From 535f545be7e0aa30fe97a9041e2cf11bc734e49a Mon Sep 17 00:00:00 2001 From: Boris Power <81998504+BorisPower@users.noreply.github.com> Date: Thu, 10 Mar 2022 18:08:53 -0800 Subject: [PATCH] Initial commit --- .gitignore | 129 + README.md | 632 + examples/Backtranslation_of_SQL_queries.py | 189 + examples/Classification.ipynb | 130 + examples/Clustering.ipynb | 262 + examples/Code_search.ipynb | 396 + examples/Customizing_embeddings.ipynb | 20994 ++++++++++ examples/Fine-tuned_classification.ipynb | 740 + examples/Get_embeddings.ipynb | 107 + examples/Obtain_dataset.ipynb | 192 + .../Recommendation_using_embeddings.ipynb | 33446 ++++++++++++++++ examples/Regression_using_embeddings.ipynb | 109 + ...emantic_text_search_using_embeddings.ipynb | 185 + examples/User_and_product_embeddings.ipynb | 174 + examples/Visualizing_embeddings_in_2D.ipynb | 142 + examples/Visualizing_embeddings_in_3D.ipynb | 268 + ...-shot_classification_with_embeddings.ipynb | 224 + .../data/geometry_English.tex | 30259 ++++++++++++++ .../data/geometry_slovenian.tex | 32857 +++++++++++++++ .../translate_latex_book.ipynb | 1163 + examples/dbpedia_samples.jsonl | 200 + examples/fine-tuned_qa/answers_with_ft.py | 150 + .../olympics-1-collect-data.ipynb | 513 + .../fine-tuned_qa/olympics-2-create-qa.ipynb | 751 + .../fine-tuned_qa/olympics-3-train-qa.ipynb | 637 + .../README.md | 50 + .../answers_functionality_example.py | 304 + .../classification_functionality_example.py | 302 + .../search_functionality_example.py | 76 + 29 files changed, 125581 insertions(+) create mode 100644 .gitignore create mode 100644 README.md create mode 100644 examples/Backtranslation_of_SQL_queries.py create mode 100644 examples/Classification.ipynb create mode 100644 examples/Clustering.ipynb create mode 100644 examples/Code_search.ipynb create mode 100644 examples/Customizing_embeddings.ipynb create mode 100644 examples/Fine-tuned_classification.ipynb create mode 100644 examples/Get_embeddings.ipynb create mode 100644 examples/Obtain_dataset.ipynb create mode 100644 examples/Recommendation_using_embeddings.ipynb create mode 100644 examples/Regression_using_embeddings.ipynb create mode 100644 examples/Semantic_text_search_using_embeddings.ipynb create mode 100644 examples/User_and_product_embeddings.ipynb create mode 100644 examples/Visualizing_embeddings_in_2D.ipynb create mode 100644 examples/Visualizing_embeddings_in_3D.ipynb create mode 100644 examples/Zero-shot_classification_with_embeddings.ipynb create mode 100644 examples/book_translation/data/geometry_English.tex create mode 100644 examples/book_translation/data/geometry_slovenian.tex create mode 100644 examples/book_translation/translate_latex_book.ipynb create mode 100644 examples/dbpedia_samples.jsonl create mode 100644 examples/fine-tuned_qa/answers_with_ft.py create mode 100644 examples/fine-tuned_qa/olympics-1-collect-data.ipynb create mode 100644 examples/fine-tuned_qa/olympics-2-create-qa.ipynb create mode 100644 examples/fine-tuned_qa/olympics-3-train-qa.ipynb create mode 100644 transition_guides_for_deprecated_API_endpoints/README.md create mode 100644 transition_guides_for_deprecated_API_endpoints/answers_functionality_example.py create mode 100644 transition_guides_for_deprecated_API_endpoints/classification_functionality_example.py create mode 100644 transition_guides_for_deprecated_API_endpoints/search_functionality_example.py diff --git a/.gitignore b/.gitignore new file mode 100644 index 0000000..b6e4761 --- /dev/null +++ b/.gitignore @@ -0,0 +1,129 @@ +# Byte-compiled / optimized / DLL files +__pycache__/ +*.py[cod] +*$py.class + +# C extensions +*.so + +# Distribution / packaging +.Python +build/ +develop-eggs/ +dist/ +downloads/ +eggs/ +.eggs/ +lib/ +lib64/ +parts/ +sdist/ +var/ +wheels/ +pip-wheel-metadata/ +share/python-wheels/ +*.egg-info/ +.installed.cfg +*.egg +MANIFEST + +# PyInstaller +# Usually these files are written by a python script from a template +# before PyInstaller builds the exe, so as to inject date/other infos into it. +*.manifest +*.spec + +# Installer logs +pip-log.txt +pip-delete-this-directory.txt + +# Unit test / coverage reports +htmlcov/ +.tox/ +.nox/ +.coverage +.coverage.* +.cache +nosetests.xml +coverage.xml +*.cover +*.py,cover +.hypothesis/ +.pytest_cache/ + +# Translations +*.mo +*.pot + +# Django stuff: +*.log +local_settings.py +db.sqlite3 +db.sqlite3-journal + +# Flask stuff: +instance/ +.webassets-cache + +# Scrapy stuff: +.scrapy + +# Sphinx documentation +docs/_build/ + +# PyBuilder +target/ + +# Jupyter Notebook +.ipynb_checkpoints + +# IPython +profile_default/ +ipython_config.py + +# pyenv +.python-version + +# pipenv +# According to pypa/pipenv#598, it is recommended to include Pipfile.lock in version control. +# However, in case of collaboration, if having platform-specific dependencies or dependencies +# having no cross-platform support, pipenv may install dependencies that don't work, or not +# install all needed dependencies. +#Pipfile.lock + +# PEP 582; used by e.g. github.com/David-OConnor/pyflow +__pypackages__/ + +# Celery stuff +celerybeat-schedule +celerybeat.pid + +# SageMath parsed files +*.sage.py + +# Environments +.env +.venv +env/ +venv/ +ENV/ +env.bak/ +venv.bak/ + +# Spyder project settings +.spyderproject +.spyproject + +# Rope project settings +.ropeproject + +# mkdocs documentation +/site + +# mypy +.mypy_cache/ +.dmypy.json +dmypy.json + +# Pyre type checker +.pyre/ diff --git a/README.md b/README.md new file mode 100644 index 0000000..ca87cac --- /dev/null +++ b/README.md @@ -0,0 +1,632 @@ +# OpenAI Cookbook + +This repository shares example code and example prompts for accomplishing common tasks with the [OpenAI API](https://openai.com/api/). + +To try these examples yourself, you’ll need an OpenAI account. [Create a free account to get started.](https://beta.openai.com/signup) + +Most code examples are written in Python, though the concepts can be applied in any language. + +In the same way that a cookbook's recipes don't span all possible meals or techniques, these examples don't span all possible use cases or methods. Use them as starting points upon which to elaborate, discover, and invent. + +## Related resources + +Beyond the code examples here, you can also learn about the [OpenAI API](https://openai.com/api/) from the following resources: + +* Try out GPT-3 in the [OpenAI Playground](https://beta.openai.com/playground) +* Read about the API in the [OpenAI Documentation](https://beta.openai.com/docs/introduction) +* Discuss the API in the [OpenAI Community Forum](https://community.openai.com/top?period=monthly) +* Look for help in the [OpenAI Help Center](https://help.openai.com/en/) +* See example prompts in the [OpenAI Examples](https://beta.openai.com/examples) + +## Examples, organized by capability + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
TextCode
Write +
  • Copywriting
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  • Blog posts
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  • Product descriptions
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  • Question generation
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  • Code completion (e.g., GitHub Copilot)
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  • Natural language software interfaces
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  • Text to code
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  • Unit tests
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    Explain +
  • Q&A about a doc
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  • Entity extraction
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  • Summarization
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  • Classification
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  • Code documentation
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  • Code explanation
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  • Docstrings
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    Edit +
  • Editing
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  • Translation
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  • Conversion between languages or styles
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  • Bug fixing
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    Compare +
  • Semantic search
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  • Recommendations
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  • Clustering
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  • Near-duplicate detection
    • +     		+
  • Code search
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  • Code clustering
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    + +## How large language models work + +[Large language models](https://openai.com/blog/better-language-models/) are functions that map text to text. Given an input string of text, a large language model tries to predict the text that will come next. + +The magic of large language models is that by being trained to minimize this prediction error over vast quantities of text, the models end up learning concepts useful for these predictions. For example, they learn concepts like: + +* how to spell +* how grammar works +* how to paraphrase +* how to answer questions +* how to hold a conversation +* how to write in many languages +* how to code +* etc. + +None of these capabilities are explicitly programmed in - they all emerge as a result of training. + +GPT-3's capabilities now power [hundreds of different software products](https://openai.com/blog/gpt-3-apps/), including productivity apps, education apps, games, and more. + +## How to control a large language model + +Of all the inputs to a large language model, by far the most influential is the text prompt. + +Large language models can be prompted to produce output in a few ways: + +* **Instruction**: Tell the model what you want +* **Completion**: Induce the model to complete the beginning of what you want +* **Demonstration**: Show the model what you want, with either: + * A few examples in the prompt + * Many hundreds or thousands of examples in a fine-tuning training dataset + +An example of each is shown below. + +### Instruction prompts + +Instruction-following models (e.g., `text-davinci-002` or any model beginning with `text-`) are specially designed to follow instructions. Write your instruction at the top of the prompt (or at the bottom, or both), and the model will do its best to follow the instruction and then stop. Instructions can be detailed, so don't be afraid to write a paragraph explicitly detailing the output you want. + +Example instruction prompt: + +```text +Extract the name of the author from the quotation below. + +“Some humans theorize that intelligent species go extinct before they can expand into outer space. If they're correct, then the hush of the night sky is the silence of the graveyard.” +― Ted Chiang, Exhalation +``` + +Output: + +```text +Ted Chiang +``` + +### Completion prompt example + +Completion-style prompts take advantage of how large language models try to write text they think is mostly likely to come next. To steer the model, try beginning a pattern or sentence that will be completed by the output you want to see. Relative to direct instructions, this mode of steering large language models can take more care and experimentation. In addition, the models won't necessarily know where to stop, so you will often need stop sequences or post-processing to cut off text generated beyond the desired output. + +Example completion prompt: + +```text +“Some humans theorize that intelligent species go extinct before they can expand into outer space. If they're correct, then the hush of the night sky is the silence of the graveyard.” +― Ted Chiang, Exhalation + +The author of this quote is +``` + +Output: + +```text + Ted Chiang +``` + +### Demonstration prompt example (few-shot learning) + +Similar to completion-style prompts, demonstrations can show the model what you want it to do. This approach is sometimes called few-shot learning, as the model learns from a few examples provided in the prompt. + +Example demonstration prompt: + +```text +Quote: +“When the reasoning mind is forced to confront the impossible again and again, it has no choice but to adapt.” +― N.K. Jemisin, The Fifth Season +Author: N.K. Jemisin + +Quote: +“Some humans theorize that intelligent species go extinct before they can expand into outer space. If they're correct, then the hush of the night sky is the silence of the graveyard.” +― Ted Chiang, Exhalation +Author: +``` + +Output: + +```text + Ted Chiang +``` + +### Fine-tuned prompt example + +With enough training examples, you can [fine-tune](https://beta.openai.com/docs/guides/fine-tuning) a custom model. In this case, instructions become unnecessary, as the model can learn the task from the training data provided. However, it can be helpful to include separator sequences (e.g., `->` or `###` or any string that doesn't commonly appear in your inputs) to tell the model when the prompt has ended and the output should begin. Without separator sequences, there is a risk that the model continues elaborating on the input text rather than starting on the answer you want to see. + +Example fine-tuned prompt (for a model that has been custom trained on similar prompt-completion pairs): + +```text +“Some humans theorize that intelligent species go extinct before they can expand into outer space. If they're correct, then the hush of the night sky is the silence of the graveyard.” +― Ted Chiang, Exhalation + +### + + +``` + +Output: + +```text + Ted Chiang +``` + +### More prompt advice + +For more prompt examples, visit [OpenAI Examples](https://beta.openai.com/examples). + +In general, the input prompt is the best lever for improving model outputs. You can try tricks like: + +* **Give more explicit instructions.** E.g., if you want the output to be a comma separated list, ask it to return a comma separated list. If you want it to say "I don't know" when the it doesn't know the answer, tell it 'Say "I don't know" if you do not know the answer.' +* **Supply better examples.** If you're demonstrating examples in your prompt, make sure that your examples are diverse and high quality. +* **Ask the model to answer as if it was an expert.** Explicitly asking the model to produce high quality output or output as if it was written by an expert can induce the model to give higher quality answers that it thinks an expert would write. E.g., "The following answer is correct, high-quality, and written by an expert." +* **Prompt the model to write down the series of steps explaining its reasoning.** E.g., prepend your answer with something like "[Let's think step by step](https://arxiv.org/pdf/2205.11916v1.pdf)." Prompting the model to give an explanation of its reasoning before its final answer can increase the likelihood that its final answer is consistent and correct. + +## Text Capabilities + +### 1. Write text + +Large language models are excellent at writing text. They can assist with: + +* Blog posts +* Email copy +* Ad copy +* Website copy +* Product descriptions +* Memos +* Storytelling +* Brainstorming +* Question generation +* etc. + +An example prompt for an instruction-following model: + +```text +Write an email to a colleague named Jill congratulating her on her promotion. The tone should be warm yet professional. Mention how you admire the work she's been putting in. Include a joke about how her pet lizard Max enjoys eating grasshoppers. Mention how you're looking forward to the team off-site next week. +``` + +Output: + +```text +Dear Jill, + +Congratulations on your promotion! I've been admiring the great work you've been putting in and I'm really happy for your success. Max the lizard must be thrilled too - I bet he's looking forward to feasting on even more grasshoppers! + +I'm really looking forward to next week's team off-site. It's going to be great to catch up with everyone and brainstorm some new ideas. + +Best, + +[Your Name] +``` + +In general, writing can work with any style of prompt. Experiment to see what works best for your use case. + +| | Advantages | Disadvantages | +| ---------------------------------------------------------- | ----------------------------------------------------------------------------- | -------------------------------------------------------------------------------- | +| Instruction-following models
    (e.g., `text-davinci-002`) | Easiest to use | Less creative; less diverse; harder to control tone, length, etc. | +| Base models
    (e.g., `davinci`) | More creative | More expensive (as including examples demonstrations in prompt will cost tokens) | +| Fine-tuned models | Can train off of many examples; cheaper than including examples in the prompt | Hard to gather training data; training makes iteration slower and more expensive | + +### 2. Explain text + +One capability of large language models is distilling information from a piece of text. This can include: + +* Answering questions about a piece of text, e.g.: + * Querying an knowledge base to help people look up things they don't know + * Querying an unfamiliar document to understand what it contains + * Querying a document with structured questions in order to extract tags, classes, entities, etc. +* Summarizing text, e.g.: + * Summarizing long documents + * Summarizing back-and-forth emails or message threads + * Summarizing detailed meeting notes with key points and next steps +* Classifying text, e.g.: + * Classifying customer feedback messages by topic or type + * Classifying documents by topic or type + * Classifying the tone or sentiment of text +* Extracting entities, e.g.: + * Extracting contact information from a customer message + * Extracting names of people or companies or products from a document + * Extracting things mentioned in customer reviews or feedback + +#### Answering questions about a piece of text + +Example prompt for answering questions about a piece of text: + +```text +Using the following text, answer the following question. If the answer is not contained within the text, say "I don't know." + +Text: +""" +Oklo Mine (sometimes Oklo Reactor or Oklo Mines), located in Oklo, Gabon on the west coast of Central Africa, is believed to be the only natural nuclear fission reactor. Oklo consists of 16 sites at which self-sustaining nuclear fission reactions are thought to have taken place approximately 1.7 billion years ago, and ran for hundreds of thousands of years. It is estimated to have averaged under 100 kW of thermal power during that time. +""" + +Question: How many natural fission reactors have ever been discovered? + +Answer: +``` + +Output: + +```text + One + ``` + +If the text you wish to ask about is longer than the token limit (~4,000 tokens for `text-davinci-002` and ~2,000 tokens for earlier models), we recommending splitting the text into smaller pieces, ranking them by relevance, and then asking the most-relevant-looking pieces. + +#### Summarization + +An example prompt for summarization: + +```text +Summarize the following text. + +Text: +""" +Two independent experiments reported their results this morning at CERN, Europe's high-energy physics laboratory near Geneva in Switzerland. Both show convincing evidence of a new boson particle weighing around 125 gigaelectronvolts, which so far fits predictions of the Higgs previously made by theoretical physicists. + +"As a layman I would say: 'I think we have it'. Would you agree?" Rolf-Dieter Heuer, CERN's director-general, asked the packed auditorium. The physicists assembled there burst into applause. +""" + +Summary: +``` + +Output: + +```text +CERN has announced the discovery of a new particle, the Higgs boson. This particle has been predicted by theoretical physicists and is a major step forward in our understanding of the universe. +``` + +#### Classification + +The best approach for classifying text depends on whether the classes are known in advance or not. + +If your classes are known in advance, classification is best done with a fine-tuned model, as demonstrated in [Fine-tuned_classification.ipynb](examples/Fine-tuned_classification.ipynb). + +If your classes are not known in advance (e.g., they are set by a user or generated on the fly), you can try zero-shot classification by either giving an instruction containing the classes or even by using embeddings to see which class label (or other classified texts) are most similar to the text ([Zero-shot_classification.ipynb](examples/Zero-shot_classification_with_embeddings.ipynb)). + +#### Entity extraction + +An example prompt for entity extraction: + +```text +From the text below, extract the following entities in the following format: +Companies: +People & titles: + +Text: +""" +In March 1981, United States v. AT&T came to trial under Assistant Attorney General William Baxter. AT&T chairman Charles L. Brown thought the company would be gutted. He realized that AT&T would lose and, in December 1981, resumed negotiations with the Justice Department. Reaching an agreement less than a month later, Brown agreed to divestiture—the best and only realistic alternative. AT&T's decision allowed it to retain its research and manufacturing arms. The decree, titled the Modification of Final Judgment, was an adjustment of the Consent Decree of 14 January 1956. Judge Harold H. Greene was given the authority over the modified decree.... + +In 1982, the U.S. government announced that AT&T would cease to exist as a monopolistic entity. On 1 January 1984, it was split into seven smaller regional companies, Bell South, Bell Atlantic, NYNEX, American Information Technologies, Southwestern Bell, US West, and Pacific Telesis, to handle regional phone services in the U.S. AT&T retains control of its long distance services, but was no longer protected from competition. +""" +``` + +Output: + +```text +Companies: United States v. AT&T, AT&T, Justice Department, Bell South, Bell Atlantic, NYNEX, American Information Technologies, Southwestern Bell, US West, Pacific Telesis +People & titles: William Baxter (Assistant Attorney General), Charles L. Brown (AT&T chairman), Harold H. Greene (Judge) +``` + +### 3. Edit text + +In addition to the [completion API endpoint](https://beta.openai.com/docs/api-reference/completions), OpenAI now offers an [edit API endpoint](https://beta.openai.com/docs/api-reference/edits) ([blog post](https://openai.com/blog/gpt-3-edit-insert/)). In contrast to completions, which only take a single text input, edits take two text inputs: the instruction and the text to be modified. + +An example edit prompt: + +Instruction input: + +```text +Fix the OCR errors +``` + +Text input: + +```text +Therewassomehostilityntheenergybehindthe researchreportedinPerceptrons....Part of ourdrivecame,aswequiteplainlyacknoweldgednourbook,fromhe facthatfundingndresearchnergywerebeingdissipatedon. . .misleadingttemptsouseconnectionistmethodsnpracticalappli-cations. +``` + +Output: + +```text +There was some hostility in the energy behind the research reported in Perceptrons....Part of our drive came, as we quite plainly acknowledged in our book, from the fact that funding and research energy were being dissipated on...misleading attempts to use connectionist methods in practical applications. +``` + +#### Translation + +Translation is another emergent capability of large language models. In 2021, [GPT-3 was used](https://arxiv.org/abs/2110.05448) to set a new state-of-the-art record in unsupervised translation on the WMT14 English-French benchmark. + +Example translation prompt using the edits endpoint: + +Instruction input: + +```text +translation into French +``` + +Text input: + +```text +That's life. +``` + +Output: + +```text +C'est la vie. +``` + +Example translation prompt using the completions endpoint: + +```text +Translate the following text from English to French. + +English: That's life. +French: +``` + +Output: + +```text + C'est la vie. +``` + +Tips for translation: + +* Performance is best on the most common languages +* We've seen better performance when the instruction is given in the final language (so if translating into French, give the instruction `Traduire le texte de l'anglais au français.` rather than `Translate the following text from English to French.`) +* Backtranslation (as described [here](https://arxiv.org/abs/2110.05448)) can also increase performance +* Text with colons and heavy punctuation can trip up the instruction-following models, especially if the instruction is using colons (e.g., `English: {english text} French:`) +* The edits endpoint has been seen to sometimes repeat the text input alongside the translation + +When it comes to translation, large language models particularly shine at combining other instructions alongside translation. For example, you can ask GPT-3 to translate Slovenian to English but keep all LaTeX typesetting commands unchanged. The following notebook details how we translated a Slovenian math book into English: + +[Translation of a Slovenian math book into English](book_translation/translate_latex_book.ipynb) + +### 4. Compare text + +The [OpenAI API embeddings endpoint](https://beta.openai.com/docs/guides/embeddings) can be used to measure similarity between pieces of text ([blog post](https://openai.com/blog/introducing-text-and-code-embeddings/)). By leveraging GPT-3's understanding of text, these embeddings [achieved state-of-the-art results](https://arxiv.org/abs/2201.10005) on benchmarks in both unsupervised learning and transfer learning settings. + +Embeddings can be used for semantic search, recommendations, cluster analysis, near-duplicate detection, and more. + +#### Semantic search + +Embeddings can be used for search either by themselves or as a feature in a larger system. + +The simplest way to use embeddings for search is as follows: + +* Before the search (precompute): + * Split your text corpus into chunks smaller than the token limit (e.g., ~2,000 tokens) + * Embed each chunk using a 'doc' model (e.g., `text-search-curie-doc-001`) + * Store those embeddings in your own database or in a vector search provider like [pinecone.io](pinecone.io) or [weaviate](weaviate.io) +* At the time of the search (live compute): + * Embed the search query using the correponding 'query' model (e.g. `text-search-curie-query-001`) + * Find the closest embeddings in your database + * Return the top results, ranked by cosine similarity + +An example of how to use embeddings for search is shown in [Semantic_search.ipynb](examples/Semantic_search.ipynb). + +In more advanced search systems, the the cosine similarity of embeddings can be used as one feature among many in ranking search results. + +#### Recommendations + +Recommendations are quite similar to search, except that instead of a free-form text query, the inputs are items in a set. And instead of using pairs of doc-query models, you can use a single symmetric similarity model (e.g., `text-similarity-curie-001`). + +An example of how to use embeddings for recommendations is shown in [Recommendations.ipynb](examples/Recommendations.ipynb). + +Similar to search, these cosine similarity scores can either be used on their own to rank items or as features in larger ranking algorithms. + +#### Customizing Embeddings + +Although OpenAI's embedding model weights cannot be fine-tuned, you can still use training data to customize embeddings to your application. + +In the following notebook, we provide an example method for customizing your embeddings using training data. The idea of the method is to train a custom matrix to multiply embedding vectors by in order to get new customized embeddings. With good training data, this custom matrix will highlight the features relevant to your training labels and suppress the rest. You can equivalently consider the matrix mulitplication as (a) a modification of the embeddings or (b) a modification of the distance function used to measure the distances between embeddings. + +* [Customizing_embeddings.ipynb](examples/Customizing_embeddings.ipynb) + +## Code Capabilities + +Large language models aren't only great at text - they can be great at code too. OpenAI's specialized code model is called [Codex](https://openai.com/blog/openai-codex/). + +Codex powers [more than 70 products](https://openai.com/blog/codex-apps/), including: + +* [GitHub Copilot](https://copilot.github.com/) (autocompletes code in VS Code and other IDEs) +* [Pygma](https://pygma.app/) (turns Figma designs into code) +* [Replit](https://replit.com/) (has an 'Explain code' button and other features) +* [Warp](https://www.warp.dev/) (a smart terminal with AI command search) +* [Machinet](https://machinet.net/) (writes Java unit test templates) + +Note that unlike instruction-following text models (e.g., `text-davinci-002`), Codex is *not* trained to follow instructions. As a result, designing good prompts can take more care. + +### 1. Write code + +An example prompt for writing code with `code-davinci-002`: + +````text +SQL tables (and columns): +* Customers(customer_id, signup_date) +* Streaming(customer_id, video_id, watch_date, watch_minutes) + +A well-written SQL query that lists customers who signed up during March 2020 and watched more than 50 hours of video in their first 30 days: +``` +```` + +Output: + +```sql +SELECT c.customer_id +FROM Customers c +JOIN Streaming s +ON c.customer_id = s.customer_id +WHERE c.signup_date BETWEEN '2020-03-01' AND '2020-03-31' +AND s.watch_date BETWEEN c.signup_date AND DATE_ADD(c.signup_date, INTERVAL 30 DAY) +GROUP BY c.customer_id +HAVING SUM(s.watch_minutes) > 50 * 60 +``` + +`code-davinci-002` is able to make inferences from variable names; for example, it infers that `watch_minutes` has units of minutes and therefore needs to be converted by a factor of 60 before being compared with 50 hours. + +### 2. Explain code + +Code explanation can be applied to many use cases: + +* Generating in-code documentation (e.g., Python docstrings, git commit messages) +* Generating out-of-code documentation (e.g., man pages) +* In an interactive code exploration tool +* Communicating program results back to users via a natural langauge interface + +An example prompt for explaining code with `code-davinci-002`: + +````text +A SQL query: +``` +SELECT c.customer_id +FROM Customers c +JOIN Streaming s +ON c.customer_id = s.customer_id +WHERE c.signup_date BETWEEN '2020-03-01' AND '2020-03-31' +AND s.watch_date BETWEEN c.signup_date AND DATE_ADD(c.signup_date, INTERVAL 30 DAY) +GROUP BY c.customer_id +HAVING SUM(s.watch_minutes) > 50 * 60 +``` + +Questions: +1. What does the SQL query do? +2. Why might someone be interested in this time period? +3. Why might a company be interested in this SQL query? + +Answers: +```` + +Output: + +```text +1. The SQL query finds all customers who signed up in March 2020 and watched more than 50 hours of content in the first 30 days after signing up. +2. The time period is interesting because it is the first month of the COVID-19 pandemic. +3. A company might be interested in this SQL query because it can help them understand how the pandemic has affected their business. +``` + +### 3. Edit code + +OpenAI's edit endpoint is particularly useful for editing code. + +Example text input to `code-davinci-edit-001`: + +```python +def tribonacci(n): + if n == 0: + return 0 + elif n == 1: + return 1 + elif n == 2: + return 1 + elif n == 3: + return 2 + else: + return tribonacci(n-1) + tribonacci(n-2) + tribonacci(n-3) +``` + +Example instruction inputs: + +```text +Add a docstring +``` + +```text +Add typing +``` + +```text +Improve the runtime +``` + +```text +Add a test +``` + +```text +Translate to JavaScript (or Rust or Lisp or any language you like) +``` + +Example output after improving the runtime and translating to JavaScript: + +```JavaScript +function tribonacci(n) { + let a = 0; + let b = 1; + let c = 1; + for (let i = 0; i < n; i++) { + [a, b, c] = [b, c, a + b + c]; + } + return a; +} +``` + +As you can see, `code-davinci-edit-001` was able to successfully reduce the function's runtime from exponential down to linear, as well as convert from Python to JavaScript. + +### 4. Compare code + +The OpenAI API also features code search embeddings, which can measure the relevance of a section of code to a text query, or the similarity between two sections of code. + +OpenAI code search embeddings significantly improved the state-of-the-art on the [CodeSearchNet](https://github.com/github/CodeSearchNet) evaluation suite, scoring 93.5% versus the previous record of 77.4%. + +Read more about OpenAI's code embeddings in the [blog post announcement](https://openai.com/blog/introducing-text-and-code-embeddings/) or [documentation](https://beta.openai.com/docs/guides/embeddings). + +Code embeddings can be useful for use cases such as: + +* Code search +* Codebase clustering & analysis + +An example of code search is shown in [Code_search.ipynb](examples/Code_search.ipynb). + +We haven't written an example of code clustering, but the idea is the same as the text clustering in [Clustering.ipynb](examples/Clustering.ipynb). diff --git a/examples/Backtranslation_of_SQL_queries.py b/examples/Backtranslation_of_SQL_queries.py new file mode 100644 index 0000000..8289a73 --- /dev/null +++ b/examples/Backtranslation_of_SQL_queries.py @@ -0,0 +1,189 @@ +from typing import List, Union + +from smokey import Smokey + +import openai + + +def get_candidates( + prompt: str, + stop: List[str], + temperature: float, + priming_prefix: str, + engine: str, + n: int = 5, +) -> List[str]: + """ + Generate N candidate completions based on the prompt, generated with a specific temperature. + + :param prompt: The prompt to start the conversation with. + :param stop: A list of tokens that indicate the end of the generation. + :param temperature: The temperature of the generation. + :param priming_prefix: The prefix to use for the priming. + :param engine: The engine to use for the generation. + :param n: The number of completions to generate. + :return: A list of completions. + """ + response = openai.Completion.create( + engine=engine, + prompt=prompt, + temperature=temperature, + max_tokens=150, + top_p=1, + frequency_penalty=0, + presence_penalty=0, + stop=stop, + n=n, + ) + responses = [priming_prefix + choice.text for choice in response.choices] + return responses + + +def rindex(lst: List, value: str) -> int: + """ + Return the index of the last occurence of a value in a list. + + :param lst: The list to search in. + :param value: The value to search for. + :return: The index of the last occurence of the value. + """ + try: + return len(lst) - lst[::-1].index(value) - 1 + except ValueError: + raise ValueError(f"Answer start token `{value}` not found in the eval template") + + +def eval_candidate( + candidate_answer: str, + original_instruction: str, + eval_template: str, + answer_start_token: str, + engine: str, +) -> float: + """ + Evaluate a candidate answer by calculating the average log probability + of the original instruction, given the candidate answer with a specific + evaluation template, aimed at reconstructing the original instruction. + + :param candidate_answer: The candidate answer to evaluate. + :param original_instruction: The original instruction. + :param eval_template: The template to use for the evaluation. + :param answer_start_token: The token to use to indicate the start of the answer. + :param engine: The engine to use for the evaluation. + :return: The evaluation of the candidate answer. + """ + response = openai.Completion.create( + engine=engine, + prompt=eval_template.format(candidate_answer, original_instruction), + temperature=0, + max_tokens=0, + top_p=1, + frequency_penalty=0, + presence_penalty=0, + logprobs=1, + echo=True, + ) + + answer_start = rindex( + response["choices"][0]["logprobs"]["tokens"], answer_start_token + ) + logprobs = response["choices"][0]["logprobs"]["token_logprobs"][answer_start + 1 :] + return sum(logprobs) / len(logprobs) + + +def backtranslation( + prompt_template: str, + additional_info: str, + instruction: str, + eval_template: str, + priming_prefix: str = "SELECT", + stop1: List[str] = ["#", ";"], + answer_start_token: str = "--", + n: int = 5, + temperature: float = 0.5, + return_all_results: bool = False, + engine: str = "davinci-codex", +) -> Union[str, List[str, float]]: + """ + Generate a number of SQL queries given a natural language instruction, + and pick the best one based on the average log probability of explaining the + candidate SQL query with the exact original instruction, when prompted for + a natural language explanation of the candidate SQL query. + + :param prompt_template: The template to use for the prompt to generate SQL. + :param additional_info: Additional information to include in the prompt + (SQL Tables, and their properties). + :param instruction: The instruction in natural language. + :param eval_template: The template to use for the evaluation. + :param priming_prefix: The prefix to use for the priming of the SQL query. + :param stop1: A list of tokens that indicate the end of the generation. + :param answer_start_token: The token to use to indicate the start of the + natural answer. + :param n: The number of candidates to generate. + :param temperature: The temperature of the generation. + :param return_all_results: Whether to return all results or just the best one. + :param engine: The engine to use for the generation and evaluation. + :return: The best SQL query, or a list of all scored generated SQL queries. + """ + prompt_template = prompt_template.format( + additional_info, instruction, priming_prefix + ) + + candidates = [] + responses = get_candidates( + prompt_template, stop1, temperature, priming_prefix, engine=engine, n=n + ) + for i in range(n): + quality = eval_candidate( + responses[i], + instruction, + eval_template, + answer_start_token, + engine=engine, + ) + candidates.append((responses[i], quality)) + + candidates.sort(key=lambda x: x[1], reverse=True) + if return_all_results: + return candidates + return candidates[0][0] + + +def main( + nl_query: str = "Return the name of each department that had more than 10 employees in June 2021", + eval_template: str = "{};\n-- Explanation of the above query in human readable format\n-- {}", + table_definitions: str = "# Employee(id, name, department_id)\n# Department(id, name, address)\n# Salary_Payments(id, employee_id, amount, date)\n", + prompt_template: str = "### Postgres SQL tables, with their properties:\n#\n{}#\n### {}\n{}", + n: int = 3, + temperature: float = 0.3, + engine: str = "davinci-codex", +): + """ + Generate a number of SQL queries given a natural language instruction, + and pick the best one based on the highest backtranslation score. + + :param nl_query: The natural language query. + :param eval_template: The template to use for the evaluation. + :param table_definitions: The definitions of the tables used in the query. + :param prompt_template: The template to use for the prompt to generate SQL. + :param n: The number of candidates to generate. + :param temperature: The temperature of the generation. + :param engine: The engine to use for the generation and evaluation. + :return: The best SQL query, or a list of all scored generated SQL queries. + """ + + result = backtranslation( + prompt_template, + table_definitions, + nl_query, + eval_template, + priming_prefix="SELECT", + temperature=temperature, + n=n, + engine=engine, + ) + print(result) + + +if __name__ == "__main__": + Smokey(main) diff --git a/examples/Classification.ipynb b/examples/Classification.ipynb new file mode 100644 index 0000000..54ad13f --- /dev/null +++ b/examples/Classification.ipynb @@ -0,0 +1,130 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Classification using the embeddings\n", + "\n", + "In the classification task we predict one of the predefined categories given an input. We will predict the score based on the embedding of the review's text, where the algorithm is correct only if it guesses the exact number of stars. We split the dataset into a training and a testing set for all the following tasks, so we can realistically evaluate performance on unseen data. The dataset is created in the [Obtain_dataset Notebook](Obtain_dataset.ipynb).\n", + "\n", + "In the following example we're predicting the number of stars in a review, from 1 to 5." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " precision recall f1-score support\n", + "\n", + " 1 0.82 0.67 0.74 21\n", + " 2 0.50 0.50 0.50 6\n", + " 3 1.00 0.46 0.63 13\n", + " 4 0.75 0.35 0.48 17\n", + " 5 0.88 1.00 0.93 143\n", + "\n", + " accuracy 0.86 200\n", + " macro avg 0.79 0.60 0.66 200\n", + "weighted avg 0.86 0.86 0.84 200\n", + "\n" + ] + } + ], + "source": [ + "import pandas as pd\n", + "import numpy as np\n", + "\n", + "from sklearn.ensemble import RandomForestClassifier\n", + "from sklearn.model_selection import train_test_split\n", + "from sklearn.metrics import classification_report, accuracy_score\n", + "\n", + "df = pd.read_csv('output/embedded_1k_reviews.csv')\n", + "df['babbage_similarity'] = df.babbage_similarity.apply(eval).apply(np.array)\n", + "\n", + "X_train, X_test, y_train, y_test = train_test_split(list(df.babbage_similarity.values), df.Score, test_size = 0.2, random_state=42)\n", + "\n", + "clf = RandomForestClassifier(n_estimators=100)\n", + "clf.fit(X_train, y_train)\n", + "preds = clf.predict(X_test)\n", + "probas = clf.predict_proba(X_test)\n", + "\n", + "report = classification_report(y_test, preds)\n", + "print(report)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can see that the model has learnt to distinguish between the categories decently. 5-star reviews show the best performance overall, and this is not too surprising, since they are the most common in the dataset." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RandomForestClassifier() - Average precision score over all classes: 0.93\n" + ] + }, + { + "data": { + "image/png": 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", + "text/plain": [ + "
    " + ] + }, + "metadata": { + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "from openai.embeddings_utils import plot_multiclass_precision_recall\n", + "\n", + "plot_multiclass_precision_recall(probas, y_test, [1,2,3,4,5], clf)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Unsurprisingly 5-star and 1-star reviews seem to be easier to predict. Perhaps with more data, the nuances between 2-4 stars could be better predicted, but there's also probably more subjectivity in how people use the inbetween scores." + ] + } + ], + "metadata": { + "interpreter": { + "hash": "be4b5d5b73a21c599de40d6deb1129796d12dc1cc33a738f7bac13269cfcafe8" + }, + "kernelspec": { + "display_name": "Python 3.7.3 64-bit ('base': conda)", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.7.3" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/Clustering.ipynb b/examples/Clustering.ipynb new file mode 100644 index 0000000..a03eec5 --- /dev/null +++ b/examples/Clustering.ipynb @@ -0,0 +1,262 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Clustering\n", + "\n", + "We use a simple k-means algorithm to demonstrate how clustering can be done. Clustering can help discover valuable, hidden groupings within the data. The dataset is created in the [Obtain_dataset Notebook](Obtain_dataset.ipynb)." + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "(1000, 2048)" + ] + }, + "execution_count": 31, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "import pandas as pd\n", + "import numpy as np\n", + "\n", + "\n", + "df = pd.read_csv('output/embedded_1k_reviews.csv')\n", + "df['text-similarity-babbage-001'] = df.babbage_similarity.apply(eval).apply(np.array)\n", + "matrix = np.vstack(df.babbage_similarity.values)\n", + "matrix.shape" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### 1. Find the clusters using K-means" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We show the simplest use of K-means. You can pick the number of clusters that fits your use case best." + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Cluster\n", + "2 2.543478\n", + "3 4.374046\n", + "0 4.709402\n", + "1 4.832099\n", + "Name: Score, dtype: float64" + ] + }, + "execution_count": 34, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "from sklearn.cluster import KMeans\n", + "\n", + "n_clusters = 4\n", + "\n", + "kmeans = KMeans(n_clusters = n_clusters,init='k-means++',random_state=42)\n", + "kmeans.fit(matrix)\n", + "labels = kmeans.labels_\n", + "df['Cluster'] = labels\n", + "\n", + "df.groupby('Cluster').Score.mean().sort_values()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "It looks like cluster 2 focused on negative reviews, while cluster 0 and 1 focused on positive reviews." + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Text(0.5, 1.0, 'Clusters identified visualized in language 2d using t-SNE')" + ] + }, + "execution_count": 40, + "metadata": {}, + "output_type": "execute_result" + }, + { + "data": { + "image/png": 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", + "text/plain": [ + "
    " + ] + }, + "metadata": { + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "from sklearn.manifold import TSNE\n", + "import matplotlib\n", + "import matplotlib.pyplot as plt\n", + "\n", + "tsne = TSNE(n_components=2, perplexity=15, random_state=42, init='random', learning_rate=200)\n", + "vis_dims2 = tsne.fit_transform(matrix)\n", + "\n", + "x = [x for x,y in vis_dims2]\n", + "y = [y for x,y in vis_dims2]\n", + "\n", + "for category, color in enumerate(['purple', 'green', 'red', 'blue']):\n", + " xs = np.array(x)[df.Cluster==category]\n", + " ys = np.array(y)[df.Cluster==category]\n", + " plt.scatter(xs, ys, color=color, alpha=0.3)\n", + "\n", + " avg_x = xs.mean()\n", + " avg_y = ys.mean()\n", + " \n", + " plt.scatter(avg_x, avg_y, marker='x', color=color, s=100)\n", + "plt.title(\"Clusters identified visualized in language 2d using t-SNE\")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Visualization of clusters in a 2d projection. The red cluster clearly represents negative reviews. The blue cluster seems quite different from the others. Let's see a few samples from each cluster." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### 2. Text samples in the clusters & naming the clusters\n", + "\n", + "Let's show random samples from each cluster. We'll use davinci-instruct-beta-v3 to name the clusters, based on a random sample of 6 reviews from that cluster." + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cluster 0 Theme: All of the customer reviews mention the great flavor of the product.\n", + "5, French Vanilla Cappuccino: Great price. Really love the the flavor. No need to add anything to \n", + "5, great coffee: A bit pricey once you add the S & H but this is one of the best flavor\n", + "5, Love It: First let me say I'm new to drinking tea. So you're not getting a well\n", + "----------------------------------------------------------------------------------------------------\n", + "Cluster 1 Theme: All three reviews mention the quality of the product.\n", + "5, Beautiful: I don't plan to grind these, have plenty other peppers for that. I go\n", + "5, Awesome: I can't find this in the stores and thought I would like it. So I bou\n", + "5, Came as expected: It was tasty and fresh. The other one I bought was old and tasted mold\n", + "----------------------------------------------------------------------------------------------------\n", + "Cluster 2 Theme: All reviews are about customer's disappointment.\n", + "1, Disappointed...: I should read the fine print, I guess. I mostly went by the picture a\n", + "5, Excellent but Price?: I first heard about this on America's Test Kitchen where it won a blin\n", + "1, Disappointed: I received the offer from Amazon and had never tried this brand before\n", + "----------------------------------------------------------------------------------------------------\n", + "Cluster 3 Theme: The reviews for these products have in common that the customers' dogs love them.\n", + "5, My Dog's Favorite Snack!: I was first introduced to this snack at my dog's training classes at p\n", + "4, Fruitables Crunchy Dog Treats: My lab goes wild for these and I am almost tempted to have a go at som\n", + "5, Happy with the product: My dog was suffering with itchy skin. He had been eating Natural Choi\n", + "----------------------------------------------------------------------------------------------------\n" + ] + } + ], + "source": [ + "import openai\n", + "\n", + "# Reading a review which belong to each group.\n", + "rev_per_cluster = 3\n", + "\n", + "for i in range(n_clusters):\n", + " print(f\"Cluster {i} Theme:\", end=\" \")\n", + " \n", + " reviews = \"\\n\".join(df[df.Cluster == i].combined.str.replace(\"Title: \", \"\").str.replace(\"\\n\\nContent: \", \": \").sample(rev_per_cluster, random_state=42).values)\n", + " response = openai.Completion.create(\n", + " engine=\"davinci-instruct-beta-v3\",\n", + " prompt=f\"What do the following customer reviews have in common?\\n\\nCustomer reviews:\\n\\\"\\\"\\\"\\n{reviews}\\n\\\"\\\"\\\"\\n\\nTheme:\",\n", + " temperature=0,\n", + " max_tokens=64,\n", + " top_p=1,\n", + " frequency_penalty=0,\n", + " presence_penalty=0\n", + " )\n", + " print(response[\"choices\"][0][\"text\"].replace('\\n',''))\n", + "\n", + " sample_cluster_rows = df[df.Cluster == i].sample(rev_per_cluster, random_state=42) \n", + " for j in range(rev_per_cluster):\n", + " print(sample_cluster_rows.Score.values[j], end=\", \")\n", + " print(sample_cluster_rows.Summary.values[j], end=\": \")\n", + " print(sample_cluster_rows.Text.str[:70].values[j])\n", + " \n", + " print(\"-\" * 100)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can see based on the average ratings per cluster, that Cluster 2 contains mostly negative reviews. Cluster 0 and 1 contain mostly positive reviews, whilst Cluster 3 appears to contain reviews about dog products." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "It's important to note that clusters will not necessarily match what you intend to use them for. A larger amount of clusters will focus on more specific patterns, whereas a small number of clusters will usually focus on largest discrepencies in the data." + ] + } + ], + "metadata": { + "interpreter": { + "hash": "be4b5d5b73a21c599de40d6deb1129796d12dc1cc33a738f7bac13269cfcafe8" + }, + "kernelspec": { + "display_name": "Python 3.7.3 64-bit ('base': conda)", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.9.9" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/Code_search.ipynb b/examples/Code_search.ipynb new file mode 100644 index 0000000..d77c7d8 --- /dev/null +++ b/examples/Code_search.ipynb @@ -0,0 +1,396 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Code search\n", + "\n", + "We index our own openai-python code repository, and show how it can be searched. We implement a simple version of file parsing and extracting of functions from python files." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total number of py files: 40\n", + "Total number of functions extracted: 64\n" + ] + } + ], + "source": [ + "import os\n", + "from glob import glob\n", + "import pandas as pd\n", + "\n", + "def get_function_name(code):\n", + " \"\"\"\n", + " Extract function name from a line beginning with \"def \"\n", + " \"\"\"\n", + " assert code.startswith(\"def \")\n", + " return code[len(\"def \"): code.index(\"(\")]\n", + "\n", + "def get_until_no_space(all_lines, i) -> str:\n", + " \"\"\"\n", + " Get all lines until a line outside the function definition is found.\n", + " \"\"\"\n", + " ret = [all_lines[i]]\n", + " for j in range(i + 1, i + 10000):\n", + " if j < len(all_lines):\n", + " if len(all_lines[j]) == 0 or all_lines[j][0] in [\" \", \"\\t\", \")\"]:\n", + " ret.append(all_lines[j])\n", + " else:\n", + " break\n", + " return \"\\n\".join(ret)\n", + "\n", + "def get_functions(filepath):\n", + " \"\"\"\n", + " Get all functions in a Python file.\n", + " \"\"\"\n", + " whole_code = open(filepath).read().replace(\"\\r\", \"\\n\")\n", + " all_lines = whole_code.split(\"\\n\")\n", + " for i, l in enumerate(all_lines):\n", + " if l.startswith(\"def \"):\n", + " code = get_until_no_space(all_lines, i)\n", + " function_name = get_function_name(code)\n", + " yield {\"code\": code, \"function_name\": function_name, \"filepath\": filepath}\n", + "\n", + "\n", + "# get user root directory\n", + "root_dir = os.path.expanduser(\"~\")\n", + "\n", + "# path to code repository directory\n", + "code_root = root_dir + \"/openai-python\"\n", + "code_files = [y for x in os.walk(code_root) for y in glob(os.path.join(x[0], '*.py'))]\n", + "print(\"Total number of py files:\", len(code_files))\n", + "all_funcs = []\n", + "for code_file in code_files:\n", + " funcs = list(get_functions(code_file))\n", + " for func in funcs:\n", + " all_funcs.append(func)\n", + "\n", + "print(\"Total number of functions extracted:\", len(all_funcs))\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "For code search models we use code-search-{model}-code to obtain embeddings for code snippets, and code-search-{model}-text to embed natural language queries." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": {}, + "outputs": [ + { + "data": { + "text/html": [ + "
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    codefunction_namefilepathcode_embedding
    0def semantic_search(engine, query, documents):...semantic_search/examples/semanticsearch/semanticsearch.py[-0.038976121693849564, -0.0031428150832653046...
    1def main():\\n parser = argparse.ArgumentPar...main/examples/semanticsearch/semanticsearch.py[-0.024289356544613838, -0.017748363316059113,...
    2def get_candidates(\\n prompt: str,\\n sto...get_candidates/examples/codex/backtranslation.py[-0.04161201789975166, -0.0169310811907053, 0....
    3def rindex(lst: List, value: str) -> int:\\n ...rindex/examples/codex/backtranslation.py[-0.027255680412054062, -0.007931121625006199,...
    4def eval_candidate(\\n candidate_answer: str...eval_candidate/examples/codex/backtranslation.py[-0.00999179296195507, -0.01640152558684349, 0...
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    " + ], + "text/plain": [ + " code function_name \\\n", + "0 def semantic_search(engine, query, documents):... semantic_search \n", + "1 def main():\\n parser = argparse.ArgumentPar... main \n", + "2 def get_candidates(\\n prompt: str,\\n sto... get_candidates \n", + "3 def rindex(lst: List, value: str) -> int:\\n ... rindex \n", + "4 def eval_candidate(\\n candidate_answer: str... eval_candidate \n", + "\n", + " filepath \\\n", + "0 /examples/semanticsearch/semanticsearch.py \n", + "1 /examples/semanticsearch/semanticsearch.py \n", + "2 /examples/codex/backtranslation.py \n", + "3 /examples/codex/backtranslation.py \n", + "4 /examples/codex/backtranslation.py \n", + "\n", + " code_embedding \n", + "0 [-0.038976121693849564, -0.0031428150832653046... \n", + "1 [-0.024289356544613838, -0.017748363316059113,... \n", + "2 [-0.04161201789975166, -0.0169310811907053, 0.... \n", + "3 [-0.027255680412054062, -0.007931121625006199,... \n", + "4 [-0.00999179296195507, -0.01640152558684349, 0... " + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "from openai.embeddings_utils import get_embedding\n", + "\n", + "df = pd.DataFrame(all_funcs)\n", + "df['code_embedding'] = df['code'].apply(lambda x: get_embedding(x, engine='code-search-babbage-code-001'))\n", + "df['filepath'] = df['filepath'].apply(lambda x: x.replace(code_root, \"\"))\n", + "df.to_csv(\"output/code_search_openai-python.csv\", index=False)\n", + "df.head()" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "/openai/tests/test_endpoints.py:test_completions_multiple_prompts score=0.681\n", + "def test_completions_multiple_prompts():\n", + " result = openai.Completion.create(\n", + " prompt=[\"This was a test\", \"This was another test\"], n=5, engine=\"ada\"\n", + " )\n", + " assert len(result.choices) == 10\n", + "\n", + "----------------------------------------------------------------------\n", + "/openai/tests/test_endpoints.py:test_completions score=0.675\n", + "def test_completions():\n", + " result = openai.Completion.create(prompt=\"This was a test\", n=5, engine=\"ada\")\n", + " assert len(result.choices) == 5\n", + "\n", + "\n", + "----------------------------------------------------------------------\n", + "/openai/tests/test_api_requestor.py:test_requestor_sets_request_id score=0.635\n", + "def test_requestor_sets_request_id(mocker: MockerFixture) -> None:\n", + " # Fake out 'requests' and confirm that the X-Request-Id header is set.\n", + "\n", + " got_headers = {}\n", + "\n", + " def fake_request(self, *args, **kwargs):\n", + " nonlocal got_headers\n", + "----------------------------------------------------------------------\n" + ] + } + ], + "source": [ + "from openai.embeddings_utils import cosine_similarity\n", + "\n", + "def search_functions(df, code_query, n=3, pprint=True, n_lines=7):\n", + " embedding = get_embedding(code_query, engine='code-search-babbage-text-001')\n", + " df['similarities'] = df.code_embedding.apply(lambda x: cosine_similarity(x, embedding))\n", + "\n", + " res = df.sort_values('similarities', ascending=False).head(n)\n", + " if pprint:\n", + " for r in res.iterrows():\n", + " print(r[1].filepath+\":\"+r[1].function_name + \" score=\" + str(round(r[1].similarities, 3)))\n", + " print(\"\\n\".join(r[1].code.split(\"\\n\")[:n_lines]))\n", + " print('-'*70)\n", + " return res\n", + "res = search_functions(df, 'Completions API tests', n=3)\n" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "/openai/validators.py:format_inferrer_validator score=0.655\n", + "def format_inferrer_validator(df):\n", + " \"\"\"\n", + " This validator will infer the likely fine-tuning format of the data, and display it to the user if it is classification.\n", + " It will also suggest to use ada and explain train/validation split benefits.\n", + " \"\"\"\n", + " ft_type = infer_task_type(df)\n", + " immediate_msg = None\n", + "----------------------------------------------------------------------\n", + "/openai/validators.py:long_examples_validator score=0.649\n", + "def long_examples_validator(df):\n", + " \"\"\"\n", + " This validator will suggest to the user to remove examples that are too long.\n", + " \"\"\"\n", + " immediate_msg = None\n", + " optional_msg = None\n", + " optional_fn = None\n", + "----------------------------------------------------------------------\n", + "/openai/validators.py:non_empty_completion_validator score=0.646\n", + "def non_empty_completion_validator(df):\n", + " \"\"\"\n", + " This validator will ensure that no completion is empty.\n", + " \"\"\"\n", + " necessary_msg = None\n", + " necessary_fn = None\n", + " immediate_msg = None\n", + "----------------------------------------------------------------------\n" + ] + } + ], + "source": [ + "res = search_functions(df, 'fine-tuning input data validation logic', n=3)" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "/openai/validators.py:common_completion_suffix_validator score=0.665\n", + "def common_completion_suffix_validator(df):\n", + " \"\"\"\n", + " This validator will suggest to add a common suffix to the completion if one doesn't already exist in case of classification or conditional generation.\n", + " \"\"\"\n", + " error_msg = None\n", + " immediate_msg = None\n", + " optional_msg = None\n", + " optional_fn = None\n", + "\n", + " ft_type = infer_task_type(df)\n", + "----------------------------------------------------------------------\n", + "/openai/validators.py:get_outfnames score=0.66\n", + "def get_outfnames(fname, split):\n", + " suffixes = [\"_train\", \"_valid\"] if split else [\"\"]\n", + " i = 0\n", + " while True:\n", + " index_suffix = f\" ({i})\" if i > 0 else \"\"\n", + " candidate_fnames = [\n", + " fname.split(\".\")[0] + \"_prepared\" + suffix + index_suffix + \".jsonl\"\n", + " for suffix in suffixes\n", + " ]\n", + " if not any(os.path.isfile(f) for f in candidate_fnames):\n", + "----------------------------------------------------------------------\n" + ] + } + ], + "source": [ + "res = search_functions(df, 'find common suffix', n=2, n_lines=10)" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "/openai/cli.py:tools_register score=0.651\n", + "def tools_register(parser):\n", + " subparsers = parser.add_subparsers(\n", + " title=\"Tools\", help=\"Convenience client side tools\"\n", + " )\n", + "\n", + " def help(args):\n", + " parser.print_help()\n", + "\n", + " parser.set_defaults(func=help)\n", + "\n", + " sub = subparsers.add_parser(\"fine_tunes.prepare_data\")\n", + " sub.add_argument(\n", + " \"-f\",\n", + " \"--file\",\n", + " required=True,\n", + " help=\"JSONL, JSON, CSV, TSV, TXT or XLSX file containing prompt-completion examples to be analyzed.\"\n", + " \"This should be the local file path.\",\n", + " )\n", + " sub.add_argument(\n", + " \"-q\",\n", + "----------------------------------------------------------------------\n" + ] + } + ], + "source": [ + "res = search_functions(df, 'Command line interface for fine-tuning', n=1, n_lines=20)" + ] + } + ], + "metadata": { + "interpreter": { + "hash": "be4b5d5b73a21c599de40d6deb1129796d12dc1cc33a738f7bac13269cfcafe8" + }, + "kernelspec": { + "display_name": "Python 3.7.3 64-bit ('base': conda)", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.7.3" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/Customizing_embeddings.ipynb b/examples/Customizing_embeddings.ipynb new file mode 100644 index 0000000..4414e9c --- /dev/null +++ b/examples/Customizing_embeddings.ipynb @@ -0,0 +1,20994 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "id": "Vq31CdSRpgkI" + }, + "source": [ + "# Customizing embeddings\n", + "\n", + "This notebook demonstrates one way to customize OpenAI embeddings to a particular task.\n", + "\n", + "The input is training data in the form of [text_1, text_2, label] where label is +1 if the pairs are similar and -1 if the pairs are dissimilar.\n", + "\n", + "The output is a matrix that you can use to multiply your embeddings. The product of this multiplication is a 'custom embedding' that will better emphasize aspects of the text relevant to your use case. In binary classification use cases, we've seen error rates drop by as much as 50%.\n", + "\n", + "In the following example, I use 1,000 sentence pairs picked from the SNLI corpus. Each pair of sentences are logically entailed (i.e., one implies the other). These pairs are our positives (label = 1). We generate synthetic negatives by combining sentences from different pairs, which are presumed to not be logically entailed (label = -1).\n", + "\n", + "For a clustering use case, you can generate positives by creating pairs from texts in the same clusters and generate negatives by creating pairs from sentences in different clusters.\n", + "\n", + "With other data sets, we have seen decent improvement with as little as ~100 training examples. Of course, performance will be better with more examples." + ] + }, + { + "cell_type": "markdown", + "metadata": { + "id": "arB38jFwpgkK" + }, + "source": [ + "# 0. Imports" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "ifvM7g4apgkK" + }, + "outputs": [], + "source": [ + "# imports\n", + "from typing import List, Tuple # for type hints\n", + "\n", + "import numpy as np # for manipulating arrays\n", + "import pandas as pd # for manipulating data in dataframes\n", + "import pickle # for saving the embeddings cache\n", + "import plotly.express as px # for plots\n", + "import random # for generating run IDs\n", + "from sklearn.model_selection import train_test_split # for splitting train & test data\n", + "import torch # for matrix optimization\n", + "\n", + "from openai.embeddings_utils import get_embedding, cosine_similarity # for embeddings\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "id": "DtBbryAapgkL" + }, + "source": [ + "## 1. Inputs\n", + "\n", + "Most inputs are here. The key things to change are where to load your datset from, where to save a cache of embeddings to, and which embedding engine you want to use.\n", + "\n", + "Depending on how your data is formatted, you'll want to rewrite the process_input_data function." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "UzxcWRCkpgkM" + }, + "outputs": [], + "source": [ + "# input parameters\n", + "embedding_cache_path = \"snli_embedding_cache.pkl\" # embeddings will be saved/loaded here\n", + "default_embedding_engine = \"babbage-similarity\" # choice of: ada, babbage, curie, davinci (curie is often a good starting point)\n", + "num_pairs_to_embed = 1000 # 1000 is arbitrary - I've gotten it to work with as little as ~100\n", + "\n", + "local_dataset_path = \"snli_train.csv\" # from: https://nlp.stanford.edu/projects/snli/\n", + "\n", + "def process_input_data(df: pd.DataFrame) -> pd.DataFrame:\n", + " # you can customize this to preprocess your own dataset\n", + " # output should be a dataframe with 3 columns: text_1, text_2, label (1 for similar, -1 for dissimilar)\n", + " df = df[df[\"label\"].isin([\"entailment\"])]\n", + " df[\"label\"] = df[\"label\"].apply(lambda x: {\"entailment\": 1, \"contradiction\": -1}[x])\n", + " df = df.rename(columns={\"sentence1\": \"text_1\", \"sentence2\": \"text_2\"})\n", + " df = df[[\"text_1\", \"text_2\", \"label\"]]\n", + " df = df.head(num_pairs_to_embed)\n", + " return df" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "id": "aBbH71hEpgkM" + }, + "source": [ + "## 2. Load and process input data" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "kAKLjYG6pgkN", + "outputId": "dc178688-e97d-4ad0-b26c-dff67b858966" + }, + "outputs": [ + { + "name": "stderr", + "output_type": "stream", + "text": [ + "/var/folders/r4/x3kdvs816995fnnph2gdpwp40000gn/T/ipykernel_46384/2274810038.py:12: SettingWithCopyWarning:\n", + "\n", + "\n", + "A value is trying to be set on a copy of a slice from a DataFrame.\n", + "Try using .loc[row_indexer,col_indexer] = value instead\n", + "\n", + "See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy\n", + "\n" + ] + }, + { + "data": { + "text/html": [ + "
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    \n", + "
    " + ], + "text/plain": [ + " text_1 \\\n", + "1 A person on a horse jumps over a broken down a... \n", + "2 A person on a horse jumps over a broken down a... \n", + "4 Children smiling and waving at camera \n", + "5 Children smiling and waving at camera \n", + "6 A boy is jumping on skateboard in the middle o... \n", + "\n", + " text_2 label \n", + "1 A person is at a diner, ordering an omelette. -1 \n", + "2 A person is outdoors, on a horse. 1 \n", + "4 There are children present 1 \n", + "5 The kids are frowning -1 \n", + "6 The boy skates down the sidewalk. -1 " + ] + }, + "execution_count": 13, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# load data\n", + "df = pd.read_csv(local_dataset_path)\n", + "\n", + "# process input data\n", + "df = process_input_data(df) # this demonstrates training data containing only positives\n", + "\n", + "# view data\n", + "df.head()" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "id": "z2F1cCoYpgkO" + }, + "source": [ + "## 3. Split data into training test sets\n", + "\n", + "Note that it's important to split data into training and test sets *before* generating synethetic negatives or positives. You don't want any text strings in the training data to show up in the test data. If there's contamination, the test metrics will look better than they'll actually be in production." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "50QmnH2qpgkO", + "outputId": "6144029b-eb29-439e-9990-7aeb28168e56" + }, + "outputs": [ + { + "name": "stderr", + "output_type": "stream", + "text": [ + "/Users/ted/.virtualenvs/openai/lib/python3.9/site-packages/pandas/core/indexing.py:1667: SettingWithCopyWarning:\n", + "\n", + "\n", + "A value is trying to be set on a copy of a slice from a DataFrame.\n", + "Try using .loc[row_indexer,col_indexer] = value instead\n", + "\n", + "See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy\n", + "\n" + ] + } + ], + "source": [ + "# split data into train and test sets\n", + "test_fraction = 0.5 # 0.5 is fairly arbitrary\n", + "random_seed = 123 # random seed is arbitrary, but is helpful in reproducibility\n", + "train_df, test_df = train_test_split(\n", + " df, test_size=test_fraction, stratify=df[\"label\"], random_state=random_seed\n", + ")\n", + "train_df.loc[:, \"dataset\"] = \"train\"\n", + "test_df.loc[:, \"dataset\"] = \"test\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "id": "MzAFkA2opgkP" + }, + "source": [ + "## 4. Generate synthetic negatives\n", + "\n", + "This is another piece of the code that you will need to modify to match your use case.\n", + "\n", + "If you have data with positives and negatives, you can skip this section.\n", + "\n", + "If you have data with only positives, you can mostly keep it as is, where it generates negatives only.\n", + "\n", + "If you have multiclass data, you will want to generate both positives and negatives. The positives can be pairs of text that share labels, and the negatives can be pairs of text that do not share labels.\n", + "\n", + "The final output should be a dataframe with text pairs, where each pair is labeled -1 or 1." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "rUYd9V0zpgkP" + }, + "outputs": [], + "source": [ + "# generate negatives\n", + "def dataframe_of_negatives(dataframe_of_positives: pd.DataFrame) -> pd.DataFrame:\n", + " \"\"\"Return dataframe of negative pairs made by combining elements of positive pairs.\"\"\"\n", + " texts = set(dataframe_of_positives[\"text_1\"].values) | set(dataframe_of_positives[\"text_2\"].values)\n", + " all_pairs = {(t1, t2) for t1 in texts for t2 in texts if t1 < t2}\n", + " positive_pairs = set(tuple(text_pair) for text_pair in dataframe_of_positives[[\"text_1\", \"text_2\"]].values)\n", + " negative_pairs = all_pairs - positive_pairs\n", + " df_of_negatives = pd.DataFrame(list(negative_pairs), columns=[\"text_1\", \"text_2\"])\n", + " df_of_negatives['label'] = -1\n", + " return df_of_negatives" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "Rkh8-J89pgkP" + }, + "outputs": [], + "source": [ + "negatives_per_positive = 1 # it will work at higher values too, but more data will be slower\n", + "# generate negatives for training dataset\n", + "train_df_negatives = dataframe_of_negatives(train_df)\n", + "train_df_negatives[\"dataset\"] = \"train\"\n", + "# generate negatives for test dataset\n", + "test_df_negatives = dataframe_of_negatives(test_df)\n", + "test_df_negatives[\"dataset\"] = \"test\"\n", + "# sample negatives and combine with positives\n", + "train_df = pd.concat([train_df, train_df_negatives.sample(n=len(train_df) * negatives_per_positive, random_state=random_seed)])\n", + "test_df = pd.concat([test_df, test_df_negatives.sample(n=len(test_df) * negatives_per_positive, random_state=random_seed)])\n", + "\n", + "df = pd.concat([train_df, test_df])" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "id": "8MVSLMSrpgkQ" + }, + "source": [ + "## 5. Calculate embeddings and cosine similarities\n", + "\n", + "Here, I create a cache to save the embeddings. This is handy so that you don't have to pay again if you want to run the code again." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "R6tWgS_ApgkQ" + }, + "outputs": [], + "source": [ + "# establish a cache of embeddings to avoid recomputing\n", + "# cache is a dict of tuples (text, engine) -> embedding\n", + "try:\n", + " with open(embedding_cache_path, \"rb\") as f:\n", + " embedding_cache = pickle.load(f)\n", + "except FileNotFoundError:\n", + " embedding_cache = {}\n", + "\n", + "\n", + "# this function will get embeddings from the cache and save them there afterward\n", + "def get_embedding_with_cache(\n", + " text: str,\n", + " engine: str = default_embedding_engine,\n", + " embedding_cache: dict = embedding_cache,\n", + " embedding_cache_path: str = embedding_cache_path,\n", + ") -> list:\n", + " if (text, engine) not in embedding_cache.keys():\n", + " # if not in cache, call API to get embedding\n", + " embedding_cache[(text, engine)] = get_embedding(text, engine)\n", + " # save embeddings cache to disk after each update\n", + " with open(embedding_cache_path, \"wb\") as embedding_cache_file:\n", + " pickle.dump(embedding_cache, embedding_cache_file)\n", + " return embedding_cache[(text, engine)]\n", + "\n", + "\n", + "# create column of embeddings\n", + "for column in [\"text_1\", \"text_2\"]:\n", + " df[f\"{column}_embedding\"] = df[column].apply(get_embedding_with_cache)\n", + "\n", + "# create column of cosine similarity between embeddings\n", + "df[\"cosine_similarity\"] = df.apply(\n", + " lambda row: cosine_similarity(row[\"text_1_embedding\"], row[\"text_2_embedding\"]),\n", + " axis=1,\n", + ")\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "id": "4pwn608LpgkQ" + }, + "source": [ + "## 6. Plot distribution of cosine similarity\n", + "\n", + "Here we measure similarity of text using cosine similarity. In our experience, most distance functions (L1, L2, cosine similarity) all work about the same. Note that our embeddings are already normalized to length 1, so cosine similarity is equivalent to dot product.\n", + "\n", + "The graphs show how much the overlap there is between the distribution of cosine similarities for similar and dissimilar pairs. If there is a high amount of overlap, that means there are some dissimilar pairs with greater cosine similarity than some similar pairs.\n", + "\n", + "The accuracy I compute is the accuracy of a simple rule that predicts 'similar (1)' if the cosine similarity is above some threshold X and otherwise predicts 'dissimilar (0)'." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "SoeDF8vqpgkQ", + "outputId": "17db817e-1702-4089-c4e8-8ca32d294930" + }, + "outputs": [ + { + "data": { + "application/vnd.plotly.v1+json": { + "config": { + "plotlyServerURL": "https://plot.ly" + }, + "data": [ + { + "alignmentgroup": "True", + "bingroup": "x", + "hovertemplate": "label=1
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"yaxis2": { + "anchor": "x2", + "domain": [ + 0.515, + 1 + ], + "matches": "y", + "title": { + "text": "count" + } + } + } + } + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "train accuracy: 88.6% ± 1.4%\n", + "test accuracy: 90.0% ± 1.3%\n" + ] + } + ], + "source": [ + "# calculate accuracy (and its standard error) of predicting label=1 if similarity>x\n", + "# x is optimized by sweeping from -1 to 1 in steps of 0.01\n", + "def accuracy_and_se(cosine_similarity:float, labeled_similarity:int) -> Tuple[float]:\n", + " accuracies = []\n", + " for threshold_thousandths in range(-1000, 1000, 1):\n", + " threshold = threshold_thousandths / 1000\n", + " total = 0\n", + " correct = 0\n", + " for cs, ls in zip(cosine_similarity, labeled_similarity):\n", + " total += 1\n", + " if cs > threshold:\n", + " prediction = 1\n", + " else:\n", + " prediction = -1\n", + " if prediction == ls:\n", + " correct += 1\n", + " accuracy = correct / total\n", + " accuracies.append(accuracy)\n", + " a = max(accuracies)\n", + " n = len(cosine_similarity)\n", + " standard_error = (a * (1 - a) / n) ** 0.5 # standard error of binomial\n", + " return a, standard_error\n", + "\n", + "# check that training and test sets are balanced\n", + "px.histogram(\n", + " df,\n", + " x=\"cosine_similarity\",\n", + " color=\"label\",\n", + " barmode=\"overlay\",\n", + " width=500,\n", + " facet_row=\"dataset\",\n", + ").show()\n", + "\n", + "for dataset in [\"train\", \"test\"]:\n", + " data = df[df[\"dataset\"] == dataset]\n", + " a, se = accuracy_and_se(data[\"cosine_similarity\"], data[\"label\"])\n", + " print(f'{dataset} accuracy: {a:0.1%} ± {1.96 * se:0.1%}')\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "id": "zHLxlnsApgkR" + }, + "source": [ + "## 7. Optimize the matrix using the training data provided" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "z52V0x8IpgkR" + }, + "outputs": [], + "source": [ + "def embedding_multiplied_by_matrix(embedding: List[float], matrix: torch.tensor) -> np.array:\n", + " embedding_tensor = torch.tensor(embedding).float()\n", + " modified_embedding = embedding_tensor @ matrix\n", + " modified_embedding = modified_embedding.detach().numpy()\n", + " return modified_embedding\n", + "\n", + "\n", + "# compute custom embeddings and new cosine similarities\n", + "def apply_matrix_to_embeddings_dataframe(matrix: torch.tensor, df: pd.DataFrame):\n", + " for column in [\"text_1_embedding\", \"text_2_embedding\"]:\n", + " df[f\"{column}_custom\"] = df[column].apply(\n", + " lambda x: embedding_multiplied_by_matrix(x, matrix)\n", + " )\n", + " df[\"cosine_similarity_custom\"] = df.apply(\n", + " lambda row: cosine_similarity(\n", + " row[\"text_1_embedding_custom\"], row[\"text_2_embedding_custom\"]\n", + " ),\n", + " axis=1,\n", + " )\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "p2ZSXu6spgkR" + }, + "outputs": [], + "source": [ + "def optimize_matrix(\n", + " modified_embedding_length: int = 2048, # in my brief experimentation, bigger was better (2048 is length of babbage encoding)\n", + " batch_size: int = 100,\n", + " max_epochs: int = 100,\n", + " learning_rate: float = 100.0, # seemed to work best when similar to batch size - feel free to try a range of values\n", + " dropout_fraction: float = 0.0, # in my testing, dropout helped by a couple percentage points (definitely not necessary)\n", + " df: pd.DataFrame = df,\n", + " print_progress: bool = True,\n", + " save_results: bool = True,\n", + ") -> torch.tensor:\n", + " \"\"\"Return matrix optimized to minimize loss on training data.\"\"\"\n", + " run_id = random.randint(0, 2 ** 31 - 1) # (range is arbitrary)\n", + " # convert from dataframe to torch tensors\n", + " # e is for embedding, s for similarity label\n", + " def tensors_from_dataframe(\n", + " df: pd.DataFrame,\n", + " embedding_column_1: str,\n", + " embedding_column_2: str,\n", + " similarity_label_column: str,\n", + " ) -> Tuple[torch.tensor]:\n", + " e1 = np.stack(np.array(df[embedding_column_1].values))\n", + " e2 = np.stack(np.array(df[embedding_column_2].values))\n", + " s = np.stack(np.array(df[similarity_label_column].astype(\"float\").values))\n", + "\n", + " e1 = torch.from_numpy(e1).float()\n", + " e2 = torch.from_numpy(e2).float()\n", + " s = torch.from_numpy(s).float()\n", + "\n", + " return e1, e2, s\n", + "\n", + " e1_train, e2_train, s_train = tensors_from_dataframe(\n", + " df[df[\"dataset\"] == \"train\"], \"text_1_embedding\", \"text_2_embedding\", \"label\"\n", + " )\n", + " e1_test, e2_test, s_test = tensors_from_dataframe(\n", + " df[df[\"dataset\"] == \"train\"], \"text_1_embedding\", \"text_2_embedding\", \"label\"\n", + " )\n", + "\n", + " # create dataset and loader\n", + " dataset = torch.utils.data.TensorDataset(e1_train, e2_train, s_train)\n", + " train_loader = torch.utils.data.DataLoader(\n", + " dataset, batch_size=batch_size, shuffle=True\n", + " )\n", + "\n", + " # define model (similarity of projected embeddings)\n", + " def model(embedding_1, embedding_2, matrix, dropout_fraction=dropout_fraction):\n", + " e1 = torch.nn.functional.dropout(embedding_1, p=dropout_fraction)\n", + " e2 = torch.nn.functional.dropout(embedding_2, p=dropout_fraction)\n", + " modified_embedding_1 = e1 @ matrix # @ is matrix multiplication\n", + " modified_embedding_2 = e2 @ matrix\n", + " similarity = torch.nn.functional.cosine_similarity(\n", + " modified_embedding_1, modified_embedding_2\n", + " )\n", + " return similarity\n", + "\n", + " # define loss function to minimize\n", + " def mse_loss(predictions, targets):\n", + " difference = predictions - targets\n", + " return torch.sum(difference * difference) / difference.numel()\n", + "\n", + " # initialize projection matrix\n", + " embedding_length = len(df[\"text_1_embedding\"].values[0])\n", + " matrix = torch.randn(\n", + " embedding_length, modified_embedding_length, requires_grad=True\n", + " )\n", + "\n", + " epochs, types, losses, accuracies, matrices = [], [], [], [], []\n", + " for epoch in range(1, 1 + max_epochs):\n", + " # iterate through training dataloader\n", + " for a, b, actual_similarity in train_loader:\n", + " # generate prediction\n", + " predicted_similarity = model(a, b, matrix)\n", + " # get loss and perform backpropagation\n", + " loss = mse_loss(predicted_similarity, actual_similarity)\n", + " loss.backward()\n", + " # update the weights\n", + " with torch.no_grad():\n", + " matrix -= matrix.grad * learning_rate\n", + " # set gradients to zero\n", + " matrix.grad.zero_()\n", + " # calculate test loss\n", + " test_predictions = model(e1_test, e2_test, matrix)\n", + " test_loss = mse_loss(test_predictions, s_test)\n", + "\n", + " # compute custom embeddings and new cosine similarities\n", + " apply_matrix_to_embeddings_dataframe(matrix, df)\n", + "\n", + " # calculate test accuracy\n", + " for dataset in [\"train\", \"test\"]:\n", + " data = df[df[\"dataset\"] == dataset]\n", + " a, se = accuracy_and_se(data[\"cosine_similarity_custom\"], data[\"label\"])\n", + "\n", + " # record results of each epoch\n", + " epochs.append(epoch)\n", + " types.append(dataset)\n", + " losses.append(loss.item() if dataset == \"train\" else test_loss.item())\n", + " accuracies.append(a)\n", + " matrices.append(matrix.detach().numpy())\n", + "\n", + " # optionally print accuracies\n", + " if print_progress is True:\n", + " print(\n", + " f\"Epoch {epoch}/{max_epochs}: {dataset} accuracy: {a:0.1%} ± {1.96 * se:0.1%}\"\n", + " )\n", + "\n", + " data = pd.DataFrame(\n", + " {\"epoch\": epochs, \"type\": types, \"loss\": losses, \"accuracy\": accuracies}\n", + " )\n", + " data[\"run_id\"] = run_id\n", + " data[\"modified_embedding_length\"] = modified_embedding_length\n", + " data[\"batch_size\"] = batch_size\n", + " data[\"max_epochs\"] = max_epochs\n", + " data[\"learning_rate\"] = learning_rate\n", + " data[\"dropout_fraction\"] = dropout_fraction\n", + " data[\n", + " \"matrix\"\n", + " ] = matrices # saving every single matrix can get big; feel free to delete/change\n", + " if save_results is True:\n", + " data.to_csv(f\"{run_id}_optimization_results.csv\", index=False)\n", + "\n", + " return data\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "nlcUW-zEpgkS", + "outputId": "4bd4bdff-628a-406f-fffe-aedbfad66446" + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Epoch 1/30: train accuracy: 89.7% ± 1.3%\n", + "Epoch 1/30: test accuracy: 91.8% ± 1.2%\n", + "Epoch 2/30: train accuracy: 91.0% ± 1.3%\n", + "Epoch 2/30: test accuracy: 92.3% ± 1.2%\n", + "Epoch 3/30: train accuracy: 91.5% ± 1.2%\n", + "Epoch 3/30: test accuracy: 93.0% ± 1.1%\n", + "Epoch 4/30: train accuracy: 92.0% ± 1.2%\n", + "Epoch 4/30: test accuracy: 93.2% ± 1.1%\n", + "Epoch 5/30: train accuracy: 92.7% ± 1.1%\n", + "Epoch 5/30: test accuracy: 93.7% ± 1.1%\n", + "Epoch 6/30: train accuracy: 92.8% ± 1.1%\n", + "Epoch 6/30: test accuracy: 94.0% ± 1.0%\n", + "Epoch 7/30: train accuracy: 93.0% ± 1.1%\n", + "Epoch 7/30: test accuracy: 94.2% ± 1.0%\n", + "Epoch 8/30: train accuracy: 93.2% ± 1.1%\n", + "Epoch 8/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 9/30: train accuracy: 93.2% ± 1.1%\n", + "Epoch 9/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 10/30: train accuracy: 93.5% ± 1.1%\n", + "Epoch 10/30: test accuracy: 94.6% ± 1.0%\n", + "Epoch 11/30: train accuracy: 93.7% ± 1.1%\n", + "Epoch 11/30: test accuracy: 94.7% ± 1.0%\n", + "Epoch 12/30: train accuracy: 93.8% ± 1.1%\n", + "Epoch 12/30: test accuracy: 94.8% ± 1.0%\n", + "Epoch 13/30: train accuracy: 93.7% ± 1.1%\n", + "Epoch 13/30: test accuracy: 94.8% ± 1.0%\n", + "Epoch 14/30: train accuracy: 93.8% ± 1.1%\n", + "Epoch 14/30: test accuracy: 94.8% ± 1.0%\n", + "Epoch 15/30: train accuracy: 93.8% ± 1.1%\n", + "Epoch 15/30: test accuracy: 94.7% ± 1.0%\n", + "Epoch 16/30: train accuracy: 93.8% ± 1.1%\n", + "Epoch 16/30: test accuracy: 94.8% ± 1.0%\n", + "Epoch 17/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 17/30: test accuracy: 94.8% ± 1.0%\n", + "Epoch 18/30: train accuracy: 93.9% ± 1.0%\n", + "Epoch 18/30: test accuracy: 94.7% ± 1.0%\n", + "Epoch 19/30: train accuracy: 93.8% ± 1.1%\n", + "Epoch 19/30: test accuracy: 94.8% ± 1.0%\n", + "Epoch 20/30: train accuracy: 93.7% ± 1.1%\n", + "Epoch 20/30: test accuracy: 94.8% ± 1.0%\n", + "Epoch 21/30: train accuracy: 93.9% ± 1.0%\n", + "Epoch 21/30: test accuracy: 94.8% ± 1.0%\n", + "Epoch 22/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 22/30: test accuracy: 94.7% ± 1.0%\n", + "Epoch 23/30: train accuracy: 93.9% ± 1.0%\n", + "Epoch 23/30: test accuracy: 94.7% ± 1.0%\n", + "Epoch 24/30: train accuracy: 93.9% ± 1.0%\n", + "Epoch 24/30: test accuracy: 94.7% ± 1.0%\n", + "Epoch 25/30: train accuracy: 93.8% ± 1.1%\n", + "Epoch 25/30: test accuracy: 94.6% ± 1.0%\n", + "Epoch 26/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 26/30: test accuracy: 94.7% ± 1.0%\n", + "Epoch 27/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 27/30: test accuracy: 94.6% ± 1.0%\n", + "Epoch 28/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 28/30: test accuracy: 94.6% ± 1.0%\n", + "Epoch 29/30: train accuracy: 93.8% ± 1.1%\n", + "Epoch 29/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 30/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 30/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 1/30: train accuracy: 89.9% ± 1.3%\n", + "Epoch 1/30: test accuracy: 91.8% ± 1.2%\n", + "Epoch 2/30: train accuracy: 91.1% ± 1.2%\n", + "Epoch 2/30: test accuracy: 92.3% ± 1.2%\n", + "Epoch 3/30: train accuracy: 91.8% ± 1.2%\n", + "Epoch 3/30: test accuracy: 92.8% ± 1.1%\n", + "Epoch 4/30: train accuracy: 92.3% ± 1.2%\n", + "Epoch 4/30: test accuracy: 93.2% ± 1.1%\n", + "Epoch 5/30: train accuracy: 92.8% ± 1.1%\n", + "Epoch 5/30: test accuracy: 93.5% ± 1.1%\n", + "Epoch 6/30: train accuracy: 93.0% ± 1.1%\n", + "Epoch 6/30: test accuracy: 93.7% ± 1.1%\n", + "Epoch 7/30: train accuracy: 93.2% ± 1.1%\n", + "Epoch 7/30: test accuracy: 94.0% ± 1.0%\n", + "Epoch 8/30: train accuracy: 93.5% ± 1.1%\n", + "Epoch 8/30: test accuracy: 94.1% ± 1.0%\n", + "Epoch 9/30: train accuracy: 93.7% ± 1.1%\n", + "Epoch 9/30: test accuracy: 94.3% ± 1.0%\n", + "Epoch 10/30: train accuracy: 93.7% ± 1.1%\n", + "Epoch 10/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 11/30: train accuracy: 93.8% ± 1.1%\n", + "Epoch 11/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 12/30: train accuracy: 93.8% ± 1.1%\n", + "Epoch 12/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 13/30: train accuracy: 93.9% ± 1.0%\n", + "Epoch 13/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 14/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 14/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 15/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 15/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 16/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 16/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 17/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 17/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 18/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 18/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 19/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 19/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 20/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 20/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 21/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 21/30: test accuracy: 94.4% ± 1.0%\n", + "Epoch 22/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 22/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 23/30: train accuracy: 93.9% ± 1.0%\n", + "Epoch 23/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 24/30: train accuracy: 93.9% ± 1.0%\n", + "Epoch 24/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 25/30: train accuracy: 93.9% ± 1.0%\n", + "Epoch 25/30: test accuracy: 94.4% ± 1.0%\n", + "Epoch 26/30: train accuracy: 93.9% ± 1.0%\n", + "Epoch 26/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 27/30: train accuracy: 93.9% ± 1.0%\n", + "Epoch 27/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 28/30: train accuracy: 93.8% ± 1.1%\n", + "Epoch 28/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 29/30: train accuracy: 93.8% ± 1.1%\n", + "Epoch 29/30: test accuracy: 94.4% ± 1.0%\n", + "Epoch 30/30: train accuracy: 93.8% ± 1.1%\n", + "Epoch 30/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 1/30: train accuracy: 89.8% ± 1.3%\n", + "Epoch 1/30: test accuracy: 92.0% ± 1.2%\n", + "Epoch 2/30: train accuracy: 91.1% ± 1.2%\n", + "Epoch 2/30: test accuracy: 92.5% ± 1.2%\n", + "Epoch 3/30: train accuracy: 91.7% ± 1.2%\n", + "Epoch 3/30: test accuracy: 92.7% ± 1.1%\n", + "Epoch 4/30: train accuracy: 92.4% ± 1.2%\n", + "Epoch 4/30: test accuracy: 93.0% ± 1.1%\n", + "Epoch 5/30: train accuracy: 92.8% ± 1.1%\n", + "Epoch 5/30: test accuracy: 93.4% ± 1.1%\n", + "Epoch 6/30: train accuracy: 93.0% ± 1.1%\n", + "Epoch 6/30: test accuracy: 93.8% ± 1.1%\n", + "Epoch 7/30: train accuracy: 93.3% ± 1.1%\n", + "Epoch 7/30: test accuracy: 94.1% ± 1.0%\n", + "Epoch 8/30: train accuracy: 93.6% ± 1.1%\n", + "Epoch 8/30: test accuracy: 94.2% ± 1.0%\n", + "Epoch 9/30: train accuracy: 93.7% ± 1.1%\n", + "Epoch 9/30: test accuracy: 94.3% ± 1.0%\n", + "Epoch 10/30: train accuracy: 93.8% ± 1.1%\n", + "Epoch 10/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 11/30: train accuracy: 93.9% ± 1.0%\n", + "Epoch 11/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 12/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 12/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 13/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 13/30: test accuracy: 94.8% ± 1.0%\n", + "Epoch 14/30: train accuracy: 94.1% ± 1.0%\n", + "Epoch 14/30: test accuracy: 94.7% ± 1.0%\n", + "Epoch 15/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 15/30: test accuracy: 94.6% ± 1.0%\n", + "Epoch 16/30: train accuracy: 94.1% ± 1.0%\n", + "Epoch 16/30: test accuracy: 94.7% ± 1.0%\n", + "Epoch 17/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 17/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 18/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 18/30: test accuracy: 94.6% ± 1.0%\n", + "Epoch 19/30: train accuracy: 94.1% ± 1.0%\n", + "Epoch 19/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 20/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 20/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 21/30: train accuracy: 94.1% ± 1.0%\n", + "Epoch 21/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 22/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 22/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 23/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 23/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 24/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 24/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 25/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 25/30: test accuracy: 94.4% ± 1.0%\n", + "Epoch 26/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 26/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 27/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 27/30: test accuracy: 94.3% ± 1.0%\n", + "Epoch 28/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 28/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 29/30: train accuracy: 94.0% ± 1.0%\n", + "Epoch 29/30: test accuracy: 94.5% ± 1.0%\n", + "Epoch 30/30: train accuracy: 93.9% ± 1.0%\n", + "Epoch 30/30: test accuracy: 94.5% ± 1.0%\n" + ] + } + ], + "source": [ + "# example hyperparameter search\n", + "# I recommend starting with max_epochs=10 while initially exploring\n", + "results = []\n", + "max_epochs = 30\n", + "dropout_fraction = 0.2\n", + "for batch_size, learning_rate in [(10, 10), (100, 100), (1000, 1000)]:\n", + " result = optimize_matrix(\n", + " batch_size=batch_size,\n", + " learning_rate=learning_rate,\n", + " max_epochs=max_epochs,\n", + " dropout_fraction=dropout_fraction,\n", + " save_results=False\n", + " )\n", + " results.append(result)\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + 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Plot the before & after, showing the results of the best matrix found during training\n", + "\n", + "The better the matrix is, the more cleanly it will separate the similar and dissimilar pairs." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "hzjoyLDOpgkS" + }, + "outputs": [], + "source": [ + "# apply result of best run to original data\n", + "best_run = runs_df.sort_values(by='accuracy', ascending=False).iloc[0]\n", + "best_matrix = best_run['matrix']\n", + "apply_matrix_to_embeddings_dataframe(best_matrix, df)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "nLnvABnXpgkS", + "outputId": "0c070faa-6e3e-4765-b082-565c72a609be" + }, + "outputs": [ + { + "data": { + "application/vnd.plotly.v1+json": { + "config": { + "plotlyServerURL": "https://plot.ly" + }, + "data": [ + { + "alignmentgroup": "True", + "bingroup": "x", + "hovertemplate": "label=1
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+ "gridcolor": "white", + "linecolor": "white", + "ticks": "", + "title": { + "standoff": 15 + }, + "zerolinecolor": "white", + "zerolinewidth": 2 + } + } + }, + "width": 500, + "xaxis": { + "anchor": "y", + "domain": [ + 0, + 0.98 + ], + "title": { + "text": "cosine_similarity_custom" + } + }, + "xaxis2": { + "anchor": "y2", + "domain": [ + 0, + 0.98 + ], + "matches": "x", + "showticklabels": false + }, + "yaxis": { + "anchor": "x", + "domain": [ + 0, + 0.485 + ], + "title": { + "text": "count" + } + }, + "yaxis2": { + "anchor": "x2", + "domain": [ + 0.515, + 1 + ], + "matches": "y", + "title": { + "text": "count" + } + } + } + } + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Test accuracy after customization: 94.5% ± 1.0%\n" + ] + } + ], + "source": [ + "# plot similarity distribution BEFORE customization\n", + "px.histogram(\n", + " df,\n", + " x=\"cosine_similarity\",\n", + " color=\"label\",\n", + " barmode=\"overlay\",\n", + " width=500,\n", + " facet_row=\"dataset\",\n", + ").show()\n", + "\n", + "test_df = df[df['dataset'] == 'test']\n", + "a, se = accuracy_and_se(test_df[\"cosine_similarity\"], test_df[\"label\"])\n", + "print(f'Test accuracy: {a:0.1%} ± {1.96 * se:0.1%}')\n", + "\n", + "# plot similarity distribution AFTER customization\n", + "px.histogram(\n", + " df,\n", + " x=\"cosine_similarity_custom\",\n", + " color=\"label\",\n", + " barmode=\"overlay\",\n", + " width=500,\n", + " facet_row=\"dataset\",\n", + ").show()\n", + "\n", + "a, se = accuracy_and_se(test_df[\"cosine_similarity_custom\"], test_df[\"label\"])\n", + "print(f'Test accuracy after customization: {a:0.1%} ± {1.96 * se:0.1%}')\n", + "\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "XO7iqiVjpgkT", + "outputId": "a100a9e0-d5aa-46ab-b8a7-4ec6f7bd1cec" + }, + "outputs": [ + { + "data": { + "text/plain": [ + "array([[-1.3517092 , 0.28163946, 0.9140799 , ..., 0.12247127,\n", + " -0.1615858 , -0.08879007],\n", + " [ 0.8833686 , -0.844946 , 2.1167974 , ..., -0.63856494,\n", + " 0.37139723, -0.5635943 ],\n", + " [ 0.2253975 , -0.58070326, -0.8530685 , ..., 0.61836195,\n", + " 1.5728145 , 0.39455906],\n", + " ...,\n", + " [ 0.00959156, 0.10840097, 0.9884689 , ..., -0.25716397,\n", + " 0.79389465, 0.16435517],\n", + " [-1.006916 , 1.0257064 , 0.32284895, ..., -0.37429097,\n", + " -0.57910526, -1.4773123 ],\n", + " [ 0.01498076, 1.5009826 , -1.0185508 , ..., 0.4380649 ,\n", + " 0.42969966, 1.0704373 ]], dtype=float32)" + ] + }, + "execution_count": 20, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "best_matrix # this is what you can multiply your embeddings by\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "id": "dPF-sczmpgkT" + }, + "outputs": [], + "source": [ + "" + ] + } + ], + "metadata": { + "interpreter": { + "hash": "365536dcbde60510dc9073d6b991cd35db2d9bac356a11f5b64279a5e6708b97" + }, + "kernelspec": { + "display_name": "Python 3.9.9 ('openai')", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.9.9" + }, + "orig_nbformat": 4, + "colab": { + "name": "customized_embeddings_example_with_synthetic_negatives.ipynb", + "provenance": [] + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} \ No newline at end of file diff --git a/examples/Fine-tuned_classification.ipynb b/examples/Fine-tuned_classification.ipynb new file mode 100644 index 0000000..60b8896 --- /dev/null +++ b/examples/Fine-tuned_classification.ipynb @@ -0,0 +1,740 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "source": [ + "# Fine tuning classification example\n", + "\n", + "We will fine-tune an ada classifier to distinguish between the two sports: Baseball and Hockey." + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 1, + "source": [ + "from sklearn.datasets import fetch_20newsgroups\n", + "import pandas as pd\n", + "import openai\n", + "\n", + "categories = ['rec.sport.baseball', 'rec.sport.hockey']\n", + "sports_dataset = fetch_20newsgroups(subset='train', shuffle=True, random_state=42, categories=categories)" + ], + "outputs": [], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + " ## Data exploration\n", + " The newsgroup dataset can be loaded using sklearn. First we will look at the data itself:" + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 2, + "source": [ + "print(sports_dataset['data'][0])" + ], + "outputs": [ + { + "output_type": "stream", + "name": "stdout", + "text": [ + "From: dougb@comm.mot.com (Doug Bank)\n", + "Subject: Re: Info needed for Cleveland tickets\n", + "Reply-To: dougb@ecs.comm.mot.com\n", + "Organization: Motorola Land Mobile Products Sector\n", + "Distribution: usa\n", + "Nntp-Posting-Host: 145.1.146.35\n", + "Lines: 17\n", + "\n", + "In article <1993Apr1.234031.4950@leland.Stanford.EDU>, bohnert@leland.Stanford.EDU (matthew bohnert) writes:\n", + "\n", + "|> I'm going to be in Cleveland Thursday, April 15 to Sunday, April 18.\n", + "|> Does anybody know if the Tribe will be in town on those dates, and\n", + "|> if so, who're they playing and if tickets are available?\n", + "\n", + "The tribe will be in town from April 16 to the 19th.\n", + "There are ALWAYS tickets available! (Though they are playing Toronto,\n", + "and many Toronto fans make the trip to Cleveland as it is easier to\n", + "get tickets in Cleveland than in Toronto. Either way, I seriously\n", + "doubt they will sell out until the end of the season.)\n", + "\n", + "-- \n", + "Doug Bank Private Systems Division\n", + "dougb@ecs.comm.mot.com Motorola Communications Sector\n", + "dougb@nwu.edu Schaumburg, Illinois\n", + "dougb@casbah.acns.nwu.edu 708-576-8207 \n", + "\n" + ] + } + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 3, + "source": [ + "sports_dataset.target_names[sports_dataset['target'][0]]\n" + ], + "outputs": [ + { + "output_type": "execute_result", + "data": { + "text/plain": [ + "'rec.sport.baseball'" + ] + }, + "metadata": {}, + "execution_count": 3 + } + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 4, + "source": [ + "len_all, len_baseball, len_hockey = len(sports_dataset.data), len([e for e in sports_dataset.target if e == 0]), len([e for e in sports_dataset.target if e == 1])\n", + "print(f\"Total examples: {len_all}, Baseball examples: {len_baseball}, Hockey examples: {len_hockey}\")" + ], + "outputs": [ + { + "output_type": "stream", + "name": "stdout", + "text": [ + "Total examples: 1197, Baseball examples: 597, Hockey examples: 600\n" + ] + } + ], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + "One sample from the baseball category can be seen above. It is an email to a mailing list. We can observe that we have 1197 examples in total, which are evenly split between the two sports." + ], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + "## Data Preparation\n", + "We transform the dataset into a pandas dataframe, with a column for prompt and completion. The prompt contains the email from the mailing list, and the completion is a name of the sport, either hockey or baseball. For demonstration purposes only and speed of fine-tuning we take only 300 examples. In a real use case the more examples the better the performance." + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 5, + "source": [ + "import pandas as pd\n", + "\n", + "labels = [sports_dataset.target_names[x].split('.')[-1] for x in sports_dataset['target']]\n", + "texts = [text.strip() for text in sports_dataset['data']]\n", + "df = pd.DataFrame(zip(texts, labels), columns = ['prompt','completion']) #[:300]\n", + "df.head()" + ], + "outputs": [ + { + "output_type": "execute_result", + "data": { + "text/plain": [ + " prompt completion\n", + "0 From: dougb@comm.mot.com (Doug Bank)\\nSubject:... baseball\n", + "1 From: gld@cunixb.cc.columbia.edu (Gary L Dare)... hockey\n", + "2 From: rudy@netcom.com (Rudy Wade)\\nSubject: Re... baseball\n", + "3 From: monack@helium.gas.uug.arizona.edu (david... hockey\n", + "4 Subject: Let it be Known\\nFrom: \n", + "\n", + "\n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + "
    promptcompletion
    0From: dougb@comm.mot.com (Doug Bank)\\nSubject:...baseball
    1From: gld@cunixb.cc.columbia.edu (Gary L Dare)...hockey
    2From: rudy@netcom.com (Rudy Wade)\\nSubject: Re...baseball
    3From: monack@helium.gas.uug.arizona.edu (david...hockey
    4Subject: Let it be Known\\nFrom: <ISSBTL@BYUVM....baseball
    \n", + "" + ] + }, + "metadata": {}, + "execution_count": 5 + } + ], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + "Both baseball and hockey are single tokens. We save the dataset as a jsonl file." + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 6, + "source": [ + "df.to_json(\"sport2.jsonl\", orient='records', lines=True)" + ], + "outputs": [], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + "### Data Preparation tool\n", + "We can now use a data preparation tool which will suggest a few improvements to our dataset before fine-tuning. Before launching the tool we update the openai library to ensure we're using the latest data preparation tool. We additionally specify `-q` which auto-accepts all suggestions." + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 7, + "source": [ + "!pip install --upgrade openai" + ], + "outputs": [], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 8, + "source": [ + "!openai tools fine_tunes.prepare_data -f sport2.jsonl -q" + ], + "outputs": [ + { + "output_type": "stream", + "name": "stdout", + "text": [ + "Analyzing...\n", + "\n", + "- Your file contains 1197 prompt-completion pairs\n", + "- Based on your data it seems like you're trying to fine-tune a model for classification\n", + "- For classification, we recommend you try one of the faster and cheaper models, such as `ada`\n", + "- For classification, you can estimate the expected model performance by keeping a held out dataset, which is not used for training\n", + "- There are 11 examples that are very long. These are rows: [134, 200, 281, 320, 404, 595, 704, 838, 1113, 1139, 1174]\n", + "For conditional generation, and for classification the examples shouldn't be longer than 2048 tokens.\n", + "- Your data does not contain a common separator at the end of your prompts. Having a separator string appended to the end of the prompt makes it clearer to the fine-tuned model where the completion should begin. See https://beta.openai.com/docs/guides/fine-tuning/preparing-your-dataset for more detail and examples. If you intend to do open-ended generation, then you should leave the prompts empty\n", + "- The completion should start with a whitespace character (` `). This tends to produce better results due to the tokenization we use. See https://beta.openai.com/docs/guides/fine-tuning/preparing-your-dataset for more details\n", + "\n", + "Based on the analysis we will perform the following actions:\n", + "- [Recommended] Remove 11 long examples [Y/n]: Y\n", + "- [Recommended] Add a suffix separator `\\n\\n###\\n\\n` to all prompts [Y/n]: Y\n", + "- [Recommended] Add a whitespace character to the beginning of the completion [Y/n]: Y\n", + "- [Recommended] Would you like to split into training and validation set? [Y/n]: Y\n", + "\n", + "\n", + "Your data will be written to a new JSONL file. Proceed [Y/n]: Y\n", + "\n", + "Wrote modified files to `sport2_prepared_train.jsonl` and `sport2_prepared_valid.jsonl`\n", + "Feel free to take a look!\n", + "\n", + "Now use that file when fine-tuning:\n", + "> openai api fine_tunes.create -t \"sport2_prepared_train.jsonl\" -v \"sport2_prepared_valid.jsonl\" --compute_classification_metrics --classification_positive_class \" baseball\"\n", + "\n", + "After you’ve fine-tuned a model, remember that your prompt has to end with the indicator string `\\n\\n###\\n\\n` for the model to start generating completions, rather than continuing with the prompt.\n", + "Once your model starts training, it'll approximately take 30.8 minutes to train a `curie` model, and less for `ada` and `babbage`. Queue will approximately take half an hour per job ahead of you.\n" + ] + } + ], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + "The tool helpfully suggests a few improvements to the dataset and splits the dataset into training and validation set.\n", + "\n", + "A suffix between a prompt and a completion is necessary to tell the model that the input text has stopped, and that it now needs to predict the class. Since we use the same separator in each example, the model is able to learn that it is meant to predict either baseball or hockey following the separator.\n", + "A whitespace prefix in completions is useful, as most word tokens are tokenized with a space prefix.\n", + "The tool also recognized that this is likely a classification task, so it suggested to split the dataset into training and validation datasets. This will allow us to easily measure expected performance on new data." + ], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + "## Fine-tuning\n", + "The tool suggests we run the following command to train the dataset. Since this is a classification task, we would like to know what the generalization performance on the provided validation set is for our classification use case. The tool suggests to add `--compute_classification_metrics --classification_positive_class \" baseball\"` in order to compute the classification metrics.\n", + "\n", + "We can simply copy the suggested command from the CLI tool. We specifically add `-m ada` to fine-tune a cheaper and faster ada model, which is usually comperable in performance to slower and more expensive models on classification use cases. " + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 9, + "source": [ + "!openai api fine_tunes.create -t \"sport2_prepared_train.jsonl\" -v \"sport2_prepared_valid.jsonl\" --compute_classification_metrics --classification_positive_class \" baseball\" -m ada" + ], + "outputs": [ + { + "output_type": "stream", + "name": "stdout", + "text": [ + "Upload progress: 100%|████████████████████| 1.52M/1.52M [00:00<00:00, 1.81Mit/s]\n", + "Uploaded file from sport2_prepared_train.jsonl: file-Dxx2xJqyjcwlhfDHpZdmCXlF\n", + "Upload progress: 100%|███████████████████████| 388k/388k [00:00<00:00, 507kit/s]\n", + "Uploaded file from sport2_prepared_valid.jsonl: file-Mvb8YAeLnGdneSAFcfiVcgcN\n", + "Created fine-tune: ft-2zaA7qi0rxJduWQpdvOvmGn3\n", + "Streaming events until fine-tuning is complete...\n", + "\n", + "(Ctrl-C will interrupt the stream, but not cancel the fine-tune)\n", + "[2021-07-30 13:15:50] Created fine-tune: ft-2zaA7qi0rxJduWQpdvOvmGn3\n", + "[2021-07-30 13:15:52] Fine-tune enqueued. Queue number: 0\n", + "[2021-07-30 13:15:56] Fine-tune started\n", + "[2021-07-30 13:18:55] Completed epoch 1/4\n", + "[2021-07-30 13:20:47] Completed epoch 2/4\n", + "[2021-07-30 13:22:40] Completed epoch 3/4\n", + "[2021-07-30 13:24:31] Completed epoch 4/4\n", + "[2021-07-30 13:26:22] Uploaded model: ada:ft-openai-2021-07-30-12-26-20\n", + "[2021-07-30 13:26:27] Uploaded result file: file-6Ki9RqLQwkChGsr9CHcr1ncg\n", + "[2021-07-30 13:26:28] Fine-tune succeeded\n", + "\n", + "Job complete! Status: succeeded 🎉\n", + "Try out your fine-tuned model:\n", + "\n", + "openai api completions.create -m ada:ft-openai-2021-07-30-12-26-20 -p \n" + ] + } + ], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + "The model is successfully trained in about ten minutes. We can see the model name is `ada:ft-openai-2021-07-30-12-26-20`, which we can use for doing inference." + ], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + "### [Advanced] Results and expected model performance\n", + "We can now download the results file to observe the expected performance on a held out validation set." + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 10, + "source": [ + "!openai api fine_tunes.results -i ft-2zaA7qi0rxJduWQpdvOvmGn3 > result.csv" + ], + "outputs": [], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 11, + "source": [ + "results = pd.read_csv('result.csv')\n", + "results[results['classification/accuracy'].notnull()].tail(1)" + ], + "outputs": [ + { + "output_type": "execute_result", + "data": { + "text/plain": [ + " step elapsed_tokens elapsed_examples training_loss \\\n", + "929 930 3027688 3720 0.044408 \n", + "\n", + " training_sequence_accuracy training_token_accuracy \\\n", + "929 1.0 1.0 \n", + "\n", + " classification/accuracy classification/precision classification/recall \\\n", + "929 0.991597 0.983471 1.0 \n", + "\n", + " classification/auroc classification/auprc classification/f1.0 \\\n", + "929 1.0 1.0 0.991667 \n", + "\n", + " validation_loss validation_sequence_accuracy validation_token_accuracy \n", + "929 NaN NaN NaN " + ], + "text/html": [ + "
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" + }, + "metadata": { + "needs_background": "light" + } + } + ], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + "## Using the model\n", + "We can now call the model to get the predictions." + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 13, + "source": [ + "test = pd.read_json('sport2_prepared_valid.jsonl', lines=True)\n", + "test.head()" + ], + "outputs": [ + { + "output_type": "execute_result", + "data": { + "text/plain": [ + " prompt completion\n", + "0 From: gld@cunixb.cc.columbia.edu (Gary L Dare)... hockey\n", + "1 From: smorris@venus.lerc.nasa.gov (Ron Morris ... hockey\n", + "2 From: golchowy@alchemy.chem.utoronto.ca (Geral... hockey\n", + "3 From: krattige@hpcc01.corp.hp.com (Kim Krattig... baseball\n", + "4 From: warped@cs.montana.edu (Doug Dolven)\\nSub... baseball" + ], + "text/html": [ + "
    \n", + "\n", + "\n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + "
    promptcompletion
    0From: gld@cunixb.cc.columbia.edu (Gary L Dare)...hockey
    1From: smorris@venus.lerc.nasa.gov (Ron Morris ...hockey
    2From: golchowy@alchemy.chem.utoronto.ca (Geral...hockey
    3From: krattige@hpcc01.corp.hp.com (Kim Krattig...baseball
    4From: warped@cs.montana.edu (Doug Dolven)\\nSub...baseball
    \n", + "
    " + ] + }, + "metadata": {}, + "execution_count": 13 + } + ], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + "We need to use the same separator following the prompt which we used during fine-tuning. In this case it is `\\n\\n###\\n\\n`. Since we're concerned with classification, we want the temperature to be as low as possible, and we only require one token completion to determine the prediction of the model." + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 14, + "source": [ + "ft_model = 'ada:ft-openai-2021-07-30-12-26-20'\n", + "res = openai.Completion.create(model=ft_model, prompt=test['prompt'][0] + '\\n\\n###\\n\\n', max_tokens=1, temperature=0)\n", + "res['choices'][0]['text']\n" + ], + "outputs": [ + { + "output_type": "execute_result", + "data": { + "text/plain": [ + "' hockey'" + ] + }, + "metadata": {}, + "execution_count": 14 + } + ], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + "To get the log probabilities, we can specify logprobs parameter on the completion request" + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 15, + "source": [ + "res = openai.Completion.create(model=ft_model, prompt=test['prompt'][0] + '\\n\\n###\\n\\n', max_tokens=1, temperature=0, logprobs=2)\n", + "res['choices'][0]['logprobs']['top_logprobs'][0]" + ], + "outputs": [ + { + "output_type": "execute_result", + "data": { + "text/plain": [ + " JSON: {\n", + " \" baseball\": -7.6311407,\n", + " \" hockey\": -0.0006307676\n", + "}" + ] + }, + "metadata": {}, + "execution_count": 15 + } + ], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + "We can see that the model predicts hockey as a lot more likely than baseball, which is the correct prediction. By requesting log_probs, we can see the prediction (log) probability for each class." + ], + "metadata": {} + }, + { + "cell_type": "markdown", + "source": [ + "### Generalization\n", + "Interestingly, our fine-tuned classifier is quite versatile. Despite being trained on emails to different mailing lists, it also successfully predicts tweets." + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 16, + "source": [ + "sample_hockey_tweet = \"\"\"Thank you to the \n", + "@Canes\n", + " and all you amazing Caniacs that have been so supportive! You guys are some of the best fans in the NHL without a doubt! Really excited to start this new chapter in my career with the \n", + "@DetroitRedWings\n", + " !!\"\"\"\n", + "res = openai.Completion.create(model=ft_model, prompt=sample_hockey_tweet + '\\n\\n###\\n\\n', max_tokens=1, temperature=0, logprobs=2)\n", + "res['choices'][0]['text']" + ], + "outputs": [ + { + "output_type": "execute_result", + "data": { + "text/plain": [ + "' hockey'" + ] + }, + "metadata": {}, + "execution_count": 16 + } + ], + "metadata": {} + }, + { + "cell_type": "code", + "execution_count": 17, + "source": [ + "sample_baseball_tweet=\"\"\"BREAKING: The Tampa Bay Rays are finalizing a deal to acquire slugger Nelson Cruz from the Minnesota Twins, sources tell ESPN.\"\"\"\n", + "res = openai.Completion.create(model=ft_model, prompt=sample_baseball_tweet + '\\n\\n###\\n\\n', max_tokens=1, temperature=0, logprobs=2)\n", + "res['choices'][0]['text']" + ], + "outputs": [ + { + "output_type": "execute_result", + "data": { + "text/plain": [ + "' baseball'" + ] + }, + "metadata": {}, + "execution_count": 17 + } + ], + "metadata": {} + } + ], + "metadata": { + "orig_nbformat": 4, + "language_info": { + "name": "python", + "version": "3.7.3", + "mimetype": "text/x-python", + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "pygments_lexer": "ipython3", + "nbconvert_exporter": "python", + "file_extension": ".py" + }, + "kernelspec": { + "name": "python3", + "display_name": "Python 3.7.3 64-bit ('base': conda)" + }, + "interpreter": { + "hash": "3b138a8faad971cc852f62bcf00f59ea0e31721743ea2c5a866ca26adf572e75" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/Get_embeddings.ipynb b/examples/Get_embeddings.ipynb new file mode 100644 index 0000000..e1b1732 --- /dev/null +++ b/examples/Get_embeddings.ipynb @@ -0,0 +1,107 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Get embeddings\n", + "\n", + "The function `get_embedding` will give us an embedding for an input text." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "12288" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "import openai\n", + "\n", + "embedding = openai.Embedding.create(input=\"Sample document text goes here\", engine=\"text-similarity-davinci-001\")['data'][0]['embedding']\n", + "len(embedding)" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "1024\n" + ] + } + ], + "source": [ + "import openai\n", + "from tenacity import retry, wait_random_exponential, stop_after_attempt\n", + "\n", + "@retry(wait=wait_random_exponential(min=1, max=20), stop=stop_after_attempt(6))\n", + "def get_embedding(text: str, engine=\"text-similarity-davinci-001\") -> List[float]:\n", + "\n", + " # replace newlines, which can negatively affect performance.\n", + " text = text.replace(\"\\n\", \" \")\n", + "\n", + " return openai.Embedding.create(input=[text], engine=engine)[\"data\"][0][\"embedding\"]\n", + "\n", + "embedding = get_embedding(\"Sample query text goes here\", engine=\"text-search-ada-query-001\")\n", + "print(len(embedding))" + ] + }, + { + "cell_type": "code", + "execution_count": 53, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "1024\n" + ] + } + ], + "source": [ + "embedding = get_embedding(\"Sample document text goes here\", engine=\"text-search-ada-doc-001\")\n", + "print(len(embedding))" + ] + } + ], + "metadata": { + "interpreter": { + "hash": "be4b5d5b73a21c599de40d6deb1129796d12dc1cc33a738f7bac13269cfcafe8" + }, + "kernelspec": { + "display_name": "Python 3.7.3 64-bit ('base': conda)", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.7.3" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/Obtain_dataset.ipynb b/examples/Obtain_dataset.ipynb new file mode 100644 index 0000000..80d1db3 --- /dev/null +++ b/examples/Obtain_dataset.ipynb @@ -0,0 +1,192 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 1. Load the dataset\n", + "\n", + "The dataset used in this example is [fine-food reviews](https://www.kaggle.com/snap/amazon-fine-food-reviews) from Amazon. The dataset contains a total of 568,454 food reviews Amazon users left up to October 2012. We will use a subset of this dataset, consisting of 1,000 most recent reviews for illustration purposes. The reviews are in English and tend to be positive or negative. Each review has a ProductId, UserId, Score, review title (Summary) and review body (Text).\n", + "\n", + "We will combine the review summary and review text into a single combined text. The model will encode this combined text and it will output a single vector embedding." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": {}, + "outputs": [ + { + "data": { + "text/html": [ + "
    \n", + "\n", + "\n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + "
    TimeProductIdUserIdScoreSummaryTextcombined
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    11303862400B001E4KFG0A3SGXH7AUHU8GW5Good Quality Dog FoodI have bought several of the Vitality canned d...Title: Good Quality Dog Food; Content: I have ...
    21346976000B00813GRG4A1D87F6ZCVE5NK1Not as AdvertisedProduct arrived labeled as Jumbo Salted Peanut...Title: Not as Advertised; Content: Product arr...
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    " + ], + "text/plain": [ + " Time ProductId UserId Score Summary \\\n", + "Id \n", + "1 1303862400 B001E4KFG0 A3SGXH7AUHU8GW 5 Good Quality Dog Food \n", + "2 1346976000 B00813GRG4 A1D87F6ZCVE5NK 1 Not as Advertised \n", + "\n", + " Text \\\n", + "Id \n", + "1 I have bought several of the Vitality canned d... \n", + "2 Product arrived labeled as Jumbo Salted Peanut... \n", + "\n", + " combined \n", + "Id \n", + "1 Title: Good Quality Dog Food; Content: I have ... \n", + "2 Title: Not as Advertised; Content: Product arr... " + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "import pandas as pd\n", + "\n", + "df = pd.read_csv('input/Reviews.csv', index_col=0)\n", + "df = df[['Time', 'ProductId', 'UserId', 'Score', 'Summary', 'Text']]\n", + "df = df.dropna()\n", + "df['combined'] = \"Title: \" + df.Summary.str.strip() + \"; Content: \" + df.Text.str.strip()\n", + "df.head(2)" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "1000" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# subsample to 1k most recent reviews and remove samples that are too long\n", + "df = df.sort_values('Time').tail(1_100)\n", + "df.drop('Time', axis=1, inplace=True)\n", + "\n", + "from transformers import GPT2TokenizerFast\n", + "tokenizer = GPT2TokenizerFast.from_pretrained(\"gpt2\")\n", + "\n", + "# remove reviews that are too long\n", + "df['n_tokens'] = df.combined.apply(lambda x: len(tokenizer.encode(x)))\n", + "df = df[df.n_tokens<2000].tail(1_000)\n", + "len(df)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### 2. Get embeddings and save them for future reuse" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": {}, + "outputs": [], + "source": [ + "from openai.embeddings_utils import get_embedding\n", + "\n", + "# This will take just under 10 minutes\n", + "df['babbage_similarity'] = df.combined.apply(lambda x: get_embedding(x, engine='text-similarity-babbage-001'))\n", + "df['babbage_search'] = df.combined.apply(lambda x: get_embedding(x, engine='text-search-babbage-doc-001'))\n", + "df.to_csv('output/embedded_1k_reviews.csv')" + ] + } + ], + "metadata": { + "interpreter": { + "hash": "be4b5d5b73a21c599de40d6deb1129796d12dc1cc33a738f7bac13269cfcafe8" + }, + "kernelspec": { + "display_name": "Python 3.7.3 64-bit ('base': conda)", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.9.9" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/Recommendation_using_embeddings.ipynb b/examples/Recommendation_using_embeddings.ipynb new file mode 100644 index 0000000..9a2cce8 --- /dev/null +++ b/examples/Recommendation_using_embeddings.ipynb @@ -0,0 +1,33446 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Recommendation using embeddings and nearest neighbor search\n", + "\n", + "Recommendations are widespread across the web.\n", + "\n", + "- 'Bought that item? Try these similar items.'\n", + "- 'Enjoy that book? Try these similar titles.'\n", + "- 'Not the help page you were looking for? Try these similar pages.'\n", + "\n", + "This notebook demonstrates how to use embeddings to find similar items to recommend. In particular, we use [AG's corpus of news articles](http://groups.di.unipi.it/~gulli/AG_corpus_of_news_articles.html) as our dataset.\n", + "\n", + "Our model will answer the question: given an article, what are the articles most similar to it?" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### 1. Imports\n", + "\n", + "First, let's import the packages and functions we'll need for later. If you don't have these, you'll need to install them. You can install them via your terminal by running `pip install {package_name}`, e.g. `pip install pandas`." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": {}, + "outputs": [], + "source": [ + "# imports\n", + "from typing import List\n", + "\n", + "import pandas as pd\n", + "import pickle\n", + "\n", + "from openai.embeddings_utils import (\n", + " get_embedding,\n", + " distances_from_embeddings,\n", + " tsne_components_from_embeddings,\n", + " chart_from_components,\n", + " indices_of_nearest_neighbors_from_distances,\n", + ")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### 2. Load data\n", + "\n", + "Next, let's load the AG news data and see what it looks like." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": {}, + "outputs": [ + { + "data": { + "text/html": [ + "
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    titledescriptionlabel_intlabel
    0World BriefingsBRITAIN: BLAIR WARNS OF CLIMATE THREAT Prime M...1World
    1Nvidia Puts a Firewall on a Motherboard (PC Wo...PC World - Upcoming chip set will include buil...4Sci/Tech
    2Olympic joy in Greek, Chinese pressNewspapers in Greece reflect a mixture of exhi...2Sports
    3U2 Can iPod with PicturesSAN JOSE, Calif. -- Apple Computer (Quote, Cha...4Sci/Tech
    4The Dream FactoryAny product, any shape, any size -- manufactur...4Sci/Tech
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    " + ], + "text/plain": [ + " title \\\n", + "0 World Briefings \n", + "1 Nvidia Puts a Firewall on a Motherboard (PC Wo... \n", + "2 Olympic joy in Greek, Chinese press \n", + "3 U2 Can iPod with Pictures \n", + "4 The Dream Factory \n", + "\n", + " description label_int label \n", + "0 BRITAIN: BLAIR WARNS OF CLIMATE THREAT Prime M... 1 World \n", + "1 PC World - Upcoming chip set will include buil... 4 Sci/Tech \n", + "2 Newspapers in Greece reflect a mixture of exhi... 2 Sports \n", + "3 SAN JOSE, Calif. -- Apple Computer (Quote, Cha... 4 Sci/Tech \n", + "4 Any product, any shape, any size -- manufactur... 4 Sci/Tech " + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# load data\n", + "dataset_path = \"https://cdn.openai.com/API/examples/data/AG_news_samples.csv\"\n", + "df = pd.read_csv(dataset_path)\n", + "\n", + "# print dataframe\n", + "n_examples = 5\n", + "df.head(n_examples)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Let's take a look at those same examples, but not truncated by ellipses." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + "Title: World Briefings\n", + "Description: BRITAIN: BLAIR WARNS OF CLIMATE THREAT Prime Minister Tony Blair urged the international community to consider global warming a dire threat and agree on a plan of action to curb the quot;alarming quot; growth of greenhouse gases.\n", + "Label: World\n", + "\n", + "Title: Nvidia Puts a Firewall on a Motherboard (PC World)\n", + "Description: PC World - Upcoming chip set will include built-in security features for your PC.\n", + "Label: Sci/Tech\n", + "\n", + "Title: Olympic joy in Greek, Chinese press\n", + "Description: Newspapers in Greece reflect a mixture of exhilaration that the Athens Olympics proved successful, and relief that they passed off without any major setback.\n", + "Label: Sports\n", + "\n", + "Title: U2 Can iPod with Pictures\n", + "Description: SAN JOSE, Calif. -- Apple Computer (Quote, Chart) unveiled a batch of new iPods, iTunes software and promos designed to keep it atop the heap of digital music players.\n", + "Label: Sci/Tech\n", + "\n", + "Title: The Dream Factory\n", + "Description: Any product, any shape, any size -- manufactured on your desktop! The future is the fabricator. By Bruce Sterling from Wired magazine.\n", + "Label: Sci/Tech\n" + ] + } + ], + "source": [ + "# print the title, description, and label of each example\n", + "for idx, row in df.head(n_examples).iterrows():\n", + " print(\"\")\n", + " print(f\"Title: {row['title']}\")\n", + " print(f\"Description: {row['description']}\")\n", + " print(f\"Label: {row['label']}\")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### 3. Build cache to save embeddings\n", + "\n", + "Before getting embeddings for these articles, let's set up a cache to save the embeddings we generate. In general, it's a good idea to save your embeddings so you can re-use them later. If you don't save them, you'll pay again each time you compute them again.\n", + "\n", + "To save you the expense of computing the embeddings needed for this demo, we've provided a pre-filled cache via the URL below. The cache is a dictionary that maps tuples of `(text, engine)` to a `list of floats` embedding. The cache is saved as a Python pickle file." + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": {}, + "outputs": [], + "source": [ + "# establish a cache of embeddings to avoid recomputing\n", + "# cache is a dict of tuples (text, engine) -> embedding, saved as a pickle file\n", + "\n", + "# set path to embedding cache\n", + "embedding_cache_path_to_load = \"https://cdn.openai.com/API/examples/data/example_embeddings_cache.pkl\"\n", + "embedding_cache_path_to_save = \"example_embeddings_cache.pkl\"\n", + "\n", + "# load the cache if it exists, and save a copy to disk\n", + "try:\n", + " embedding_cache = pd.read_pickle(embedding_cache_path_to_load)\n", + "except FileNotFoundError:\n", + " embedding_cache = {}\n", + "with open(embedding_cache_path_to_save, \"wb\") as embedding_cache_file:\n", + " pickle.dump(embedding_cache, embedding_cache_file)\n", + "\n", + "# define a function to retrieve embeddings from the cache if present, and otherwise request via the API\n", + "def embedding_from_string(\n", + " string: str,\n", + " engine: str = \"text-similarity-babbage-001\",\n", + " embedding_cache=embedding_cache\n", + ") -> List:\n", + " \"\"\"Return embedding of given string, using a cache to avoid recomputing.\"\"\"\n", + " if (string, engine) not in embedding_cache.keys():\n", + " embedding_cache[(string, engine)] = get_embedding(string, engine)\n", + " print('NOT FOUND')\n", + " with open(embedding_cache_path_to_save, \"wb\") as embedding_cache_file:\n", + " pickle.dump(embedding_cache, embedding_cache_file)\n", + " return embedding_cache[(string, engine)]" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Let's check that it works by getting an embedding." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + "Example string: BRITAIN: BLAIR WARNS OF CLIMATE THREAT Prime Minister Tony Blair urged the international community to consider global warming a dire threat and agree on a plan of action to curb the quot;alarming quot; growth of greenhouse gases.\n", + "\n", + "Example embedding: [-0.029093433171510696, 0.007570988964289427, -0.011933144181966782, -0.016499919816851616, 0.026675179600715637, -0.016704540699720383, -0.019439026713371277, 0.015421006828546524, -0.009700911119580269, -0.02580088935792446]...\n" + ] + } + ], + "source": [ + "# as an example, take the first description from the dataset\n", + "example_string = df[\"description\"].values[0]\n", + "print(f\"\\nExample string: {example_string}\")\n", + "\n", + "# print the first 10 dimensions of the embedding\n", + "example_embedding = embedding_from_string(example_string, engine=\"text-similarity-babbage-001\")\n", + "print(f\"\\nExample embedding: {example_embedding[:10]}...\")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### 4. Recommend similar articles based on embeddings\n", + "\n", + "To find similar articles, let's follow a three-step plan:\n", + "1. Get the similarity embeddings of all the article descriptions\n", + "2. Calculate the distance between a source title and all other articles\n", + "3. Print out the other articles closest to the source title" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": {}, + "outputs": [], + "source": [ + "def print_recommendations_from_strings(\n", + " strings: List[str],\n", + " index_of_source_string: int,\n", + " k_nearest_neighbors: int = 1,\n", + " engine=\"text-similarity-babbage-001\",\n", + ") -> List[int]:\n", + " \"\"\"Print out the k nearest neighbors of a given string.\"\"\"\n", + " # get embeddings for all strings\n", + " embeddings = [embedding_from_string(string, engine=engine) for string in strings]\n", + " # get the embedding of the source string\n", + " query_embedding = embeddings[index_of_source_string]\n", + " # get distances between the source embedding and other embeddings (function from embeddings_utils.py)\n", + " distances = distances_from_embeddings(query_embedding, embeddings, distance_metric=\"cosine\")\n", + " # get indices of nearest neighbors (function from embeddings_utils.py)\n", + " indices_of_nearest_neighbors = indices_of_nearest_neighbors_from_distances(distances)\n", + "\n", + " # print out source string\n", + " query_string = strings[index_of_source_string]\n", + " print(f\"Source string: {query_string}\")\n", + " # print out its k nearest neighbors\n", + " k_counter = 0\n", + " for i in indices_of_nearest_neighbors:\n", + " # skip any strings that are identical matches to the starting string\n", + " if query_string == strings[i]:\n", + " continue\n", + " # stop after printing out k articles\n", + " if k_counter >= k_nearest_neighbors:\n", + " break\n", + " k_counter += 1\n", + "\n", + " # print out the similar strings and their distances\n", + " print(\n", + " f\"\"\"\n", + " --- Recommendation #{k_counter} (nearest neighbor {k_counter} of {k_nearest_neighbors}) ---\n", + " String: {strings[i]}\n", + " Distance: {distances[i]:0.3f}\"\"\"\n", + " )\n", + "\n", + " return indices_of_nearest_neighbors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### 5. Example recommendations\n", + "\n", + "Let's look for articles similar to first one, which was about Tony Blair." + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Source string: BRITAIN: BLAIR WARNS OF CLIMATE THREAT Prime Minister Tony Blair urged the international community to consider global warming a dire threat and agree on a plan of action to curb the quot;alarming quot; growth of greenhouse gases.\n", + "\n", + " --- Recommendation #1 (nearest neighbor 1 of 5) ---\n", + " String: THE re-election of British Prime Minister Tony Blair would be seen as an endorsement of the military action in Iraq, Prime Minister John Howard said today.\n", + " Distance: 0.164\n", + "\n", + " --- Recommendation #2 (nearest neighbor 2 of 5) ---\n", + " String: Israel is prepared to back a Middle East conference convened by Tony Blair early next year despite having expressed fears that the British plans were over-ambitious and designed \n", + " Distance: 0.169\n", + "\n", + " --- Recommendation #3 (nearest neighbor 3 of 5) ---\n", + " String: WASHINGTON (Reuters) - President Bush on Friday set a four-year goal of seeing a Palestinian state established and he and British Prime Minister Tony Blair vowed to mobilize international support to help make it happen now that Yasser Arafat is dead.\n", + " Distance: 0.174\n", + "\n", + " --- Recommendation #4 (nearest neighbor 4 of 5) ---\n", + " String: AP - President Bush declared Friday that charges of voter fraud have cast doubt on the Ukrainian election, and warned that any European-negotiated pact on Iran's nuclear program must ensure the world can verify Tehran's compliance.\n", + " Distance: 0.179\n", + "\n", + " --- Recommendation #5 (nearest neighbor 5 of 5) ---\n", + " String: AFP - A battle group of British troops rolled out of southern Iraq on a US-requested mission to deadlier areas near Baghdad, in a major political gamble for British Prime Minister Tony Blair.\n", + " Distance: 0.182\n" + ] + } + ], + "source": [ + "article_descriptions = df[\"description\"].tolist()\n", + "\n", + "tony_blair_articles = print_recommendations_from_strings(\n", + " strings=article_descriptions, # let's base similarity off of the article description\n", + " index_of_source_string=0, # let's look at articles similar to the first one about Tony Blair\n", + " k_nearest_neighbors=5, # let's look at the 5 most similar articles\n", + ")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Pretty good! All 5 of the recommendations look similar to the original article about Tony Blair. Interestingly, note that #4 doesn't mention the words Tony Blair, but is nonetheless recommended by the model, presumably because the model understands that Tony Blair tends to be related to President Bush or European pacts over Iran's nuclear program. This illustrates the potential power of using embeddings rather than basic string matching; our models understand what topics are related to one another, even when their words don't overlap." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Let's see how our recommender does on the second example article about NVIDIA's new chipset with more security." + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Source string: PC World - Upcoming chip set will include built-in security features for your PC.\n", + "\n", + " --- Recommendation #1 (nearest neighbor 1 of 5) ---\n", + " String: PC World - Updated antivirus software for businesses adds intrusion prevention features.\n", + " Distance: 0.108\n", + "\n", + " --- Recommendation #2 (nearest neighbor 2 of 5) ---\n", + " String: PC World - Send your video throughout your house--wirelessly--with new gateways and media adapters.\n", + " Distance: 0.160\n", + "\n", + " --- Recommendation #3 (nearest neighbor 3 of 5) ---\n", + " String: PC World - The one-time World Class Product of the Year PDA gets a much-needed upgrade.\n", + " Distance: 0.161\n", + "\n", + " --- Recommendation #4 (nearest neighbor 4 of 5) ---\n", + " String: PC World - Symantec, McAfee hope raising virus-definition fees will move users to\\ suites.\n", + " Distance: 0.166\n", + "\n", + " --- Recommendation #5 (nearest neighbor 5 of 5) ---\n", + " String: Ziff Davis - The company this week will unveil more programs and technologies designed to ease users of its high-end servers onto its Integrity line, which uses Intel's 64-bit Itanium processor.\n", + " Distance: 0.193\n" + ] + } + ], + "source": [ + "chipset_security_articles = print_recommendations_from_strings(\n", + " strings=article_descriptions, # let's base similarity off of the article description\n", + " index_of_source_string=1, # let's look at articles similar to the second one about a more secure chipset\n", + " k_nearest_neighbors=5, # let's look at the 5 most similar articles\n", + ")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "From the printed distances, you can see that the #1 recommendation is much closer than all the others (0.108 vs 0.160+). And the #1 recommendation looks very similar to the starting article - it's another article from PC World about increasing computer security. Pretty good! " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Appendix: Using embeddings in more sophisticated recommenders\n", + "\n", + "A more sophisticated way to build a recommender system is to train a machine learning model that takes in tens or hundreds of signals, such as item popularity or user click data. Even in this system, embeddings can be a very useful signal into the recommender, especially for items that are being 'cold started' with no user data yet (e.g., a brand new product added to the catalog without any clicks yet)." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Appendix: Using embeddings to visualize similar articles" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "To get a sense of what our nearest neighbor recommender is doing, let's visualize the article embeddings. Although we can't plot the 2048 dimensions of each embedding vector, we can use techniques like [t-SNE](https://en.wikipedia.org/wiki/T-distributed_stochastic_neighbor_embedding) or [PCA](https://en.wikipedia.org/wiki/Principal_component_analysis) to compress the embeddings down into 2 or 3 dimensions, which we can chart.\n", + "\n", + "Before visualizing the nearest neighbors, let's visualize all of the article descriptions using t-SNE. Note that t-SNE is not deterministic, meaning that results may vary from run to run." + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": {}, + "outputs": [ + { + "name": "stderr", + "output_type": "stream", + "text": [ + "/Users/ted/.virtualenvs/openai/lib/python3.9/site-packages/sklearn/manifold/_t_sne.py:982: FutureWarning: The PCA initialization in TSNE will change to have the standard deviation of PC1 equal to 1e-4 in 1.2. This will ensure better convergence.\n", + " warnings.warn(\n" + ] + }, + { + "data": { + "application/vnd.plotly.v1+json": { + "config": { + "plotlyServerURL": "https://plot.ly" + }, + "data": [ + { + "customdata": [ + [ + "BRITAIN: BLAIR WARNS OF
    CLIMATE THREAT Prime Minister
    Tony Blair urged the
    international community to
    consider global warming a dire
    threat and agree on a plan of
    action to curb the
    quot;alarming quot; growth of
    greenhouse gases." + ], + [ + "KABUL, Sept 22 (AFP): Three US
    soldiers were killed and 14
    wounded in a series of fierce
    clashes with suspected Taliban
    fighters in south and eastern
    Afghanistan this week, the US
    military said Wednesday." + ], + [ + "AUSTRALIAN journalist John
    Martinkus is lucky to be alive
    after spending 24 hours in the
    hands of Iraqi militants at
    the weekend. Martinkus was in
    Baghdad working for the SBS
    Dateline TV current affairs
    program" + ], + [ + " GAZA (Reuters) - An Israeli
    helicopter fired a missile
    into a town in the southern
    Gaza Strip late on Wednesday,
    witnesses said, hours after a
    Palestinian suicide bomber
    blew herself up in Jerusalem,
    killing two Israeli border
    policemen." + ], + [ + "RIYADH, Saudi Arabia -- Saudi
    police are seeking two young
    men in the killing of a Briton
    in a Riyadh parking lot, the
    Interior Ministry said today,
    and the British ambassador
    called it a terrorist attack." + ], + [ + "A gas explosion at a coal mine
    in northern China killed 33
    workers in the 10th deadly
    mine blast reported in three
    months. The explosion occurred
    yesterday at 4:20 pm at Nanlou
    township" + ], + [ + "Reuters - Palestinian leader
    Mahmoud Abbas called\\Israel
    \"the Zionist enemy\" Tuesday,
    unprecedented language for\\the
    relative moderate who is
    expected to succeed Yasser
    Arafat." + ], + [ + "Nasser al-Qidwa, Palestinian
    representative at the United
    Nations and nephew of late
    leader Yasser Arafat, handed
    Arafat #39;s death report to
    the Palestinian National
    Authority (PNA) on Saturday." + ], + [ + "CAIRO, Egypt - France's
    foreign minister appealed
    Monday for the release of two
    French journalists abducted in
    Baghdad, saying the French
    respect all religions. He did
    not rule out traveling to
    Baghdad..." + ], + [ + "United Arab Emirates President
    and ruler of Abu Dhabi Sheik
    Zayed bin Sultan al-Nayhan
    died Tuesday, official
    television reports. He was 86." + ], + [ + "PALESTINIAN leader Yasser
    Arafat today issued an urgent
    call for the immediate release
    of two French journalists
    taken hostage in Iraq." + ], + [ + "The al-Qaida terrorist network
    spent less than \\$50,000 on
    each of its major attacks
    except for the Sept. 11, 2001,
    suicide hijackings, and one of
    its hallmarks is using" + ], + [ + "A FATHER who scaled the walls
    of a Cardiff court dressed as
    superhero Robin said the
    Buckingham Palace protester
    posed no threat. Fathers 4
    Justice activist Jim Gibson,
    who earlier this year staged
    an eye-catching" + ], + [ + "Julia Gillard has reportedly
    bowed out of the race to
    become shadow treasurer,
    taking enormous pressure off
    Opposition Leader Mark Latham." + ], + [ + "AFP - Maybe it's something to
    do with the fact that the
    playing area is so vast that
    you need a good pair of
    binoculars to see the action
    if it's not taking place right
    in front of the stands." + ], + [ + "Egypt #39;s release of accused
    Israeli spy Azzam Azzam in an
    apparent swap for six Egyptian
    students held on suspicion of
    terrorism is expected to melt
    the ice and perhaps result" + ], + [ + "GAZA CITY, Gaza Strip: Hamas
    militants killed an Israeli
    soldier and wounded four with
    an explosion in a booby-
    trapped chicken coop on
    Tuesday, in what the Islamic
    group said was an elaborate
    scheme to lure troops to the
    area with the help of a double" + ], + [ + "AP - The 300 men filling out
    forms in the offices of an
    Iranian aid group were offered
    three choices: Train for
    suicide attacks against U.S.
    troops in Iraq, for suicide
    attacks against Israelis or to
    assassinate British author
    Salman Rushdie." + ], + [ + "ATHENS, Greece - Gail Devers,
    the most talented yet star-
    crossed hurdler of her
    generation, was unable to
    complete even one hurdle in
    100-meter event Sunday -
    failing once again to win an
    Olympic hurdling medal.
    Devers, 37, who has three
    world championships in the
    hurdles but has always flopped
    at the Olympics, pulled up
    short and screamed as she slid
    under the first hurdle..." + ], + [ + " NAIROBI (Reuters) - The
    Sudanese government and its
    southern rebel opponents have
    agreed to sign a pledge in the
    Kenyan capital on Friday to
    formally end a brutal 21-year-
    old civil war, with U.N.
    Security Council ambassadors
    as witnesses." + ], + [ + "AP - Former Guatemalan
    President Alfonso Portillo
    #151; suspected of corruption
    at home #151; is living and
    working part-time in the same
    Mexican city he fled two
    decades ago to avoid arrest on
    murder charges, his close
    associates told The Associated
    Press on Sunday." + ], + [ + "washingtonpost.com - BRUSSELS,
    Aug. 26 -- The United States
    will have to wait until next
    year to see its fight with the
    European Union over biotech
    foods resolved, as the World
    Trade Organization agreed to
    an E.U. request to bring
    scientists into the debate,
    officials said Thursday." + ], + [ + "Insisting that Hurriyat
    Conference is the real
    representative of Kashmiris,
    Pakistan has claimed that
    India is not ready to accept
    ground realities in Kashmir." + ], + [ + "VIENNA -- After two years of
    investigating Iran's atomic
    program, the UN nuclear
    watchdog still cannot rule out
    that Tehran has a secret atom
    bomb project as Washington
    insists, the agency's chief
    said yesterday." + ], + [ + "AFP - US Secretary of State
    Colin Powell wrapped up a
    three-nation tour of Asia
    after winning pledges from
    Japan, China and South Korea
    to press North Korea to resume
    stalled talks on its nuclear
    weapons programs." + ], + [ + "CAIRO, Egypt An Egyptian
    company says one of its four
    workers who had been kidnapped
    in Iraq has been freed. It
    says it can #39;t give the
    status of the others being
    held hostage but says it is
    quot;doing its best to secure
    quot; their release." + ], + [ + "AFP - Hosts India braced
    themselves for a harrowing
    chase on a wearing wicket in
    the first Test after Australia
    declined to enforce the
    follow-on here." + ], + [ + "Prime Minister Paul Martin of
    Canada urged Haitian leaders
    on Sunday to allow the
    political party of the deposed
    president, Jean-Bertrand
    Aristide, to take part in new
    elections." + ], + [ + "Hostage takers holding up to
    240 people at a school in
    southern Russia have refused
    to talk with a top Islamic
    leader and demanded to meet
    with regional leaders instead,
    ITAR-TASS reported on
    Wednesday." + ], + [ + "Three children from a care
    home are missing on the
    Lancashire moors after they
    are separated from a group." + ], + [ + "Diabetics should test their
    blood sugar levels more
    regularly to reduce the risk
    of cardiovascular disease, a
    study says." + ], + [ + "Iraq's interim Prime Minister
    Ayad Allawi announced that
    proceedings would begin
    against former Baath Party
    leaders." + ], + [ + "A toxic batch of home-brewed
    alcohol has killed 31 people
    in several towns in central
    Pakistan, police and hospital
    officials say." + ], + [ + " BEIJING (Reuters) - North
    Korea is committed to holding
    six-party talks aimed at
    resolving the crisis over its
    nuclear weapons program, but
    has not indicated when, a top
    British official said on
    Tuesday." + ], + [ + " BAGHDAD (Reuters) - Iraq's
    interim government extended
    the closure of Baghdad
    international airport
    indefinitely on Saturday
    under emergency rule imposed
    ahead of this week's U.S.-led
    offensive on Falluja." + ], + [ + "Rivaling Bush vs. Kerry for
    bitterness, doctors and trial
    lawyers are squaring off this
    fall in an unprecedented four-
    state struggle over limiting
    malpractice awards..." + ], + [ + "AP - Hundreds of tribesmen
    gathered Tuesday near the area
    where suspected al-Qaida-
    linked militants are holding
    two Chinese engineers and
    demanding safe passage to
    their reputed leader, a former
    U.S. prisoner from Guantanamo
    Bay, Cuba, officials and
    residents said." + ], + [ + "In an alarming development,
    high-precision equipment and
    materials which could be used
    for making nuclear bombs have
    disappeared from some Iraqi
    facilities, the United Nations
    watchdog agency has said." + ], + [ + "A US airman dies and two are
    hurt as a helicopter crashes
    due to technical problems in
    western Afghanistan." + ], + [ + "Jacques Chirac has ruled out
    any withdrawal of French
    troops from Ivory Coast,
    despite unrest and anti-French
    attacks, which have forced the
    evacuation of thousands of
    Westerners." + ], + [ + "Japanese Prime Minister
    Junichiro Koizumi reshuffled
    his cabinet yesterday,
    replacing several top
    ministers in an effort to
    boost his popularity,
    consolidate political support
    and quicken the pace of
    reforms in the world #39;s
    second-largest economy." + ], + [ + "TBILISI (Reuters) - At least
    two Georgian soldiers were
    killed and five wounded in
    artillery fire with
    separatists in the breakaway
    region of South Ossetia,
    Georgian officials said on
    Wednesday." + ], + [ + "Laksamana.Net - Two Indonesian
    female migrant workers freed
    by militants in Iraq are
    expected to arrive home within
    a day or two, the Foreign
    Affairs Ministry said
    Wednesday (6/10/04)." + ], + [ + "A bus was hijacked today and
    shots were fired at police who
    surrounded it on the outskirts
    of Athens. Police did not know
    how many passengers were
    aboard the bus." + ], + [ + "AP - President Bashar Assad
    shuffled his Cabinet on
    Monday, just weeks after the
    United States and the United
    Nations challenged Syria over
    its military presence in
    Lebanon and the security
    situation along its border
    with Iraq." + ], + [ + "AP - President Vladimir Putin
    has signed a bill confirming
    Russia's ratification of the
    Kyoto Protocol, the Kremlin
    said Friday, clearing the way
    for the global climate pact to
    come into force early next
    year." + ], + [ + "AP - The authenticity of newly
    unearthed memos stating that
    George W. Bush failed to meet
    standards of the Texas Air
    National Guard during the
    Vietnam War was questioned
    Thursday by the son of the
    late officer who reportedly
    wrote the memos." + ], + [ + "Canadian Press - OAKVILLE,
    Ont. (CP) - The body of a
    missing autistic man was
    pulled from a creek Monday,
    just metres from where a key
    piece of evidence was
    uncovered but originally
    overlooked because searchers
    had the wrong information." + ], + [ + "AFP - German Chancellor
    Gerhard Schroeder arrived in
    Libya for an official visit
    during which he is to hold
    talks with Libyan leader
    Moamer Kadhafi." + ], + [ + "The government will examine
    claims 100,000 Iraqi civilians
    have been killed since the US-
    led invasion, Jack Straw says." + ], + [ + "Eton College and Clarence
    House joined forces yesterday
    to deny allegations due to be
    made at an employment tribunal
    today by a former art teacher
    that she improperly helped
    Prince Harry secure an A-level
    pass in art two years ago." + ], + [ + "AFP - Great Britain's chances
    of qualifying for the World
    Group of the Davis Cup were
    evaporating rapidly after
    Austria moved into a 2-1 lead
    following the doubles." + ], + [ + "Asia-Pacific leaders meet in
    Australia to discuss how to
    keep nuclear weapons out of
    the hands of extremists." + ], + [ + " TALL AFAR, Iraq -- A three-
    foot-high coil of razor wire,
    21-ton armored vehicles and
    American soldiers with black
    M-4 assault rifles stood
    between tens of thousands of
    people and their homes last
    week." + ], + [ + "LAKE GEORGE, N.Y. - Even
    though he's facing double hip
    replacement surgery, Bill
    Smith is more than happy to
    struggle out the door each
    morning, limp past his brand
    new P.T..." + ], + [ + " JERUSALEM (Reuters) - Israeli
    Prime Minister Ariel Sharon
    poured cold water on Tuesday
    on recent international
    efforts to restart stalled
    peace talks with Syria, saying
    there was \"no possibility\" of
    returning to previous
    discussions." + ], + [ + "Dutch smugness was slapped
    hard during the past
    fortnight. The rude awakening
    began with the barbaric
    slaying of controversial
    filmmaker Theo van Gogh on
    November 2. Then followed a
    reciprocal cycle of some" + ], + [ + "pee writes quot;A passenger
    on a commuter plane in
    northern Norway attacked both
    pilots and at least one
    passenger with an axe as the
    aircraft was coming in to
    land." + ], + [ + "Prime Minister Ariel Sharon
    pledged Sunday to escalate a
    broad Israeli offensive in
    northern Gaza, saying troops
    will remain until Palestinian
    rocket attacks are halted.
    Israeli officials said the
    offensive -- in which 58
    Palestinians and three
    Israelis have been killed --
    will help clear the way for an
    Israeli withdrawal." + ], + [ + "NEW YORK - Wall Street
    professionals know to keep
    their expectations in check in
    September, historically the
    worst month of the year for
    stocks. As summertime draws to
    a close, money managers are
    getting back to business,
    cleaning house, and often
    sending the market lower in
    the process..." + ], + [ + "A group linked to al Qaeda
    ally Abu Musab al-Zarqawi said
    it had tried to kill Iraq
    #39;s environment minister on
    Tuesday and warned it would
    not miss next time, according
    to an Internet statement." + ], + [ + "The Israeli military killed
    four Palestinian militants on
    Wednesday as troops in tanks
    and armored vehicles pushed
    into another town in the
    northern Gaza Strip, extending" + ], + [ + "KIRKUK, Iraq - A suicide
    attacker detonated a car bomb
    Saturday outside a police
    academy in the northern Iraqi
    city of Kirkuk as hundreds of
    trainees and civilians were
    leaving for the day, killing
    at least 20 people and
    wounding 36, authorities said.
    Separately, U.S and Iraqi
    forces clashed with insurgents
    in another part of northern
    Iraq after launching an
    operation to destroy an
    alleged militant cell in the
    town of Tal Afar, the U.S..." + ], + [ + "AP - Many states are facing
    legal challenges over possible
    voting problems Nov. 2. A look
    at some of the developments
    Thursday:" + ], + [ + "Israeli troops withdrew from
    the southern Gaza Strip town
    of Khan Yunis on Tuesday
    morning, following a 30-hour
    operation that left 17
    Palestinians dead." + ], + [ + "PM-designate Omar Karameh
    forms a new 30-member cabinet
    which includes women for the
    first time." + ], + [ + "Bahrain #39;s king pardoned a
    human rights activist who
    convicted of inciting hatred
    of the government and
    sentenced to one year in
    prison Sunday in a case linked
    to criticism of the prime
    minister." + ], + [ + "Leaders from 38 Asian and
    European nations are gathering
    in Vietnam for a summit of the
    Asia-Europe Meeting, know as
    ASEM. One thousand delegates
    are to discuss global trade
    and regional politics during
    the two-day forum." + ], + [ + "A US soldier has pleaded
    guilty to murdering a wounded
    16-year-old Iraqi boy. Staff
    Sergeant Johnny Horne was
    convicted Friday of the
    unpremeditated murder" + ], + [ + "Guinea-Bissau #39;s army chief
    of staff and former interim
    president, General Verissimo
    Correia Seabra, was killed
    Wednesday during unrest by
    mutinous soldiers in the
    former Portuguese" + ], + [ + "31 October 2004 -- Exit polls
    show that Prime Minister
    Viktor Yanukovich and
    challenger Viktor Yushchenko
    finished on top in Ukraine
    #39;s presidential election
    today and will face each other
    in a run-off next month." + ], + [ + "Rock singer Bono pledges to
    spend the rest of his life
    trying to eradicate extreme
    poverty around the world." + ], + [ + "AP - Just when tourists
    thought it was safe to go back
    to the Princess Diana memorial
    fountain, the mud has struck." + ], + [ + "AP - Three times a week, The
    Associated Press picks an
    issue and asks President Bush
    and Democratic presidential
    candidate John Kerry a
    question about it. Today's
    question and their responses:" + ], + [ + "In an apparent damage control
    exercise, Russian President
    Vladimir Putin on Saturday
    said he favored veto rights
    for India as new permanent
    member of the UN Security
    Council." + ], + [ + "AP - Nigeria's Senate has
    ordered a subsidiary of
    petroleum giant Royal/Dutch
    Shell to pay a Nigerian ethnic
    group #36;1.5 billion for oil
    spills in their homelands, but
    the legislative body can't
    enforce the resolution, an
    official said Wednesday." + ], + [ + "Australian troops in Baghdad
    came under attack today for
    the first time since the end
    of the Iraq war when a car
    bomb exploded injuring three
    soldiers and damaging an
    Australian armoured convoy." + ], + [ + "Pakistans decision to refuse
    the International Atomic
    Energy Agency to have direct
    access to Dr AQ Khan is
    correct on both legal and
    political counts." + ], + [ + "MANILA, 4 December 2004 - With
    floods receding, rescuers
    raced to deliver food to
    famished survivors in
    northeastern Philippine
    villages isolated by back-to-
    back storms that left more
    than 650 people dead and
    almost 400 missing." + ], + [ + "Talks on where to build the
    world #39;s first nuclear
    fusion reactor ended without a
    deal on Tuesday but the
    European Union said Japan and
    the United States no longer
    firmly opposed its bid to put
    the plant in France." + ], + [ + "CLEVELAND - The White House
    said Vice President Dick
    Cheney faces a \"master
    litigator\" when he debates
    Sen. John Edwards Tuesday
    night, a backhanded compliment
    issued as the Republican
    administration defended itself
    against criticism that it has
    not acknowledged errors in
    waging war in Iraq..." + ], + [ + "SEOUL (Reuters) - The chairman
    of South Korea #39;s ruling
    Uri Party resigned on Thursday
    after saying his father had
    served as a military police
    officer during Japan #39;s
    1910-1945 colonial rule on the
    peninsula." + ], + [ + "ALERE, Uganda -- Kasmiro
    Bongonyinge remembers sitting
    up suddenly in his bed. It was
    just after sunrise on a summer
    morning two years ago, and the
    old man, 87 years old and
    blind, knew something was
    wrong." + ], + [ + "JAKARTA - Official results
    have confirmed former army
    general Susilo Bambang
    Yudhoyono as the winner of
    Indonesia #39;s first direct
    presidential election, while
    incumbent Megawati
    Sukarnoputri urged her nation
    Thursday to wait for the
    official announcement" + ], + [ + "Reuters - A ragged band of
    children\\emerges ghost-like
    from mists in Ethiopia's
    highlands,\\thrusting bunches
    of carrots at a car full of
    foreigners." + ], + [ + "AP - A U.N. human rights
    expert criticized the U.S.-led
    coalition forces in
    Afghanistan for violating
    international law by allegedly
    beating Afghans to death and
    forcing some to remove their
    clothes or wear hoods." + ], + [ + " JERUSALEM (Reuters) - Israeli
    Prime Minister Ariel Sharon
    said on Thursday Yasser
    Arafat's death could be a
    turning point for peacemaking
    but he would pursue a
    unilateral plan that would
    strip Palestinians of some
    land they want for a state." + ], + [ + " AL-ASAD AIRBASE, Iraq
    (Reuters) - Defense Secretary
    Donald Rumsfeld swept into an
    airbase in Iraq's western
    desert Sunday to make a
    first-hand evaluation of
    operations to quell a raging
    Iraqi insurgency in his first
    such visit in five months." + ], + [ + "WASHINGTON - Democrat John
    Kerry accused President Bush
    on Monday of sending U.S.
    troops to the \"wrong war in
    the wrong place at the wrong
    time\" and said he'd try to
    bring them all home in four
    years..." + ], + [ + "More lorry drivers are
    bringing supplies to Nepal's
    capital in defiance of an
    indefinite blockade by Maoist
    rebels." + ], + [ + " BEIJING (Reuters) - Floods
    and landslides have killed 76
    people in southwest China in
    the past four days and washed
    away homes and roads, knocked
    down power lines and cut off
    at least one city, state
    media said on Monday." + ], + [ + "AP - Victims of the Sept. 11
    attacks were mourned worldwide
    Saturday, but in the Middle
    East, amid sympathy for the
    dead, Arabs said Washington's
    support for Israel and the war
    on terror launched in the
    aftermath of the World Trade
    Center's collapse have only
    fueled anger and violence." + ], + [ + "SEATTLE - Ichiro Suzuki set
    the major league record for
    hits in a season with 258,
    breaking George Sisler's
    84-year-old mark with a pair
    of singles Friday night. The
    Seattle star chopped a leadoff
    single in the first inning,
    then made history with a
    grounder up the middle in the
    third..." + ], + [ + "The intruder who entered
    British Queen Elizabeth II
    #39;s official Scottish
    residence and caused a
    security scare was a reporter
    from the London-based Sunday
    Times newspaper, local media
    reported Friday." + ], + [ + "Canadian Press - FREDERICTON
    (CP) - A New Brunswick truck
    driver arrested in Ontario
    this week has been accused by
    police of stealing 50,000 cans
    of Moosehead beer." + ], + [ + "Chinese authorities detained a
    prominent, U.S.-based Buddhist
    leader in connection with his
    plans to reopen an ancient
    temple complex in the Chinese
    province of Inner Mongolia
    last week and have forced
    dozens of his American
    followers to leave the region,
    local officials said
    Wednesday." + ], + [ + "Adorned with Turkish and EU
    flags, Turkey #39;s newspapers
    hailed Thursday an official EU
    report recommending the
    country start talks to join
    the bloc, while largely
    ignoring the stringent
    conditions attached to the
    announcement." + ], + [ + "Thailand's prime minister
    visits the southern town where
    scores of Muslims died in army
    custody after a rally." + ], + [ + "Beijing: At least 170 miners
    were trapped underground after
    a gas explosion on Sunday
    ignited a fire in a coalmine
    in north-west China #39;s
    Shaanxi province, reports
    said." + ], + [ + "SAMARRA (Iraq): With renewe d
    wave of skirmishes between the
    Iraqi insurgents and the US-
    led coalition marines, several
    people including top police
    officers were put to death on
    Saturday." + ], + [ + "AFP - Like most US Latinos,
    members of the extended
    Rodriguez family say they will
    cast their votes for Democrat
    John Kerry in next month's
    presidential polls." + ], + [ + "FALLUJAH, Iraq -- Four Iraqi
    fighters huddled in a trench,
    firing rocket-propelled
    grenades at Lieutenant Eric
    Gregory's Bradley Fighting
    Vehicle and the US tanks and
    Humvees that were lumbering
    through tight streets between
    boxlike beige houses." + ], + [ + "AP - Several thousand
    Christians who packed a
    cathedral compound in the
    Egyptian capital hurled stones
    at riot police Wednesday to
    protest a woman's alleged
    forced conversion to Islam. At
    least 30 people were injured." + ], + [ + "A group of Saudi religious
    scholars have signed an open
    letter urging Iraqis to
    support jihad against US-led
    forces. quot;Fighting the
    occupiers is a duty for all
    those who are able, quot; they
    said in a statement posted on
    the internet at the weekend." + ], + [ + "Mountaineers retrieve three
    bodies believed to have been
    buried for 22 years on an
    Indian glacier." + ], + [ + "President Thabo Mbeki met with
    Ivory Coast Prime Minister
    Seydou Diarra for three hours
    yesterday as part of talks
    aimed at bringing peace to the
    conflict-wracked Ivory Coast." + ], + [ + " KATHMANDU (Reuters) - Nepal's
    Maoist rebels have
    temporarily suspended a
    crippling economic blockade of
    the capital from Wednesday,
    saying the move was in
    response to popular appeals." + ], + [ + "Reuters - An Algerian
    suspected of being a leader\\of
    the Madrid train bombers has
    been identified as one of
    seven\\people who blew
    themselves up in April to
    avoid arrest, Spain's\\Interior
    Ministry said on Friday." + ], + [ + "KABUL: An Afghan man was found
    guilty on Saturday of killing
    four journalists in 2001,
    including two from Reuters,
    and sentenced to death." + ], + [ + "Yasser Arafat, the leader for
    decades of a fight for
    Palestinian independence from
    Israel, has died at a military
    hospital in Paris, according
    to news reports." + ], + [ + " JABALYA, Gaza Strip (Reuters)
    - Israel pulled most of its
    forces out of the northern
    Gaza Strip Saturday after a
    four-day incursion it said
    was staged to halt Palestinian
    rocket attacks on southern
    Israeli towns." + ], + [ + "THE Turkish embassy in Baghdad
    was investigating a television
    report that two Turkish
    hostages had been killed in
    Iraq, but no confirmation was
    available so far, a senior
    Turkish diplomat said today." + ], + [ + "Reuters - Thousands of
    supporters of
    Ukraine's\\opposition leader,
    Viktor Yushchenko, celebrated
    on the streets\\in the early
    hours on Monday after an exit
    poll showed him\\winner of a
    bitterly fought presidential
    election." + ], + [ + "LONDON : The United States
    faced rare criticism over
    human rights from close ally
    Britain, with an official
    British government report
    taking Washington to task over
    concerns about Iraq and the
    Guantanamo Bay jail." + ], + [ + "LONDON - A bomb threat that
    mentioned Iraq forced a New
    York-bound Greek airliner to
    make an emergency landing
    Sunday at London's Stansted
    Airport escorted by military
    jets, authorities said. An
    airport spokeswoman said an
    Athens newspaper had received
    a phone call saying there was
    a bomb on board the Olympic
    Airlines plane..." + ], + [ + "ATHENS, Greece - Sheryl
    Swoopes made three big plays
    at the end - two baskets and
    another on defense - to help
    the United States squeeze out
    a 66-62 semifinal victory over
    Russia on Friday. Now, only
    one game stands between the
    U.S..." + ], + [ + "Scientists are developing a
    device which could improve the
    lives of kidney dialysis
    patients." + ], + [ + "KABUL, Afghanistan The Afghan
    government is blaming drug
    smugglers for yesterday #39;s
    attack on the leading vice
    presidential candidate ." + ], + [ + "One of the leading figures in
    the Greek Orthodox Church, the
    Patriarch of Alexandria Peter
    VII, has been killed in a
    helicopter crash in the Aegean
    Sea." + ], + [ + "CANBERRA, Australia -- The
    sweat-stained felt hats worn
    by Australian cowboys, as much
    a part of the Outback as
    kangaroos and sun-baked soil,
    may be heading for the history
    books. They fail modern
    industrial safety standards." + ], + [ + "A London-to-Washington flight
    is diverted after a security
    alert involving the singer
    formerly known as Cat Stevens." + ], + [ + "AP - President Bush declared
    Friday that charges of voter
    fraud have cast doubt on the
    Ukrainian election, and warned
    that any European-negotiated
    pact on Iran's nuclear program
    must ensure the world can
    verify Tehran's compliance." + ], + [ + "TheSpaceShipOne team is handed
    the \\$10m cheque and trophy it
    won for claiming the Ansari
    X-Prize." + ], + [ + "Security officials have
    identified six of the
    militants who seized a school
    in southern Russia as being
    from Chechnya, drawing a
    strong connection to the
    Chechen insurgents who have
    been fighting Russian forces
    for years." + ], + [ + "SEOUL -- North Korea set three
    conditions yesterday to be met
    before it would consider
    returning to six-party talks
    on its nuclear programs." + ], + [ + "US-backed Iraqi commandos were
    poised Friday to storm rebel
    strongholds in the northern
    city of Mosul, as US military
    commanders said they had
    quot;broken the back quot; of
    the insurgency with their
    assault on the former rebel
    bastion of Fallujah." + ], + [ + "JERUSALEM (Reuters) - Prime
    Minister Ariel Sharon, facing
    a party mutiny over his plan
    to quit the Gaza Strip, has
    approved 1,000 more Israeli
    settler homes in the West Bank
    in a move that drew a cautious
    response on Tuesday from ..." + ], + [ + "GHAZNI, Afghanistan, 6 October
    2004 - Wartime security was
    rolled out for Afghanistans
    interim President Hamid Karzai
    as he addressed his first
    election campaign rally
    outside the capital yesterday
    amid spiraling violence." + ], + [ + "China has confirmed that it
    found a deadly strain of bird
    flu in pigs as early as two
    years ago. China #39;s
    Agriculture Ministry said two
    cases had been discovered, but
    it did not say exactly where
    the samples had been taken." + ], + [ + "AP - Ten years after the Irish
    Republican Army's momentous
    cease-fire, negotiations
    resumed Wednesday in hope of
    reviving a Catholic-Protestant
    administration, an elusive
    goal of Northern Ireland's
    hard-fought peace process." + ], + [ + " SANTO DOMINGO, Dominican
    Republic, Sept. 18 -- Tropical
    Storm Jeanne headed for the
    Bahamas on Saturday after an
    assault on the Dominican
    Republic that killed 10
    people, destroyed hundreds of
    houses and forced thousands
    from their homes." + ], + [ + "An explosion tore apart a car
    in Gaza City Monday, killing
    at least one person,
    Palestinian witnesses said.
    They said Israeli warplanes
    were circling overhead at the
    time of the blast, indicating
    a possible missile strike." + ], + [ + "Beijing, Oct. 25 (PTI): China
    and the US today agreed to
    work jointly to re-energise
    the six-party talks mechanism
    aimed at dismantling North
    Korea #39;s nuclear programmes
    while Washington urged Beijing
    to resume" + ], + [ + "AFP - Sporadic gunfire and
    shelling took place overnight
    in the disputed Georgian
    region of South Ossetia in
    violation of a fragile
    ceasefire, wounding seven
    Georgian servicemen." + ], + [ + " FALLUJA, Iraq (Reuters) -
    U.S. forces hit Iraq's rebel
    stronghold of Falluja with the
    fiercest air and ground
    bombardment in months, as
    insurgents struck back on
    Saturday with attacks that
    killed up to 37 people in
    Samarra." + ], + [ + " NAJAF, Iraq (Reuters) - The
    fate of a radical Shi'ite
    rebellion in the holy city of
    Najaf was uncertain Friday
    amid disputed reports that
    Iraqi police had gained
    control of the Imam Ali
    Mosque." + ], + [ + "Until this week, only a few
    things about the strange,
    long-ago disappearance of
    Charles Robert Jenkins were
    known beyond a doubt. In the
    bitter cold of Jan. 5, 1965,
    the 24-year-old US Army
    sergeant was leading" + ], + [ + "The United States on Tuesday
    modified slightly a threat of
    sanctions on Sudan #39;s oil
    industry in a revised text of
    its UN resolution on
    atrocities in the country
    #39;s Darfur region." + ], + [ + "AP - France intensified
    efforts Tuesday to save the
    lives of two journalists held
    hostage in Iraq, and the Arab
    League said the militants'
    deadline for France to revoke
    a ban on Islamic headscarves
    in schools had been extended." + ], + [ + "At least 12 people die in an
    explosion at a fuel pipeline
    on the outskirts of Nigeria's
    biggest city, Lagos." + ], + [ + "Volkswagen demanded a two-year
    wage freeze for the
    170,000-strong workforce at
    Europe #39;s biggest car maker
    yesterday, provoking union
    warnings of imminent conflict
    at key pay and conditions
    negotiations." + ], + [ + "Citing security concerns, the
    U.S. Embassy on Thursday
    banned its employees from
    using the highway linking the
    embassy area to the
    international airport, a
    10-mile stretch of road
    plagued by frequent suicide
    car-bomb attacks." + ], + [ + "AP - Tom Daschle bade his
    fellow Senate Democrats
    farewell Tuesday with a plea
    that they seek common ground
    with Republicans yet continue
    to fight for the less
    fortunate." + ], + [ + "AP - Police defused a bomb in
    a town near Prime Minister
    Silvio Berlusconi's villa on
    the island of Sardinia on
    Wednesday shortly after
    British Prime Minister Tony
    Blair finished a visit there
    with the Italian leader." + ], + [ + "The coffin of Yasser Arafat,
    draped with the Palestinian
    flag, was bound for Ramallah
    in the West Bank Friday,
    following a formal funeral on
    a military compound near
    Cairo." + ], + [ + "US Ambassador to the United
    Nations John Danforth resigned
    on Thursday after serving in
    the post for less than six
    months. Danforth, 68, said in
    a letter released Thursday" + ], + [ + "ISLAMABAD, Pakistan -- Photos
    were published yesterday in
    newspapers across Pakistan of
    six terror suspects, including
    a senior Al Qaeda operative,
    the government says were
    behind attempts to assassinate
    the nation's president." + ], + [ + " ATHENS (Reuters) - The Athens
    Paralympics canceled
    celebrations at its closing
    ceremony after seven
    schoolchildren traveling to
    watch the event died in a bus
    crash on Monday." + ], + [ + "DUBAI : An Islamist group has
    threatened to kill two Italian
    women held hostage in Iraq if
    Rome does not withdraw its
    troops from the war-torn
    country within 24 hours,
    according to an internet
    statement." + ], + [ + "A heavy quake rocked Indonesia
    #39;s Papua province killing
    at least 11 people and
    wounding 75. The quake
    destroyed 150 buildings,
    including churches, mosques
    and schools." + ], + [ + "Reuters - A small group of
    suspected\\gunmen stormed
    Uganda's Water Ministry
    Wednesday and took
    three\\people hostage to
    protest against proposals to
    allow President\\Yoweri
    Museveni for a third
    term.\\Police and soldiers with
    assault rifles cordoned off
    the\\three-story building, just
    328 feet from Uganda's
    parliament\\building in the
    capital Kampala." + ], + [ + "Venezuela suggested Friday
    that exiles living in Florida
    may have masterminded the
    assassination of a prosecutor
    investigating a short-lived
    coup against leftist President
    Hugo Chvez" + ], + [ + "Facing a popular outcry at
    home and stern warnings from
    Europe, the Turkish government
    discreetly stepped back
    Tuesday from a plan to
    introduce a motion into a
    crucial penal reform bill to
    make adultery a crime
    punishable by prison." + ], + [ + "North-west Norfolk MP Henry
    Bellingham has called for the
    release of an old college
    friend accused of plotting a
    coup in Equatorial Guinea." + ], + [ + "AFP - Want to buy a castle?
    Head for the former East
    Germany." + ], + [ + "AFP - Steven Gerrard has moved
    to allay Liverpool fans' fears
    that he could be out until
    Christmas after breaking a
    metatarsal bone in his left
    foot." + ], + [ + "MINSK - Legislative elections
    in Belarus held at the same
    time as a referendum on
    whether President Alexander
    Lukashenko should be allowed
    to seek a third term fell
    significantly short of
    democratic standards, foreign
    observers said here Monday." + ], + [ + "An Olympic sailor is charged
    with the manslaughter of a
    Briton who died after being
    hit by a car in Athens." + ], + [ + "AP - Secretary of State Colin
    Powell on Friday praised the
    peace deal that ended fighting
    in Iraq's holy city of Najaf
    and said the presence of U.S.
    forces in the area helped make
    it possible." + ], + [ + "26 August 2004 -- Iraq #39;s
    top Shi #39;ite cleric, Grand
    Ayatollah Ali al-Sistani,
    arrived in the city of Al-
    Najaf today in a bid to end a
    weeks-long conflict between US
    forces and militiamen loyal to
    Shi #39;ite cleric Muqtada al-
    Sadr." + ], + [ + "PARIS : French trade unions
    called on workers at France
    Telecom to stage a 24-hour
    strike September 7 to protest
    government plans to privatize
    the public telecommunications
    operator, union sources said." + ], + [ + "The Indonesian tourism
    industry has so far not been
    affected by last week #39;s
    bombing outside the Australian
    embassy in Jakarta and
    officials said they do not
    expect a significant drop in
    visitor numbers as a result of
    the attack." + ], + [ + "MARK Thatcher will have to
    wait until at least next April
    to face trial on allegations
    he helped bankroll a coup
    attempt in oil-rich Equatorial
    Guinea." + ], + [ + "NEW YORK - A drop in oil
    prices and upbeat outlooks
    from Wal-Mart and Lowe's
    helped send stocks sharply
    higher Monday on Wall Street,
    with the swing exaggerated by
    thin late summer trading. The
    Dow Jones industrials surged
    nearly 130 points..." + ], + [ + "ROSTOV-ON-DON, Russia --
    Hundreds of protesters
    ransacked and occupied the
    regional administration
    building in a southern Russian
    province Tuesday, demanding
    the resignation of the region
    #39;s president, whose former
    son-in-law has been linked to
    a multiple" + ], + [ + "AFP - Iraqi Foreign Minister
    Hoshyar Zebari arrived
    unexpectedly in the holy city
    of Mecca Wednesday where he
    met Crown Prince Abdullah bin
    Abdul Aziz, the official SPA
    news agency reported." + ], + [ + "Haitian police and UN troops
    moved into a slum neighborhood
    on Sunday and cleared street
    barricades that paralyzed a
    part of the capital." + ], + [ + "withdrawal of troops and
    settlers from occupied Gaza
    next year. Militants seek to
    claim any pullout as a
    victory. quot;Islamic Jihad
    will not be broken by this
    martyrdom, quot; said Khaled
    al-Batsh, a senior political
    leader in Gaza." + ], + [ + "The U.S. military has found
    nearly 20 houses where
    intelligence officers believe
    hostages were tortured or
    killed in this city, including
    the house with the cage that
    held a British contractor who
    was beheaded last month." + ], + [ + "AFP - Opponents of the Lao
    government may be plotting
    bomb attacks in Vientiane and
    other areas of Laos timed to
    coincide with a summit of
    Southeast Asian leaders the
    country is hosting next month,
    the United States said." + ], + [ + "AP - Russia agreed Thursday to
    send warships to help NATO
    naval patrols that monitor
    suspicious vessels in the
    Mediterranean, part of a push
    for closer counterterrorism
    cooperation between Moscow and
    the western alliance." + ], + [ + "A military plane crashed in
    the mountains near Caracas,
    killing all 16 persons on
    board, including two high-
    ranking military officers,
    officials said." + ], + [ + "A voice recording said to be
    that of suspected Al Qaeda
    commander Abu Mussab al-
    Zarqawi, claims Iraq #39;s
    Prime Minister Iyad Allawi is
    the militant network #39;s
    number one target." + ], + [ + "BEIJING -- More than a year
    after becoming China's
    president, Hu Jintao was
    handed the full reins of power
    yesterday when his
    predecessor, Jiang Zemin, gave
    up the nation's most powerful
    military post." + ], + [ + "AP - Greenpeace activists
    scaled the walls of Ford Motor
    Co.'s Norwegian headquarters
    Tuesday to protest plans to
    destroy hundreds of non-
    polluting electric cars." + ], + [ + "AFP - The chances of Rupert
    Murdoch's News Corp relocating
    from Australia to the United
    States have increased after
    one of its biggest
    institutional investors has
    chosen to abstain from a vote
    next week on the move." + ], + [ + "AFP - An Indian minister said
    a school text-book used in the
    violence-prone western state
    of Gujarat portrayed Adolf
    Hitler as a role model." + ], + [ + "DOVER, N.H. (AP) -- Democrat
    John Kerry is seizing on the
    Bush administration's failure
    to secure hundreds of tons of
    explosives now missing in
    Iraq." + ], + [ + "United Nations officials
    report security breaches in
    internally displaced people
    and refugee camps in Sudan
    #39;s embattled Darfur region
    and neighboring Chad." + ], + [ + "KINGSTON, Jamaica - Hurricane
    Ivan's deadly winds and
    monstrous waves bore down on
    Jamaica on Friday, threatening
    a direct hit on its densely
    populated capital after
    ravaging Grenada and killing
    at least 33 people. The
    Jamaican government ordered
    the evacuation of half a
    million people from coastal
    areas, where rains on Ivan's
    outer edges were already
    flooding roads..." + ], + [ + "North Korea has denounced as
    quot;wicked terrorists quot;
    the South Korean officials who
    orchestrated last month #39;s
    airlift to Seoul of 468 North
    Korean defectors." + ], + [ + "The Black Watch regiment has
    returned to its base in Basra
    in southern Iraq after a
    month-long mission standing in
    for US troops in a more
    violent part of the country,
    the Ministry of Defence says." + ], + [ + "AP - A senior Congolese
    official said Tuesday his
    nation had been invaded by
    neighboring Rwanda, and U.N.
    officials said they were
    investigating claims of
    Rwandan forces clashing with
    militias in the east." + ], + [ + "UNITED NATIONS - The United
    Nations #39; nuclear agency
    says it is concerned about the
    disappearance of equipment and
    materials from Iraq that could
    be used to make nuclear
    weapons." + ], + [ + " BRUSSELS (Reuters) - The EU's
    historic deal with Turkey to
    open entry talks with the vast
    Muslim country was hailed by
    supporters as a bridge builder
    between Europe and the Islamic
    world." + ], + [ + "Iraqi President Ghazi al-
    Yawar, who was due in Paris on
    Sunday to start a European
    tour, has postponed his visit
    to France due to the ongoing
    hostage drama involving two
    French journalists, Arab
    diplomats said Friday." + ], + [ + " SAO PAULO, Brazil (Reuters) -
    President Luiz Inacio Lula da
    Silva's Workers' Party (PT)
    won the mayoralty of six state
    capitals in Sunday's municipal
    vote but was forced into a
    run-off to defend its hold on
    the race's biggest prize, the
    city of Sao Paulo." + ], + [ + "ATHENS, Greece - They are
    America's newest golden girls
    - powerful and just a shade
    from perfection. The U.S..." + ], + [ + "AMMAN, Sept. 15. - The owner
    of a Jordanian truck company
    announced today that he had
    ordered its Iraq operations
    stopped in a bid to save the
    life of a driver held hostage
    by a militant group." + ], + [ + "Israel is prepared to back a
    Middle East conference
    convened by Tony Blair early
    next year despite having
    expressed fears that the
    British plans were over-
    ambitious and designed" + ], + [ + "AP - U.S. State Department
    officials learned that seven
    American children had been
    abandoned at a Nigerian
    orphanage but waited more than
    a week to check on the youths,
    who were suffering from
    malnutrition, malaria and
    typhoid, a newspaper reported
    Saturday." + ], + [ + "\\Angry mobs in Ivory Coast's
    main city, Abidjan, marched on
    the airport, hours after it
    came under French control." + ], + [ + "Several workers are believed
    to have been killed and others
    injured after a contruction
    site collapsed at Dubai
    airport. The workers were
    trapped under rubble at the
    site of a \\$4." + ], + [ + "Talks between Sudan #39;s
    government and two rebel
    groups to resolve the nearly
    two-year battle resume Friday.
    By Abraham McLaughlin Staff
    writer of The Christian
    Science Monitor." + ], + [ + "Stansted airport is the
    designated emergency landing
    ground for planes in British
    airspace hit by in-flight
    security alerts. Emergency
    services at Stansted have
    successfully dealt" + ], + [ + "The massive military operation
    to retake Fallujah has been
    quot;accomplished quot;, a
    senior Iraqi official said.
    Fierce fighting continued in
    the war-torn city where
    pockets of resistance were
    still holding out against US
    forces." + ], + [ + "There are some signs of
    progress in resolving the
    Nigerian conflict that is
    riling global oil markets. The
    leader of militia fighters
    threatening to widen a battle
    for control of Nigeria #39;s
    oil-rich south has" + ], + [ + "A strong earthquake hit Taiwan
    on Monday, shaking buildings
    in the capital Taipei for
    several seconds. No casualties
    were reported." + ], + [ + "A policeman ran amok at a
    security camp in Indian-
    controlled Kashmir after an
    argument and shot dead seven
    colleagues before he was
    gunned down, police said on
    Sunday." + ], + [ + "New York police have developed
    a pre-emptive strike policy,
    cutting off demonstrations
    before they grow large." + ], + [ + "Bulgaria has started its first
    co-mission with the EU in
    Bosnia and Herzegovina, along
    with some 30 countries,
    including Canada and Turkey." + ], + [ + "AP - The pileup of events in
    the city next week, including
    the Republican National
    Convention, will add to the
    security challenge for the New
    York Police Department, but
    commissioner Ray Kelly says,
    \"With a big, experienced
    police force, we can do it.\"" + ], + [ + "LONDON, Dec 11 (IranMania) -
    Iraqi Vice-President Ibrahim
    al-Jaafari refused to believe
    in remarks published Friday
    that Iran was attempting to
    influence Iraqi polls with the
    aim of creating a
    quot;crescent quot; dominated
    by Shiites in the region." + ], + [ + "The late Princess Dianas
    former bodyguard, Ken Wharfe,
    dismisses her suspicions that
    one of her lovers was bumped
    off. Princess Diana had an
    affair with Barry Mannakee, a
    policeman who was assigned to
    protect her." + ], + [ + "Long considered beyond the
    reach of mainland mores, the
    Florida city is trying to
    limit blatant displays of
    sexual behavior." + ], + [ + "Senator John Kerry said today
    that the war in Iraq was a
    \"profound diversion\" from the
    war on terror and Osama bin
    Laden." + ], + [ + "A group claiming to have
    captured two Indonesian women
    in Iraq has said it will
    release them if Jakarta frees
    Muslim cleric Abu Bakar Bashir
    being held for alleged
    terrorist links." + ], + [ + "Indonesian police said
    yesterday that DNA tests had
    identified a suicide bomber
    involved in a deadly attack
    this month on the Australian
    embassy in Jakarta." + ], + [ + "NEW YORK - Wal-Mart Stores
    Inc.'s warning of
    disappointing sales sent
    stocks fluctuating Monday as
    investors' concerns about a
    slowing economy offset their
    relief over a drop in oil
    prices. October contracts
    for a barrel of light crude
    were quoted at \\$46.48, down
    24 cents, on the New York
    Mercantile Exchange..." + ], + [ + "Iraq #39;s top Shi #39;ite
    cleric made a sudden return to
    the country on Wednesday and
    said he had a plan to end an
    uprising in the quot;burning
    city quot; of Najaf, where
    fighting is creeping ever
    closer to its holiest shrine." + ], + [ + "KABUL, Afghanistan Aug. 22,
    2004 - US soldiers sprayed a
    pickup truck with bullets
    after it failed to stop at a
    roadblock in central
    Afghanistan, killing two women
    and a man and critically
    wounding two other" + ], + [ + "SYDNEY -- Prime Minister John
    Howard of Australia, a key US
    ally and supporter of the Iraq
    war, celebrated his election
    win over opposition Labor
    after voters enjoying the
    fruits of a strong economy
    gave him another term." + ], + [ + "BAGHDAD, Iraq - Two rockets
    hit a downtown Baghdad hotel
    housing foreigners and
    journalists Thursday, and
    gunfire erupted in the
    neighborhood across the Tigris
    River from the U.S. Embassy
    compound..." + ], + [ + "The Prevention of Terrorism
    Act 2002 (Pota) polarised the
    country, not just by the
    manner in which it was pushed
    through by the NDA government
    through a joint session of
    Parliament but by the shabby
    and often biased manner in
    which it was enforced." + ], + [ + "The US military says marines
    in Fallujah shot and killed an
    insurgent who engaged them as
    he was faking being dead, a
    week after footage of a marine
    killing an apparently unarmed
    and wounded Iraqi caused a
    stir in the region." + ], + [ + "Description: NPR #39;s Alex
    Chadwick talks to Colin Brown,
    deputy political editor for
    the United Kingdom #39;s
    Independent newspaper,
    currently covering the British
    Labour Party Conference." + ], + [ + "Hamas vowed revenge yesterday
    after an Israeli airstrike in
    Gaza killed one of its senior
    commanders - the latest
    assassination to have weakened
    the militant group." + ], + [ + "A senior member of the
    Palestinian resistance group
    Hamas has been released from
    an Israeli prison after
    completing a two-year
    sentence." + ], + [ + "MPs have announced a new
    inquiry into family courts and
    whether parents are treated
    fairly over issues such as
    custody or contact with their
    children." + ], + [ + "Canadian Press - MELBOURNE,
    Australia (AP) - A 36-year-old
    businesswoman was believed to
    be the first woman to walk
    around Australia on Friday
    after striding into her
    hometown of Melbourne to
    complete her 16,700-kilometre
    trek in 365 days." + ], + [ + "Most remaining Pakistani
    prisoners held at the US
    Guantanamo Bay prison camp are
    freed, officials say." + ], + [ + "French police are
    investigating an arson-caused
    fire at a Jewish Social Center
    that might have killed dozens
    without the quick response of
    firefighters." + ], + [ + "Rodney King, whose videotaped
    beating led to riots in Los
    Angeles in 1992, is out of
    jail now and talking frankly
    for the first time about the
    riots, himself and the
    American way of life." + ], + [ + "AFP - Radical Islamic cleric
    Abu Hamza al-Masri was set to
    learn Thursday whether he
    would be charged under
    Britain's anti-terrorism law,
    thus delaying his possible
    extradition to the United
    States to face terrorism-
    related charges." + ], + [ + "Louisen Louis, 30, walked
    Monday in the middle of a
    street that resembled a small
    river with brown rivulets and
    waves. He wore sandals and had
    a cut on one of his big toes." + ], + [ + "A car bomb exploded outside
    the main hospital in Chechny
    #39;s capital, Grozny, on
    Sunday, injuring 17 people in
    an attack apparently targeting
    members of a Chechen security
    force bringing in wounded from
    an earlier explosion" + ], + [ + "AP - Gay marriage is emerging
    as a big enough issue in
    several states to influence
    races both for Congress and
    the presidency." + ], + [ + "More than 30 aid workers have
    been airlifted to safety from
    a town in Sudan #39;s troubled
    Darfur region after fighting
    broke out and their base was
    bombed, a British charity
    says." + ], + [ + "It #39;s the mildest of mild
    winters down here in the south
    of Italy and, last weekend at
    Bcoli, a pretty suburb by the
    seaside west of Naples, the
    customers of Pizzeria quot;Da
    Enrico quot; were making the
    most of it." + ], + [ + "WASHINGTON - A spotty job
    market and stagnant paychecks
    cloud this Labor Day holiday
    for many workers, highlighting
    the importance of pocketbook
    issues in the presidential
    election. \"Working harder
    and enjoying it less,\" said
    economist Ken Mayland,
    president of ClearView
    Economics, summing up the
    state of working America..." + ], + [ + "Canadian Press - MONTREAL (CP)
    - A 19-year-old man charged in
    a firebombing at a Jewish
    elementary school pleaded
    guilty Thursday to arson." + ], + [ + " quot;Resuming uranium
    enrichment is not in our
    agenda. We are still committed
    to the suspension, quot;
    Foreign Ministry spokesman
    Hamid Reza." + ], + [ + "The U.S. military presence in
    Iraq will grow to 150,000
    troops by next month, the
    highest level since the
    invasion last year." + ], + [ + "UPDATE, SUN 9PM: More than a
    million people have left their
    homes in Cuba, as Hurricane
    Ivan approaches. The ferocious
    storm is headed that way,
    after ripping through the
    Cayman Islands, tearing off
    roofs, flooding homes and
    causing general havoc." + ], + [ + "German Chancellor Gerhard
    Schroeder said Sunday that
    there was quot;no problem
    quot; with Germany #39;s
    support to the start of
    negotiations on Turkey #39;s
    entrance into EU." + ], + [ + "MANILA Fernando Poe Jr., the
    popular actor who challenged
    President Gloria Macapagal
    Arroyo in the presidential
    elections this year, died
    early Tuesday." + ], + [ + "AMSTERDAM, NETHERLANDS - A
    Dutch filmmaker who outraged
    members of the Muslim
    community by making a film
    critical of the mistreatment
    of women in Islamic society
    was gunned down and stabbed to
    death Tuesday on an Amsterdam
    street." + ], + [ + "Zimbabwe #39;s most persecuted
    white MP began a year of hard
    labour last night after
    parliament voted to jail him
    for shoving the Justice
    Minister during a debate over
    land seizures." + ], + [ + "A smashing blow is being dealt
    to thousands of future
    pensioners by a law that has
    just been brought into force
    by the Federal Government." + ], + [ + "AP - An Israeli helicopter
    fired two missiles in Gaza
    City after nightfall
    Wednesday, one at a building
    in the Zeitoun neighborhood,
    witnesses said, setting a
    fire." + ], + [ + "The Philippines put the toll
    at more than 1,000 dead or
    missing in four storms in two
    weeks but, even with a break
    in the weather on Saturday" + ], + [ + "Reuters - Four explosions were
    reported at petrol\\stations in
    the Madrid area on Friday,
    Spanish radio stations\\said,
    following a phone warning in
    the name of the armed
    Basque\\separatist group ETA to
    a Basque newspaper." + ], + [ + "WEST PALM BEACH, Fla. -
    Hurricane Jeanne got stronger,
    bigger and faster as it
    battered the Bahamas and bore
    down on Florida Saturday,
    sending huge waves crashing
    onto beaches and forcing
    thousands into shelters just
    weeks after Frances ravaged
    this area..." + ], + [ + "NEW YORK - Elena Dementieva
    shook off a subpar serve that
    produced 15 double-faults, an
    aching left thigh and an upset
    stomach to advance to the
    semifinals at the U.S. Open
    with a 4-6, 6-4, 7-6 (1)
    victory Tuesday over Amelie
    Mauresmo..." + ], + [ + "Prime Minister Dr Manmohan
    Singh inaugurated a research
    centre in the Capital on
    Thursday to mark 400 years of
    compilation of Sikh holy book
    the Guru Granth Sahib." + ], + [ + "THE re-election of British
    Prime Minister Tony Blair
    would be seen as an
    endorsement of the military
    action in Iraq, Prime Minister
    John Howard said today." + ], + [ + "President George W. Bush
    pledged Friday to spend some
    of the political capital from
    his re-election trying to
    secure a lasting Middle East
    peace, and he envisioned the
    establishment" + ], + [ + "NEW DELHI - A bomb exploded
    during an Independence Day
    parade in India's remote
    northeast on Sunday, killing
    at least 15 people, officials
    said, just an hour after Prime
    Minister Manmohan Singh
    pledged to fight terrorism.
    The outlawed United Liberation
    Front of Asom was suspected of
    being behind the attack in
    Assam state and a second one
    later in the area, said Assam
    Inspector General of Police
    Khagen Sharma..." + ], + [ + "A UN envoy to Sudan will visit
    Darfur tomorrow to check on
    the government #39;s claim
    that some 70,000 people
    displaced by conflict there
    have voluntarily returned to
    their homes, a spokesman said." + ], + [ + "AP - Most of the presidential
    election provisional ballots
    rejected so far in Ohio came
    from people who were not even
    registered to vote, election
    officials said after spending
    nearly two weeks poring over
    thousands of disputed votes." + ], + [ + "AP - Rival inmates fought each
    other with knives and sticks
    Wednesday at a San Salvador
    prison, leaving at least 31
    people dead and two dozen
    injured, officials said." + ], + [ + "BAGHDAD - Two Egyptian
    employees of a mobile phone
    company were seized when
    gunmen stormed into their
    Baghdad office, the latest in
    a series of kidnappings in the
    country." + ], + [ + "BRISBANE, Australia - The body
    of a whale resembling a giant
    dolphin that washed up on an
    eastern Australian beach has
    intrigued local scientists,
    who agreed Wednesday that it
    is rare but are not sure just
    how rare." + ], + [ + "President Bush aims to
    highlight American drug-
    fighting aid in Colombia and
    boost a conservative Latin
    American leader with a stop in
    the Andean nation where
    thousands of security forces
    are deployed to safeguard his
    brief stay." + ], + [ + "Dubai - Former Palestinian
    security minister Mohammed
    Dahlan said on Monday that a
    quot;gang of mercenaries quot;
    known to the Palestinian
    police were behind the
    shooting that resulted in two
    deaths in a mourning tent for
    Yasser Arafat in Gaza." + ], + [ + "A Frenchman working for Thales
    SA, Europe #39;s biggest maker
    of military electronics, was
    shot dead while driving home
    at night in the Saudi Arabian
    city of Jeddah." + ], + [ + "China will take tough measures
    this winter to improve the
    country #39;s coal mine safety
    and prevent accidents. State
    Councilor Hua Jianmin said
    Thursday the industry should
    take" + ], + [ + "BAGHDAD (Iraq): As the
    intensity of skirmishes
    swelled on the soils of Iraq,
    dozens of people were put to
    death with toxic shots by the
    US helicopter gunship, which
    targeted the civilians,
    milling around a burning
    American vehicle in a Baghdad
    street on" + ], + [ + "Reuters - A key Iranian
    nuclear facility which
    the\\U.N.'s nuclear watchdog
    has urged Tehran to shut down
    is\\nearing completion, a
    senior Iranian nuclear
    official said on\\Sunday." + ], + [ + "Spain's Football Federation
    launches an investigation into
    racist comments made by
    national coach Luis Aragones." + ], + [ + "Bricks and plaster blew inward
    from the wall, as the windows
    all shattered and I fell to
    the floorwhether from the
    shock wave, or just fright, it
    wasn #39;t clear." + ], + [ + "Surfersvillage Global Surf
    News, 13 September 2004: - -
    Hurricane Ivan, one of the
    most powerful storms to ever
    hit the Caribbean, killed at
    least 16 people in Jamaica,
    where it wrecked houses and
    washed away roads on Saturday,
    but appears to have spared" + ], + [ + "LONDON, England -- A US
    scientist is reported to have
    observed a surprising jump in
    the amount of carbon dioxide,
    the main greenhouse gas." + ], + [ + "Zimbabwe #39;s ruling Zanu-PF
    old guard has emerged on top
    after a bitter power struggle
    in the deeply divided party
    during its five-yearly
    congress, which ended
    yesterday." + ], + [ + "Reuters - Thousands of
    demonstrators pressing
    to\\install Ukraine's
    opposition leader as president
    after a\\disputed election
    launched fresh street rallies
    in the capital\\for the third
    day Wednesday." + ], + [ + "Michael Jackson wishes he had
    fought previous child
    molestation claims instead of
    trying to \"buy peace\", his
    lawyer says." + ], + [ + "North Korea says it will not
    abandon its weapons programme
    after the South admitted
    nuclear activities." + ], + [ + "While there is growing
    attention to ongoing genocide
    in Darfur, this has not
    translated into either a
    meaningful international
    response or an accurate
    rendering of the scale and
    evident course of the
    catastrophe." + ], + [ + "THE prosecution on terrorism
    charges of extremist Islamic
    cleric and accused Jemaah
    Islamiah leader Abu Bakar
    Bashir will rely heavily on
    the potentially tainted
    testimony of at least two
    convicted Bali bombers, his
    lawyers have said." + ], + [ + "Clashes between US troops and
    Sadr militiamen escalated
    Thursday, as the US surrounded
    Najaf for possible siege." + ], + [ + "AFP - A battle group of
    British troops rolled out of
    southern Iraq on a US-
    requested mission to deadlier
    areas near Baghdad, in a major
    political gamble for British
    Prime Minister Tony Blair." + ], + [ + "over half the children in the
    world - suffer extreme
    deprivation because of war,
    HIV/AIDS or poverty, according
    to a report released yesterday
    by the United Nations Children
    #39;s Fund." + ], + [ + "Reuters - Philippine rescue
    teams\\evacuated thousands of
    people from the worst flooding
    in the\\central Luzon region
    since the 1970s as hungry
    victims hunted\\rats and birds
    for food." + ], + [ + "The Afghan president expresses
    deep concern after a bomb
    attack which left at least
    seven people dead." + ], + [ + "Gardez (Afghanistan), Sept. 16
    (Reuters): Afghan President
    Hamid Karzai escaped an
    assassination bid today when a
    rocket was fired at his US
    military helicopter as it was
    landing in the southeastern
    town of Gardez." + ], + [ + "The Jets came up with four
    turnovers by Dolphins
    quarterback Jay Fiedler in the
    second half, including an
    interception returned 66 yards
    for a touchdown." + ], + [ + "DUBLIN -- Prime Minister
    Bertie Ahern urged Irish
    Republican Army commanders
    yesterday to meet what he
    acknowledged was ''a heavy
    burden quot;: disarming and
    disbanding their organization
    in support of Northern
    Ireland's 1998 peace accord." + ], + [ + "While reproductive planning
    and women #39;s equality have
    improved substantially over
    the past decade, says a United
    Nations report, world
    population will increase from
    6.4 billion today to 8.9
    billion by 2050, with the 50
    poorest countries tripling in" + ], + [ + "BAR's Anthony Davidson and
    Jenson Button set the pace at
    the first Chinese Grand Prix." + ], + [ + "WASHINGTON - Contradicting the
    main argument for a war that
    has cost more than 1,000
    American lives, the top U.S.
    arms inspector reported
    Wednesday that he found no
    evidence that Iraq produced
    any weapons of mass
    destruction after 1991..." + ], + [ + "AFP - Style mavens will be
    scanning the catwalks in Paris
    this week for next spring's
    must-have handbag, as a
    sweeping exhibition at the
    French capital's fashion and
    textile museum reveals the bag
    in all its forms." + ], + [ + "Canadian Press - SAINT-
    QUENTIN, N.B. (CP) - A major
    highway in northern New
    Brunswick remained closed to
    almost all traffic Monday, as
    local residents protested
    planned health care cuts." + ], + [ + " NAJAF, Iraq (Reuters) - A
    radical Iraqi cleric leading a
    Shi'ite uprising agreed on
    Wednesday to disarm his
    militia and leave one of the
    country's holiest Islamic
    shrines after warnings of an
    onslaught by government
    forces." + ], + [ + "Saudi security forces have
    killed a wanted militant near
    the scene of a deadly shootout
    Thursday. Officials say the
    militant was killed in a
    gunbattle Friday in the
    northern town of Buraida,
    hours after one" + ], + [ + "Two South Africans acquitted
    by a Zimbabwean court of
    charges related to the alleged
    coup plot in Equatorial Guinea
    are to be questioned today by
    the South African authorities." + ], + [ + "MOSCOW (CP) - Russia mourned
    89 victims of a double air
    disaster today as debate
    intensified over whether the
    two passenger liners could
    have plunged almost
    simultaneously from the sky by
    accident." + ], + [ + "Australia #39;s prime minister
    says a body found in Fallujah
    is likely that of kidnapped
    aid worker Margaret Hassan.
    John Howard told Parliament a
    videotape of an Iraqi
    terrorist group executing a
    Western woman appears to have
    been genuine." + ], + [ + "AP - Their first debate less
    than a week away, President
    Bush and Democrat John Kerry
    kept their public schedules
    clear on Saturday and began to
    focus on their prime-time
    showdown." + ], + [ + "PARIS Getting to the bottom of
    what killed Yassar Arafat
    could shape up to be an ugly
    family tug-of-war. Arafat
    #39;s half-brother and nephew
    want copies of Arafat #39;s
    medical records from the
    suburban Paris hospital" + ], + [ + " THE HAGUE (Reuters) - Former
    Yugoslav President Slobodan
    Milosevic condemned his war
    crimes trial as a \"pure farce\"
    on Wednesday in a defiant
    finish to his opening defense
    statement against charges of
    ethnic cleansing in the
    Balkans." + ], + [ + " GUWAHATI, India (Reuters) -
    People braved a steady drizzle
    to come out to vote in a
    remote northeast Indian state
    on Thursday, as troops
    guarded polling stations in an
    election being held under the
    shadow of violence." + ], + [ + "AFP - Three of the nine
    Canadian sailors injured when
    their newly-delivered,
    British-built submarine caught
    fire in the North Atlantic
    were airlifted Wednesday to
    hospital in northwest Ireland,
    officials said." + ], + [ + "BAGHDAD, Iraq - A series of
    strong explosions shook
    central Baghdad near dawn
    Sunday, and columns of thick
    black smoke rose from the
    Green Zone where U.S. and
    Iraqi government offices are
    located..." + ], + [ + "Sven-Goran Eriksson may gamble
    by playing goalkeeper Paul
    Robinson and striker Jermain
    Defoe in Poland." + ], + [ + "Foreign Secretary Jack Straw
    has flown to Khartoum on a
    mission to pile the pressure
    on the Sudanese government to
    tackle the humanitarian
    catastrophe in Darfur." + ], + [ + "Reuters - A senior U.S.
    official said on
    Wednesday\\deals should not be
    done with hostage-takers ahead
    of the\\latest deadline set by
    Afghan Islamic militants who
    have\\threatened to kill three
    kidnapped U.N. workers." + ], + [ + "Sinn Fein leader Gerry Adams
    has put the pressure for the
    success or failure of the
    Northern Ireland assembly
    talks firmly on the shoulders
    of Ian Paisley." + ], + [ + " JAKARTA (Reuters) - President
    Megawati Sukarnoputri urged
    Indonesians on Thursday to
    accept the results of the
    country's first direct
    election of a leader, but
    stopped short of conceding
    defeat." + ], + [ + "ISLAMABAD: Pakistan early
    Monday test-fired its
    indigenously developed short-
    range nuclear-capable Ghaznavi
    missile, the Inter Services
    Public Relations (ISPR) said
    in a statement." + ], + [ + "The trial of a man accused of
    murdering York backpacker
    Caroline Stuttle begins in
    Australia." + ], + [ + "BRUSSELS: The EU sought
    Wednesday to keep pressure on
    Turkey over its bid to start
    talks on joining the bloc, as
    last-minute haggling seemed
    set to go down to the wire at
    a summit poised to give a
    green light to Ankara." + ], + [ + "AP - J. Cofer Black, the State
    Department official in charge
    of counterterrorism, is
    leaving government in the next
    few weeks." + ], + [ + "AFP - The United States
    presented a draft UN
    resolution that steps up the
    pressure on Sudan over the
    crisis in Darfur, including
    possible international
    sanctions against its oil
    sector." + ], + [ + "AFP - At least 33 people were
    killed and dozens others
    wounded when two bombs ripped
    through a congregation of
    Sunni Muslims in Pakistan's
    central city of Multan, police
    said." + ], + [ + "AFP - A series of torchlight
    rallies and vigils were held
    after darkness fell on this
    central Indian city as victims
    and activists jointly
    commemorated a night of horror
    20 years ago when lethal gas
    leaked from a pesticide plant
    and killed thousands." + ], + [ + "A Zimbabwe court Friday
    convicted a British man
    accused of leading a coup plot
    against the government of oil-
    rich Equatorial Guinea on
    weapons charges, but acquitted
    most of the 69 other men held
    with him." + ], + [ + "Canadian Press - TORONTO (CP)
    - The fatal stabbing of a
    young man trying to eject
    unwanted party guests from his
    family home, the third such
    knifing in just weeks, has
    police worried about a
    potentially fatal holiday
    recipe: teens, alcohol and
    knives." + ], + [ + "MOSCOW - A female suicide
    bomber set off a shrapnel-
    filled explosive device
    outside a busy Moscow subway
    station on Tuesday night,
    officials said, killing 10
    people and injuring more than
    50." + ], + [ + "ABIDJAN (AFP) - Two Ivory
    Coast military aircraft
    carried out a second raid on
    Bouake, the stronghold of the
    former rebel New Forces (FN)
    in the divided west African
    country, a French military
    source told AFP." + ], + [ + "AFP - A state of civil
    emergency in the rebellion-hit
    Indonesian province of Aceh
    has been formally extended by
    six month, as the country's
    president pledged to end
    violence there without foreign
    help." + ], + [ + "US Secretary of State Colin
    Powell on Monday said he had
    spoken to both Indian Foreign
    Minister K Natwar Singh and
    his Pakistani counterpart
    Khurshid Mahmud Kasuri late
    last week before the two met
    in New Delhi this week for
    talks." + ], + [ + "NEW YORK - Victor Diaz hit a
    tying, three-run homer with
    two outs in the ninth inning,
    and Craig Brazell's first
    major league home run in the
    11th gave the New York Mets a
    stunning 4-3 victory over the
    Chicago Cubs on Saturday.
    The Cubs had much on the
    line..." + ], + [ + "AFP - At least 54 people have
    died and more than a million
    have fled their homes as
    torrential rains lashed parts
    of India and Bangladesh,
    officials said." + ], + [ + "LOS ANGELES - California has
    adopted the world's first
    rules to reduce greenhouse
    emissions for autos, taking
    what supporters see as a
    dramatic step toward cleaning
    up the environment but also
    ensuring higher costs for
    drivers. The rules may lead
    to sweeping changes in
    vehicles nationwide,
    especially if other states opt
    to follow California's
    example..." + ], + [ + "AFP - Republican and
    Democratic leaders each
    declared victory after the
    first head-to-head sparring
    match between President George
    W. Bush and Democratic
    presidential hopeful John
    Kerry." + ], + [ + "Last night in New York, the UN
    secretary-general was given a
    standing ovation - a robust
    response to a series of
    attacks in past weeks." + ], + [ + "JOHANNESBURG -- Meeting in
    Nigeria four years ago,
    African leaders set a goal
    that 60 percent of children
    and pregnant women in malaria-
    affected areas around the
    continent would be sleeping
    under bed nets by the end of
    2005." + ], + [ + "AP - Duke Bainum outspent Mufi
    Hannemann in Honolulu's most
    expensive mayoral race, but
    apparently failed to garner
    enough votes in Saturday's
    primary to claim the office
    outright." + ], + [ + "POLITICIANS and aid agencies
    yesterday stressed the
    importance of the media in
    keeping the spotlight on the
    appalling human rights abuses
    taking place in the Darfur
    region of Sudan." + ], + [ + "\\Children who have a poor diet
    are more likely to become
    aggressive and anti-social, US
    researchers believe." + ], + [ + "Canadian Press - OTTAWA (CP) -
    Contrary to Immigration
    Department claims, there is no
    shortage of native-borne
    exotic dancers in Canada, says
    a University of Toronto law
    professor who has studied the
    strip club business." + ], + [ + "The European Union presidency
    yesterday expressed optimism
    that a deal could be struck
    over Turkey #39;s refusal to
    recognize Cyprus in the lead-
    up to next weekend #39;s EU
    summit, which will decide
    whether to give Ankara a date
    for the start of accession
    talks." + ], + [ + " WASHINGTON (Reuters) -
    President Bush on Friday set a
    four-year goal of seeing a
    Palestinian state established
    and he and British Prime
    Minister Tony Blair vowed to
    mobilize international
    support to help make it happen
    now that Yasser Arafat is
    dead." + ], + [ + " KHARTOUM (Reuters) - Sudan on
    Saturday questioned U.N.
    estimates that up to 70,000
    people have died from hunger
    and disease in its remote
    Darfur region since a
    rebellion began 20 months
    ago." + ], + [ + "AFP - At least four Georgian
    soldiers were killed and five
    wounded in overnight clashes
    in Georgia's separatist, pro-
    Russian region of South
    Ossetia, Georgian officers
    near the frontline with
    Ossetian forces said early
    Thursday." + ], + [ + "The European Commission is
    expected later this week to
    recommend EU membership talks
    with Turkey. Meanwhile, German
    Chancellor Gerhard Schroeder
    and Turkish Prime Minister
    Tayyip Erdogan are
    anticipating a quot;positive
    report." + ], + [ + "The Canadian government
    signalled its intention
    yesterday to reintroduce
    legislation to decriminalise
    the possession of small
    amounts of marijuana." + ], + [ + "A screensaver targeting spam-
    related websites appears to
    have been too successful." + ], + [ + " ABIDJAN (Reuters) - Ivory
    Coast warplanes killed nine
    French soldiers on Saturday in
    a bombing raid during the
    fiercest clashes with rebels
    for 18 months and France hit
    back by destroying most of
    the West African country's
    small airforce." + ], + [ + "GAZA CITY, Gaza Strip --
    Islamic militant groups behind
    many suicide bombings
    dismissed yesterday a call
    from Mahmoud Abbas, the
    interim Palestinian leader, to
    halt attacks in the run-up to
    a Jan. 9 election to replace
    Yasser Arafat." + ], + [ + "Secretary of State Colin
    Powell is wrapping up an East
    Asia trip focused on prodding
    North Korea to resume talks
    aimed at ending its nuclear-
    weapons program." + ], + [ + "The risk of intestinal damage
    from common painkillers may be
    higher than thought, research
    suggests." + ], + [ + "AN earthquake measuring 7.3 on
    the Richter Scale hit western
    Japan this morning, just hours
    after another strong quake
    rocked the area." + ], + [ + "Britain #39;s Prince Harry,
    struggling to shed a growing
    quot;wild child quot; image,
    won #39;t apologize to a
    photographer he scuffled with
    outside an exclusive London
    nightclub, a royal spokesman
    said on Saturday." + ], + [ + "Pakistan #39;s interim Prime
    Minister Chaudhry Shaujaat
    Hussain has announced his
    resignation, paving the way
    for his successor Shauket
    Aziz." + ], + [ + "A previously unknown group
    calling itself Jamaat Ansar
    al-Jihad al-Islamiya says it
    set fire to a Jewish soup
    kitchen in Paris, according to
    an Internet statement." + ], + [ + "The deadliest attack on
    Americans in Iraq since May
    came as Iraqi officials
    announced that Saddam
    Hussein's deputy had not been
    captured on Sunday." + ], + [ + "A fundraising concert will be
    held in London in memory of
    the hundreds of victims of the
    Beslan school siege." + ], + [ + "AFP - The landmark trial of a
    Rwandan Roman Catholic priest
    accused of supervising the
    massacre of 2,000 of his Tutsi
    parishioners during the
    central African country's 1994
    genocide opens at a UN court
    in Tanzania." + ], + [ + "The Irish government has
    stepped up its efforts to free
    the British hostage in Iraq,
    Ken Bigley, whose mother is
    from Ireland, by talking to
    diplomats from Iran and
    Jordan." + ], + [ + "AP - Republican Sen. Lincoln
    Chafee, who flirted with
    changing political parties in
    the wake of President Bush's
    re-election victory, says he
    will stay in the GOP." + ], + [ + "South Korean President Roh
    Moo-hyun pays a surprise visit
    to troops in Iraq, after his
    government decided to extend
    their mandate." + ], + [ + "As the European Union
    approaches a contentious
    decision - whether to let
    Turkey join the club - the
    Continent #39;s rulers seem to
    have left their citizens
    behind." + ], + [ + " BAGHDAD (Reuters) - A
    deadline set by militants who
    have threatened to kill two
    Americans and a Briton seized
    in Iraq was due to expire
    Monday, and more than two
    dozen other hostages were
    also facing death unless rebel
    demands were met." + ], + [ + "Some Venezuelan television
    channels began altering their
    programs Thursday, citing
    fears of penalties under a new
    law restricting violence and
    sexual content over the
    airwaves." + ], + [ + "afrol News, 4 October -
    Hundred years of conservation
    efforts have lifted the
    southern black rhino
    population from about hundred
    to 11,000 animals." + ], + [ + "THE death of Philippine movie
    star and defeated presidential
    candidate Fernando Poe has
    drawn some political backlash,
    with some people seeking to
    use his sudden demise as a
    platform to attack President
    Gloria Arroyo." + ], + [ + "VIENTIANE, Laos China moved
    yet another step closer in
    cementing its economic and
    diplomatic relationships with
    Southeast Asia today when
    Prime Minister Wen Jiabao
    signed a trade accord at a
    regional summit that calls for
    zero tariffs on a wide range
    of" + ], + [ + "British judges in London
    Tuesday ordered radical Muslim
    imam Abu Hamza to stand trial
    for soliciting murder and
    inciting racial hatred." + ], + [ + "SARASOTA, Fla. - The
    devastation brought on by
    Hurricane Charley has been
    especially painful for an
    elderly population that is
    among the largest in the
    nation..." + ], + [ + " quot;He is charged for having
    a part in the Bali incident,
    quot; state prosecutor Andi
    Herman told Reuters on
    Saturday. bombing attack at
    the US-run JW." + ], + [ + "The United Nations called on
    Monday for an immediate
    ceasefire in eastern Congo as
    fighting between rival army
    factions flared for a third
    day." + ], + [ + "Allowing dozens of casinos to
    be built in the UK would bring
    investment and thousands of
    jobs, Tony Blair says." + ], + [ + "The shock here was not just
    from the awful fact itself,
    that two vibrant young Italian
    women were kidnapped in Iraq,
    dragged from their office by
    attackers who, it seems, knew
    their names." + ], + [ + "Tehran/Vianna, Sept. 19 (NNN):
    Iran on Sunday rejected the
    International Atomic Energy
    Agency (IAEA) call to suspend
    of all its nuclear activities,
    saying that it will not agree
    to halt uranium enrichment." + ], + [ + "A 15-year-old Argentine
    student opened fire at his
    classmates on Tuesday in a
    middle school in the south of
    the Buenos Aires province,
    leaving at least four dead and
    five others wounded, police
    said." + ], + [ + "NEW YORK - A cable pay-per-
    view company has decided not
    to show a three-hour election
    eve special with filmmaker
    Michael Moore that included a
    showing of his documentary
    \"Fahrenheit 9/11,\" which is
    sharply critical of President
    Bush. The company, iN
    DEMAND, said Friday that its
    decision is due to \"legitimate
    business and legal concerns.\"
    A spokesman would not
    elaborate..." + ], + [ + "Democracy candidates picked up
    at least one more seat in
    parliament, according to exit
    polls." + ], + [ + "The IOC wants suspended
    Olympic member Ivan Slavkov to
    be thrown out of the
    organisation." + ], + [ + " BANGKOK (Reuters) - The
    ouster of Myanmar's prime
    minister, architect of a
    tentative \"roadmap to
    democracy,\" has dashed faint
    hopes for an end to military
    rule and leaves Southeast
    Asia's policy of constructive
    engagement in tatters." + ], + [ + "The anguish of hostage Kenneth
    Bigley in Iraq hangs over
    Prime Minister Tony Blair
    today as he faces the twin
    test of a local election and a
    debate by his Labour Party
    about the divisive war." + ], + [ + " quot;There were 16 people
    travelling aboard. ... It
    crashed into a mountain, quot;
    Col. Antonio Rivero, head of
    the Civil Protection service,
    told." + ], + [ + "President Bush is reveling in
    winning the popular vote and
    feels he can no longer be
    considered a one-term accident
    of history." + ], + [ + "Canadian Press - TORONTO (CP)
    - Thousands of Royal Bank
    clerks are being asked to
    display rainbow stickers at
    their desks and cubicles to
    promote a safe work
    environment for gays,
    lesbians, and bisexuals." + ], + [ + " BAGHDAD (Reuters) - A car
    bomb killed two American
    soldiers and wounded eight
    when it exploded in Baghdad on
    Saturday, the U.S. military
    said in a statement." + ], + [ + "Sudanese authorities have
    moved hundreds of pro-
    government fighters from the
    crisis-torn Darfur region to
    other parts of the country to
    keep them out of sight of
    foreign military observers
    demanding the militia #39;s
    disarmament, a rebel leader
    charged" + ], + [ + "BAGHDAD, Iraq - A car bomb
    Tuesday ripped through a busy
    market near a Baghdad police
    headquarters where Iraqis were
    waiting to apply for jobs on
    the force, and gunmen opened
    fire on a van carrying police
    home from work in Baqouba,
    killing at least 59 people
    total and wounding at least
    114. The attacks were the
    latest attempts by militants
    to wreck the building of a
    strong Iraqi security force, a
    keystone of the U.S..." + ], + [ + "The Israeli prime minister
    said today that he wanted to
    begin withdrawing settlers
    from Gaza next May or June." + ], + [ + "Queen Elizabeth II stopped
    short of apologizing for the
    Allies #39; bombing of Dresden
    during her first state visit
    to Germany in 12 years and
    instead acknowledged quot;the
    appalling suffering of war on
    both sides." + ], + [ + "AP - Fugitive Taliban leader
    Mullah Mohammed Omar has
    fallen out with some of his
    lieutenants, who blame him for
    the rebels' failure to disrupt
    the landmark Afghan
    presidential election, the
    U.S. military said Wednesday." + ], + [ + "HAVANA -- Cuban President
    Fidel Castro's advancing age
    -- and ultimately his
    mortality -- were brought home
    yesterday, a day after he
    fractured a knee and arm when
    he tripped and fell at a
    public event." + ], + [ + " BRASILIA, Brazil (Reuters) -
    The United States and Brazil
    predicted on Tuesday Latin
    America's largest country
    would resolve a dispute with
    the U.N. nuclear watchdog over
    inspections of a uranium
    enrichment plant." + ], + [ + "Two bombs exploded today near
    a tea shop and wounded 20
    people in southern Thailand,
    police said, as violence
    continued unabated in the
    Muslim-majority region where
    residents are seething over
    the deaths of 78 detainees
    while in military custody." + ], + [ + "Prime Minister Junichiro
    Koizumi, back in Tokyo after
    an 11-day diplomatic mission
    to the Americas, hunkered down
    with senior ruling party
    officials on Friday to focus
    on a major reshuffle of
    cabinet and top party posts." + ], + [ + "NEW YORK - A sluggish gross
    domestic product reading was
    nonetheless better than
    expected, prompting investors
    to send stocks slightly higher
    Friday on hopes that the
    economic slowdown would not be
    as bad as first thought.
    The 2.8 percent GDP growth in
    the second quarter, a revision
    from the 3 percent preliminary
    figure reported in July, is a
    far cry from the 4.5 percent
    growth in the first quarter..." + ], + [ + " SEOUL (Reuters) - A huge
    explosion in North Korea last
    week may have been due to a
    combination of demolition work
    for a power plant and
    atmospheric clouds, South
    Korea's spy agency said on
    Wednesday." + ], + [ + "AP - Sudan's U.N. ambassador
    challenged the United States
    to send troops to the Darfur
    region if it really believes a
    genocide is taking place as
    the U.S. Congress and
    President Bush's
    administration have
    determined." + ], + [ + "AFP - A junior British
    minister said that people
    living in camps after fleeing
    their villages because of
    conflict in Sudan's Darfur
    region lived in fear of
    leaving their temporary homes
    despite a greater presence now
    of aid workers." + ], + [ + "US Deputy Secretary of State
    Richard Armitage (L) shakes
    hands with Pakistani Foreign
    Minister Khurshid Kasuri prior
    to their meeting in Islamabad,
    09 November 2004." + ], + [ + "AP - Democrat John Edwards
    kept up a long-distance debate
    over his \"two Americas\"
    campaign theme with Vice
    President Dick Cheney on
    Tuesday, saying it was no
    illusion to thousands of laid-
    off workers in Ohio." + ], + [ + "A group of 76 Eritreans on a
    repatriation flight from Libya
    Friday forced their plane to
    change course and land in the
    Sudanese capital, Khartoum,
    where they" + ], + [ + "AP - They say that opposites
    attract, and in the case of
    Sen. John Kerry and Teresa
    Heinz Kerry, that may be true
    #151; at least in their public
    personas." + ], + [ + "AP - Impoverished North Korea
    might resort to selling
    weapons-grade plutonium to
    terrorists for much-needed
    cash, and that would be
    \"disastrous for the world,\"
    the top U.S. military
    commander in South Korea said
    Friday." + ], + [ + "Dubai: Osama bin Laden on
    Thursday called on his
    fighters to strike Gulf oil
    supplies and warned Saudi
    leaders they risked a popular
    uprising in an audio message
    said to be by the Western
    world #39;s most wanted terror
    mastermind." + ], + [ + "BEIJING -- Chinese authorities
    have arrested or reprimanded
    more than 750 officials in
    recent months in connection
    with billions of dollars in
    financial irregularities
    ranging from unpaid taxes to
    embezzlement, according to a
    report made public yesterday." + ], + [ + "Canadian Press - GAUHATI,
    India (AP) - Residents of
    northeastern India were
    bracing for more violence
    Friday, a day after bombs
    ripped through two buses and a
    grenade was hurled into a
    crowded market in attacks that
    killed four people and wounded
    54." + ], + [ + "Some 28 people are charged
    with trying to overthrow
    Sudan's President Bashir,
    reports say." + ], + [ + "Hong Kong #39;s Beijing-backed
    chief executive yesterday
    ruled out any early moves to
    pass a controversial national
    security law which last year
    sparked a street protest by
    half a million people." + ], + [ + "Beer consumption has doubled
    over the past five years,
    prompting legislators to
    implement new rules." + ], + [ + "At least 79 people were killed
    and 74 were missing after some
    of the worst rainstorms in
    recent years triggered
    landslides and flash floods in
    southwest China, disaster
    relief officials said
    yesterday." + ], + [ + "BANGKOK Thai military aircraft
    dropped 100 million paper
    birds over southern Thailand
    on Sunday in a gesture
    intended to promote peace in
    mainly Muslim provinces, where
    more than 500 people have died
    this year in attacks by
    separatist militants and" + ], + [ + "Insurgents threatened on
    Saturday to cut the throats of
    two Americans and a Briton
    seized in Baghdad, and
    launched a suicide car bomb
    attack on Iraqi security
    forces in Kirkuk that killed
    at least 23 people." + ], + [ + " BEIJING (Reuters) - Secretary
    of State Colin Powell will
    urge China Monday to exert its
    influence over North Korea to
    resume six-party negotiations
    on scrapping its suspected
    nuclear weapons program." + ], + [ + "Reuters - An Israeli missile
    strike killed at least\\two
    Hamas gunmen in Gaza City
    Monday, a day after a
    top\\commander of the Islamic
    militant group was killed in a
    similar\\attack, Palestinian
    witnesses and security sources
    said." + ], + [ + "AP - The Pentagon has restored
    access to a Web site that
    assists soldiers and other
    Americans living overseas in
    voting, after receiving
    complaints that its security
    measures were preventing
    legitimate voters from using
    it." + ], + [ + "Montreal - The British-built
    Canadian submarine HMCS
    Chicoutimi, crippled since a
    fire at sea that killed one of
    its crew, is under tow and is
    expected in Faslane, Scotland,
    early next week, Canadian
    naval commander Tyrone Pile
    said on Friday." + ], + [ + "Thirty-five Pakistanis freed
    from the US Guantanamo Bay
    prison camp arrived home on
    Saturday and were taken
    straight to prison for further
    interrogation, the interior
    minister said." + ], + [ + "Witnesses in the trial of a US
    soldier charged with abusing
    prisoners at Abu Ghraib have
    told the court that the CIA
    sometimes directed abuse and
    orders were received from
    military command to toughen
    interrogations." + ], + [ + "AP - Italian and Lebanese
    authorities have arrested 10
    suspected terrorists who
    planned to blow up the Italian
    Embassy in Beirut, an Italian
    news agency and the Defense
    Ministry said Tuesday." + ], + [ + "CORAL GABLES, Fla. -- John F.
    Kerry and President Bush
    clashed sharply over the war
    in Iraq last night during the
    first debate of the
    presidential campaign season,
    with the senator from
    Massachusetts accusing" + ], + [ + "GONAIVES, Haiti -- While
    desperately hungry flood
    victims wander the streets of
    Gonaives searching for help,
    tons of food aid is stacking
    up in a warehouse guarded by
    United Nations peacekeepers." + ], + [ + "AFP - The resignation of
    disgraced Fiji Vice-President
    Ratu Jope Seniloli failed to
    quell anger among opposition
    leaders and the military over
    his surprise release from
    prison after serving just
    three months of a four-year
    jail term for his role in a
    failed 2000 coup." + ], + [ + "Sourav Ganguly files an appeal
    against a two-match ban
    imposed for time-wasting." + ], + [ + "Greek police surrounded a bus
    full of passengers seized by
    armed hijackers along a
    highway from an Athens suburb
    Wednesday, police said." + ], + [ + "BAGHDAD: A suicide car bomber
    attacked a police convoy in
    Baghdad yesterday as guerillas
    kept pressure on Iraq #39;s
    security forces despite a
    bloody rout of insurgents in
    Fallujah." + ], + [ + "Fixtures and fittings from
    Damien Hirst's restaurant
    Pharmacy sell for 11.1, 8m
    more than expected." + ], + [ + "Five years after catastrophic
    floods and mudslides killed
    thousands along Venezuela's
    Caribbean coast, survivors in
    this town still see the signs
    of destruction - shattered
    concrete walls and tall weeds
    growing atop streets covered
    in dried mud." + ], + [ + "The UN nuclear watchdog
    confirmed Monday that nearly
    400 tons of powerful
    explosives that could be used
    in conventional or nuclear
    missiles disappeared from an
    unguarded military
    installation in Iraq." + ], + [ + "Militants in Iraq have killed
    the second of two US civilians
    they were holding hostage,
    according to a statement on an
    Islamist website." + ], + [ + "More than 4,000 American and
    Iraqi soldiers mounted a
    military assault on the
    insurgent-held city of
    Samarra." + ], + [ + " MEXICO CITY (Reuters) -
    Mexican President Vicente Fox
    said on Monday he hoped
    President Bush's re-election
    meant a bilateral accord on
    migration would be reached
    before his own term runs out
    at the end of 2006." + ], + [ + " BRUSSELS (Reuters) - French
    President Jacques Chirac said
    on Friday he had not refused
    to meet Iraqi interim Prime
    Minister Iyad Allawi after
    reports he was snubbing the
    Iraqi leader by leaving an EU
    meeting in Brussels early." + ], + [ + "NEW YORK - Former President
    Bill Clinton was in good
    spirits Saturday, walking
    around his hospital room in
    street clothes and buoyed by
    thousands of get-well messages
    as he awaited heart bypass
    surgery early this coming
    week, people close to the
    family said. Clinton was
    expected to undergo surgery as
    early as Monday but probably
    Tuesday, said Democratic Party
    Chairman Terry McAuliffe, who
    said the former president was
    \"upbeat\" when he spoke to him
    by phone Friday..." + ], + [ + "Poland will cut its troops in
    Iraq early next year and won
    #39;t stay in the country
    quot;an hour longer quot; than
    needed, the country #39;s
    prime minister said Friday." + ], + [ + "A car bomb exploded at a US
    military convoy in the
    northern Iraqi city of Mosul,
    causing several casualties,
    the army and Iraqi officials
    said." + ], + [ + "ATHENS, Greece - Michael
    Phelps took care of qualifying
    for the Olympic 200-meter
    freestyle semifinals Sunday,
    and then found out he had been
    added to the American team for
    the evening's 400 freestyle
    relay final. Phelps' rivals
    Ian Thorpe and Pieter van den
    Hoogenband and teammate Klete
    Keller were faster than the
    teenager in the 200 free
    preliminaries..." + ], + [ + " JERUSALEM (Reuters) - Israeli
    Prime Minister Ariel Sharon
    intends to present a timetable
    for withdrawal from Gaza to
    lawmakers from his Likud Party
    Tuesday despite a mutiny in
    the right-wing bloc over the
    plan." + ], + [ + " KABUL (Reuters) - Three U.N.
    workers held by militants in
    Afghanistan were in their
    third week of captivity on
    Friday after calls from both
    sides for the crisis to be
    resolved ahead of this
    weekend's Muslim festival of
    Eid al-Fitr." + ], + [ + "ROME, Oct 29 (AFP) - French
    President Jacques Chirac urged
    the head of the incoming
    European Commission Friday to
    take quot;the appropriate
    decisions quot; to resolve a
    row over his EU executive team
    which has left the EU in
    limbo." + ], + [ + "Russia's parliament will
    launch an inquiry into a
    school siege that killed over
    300 hostages, President
    Vladimir Putin said on Friday,
    but analysts doubt it will
    satisfy those who blame the
    carnage on security services." + ], + [ + "Tony Blair talks to business
    leaders about new proposals
    for a major shake-up of the
    English exam system." + ], + [ + "KUALA LUMPUR, Malaysia A
    Malaysian woman has claimed a
    new world record after living
    with over six-thousand
    scorpions for 36 days
    straight." + ], + [ + "could take drastic steps if
    the talks did not proceed as
    Tehran wants. Mehr news agency
    quoted him as saying on
    Wednesday. that could be used" + ], + [ + "Israeli authorities have
    launched an investigation into
    death threats against Israeli
    Prime Minister Ariel Sharon
    and other officials supporting
    his disengagement plan from
    Gaza and parts of the West
    Bank, Jerusalem police said
    Tuesday." + ], + [ + "It was a carefully scripted
    moment when Russian President
    Vladimir Putin began quoting
    Taras Shevchenko, this country
    #39;s 19th-century bard,
    during a live television" + ], + [ + "China says it welcomes Russia
    #39;s ratification of the
    Kyoto Protocol that aims to
    stem global warming by
    reducing greenhouse-gas
    emissions." + ], + [ + "The Metropolitan
    Transportation Authority is
    proposing a tax increase to
    raise \\$900 million a year to
    help pay for a five-year
    rebuilding program." + ], + [ + "AFP - The Canadian armed
    forces chief of staff on was
    elected to take over as head
    of NATO's Military Committee,
    the alliance's highest
    military authority, military
    and diplomatic sources said." + ], + [ + "AFP - Two proposed resolutions
    condemning widespread rights
    abuses in Sudan and Zimbabwe
    failed to pass a UN committee,
    mired in debate between
    African and western nations." + ], + [ + "The return of noted reformer
    Nabil Amr to Palestinian
    politics comes at a crucial
    juncture." + ], + [ + "AP - After two debates, voters
    have seen President Bush look
    peevish and heard him pass the
    buck. They've watched Sen.
    John Kerry deny he's a flip-
    flopper and then argue that
    Saddam Hussein was a threat
    #151; and wasn't. It's no
    wonder so few minds have
    changed." + ], + [ + "The United States says the
    Lebanese parliament #39;s
    decision Friday to extend the
    term of pro-Syrian President
    Emile Lahoud made a
    quot;crude mockery of
    democratic principles." + ], + [ + "Canadian Press - OTTAWA (CP) -
    The prime minister says he's
    been assured by President
    George W. Bush that the U.S.
    missile defence plan does not
    necessarily involve the
    weaponization of space." + ], + [ + "Fifty-six miners are dead and
    another 92 are still stranded
    underground after a gas blast
    at the Daping coal mine in
    Central China #39;s Henan
    Province, safety officials
    said Thursday." + ], + [ + "BEIRUT, Lebanon - Three
    Lebanese men and their Iraqi
    driver have been kidnapped in
    Iraq, the Lebanese Foreign
    Ministry said Sunday, as
    Iraq's prime minister said his
    government was working for the
    release of two Americans and a
    Briton also being held
    hostage. Gunmen snatched
    the three Lebanese, who worked
    for a travel agency with a
    branch in Baghdad, as they
    drove on the highway between
    the capital and Fallujah on
    Friday night, a ministry
    official said..." + ], + [ + "PORTLAND, Ore. - It's been
    almost a year since singer-
    songwriter Elliott Smith
    committed suicide, and fans
    and friends will be looking
    for answers as the posthumous
    \"From a Basement on the Hill\"
    is released..." + ], + [ + " GAZA (Reuters) - Israel
    expanded its military
    offensive in northern Gaza,
    launching two air strikes
    early on Monday that killed
    at least three Palestinians
    and wounded two, including a
    senior Hamas leader, witnesses
    and medics said." + ], + [ + "Reuters - Sudan's government
    resumed\\talks with rebels in
    the oil-producing south on
    Thursday while\\the United
    Nations set up a panel to
    investigate charges
    of\\genocide in the west of
    Africa's largest country." + ], + [ + "NEW YORK - Optimism that the
    embattled technology sector
    was due for a recovery sent
    stocks modestly higher Monday
    despite a new revenue warning
    from semiconductor company
    Broadcom Inc. While
    Broadcom, which makes chips
    for television set-top boxes
    and other electronics, said
    high inventories resulted in
    delayed shipments, investors
    were encouraged as it said
    future quarters looked
    brighter..." + ], + [ + "Carlos Barrios Orta squeezed
    himself into his rubber diving
    suit, pulled on an 18-pound
    helmet that made him look like
    an astronaut, then lowered
    himself into the sewer. He
    disappeared into the filthy
    water, which looked like some
    cauldron of rancid beef stew,
    until the only sign of him was
    air bubbles breaking the
    surface." + ], + [ + "The mutilated body of a woman
    of about sixty years,
    amputated of arms and legs and
    with the sliced throat, has
    been discovered this morning
    in a street of the south of
    Faluya by the Marines,
    according to a statement by
    AFP photographer." + ], + [ + "(SH) - In Afghanistan, Hamid
    Karzai defeated a raft of
    candidates to win his historic
    election. In Iraq, more than
    200 political parties have
    registered for next month
    #39;s elections." + ], + [ + "Margaret Hassan, who works for
    charity Care International,
    was taken hostage while on her
    way to work in Baghdad. Here
    are the main events since her
    kidnapping." + ], + [ + "Reuters - The United States
    modified its\\call for U.N.
    sanctions against Sudan on
    Tuesday but still kept\\up the
    threat of punitive measures if
    Khartoum did not
    stop\\atrocities in its Darfur
    region." + ], + [ + "Human rights and environmental
    activists have hailed the
    award of the 2004 Nobel Peace
    Prize to Wangari Maathai of
    Kenya as fitting recognition
    of the growing role of civil
    society" + ], + [ + "An Al Qaeda-linked militant
    group beheaded an American
    hostage in Iraq and threatened
    last night to kill another two
    Westerners in 24 hours unless
    women prisoners were freed
    from Iraqi jails." + ], + [ + "Argentina, Denmark, Greece,
    Japan and Tanzania on Friday
    won coveted two-year terms on
    the UN Security Council at a
    time when pressure is mounting
    to expand the powerful
    15-nation body." + ], + [ + "ISLAMABAD: Pakistan and India
    agreed on Friday to re-open
    the Khokhropar-Munabao railway
    link, which was severed nearly
    40 years ago." + ], + [ + "NEW YORK - Stocks moved higher
    Friday as a stronger than
    expected retail sales report
    showed that higher oil prices
    aren't scaring consumers away
    from spending. Federal Reserve
    Chairman Alan Greenspan's
    positive comments on oil
    prices also encouraged
    investors..." + ], + [ + "Saddam Hussein lives in an
    air-conditioned 10-by-13 foot
    cell on the grounds of one of
    his former palaces, tending
    plants and proclaiming himself
    Iraq's lawful ruler." + ], + [ + "AP - Brazilian U.N.
    peacekeepers will remain in
    Haiti until presidential
    elections are held in that
    Caribbean nation sometime next
    year, President Luiz Inacio
    Lula da Silva said Monday." + ], + [ + "Iraq is quot;working 24 hours
    a day to ... stop the
    terrorists, quot; interim
    Iraqi Prime Minister Ayad
    Allawi said Monday. Iraqis are
    pushing ahead with reforms and
    improvements, Allawi told" + ], + [ + "AP - Musicians, composers and
    authors were among the more
    than two dozen people
    Wednesday honored with
    National Medal of Arts and
    National Humanities awards at
    the White House." + ], + [ + "Wynton Marsalis, the trumpet-
    playing star and artistic
    director of Jazz at Lincoln
    Center, has become an entity
    above and around the daily
    jazz world." + ], + [ + "BUENOS AIRES: Pakistan has
    ratified the Kyoto Protocol on
    Climatic Change, Environment
    Minister Malik Khan said on
    Thursday. He said that
    Islamabad has notified UN
    authorities of ratification,
    which formally comes into
    effect in February 2005." + ], + [ + "Staff Sgt. Johnny Horne, Jr.,
    and Staff Sgt. Cardenas Alban
    have been charged with murder
    in the death of an Iraqi, the
    1st Cavalry Division announced
    Monday." + ], + [ + "President Vladimir Putin makes
    a speech as he hosts Russia
    #39;s Olympic athletes at a
    Kremlin banquet in Moscow,
    Thursday, Nov. 4, 2004." + ], + [ + "AFP - Hundreds of Buddhists in
    southern Russia marched
    through snow to see and hear
    the Dalai Lama as he continued
    a long-awaited visit to the
    country in spite of Chinese
    protests." + ], + [ + "British officials were on
    diplomatic tenterhooks as they
    awaited the arrival on
    Thursday of French President
    Jacques Chirac for a two-day
    state visit to Britain." + ], + [ + " SYDNEY (Reuters) - A group of
    women on Pitcairn Island in
    the South Pacific are standing
    by their men, who face
    underage sex charges, saying
    having sex at age 12 is a
    tradition dating back to 18th
    century mutineers who settled
    on the island." + ], + [ + "Two aircraft are flying out
    from the UK on Sunday to
    deliver vital aid and supplies
    to Haiti, which has been
    devastated by tropical storm
    Jeanne." + ], + [ + "A scare triggered by a
    vibrating sex toy shut down a
    major Australian regional
    airport for almost an hour
    Monday, police said. The
    vibrating object was
    discovered Monday morning" + ], + [ + "AP - An Indiana congressional
    candidate abruptly walked off
    the set of a debate because
    she got stage fright." + ], + [ + "Reuters - Having reached out
    to Kashmiris during a two-day
    visit to the region, the prime
    minister heads this weekend to
    the country's volatile
    northeast, where public anger
    is high over alleged abuses by
    Indian soldiers." + ], + [ + "SAN JUAN IXTAYOPAN, Mexico --
    A pair of wooden crosses
    outside the elementary school
    are all that mark the macabre
    site where, just weeks ago, an
    angry mob captured two federal
    police officers, beat them
    unconscious, and set them on
    fire." + ], + [ + "DUBLIN : An Olympic Airlines
    plane diverted to Ireland
    following a bomb alert, the
    second faced by the Greek
    carrier in four days, resumed
    its journey to New York after
    no device was found on board,
    airport officials said." + ], + [ + " WASHINGTON (Reuters) - The
    United States said on Friday
    it is preparing a new U.N.
    resolution on Darfur and that
    Secretary of State Colin
    Powell might address next week
    whether the violence in
    western Sudan constitutes
    genocide." + ], + [ + "NAJAF, Iraq : Iraq #39;s top
    Shiite Muslim clerics, back in
    control of Najaf #39;s Imam
    Ali shrine after a four-month
    militia occupation, met amid
    the ruins of the city as life
    spluttered back to normality." + ], + [ + "Reuters - House of
    Representatives
    Majority\\Leader Tom DeLay,
    admonished twice in six days
    by his chamber's\\ethics
    committee, withstood calls on
    Thursday by rival\\Democrats
    and citizen groups that he
    step aside." + ], + [ + "AFP - Australia has accounted
    for all its nationals known to
    be working in Iraq following a
    claim by a radical Islamic
    group to have kidnapped two
    Australians, Foreign Affairs
    Minister Alexander Downer
    said." + ], + [ + "The hurricane appeared to be
    strengthening and forecasters
    warned of an increased threat
    of serious flooding and wind
    damage." + ], + [ + "Four homeless men were
    bludgeoned to death and six
    were in critical condition on
    Friday following early morning
    attacks by unknown assailants
    in downtown streets of Sao
    Paulo, a police spokesman
    said." + ], + [ + "Phone companies are not doing
    enough to warn customers about
    internet \"rogue-dialling\"
    scams, watchdog Icstis warns." + ], + [ + "India's main opposition party
    takes action against senior
    party member Uma Bharti after
    a public row." + ], + [ + "The Russian military yesterday
    extended its offer of a \\$10
    million reward for information
    leading to the capture of two
    separatist leaders who, the
    Kremlin claims, were behind
    the Beslan massacre." + ], + [ + "Israels Shin Bet security
    service has tightened
    protection of the prime
    minister, MPs and parliament
    ahead of next weeks crucial
    vote on a Gaza withdrawal." + ], + [ + "Reuters - Iraq's interim
    defense minister
    accused\\neighbors Iran and
    Syria on Wednesday of aiding
    al Qaeda\\Islamist Abu Musab
    al-Zarqawi and former agents
    of Saddam\\Hussein to promote a
    \"terrorist\" insurgency in
    Iraq." + ], + [ + "AP - The Senate race in
    Kentucky stayed at fever pitch
    on Thursday as Democratic
    challenger Daniel Mongiardo
    stressed his opposition to gay
    marriage while accusing
    Republican incumbent Jim
    Bunning of fueling personal
    attacks that seemed to suggest
    Mongiardo is gay." + ], + [ + " PORT LOUIS, Aug. 17
    (Xinhuanet) -- Southern
    African countries Tuesday
    pledged better trade and
    investment relations with
    China as well as India in the
    final communique released at
    the end of their two-day
    summit." + ], + [ + "BAGHDAD - A militant group has
    released a video saying it
    kidnapped a missing journalist
    in Iraq and would kill him
    unless US forces left Najaf
    within 48 hours." + ], + [ + "18 August 2004 -- There has
    been renewed fighting in the
    Iraqi city of Al-Najaf between
    US and Iraqi troops and Shi
    #39;a militiamen loyal to
    radical cleric Muqtada al-
    Sadr." + ], + [ + "Australia tighten their grip
    on the third Test and the
    series after dominating India
    on day two in Nagpur." + ], + [ + " SEOUL (Reuters) - North
    Korea's two-year-old nuclear
    crisis has taxed the world's
    patience, the chief United
    Nations nuclear regulator
    said on Wednesday, urging
    communist Pyongyang to return
    to its disarmament treaty
    obligations." + ], + [ + "GROZNY, Russia - The Russian
    government's choice for
    president of war-ravaged
    Chechnya appeared to be the
    victor Sunday in an election
    tainted by charges of fraud
    and shadowed by last week's
    terrorist destruction of two
    airliners. Little more than
    two hours after polls closed,
    acting Chechen president
    Sergei Abramov said
    preliminary results showed
    Maj..." + ], + [ + "BAGHDAD, Sept 5 (AFP) - Izzat
    Ibrahim al-Duri, Saddam
    Hussein #39;s deputy whose
    capture was announced Sunday,
    is 62 and riddled with cancer,
    but was public enemy number
    two in Iraq for the world
    #39;s most powerful military." + ], + [ + "AP - An explosion targeted the
    Baghdad governor's convoy as
    he was traveling through the
    capital Tuesday, killing two
    people but leaving him
    uninjured, the Interior
    Ministry said." + ], + [ + "GENEVA: Rescuers have found
    the bodies of five Swiss
    firemen who died after the
    ceiling of an underground car
    park collapsed during a fire,
    a police spokesman said last
    night." + ], + [ + "John Kerry has held 10 \"front
    porch visit\" events an actual
    front porch is optional where
    perhaps 100 people ask
    questions in a low-key
    campaigning style." + ], + [ + "The right-win opposition
    Conservative Party and Liberal
    Center Union won 43 seats in
    the 141-member Lithuanian
    parliament, after more than 99
    percent of the votes were
    counted" + ], + [ + "KHARTOUM, Aug 18 (Reuters) -
    The United Nations said on
    Wednesday it was very
    concerned by Sudan #39;s lack
    of practical progress in
    bringing security to Darfur,
    where more than a million
    people have fled their homes
    for fear of militia ..." + ], + [ + "AFP - With less than two
    months until the November 2
    election, President George W.
    Bush is working to shore up
    support among his staunchest
    supporters even as he courts
    undecided voters." + ], + [ + "Chile's government says it
    will build a prison for
    officers convicted of human
    rights abuses in the Pinochet
    era." + ], + [ + "Palestinian leader Mahmoud
    Abbas reiterated calls for his
    people to drop their weapons
    in the struggle for a state. a
    clear change of strategy for
    peace with Israel after Yasser
    Arafat #39;s death." + ], + [ + " BAGHDAD (Reuters) - Iraq's
    U.S.-backed government said on
    Tuesday that \"major neglect\"
    by its American-led military
    allies led to a massacre of 49
    army recruits at the weekend." + ], + [ + "BEIJING -- Police have
    detained a man accused of
    slashing as many as nine boys
    to death as they slept in
    their high school dormitory in
    central China, state media
    reported today." + ], + [ + "Israeli Prime Minister Ariel
    Sharon #39;s Likud party
    agreed on Thursday to a
    possible alliance with
    opposition Labour in a vote
    that averted a snap election
    and strengthened his Gaza
    withdrawal plan." + ], + [ + "While the American forces are
    preparing to launch a large-
    scale attack against Falluja
    and al-Ramadi, one of the
    chieftains of Falluja said
    that contacts are still
    continuous yesterday between
    members representing the city
    #39;s delegation and the
    interim" + ], + [ + "UN Security Council
    ambassadors were still
    quibbling over how best to
    pressure Sudan and rebels to
    end two different wars in the
    country even as they left for
    Kenya on Tuesday for a meeting
    on the crisis." + ], + [ + "HENDALA, Sri Lanka -- Day
    after day, locked in a cement
    room somewhere in Iraq, the
    hooded men beat him. They told
    him he would be beheaded.
    ''Ameriqi! quot; they shouted,
    even though he comes from this
    poor Sri Lankan fishing
    village." + ], + [ + "THE kidnappers of British aid
    worker Margaret Hassan last
    night threatened to turn her
    over to the group which
    beheaded Ken Bigley if the
    British Government refuses to
    pull its troops out of Iraq." + ], + [ + "<p></p><p>
    BOGOTA, Colombia (Reuters) -
    Ten Colombian police
    officerswere killed on Tuesday
    in an ambush by the National
    LiberationArmy, or ELN, in the
    worst attack by the Marxist
    group inyears, authorities
    told Reuters.</p>" + ], + [ + "AFP - US President George W.
    Bush called his Philippines
    counterpart Gloria Arroyo and
    said their countries should
    keep strong ties, a spokesman
    said after a spat over
    Arroyo's handling of an Iraq
    kidnapping." + ], + [ + "A scathing judgment from the
    UK #39;s highest court
    condemning the UK government
    #39;s indefinite detention of
    foreign terror suspects as a
    threat to the life of the
    nation, left anti-terrorist
    laws in the UK in tatters on
    Thursday." + ], + [ + "Reuters - Australia's
    conservative Prime
    Minister\\John Howard, handed
    the most powerful mandate in a
    generation,\\got down to work
    on Monday with reform of
    telecommunications,\\labor and
    media laws high on his agenda." + ], + [ + " TOKYO (Reuters) - Typhoon
    Megi killed one person as it
    slammed ashore in northern
    Japan on Friday, bringing the
    death toll to at least 13 and
    cutting power to thousands of
    homes before heading out into
    the Pacific." + ], + [ + "Cairo: Egyptian President
    Hosni Mubarak held telephone
    talks with Palestinian leader
    Yasser Arafat about Israel
    #39;s plan to pull troops and
    the 8,000 Jewish settlers out
    of the Gaza Strip next year,
    the Egyptian news agency Mena
    said." + ], + [ + "The first American military
    intelligence soldier to be
    court-martialed over the Abu
    Ghraib abuse scandal was
    sentenced Saturday to eight
    months in jail, a reduction in
    rank and a bad-conduct
    discharge." + ], + [ + " TOKYO (Reuters) - Japan will
    seek an explanation at weekend
    talks with North Korea on
    activity indicating Pyongyang
    may be preparing a missile
    test, although Tokyo does not
    think a launch is imminent,
    Japan's top government
    spokesman said." + ], + [ + " BEIJING (Reuters) - Iran will
    never be prepared to
    dismantle its nuclear program
    entirely but remains committed
    to the non-proliferation
    treaty (NPT), its chief
    delegate to the International
    Atomic Energy Agency said on
    Wednesday." + ], + [ + "<p></p><p>
    SANTIAGO, Chile (Reuters) -
    President Bush on
    Saturdayreached into a throng
    of squabbling bodyguards and
    pulled aSecret Service agent
    away from Chilean security
    officers afterthey stopped the
    U.S. agent from accompanying
    the president ata
    dinner.</p>" + ], + [ + "Iranian deputy foreign
    minister Gholamali Khoshrou
    denied Tuesday that his
    country #39;s top leaders were
    at odds over whether nuclear
    weapons were un-Islamic,
    insisting that it will
    quot;never quot; make the
    bomb." + ], + [ + " quot;Israel mercenaries
    assisting the Ivory Coast army
    operated unmanned aircraft
    that aided aerial bombings of
    a French base in the country,
    quot; claimed" + ], + [ + "AFP - SAP, the world's leading
    maker of business software,
    may need an extra year to
    achieve its medium-term profit
    target of an operating margin
    of 30 percent, its chief
    financial officer said." + ], + [ + "AFP - The airline Swiss said
    it had managed to cut its
    first-half net loss by about
    90 percent but warned that
    spiralling fuel costs were
    hampering a turnaround despite
    increasing passenger travel." + ], + [ + "Vulnerable youngsters expelled
    from schools in England are
    being let down by the system,
    say inspectors." + ], + [ + "Yasser Arafat was undergoing
    tests for possible leukaemia
    at a military hospital outside
    Paris last night after being
    airlifted from his Ramallah
    headquarters to an anxious
    farewell from Palestinian
    well-wishers." + ], + [ + "Nepal #39;s main opposition
    party urged the government on
    Monday to call a unilateral
    ceasefire with Maoist rebels
    and seek peace talks to end a
    road blockade that has cut the
    capital off from the rest of
    the country." + ], + [ + "BOSTON - The New York Yankees
    and Boston were tied 4-4 after
    13 innings Monday night with
    the Red Sox trying to stay
    alive in the AL championship
    series. Boston tied the
    game with two runs in the
    eighth inning on David Ortiz's
    solo homer, a walk to Kevin
    Millar, a single by Trot Nixon
    and a sacrifice fly by Jason
    Varitek..." + ], + [ + "Reuters - The country may be
    more\\or less evenly divided
    along partisan lines when it
    comes to\\the presidential
    race, but the Republican Party
    prevailed in\\the Nielsen
    polling of this summer's
    nominating conventions." + ], + [ + " In Vice President Cheney's
    final push before next
    Tuesday's election, talk of
    nuclear annihilation and
    escalating war rhetoric have
    blended with balloon drops,
    confetti cannons and the other
    trappings of modern
    campaigning with such ferocity
    that it is sometimes tough to
    tell just who the enemy is." + ] + ], + "hovertemplate": "label=World
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    will include built-in security
    features for your PC." + ], + [ + "SAN JOSE, Calif. -- Apple
    Computer (Quote, Chart)
    unveiled a batch of new iPods,
    iTunes software and promos
    designed to keep it atop the
    heap of digital music players." + ], + [ + "Any product, any shape, any
    size -- manufactured on your
    desktop! The future is the
    fabricator. By Bruce Sterling
    from Wired magazine." + ], + [ + "The Microsoft CEO says one way
    to stem growing piracy of
    Windows and Office in emerging
    markets is to offer low-cost
    computers." + ], + [ + "PalmSource surprised the
    mobile vendor community today
    with the announcement that it
    will acquire China MobileSoft
    (CMS), ostensibly to leverage
    that company's expertise in
    building a mobile version of
    the Linux operating system." + ], + [ + "JEFFERSON CITY - An election
    security expert has raised
    questions about Missouri
    Secretary of State Matt Blunt
    #39;s plan to let soldiers at
    remote duty stations or in
    combat areas cast their
    ballots with the help of
    e-mail." + ], + [ + "A new, optional log-on service
    from America Online that makes
    it harder for scammers to
    access a persons online
    account will not be available
    for Macintosh" + ], + [ + "It #39;s official. Microsoft
    recently stated
    definitivelyand contrary to
    rumorsthat there will be no
    new versions of Internet
    Explorer for users of older
    versions of Windows." + ], + [ + "But fresh antitrust suit is in
    the envelope, says Novell" + ], + [ + "Chips that help a computer's
    main microprocessors perform
    specific types of math
    problems are becoming a big
    business once again.\\" + ], + [ + "A rocket carrying two Russian
    cosmonauts and an American
    astronaut to the international
    space station streaked into
    orbit on Thursday, the latest
    flight of a Russian space
    vehicle to fill in for
    grounded US shuttles." + ], + [ + "OCTOBER 12, 2004
    (COMPUTERWORLD) - Microsoft
    Corp. #39;s monthly patch
    release cycle may be making it
    more predictable for users to
    deploy software updates, but
    there appears to be little
    letup in the number of holes
    being discovered in the
    company #39;s software" + ], + [ + "Wearable tech goes mainstream
    as the Gap introduces jacket
    with built-in radio.\\" + ], + [ + "Funding round of \\$105 million
    brings broadband Internet
    telephony company's total haul
    to \\$208 million." + ], + [ + "washingtonpost.com - Anthony
    Casalena was 17 when he got
    his first job as a programmer
    for a start-up called
    HyperOffice, a software
    company that makes e-mail and
    contact management
    applications for the Web.
    Hired as an intern, he became
    a regular programmer after two
    weeks and rewrote the main
    product line." + ], + [ + "The fossilized neck bones of a
    230-million-year-old sea
    creature have features
    suggesting that the animal
    #39;s snakelike throat could
    flare open and create suction
    that would pull in prey." + ], + [ + "IBM late on Tuesday announced
    the biggest update of its
    popular WebSphere business
    software in two years, adding
    features such as automatically
    detecting and fixing problems." + ], + [ + "AP - British entrepreneur
    Richard Branson said Monday
    that his company plans to
    launch commercial space
    flights over the next few
    years." + ], + [ + "A group of Internet security
    and instant messaging
    providers have teamed up to
    detect and thwart the growing
    threat of IM and peer-to-peer
    viruses and worms, they said
    this week." + ], + [ + "His first space flight was in
    1965 when he piloted the first
    manned Gemini mission. Later
    he made two trips to the moon
    -- orbiting during a 1969
    flight and then walking on the
    lunar surface during a mission
    in 1972." + ], + [ + "By RACHEL KONRAD SAN
    FRANCISCO (AP) -- EBay Inc.,
    which has been aggressively
    expanding in Asia, plans to
    increase its stake in South
    Korea's largest online auction
    company. The Internet
    auction giant said Tuesday
    night that it would purchase
    nearly 3 million publicly held
    shares of Internet Auction
    Co..." + ], + [ + "AP - China's biggest computer
    maker, Lenovo Group, said
    Wednesday it has acquired a
    majority stake in
    International Business
    Machines Corp.'s personal
    computer business for
    #36;1.75 billion, one of the
    biggest Chinese overseas
    acquisitions ever." + ], + [ + "Big evolutionary insights
    sometimes come in little
    packages. Witness the
    startling discovery, in a cave
    on the eastern Indonesian
    island of Flores, of the
    partial skeleton of a half-
    size Homo species that" + ], + [ + "Luke Skywalker and Darth Vader
    may get all the glory, but a
    new Star Wars video game
    finally gives credit to the
    everyday grunts who couldn
    #39;t summon the Force for
    help." + ], + [ + " LOS ANGELES (Reuters) -
    Federal authorities raided
    three Washington, D.C.-area
    video game stores and arrested
    two people for modifying
    video game consoles to play
    pirated video games, a video
    game industry group said on
    Wednesday." + ], + [ + "Thornbugs communicate by
    vibrating the branches they
    live on. Now scientists are
    discovering just what the bugs
    are \"saying.\"" + ], + [ + "Federal prosecutors cracked a
    global cartel that had
    illegally fixed prices of
    memory chips in personal
    computers and servers for
    three years." + ], + [ + "AP - Oracle Corp. CEO Larry
    Ellison reiterated his
    determination to prevail in a
    long-running takeover battle
    with rival business software
    maker PeopleSoft Inc.,
    predicting the proposed deal
    will create a more competitive
    company with improved customer
    service." + ], + [ + "AP - Scientists warned Tuesday
    that a long-term increase in
    global temperature of 3.5
    degrees could threaten Latin
    American water supplies,
    reduce food yields in Asia and
    result in a rise in extreme
    weather conditions in the
    Caribbean." + ], + [ + "AP - Further proof New York's
    real estate market is
    inflated: The city plans to
    sell space on top of lampposts
    to wireless phone companies
    for #36;21.6 million a year." + ], + [ + "The rollout of new servers and
    networking switches in stores
    is part of a five-year
    agreement supporting 7-Eleven
    #39;s Retail Information
    System." + ], + [ + "Top Hollywood studios have
    launched a wave of court cases
    against internet users who
    illegally download film files.
    The Motion Picture Association
    of America, which acts as the
    representative of major film" + ], + [ + "AUSTRALIANS went into a
    television-buying frenzy the
    run-up to the Athens Olympics,
    suggesting that as a nation we
    could easily have scored a
    gold medal for TV purchasing." + ], + [ + "The next time you are in your
    bedroom with your PC plus
    Webcam switched on, don #39;t
    think that your privacy is all
    intact. If you have a Webcam
    plugged into an infected
    computer, there is a
    possibility that" + ], + [ + "AP - Shawn Fanning's Napster
    software enabled countless
    music fans to swap songs on
    the Internet for free, turning
    him into the recording
    industry's enemy No. 1 in the
    process." + ], + [ + "Reuters - Walt Disney Co.
    is\\increasing investment in
    video games for hand-held
    devices and\\plans to look for
    acquisitions of small game
    publishers and\\developers,
    Disney consumer products
    Chairman Andy Mooney said\\on
    Monday." + ], + [ + "BANGKOK - A UN conference last
    week banned commercial trade
    in the rare Irrawaddy dolphin,
    a move environmentalists said
    was needed to save the
    threatened species." + ], + [ + "PC World - Updated antivirus
    software for businesses adds
    intrusion prevention features." + ], + [ + "Reuters - Online media company
    Yahoo Inc.\\ late on Monday
    rolled out tests of redesigned
    start\\pages for its popular
    Yahoo.com and My.Yahoo.com
    sites." + ], + [ + "The Sun may have captured
    thousands or even millions of
    asteroids from another
    planetary system during an
    encounter more than four
    billion years ago, astronomers
    are reporting." + ], + [ + "Thumb through the book - then
    buy a clean copy from Amazon" + ], + [ + "Reuters - High-definition
    television can\\show the sweat
    beading on an athlete's brow,
    but the cost of\\all the
    necessary electronic equipment
    can get a shopper's own\\pulse
    racing." + ], + [ + "MacCentral - Apple Computer
    Inc. on Monday released an
    update for Apple Remote
    Desktop (ARD), the company's
    software solution to assist
    Mac system administrators and
    computer managers with asset
    management, software
    distribution and help desk
    support. ARD 2.1 includes
    several enhancements and bug
    fixes." + ], + [ + "The fastest-swiveling space
    science observatory ever built
    rocketed into orbit on
    Saturday to scan the universe
    for celestial explosions." + ], + [ + "Copernic Unleashes Desktop
    Search Tool\\\\Copernic
    Technologies Inc. today
    announced Copernic Desktop
    Search(TM) (CDS(TM)), \"The
    Search Engine For Your PC
    (TM).\" Copernic has used the
    experience gained from over 30
    million downloads of its
    Windows-based Web search
    software to develop CDS, a
    desktop search product that
    users are saying is far ..." + ], + [ + "The DVD Forum moved a step
    further toward the advent of
    HD DVD media and drives with
    the approval of key physical
    specifications at a meeting of
    the organisations steering
    committee last week." + ], + [ + "Often pigeonholed as just a
    seller of televisions and DVD
    players, Royal Philips
    Electronics said third-quarter
    profit surged despite a slide
    into the red by its consumer
    electronics division." + ], + [ + "AP - Google, the Internet
    search engine, has done
    something that law enforcement
    officials and their computer
    tools could not: Identify a
    man who died in an apparent
    hit-and-run accident 11 years
    ago in this small town outside
    Yakima." + ], + [ + "We are all used to IE getting
    a monthly new security bug
    found, but Winamp? In fact,
    this is not the first security
    flaw found in the application." + ], + [ + "The Apache Software Foundation
    and the Debian Project said
    they won't support the Sender
    ID e-mail authentication
    standard in their products." + ], + [ + "USATODAY.com - The newly
    restored THX 1138 arrives on
    DVD today with Star Wars
    creator George Lucas' vision
    of a Brave New World-like
    future." + ], + [ + "The chances of scientists
    making any one of five
    discoveries by 2010 have been
    hugely underestimated,
    according to bookmakers." + ], + [ + "Deepening its commitment to
    help corporate users create
    SOAs (service-oriented
    architectures) through the use
    of Web services, IBM's Global
    Services unit on Thursday
    announced the formation of an
    SOA Management Practice." + ], + [ + "Toshiba Corp. #39;s new
    desktop-replacement multimedia
    notebooks, introduced on
    Tuesday, are further evidence
    that US consumers still have
    yet to embrace the mobility
    offered by Intel Corp." + ], + [ + "Aregular Amazon customer,
    Yvette Thompson has found
    shopping online to be mostly
    convenient and trouble-free.
    But last month, after ordering
    two CDs on Amazon.com, the
    Silver Spring reader
    discovered on her bank
    statement that she was double-
    charged for the \\$26.98 order.
    And there was a \\$25 charge
    that was a mystery." + ], + [ + "Finnish mobile giant Nokia has
    described its new Preminet
    solution, which it launched
    Monday (Oct. 25), as a
    quot;major worldwide
    initiative." + ], + [ + " SAN FRANCISCO (Reuters) - At
    virtually every turn, Intel
    Corp. executives are heaping
    praise on an emerging long-
    range wireless technology
    known as WiMAX, which can
    blanket entire cities with
    high-speed Internet access." + ], + [ + "Phishing is one of the
    fastest-growing forms of
    personal fraud in the world.
    While consumers are the most
    obvious victims, the damage
    spreads far wider--hurting
    companies #39; finances and
    reputations and potentially" + ], + [ + "Reuters - The U.S. Interior
    Department on\\Friday gave
    final approval to a plan by
    ConocoPhillips and\\partner
    Anadarko Petroleum Corp. to
    develop five tracts around\\the
    oil-rich Alpine field on
    Alaska's North Slope." + ], + [ + "Atari has opened the initial
    sign-up phase of the closed
    beta for its Dungeons amp;
    Dragons real-time-strategy
    title, Dragonshard." + ], + [ + "Big Blue adds features, beefs
    up training efforts in China;
    rival Unisys debuts new line
    and pricing plan." + ], + [ + "South Korea's Hynix
    Semiconductor Inc. and Swiss-
    based STMicroelectronics NV
    signed a joint-venture
    agreement on Tuesday to
    construct a memory chip
    manufacturing plant in Wuxi,
    about 100 kilometers west of
    Shanghai, in China." + ], + [ + "Well, Intel did say -
    dismissively of course - that
    wasn #39;t going to try to
    match AMD #39;s little dual-
    core Opteron demo coup of last
    week and show off a dual-core
    Xeon at the Intel Developer
    Forum this week and - as good
    as its word - it didn #39;t." + ], + [ + "IT #39;S BEEN a heck of an
    interesting two days here in
    Iceland. I #39;ve seen some
    interesting technology, heard
    some inventive speeches and
    met some people with different
    ideas." + ], + [ + "Reuters - Two clients of
    Germany's Postbank\\(DPBGn.DE)
    fell for an e-mail fraud that
    led them to reveal\\money
    transfer codes to a bogus Web
    site -- the first case of\\this
    scam in German, prosecutors
    said on Thursday." + ], + [ + "US spending on information
    technology goods, services,
    and staff will grow seven
    percent in 2005 and continue
    at a similar pace through
    2008, according to a study
    released Monday by Forrester
    Research." + ], + [ + "Look, Ma, no hands! The U.S.
    space agency's latest
    spacecraft can run an entire
    mission by itself. By Amit
    Asaravala." + ], + [ + "Joining America Online,
    EarthLink and Yahoo against
    spamming, Microsoft Corp.
    today announced the filing of
    three new anti-Spam lawsuits
    under the CAN-SPAM federal law
    as part of its initiative in
    solving the Spam problem for
    Internet users worldwide." + ], + [ + "A nationwide inspection shows
    Internet users are not as safe
    online as they believe. The
    inspections found most
    consumers have no firewall
    protection, outdated antivirus
    software and dozens of spyware
    programs secretly running on
    their computers." + ], + [ + "More than three out of four
    (76 percent) consumers are
    experiencing an increase in
    spoofing and phishing
    incidents, and 35 percent
    receive fake e-mails at least
    once a week, according to a
    recent national study." + ], + [ + "AP - Deep in the Atlantic
    Ocean, undersea explorers are
    living a safer life thanks to
    germ-fighting clothing made in
    Kinston." + ], + [ + "Computer Associates Monday
    announced the general
    availability of three
    Unicenter performance
    management products for
    mainframe data management." + ], + [ + "Reuters - The European Union
    approved on\\Wednesday the
    first biotech seeds for
    planting and sale across\\EU
    territory, flying in the face
    of widespread
    consumer\\resistance to
    genetically modified (GMO)
    crops and foods." + ], + [ + "p2pnet.net News:- A Microsoft
    UK quot;WEIGHING THE COST OF
    LINUX VS. WINDOWS? LET #39;S
    REVIEW THE FACTS quot;
    magazine ad has been nailed as
    misleading by Britain #39;s
    Advertising Standards
    Authority (ASA)." + ], + [ + "The retail sector overall may
    be reporting a sluggish start
    to the season, but holiday
    shoppers are scooping up tech
    goods at a brisk pace -- and
    they're scouring the Web for
    bargains more than ever.
    <FONT face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\" color=\"#666666\">&
    lt;B>-washingtonpost.com<
    ;/B></FONT>" + ], + [ + "The team behind the Beagle 2
    mission has unveiled its
    design for a successor to the
    British Mars lander." + ], + [ + "Survey points to popularity in
    Europe, the Middle East and
    Asia of receivers that skip
    the pay TV and focus on free
    programming." + ], + [ + "Linux publisher Red Hat Inc.
    said Tuesday that information-
    technology consulting firm
    Unisys Corp. will begin
    offering a business version of
    the company #39;s open-source
    operating system on its
    servers." + ], + [ + "IBM's p5-575, a specialized
    server geared for high-
    performance computing, has
    eight 1.9GHz Power5
    processors." + ], + [ + "Reuters - British police said
    on Monday they had\\charged a
    man with sending hoax emails
    to relatives of people\\missing
    since the Asian tsunami,
    saying their loved ones
    had\\been confirmed dead." + ], + [ + "LOS ANGELES - On Sept. 1,
    former secretary of
    Agriculture Dan Glickman
    replaced the legendary Jack
    Valenti as president and CEO
    of Hollywood #39;s trade
    group, the Motion Picture
    Association of America." + ], + [ + "The director of the National
    Hurricane Center stays calm in
    the midst of a storm, but
    wants everyone in hurricane-
    prone areas to get the message
    from his media advisories:
    Respect the storm's power and
    make proper response plans." + ], + [ + "SAN FRANCISCO Several
    California cities and
    counties, including Los
    Angeles and San Francisco, are
    suing Microsoft for what could
    amount to billions of dollars." + ], + [ + "Google plans to release a
    version of its desktop search
    tool for computers that run
    Apple Computer #39;s Mac
    operating system, Google #39;s
    chief executive, Eric Schmidt,
    said Friday." + ], + [ + "AMD : sicurezza e prestazioni
    ottimali con il nuovo
    processore mobile per notebook
    leggeri e sottili; Acer Inc.
    preme sull #39;acceleratore
    con il nuovo notebook a
    marchio Ferrari." + ], + [ + "Firefox use around the world
    climbed 34 percent in the last
    month, according to a report
    published by Web analytics
    company WebSideStory Monday." + ], + [ + "If a plastic card that gives
    you credit for something you
    don't want isn't your idea of
    a great gift, you can put it
    up for sale or swap." + ], + [ + "WASHINGTON Aug. 17, 2004
    Scientists are planning to
    take the pulse of the planet
    and more in an effort to
    improve weather forecasts,
    predict energy needs months in
    advance, anticipate disease
    outbreaks and even tell
    fishermen where the catch will
    be ..." + ], + [ + "New communications technology
    could spawn future products.
    Could your purse tell you to
    bring an umbrella?" + ], + [ + "SAP has won a \\$35 million
    contract to install its human
    resources software for the US
    Postal Service. The NetWeaver-
    based system will replace the
    Post Office #39;s current
    25-year-old legacy application" + ], + [ + "NewsFactor - For years,
    companies large and small have
    been convinced that if they
    want the sophisticated
    functionality of enterprise-
    class software like ERP and
    CRM systems, they must buy
    pre-packaged applications.
    And, to a large extent, that
    remains true." + ], + [ + "Following in the footsteps of
    the RIAA, the MPAA (Motion
    Picture Association of
    America) announced that they
    have began filing lawsuits
    against people who use peer-
    to-peer software to download
    copyrighted movies off the
    Internet." + ], + [ + "MacCentral - At a special
    music event featuring Bono and
    The Edge from rock group U2
    held on Tuesday, Apple took
    the wraps off the iPod Photo,
    a color iPod available in 40GB
    or 60GB storage capacities.
    The company also introduced
    the iPod U2, a special edition
    of Apple's 20GB player clad in
    black, equipped with a red
    Click Wheel and featuring
    engraved U2 band member
    signatures. The iPod Photo is
    available immediately, and
    Apple expects the iPod U2 to
    ship in mid-November." + ], + [ + "Reuters - Oracle Corp is
    likely to win clearance\\from
    the European Commission for
    its hostile #36;7.7
    billion\\takeover of rival
    software firm PeopleSoft Inc.,
    a source close\\to the
    situation said on Friday." + ], + [ + "IBM (Quote, Chart) said it
    would spend a quarter of a
    billion dollars over the next
    year and a half to grow its
    RFID (define) business." + ], + [ + "SPACE.com - NASA released one
    of the best pictures ever made
    of Saturn's moon Titan as the
    Cassini spacecraft begins a
    close-up inspection of the
    satellite today. Cassini is
    making the nearest flyby ever
    of the smog-shrouded moon." + ], + [ + "ANNAPOLIS ROYAL, NS - Nova
    Scotia Power officials
    continued to keep the sluice
    gates open at one of the
    utility #39;s hydroelectric
    plants Wednesday in hopes a
    wayward whale would leave the
    area and head for the open
    waters of the Bay of Fundy." + ], + [ + "In fact, Larry Ellison
    compares himself to the
    warlord, according to
    PeopleSoft's former CEO,
    defending previous remarks he
    made." + ], + [ + "You can empty your pockets of
    change, take off your belt and
    shoes and stick your keys in
    the little tray. But if you've
    had radiation therapy
    recently, you still might set
    off Homeland Security alarms." + ], + [ + "SEOUL, Oct 19 Asia Pulse -
    LG.Philips LCD Co.
    (KSE:034220), the world #39;s
    second-largest maker of liquid
    crystal display (LCD), said
    Tuesday it has developed the
    world #39;s largest organic
    light emitting diode" + ], + [ + "Security experts warn of
    banner ads with a bad attitude
    --and a link to malicious
    code. Also: Phishers, be gone." + ], + [ + "com October 15, 2004, 5:11 AM
    PT. Wood paneling and chrome
    made your dad #39;s station
    wagon look like a million
    bucks, and they might also be
    just the ticket for Microsoft
    #39;s fledgling" + ], + [ + "Computer Associates
    International yesterday
    reported a 6 increase in
    revenue during its second
    fiscal quarter, but posted a
    \\$94 million loss after paying
    to settle government
    investigations into the
    company, it said yesterday." + ], + [ + "PC World - Send your video
    throughout your house--
    wirelessly--with new gateways
    and media adapters." + ], + [ + "Links to this week's topics
    from search engine forums
    across the web: New MSN Search
    Goes LIVE in Beta - Microsoft
    To Launch New Search Engine -
    Google Launches 'Google
    Advertising Professionals' -
    Organic vs Paid Traffic ROI? -
    Making Money With AdWords? -
    Link Building 101" + ], + [ + "Eight conservation groups are
    fighting the US government
    over a plan to poison
    thousands of prairie dogs in
    the grasslands of South
    Dakota, saying wildlife should
    not take a backseat to
    ranching interests." + ], + [ + "Siding with chip makers,
    Microsoft said it won't charge
    double for its per-processor
    licenses when dual-core chips
    come to market next year." + ], + [ + "IBM and the Spanish government
    have introduced a new
    supercomputer they hope will
    be the most powerful in
    Europe, and one of the 10 most
    powerful in the world." + ], + [ + "Mozilla's new web browser is
    smart, fast and user-friendly
    while offering a slew of
    advanced, customizable
    functions. By Michelle Delio." + ], + [ + "originally offered on notebook
    PCs -- to its Opteron 32- and
    64-bit x86 processors for
    server applications. The
    technology will help servers
    to run" + ], + [ + "MacCentral - After Apple
    unveiled the iMac G5 in Paris
    this week, Vice President of
    Hardware Product Marketing
    Greg Joswiak gave Macworld
    editors a guided tour of the
    desktop's new design. Among
    the topics of conversation:
    the iMac's cooling system, why
    pre-installed Bluetooth
    functionality and FireWire 800
    were left out, and how this
    new model fits in with Apple's
    objectives." + ], + [ + "We #39;ve known about
    quot;strained silicon quot;
    for a while--but now there
    #39;s a better way to do it.
    Straining silicon improves
    chip performance." + ], + [ + "This week, Sir Richard Branson
    announced his new company,
    Virgin Galactic, has the
    rights to the first commercial
    flights into space." + ], + [ + "71-inch HDTV comes with a home
    stereo system and components
    painted in 24-karat gold." + ], + [ + "MacCentral - RealNetworks Inc.
    said on Tuesday that it has
    sold more than a million songs
    at its online music store
    since slashing prices last
    week as part of a limited-time
    sale aimed at growing the user
    base of its new digital media
    software." + ], + [ + "With the presidential election
    less than six weeks away,
    activists and security experts
    are ratcheting up concern over
    the use of touch-screen
    machines to cast votes.
    <FONT face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\"\\ color=\"#666666\">
    <B>-washingtonpost.com&l
    t;/B></FONT>" + ], + [ + "NEW YORK, September 14 (New
    Ratings) - Yahoo! Inc
    (YHOO.NAS) has agreed to
    acquire Musicmatch Inc, a
    privately held digital music
    software company, for about
    \\$160 million in cash." + ], + [ + "Gov. Rod Blagojevich plans to
    propose a ban Thursday on the
    sale of violent and sexually
    explicit video games to
    minors, something other states
    have tried with little
    success." + ], + [ + "A cable channel plans to
    resurrect each of the 1,230
    regular-season games listed on
    the league's defunct 2004-2005
    schedule by setting them in
    motion on a video game
    console." + ], + [ + "SiliconValley.com - When
    \"Halo\" became a smash video
    game hit following Microsoft's
    launch of the Xbox console in
    2001, it was a no-brainer that
    there would be a sequel to the
    science fiction shoot-em-up." + ], + [ + "PARIS -- The city of Paris
    intends to reduce its
    dependence on software
    suppliers with \"de facto
    monopolies,\" but considers an
    immediate switch of its 17,000
    desktops to open source
    software too costly, it said
    Wednesday." + ], + [ + "The vast majority of consumers
    are unaware that an Apple iPod
    digital music player only
    plays proprietary iTunes
    files, while a smaller
    majority agree that it is
    within RealNetworks #39;
    rights to develop a program
    that will make its music files
    compatible" + ], + [ + " PROVIDENCE, R.I. (Reuters) -
    You change the oil in your car
    every 5,000 miles or so. You
    clean your house every week or
    two. Your PC needs regular
    maintenance as well --
    especially if you're using
    Windows and you spend a lot of
    time on the Internet." + ], + [ + "The Motley Fool - Here's
    something you don't see every
    day -- the continuing brouhaha
    between Oracle (Nasdaq: ORCL -
    News) and PeopleSoft (Nasdaq:
    PSFT - News) being a notable
    exception. South Africa's
    Harmony Gold Mining Company
    (NYSE: HMY - News) has
    announced a hostile takeover
    bid to acquire fellow South
    African miner Gold Fields
    Limited (NYSE: GFI - News).
    The transaction, if it takes
    place, would be an all-stock
    acquisition, with Harmony
    issuing 1.275 new shares in
    payment for each share of Gold
    Fields. The deal would value
    Gold Fields at more than
    #36;8 billion. ..." + ], + [ + "SPACE.com - NASA's Mars \\rover
    Opportunity nbsp;will back its
    \\way out of a nbsp;crater it
    has spent four months
    exploring after reaching
    terrain nbsp;that appears \\too
    treacherous to tread. nbsp;" + ], + [ + "Sony Corp. announced Tuesday a
    new 20 gigabyte digital music
    player with MP3 support that
    will be available in Great
    Britain and Japan before
    Christmas and elsewhere in
    Europe in early 2005." + ], + [ + "TOKYO (AP) -- The electronics
    and entertainment giant Sony
    Corp. (SNE) is talking with
    Wal-Mart Stores Inc..." + ], + [ + "After an unprecedented span of
    just five days, SpaceShipOne
    is ready for a return trip to
    space on Monday, its final
    flight to clinch a \\$10
    million prize." + ], + [ + "Hi-tech monitoring of
    livestock at pig farms could
    help improve the animal growth
    process and reduce costs." + ], + [ + "Cable amp; Wireless plc
    (NYSE: CWP - message board) is
    significantly ramping up its
    investment in local loop
    unbundling (LLU) in the UK,
    and it plans to spend up to 85
    million (\\$152." + ], + [ + "AFP - Microsoft said that it
    had launched a new desktop
    search tool that allows
    personal computer users to
    find documents or messages on
    their PCs." + ], + [ + "The three largest computer
    makers spearheaded a program
    today designed to standardize
    working conditions for their
    non-US workers." + ], + [ + "Description: Illinois Gov. Rod
    Blagojevich is backing state
    legislation that would ban
    sales or rentals of video
    games with graphic sexual or
    violent content to children
    under 18." + ], + [ + "Are you bidding on keywords
    through Overture's Precision
    Match, Google's AdWords or
    another pay-for-placement
    service? If so, you're
    eligible to participate in
    their contextual advertising
    programs." + ], + [ + "Nobel Laureate Wilkins, 87,
    played an important role in
    the discovery of the double
    helix structure of DNA, the
    molecule that carries our
    quot;life code quot;,
    Kazinform refers to BBC News." + ], + [ + "A number of signs point to
    increasing demand for tech
    workers, but not all the
    clouds have been driven away." + ], + [ + "Microsoft Corp. (MSFT.O:
    Quote, Profile, Research)
    filed nine new lawsuits
    against spammers who send
    unsolicited e-mail, including
    an e-mail marketing Web
    hosting company, the world
    #39;s largest software maker
    said on Thursday." + ], + [ + "The rocket plane SpaceShipOne
    is just one flight away from
    claiming the Ansari X-Prize, a
    \\$10m award designed to kick-
    start private space travel." + ], + [ + "Reuters - Three American
    scientists won the\\2004 Nobel
    physics prize on Tuesday for
    showing how tiny
    quark\\particles interact,
    helping to explain everything
    from how a\\coin spins to how
    the universe was built." + ], + [ + "AP - Sharp Electronics Corp.
    plans to stop selling its
    Linux-based handheld computer
    in the United States, another
    sign of the slowing market for
    personal digital assistants." + ], + [ + "Reuters - Glaciers once held
    up by a floating\\ice shelf off
    Antarctica are now sliding off
    into the sea --\\and they are
    going fast, scientists said on
    Tuesday." + ], + [ + "AP - Warning lights flashed
    atop four police cars as the
    caravan wound its way up the
    driveway in a procession fit
    for a presidential candidate.
    At long last, Azy and Indah
    had arrived. They even flew
    through a hurricane to get
    here." + ], + [ + "\\Female undergraduates work
    harder and are more open-
    minded than males, leading to
    better results, say
    scientists." + ], + [ + "The United Nations annual
    World Robotics Survey predicts
    the use of robots around the
    home will surge seven-fold by
    2007. The boom is expected to
    be seen in robots that can mow
    lawns and vacuum floors, among
    other chores." + ], + [ + "Want to dive deep -- really
    deep -- into the technical
    literature about search
    engines? Here's a road map to
    some of the best web
    information retrieval
    resources available online." + ], + [ + "Reuters - Ancel Keys, a
    pioneer in public health\\best
    known for identifying the
    connection between
    a\\cholesterol-rich diet and
    heart disease, has died." + ], + [ + "Reuters - T-Mobile USA, the
    U.S. wireless unit\\of Deutsche
    Telekom AG (DTEGn.DE), does
    not expect to offer\\broadband
    mobile data services for at
    least the next two years,\\its
    chief executive said on
    Thursday." + ], + [ + "Verizon Communications is
    stepping further into video as
    a way to compete against cable
    companies." + ], + [ + "Oracle Corp. plans to release
    the latest version of its CRM
    (customer relationship
    management) applications
    within the next two months, as
    part of an ongoing update of
    its E-Business Suite." + ], + [ + "Hosted CRM service provider
    Salesforce.com took another
    step forward last week in its
    strategy to build an online
    ecosystem of vendors that
    offer software as a service." + ], + [ + "washingtonpost.com -
    Technology giants IBM and
    Hewlett-Packard are injecting
    hundreds of millions of
    dollars into radio-frequency
    identification technology,
    which aims to advance the
    tracking of items from ho-hum
    bar codes to smart tags packed
    with data." + ], + [ + "The Norfolk Broads are on
    their way to getting a clean
    bill of ecological health
    after a century of stagnation." + ], + [ + "For the second time this year,
    an alliance of major Internet
    providers - including Atlanta-
    based EarthLink -iled a
    coordinated group of lawsuits
    aimed at stemming the flood of
    online junk mail." + ], + [ + "Via Technologies has released
    a version of the open-source
    Xine media player that is
    designed to take advantage of
    hardware digital video
    acceleration capabilities in
    two of the company #39;s PC
    chipsets, the CN400 and
    CLE266." + ], + [ + "SCO Group has a plan to keep
    itself fit enough to continue
    its legal battles against
    Linux and to develop its Unix-
    on-Intel operating systems." + ], + [ + "New software helps corporate
    travel managers track down
    business travelers who
    overspend. But it also poses a
    dilemma for honest travelers
    who are only trying to save
    money." + ], + [ + "NATO Secretary-General Jaap de
    Hoop Scheffer has called a
    meeting of NATO states and
    Russia on Tuesday to discuss
    the siege of a school by
    Chechen separatists in which
    more than 335 people died, a
    NATO spokesman said." + ], + [ + "AFP - Senior executives at
    business software group
    PeopleSoft unanimously
    recommended that its
    shareholders reject a 8.8
    billion dollar takeover bid
    from Oracle Corp, PeopleSoft
    said in a statement Wednesday." + ], + [ + "Reuters - Neolithic people in
    China may have\\been the first
    in the world to make wine,
    according to\\scientists who
    have found the earliest
    evidence of winemaking\\from
    pottery shards dating from
    7,000 BC in northern China." + ], + [ + " TOKYO (Reuters) - Electronics
    conglomerate Sony Corp.
    unveiled eight new flat-screen
    televisions on Thursday in a
    product push it hopes will
    help it secure a leading 35
    percent of the domestic
    market in the key month of
    December." + ], + [ + "Donald Halsted, one target of
    a class-action suit alleging
    financial improprieties at
    bankrupt Polaroid, officially
    becomes CFO." + ], + [ + "The 7710 model features a
    touch screen, pen input, a
    digital camera, an Internet
    browser, a radio, video
    playback and streaming and
    recording capabilities, the
    company said." + ], + [ + "A top Red Hat executive has
    attacked the open-source
    credentials of its sometime
    business partner Sun
    Microsystems. In a Web log
    posting Thursday, Michael
    Tiemann, Red Hat #39;s vice
    president of open-source
    affairs" + ], + [ + "MySQL developers turn to an
    unlikely source for database
    tool: Microsoft. Also: SGI
    visualizes Linux, and the
    return of Java veteran Kim
    Polese." + ], + [ + "Dell Inc. said its profit
    surged 25 percent in the third
    quarter as the world's largest
    personal computer maker posted
    record sales due to rising
    technology spending in the
    corporate and government
    sectors in the United States
    and abroad." + ], + [ + "Reuters - SBC Communications
    said on Monday it\\would offer
    a television set-top box that
    can handle music,\\photos and
    Internet downloads, part of
    SBC's efforts to expand\\into
    home entertainment." + ], + [ + " WASHINGTON (Reuters) - Hopes
    -- and worries -- that U.S.
    regulators will soon end the
    ban on using wireless phones
    during U.S. commercial flights
    are likely at least a year or
    two early, government
    officials and analysts say." + ], + [ + "After a year of pilots and
    trials, Siebel Systems jumped
    with both feet into the SMB
    market Tuesday, announcing a
    new approach to offer Siebel
    Professional CRM applications
    to SMBs (small and midsize
    businesses) -- companies with
    revenues up to about \\$500
    million." + ], + [ + "Intel won't release a 4-GHz
    version of its flagship
    Pentium 4 product, having
    decided instead to realign its
    engineers around the company's
    new design priorities, an
    Intel spokesman said today.
    \\\\" + ], + [ + "AP - A Soyuz spacecraft
    carrying two Russians and an
    American rocketed closer
    Friday to its docking with the
    international space station,
    where the current three-man
    crew made final departure
    preparations." + ], + [ + "Scientists have manipulated
    carbon atoms to create a
    material that could be used to
    create light-based, versus
    electronic, switches. The
    material could lead to a
    supercharged Internet based
    entirely on light, scientists
    say." + ], + [ + "LOS ANGELES (Reuters)
    Qualcomm has dropped an \\$18
    million claim for monetary
    damages from rival Texas
    Instruments for publicly
    discussing terms of a
    licensing pact, a TI
    spokeswoman confirmed Tuesday." + ], + [ + "Hewlett-Packard is the latest
    IT vendor to try blogging. But
    analysts wonder if the weblog
    trend is the 21st century
    equivalent of CB radios, which
    made a big splash in the 1970s
    before fading." + ], + [ + "The scientists behind Dolly
    the sheep apply for a license
    to clone human embryos. They
    want to take stem cells from
    the embryos to study Lou
    Gehrig's disease." + ], + [ + "Some of the silly tunes
    Japanese pay to download to
    use as the ring tone for their
    mobile phones sure have their
    knockers, but it #39;s for
    precisely that reason that a
    well-known counselor is raking
    it in at the moment, according
    to Shukan Gendai (10/2)." + ], + [ + "IBM announced yesterday that
    it will invest US\\$250 million
    (S\\$425 million) over the next
    five years and employ 1,000
    people in a new business unit
    to support products and
    services related to sensor
    networks." + ], + [ + "AP - Microsoft Corp. goes into
    round two Friday of its battle
    to get the European Union's
    sweeping antitrust ruling
    lifted having told a judge
    that it had been prepared
    during settlement talks to
    share more software code with
    its rivals than the EU
    ultimately demanded." + ], + [ + "AP - Sears, Roebuck and Co.,
    which has successfully sold
    its tools and appliances on
    the Web, is counting on having
    the same magic with bedspreads
    and sweaters, thanks in part
    to expertise gained by its
    purchase of Lands' End Inc." + ], + [ + "Austin police are working with
    overseas officials to bring
    charges against an English man
    for sexual assault of a child,
    a second-degree felony." + ], + [ + "San Francisco
    developer/publisher lands
    coveted Paramount sci-fi
    license, \\$6.5 million in
    funding on same day. Although
    it is less than two years old,
    Perpetual Entertainment has
    acquired one of the most
    coveted sci-fi licenses on the
    market." + ], + [ + "By KELLY WIESE JEFFERSON
    CITY, Mo. (AP) -- Missouri
    will allow members of the
    military stationed overseas to
    return absentee ballots via
    e-mail, raising concerns from
    Internet security experts
    about fraud and ballot
    secrecy..." + ], + [ + "Avis Europe PLC has dumped a
    new ERP system based on
    software from PeopleSoft Inc.
    before it was even rolled out,
    citing substantial delays and
    higher-than-expected costs." + ], + [ + "Yahoo #39;s (Quote, Chart)
    public embrace of the RSS
    content syndication format
    took a major leap forward with
    the release of a revamped My
    Yahoo portal seeking to
    introduce the technology to
    mainstream consumers." + ], + [ + "TOKYO (AP) -- Japanese
    electronics and entertainment
    giant Sony Corp. (SNE) plans
    to begin selling a camcorder
    designed for consumers that
    takes video at digital high-
    definition quality and is
    being priced at about
    \\$3,600..." + ], + [ + "In a meeting of the cinematic
    with the scientific, Hollywood
    helicopter stunt pilots will
    try to snatch a returning NASA
    space probe out of the air
    before it hits the ground." + ], + [ + "Legend has it (incorrectly, it
    seems) that infamous bank
    robber Willie Sutton, when
    asked why banks were his
    favorite target, responded,
    quot;Because that #39;s where
    the money is." + ], + [ + "Pfizer, GlaxoSmithKline and
    Purdue Pharma are the first
    drugmakers willing to take the
    plunge and use radio frequency
    identification technology to
    protect their US drug supply
    chains from counterfeiters." + ], + [ + "Search any fee-based digital
    music service for the best-
    loved musical artists of the
    20th century and most of the
    expected names show up." + ], + [ + "America Online Inc. is
    packaging new features to
    combat viruses, spam and
    spyware in response to growing
    online security threats.
    Subscribers will be able to
    get the free tools" + ], + [ + "CAPE CANAVERAL-- NASA aims to
    launch its first post-Columbia
    shuttle mission during a
    shortened nine-day window
    March, and failure to do so
    likely would delay a planned
    return to flight until at
    least May." + ], + [ + "The Chinese city of Beijing
    has cancelled an order for
    Microsoft software, apparently
    bowing to protectionist
    sentiment. The deal has come
    under fire in China, which is
    trying to build a domestic
    software industry." + ], + [ + "Apple says it will deliver its
    iTunes music service to more
    European countries next month.
    Corroborating several reports
    in recent months, Reuters is
    reporting today that Apple
    Computer is planning the next" + ], + [ + "Reuters - Motorola Inc., the
    world's\\second-largest mobile
    phone maker, said on Tuesday
    it expects\\to sustain strong
    sales growth in the second
    half of 2004\\thanks to new
    handsets with innovative
    designs and features." + ], + [ + "PRAGUE, Czech Republic --
    Eugene Cernan, the last man to
    walk on the moon during the
    final Apollo landing, said
    Thursday he doesn't expect
    space tourism to become
    reality in the near future,
    despite a strong demand.
    Cernan, now 70, who was
    commander of NASA's Apollo 17
    mission and set foot on the
    lunar surface in December 1972
    during his third space flight,
    acknowledged that \"there are
    many people interested in
    space tourism.\" But the
    former astronaut said he
    believed \"we are a long way
    away from the day when we can
    send a bus of tourists to the
    moon.\" He spoke to reporters
    before being awarded a medal
    by the Czech Academy of
    Sciences for his contribution
    to science..." + ], + [ + "Never shy about entering a
    market late, Microsoft Corp.
    is planning to open the
    virtual doors of its long-
    planned Internet music store
    next week. <FONT
    face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\"\\ color=\"#666666\">
    <B>-Leslie
    Walker</B></FONT>" + ], + [ + "AP - On his first birthday
    Thursday, giant panda cub Mei
    Sheng delighted visitors by
    playing for the first time in
    snow delivered to him at the
    San Diego Zoo. He also sat on
    his ice cake, wrestled with
    his mom, got his coat
    incredibly dirty, and didn't
    read any of the more than 700
    birthday wishes sent him via
    e-mail from as far away as
    Ireland and Argentina." + ], + [ + "AP - Researchers put a
    satellite tracking device on a
    15-foot shark that appeared to
    be lost in shallow water off
    Cape Cod, the first time a
    great white has been tagged
    that way in the Atlantic." + ], + [ + "The Windows Future Storage
    (WinFS) technology that got
    cut out of Windows
    quot;Longhorn quot; is in
    serious trouble, and not just
    the hot water a feature might
    encounter for missing its
    intended production vehicle." + ], + [ + "washingtonpost.com - Let the
    games begin. Not the Olympics
    again, but the all-out battle
    between video game giants Sony
    Corp. and Nintendo Co. Ltd.
    The two Japanese companies are
    rolling out new gaming
    consoles, but Nintendo has
    beaten Sony to the punch by
    announcing an earlier launch
    date for its new hand-held
    game player." + ], + [ + "AP - Echoing what NASA
    officials said a day earlier,
    a Russian space official on
    Friday said the two-man crew
    on the international space
    station could be forced to
    return to Earth if a planned
    resupply flight cannot reach
    them with food supplies later
    this month." + ], + [ + "InfoWorld - SANTA CLARA,
    CALIF. -- Accommodating large
    patch sets in Linux is
    expected to mean forking off
    of the 2.7 version of the
    platform to accommodate these
    changes, according to Andrew
    Morton, lead maintainer of the
    Linux kernel for Open Source
    Development Labs (OSDL)." + ], + [ + "AMSTERDAM The mobile phone
    giants Vodafone and Nokia
    teamed up on Thursday to
    simplify cellphone software
    written with the Java computer
    language." + ], + [ + "The overall Linux market is
    far larger than previous
    estimates show, a new study
    says. In an analysis of the
    Linux market released late
    Tuesday, market research firm
    IDC estimated that the Linux
    market -- including" + ], + [ + "By PAUL ELIAS SAN FRANCISCO
    (AP) -- Several California
    cities and counties, including
    San Francisco and Los Angeles,
    sued Microsoft Corp. (MSFT) on
    Friday, accusing the software
    giant of illegally charging
    inflated prices for its
    products because of monopoly
    control of the personal
    computer operating system
    market..." + ], + [ + "Public opinion of the database
    giant sinks to 12-year low, a
    new report indicates." + ], + [ + "For spammers, it #39;s been a
    summer of love. Two newly
    issued reports tracking the
    circulation of unsolicited
    e-mails say pornographic spam
    dominated this summer, nearly
    all of it originating from
    Internet addresses in North
    America." + ], + [ + "Microsoft portrayed its
    Longhorn decision as a
    necessary winnowing to hit the
    2006 timetable. The
    announcement on Friday,
    Microsoft executives insisted,
    did not point to a setback in
    software" + ], + [ + "A problem in the Service Pack
    2 update for Windows XP may
    keep owners of AMD-based
    computers from using the long-
    awaited security package,
    according to Microsoft." + ], + [ + "Five years ago, running a
    telephone company was an
    immensely profitable
    proposition. Since then, those
    profits have inexorably
    declined, and now that decline
    has taken another gut-
    wrenching dip." + ], + [ + "NEW YORK - The litigious
    Recording Industry Association
    of America (RIAA) is involved
    in another legal dispute with
    a P-to-P (peer-to-peer)
    technology maker, but this
    time, the RIAA is on defense.
    Altnet Inc. filed a lawsuit
    Wednesday accusing the RIAA
    and several of its partners of
    infringing an Altnet patent
    covering technology for
    identifying requested files on
    a P-to-P network." + ], + [ + "For two weeks before MTV
    debuted U2 #39;s video for the
    new single quot;Vertigo,
    quot; fans had a chance to see
    the band perform the song on
    TV -- in an iPod commercial." + ], + [ + "Oct. 26, 2004 - The US-
    European spacecraft Cassini-
    Huygens on Tuesday made a
    historic flyby of Titan,
    Saturn #39;s largest moon,
    passing so low as to almost
    touch the fringes of its
    atmosphere." + ], + [ + "Reuters - A volcano in central
    Japan sent smoke and\\ash high
    into the sky and spat out
    molten rock as it erupted\\for
    a fourth straight day on
    Friday, but experts said the
    peak\\appeared to be quieting
    slightly." + ], + [ + "Shares of Google Inc. made
    their market debut on Thursday
    and quickly traded up 19
    percent at \\$101.28. The Web
    search company #39;s initial
    public offering priced at \\$85" + ], + [ + "Reuters - Global warming is
    melting\\Ecuador's cherished
    mountain glaciers and could
    cause several\\of them to
    disappear over the next two
    decades, Ecuadorean and\\French
    scientists said on Wednesday." + ], + [ + "Rather than tell you, Dan
    Kranzler chooses instead to
    show you how he turned Mforma
    into a worldwide publisher of
    video games, ringtones and
    other hot downloads for mobile
    phones." + ], + [ + "Not being part of a culture
    with a highly developed
    language, could limit your
    thoughts, at least as far as
    numbers are concerned, reveals
    a new study conducted by a
    psychologist at the Columbia
    University in New York." + ], + [ + "CAMBRIDGE, Mass. A native of
    Red Oak, Iowa, who was a
    pioneer in astronomy who
    proposed the quot;dirty
    snowball quot; theory for the
    substance of comets, has died." + ], + [ + "A Portuguese-sounding version
    of the virus has appeared in
    the wild. Be wary of mail from
    Manaus." + ], + [ + "Reuters - Hunters soon may be
    able to sit at\\their computers
    and blast away at animals on a
    Texas ranch via\\the Internet,
    a prospect that has state
    wildlife officials up\\in arms." + ], + [ + "The Bedminster-based company
    yesterday said it was pushing
    into 21 new markets with the
    service, AT amp;T CallVantage,
    and extending an introductory
    rate offer until Sept. 30. In
    addition, the company is
    offering in-home installation
    of up to five ..." + ], + [ + "Samsung Electronics Co., Ltd.
    has developed a new LCD
    (liquid crystal display)
    technology that builds a touch
    screen into the display, a
    development that could lead to
    thinner and cheaper display
    panels for mobile phones, the
    company said Tuesday." + ], + [ + "The message against illegally
    copying CDs for uses such as
    in file-sharing over the
    Internet has widely sunk in,
    said the company in it #39;s
    recent announcement to drop
    the Copy-Control program." + ], + [ + "Reuters - California will
    become hotter and\\drier by the
    end of the century, menacing
    the valuable wine and\\dairy
    industries, even if dramatic
    steps are taken to curb\\global
    warming, researchers said on
    Monday." + ], + [ + "IBM said Monday that it won a
    500 million (AUD\\$1.25
    billion), seven-year services
    contract to help move UK bank
    Lloyds TBS from its
    traditional voice
    infrastructure to a converged
    voice and data network." + ], + [ + "Space shuttle astronauts will
    fly next year without the
    ability to repair in orbit the
    type of damage that destroyed
    the Columbia vehicle in
    February 2003." + ], + [ + "By cutting WinFS from Longhorn
    and indefinitely delaying the
    storage system, Microsoft
    Corp. has also again delayed
    the Microsoft Business
    Framework (MBF), a new Windows
    programming layer that is
    closely tied to WinFS." + ], + [ + "Samsung's new SPH-V5400 mobile
    phone sports a built-in
    1-inch, 1.5-gigabyte hard disk
    that can store about 15 times
    more data than conventional
    handsets, Samsung said." + ], + [ + "Vendor says it #39;s
    developing standards-based
    servers in various form
    factors for the telecom
    market. By Darrell Dunn.
    Hewlett-Packard on Thursday
    unveiled plans to create a
    portfolio of products and
    services" + ], + [ + "Ziff Davis - The company this
    week will unveil more programs
    and technologies designed to
    ease users of its high-end
    servers onto its Integrity
    line, which uses Intel's
    64-bit Itanium processor." + ], + [ + "The Mac maker says it will
    replace about 28,000 batteries
    in one model of PowerBook G4
    and tells people to stop using
    the notebook." + ], + [ + "Millions of casual US anglers
    are having are larger than
    appreciated impact on sea fish
    stocks, scientists claim." + ], + [ + "Microsoft Xbox Live traffic on
    service provider networks
    quadrupled following the
    November 9th launch of Halo-II
    -- which set entertainment
    industry records by selling
    2.4-million units in the US
    and Canada on the first day of
    availability, driving cash" + ], + [ + "Lawyers in a California class
    action suit against Microsoft
    will get less than half the
    payout they had hoped for. A
    judge in San Francisco ruled
    that the attorneys will
    collect only \\$112." + ], + [ + "Google Browser May Become
    Reality\\\\There has been much
    fanfare in the Mozilla fan
    camps about the possibility of
    Google using Mozilla browser
    technology to produce a
    GBrowser - the Google Browser.
    Over the past two weeks, the
    news and speculation has
    escalated to the point where
    even Google itself is ..." + ], + [ + "Hewlett-Packard has joined
    with Brocade to integrate
    Brocade #39;s storage area
    network switching technology
    into HP Bladesystem servers to
    reduce the amount of fabric
    infrastructure needed in a
    datacentre." + ], + [ + "The US Senate Commerce
    Committee on Wednesday
    approved a measure that would
    provide up to \\$1 billion to
    ensure consumers can still
    watch television when
    broadcasters switch to new,
    crisp digital signals." + ], + [ + "Reuters - Internet stocks
    are\\as volatile as ever, with
    growth-starved investors
    flocking to\\the sector in the
    hope they've bought shares in
    the next online\\blue chip." + ], + [ + "NASA has released an inventory
    of the scientific devices to
    be put on board the Mars
    Science Laboratory rover
    scheduled to land on the
    surface of Mars in 2009, NASAs
    news release reads." + ], + [ + "The U.S. Congress needs to
    invest more in the U.S.
    education system and do more
    to encourage broadband
    adoption, the chief executive
    of Cisco said Wednesday.<p&
    gt;ADVERTISEMENT</p><
    p><img src=\"http://ad.do
    ubleclick.net/ad/idg.us.ifw.ge
    neral/sbcspotrssfeed;sz=1x1;or
    d=200301151450?\" width=\"1\"
    height=\"1\"
    border=\"0\"/><a href=\"htt
    p://ad.doubleclick.net/clk;922
    8975;9651165;a?http://www.info
    world.com/spotlights/sbc/main.
    html?lpid0103035400730000idlp\"
    >SBC Case Study: Crate Ba
    rrel</a><br/>What
    sold them on improving their
    network? A system that could
    cut management costs from the
    get-go. Find out
    more.</p>" + ], + [ + "Thirty-two countries and
    regions will participate the
    Fifth China International
    Aviation and Aerospace
    Exhibition, opening Nov. 1 in
    Zhuhai, a city in south China
    #39;s Guangdong Province." + ], + [ + "p2pnet.net News:- Virgin
    Electronics has joined the mp3
    race with a \\$250, five gig
    player which also handles
    Microsoft #39;s WMA format." + ], + [ + "Microsoft has suspended the
    beta testing of the next
    version of its MSN Messenger
    client because of a potential
    security problem, a company
    spokeswoman said Wednesday." + ], + [ + "AP - Business software maker
    Oracle Corp. attacked the
    credibility and motives of
    PeopleSoft Inc.'s board of
    directors Monday, hoping to
    rally investor support as the
    17-month takeover battle
    between the bitter business
    software rivals nears a
    climactic showdown." + ], + [ + "A new worm has been discovered
    in the wild that #39;s not
    just settling for invading
    users #39; PCs--it wants to
    invade their homes too." + ], + [ + "Researchers have for the first
    time established the existence
    of odd-parity superconductors,
    materials that can carry
    electric current without any
    resistance." + ], + [ + "BRUSSELS, Belgium (AP) --
    European antitrust regulators
    said Monday they have extended
    their review of a deal between
    Microsoft Corp. (MSFT) and
    Time Warner Inc..." + ], + [ + "Fans who can't get enough of
    \"The Apprentice\" can visit a
    new companion Web site each
    week and watch an extra 40
    minutes of video not broadcast
    on the Thursday
    show.<br><FONT
    face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\"\\ color=\"#666666\">
    <B>-Leslie
    Walker</b></font>" + ], + [ + "An adult Web site publisher is
    suing Google, saying the
    search engine company made it
    easier for users to see the
    site #39;s copyrighted nude
    photographs without paying or
    gaining access through the
    proper channels." + ], + [ + "A Washington-based public
    opinion firm has released the
    results of an election day
    survey of Nevada voters
    showing 81 support for the
    issuance of paper receipts
    when votes are cast
    electronically." + ], + [ + "NAPSTER creator SHAWN FANNING
    has revealed his plans for a
    new licensed file-sharing
    service with an almost
    unlimited selection of tracks." + ], + [ + "Two separate studies by U.S.
    researchers find that super
    drug-resistant strains of
    tuberculosis are at the
    tipping point of a global
    epidemic, and only small
    changes could help them spread
    quickly." + ], + [ + "Dell cut prices on some
    servers and PCs by as much as
    22 percent because it #39;s
    paying less for parts. The
    company will pass the savings
    on components such as memory
    and liquid crystal displays" + ], + [ + "WASHINGTON: The European-
    American Cassini-Huygens space
    probe has detected traces of
    ice flowing on the surface of
    Saturn #39;s largest moon,
    Titan, suggesting the
    existence of an ice volcano,
    NASA said Tuesday." + ], + [ + "All ISS systems continue to
    function nominally, except
    those noted previously or
    below. Day 7 of joint
    Exp.9/Exp.10 operations and
    last full day before 8S
    undocking." + ], + [ + " quot;Magic can happen. quot;
    Sirius Satellite Radio
    (nasdaq: SIRI - news - people
    ) may have signed Howard Stern
    and the men #39;s NCAA
    basketball tournaments, but XM
    Satellite Radio (nasdaq: XMSR
    - news - people ) has its
    sights on your cell phone." + ], + [ + "Trick-or-treaters can expect
    an early Halloween treat on
    Wednesday night, when a total
    lunar eclipse makes the moon
    look like a glowing pumpkin." + ], + [ + "Sifting through millions of
    documents to locate a valuable
    few is tedious enough, but
    what happens when those files
    are scattered across different
    repositories?" + ], + [ + "NASA #39;s Mars rovers have
    uncovered more tantalizing
    evidence of a watery past on
    the Red Planet, scientists
    said Wednesday. And the
    rovers, Spirit and
    Opportunity, are continuing to
    do their jobs months after
    they were expected to ..." + ], + [ + "Intel Chief Technology Officer
    Pat Gelsinger said on
    Thursday, Sept. 9, that the
    Internet needed to be upgraded
    in order to deal with problems
    that will become real issues
    soon." + ], + [ + " LONDON (Reuters) - Television
    junkies of the world, get
    ready for \"Friends,\" \"Big
    Brother\" and \"The Simpsons\" to
    phone home." + ], + [ + "I #39;M FEELING a little bit
    better about the hundreds of
    junk e-mails I get every day
    now that I #39;ve read that
    someone else has much bigger
    e-mail troubles." + ], + [ + "Microsoft's antispam Sender ID
    technology continues to get
    the cold shoulder. Now AOL
    adds its voice to a growing
    chorus of businesses and
    organizations shunning the
    proprietary e-mail
    authentication system." + ], + [ + "Through the World Community
    Grid, your computer could help
    address the world's health and
    social problems." + ], + [ + "A planned component for
    Microsoft #39;s next version
    of Windows is causing
    consternation among antivirus
    experts, who say that the new
    module, a scripting platform
    called Microsoft Shell, could
    give birth to a whole new
    generation of viruses and
    remotely" + ], + [ + "SAN JOSE, California Yahoo
    will likely have a tough time
    getting American courts to
    intervene in a dispute over
    the sale of Nazi memorabilia
    in France after a US appeals
    court ruling." + ], + [ + "eBay Style director Constance
    White joins Post fashion
    editor Robin Givhan and host
    Janet Bennett to discuss how
    to find trends and bargains
    and pull together a wardrobe
    online." + ], + [ + "With a sudden shudder, the
    ground collapsed and the pipe
    pushed upward, buckling into a
    humped shape as Cornell
    University scientists produced
    the first simulated earthquake" + ], + [ + "Microsoft (Quote, Chart) has
    fired another salvo in its
    ongoing spam battle, this time
    against porn peddlers who don
    #39;t keep their smut inside
    the digital equivalent of a
    quot;Brown Paper Wrapper." + ], + [ + " SEATTLE (Reuters) - The next
    version of the Windows
    operating system, Microsoft
    Corp.'s <A HREF=\"http://www
    .reuters.co.uk/financeQuoteLoo
    kup.jhtml?ticker=MSFT.O
    qtype=sym infotype=info
    qcat=news\">MSFT.O</A>
    flagship product, will ship
    in 2006, the world's largest
    software maker said on
    Friday." + ], + [ + "Fossil remains of the oldest
    and smallest known ancestor of
    Tyrannosaurus rex, the world
    #39;s favorite ferocious
    dinosaur, have been discovered
    in China with evidence that
    its body was cloaked in downy
    quot;protofeathers." + ], + [ + "People fishing for sport are
    doing far more damage to US
    marine fish stocks than anyone
    thought, accounting for nearly
    a quarter of the" + ], + [ + "This particular index is
    produced by the University of
    Michigan Business School, in
    partnership with the American
    Society for Quality and CFI
    Group, and is supported in
    part by ForeSee Results" + ], + [ + "PC World - Symantec, McAfee
    hope raising virus-definition
    fees will move users to\\
    suites." + ], + [ + "By byron kho. A consortium of
    movie and record companies
    joined forces on Friday to
    request that the US Supreme
    Court take another look at
    peer-to-peer file-sharing
    programs." + ], + [ + "LONDON - Wild capuchin monkeys
    can understand cause and
    effect well enough to use
    rocks to dig for food,
    scientists have found.
    Capuchin monkeys often use
    tools and solve problems in
    captivity and sometimes" + ], + [ + "The key to hidden treasure
    lies in your handheld GPS
    unit. GPS-based \"geocaching\"
    is a high-tech sport being
    played by thousands of people
    across the globe." + ], + [ + "The U.S. information tech
    sector lost 403,300 jobs
    between March 2001 and April
    2004, and the market for tech
    workers remains bleak,
    according to a new report." + ], + [ + "com September 30, 2004, 11:11
    AM PT. SanDisk announced
    Thursday increased capacities
    for several different flash
    memory cards. The Sunnyvale,
    Calif." + ], + [ + "roundup Plus: Tech firms rally
    against copyright bill...Apple
    .Mac customers suffer e-mail
    glitches...Alvarion expands
    wireless broadband in China." + ], + [ + "Red Hat is acquiring security
    and authentication tools from
    Netscape Security Solutions to
    bolster its software arsenal.
    Red Hat #39;s CEO and chairman
    Matthew Szulik spoke about the
    future strategy of the Linux
    supplier." + ], + [ + "Global Web portal Yahoo! Inc.
    Wednesday night made available
    a beta version of a new search
    service for videos. Called
    Yahoo! Video Search, the
    search engine crawls the Web
    for different types of media
    files" + ], + [ + "Interactive posters at 25
    underground stations are
    helping Londoners travel
    safely over Christmas." + ], + [ + "According to Swiss
    authorities, history was made
    Sunday when 2723 people in
    four communities in canton
    Geneva, Switzerland, voted
    online in a national federal
    referendum." + ], + [ + " Nextel was the big story in
    telecommunications yesterday,
    thanks to the Reston company's
    mega-merger with Sprint, but
    the future of wireless may be
    percolating in dozens of
    Washington area start-ups." + ], + [ + "I have been anticipating this
    day like a child waits for
    Christmas. Today, PalmOne
    introduces the Treo 650, the
    answer to my quot;what smart
    phone will I buy?" + ], + [ + "Wikipedia has surprised Web
    watchers by growing fast and
    maturing into one of the most
    popular reference sites." + ], + [ + "It only takes 20 minutes on
    the Internet for an
    unprotected computer running
    Microsoft Windows to be taken
    over by a hacker. Any personal
    or financial information
    stored" + ], + [ + "Prices for flash memory cards
    -- the little modules used by
    digital cameras, handheld
    organizers, MP3 players and
    cell phones to store pictures,
    music and other data -- are
    headed down -- way down. Past
    trends suggest that prices
    will drop 35 percent a year,
    but industry analysts think
    that rate will be more like 40
    or 50 percent this year and
    next, due to more
    manufacturers entering the
    market." + ], + [ + "While the US software giant
    Microsoft has achieved almost
    sweeping victories in
    government procurement
    projects in several Chinese
    provinces and municipalities,
    the process" + ], + [ + "Microsoft Corp. has delayed
    automated distribution of a
    major security upgrade to its
    Windows XP Professional
    operating system, citing a
    desire to give companies more
    time to test it." + ], + [ + "Gateway computers will be more
    widely available at Office
    Depot, in the PC maker #39;s
    latest move to broaden
    distribution at retail stores
    since acquiring rival
    eMachines this year." + ], + [ + "Intel Corp. is refreshing its
    64-bit Itanium 2 processor
    line with six new chips based
    on the Madison core. The new
    processors represent the last
    single-core Itanium chips that
    the Santa Clara, Calif." + ], + [ + "For the first time, broadband
    connections are reaching more
    than half (51 percent) of the
    American online population at
    home, according to measurement
    taken in July by
    Nielsen/NetRatings, a
    Milpitas-based Internet
    audience measurement and
    research ..." + ], + [ + "Consider the New World of
    Information - stuff that,
    unlike the paper days of the
    past, doesn't always
    physically exist. You've got
    notes, scrawlings and
    snippets, Web graphics, photos
    and sounds. Stuff needs to be
    cut, pasted, highlighted,
    annotated, crossed out,
    dragged away. And, as Ross
    Perot used to say (or maybe it
    was Dana Carvey impersonating
    him), don't forget the graphs
    and charts." + ], + [ + "Major Hollywood studios on
    Tuesday announced scores of
    lawsuits against computer
    server operators worldwide,
    including eDonkey, BitTorrent
    and DirectConnect networks,
    for allowing trading of
    pirated movies." + ], + [ + "But will Wi-Fi, high-
    definition broadcasts, mobile
    messaging and other
    enhancements improve the game,
    or wreck it?\\<br />
    Photos of tech-friendly parks\\" + ], + [ + "On-demand viewing isn't just
    for TiVo owners anymore.
    Television over internet
    protocol, or TVIP, offers
    custom programming over
    standard copper wires." + ], + [ + "PalmSource #39;s European
    developer conference is going
    on now in Germany, and this
    company is using this
    opportunity to show off Palm
    OS Cobalt 6.1, the latest
    version of its operating
    system." + ], + [ + "Speaking to members of the
    Massachusetts Software
    Council, Microsoft CEO Steve
    Ballmer touted a bright future
    for technology but warned his
    listeners to think twice
    before adopting open-source
    products like Linux." + ], + [ + "MIAMI - The Trillian instant
    messaging (IM) application
    will feature several
    enhancements in its upcoming
    version 3.0, including new
    video and audio chat
    capabilities, enhanced IM
    session logs and integration
    with the Wikipedia online
    encyclopedia, according to
    information posted Friday on
    the product developer's Web
    site." + ], + [ + "Reuters - Online DVD rental
    service Netflix Inc.\\and TiVo
    Inc., maker of a digital video
    recorder, on Thursday\\said
    they have agreed to develop a
    joint entertainment\\offering,
    driving shares of both
    companies higher." + ], + [ + "THIS YULE is all about console
    supply and there #39;s
    precious little units around,
    it has emerged. Nintendo has
    announced that it is going to
    ship another 400,000 units of
    its DS console to the United
    States to meet the shortfall
    there." + ], + [ + "Annual global semiconductor
    sales growth will probably
    fall by half in 2005 and
    memory chip sales could
    collapse as a supply glut saps
    prices, world-leading memory
    chip maker Samsung Electronics
    said on Monday." + ], + [ + "NEW YORK - Traditional phone
    systems may be going the way
    of the Pony Express. Voice-
    over-Internet Protocol,
    technology that allows users
    to make and receive phone
    calls using the Internet, is
    giving the old circuit-
    switched system a run for its
    money." + ], + [ + "AP - The first U.S. cases of
    the fungus soybean rust, which
    hinders plant growth and
    drastically cuts crop
    production, were found at two
    research sites in Louisiana,
    officials said Wednesday." + ], + [ + " quot;There #39;s no way
    anyone would hire them to
    fight viruses, quot; said
    Sophos security analyst Gregg
    Mastoras. quot;For one, no
    security firm could maintain
    its reputation by employing
    hackers." + ], + [ + "Symantec has revoked its
    decision to blacklist a
    program that allows Web
    surfers in China to browse
    government-blocked Web sites.
    The move follows reports that
    the firm labelled the Freegate
    program, which" + ], + [ + "Microsoft has signed a pact to
    work with the United Nations
    Educational, Scientific and
    Cultural Organization (UNESCO)
    to increase computer use,
    Internet access and teacher
    training in developing
    countries." + ], + [ + "roundup Plus: Microsoft tests
    Windows Marketplace...Nortel
    delays financials
    again...Microsoft updates
    SharePoint." + ], + [ + "OPEN SOURCE champion Microsoft
    is expanding its programme to
    give government organisations
    some of its source code. In a
    communique from the lair of
    the Vole, in Redmond,
    spinsters have said that
    Microsoft" + ], + [ + "WASHINGTON Can you always tell
    when somebody #39;s lying? If
    so, you might be a wizard of
    the fib. A California
    psychology professor says
    there #39;s a tiny subculture
    of people that can pick out a
    lie nearly every time they
    hear one." + ], + [ + "Type design was once the
    province of skilled artisans.
    With the help of new computer
    programs, neophytes have
    flooded the Internet with
    their creations." + ], + [ + "RCN Inc., co-owner of
    Starpower Communications LLC,
    the Washington area
    television, telephone and
    Internet provider, filed a
    plan of reorganization
    yesterday that it said puts
    the company" + ], + [ + "<strong>Letters</stro
    ng> Reports of demise
    premature" + ], + [ + "Intel, the world #39;s largest
    chip maker, scrapped a plan
    Thursday to enter the digital
    television chip business,
    marking a major retreat from
    its push into consumer
    electronics." + ], + [ + "The US is the originator of
    over 42 of the worlds
    unsolicited commercial e-mail,
    making it the worst offender
    in a league table of the top
    12 spam producing countries
    published yesterday by anti-
    virus firm Sophos." + ], + [ + "Intel is drawing the curtain
    on some of its future research
    projects to continue making
    transistors smaller, faster,
    and less power-hungry out as
    far as 2020." + ], + [ + "Plus: Experts fear Check 21
    could lead to massive bank
    fraud." + ], + [ + "By SIOBHAN McDONOUGH
    WASHINGTON (AP) -- Fewer
    American youths are using
    marijuana, LSD and Ecstasy,
    but more are abusing
    prescription drugs, says a
    government report released
    Thursday. The 2003 National
    Survey on Drug Use and Health
    also found youths and young
    adults are more aware of the
    risks of using pot once a
    month or more frequently..." + ], + [ + "A TNO engineer prepares to
    start capturing images for the
    world's biggest digital photo." + ], + [ + "PalmOne is aiming to sharpen
    up its image with the launch
    of the Treo 650 on Monday. As
    previously reported, the smart
    phone update has a higher-
    resolution screen and a faster
    processor than the previous
    top-of-the-line model, the
    Treo 600." + ], + [ + "kinrowan writes quot;MIT,
    inventor of Kerberos, has
    announced a pair of
    vulnerabities in the software
    that will allow an attacker to
    either execute a DOS attack or
    execute code on the machine." + ], + [ + "Reuters - Former Pink Floyd
    mainman Roger\\Waters released
    two new songs, both inspired
    by the U.S.-led\\invasion of
    Iraq, via online download
    outlets Tuesday." + ], + [ + "NASA has been working on a
    test flight project for the
    last few years involving
    hypersonic flight. Hypersonic
    flight is fight at speeds
    greater than Mach 5, or five
    times the speed of sound." + ], + [ + "AP - Microsoft Corp. announced
    Wednesday that it would offer
    a low-cost, localized version
    of its Windows XP operating
    system in India to tap the
    large market potential in this
    country of 1 billion people,
    most of whom do not speak
    English." + ], + [ + "AP - Utah State University has
    secured a #36;40 million
    contract with NASA to build an
    orbiting telescope that will
    examine galaxies and try to
    find new stars." + ], + [ + "I spend anywhere from three to
    eight hours every week
    sweating along with a motley
    crew of local misfits,
    shelving, sorting, and hauling
    ton after ton of written
    matter in a rowhouse basement
    in Baltimore. We have no heat
    nor air conditioning, but
    still, every week, we come and
    work. Volunteer night is
    Wednesday, but many of us also
    work on the weekends, when
    we're open to the public.
    There are times when we're
    freezing and we have to wear
    coats and gloves inside,
    making handling books somewhat
    tricky; other times, we're all
    soaked with sweat, since it's
    90 degrees out and the
    basement is thick with bodies.
    One learns to forget about
    personal space when working at
    The Book Thing, since you can
    scarcely breathe without
    bumping into someone, and we
    are all so accustomed to
    having to scrape by each other
    that most of us no longer
    bother to say \"excuse me\"
    unless some particularly
    dramatic brushing occurs." + ], + [ + "Alarmed by software glitches,
    security threats and computer
    crashes with ATM-like voting
    machines, officials from
    Washington, D.C., to
    California are considering an
    alternative from an unlikely
    place: Nevada." + ], + [ + "More Americans are quitting
    their jobs and taking the risk
    of starting a business despite
    a still-lackluster job market." + ], + [ + "SPACE.com - With nbsp;food
    stores nbsp;running low, the
    two \\astronauts living aboard
    the International Space
    Station (ISS) are cutting back
    \\their meal intake and
    awaiting a critical cargo
    nbsp;delivery expected to
    arrive \\on Dec. 25." + ], + [ + "Virgin Electronics hopes its
    slim Virgin Player, which
    debuts today and is smaller
    than a deck of cards, will
    rise as a lead competitor to
    Apple's iPod." + ], + [ + "Researchers train a monkey to
    feed itself by guiding a
    mechanical arm with its mind.
    It could be a big step forward
    for prosthetics. By David
    Cohn." + ], + [ + "AP - Scientists in 17
    countries will scout waterways
    to locate and study the
    world's largest freshwater
    fish species, many of which
    are declining in numbers,
    hoping to learn how to better
    protect them, researchers
    announced Thursday." + ], + [ + "AP - Google Inc.'s plans to
    move ahead with its initial
    public stock offering ran into
    a roadblock when the
    Securities and Exchange
    Commission didn't approve the
    Internet search giant's
    regulatory paperwork as
    requested." + ], + [ + "Citing security risks, a state
    university is urging students
    to drop Internet Explorer in
    favor of alternative Web
    browsers such as Firefox and
    Safari." + ], + [ + "Despite being ranked eleventh
    in the world in broadband
    penetration, the United States
    is rolling out high-speed
    services on a quot;reasonable
    and timely basis to all
    Americans, quot; according to
    a new report narrowly approved
    today by the Federal
    Communications" + ], + [ + "NewsFactor - Taking an
    innovative approach to the
    marketing of high-performance
    \\computing, Sun Microsystems
    (Nasdaq: SUNW) is offering its
    N1 Grid program in a pay-for-
    use pricing model that mimics
    the way such commodities as
    electricity and wireless phone
    plans are sold." + ], + [ + "Reuters - Madonna and m-Qube
    have\\made it possible for the
    star's North American fans to
    download\\polyphonic ring tones
    and other licensed mobile
    content from\\her official Web
    site, across most major
    carriers and without\\the need
    for a credit card." + ], + [ + "The chipmaker is back on a
    buying spree, having scooped
    up five other companies this
    year." + ], + [ + "The company hopes to lure
    software partners by promising
    to save them from
    infrastructure headaches." + ], + [ + "Call it the Maximus factor.
    Archaeologists working at the
    site of an old Roman temple
    near Guy #39;s hospital in
    London have uncovered a pot of
    cosmetic cream dating back to
    AD2." + ], + [ + "The deal comes as Cisco pushes
    to develop a market for CRS-1,
    a new line of routers aimed at
    telephone, wireless and cable
    companies." + ], + [ + "SEPTEMBER 14, 2004 (IDG NEWS
    SERVICE) - Sun Microsystems
    Inc. and Microsoft Corp. next
    month plan to provide more
    details on the work they are
    doing to make their products
    interoperable, a Sun executive
    said yesterday." + ], + [ + "AP - Astronomy buffs and
    amateur stargazers turned out
    to watch a total lunar eclipse
    Wednesday night #151; the
    last one Earth will get for
    nearly two and a half years." + ], + [ + "The compact disc has at least
    another five years as the most
    popular music format before
    online downloads chip away at
    its dominance, a new study
    said on Tuesday." + ], + [ + "Does Your Site Need a Custom
    Search Engine
    Toolbar?\\\\Today's surfers
    aren't always too comfortable
    installing software on their
    computers. Especially free
    software that they don't
    necessarily understand. With
    all the horror stories of
    viruses, spyware, and adware
    that make the front page these
    days, it's no wonder. So is
    there ..." + ], + [ + "Spammers aren't ducking
    antispam laws by operating
    offshore, they're just
    ignoring it." + ], + [ + "AP - Biologists plan to use
    large nets and traps this week
    in Chicago's Burnham Harbor to
    search for the northern
    snakehead #151; a type of
    fish known for its voracious
    appetite and ability to wreak
    havoc on freshwater
    ecosystems." + ], + [ + "BAE Systems PLC and Northrop
    Grumman Corp. won \\$45 million
    contracts yesterday to develop
    prototypes of anti-missile
    technology that could protect
    commercial aircraft from
    shoulder-fired missiles." + ], + [ + "Users of Microsoft #39;s
    Hotmail, most of whom are
    accustomed to getting regular
    sales pitches for premium
    e-mail accounts, got a
    pleasant surprise in their
    inboxes this week: extra
    storage for free." + ], + [ + "IT services provider
    Electronic Data Systems
    yesterday reported a net loss
    of \\$153 million for the third
    quarter, with earnings hit in
    part by an asset impairment
    charge of \\$375 million
    connected with EDS's N/MCI
    project." + ], + [ + "International Business
    Machines Corp. said Wednesday
    it had agreed to settle most
    of the issues in a suit over
    changes in its pension plans
    on terms that allow the
    company to continue to appeal
    a key question while capping
    its liability at \\$1.7
    billion. <BR><FONT
    face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\"\\ color=\"#666666\">
    <B>-The Washington
    Post</B></FONT>" + ], + [ + "Edward Kozel, Cisco's former
    chief technology officer,
    joins the board of Linux
    seller Red Hat." + ], + [ + "Reuters - International
    Business Machines\\Corp. late
    on Wednesday rolled out a new
    version of its\\database
    software aimed at users of
    Linux and Unix
    operating\\systems that it
    hopes will help the company
    take away market\\share from
    market leader Oracle Corp. ." + ], + [ + "NASA #39;s Mars Odyssey
    mission, originally scheduled
    to end on Tuesday, has been
    granted a stay of execution
    until at least September 2006,
    reveal NASA scientists." + ], + [ + "BusinessWeek Online - The
    jubilation that swept East
    Germany after the fall of the
    Berlin Wall in 1989 long ago
    gave way to the sober reality
    of globalization and market
    forces. Now a decade of
    resentment seems to be boiling
    over. In Eastern cities such
    as Leipzig or Chemnitz,
    thousands have taken to the
    streets since July to protest
    cuts in unemployment benefits,
    the main source of livelihood
    for 1.6 million East Germans.
    Discontent among
    reunification's losers fueled
    big gains by the far left and
    far right in Brandenburg and
    Saxony state elections Sept.
    19. ..." + ], + [ + "In a discovery sure to set off
    a firestorm of debate over
    human migration to the western
    hemisphere, archaeologists in
    South Carolina say they have
    uncovered evidence that people
    lived in eastern North America
    at least 50,000 years ago -
    far earlier than" + ], + [ + "AP - Former president Bill
    Clinton on Monday helped
    launch a new Internet search
    company backed by the Chinese
    government which says its
    technology uses artificial
    intelligence to produce better
    results than Google Inc." + ], + [ + "Sun Microsystems plans to
    release its Sun Studio 10
    development tool in the fourth
    quarter of this year,
    featuring support for 64-bit
    applications running on AMD
    Opteron and Nocona processors,
    Sun officials said on Tuesday." + ], + [ + "Most IT Managers won #39;t
    question the importance of
    security, but this priority
    has been sliding between the
    third and fourth most
    important focus for companies." + ], + [ + "AP - The Federal Trade
    Commission on Thursday filed
    the first case in the country
    against software companies
    accused of infecting computers
    with intrusive \"spyware\" and
    then trying to sell people the
    solution." + ], + [ + "While shares of Apple have
    climbed more than 10 percent
    this week, reaching 52-week
    highs, two research firms told
    investors Friday they continue
    to remain highly bullish about
    the stock." + ], + [ + "States are now barred from
    imposing telecommunications
    regulations on Net phone
    providers." + ], + [ + "Strong sales of new mobile
    phone models boosts profits at
    Carphone Warehouse but the
    retailer's shares fall on
    concerns at a decline in
    profits from pre-paid phones." + ], + [ + " NEW YORK (Reuters) - IBM and
    top scientific research
    organizations are joining
    forces in a humanitarian
    effort to tap the unused
    power of millions of computers
    and help solve complex social
    problems." + ], + [ + "AP - Grizzly and black bears
    killed a majority of elk
    calves in northern Yellowstone
    National Park for the second
    year in a row, preliminary
    study results show." + ], + [ + "PC World - The one-time World
    Class Product of the Year PDA
    gets a much-needed upgrade." + ], + [ + "As part of its much-touted new
    MSN Music offering, Microsoft
    Corp. (MSFT) is testing a Web-
    based radio service that
    mimics nearly 1,000 local
    radio stations, allowing users
    to hear a version of their
    favorite radio station with
    far fewer interruptions." + ], + [ + "Ziff Davis - A quick
    resolution to the Mambo open-
    source copyright dispute seems
    unlikely now that one of the
    parties has rejected an offer
    for mediation." + ], + [ + "Students take note - endless
    journeys to the library could
    become a thing of the past
    thanks to a new multimillion-
    pound scheme to make classic
    texts available at the click
    of a mouse." + ], + [ + "Moscow - The next crew of the
    International Space Station
    (ISS) is to contribute to the
    Russian search for a vaccine
    against Aids, Russian
    cosmonaut Salijan Sharipovthe
    said on Thursday." + ], + [ + "Locked away in fossils is
    evidence of a sudden solar
    cooling. Kate Ravilious meets
    the experts who say it could
    explain a 3,000-year-old mass
    migration - and today #39;s
    global warming." + ], + [ + "By ANICK JESDANUN NEW YORK
    (AP) -- Taran Rampersad didn't
    complain when he failed to
    find anything on his hometown
    in the online encyclopedia
    Wikipedia. Instead, he simply
    wrote his own entry for San
    Fernando, Trinidad and
    Tobago..." + ], + [ + "How do you top a battle
    between Marines and an alien
    religious cult fighting to the
    death on a giant corona in
    outer space? The next logical
    step is to take that battle to
    Earth, which is exactly what
    Microsoft" + ], + [ + "Four U.S. agencies yesterday
    announced a coordinated attack
    to stem the global trade in
    counterfeit merchandise and
    pirated music and movies, an
    underground industry that law-
    enforcement officials estimate
    to be worth \\$500 billion each
    year." + ], + [ + "Half of all U.S. Secret
    Service agents are dedicated
    to protecting President
    Washington #151;and all the
    other Presidents on U.S.
    currency #151;from
    counterfeiters." + ], + [ + "Maxime Faget conceived and
    proposed the development of
    the one-man spacecraft used in
    Project Mercury, which put the
    first American astronauts into
    suborbital flight, then
    orbital flight" + ], + [ + "The patch fixes a flaw in the
    e-mail server software that
    could be used to get access to
    in-boxes and information." + ], + [ + "AP - Being the biggest dog may
    pay off at feeding time, but
    species that grow too large
    may be more vulnerable to
    extinction, new research
    suggests." + ], + [ + "Newest Efficeon processor also
    offers higher frequency using
    less power." + ], + [ + "AP - In what it calls a first
    in international air travel,
    Finnair says it will let its
    frequent fliers check in using
    text messages on mobile
    phones." + ], + [ + "Search Engine for Programming
    Code\\\\An article at Newsforge
    pointed me to Koders (
    http://www.koders.com ) a
    search engine for finding
    programming code. Nifty.\\\\The
    front page allows you to
    specify keywords, sixteen
    languages (from ASP to VB.NET)
    and sixteen licenses (from AFL
    to ZPL -- fortunately there's
    an information page to ..." + ], + [ + " #147;Apple once again was the
    star of the show at the annual
    MacUser awards, #148; reports
    MacUser, #147;taking away six
    Maxine statues including
    Product of the Year for the
    iTunes Music Store. #148;
    Other Apple award winners:
    Power Mac G5, AirPort Express,
    20-inch Apple Cinema Display,
    GarageBand and DVD Studio Pro
    3. Nov 22" + ], + [ + "AP - A sweeping wildlife
    preserve in southwestern
    Arizona is among the nation's
    10 most endangered refuges,
    due in large part to illegal
    drug and immigrant traffic and
    Border Patrol operations, a
    conservation group said
    Friday." + ], + [ + "SEPTEMBER 20, 2004
    (COMPUTERWORLD) - At
    PeopleSoft Inc. #39;s Connect
    2004 conference in San
    Francisco this week, the
    software vendor is expected to
    face questions from users
    about its ability to fend off
    Oracle Corp." + ], + [ + "PBS's Charlie Rose quizzes Sun
    co-founder Bill Joy,
    Monster.com chief Jeff Taylor,
    and venture capitalist John
    Doerr." + ], + [ + "Skype Technologies is teaming
    with Siemens to offer cordless
    phone users the ability to
    make Internet telephony calls,
    in addition to traditional
    calls, from their handsets." + ], + [ + "AP - After years of
    resistance, the U.S. trucking
    industry says it will not try
    to impede or delay a new
    federal rule aimed at cutting
    diesel pollution." + ], + [ + "Umesh Patel, a 36-year old
    software engineer from
    California, debated until the
    last minute." + ], + [ + "AP - From now until the start
    of winter, male tarantulas are
    roaming around, searching for
    female mates, an ideal time to
    find out where the spiders
    flourish in Arkansas." + ], + [ + "NewsFactor - While a U.S.
    District Court continues to
    weigh the legality of
    \\Oracle's (Nasdaq: ORCL)
    attempted takeover of
    \\PeopleSoft (Nasdaq: PSFT),
    Oracle has taken the necessary
    steps to ensure the offer does
    not die on the vine with
    PeopleSoft's shareholders." + ], + [ + "Analysts said the smartphone
    enhancements hold the
    potential to bring PalmSource
    into an expanding market that
    still has room despite early
    inroads by Symbian and
    Microsoft." + ], + [ + "TOKYO (AFP) - Japan #39;s
    Fujitsu and Cisco Systems of
    the US said they have agreed
    to form a strategic alliance
    focusing on routers and
    switches that will enable
    businesses to build advanced
    Internet Protocol (IP)
    networks." + ], + [ + "It #39;s a new internet
    browser. The first full
    version, Firefox 1.0, was
    launched earlier this month.
    In the sense it #39;s a
    browser, yes, but the
    differences are larger than
    the similarities." + ], + [ + "Microsoft will accelerate SP2
    distribution to meet goal of
    100 million downloads in two
    months." + ], + [ + "International Business
    Machines Corp., taken to court
    by workers over changes it
    made to its traditional
    pension plan, has decided to
    stop offering any such plan to
    new employees." + ], + [ + "NEW YORK - With four of its
    executives pleading guilty to
    price-fixing charges today,
    Infineon will have a hard time
    arguing that it didn #39;t fix
    prices in its ongoing
    litigation with Rambus." + ], + [ + "A report indicates that many
    giants of the industry have
    been able to capture lasting
    feelings of customer loyalty." + ], + [ + "It is the news that Internet
    users do not want to hear: the
    worldwide web is in danger of
    collapsing around us. Patrick
    Gelsinger, the chief
    technology officer for
    computer chip maker Intel,
    told a conference" + ], + [ + "Cisco Systems CEO John
    Chambers said today that his
    company plans to offer twice
    as many new products this year
    as ever before." + ], + [ + "Every day, somewhere in the
    universe, there #39;s an
    explosion that puts the power
    of the Sun in the shade. Steve
    Connor investigates the riddle
    of gamma-ray bursts." + ], + [ + "Ten months after NASA #39;s
    twin rovers landed on Mars,
    scientists reported this week
    that both robotic vehicles are
    still navigating their rock-
    studded landscapes with all
    instruments operating" + ], + [ + "British scientists say they
    have found a new, greener way
    to power cars and homes using
    sunflower oil, a commodity
    more commonly used for cooking
    fries." + ], + [ + "The United States #39; top
    computer-security official has
    resigned after a little more
    than a year on the job, the US
    Department of Homeland
    Security said on Friday." + ], + [ + " SAN FRANCISCO (Reuters) -
    Intel Corp. <A HREF=\"http:/
    /www.reuters.co.uk/financeQuot
    eLookup.jhtml?ticker=INTC.O
    qtype=sym infotype=info
    qcat=news\">INTC.O</A>
    on Thursday outlined its
    vision of the Internet of the
    future, one in which millions
    of computer servers would
    analyze and direct network
    traffic to make the Web safer
    and more efficient." + ], + [ + "Check out new gadgets as
    holiday season approaches." + ], + [ + "TOKYO, Sep 08, 2004 (AFX-UK
    via COMTEX) -- Sony Corp will
    launch its popular projection
    televisions with large liquid
    crystal display (LCD) screens
    in China early next year, a
    company spokeswoman said." + ], + [ + "AFP - Outgoing EU competition
    commissioner Mario Monti wants
    to reach a decision on
    Oracle's proposed takeover of
    business software rival
    PeopleSoft by the end of next
    month, an official said." + ], + [ + " WASHINGTON (Reuters) -
    Satellite companies would be
    able to retransmit
    broadcasters' television
    signals for another five
    years but would have to offer
    those signals on a single
    dish, under legislation
    approved by Congress on
    Saturday." + ], + [ + "Sven Jaschan, who may face
    five years in prison for
    spreading the Netsky and
    Sasser worms, is now working
    in IT security. Photo: AFP." + ], + [ + "TechWeb - An Indian Institute
    of Technology professor--and
    open-source evangelist--
    discusses the role of Linux
    and open source in India." + ], + [ + "Here are some of the latest
    health and medical news
    developments, compiled by
    editors of HealthDay: -----
    Contaminated Fish in Many U.S.
    Lakes and Rivers Fish
    that may be contaminated with
    dioxin, mercury, PCBs and
    pesticides are swimming in
    more than one-third of the
    United States' lakes and
    nearly one-quarter of its
    rivers, according to a list of
    advisories released by the
    Environmental Protection
    Agency..." + ], + [ + "Bill Gates is giving his big
    speech right now at Microsofts
    big Digital Entertainment
    Anywhere event in Los Angeles.
    Well have a proper report for
    you soon, but in the meantime
    we figured wed" + ], + [ + "Spinal and non-spinal
    fractures are reduced by
    almost a third in women age 80
    or older who take a drug
    called strontium ranelate,
    European investigators
    announced" + ], + [ + "BERLIN: Apple Computer Inc is
    planning the next wave of
    expansion for its popular
    iTunes online music store with
    a multi-country European
    launch in October, the
    services chief architect said
    on Wednesday." + ], + [ + "Calls to 13 other countries
    will be blocked to thwart
    auto-dialer software." + ], + [ + "com September 27, 2004, 5:00
    AM PT. This fourth priority
    #39;s main focus has been
    improving or obtaining CRM and
    ERP software for the past year
    and a half." + ], + [ + "SPACE.com - The Zero Gravity
    Corporation \\ has been given
    the thumbs up by the Federal
    Aviation Administration (FAA)
    to \\ conduct quot;weightless
    flights quot; for the general
    public, providing the \\
    sensation of floating in
    space." + ], + [ + "A 32-year-old woman in Belgium
    has become the first woman
    ever to give birth after
    having ovarian tissue removed,
    frozen and then implanted back
    in her body." + ], + [ + "No sweat, says the new CEO,
    who's imported from IBM. He
    also knows this will be the
    challenge of a lifetime." + ], + [ + "AP - Their discoveries may be
    hard for many to comprehend,
    but this year's Nobel Prize
    winners still have to explain
    what they did and how they did
    it." + ], + [ + "More than 15 million homes in
    the UK will be able to get on-
    demand movies by 2008, say
    analysts." + ], + [ + "A well-researched study by the
    National Academy of Sciences
    makes a persuasive case for
    refurbishing the Hubble Space
    Telescope. NASA, which is
    reluctant to take this
    mission, should rethink its
    position." + ], + [ + "LONDON -- In a deal that
    appears to buck the growing
    trend among governments to
    adopt open-source
    alternatives, the U.K.'s
    Office of Government Commerce
    (OGC) is negotiating a renewal
    of a three-year agreement with
    Microsoft Corp." + ], + [ + "WASHINGTON - A Pennsylvania
    law requiring Internet service
    providers (ISPs) to block Web
    sites the state's prosecuting
    attorneys deem to be child
    pornography has been reversed
    by a U.S. federal court, with
    the judge ruling the law
    violated free speech rights." + ], + [ + "The Microsoft Corp. chairman
    receives four million e-mails
    a day, but practically an
    entire department at the
    company he founded is
    dedicated to ensuring that
    nothing unwanted gets into his
    inbox, the company #39;s chief
    executive said Thursday." + ], + [ + "The 7100t has a mobile phone,
    e-mail, instant messaging, Web
    browsing and functions as an
    organiser. The device looks
    like a mobile phone and has
    the features of the other
    BlackBerry models, with a
    large screen" + ], + [ + "Apple Computer has unveiled
    two new versions of its hugely
    successful iPod: the iPod
    Photo and the U2 iPod. Apple
    also has expanded" + ], + [ + "AP - This year's hurricanes
    spread citrus canker to at
    least 11,000 trees in
    Charlotte County, one of the
    largest outbreaks of the
    fruit-damaging infection to
    ever affect Florida's citrus
    industry, state officials
    said." + ], + [ + "IBM moved back into the iSCSI
    (Internet SCSI) market Friday
    with a new array priced at
    US\\$3,000 and aimed at the
    small and midsize business
    market." + ], + [ + "American Technology Research
    analyst Shaw Wu has initiated
    coverage of Apple Computer
    (AAPL) with a #39;buy #39;
    recommendation, and a 12-month
    target of US\\$78 per share." + ], + [ + "washingtonpost.com - Cell
    phone shoppers looking for new
    deals and features didn't find
    them yesterday, a day after
    Sprint Corp. and Nextel
    Communications Inc. announced
    a merger that the phone
    companies promised would shake
    up the wireless business." + ], + [ + "JAPANESE GIANT Sharp has
    pulled the plug on its Linux-
    based PDAs in the United
    States as no-one seems to want
    them. The company said that it
    will continue to sell them in
    Japan where they sell like hot
    cakes" + ], + [ + "New NavOne offers handy PDA
    and Pocket PC connectivity,
    but fails to impress on
    everything else." + ], + [ + "The Marvel deal calls for
    Mforma to work with the comic
    book giant to develop a
    variety of mobile applications
    based on Marvel content." + ], + [ + "washingtonpost.com - The
    Portable Media Center -- a
    new, Microsoft-conceived
    handheld device that presents
    video and photos as well as
    music -- would be a decent
    idea if there weren't such a
    thing as lampposts. Or street
    signs. Or trees. Or other
    cars." + ], + [ + "The Air Force Reserve #39;s
    Hurricane Hunters, those
    fearless crews who dive into
    the eyewalls of hurricanes to
    relay critical data on
    tropical systems, were chased
    from their base on the
    Mississippi Gulf Coast by
    Hurricane Ivan." + ], + [ + "Embargo or not, Fidel Castro's
    socialist paradise has quietly
    become a pharmaceutical
    powerhouse. (They're still
    working on the capitalism
    thing.) By Douglas Starr from
    Wired magazine." + ], + [ + "A new worm can spy on users by
    hijacking their Web cameras, a
    security firm warned Monday.
    The Rbot.gr worm -- the latest
    in a long line of similar
    worms; one security firm
    estimates that more than 4,000
    variations" + ], + [ + "The separation of PalmOne and
    PalmSource will be complete
    with Eric Benhamou's
    resignation as the latter's
    chairman." + ], + [ + "\\\\\"(CNN) -- A longtime
    associate of al Qaeda leader
    Osama bin Laden surrendered
    to\\Saudi Arabian officials
    Tuesday, a Saudi Interior
    Ministry official said.\"\\\\\"But
    it is unclear what role, if
    any, Khaled al-Harbi may have
    had in any terror\\attacks
    because no public charges have
    been filed against him.\"\\\\\"The
    Saudi government -- in a
    statement released by its
    embassy in Washington
    --\\called al-Harbi's surrender
    \"the latest direct result\" of
    its limited, one-month\\offer
    of leniency to terror
    suspects.\"\\\\This is great! I
    hope this really starts to pay
    off. Creative solutions
    to\\terrorism that don't
    involve violence. \\\\How
    refreshing! \\\\Are you paying
    attention Bush
    administration?\\\\" + ], + [ + "AT T Corp. is cutting 7,400
    more jobs and slashing the
    book value of its assets by
    \\$11.4 billion, drastic moves
    prompted by the company's plan
    to retreat from the
    traditional consumer telephone
    business following a lost
    court battle." + ], + [ + "Celerons form the basis of
    Intel #39;s entry-level
    platform which includes
    integrated/value motherboards
    as well. The Celeron 335D is
    the fastest - until the
    Celeron 340D - 2.93GHz -
    becomes mainstream - of the" + ], + [ + "The entertainment industry
    asks the Supreme Court to
    reverse the Grokster decision,
    which held that peer-to-peer
    networks are not liable for
    copyright abuses of their
    users. By Michael Grebb." + ], + [ + "Hewlett-Packard will shell out
    \\$16.1 billion for chips in
    2005, but Dell's wallet is
    wide open, too." + ], + [ + "A remote attacker could take
    complete control over
    computers running many
    versions of Microsoft software
    by inserting malicious code in
    a JPEG image that executes
    through an unchecked buffer" + ], + [ + "The lawsuit claims the
    companies use a patented
    Honeywell technology for
    brightening images and
    reducing interference on
    displays." + ], + [ + "The co-president of Oracle
    testified that her company was
    serious about its takeover
    offer for PeopleSoft and was
    not trying to scare off its
    customers." + ], + [ + "An extremely rare Hawaiian
    bird dies in captivity,
    possibly marking the
    extinction of its entire
    species only 31 years after it
    was first discovered." + ], + [ + "Does Geico's trademark lawsuit
    against Google have merit? How
    will the case be argued?
    What's the likely outcome of
    the trial? A mock court of
    trademark experts weighs in
    with their verdict." + ], + [ + "Treo 650 boasts a high-res
    display, an improved keyboard
    and camera, a removable
    battery, and more. PalmOne
    this week is announcing the
    Treo 650, a hybrid PDA/cell-
    phone device that addresses
    many of the shortcomings" + ], + [ + "International Business
    Machines Corp.'s possible exit
    from the personal computer
    business would be the latest
    move in what amounts to a long
    goodbye from a field it
    pioneered and revolutionized." + ], + [ + "Leipzig Game Convention in
    Germany, the stage for price-
    slash revelations. Sony has
    announced that it #39;s
    slashing the cost of PS2 in
    the UK and Europe to 104.99
    GBP." + ], + [ + "In yet another devastating
    body blow to the company,
    Intel (Nasdaq: INTC) announced
    it would be canceling its
    4-GHz Pentium chip. The
    semiconductor bellwether said
    it was switching" + ], + [ + "Seagate #39;s native SATA
    interface technology with
    Native Command Queuing (NCQ)
    allows the Barracuda 7200.8 to
    match the performance of
    10,000-rpm SATA drives without
    sacrificing capacity" + ], + [ + "You #39;re probably already
    familiar with one of the most
    common questions we hear at
    iPodlounge: quot;how can I
    load my iPod up with free
    music?" + ], + [ + "Vice chairman of the office of
    the chief executive officer in
    Novell Inc, Chris Stone is
    leaving the company to pursue
    other opportunities in life." + ], + [ + "washingtonpost.com - Microsoft
    is going to Tinseltown today
    to announce plans for its
    revamped Windows XP Media
    Center, part of an aggressive
    push to get ahead in the
    digital entertainment race." + ], + [ + "AP - Most of the turkeys
    gracing the nation's dinner
    tables Thursday have been
    selectively bred for their
    white meat for so many
    generations that simply
    walking can be a problem for
    many of the big-breasted birds
    and sex is no longer possible." + ], + [ + " SEATTLE (Reuters) - Microsoft
    Corp. <A HREF=\"http://www.r
    euters.co.uk/financeQuoteLooku
    p.jhtml?ticker=MSFT.O
    qtype=sym infotype=info
    qcat=news\">MSFT.O</A>
    is making a renewed push this
    week to get its software into
    living rooms with the launch
    of a new version of its
    Windows XP Media Center, a
    personal computer designed for
    viewing movies, listening to
    music and scrolling through
    digital pictures." + ], + [ + "The new software is designed
    to simplify the process of
    knitting together back-office
    business applications." + ], + [ + "FRANKFURT, GERMANY -- The
    German subsidiaries of
    Hewlett-Packard Co. (HP) and
    Novell Inc. are teaming to
    offer Linux-based products to
    the country's huge public
    sector." + ], + [ + "SiliconValley.com - \"I'm
    back,\" declared Apple
    Computer's Steve Jobs on
    Thursday morning in his first
    public appearance before
    reporters since cancer surgery
    in late July." + ], + [ + "Health India: London, Nov 4 :
    Cosmetic face cream used by
    fashionable Roman women was
    discovered at an ongoing
    archaeological dig in London,
    in a metal container, complete
    with the lid and contents." + ], + [ + "MacCentral - Microsoft's
    Macintosh Business Unit on
    Tuesday issued a patch for
    Virtual PC 7 that fixes a
    problem that occurred when
    running the software on Power
    Mac G5 computers with more
    than 2GB of RAM installed.
    Previously, Virtual PC 7 would
    not run on those computers,
    causing a fatal error that
    crashed the application.
    Microsoft also noted that
    Virtual PC 7.0.1 also offers
    stability improvements,
    although it wasn't more
    specific than that -- some
    users have reported problems
    with using USB devices and
    other issues." + ], + [ + "Don't bother buying Star Wars:
    Battlefront if you're looking
    for a first-class shooter --
    there are far better games out
    there. But if you're a Star
    Wars freak and need a fix,
    this title will suffice. Lore
    Sjberg reviews Battlefront." + ], + [ + "More than by brain size or
    tool-making ability, the human
    species was set apart from its
    ancestors by the ability to
    jog mile after lung-stabbing
    mile with greater endurance
    than any other primate,
    according to research
    published today in the journal" + ], + [ + "Woodland Hills-based Brilliant
    Digital Entertainment and its
    subsidiary Altnet announced
    yesterday that they have filed
    a patent infringement suit
    against the Recording Industry
    Association of America (RIAA)." + ], + [ + "Sony Ericsson and Cingular
    provide Z500a phones and
    service for military families
    to stay connected on today
    #39;s quot;Dr. Phil Show
    quot;." + ], + [ + "Officials at EarthLink #39;s
    (Quote, Chart) R amp;D
    facility have quietly released
    a proof-of-concept file-
    sharing application based on
    the Session Initiated Protocol
    (define)." + ], + [ + " quot;It #39;s your mail,
    quot; the Google Web site
    said. quot;You should be able
    to choose how and where you
    read it. You can even switch
    to other e-mail services
    without having to worry about
    losing access to your
    messages." + ], + [ + "The world #39;s only captive
    great white shark made history
    this week when she ate several
    salmon fillets, marking the
    first time that a white shark
    in captivity" + ], + [ + "Communications aggregator
    iPass said Monday that it is
    adding in-flight Internet
    access to its access
    portfolio. Specifically, iPass
    said it will add access
    offered by Connexion by Boeing
    to its list of access
    providers." + ], + [ + "Company launches free test
    version of service that
    fosters popular Internet
    activity." + ], + [ + "Outdated computer systems are
    hampering the work of
    inspectors, says the UN
    nuclear agency." + ], + [ + "MacCentral - You Software Inc.
    announced on Tuesday the
    availability of You Control:
    iTunes, a free\\download that
    places iTunes controls in the
    Mac OS X menu bar.
    Without\\leaving the current
    application, you can pause,
    play, rewind or skip songs,\\as
    well as control iTunes' volume
    and even browse your entire
    music library\\by album, artist
    or genre. Each time a new song
    plays, You Control:
    iTunes\\also pops up a window
    that displays the artist and
    song name and the
    album\\artwork, if it's in the
    library. System requirements
    call for Mac OS X\\v10.2.6 and
    10MB free hard drive space.
    ..." + ], + [ + "The US Secret Service Thursday
    announced arrests in eight
    states and six foreign
    countries of 28 suspected
    cybercrime gangsters on
    charges of identity theft,
    computer fraud, credit-card
    fraud, and conspiracy." + ] + ], + "hovertemplate": "label=Sci/Tech
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    mixture of exhilaration that
    the Athens Olympics proved
    successful, and relief that
    they passed off without any
    major setback." + ], + [ + "Loudon, NH -- As the rain
    began falling late Friday
    afternoon at New Hampshire
    International Speedway, the
    rich in the Nextel Cup garage
    got richer." + ], + [ + "AP - Ottawa Senators right
    wing Marian Hossa is switching
    European teams during the NHL
    lockout." + ], + [ + "Chelsea terminated Romania
    striker Adrian Mutu #39;s
    contract, citing gross
    misconduct after the player
    failed a doping test for
    cocaine and admitted taking
    the drug, the English soccer
    club said in a statement." + ], + [ + "The success of Paris #39; bid
    for Olympic Games 2012 would
    bring an exceptional
    development for France for at
    least 6 years, said Jean-
    Francois Lamour, French
    minister for Youth and Sports
    on Tuesday." + ], + [ + "Madison, WI (Sports Network) -
    Anthony Davis ran for 122
    yards and two touchdowns to
    lead No. 6 Wisconsin over
    Northwestern, 24-12, to
    celebrate Homecoming weekend
    at Camp Randall Stadium." + ], + [ + "LaVar Arrington participated
    in his first practice since
    Oct. 25, when he aggravated a
    knee injury that sidelined him
    for 10 games." + ], + [ + " NEW YORK (Reuters) - Billie
    Jean King cut her final tie
    with the U.S. Fed Cup team
    Tuesday when she retired as
    coach." + ], + [ + "The Miami Dolphins arrived for
    their final exhibition game
    last night in New Orleans
    short nine players." + ], + [ + "AP - From LSU at No. 1 to Ohio
    State at No. 10, The AP
    women's basketball poll had no
    changes Monday." + ], + [ + "nother stage victory for race
    leader Petter Solberg, his
    fifth since the start of the
    rally. The Subaru driver is
    not pulling away at a fast
    pace from Gronholm and Loeb
    but the gap is nonetheless
    increasing steadily." + ], + [ + "April 1980 -- Players strike
    the last eight days of spring
    training. Ninety-two
    exhibition games are canceled.
    June 1981 -- Players stage
    first midseason strike in
    history." + ], + [ + "On Sunday, the day after Ohio
    State dropped to 0-3 in the
    Big Ten with a 33-7 loss at
    Iowa, the Columbus Dispatch
    ran a single word above its
    game story on the Buckeyes:
    quot;Embarrassing." + ], + [ + "BOSTON - Curt Schilling got
    his 20th win on the eve of
    Boston #39;s big series with
    the New York Yankees. Now he
    wants much more. quot;In a
    couple of weeks, hopefully, it
    will get a lot better, quot;
    he said after becoming" + ], + [ + "Shooed the ghosts of the
    Bambino and the Iron Horse and
    the Yankee Clipper and the
    Mighty Mick, on his 73rd
    birthday, no less, and turned
    Yankee Stadium into a morgue." + ], + [ + "Gold medal-winning Marlon
    Devonish says the men #39;s
    4x100m Olympic relay triumph
    puts British sprinting back on
    the map. Devonish, Darren
    Campbell, Jason Gardener and
    Mark Lewis-Francis edged out
    the American" + ], + [ + "he difficult road conditions
    created a few incidents in the
    first run of the Epynt stage.
    Francois Duval takes his
    second stage victory since the
    start of the rally, nine
    tenths better than Sebastien
    Loeb #39;s performance in
    second position." + ], + [ + "Tallahassee, FL (Sports
    Network) - Florida State head
    coach Bobby Bowden has
    suspended senior wide receiver
    Craphonso Thorpe for the
    Seminoles #39; game against
    fellow Atlantic Coast
    Conference member Duke on
    Saturday." + ], + [ + "Former champion Lleyton Hewitt
    bristled, battled and
    eventually blossomed as he
    took another step towards a
    second US Open title with a
    second-round victory over
    Moroccan Hicham Arazi on
    Friday." + ], + [ + "As the Mets round out their
    search for a new manager, the
    club is giving a last-minute
    nod to its past. Wally
    Backman, an infielder for the
    Mets from 1980-88 who played
    second base on the 1986" + ], + [ + "AMSTERDAM, Aug 18 (Reuters) -
    Midfielder Edgar Davids #39;s
    leadership qualities and
    never-say-die attitude have
    earned him the captaincy of
    the Netherlands under new
    coach Marco van Basten." + ], + [ + "COUNTY KILKENNY, Ireland (PA)
    -- Hurricane Jeanne has led to
    world No. 1 Vijay Singh
    pulling out of this week #39;s
    WGC-American Express
    Championship at Mount Juliet." + ], + [ + "Green Bay, WI (Sports Network)
    - The Green Bay Packers will
    be without the services of Pro
    Bowl center Mike Flanagan for
    the remainder of the season,
    as he will undergo left knee
    surgery." + ], + [ + "COLUMBUS, Ohio -- NCAA
    investigators will return to
    Ohio State University Monday
    to take another look at the
    football program after the
    latest round of allegations
    made by former players,
    according to the Akron Beacon
    Journal." + ], + [ + "Manchester United gave Alex
    Ferguson a 1,000th game
    anniversary present by
    reaching the last 16 of the
    Champions League yesterday,
    while four-time winners Bayern
    Munich romped into the second
    round with a 5-1 beating of
    Maccabi Tel Aviv." + ], + [ + "The last time the Angels
    played the Texas Rangers, they
    dropped two consecutive
    shutouts at home in their most
    agonizing lost weekend of the
    season." + ], + [ + "SEATTLE - Chasing a nearly
    forgotten ghost of the game,
    Ichiro Suzuki broke one of
    baseball #39;s oldest records
    Friday night, smoking a single
    up the middle for his 258th
    hit of the year and breaking
    George Sisler #39;s record for
    the most hits in a season" + ], + [ + "Grace Park, her caddie - and
    fans - were poking around in
    the desert brush alongside the
    18th fairway desperately
    looking for her ball." + ], + [ + "Washington Redskins kicker
    John Hall strained his right
    groin during practice
    Thursday, his third leg injury
    of the season. Hall will be
    held out of practice Friday
    and is questionable for Sunday
    #39;s game against the Chicago
    Bears." + ], + [ + "The Florida Gators and
    Arkansas Razorbacks meet for
    just the sixth time Saturday.
    The Gators hold a 4-1
    advantage in the previous five
    meetings, including last year
    #39;s 33-28 win." + ], + [ + "AP - 1941 #151; Brooklyn
    catcher Mickey Owen dropped a
    third strike on Tommy Henrich
    of what would have been the
    last out of a Dodgers victory
    against the New York Yankees.
    Given the second chance, the
    Yankees scored four runs for a
    7-4 victory and a 3-1 lead in
    the World Series, which they
    ended up winning." + ], + [ + "Braves shortstop Rafael Furcal
    was arrested on drunken
    driving charges Friday, his
    second D.U.I. arrest in four
    years." + ], + [ + " ATHENS (Reuters) - Natalie
    Coughlin's run of bad luck
    finally took a turn for the
    better when she won the gold
    medal in the women's 100
    meters backstroke at the
    Athens Olympics Monday." + ], + [ + "AP - Pedro Feliz hit a
    tiebreaking grand slam with
    two outs in the eighth inning
    for his fourth hit of the day,
    and the Giants helped their
    playoff chances with a 9-5
    victory over the Los Angeles
    Dodgers on Saturday." + ], + [ + "LEVERKUSEN/ROME, Dec 7 (SW) -
    Dynamo Kiev, Bayer Leverkusen,
    and Real Madrid all have a
    good chance of qualifying for
    the Champions League Round of
    16 if they can get the right
    results in Group F on
    Wednesday night." + ], + [ + "Ed Hinkel made a diving,
    fingertip catch for a key
    touchdown and No. 16 Iowa
    stiffened on defense when it
    needed to most to beat Iowa
    State 17-10 Saturday." + ], + [ + "During last Sunday #39;s
    Nextel Cup race, amid the
    ongoing furor over Dale
    Earnhardt Jr. #39;s salty
    language, NBC ran a commercial
    for a show coming on later
    that night called quot;Law
    amp; Order: Criminal Intent." + ], + [ + "AP - After playing in hail,
    fog and chill, top-ranked
    Southern California finishes
    its season in familiar
    comfort. The Trojans (9-0, 6-0
    Pacific-10) have two games at
    home #151; against Arizona on
    Saturday and Notre Dame on
    Nov. 27 #151; before their
    rivalry game at UCLA." + ], + [ + "The remnants of Hurricane
    Jeanne rained out Monday's
    game between the Mets and the
    Atlanta Braves. It will be
    made up Tuesday as part of a
    doubleheader." + ], + [ + "AP - NASCAR is not expecting
    any immediate changes to its
    top-tier racing series
    following the merger between
    telecommunications giant
    Sprint Corp. and Nextel
    Communications Inc." + ], + [ + "Like wide-open races? You
    #39;ll love the Big 12 North.
    Here #39;s a quick morning
    line of the Big 12 North as it
    opens conference play this
    weekend." + ], + [ + "Taquan Dean scored 22 points,
    Francisco Garcia added 19 and
    No. 13 Louisville withstood a
    late rally to beat Florida
    74-70 Saturday." + ], + [ + "NEW YORK -- This was all about
    Athens, about redemption for
    one and validation for the
    other. Britain's Paula
    Radcliffe, the fastest female
    marathoner in history, failed
    to finish either of her
    Olympic races last summer.
    South Africa's Hendrik Ramaala
    was a five-ringed dropout,
    too, reinforcing his
    reputation as a man who could
    go only half the distance." + ], + [ + "LONDON -- Ernie Els has set
    his sights on an improved
    putting display this week at
    the World Golf Championships
    #39; NEC Invitational in
    Akron, Ohio, after the
    disappointment of tying for
    fourth place at the PGA
    Championship last Sunday." + ], + [ + "Fiji #39;s Vijay Singh
    replaced Tiger Woods as the
    world #39;s No.1 ranked golfer
    today by winning the PGA
    Deutsche Bank Championship." + ], + [ + "LEIPZIG, Germany : Jurgen
    Klinsmann enjoyed his first
    home win as German manager
    with his team defeating ten-
    man Cameroon 3-0 in a friendly
    match." + ], + [ + "AP - Kevin Brown had a chance
    to claim a place in Yankees
    postseason history with his
    start in Game 7 of the AL
    championship series." + ], + [ + "HOMESTEAD, Fla. -- Kurt Busch
    got his first big break in
    NASCAR by winning a 1999
    talent audition nicknamed
    quot;The Gong Show. quot; He
    was selected from dozens of
    unknown, young race drivers to
    work for one of the sport
    #39;s most famous team owners,
    Jack Roush." + ], + [ + "Zurich, Switzerland (Sports
    Network) - Former world No. 1
    Venus Williams advanced on
    Thursday and will now meet
    Wimbledon champion Maria
    Sharapova in the quarterfinals
    at the \\$1." + ], + [ + "INDIA #39;S cricket chiefs
    began a frenetic search today
    for a broadcaster to show next
    month #39;s home series
    against world champion
    Australia after cancelling a
    controversial \\$US308 million
    (\\$440 million) television
    deal." + ], + [ + "The International Olympic
    Committee (IOC) has urged
    Beijing to ensure the city is
    ready to host the 2008 Games
    well in advance, an official
    said on Wednesday." + ], + [ + "Virginia Tech scores 24 points
    off four first-half turnovers
    in a 55-6 wipeout of Maryland
    on Thursday to remain alone
    atop the ACC." + ], + [ + "AP - Martina Navratilova's
    long, illustrious career will
    end without an Olympic medal.
    The 47-year-old Navratilova
    and Lisa Raymond lost 6-4,
    4-6, 6-4 on Thursday night to
    fifth-seeded Shinobu Asagoe
    and Ai Sugiyama of Japan in
    the quarterfinals, one step
    shy of the medal round." + ], + [ + "PSV Eindhoven re-established
    their five-point lead at the
    top of the Dutch Eredivisie
    today with a 2-0 win at
    Vitesse Arnhem. Former
    Sheffield Wednesday striker
    Gerald Sibon put PSV ahead in
    the 56th minute" + ], + [ + "TODAY AUTO RACING 3 p.m. --
    NASCAR Nextel Cup Sylvania 300
    qualifying at N.H.
    International Speedway,
    Loudon, N.H., TNT PRO BASEBALL
    7 p.m. -- Red Sox at New York
    Yankees, Ch. 38, WEEI (850)
    (on cable systems where Ch. 38
    is not available, the game
    will air on NESN); Chicago
    Cubs at Cincinnati, ESPN 6
    p.m. -- Eastern League finals:
    ..." + ], + [ + "MUMBAI, SEPTEMBER 21: The
    Board of Control for Cricket
    in India (BCCI) today informed
    the Bombay High Court that it
    was cancelling the entire
    tender process regarding
    cricket telecast rights as
    also the conditional deal with
    Zee TV." + ], + [ + "JEJU, South Korea : Grace Park
    of South Korea won an LPGA
    Tour tournament, firing a
    seven-under-par 65 in the
    final round of the
    1.35-million dollar CJ Nine
    Bridges Classic." + ], + [ + "AP - The NHL will lock out its
    players Thursday, starting a
    work stoppage that threatens
    to keep the sport off the ice
    for the entire 2004-05 season." + ], + [ + "When Paula Radcliffe dropped
    out of the Olympic marathon
    miles from the finish, she
    sobbed uncontrollably.
    Margaret Okayo knew the
    feeling." + ], + [ + "First baseman Richie Sexson
    agrees to a four-year contract
    with the Seattle Mariners on
    Wednesday." + ], + [ + "Notes and quotes from various
    drivers following California
    Speedway #39;s Pop Secret 500.
    Jeff Gordon slipped to second
    in points following an engine
    failure while Jimmie Johnson
    moved back into first." + ], + [ + " MEMPHIS, Tenn. (Sports
    Network) - The Memphis
    Grizzlies signed All-Star
    forward Pau Gasol to a multi-
    year contract on Friday.
    Terms of the deal were not
    announced." + ], + [ + "Andre Agassi brushed past
    Jonas Bjorkman 6-3 6-4 at the
    Stockholm Open on Thursday to
    set up a quarter-final meeting
    with Spanish sixth seed
    Fernando Verdasco." + ], + [ + " BOSTON (Reuters) - Boston was
    tingling with anticipation on
    Saturday as the Red Sox
    prepared to host Game One of
    the World Series against the
    St. Louis Cardinals and take a
    step toward ridding
    themselves of a hex that has
    hung over the team for eight
    decades." + ], + [ + "FOR the first time since his
    appointment as Newcastle
    United manager, Graeme Souness
    has been required to display
    the strong-arm disciplinary
    qualities that, to Newcastle
    directors, made" + ], + [ + "WHY IT HAPPENED Tom Coughlin
    won his first game as Giants
    coach and immediately
    announced a fine amnesty for
    all Giants. Just kidding." + ], + [ + "A second-place finish in his
    first tournament since getting
    married was good enough to
    take Tiger Woods from third to
    second in the world rankings." + ], + [ + " COLORADO SPRINGS, Colorado
    (Reuters) - World 400 meters
    champion Jerome Young has been
    given a lifetime ban from
    athletics for a second doping
    offense, the U.S. Anti-Doping
    Agency (USADA) said Wednesday." + ], + [ + "LONDON - In two years, Arsenal
    will play their home matches
    in the Emirates stadium. That
    is what their new stadium at
    Ashburton Grove will be called
    after the Premiership
    champions yesterday agreed to
    the" + ], + [ + "KNOXVILLE, Tenn. -- Jason
    Campbell threw for 252 yards
    and two touchdowns, and No. 8
    Auburn proved itself as a
    national title contender by
    overwhelming No. 10 Tennessee,
    34-10, last night." + ], + [ + "WASHINGTON -- Another
    Revolution season concluded
    with an overtime elimination.
    Last night, the Revolution
    thrice rallied from deficits
    for a 3-3 tie with D.C. United
    in the Eastern Conference
    final, then lost in the first-
    ever Major League Soccer match
    decided by penalty kicks." + ], + [ + "Dwyane Wade calls himself a
    quot;sidekick, quot; gladly
    accepting the role Kobe Bryant
    never wanted in Los Angeles.
    And not only does second-year
    Heat point" + ], + [ + "World number one golfer Vijay
    Singh of Fiji has captured his
    eighth PGA Tour event of the
    year with a win at the 84
    Lumber Classic in Farmington,
    Pennsylvania." + ], + [ + "The noise was deafening and
    potentially unsettling in the
    minutes before the start of
    the men #39;s Olympic
    200-meter final. The Olympic
    Stadium crowd chanted
    quot;Hellas!" + ], + [ + "Brazilian forward Ronaldinho
    scored a sensational goal for
    Barcelona against Milan,
    making for a last-gasp 2-1.
    The 24-year-old #39;s
    brilliant left-foot hit at the
    Nou Camp Wednesday night sent
    Barcelona atop of Group F." + ], + [ + "Bee Staff Writer. SANTA CLARA
    - Andre Carter #39;s back
    injury has kept him out of the
    49ers #39; lineup since Week
    1. It #39;s also interfering
    with him rooting on his alma
    mater in person." + ], + [ + "AP - Kellen Winslow Jr. ended
    his second NFL holdout Friday." + ], + [ + "HOUSTON - With only a few
    seconds left in a certain
    victory for Miami, Peyton
    Manning threw a meaningless
    6-yard touchdown pass to
    Marvin Harrison to cut the
    score to 24-15." + ], + [ + "DEADLY SCORER: Manchester
    United #39;s Wayne Rooney
    celebrating his three goals
    against Fenerbahce this week
    at Old Trafford. (AP)." + ], + [ + "You can feel the confidence
    level, not just with Team
    Canada but with all of Canada.
    There #39;s every expectation,
    from one end of the bench to
    the other, that Martin Brodeur
    is going to hold the fort." + ], + [ + "Heading into the first game of
    a new season, every team has
    question marks. But in 2004,
    the Denver Broncos seemed to
    have more than normal." + ], + [ + "Anaheim, Calif. - There is a
    decidedly right lean to the
    three-time champions of the
    American League Central. In a
    span of 26 days, the Minnesota
    Twins lost the left side of
    their infield to free agency." + ], + [ + "With the NFL trading deadline
    set for 4 p.m. Tuesday,
    Patriots coach Bill Belichick
    said there didn't seem to be
    much happening on the trade
    front around the league." + ], + [ + "AP - David Beckham broke his
    rib moments after scoring
    England's second goal in
    Saturday's 2-0 win over Wales
    in a World Cup qualifying
    game." + ], + [ + "Nothing changed at the top of
    Serie A as all top teams won
    their games to keep the
    distance between one another
    unaltered. Juventus came back
    from behind against Lazio to
    win thanks to another goal by
    Ibrahimovic" + ], + [ + "RICHMOND, Va. Jeremy Mayfield
    won his first race in over
    four years, taking the
    Chevrolet 400 at Richmond
    International Raceway after
    leader Kurt Busch ran out of
    gas eight laps from the
    finish." + ], + [ + "AP - Track star Marion Jones
    filed a defamation lawsuit
    Wednesday against the man
    whose company is at the center
    of a federal investigation
    into illegal steroid use among
    some of the nation's top
    athletes." + ], + [ + "England #39;s players hit out
    at cricket #39;s authorities
    tonight and claimed they had
    been used as quot;political
    pawns quot; after the Zimbabwe
    government produced a
    spectacular U-turn to ensure
    the controversial one-day
    series will go ahead." + ], + [ + "AP - The Japanese won the
    pregame home run derby. Then
    the game started and the major
    league All-Stars put their
    bats to work. Back-to-back
    home runs by Moises Alou and
    Vernon Wells in the fourth
    inning and by Johnny Estrada
    and Brad Wilkerson in the
    ninth powered the major
    leaguers past the Japanese
    stars 7-3 Sunday for a 3-0
    lead in the eight-game series." + ], + [ + "With Chelsea losing their
    unbeaten record and Manchester
    United failing yet again to
    win, William Hill now make
    Arsenal red-hot 2/5 favourites
    to retain the title." + ], + [ + "Theresa special bookcase in Al
    Grohs office completely full
    of game plans from his days in
    the NFL. Green ones are from
    the Jets." + ], + [ + "Newcastle ensured their place
    as top seeds in Friday #39;s
    third round UEFA Cup draw
    after holding Sporting Lisbon
    to a 1-1 draw at St James #39;
    Park." + ], + [ + "CBC SPORTS ONLINE - Bode
    Miller continued his
    impressive 2004-05 World Cup
    skiing season by winning a
    night slalom race in
    Sestriere, Italy on Monday." + ], + [ + "Damien Rhodes scored on a
    2-yard run in the second
    overtime, then Syracuse's
    defense stopped Pittsburgh on
    fourth and 1, sending the
    Orange to a 38-31 victory
    yesterday in Syracuse, N.Y." + ], + [ + "AP - Anthony Harris scored 18
    of his career-high 23 points
    in the second half to help
    Miami upset No. 19 Florida
    72-65 Saturday and give first-
    year coach Frank Haith his
    biggest victory." + ], + [ + "AP - The five cities looking
    to host the 2012 Summer Games
    submitted bids to the
    International Olympic
    Committee on Monday, entering
    the final stage of a long
    process in hopes of landing
    one of the biggest prizes in
    sports." + ], + [ + "The FIA has already cancelled
    todays activities at Suzuka as
    Super Typhoon Ma-On heads
    towards the 5.807km circuit.
    Saturday practice has been
    cancelled altogether while
    pre-qualifying and final
    qualifying" + ], + [ + " GRAND PRAIRIE, Texas
    (Reuters) - Betting on horses
    was banned in Texas until as
    recently as 1987. Times have
    changed rapidly since.
    Saturday, Lone Star Park race
    track hosts the \\$14 million
    Breeders Cup, global racing's
    end-of-season extravaganza." + ], + [ + "It might be a stay of
    execution for Coach P, or it
    might just be a Christmas
    miracle come early. SU #39;s
    upset win over BC has given
    hope to the Orange playing in
    a post season Bowl game." + ], + [ + "PHIL Neville insists
    Manchester United don #39;t
    fear anyone in the Champions
    League last 16 and declared:
    quot;Bring on the Italians." + ], + [ + "TORONTO -- Toronto Raptors
    point guard Alvin Williams
    will miss the rest of the
    season after undergoing
    surgery on his right knee
    Monday." + ], + [ + "AP - Tennessee's two freshmen
    quarterbacks have Volunteers
    fans fantasizing about the
    next four years. Brent
    Schaeffer and Erik Ainge
    surprised many with the nearly
    seamless way they rotated
    throughout a 42-17 victory
    over UNLV on Sunday night." + ], + [ + "MADRID, Aug 18 (Reuters) -
    Portugal captain Luis Figo
    said on Wednesday he was
    taking an indefinite break
    from international football,
    but would not confirm whether
    his decision was final." + ], + [ + "AP - Shaun Rogers is in the
    backfield as often as some
    running backs. Whether teams
    dare to block Detroit's star
    defensive tackle with one
    player or follow the trend of
    double-teaming him, he often
    rips through offensive lines
    with a rare combination of
    size, speed, strength and
    nimble footwork." + ], + [ + "The Lions and Eagles entered
    Sunday #39;s game at Ford
    Field in the same place --
    atop their respective
    divisions -- and with
    identical 2-0 records." + ], + [ + "After Marcos Moreno threw four
    more interceptions in last
    week's 14-13 overtime loss at
    N.C. A T, Bison Coach Ray
    Petty will start Antoine
    Hartfield against Norfolk
    State on Saturday." + ], + [ + "SOUTH WILLIAMSPORT, Pa. --
    Looking ahead to the US
    championship game almost cost
    Conejo Valley in the
    semifinals of the Little
    League World Series." + ], + [ + "The Cubs didn #39;t need to
    fly anywhere near Florida to
    be in the eye of the storm.
    For a team that is going on
    100 years since last winning a
    championship, the only thing
    they never are at a loss for
    is controversy." + ], + [ + "KETTERING, Ohio Oct. 12, 2004
    - Cincinnati Bengals defensive
    end Justin Smith pleaded not
    guilty to a driving under the
    influence charge." + ], + [ + "MINNEAPOLIS -- For much of the
    2004 season, Twins pitcher
    Johan Santana didn #39;t just
    beat opposing hitters. Often,
    he overwhelmed and owned them
    in impressive fashion." + ], + [ + "At a charity auction in New
    Jersey last weekend, baseball
    memorabilia dealer Warren
    Heller was approached by a man
    with an unusual but topical
    request." + ], + [ + "INTER Milan coach Roberto
    Mancini believes the club
    #39;s lavish (northern) summer
    signings will enable them to
    mount a serious Serie A
    challenge this season." + ], + [ + "AP - Brad Ott shot an 8-under
    64 on Sunday to win the
    Nationwide Tour's Price Cutter
    Charity Championship for his
    first Nationwide victory." + ], + [ + "AP - New York Jets running
    back Curtis Martin passed Eric
    Dickerson and Jerome Bettis on
    the NFL career rushing list
    Sunday against the St. Louis
    Rams, moving to fourth all-
    time." + ], + [ + "Instead of standing for ante
    meridian and post meridian,
    though, fans will remember the
    time periods of pre-Mia and
    after-Mia. After playing for
    18 years and shattering nearly
    every record" + ], + [ + "Stephon Marbury, concerned
    about his lousy shooting in
    Athens, used an off day to go
    to the gym and work on his
    shot. By finding his range, he
    saved the United States #39;
    hopes for a basketball gold
    medal." + ], + [ + "AP - Randy Moss is expected to
    play a meaningful role for the
    Minnesota Vikings this weekend
    against the Giants, even
    without a fully healed right
    hamstring." + ], + [ + "Pros: Fits the recent profile
    (44, past PGA champion, fiery
    Ryder Cup player); the job is
    his if he wants it. Cons:
    Might be too young to be
    willing to burn two years of
    play on tour." + ], + [ + "Elmer Santos scored in the
    second half, lifting East
    Boston to a 1-0 win over
    Brighton yesterday afternoon
    and giving the Jets an early
    leg up in what is shaping up
    to be a tight Boston City
    League race." + ], + [ + "Saints special teams captain
    Steve Gleason expects to be
    fined by the league after
    being ejected from Sunday's
    game against the Carolina
    Panthers for throwing a punch." + ], + [ + "Play has begun in the
    Australian Masters at
    Huntingdale in Melbourne with
    around half the field of 120
    players completing their first
    rounds." + ], + [ + "LOUISVILLE, Ky. - Louisville
    men #39;s basketball head
    coach Rick Pitino and senior
    forward Ellis Myles met with
    members of the media on Friday
    to preview the Cardinals #39;
    home game against rival
    Kentucky on Satursday." + ], + [ + "AP - Sounds like David
    Letterman is as big a \"Pops\"
    fan as most everyone else." + ], + [ + "Arsenal was held to a 1-1 tie
    by struggling West Bromwich
    Albion on Saturday, failing to
    pick up a Premier League
    victory when Rob Earnshaw
    scored with 11 minutes left." + ], + [ + "Baseball #39;s executive vice
    president Sandy Alderson
    insisted last month that the
    Cubs, disciplined for an
    assortment of run-ins with
    umpires, would not be targeted
    the rest of the season by
    umpires who might hold a
    grudge." + ], + [ + "As Superman and Batman would
    no doubt reflect during their
    cigarette breaks, the really
    draining thing about being a
    hero was that you have to keep
    riding to the rescue." + ], + [ + "AP - Eric Hinske and Vernon
    Wells homered, and the Toronto
    Blue Jays completed a three-
    game sweep of the Baltimore
    Orioles with an 8-5 victory
    Sunday." + ], + [ + " NEW YORK (Reuters) - Todd
    Walker homered, had three hits
    and drove in four runs to lead
    the Chicago Cubs to a 12-5 win
    over the Cincinnati Reds in
    National League play at
    Wrigley Field on Monday." + ], + [ + "MIAMI (Sports Network) -
    Shaquille O #39;Neal made his
    home debut, but once again it
    was Dwyane Wade stealing the
    show with 28 points as the
    Miami Heat downed the
    Cleveland Cavaliers, 92-86, in
    front of a record crowd at
    AmericanAirlines Arena." + ], + [ + "AP - The San Diego Chargers
    looked sharp #151; and played
    the same way. Wearing their
    powder-blue throwback jerseys
    and white helmets from the
    1960s, the Chargers did almost
    everything right in beating
    the Jacksonville Jaguars 34-21
    on Sunday." + ], + [ + "NERVES - no problem. That
    #39;s the verdict of Jose
    Mourinho today after his
    Chelsea side gave a resolute
    display of character at
    Highbury." + ], + [ + "AP - The latest low point in
    Ron Zook's tenure at Florida
    even has the coach wondering
    what went wrong. Meanwhile,
    Sylvester Croom's first big
    win at Mississippi State has
    given the Bulldogs and their
    fans a reason to believe in
    their first-year leader.
    Jerious Norwood's 37-yard
    touchdown run with 32 seconds
    remaining lifted the Bulldogs
    to a 38-31 upset of the 20th-
    ranked Gators on Saturday." + ], + [ + "Someone forgot to inform the
    US Olympic basketball team
    that it was sent to Athens to
    try to win a gold medal, not
    to embarrass its country." + ], + [ + "Ten-man Paris St Germain
    clinched their seventh
    consecutive victory over arch-
    rivals Olympique Marseille
    with a 2-1 triumph in Ligue 1
    on Sunday thanks to a second-
    half winner by substitute
    Edouard Cisse." + ], + [ + "Roy Oswalt wasn #39;t
    surprised to hear the Astros
    were flying Sunday night
    through the remnants of a
    tropical depression that
    dumped several inches of rain
    in Louisiana and could bring
    showers today in Atlanta." + ], + [ + "AP - This hardly seemed
    possible when Pitt needed
    frantic rallies to overcome
    Division I-AA Furman or Big
    East cellar dweller Temple. Or
    when the Panthers could barely
    move the ball against Ohio
    #151; not Ohio State, but Ohio
    U." + ], + [ + "Everyone is moaning about the
    fallout from last weekend but
    they keep on reporting the
    aftermath. #39;The fall-out
    from the so-called
    quot;Battle of Old Trafford
    quot; continues to settle over
    the nation and the debate" + ], + [ + "American Lindsay Davenport
    regained the No. 1 ranking in
    the world for the first time
    since early 2002 by defeating
    Dinara Safina of Russia 6-4,
    6-2 in the second round of the
    Kremlin Cup on Thursday." + ], + [ + "Freshman Jeremy Ito kicked
    four field goals and Ryan
    Neill scored on a 31-yard
    interception return to lead
    improving Rutgers to a 19-14
    victory on Saturday over
    visiting Michigan State." + ], + [ + "Third-seeded Guillermo Canas
    defeated Guillermo Garcia-
    Lopez of Spain 7-6 (1), 6-3
    Monday on the first day of the
    Shanghai Open on Monday." + ], + [ + "But to play as feebly as it
    did for about 35 minutes last
    night in Game 1 of the WNBA
    Finals and lose by only four
    points -- on the road, no less
    -- has to be the best
    confidence builder since Cindy
    St." + ], + [ + "With yesterday #39;s report on
    its athletic department
    violations completed, the
    University of Washington says
    it is pleased to be able to
    move forward." + ], + [ + "Sammy Sosa was fined \\$87,400
    -- one day's salary -- for
    arriving late to the Cubs'
    regular-season finale at
    Wrigley Field and leaving the
    game early. The slugger's
    agent, Adam Katz , said
    yesterday Sosa most likely
    will file a grievance. Sosa
    arrived 70 minutes before
    Sunday's first pitch, and he
    apparently left 15 minutes
    after the game started without
    ..." + ], + [ + "Is the Oklahoma defense a
    notch below its predecessors?
    Is Texas #39; offense a step-
    ahead? Why is Texas Tech
    feeling good about itself
    despite its recent loss?" + ], + [ + "New York gets 57 combined
    points from its starting
    backcourt of Jamal Crawford
    and Stephon Marbury and tops
    Denver, 107-96." + ], + [ + "AP - Shawn Marion had a
    season-high 33 points and 15
    rebounds, leading the Phoenix
    Suns on a fourth-quarter
    comeback despite the absence
    of Steve Nash in a 95-86 win
    over the New Orleans Hornets
    on Friday night." + ], + [ + "Messina upset defending
    champion AC Milan 2-1
    Wednesday, while Juventus won
    its third straight game to
    stay alone atop the Italian
    league standings." + ], + [ + "AP - Padraig Harrington
    rallied to a three-stroke
    victory in the German Masters
    on a windy Sunday, closing
    with a 2-under-par 70 and
    giving his game a big boost
    before the Ryder Cup." + ], + [ + "Ironically it was the first
    regular season game for the
    Carolina Panthers that not
    only began the history of the
    franchise, but also saw the
    beginning of a rivalry that
    goes on to this day." + ], + [ + "Baltimore Ravens running back
    Jamal Lewis did not appear at
    his arraignment Friday, but
    his lawyers entered a not
    guilty plea on charges in an
    expanded drug conspiracy
    indictment." + ], + [ + "After serving a five-game
    suspension, Milton Bradley
    worked out with the Dodgers as
    they prepared for Tuesday's
    opener against the St. Louis
    Cardinals." + ], + [ + "AP - Prime Time won't be
    playing in prime time this
    time. Deion Sanders was on the
    inactive list and missed a
    chance to strut his stuff on
    \"Monday Night Football.\"" + ], + [ + "The man who delivered the
    knockout punch was picked up
    from the Seattle scrap heap
    just after the All-Star Game.
    Before that, John Olerud
    certainly hadn't figured on
    facing Pedro Martinez in
    Yankee Stadium in October." + ], + [ + "LONDON : A consortium,
    including former world
    champion Nigel Mansell, claims
    it has agreed terms to ensure
    Silverstone remains one of the
    venues for the 2005 Formula
    One world championship." + ], + [ + " BATON ROUGE, La. (Sports
    Network) - LSU has named Les
    Miles its new head football
    coach, replacing Nick Saban." + ], + [ + "AP - The Chicago Blackhawks
    re-signed goaltender Michael
    Leighton to a one-year
    contract Wednesday." + ], + [ + "ATHENS -- She won her first
    Olympic gold medal in kayaking
    when she was 18, the youngest
    paddler to do so in Games
    history. Yesterday, at 42,
    Germany #39;s golden girl
    Birgit Fischer won her eighth
    Olympic gold in the four-woman
    500-metre kayak race." + ], + [ + "England boss Sven Goran
    Eriksson has defended
    goalkeeper David James after
    last night #39;s 2-2 draw in
    Austria. James allowed Andreas
    Ivanschitz #39;s shot to slip
    through his fingers to
    complete Austria comeback from
    two goals down." + ], + [ + "The quot;future quot; is
    getting a chance to revive the
    presently struggling New York
    Giants. Two other teams also
    decided it was time for a
    change at quarterback, but the
    Buffalo Bills are not one of
    them." + ], + [ + "Flushing Meadows, NY (Sports
    Network) - The men #39;s
    semifinals at the 2004 US Open
    will be staged on Saturday,
    with three of the tournament
    #39;s top-five seeds ready for
    action at the USTA National
    Tennis Center." + ], + [ + "Given nearly a week to examine
    the security issues raised by
    the now-infamous brawl between
    players and fans in Auburn
    Hills, Mich., Nov. 19, the
    Celtics returned to the
    FleetCenter last night with
    two losses and few concerns
    about their on-court safety." + ], + [ + "Arsenals Thierry Henry today
    missed out on the European
    Footballer of the Year award
    as Andriy Shevchenko took the
    honour. AC Milan frontman
    Shevchenko held off
    competition from Barcelona
    pair Deco and" + ], + [ + "Neil Mellor #39;s sensational
    late winner for Liverpool
    against Arsenal on Sunday has
    earned the back-up striker the
    chance to salvage a career
    that had appeared to be
    drifting irrevocably towards
    the lower divisions." + ], + [ + "ABOUT 70,000 people were
    forced to evacuate Real Madrid
    #39;s Santiago Bernabeu
    stadium minutes before the end
    of a Primera Liga match
    yesterday after a bomb threat
    in the name of ETA Basque
    separatist guerillas." + ], + [ + "The team learned on Monday
    that full-back Jon Ritchie
    will miss the rest of the
    season with a torn anterior
    cruciate ligament in his left
    knee." + ], + [ + "Venus Williams barely kept
    alive her hopes of qualifying
    for next week #39;s WTA Tour
    Championships. Williams,
    seeded fifth, survived a
    third-set tiebreaker to
    outlast Yuilana Fedak of the
    Ukraine, 6-4 2-6 7-6" + ], + [ + " SYDNEY (Reuters) - Arnold
    Palmer has taken a swing at
    America's top players,
    criticizing their increasing
    reluctance to travel abroad
    to play in tournaments." + ], + [ + "Although he was well-beaten by
    Retief Goosen in Sunday #39;s
    final round of The Tour
    Championship in Atlanta, there
    has been some compensation for
    the former world number one,
    Tiger Woods." + ], + [ + "WAYNE Rooney and Henrik
    Larsson are among the players
    nominated for FIFAs
    prestigious World Player of
    the Year award. Rooney is one
    of four Manchester United
    players on a list which is
    heavily influenced by the
    Premiership." + ], + [ + "It didn #39;t look good when
    it happened on the field, and
    it looked worse in the
    clubhouse. Angels second
    baseman Adam Kennedy left the
    Angels #39; 5-2 win over the
    Seattle Mariners" + ], + [ + "Freshman Darius Walker ran for
    115 yards and scored two
    touchdowns, helping revive an
    Irish offense that had managed
    just one touchdown in the
    season's first six quarters." + ], + [ + "AP - NBC is adding a 5-second
    delay to its NASCAR telecasts
    after Dale Earnhardt Jr. used
    a vulgarity during a postrace
    TV interview last weekend." + ], + [ + " BOSTON (Sports Network) - The
    New York Yankees will start
    Orlando \"El Duque\" Hernandez
    in Game 4 of the American
    League Championship Series on
    Saturday against the Boston
    Red Sox." + ], + [ + "The future of Sven-Goran
    Eriksson as England coach is
    the subject of intense
    discussion after the draw in
    Austria. Has the Swede lost
    the confidence of the nation
    or does he remain the best man
    for the job?" + ], + [ + "Spain begin their third final
    in five seasons at the Olympic
    stadium hoping to secure their
    second title since their first
    in Barcelona against Australia
    in 2000." + ], + [ + "Second-seeded David Nalbandian
    of Argentina lost at the Japan
    Open on Thursday, beaten by
    Gilles Muller of Luxembourg
    7-6 (4), 3-6, 6-4 in the third
    round." + ], + [ + "Thursday #39;s unexpected
    resignation of Memphis
    Grizzlies coach Hubie Brown
    left a lot of questions
    unanswered. In his unique way
    of putting things, the
    71-year-old Brown seemed to
    indicate he was burned out and
    had some health concerns." + ], + [ + "RUDI Voller had quit as coach
    of Roma after a 3-1 defeat
    away to Bologna, the Serie A
    club said today. Under the
    former Germany coach, Roma had
    taken just four league points
    from a possible 12." + ], + [ + "ONE by one, the players #39;
    faces had flashed up on the
    giant Ibrox screens offering
    season #39;s greetings to the
    Rangers fans. But the main
    presents were reserved for
    Auxerre." + ], + [ + "South Korea #39;s Grace Park
    shot a seven-under-par 65 to
    triumph at the CJ Nine Bridges
    Classic on Sunday. Park #39;s
    victory made up her final-
    round collapse at the Samsung
    World Championship two weeks
    ago." + ], + [ + "Titans QB Steve McNair was
    released from a Nashville
    hospital after a two-night
    stay for treatment of a
    bruised sternum. McNair was
    injured during the fourth
    quarter of the Titans #39;
    15-12 loss to Jacksonville on
    Sunday." + ], + [ + "Keith Miller, Australia #39;s
    most prolific all-rounder in
    Test cricket, died today at a
    nursing home, Cricket
    Australia said. He was 84." + ], + [ + "TORONTO Former Toronto pitcher
    John Cerutti (seh-ROO
    #39;-tee) was found dead in
    his hotel room today,
    according to the team. He was
    44." + ], + [ + "Defense: Ken Lucas. His
    biggest play was his first
    one. The fourth-year
    cornerback intercepted a Ken
    Dorsey pass that kissed off
    the hands of wide receiver
    Rashaun Woods and returned it
    25 yards to set up the
    Seahawks #39; first score." + ], + [ + "The powerful St. Louis trio of
    Albert Pujols, Scott Rolen and
    Jim Edmonds is 4 for 23 with
    one RBI in the series and with
    runners on base, they are 1
    for 13." + ], + [ + "The Jets signed 33-year-old
    cornerback Terrell Buckley,
    who was released by New
    England on Sunday, after
    putting nickel back Ray
    Mickens on season-ending
    injured reserve yesterday with
    a torn ACL in his left knee." + ], + [ + "WEST INDIES thrilling victory
    yesterday in the International
    Cricket Council Champions
    Trophy meant the world to the
    five million people of the
    Caribbean." + ], + [ + "Scotland manager Berti Vogts
    insists he is expecting
    nothing but victory against
    Moldova on Wednesday. The game
    certainly is a must-win affair
    if the Scots are to have any
    chance of qualifying for the
    2006 World Cup finals." + ], + [ + " INDIANAPOLIS (Reuters) -
    Jenny Thompson will take the
    spotlight from injured U.S.
    team mate Michael Phelps at
    the world short course
    championships Saturday as she
    brings down the curtain on a
    spectacular swimming career." + ], + [ + "Martin Brodeur made 27 saves,
    and Brad Richards, Kris Draper
    and Joe Sakic scored goals to
    help Canada beat Russia 3-1
    last night in the World Cup of
    Hockey, giving the Canadians a
    3-0 record in round-robin
    play." + ], + [ + "ST. LOUIS -- Mike Martz #39;s
    week of anger was no empty
    display. He saw the defending
    NFC West champions slipping
    and thought taking potshots at
    his players might be his best
    shot at turning things around." + ], + [ + "Romanian soccer star Adrian
    Mutu as he arrives at the
    British Football Association
    in London, ahead of his
    disciplinary hearing, Thursday
    Nov. 4, 2004." + ], + [ + "Australia completed an
    emphatic Test series sweep
    over New Zealand with a
    213-run win Tuesday, prompting
    a caution from Black Caps
    skipper Stephen Fleming for
    other cricket captains around
    the globe." + ], + [ + "Liverpool, England (Sports
    Network) - Former English
    international and Liverpool
    great Emlyn Hughes passed away
    Tuesday from a brain tumor." + ], + [ + "JACKSONVILLE, Fla. -- They
    were singing in the Colts #39;
    locker room today, singing
    like a bunch of wounded
    songbirds. Never mind that
    Marcus Pollard, Dallas Clark
    and Ben Hartsock won #39;t be
    recording a remake of Kenny
    Chesney #39;s song,
    quot;Young, quot; any time
    soon." + ], + [ + "As the close-knit NASCAR
    community mourns the loss of
    team owner Rick Hendrick #39;s
    son, brother, twin nieces and
    six others in a plane crash
    Sunday, perhaps no one outside
    of the immediate family
    grieves more deeply than
    Darrell Waltrip." + ], + [ + "AP - Purdue quarterback Kyle
    Orton has no trouble
    remembering how he felt after
    last year's game at Michigan." + ], + [ + "Brown is a second year player
    from Memphis and has spent the
    2004 season on the Steelers
    #39; practice squad. He played
    in two games last year." + ], + [ + "Barret Jackman, the last of
    the Blues left to sign before
    the league #39;s probable
    lockout on Wednesday,
    finalized a deal Monday that
    is rare in the current
    economic climate but fitting
    for him." + ], + [ + "AP - In the tumult of the
    visitors' clubhouse at Yankee
    Stadium, champagne pouring all
    around him, Theo Epstein held
    a beer. \"I came in and there
    was no champagne left,\" he
    said this week. \"I said, 'I'll
    have champagne if we win it
    all.'\" Get ready to pour a
    glass of bubbly for Epstein.
    No I.D. necessary." + ], + [ + "Barcelona held on from an
    early Deco goal to edge game
    local rivals Espanyol 1-0 and
    carve out a five point
    tabletop cushion. Earlier,
    Ronaldo rescued a point for
    Real Madrid, who continued
    their middling form with a 1-1
    draw at Real Betis." + ], + [ + "MONTREAL (CP) - The Expos may
    be history, but their demise
    has heated up the market for
    team memorabilia. Vintage
    1970s and 1980s shirts are
    already sold out, but
    everything from caps, beer
    glasses and key-chains to
    dolls of mascot Youppi!" + ], + [ + "A 76th minute goal from
    European Footballer of the
    Year Pavel Nedved gave
    Juventus a 1-0 win over Bayern
    Munich on Tuesday handing the
    Italians clear control at the
    top of Champions League Group
    C." + ], + [ + "ANN ARBOR, Mich. -- Some NHL
    players who took part in a
    charity hockey game at the
    University of Michigan on
    Thursday were hopeful the news
    that the NHL and the players
    association will resume talks
    next week" + ], + [ + "BOULDER, Colo. -- Vernand
    Morency ran for 165 yards and
    two touchdowns and Donovan
    Woods threw for three more
    scores, lifting No. 22
    Oklahoma State to a 42-14
    victory over Colorado
    yesterday." + ], + [ + "PULLMAN - Last week, in
    studying USC game film, Cougar
    coaches thought they found a
    chink in the national
    champions armor. And not just
    any chink - one with the
    potential, right from the get
    go, to" + ], + [ + "AP - Matt Leinart was quite a
    baseball prospect growing up,
    showing so much promise as a
    left-handed pitcher that
    scouts took notice before high
    school." + ], + [ + "LSU will stick with a two-
    quarterback rotation Saturday
    at Auburn, according to Tigers
    coach Nick Saban, who seemed
    to have some fun telling the
    media what he will and won
    #39;t discuss Monday." + ], + [ + "Seattle -- - Not so long ago,
    the 49ers were inflicting on
    other teams the kind of pain
    and embarrassment they felt in
    their 34-0 loss to the
    Seahawks on Sunday." + ], + [ + "London - Manchester City held
    fierce crosstown rivals
    Manchester United to a 0-0
    draw on Sunday, keeping the
    Red Devils eleven points
    behind leaders Chelsea." + ], + [ + "New Ole Miss head coach Ed
    Orgeron, speaking for the
    first time since his hiring,
    made clear the goal of his
    football program. quot;The
    goal of this program will be
    to go to the Sugar Bowl, quot;
    Orgeron said." + ], + [ + "It is much too easy to call
    Pedro Martinez the selfish
    one, to say he is walking out
    on the Red Sox, his baseball
    family, for the extra year of
    the Mets #39; crazy money." + ], + [ + "It is impossible for young
    tennis players today to know
    what it was like to be Althea
    Gibson and not to be able to
    quot;walk in the front door,
    quot; Garrison said." + ], + [ + "Here #39;s an obvious word of
    advice to Florida athletic
    director Jeremy Foley as he
    kicks off another search for
    the Gators football coach: Get
    Steve Spurrier on board." + ], + [ + "Amid the stormy gloom in
    Gotham, the rain-idled Yankees
    last night had plenty of time
    to gather in front of their
    televisions and watch the Red
    Sox Express roar toward them.
    The national telecast might
    have been enough to send a
    jittery Boss Steinbrenner
    searching his Bartlett's
    Familiar Quotations for some
    quot;Little Engine That Could
    quot; metaphor." + ], + [ + "FULHAM fans would have been
    singing the late Elvis #39;
    hit #39;The wonder of you
    #39; to their player Elvis
    Hammond. If not for Frank
    Lampard spoiling the party,
    with his dedication to his
    late grandfather." + ], + [ + "A bird #39;s eye view of the
    circuit at Shanghai shows what
    an event Sunday #39;s Chinese
    Grand Prix will be. The course
    is arguably one of the best
    there is, and so it should be
    considering the amount of
    money that has been spent on
    it." + ], + [ + "AP - Sirius Satellite Radio
    signed a deal to air the men's
    NCAA basketball tournament
    through 2007, the latest move
    made in an attempt to draw
    customers through sports
    programming." + ], + [ + "(Sports Network) - The
    inconsistent San Diego Padres
    will try for consecutive wins
    for the first time since
    August 28-29 tonight, when
    they begin a huge four-game
    set against the Los Angeles
    Dodgers at Dodger Stadium." + ], + [ + " NEW YORK (Reuters) - Top seed
    Roger Federer survived a
    stirring comeback from twice
    champion Andre Agassi to reach
    the semifinals of the U.S.
    Open for the first time on
    Thursday, squeezing through
    6-3, 2-6, 7-5, 3-6, 6-3." + ], + [ + "SAN FRANCISCO - What Babe Ruth
    was to the first half of the
    20th century and Hank Aaron
    was to the second, Barry Bonds
    has become for the home run
    generation." + ], + [ + "Birgit Fischer settled for
    silver, leaving the 42-year-
    old Olympian with two medals
    in two days against decidedly
    younger competition." + ], + [ + "There was no mystery ... no
    secret strategy ... no baited
    trap that snapped shut and
    changed the course of history
    #39;s most lucrative non-
    heavyweight fight." + ], + [ + "Notre Dame accepted an
    invitation Sunday to play in
    the Insight Bowl in Phoenix
    against a Pac-10 team on Dec.
    28. The Irish (6-5) accepted
    the bid a day after losing to
    Southern California" + ], + [ + "Greg Anderson has so dominated
    Pro Stock this season that his
    championship quest has evolved
    into a pursuit of NHRA
    history. By Bob Hesser, Racers
    Edge Photography." + ], + [ + "Michael Powell, chairman of
    the FCC, said Wednesday he was
    disappointed with ABC for
    airing a sexually suggestive
    opening to \"Monday Night
    Football.\"" + ], + [ + "Former Washington football
    coach Rick Neuheisel looked
    forward to a return to
    coaching Wednesday after being
    cleared by the NCAA of
    wrongdoing related to his
    gambling on basketball games." + ], + [ + "AP - Rutgers basketball player
    Shalicia Hurns was suspended
    from the team after pleading
    guilty to punching and tying
    up her roommate during a
    dispute over painkilling
    drugs." + ], + [ + "This was the event Michael
    Phelps didn't really need to
    compete in if his goal was to
    win eight golds. He probably
    would have had a better chance
    somewhere else." + ], + [ + "Gabe Kapler became the first
    player to leave the World
    Series champion Boston Red
    Sox, agreeing to a one-year
    contract with the Yomiuri
    Giants in Tokyo." + ], + [ + "BALI, Indonesia - Svetlana
    Kuznetsova, fresh off her
    championship at the US Open,
    defeated Australian qualifier
    Samantha Stosur 6-4, 6-4
    Thursday to reach the
    quarterfinals of the Wismilak
    International." + ], + [ + "AP - Just like the old days in
    Dallas, Emmitt Smith made life
    miserable for the New York
    Giants on Sunday." + ], + [ + "TAMPA, Fla. - Chris Simms
    first NFL start lasted 19
    plays, and it might be a while
    before he plays again for the
    Tampa Bay Buccaneers." + ], + [ + "Jarno Trulli made the most of
    the conditions in qualifying
    to claim pole ahead of Michael
    Schumacher, while Fernando
    finished third." + ], + [ + "AP - Authorities are
    investigating whether bettors
    at New York's top thoroughbred
    tracks were properly informed
    when jockeys came in
    overweight at races, a source
    familiar with the probe told
    The Associated Press." + ], + [ + "Michael Owen scored his first
    goal for Real Madrid in a 1-0
    home victory over Dynamo Kiev
    in the Champions League. The
    England striker toe-poked home
    Ronaldo #39;s cross in the
    35th minute to join the
    Russians" + ], + [ + "Jason Giambi has returned to
    the New York Yankees'
    clubhouse but is still
    clueless as to when he will be
    able to play again." + ], + [ + "Seventy-five National Hockey
    League players met with union
    leaders yesterday to get an
    update on a lockout that shows
    no sign of ending." + ], + [ + "Howard, at 6-4 overall and 3-3
    in the Mid-Eastern Athletic
    Conference, can clinch a
    winning record in the MEAC
    with a win over Delaware State
    on Saturday." + ], + [ + " ATHENS (Reuters) - Top-ranked
    Argentina booked their berth
    in the women's hockey semi-
    finals at the Athens Olympics
    on Friday but defending
    champions Australia now face
    an obstacle course to qualify
    for the medal matches." + ], + [ + "The Notre Dame message boards
    are no longer discussing
    whether Tyrone Willingham
    should be fired. Theyre
    already arguing about whether
    the next coach should be Barry
    Alvarez or Steve Spurrier." + ], + [ + " THOMASTOWN, Ireland (Reuters)
    - World number three Ernie
    Els overcame difficult weather
    conditions to fire a sparkling
    eight-under-par 64 and move
    two shots clear after two
    rounds of the WGC-American
    Express Championship Friday." + ], + [ + "Lamar Odom supplemented 20
    points with 13 rebounds and
    Kobe Bryant added 19 points to
    overcome a big night from Yao
    Ming as the Los Angeles Lakers
    ground out an 84-79 win over
    the Rockets in Houston
    Saturday." + ], + [ + "It rained Sunday, of course,
    and but another soppy, sloppy
    gray day at Westside Tennis
    Club did nothing to deter
    Roger Federer from his
    appointed rounds." + ], + [ + "Amelie Mauresmo was handed a
    place in the Advanta
    Championships final after
    Maria Sharapova withdrew from
    their semi-final because of
    injury." + ], + [ + " EAST RUTHERFORD, New Jersey
    (Sports Network) - Retired NBA
    center and seven-time All-
    Star Alonzo Mourning is going
    to give his playing career
    one more shot." + ], + [ + "Jordan have confirmed that
    Timo Glock will replace
    Giorgio Pantano for this
    weekend #39;s Chinese GP as
    the team has terminated its
    contract with Pantano." + ], + [ + "The continuing heartache of
    Wake Forest #39;s ACC football
    season was best described by
    fifth-ranked Florida State
    coach Bobby Bowden, after his
    Seminoles had edged the
    Deacons 20-17 Saturday at
    Groves Stadium." + ], + [ + "THE glory days have returned
    to White Hart Lane. When Spurs
    new first-team coach Martin
    Jol promised a return to the
    traditions of the 1960s,
    nobody could have believed he
    was so determined to act so
    quickly and so literally." + ], + [ + "AP - When Paula Radcliffe
    dropped out of the Olympic
    marathon miles from the
    finish, she sobbed
    uncontrollably. Margaret Okayo
    knew the feeling. Okayo pulled
    out of the marathon at the
    15th mile with a left leg
    injury, and she cried, too.
    When she watched Radcliffe
    quit, Okayo thought, \"Let's
    cry together.\"" + ], + [ + " ATLANTA (Sports Network) -
    The Atlanta Hawks signed free
    agent Kevin Willis on
    Wednesday, nearly a decade
    after the veteran big man
    ended an 11- year stint with
    the team." + ], + [ + "When his right-front tire went
    flying off early in the Ford
    400, the final race of the
    NASCAR Nextel Cup Series
    season, Kurt Busch, it seemed,
    was destined to spend his
    offseason" + ], + [ + "Brandon Backe and Woody
    Williams pitched well last
    night even though neither
    earned a win. But their
    outings will show up in the
    record books." + ], + [ + "The Football Association today
    decided not to charge David
    Beckham with bringing the game
    into disrepute. The FA made
    the surprise announcement
    after their compliance unit
    ruled" + ], + [ + " CRANS-SUR-SIERRE, Switzerland
    (Reuters) - World number
    three Ernie Els says he feels
    a failure after narrowly
    missing out on three of the
    year's four major
    championships." + ], + [ + "Striker Bonaventure Kalou
    netted twice to send AJ
    Auxerre through to the first
    knockout round of the UEFA Cup
    at the expense of Rangers on
    Wednesday." + ], + [ + "THIS weekend sees the
    quot;other quot; showdown
    between New York and New
    England as the Jets and
    Patriots clash in a battle of
    the unbeaten teams." + ], + [ + "Nicolas Anelka is fit for
    Manchester City #39;s
    Premiership encounter against
    Tottenham at Eastlands, but
    the 13million striker will
    have to be content with a
    place on the bench." + ], + [ + " PITTSBURGH (Reuters) - Ben
    Roethlisberger passed for 183
    yards and two touchdowns,
    Hines Ward scored twice and
    the Pittsburgh Steelers
    rolled to a convincing 27-3
    victory over Philadelphia on
    Sunday for their second
    straight win against an
    undefeated opponent." + ], + [ + " ATHENS (Reuters) - The U.S.
    men's basketball team got
    their first comfortable win
    at the Olympic basketball
    tournament Monday, routing
    winless Angola 89-53 in their
    final preliminary round game." + ], + [ + "Often, the older a pitcher
    becomes, the less effective he
    is on the mound. Roger Clemens
    apparently didn #39;t get that
    memo. On Tuesday, the 42-year-
    old Clemens won an
    unprecedented" + ], + [ + "PSV Eindhoven faces Arsenal at
    Highbury tomorrow night on the
    back of a free-scoring start
    to the season. Despite losing
    Mateja Kezman to Chelsea in
    the summer, the Dutch side has
    scored 12 goals in the first" + ], + [ + "PLAYER OF THE GAME: Playing
    with a broken nose, Seattle
    point guard Sue Bird set a
    WNBA playoff record for
    assists with 14, also pumping
    in 10 points as the Storm
    claimed the Western Conference
    title last night." + ], + [ + "Frankfurt - World Cup winners
    Brazil were on Monday drawn to
    meet European champions
    Greece, Gold Cup winners
    Mexico and Asian champions
    Japan at the 2005
    Confederations Cup." + ], + [ + "Third baseman Vinny Castilla
    said he fits fine with the
    Colorado youth movement, even
    though he #39;ll turn 38 next
    season and the Rockies are
    coming off the second-worst" + ], + [ + "(Sports Network) - Two of the
    top teams in the American
    League tangle in a possible
    American League Division
    Series preview tonight, as the
    West-leading Oakland Athletics
    host the wild card-leading
    Boston Red Sox for the first
    of a three-game set at the" + ], + [ + "Inverness Caledonian Thistle
    appointed Craig Brewster as
    its new manager-player
    Thursday although he #39;s
    unable to play for the team
    until January." + ], + [ + "ATHENS, Greece -- Alan Shearer
    converted an 87th-minute
    penalty to give Newcastle a
    1-0 win over Panionios in
    their UEFA Cup Group D match." + ], + [ + "Derek Jeter turned a season
    that started with a terrible
    slump into one of the best in
    his accomplished 10-year
    career. quot;I don #39;t
    think there is any question,
    quot; the New York Yankees
    manager said." + ], + [ + "China's Guo Jingjing easily
    won the women's 3-meter
    springboard last night, and Wu
    Minxia made it a 1-2 finish
    for the world's diving
    superpower, taking the silver." + ], + [ + "GREEN BAY, Wisconsin (Ticker)
    -- Brett Favre will be hoping
    his 200th consecutive start
    turns out better than his last
    two have against the St." + ], + [ + "When an NFL team opens with a
    prolonged winning streak,
    former Miami Dolphins coach
    Don Shula and his players from
    the 17-0 team of 1972 root
    unabashedly for the next
    opponent." + ], + [ + "MIANNE Bagger, the transsexual
    golfer who prompted a change
    in the rules to allow her to
    compete on the professional
    circuit, made history
    yesterday by qualifying to
    play full-time on the Ladies
    European Tour." + ], + [ + "Great Britain #39;s gold medal
    tally now stands at five after
    Leslie Law was handed the
    individual three day eventing
    title - in a courtroom." + ], + [ + "Mayor Tom Menino must be
    proud. His Boston Red Sox just
    won their first World Series
    in 86 years and his Hyde Park
    Blue Stars yesterday clinched
    their first Super Bowl berth
    in 32 years, defeating
    O'Bryant, 14-0. Who would have
    thought?" + ], + [ + "South Korea have appealed to
    sport #39;s supreme legal body
    in an attempt to award Yang
    Tae-young the Olympic
    gymnastics all-round gold
    medal after a scoring error
    robbed him of the title in
    Athens." + ], + [ + "AP - Johan Santana had an
    early lead and was well on his
    way to his 10th straight win
    when the rain started to fall." + ], + [ + "ATHENS-In one of the biggest
    shocks in Olympic judo
    history, defending champion
    Kosei Inoue was defeated by
    Dutchman Elco van der Geest in
    the men #39;s 100-kilogram
    category Thursday." + ], + [ + "NBC is adding a 5-second delay
    to its Nascar telecasts after
    Dale Earnhardt Jr. used a
    vulgarity during a postrace
    interview last weekend." + ], + [ + "San Francisco Giants
    outfielder Barry Bonds, who
    became the third player in
    Major League Baseball history
    to hit 700 career home runs,
    won the National League Most
    Valuable Player Award" + ], + [ + "Portsmouth chairman Milan
    Mandaric said on Tuesday that
    Harry Redknapp, who resigned
    as manager last week, was
    innocent of any wrong-doing
    over agent or transfer fees." + ], + [ + "This record is for all the
    little guys, for all the
    players who have to leg out
    every hit instead of taking a
    relaxing trot around the
    bases, for all the batters
    whose muscles aren #39;t" + ], + [ + "Charlie Hodgson #39;s record-
    equalling performance against
    South Africa was praised by
    coach Andy Robinson after the
    Sale flyhalf scored 27 points
    in England #39;s 32-16 victory
    here at Twickenham on
    Saturday." + ], + [ + "BASEBALL Atlanta (NL):
    Optioned P Roman Colon to
    Greenville (Southern);
    recalled OF Dewayne Wise from
    Richmond (IL). Boston (AL):
    Purchased C Sandy Martinez
    from Cleveland (AL) and
    assigned him to Pawtucket
    (IL). Cleveland (AL): Recalled
    OF Ryan Ludwick from Buffalo
    (IL). Chicago (NL): Acquired
    OF Ben Grieve from Milwaukee
    (NL) for player to be named
    and cash; acquired C Mike ..." + ], + [ + "BRONX, New York (Ticker) --
    Kelvim Escobar was the latest
    Anaheim Angels #39; pitcher to
    subdue the New York Yankees.
    Escobar pitched seven strong
    innings and Bengie Molina tied
    a career-high with four hits,
    including" + ], + [ + "Sep 08 - Vijay Singh revelled
    in his status as the new world
    number one after winning the
    Deutsche Bank Championship by
    three shots in Boston on
    Monday." + ], + [ + "Many people in golf are asking
    that today. He certainly wasn
    #39;t A-list and he wasn #39;t
    Larry Nelson either. But you
    couldn #39;t find a more solid
    guy to lead the United States
    into Ireland for the 2006
    Ryder Cup Matches." + ], + [ + "Australian Stuart Appleby, who
    was the joint second-round
    leader, returned a two-over 74
    to drop to third at three-
    under while American Chris
    DiMarco moved into fourth with
    a round of 69." + ], + [ + "With a doubleheader sweep of
    the Minnesota Twins, the New
    York Yankees moved to the
    verge of clinching their
    seventh straight AL East
    title." + ], + [ + "Athens, Greece (Sports
    Network) - The first official
    track event took place this
    morning and Italy #39;s Ivano
    Brugnetti won the men #39;s
    20km walk at the Summer
    Olympics in Athens." + ], + [ + "Champions Arsenal opened a
    five-point lead at the top of
    the Premier League after a 4-0
    thrashing of Charlton Athletic
    at Highbury Saturday." + ], + [ + "The Redskins and Browns have
    traded field goals and are
    tied, 3-3, in the first
    quarter in Cleveland." + ], + [ + "Forget September call-ups. The
    Red Sox may tap their minor
    league system for an extra
    player or two when the rules
    allow them to expand their
    25-man roster Wednesday, but
    any help from the farm is
    likely to pale against the
    abundance of talent they gain
    from the return of numerous
    players, including Trot Nixon
    , from the disabled list." + ], + [ + "THOUSAND OAKS -- Anonymity is
    only a problem if you want it
    to be, and it is obvious Vijay
    Singh doesn #39;t want it to
    be. Let others chase fame." + ], + [ + "David Coulthard #39;s season-
    long search for a Formula One
    drive next year is almost
    over. Negotiations between Red
    Bull Racing and Coulthard, who
    tested for the Austrian team
    for the first time" + ], + [ + "AP - Southern California
    tailback LenDale White
    remembers Justin Holland from
    high school. The Colorado
    State quarterback made quite
    an impression." + ], + [ + "TIM HENMAN last night admitted
    all of his energy has been
    drained away as he bowed out
    of the Madrid Masters. The top
    seed, who had a blood test on
    Wednesday to get to the bottom
    of his fatigue, went down" + ], + [ + "ATHENS -- US sailors needed a
    big day to bring home gold and
    bronze medals from the sailing
    finale here yesterday. But
    rolling the dice on windshifts
    and starting tactics backfired
    both in Star and Tornado
    classes, and the Americans had
    to settle for a single silver
    medal." + ], + [ + "AP - Cavaliers forward Luke
    Jackson was activated
    Wednesday after missing five
    games because of tendinitis in
    his right knee. Cleveland also
    placed forward Sasha Pavlovic
    on the injured list." + ], + [ + "The Brisbane Lions #39;
    football manager stepped out
    of the changerooms just before
    six o #39;clock last night and
    handed one of the milling
    supporters a six-pack of beer." + ], + [ + "Cavaliers owner Gordon Gund is
    in quot;serious quot;
    negotiations to sell the NBA
    franchise, which has enjoyed a
    dramatic financial turnaround
    since the arrival of star
    LeBron James." + ], + [ + "After two days of gloom, China
    was back on the winning rails
    on Thursday with Liu Chunhong
    winning a weightlifting title
    on her record-shattering binge
    and its shuttlers contributing
    two golds in the cliff-hanging
    finals." + ], + [ + "One question that arises
    whenever a player is linked to
    steroids is, \"What would he
    have done without them?\"
    Baseball history whispers an
    answer." + ], + [ + "The second round of the
    Canadian Open golf tournament
    continues Saturday Glenn Abbey
    Golf Club in Oakville,
    Ontario, after play was
    suspended late Friday due to
    darkness." + ], + [ + "A massive plan to attract the
    2012 Summer Olympics to New
    York, touting the city's
    diversity, financial and media
    power, was revealed Wednesday." + ], + [ + "NICK Heidfeld #39;s test with
    Williams has been brought
    forward after BAR blocked
    plans for Anthony Davidson to
    drive its Formula One rival
    #39;s car." + ], + [ + "Grace Park closed with an
    eagle and two birdies for a
    7-under-par 65 and a two-
    stroke lead after three rounds
    of the Wachovia LPGA Classic
    on Saturday." + ], + [ + "Carlos Beltran drives in five
    runs to carry the Astros to a
    12-3 rout of the Braves in
    Game 5 of their first-round NL
    playoff series." + ], + [ + "Since Lennox Lewis #39;s
    retirement, the heavyweight
    division has been knocked for
    having more quantity than
    quality. Eight heavyweights on
    Saturday night #39;s card at
    Madison Square Garden hope to
    change that perception, at
    least for one night." + ], + [ + "Tim Duncan had 17 points and
    10 rebounds, helping the San
    Antonio Spurs to a 99-81
    victory over the New York
    Kicks. This was the Spurs
    fourth straight win this
    season." + ], + [ + "Nagpur: India suffered a
    double blow even before the
    first ball was bowled in the
    crucial third cricket Test
    against Australia on Tuesday
    when captain Sourav Ganguly
    and off spinner Harbhajan
    Singh were ruled out of the
    match." + ], + [ + "AP - Former New York Yankees
    hitting coach Rick Down was
    hired for the same job by the
    Mets on Friday, reuniting him
    with new manager Willie
    Randolph." + ], + [ + "FILDERSTADT (Germany) - Amelie
    Mauresmo and Lindsay Davenport
    took their battle for the No.
    1 ranking and Porsche Grand
    Prix title into the semi-
    finals with straight-sets
    victories on Friday." + ], + [ + "Carter returned, but it was
    running back Curtis Martin and
    the offensive line that put
    the Jets ahead. Martin rushed
    for all but 10 yards of a
    45-yard drive that stalled at
    the Cardinals 10." + ], + [ + "DENVER (Ticker) -- Jake
    Plummer more than made up for
    a lack of a running game.
    Plummer passed for 294 yards
    and two touchdowns as the
    Denver Broncos posted a 23-13
    victory over the San Diego
    Chargers in a battle of AFC
    West Division rivals." + ], + [ + "It took all of about five
    minutes of an introductory
    press conference Wednesday at
    Heritage Hall for USC
    basketball to gain something
    it never really had before." + ], + [ + "AP - The Boston Red Sox looked
    at the out-of-town scoreboard
    and could hardly believe what
    they saw. The New York Yankees
    were trailing big at home
    against the Cleveland Indians
    in what would be the worst
    loss in the 101-year history
    of the storied franchise." + ], + [ + "The Red Sox will either
    complete an amazing comeback
    as the first team to rebound
    from a 3-0 deficit in
    postseason history, or the
    Yankees will stop them." + ], + [ + "The Red Sox have reached
    agreement with free agent
    pitcher Matt Clement yesterday
    on a three-year deal that will
    pay him around \\$25 million,
    his agent confirmed yesterday." + ], + [ + "HEN Manny Ramirez and David
    Ortiz hit consecutive home
    runs Sunday night in Chicago
    to put the Red Sox ahead,
    there was dancing in the
    streets in Boston." + ], + [ + "MIAMI -- Bryan Randall grabbed
    a set of Mardi Gras beads and
    waved them aloft, while his
    teammates exalted in the
    prospect of a trip to New
    Orleans." + ], + [ + "TORONTO (CP) - With an injured
    Vince Carter on the bench, the
    Toronto Raptors dropped their
    sixth straight game Friday,
    101-87 to the Denver Nuggets." + ], + [ + "While not quite a return to
    glory, Monday represents the
    Redskins' return to the
    national consciousness." + ], + [ + "Vijay Singh has won the US PGA
    Tour player of the year award
    for the first time, ending
    Tiger Woods #39;s five-year
    hold on the honour." + ], + [ + "England got strikes from
    sparkling debut starter
    Jermain Defoe and Michael Owen
    to defeat Poland in a Uefa
    World Cup qualifier in
    Chorzow." + ], + [ + "Titleholder Ernie Els moved
    within sight of a record sixth
    World Match Play title on
    Saturday by solving a putting
    problem to overcome injured
    Irishman Padraig Harrington 5
    and 4." + ], + [ + "If the Washington Nationals
    never win a pennant, they have
    no reason to ever doubt that
    DC loves them. Yesterday, the
    District City Council
    tentatively approved a tab for
    a publicly financed ballpark
    that could amount to as much
    as \\$630 million." + ], + [ + "Defensive back Brandon
    Johnson, who had two
    interceptions for Tennessee at
    Mississippi, was suspended
    indefinitely Monday for
    violation of team rules." + ], + [ + "Points leader Kurt Busch spun
    out and ran out of fuel, and
    his misfortune was one of the
    reasons crew chief Jimmy
    Fennig elected not to pit with
    20 laps to go." + ], + [ + "(CP) - The NHL all-star game
    hasn #39;t been cancelled
    after all. It #39;s just been
    moved to Russia. The agent for
    New York Rangers winger
    Jaromir Jagr confirmed Monday
    that the Czech star had joined
    Omsk Avangard" + ], + [ + "HERE in the land of myth, that
    familiar god of sports --
    karma -- threw a bolt of
    lightning into the Olympic
    stadium yesterday. Marion
    Jones lunged desperately with
    her baton in the 4 x 100m
    relay final, but couldn #39;t
    reach her target." + ], + [ + "AP - Kenny Rogers lost at the
    Coliseum for the first time in
    more than 10 years, with Bobby
    Crosby's three-run double in
    the fifth inning leading the
    Athletics to a 5-4 win over
    the Texas Rangers on Thursday." + ], + [ + "Dambulla, Sri Lanka - Kumar
    Sangakkara and Avishka
    Gunawardene slammed impressive
    half-centuries to help an
    under-strength Sri Lanka crush
    South Africa by seven wickets
    in the fourth one-day
    international here on
    Saturday." + ], + [ + "Fresh off being the worst team
    in baseball, the Arizona
    Diamondbacks set a new record
    this week: fastest team to
    both hire and fire a manager." + ], + [ + "Nebraska head coach Bill
    Callahan runs off the field at
    halftime of the game against
    Baylor in Lincoln, Neb.,
    Saturday, Oct. 16, 2004." + ], + [ + " EAST RUTHERFORD, N.J. (Sports
    Network) - The Toronto
    Raptors have traded All-Star
    swingman Vince Carter to the
    New Jersey Nets in exchange
    for center Alonzo Mourning,
    forward Eric Williams,
    center/forward Aaron Williams
    and two first- round draft
    picks." + ], + [ + "What riot? quot;An Argentine
    friend of mine was a little
    derisive of the Pacers-Pistons
    eruption, quot; says reader
    Mike Gaynes. quot;He snorted,
    #39;Is that what Americans
    call a riot?" + ], + [ + "All season, Chris Barnicle
    seemed prepared for just about
    everything, but the Newton
    North senior was not ready for
    the hot weather he encountered
    yesterday in San Diego at the
    Footlocker Cross-Country
    National Championships. Racing
    in humid conditions with
    temperatures in the 70s, the
    Massachusetts Division 1 state
    champion finished sixth in 15
    minutes 34 seconds in the
    5-kilometer race. ..." + ], + [ + "AP - Coach Tyrone Willingham
    was fired by Notre Dame on
    Tuesday after three seasons in
    which he failed to return one
    of the nation's most storied
    football programs to
    prominence." + ], + [ + "COLLEGE PARK, Md. -- Joel
    Statham completed 18 of 25
    passes for 268 yards and two
    touchdowns in No. 23
    Maryland's 45-22 victory over
    Temple last night, the
    Terrapins' 12th straight win
    at Byrd Stadium." + ], + [ + "Manchester United boss Sir
    Alex Ferguson wants the FA to
    punish Arsenal good guy Dennis
    Bergkamp for taking a swing at
    Alan Smith last Sunday." + ], + [ + "Published reports say Barry
    Bonds has testified that he
    used a clear substance and a
    cream given to him by a
    trainer who was indicted in a
    steroid-distribution ring." + ], + [ + " ATHENS (Reuters) - Christos
    Angourakis added his name to
    Greece's list of Paralympic
    medal winners when he claimed
    a bronze in the T53 shot put
    competition Thursday." + ], + [ + "Jay Fiedler threw for one
    touchdown, Sage Rosenfels
    threw for another and the
    Miami Dolphins got a victory
    in a game they did not want to
    play, beating the New Orleans
    Saints 20-19 Friday night." + ], + [ + " NEW YORK (Reuters) - Terrell
    Owens scored three touchdowns
    and the Philadelphia Eagles
    amassed 35 first-half points
    on the way to a 49-21
    drubbing of the Dallas Cowboys
    in Irving, Texas, Monday." + ], + [ + "A late strike by Salomon Kalou
    sealed a 2-1 win for Feyenoord
    over NEC Nijmegen, while
    second placed AZ Alkmaar
    defeated ADO Den Haag 2-0 in
    the Dutch first division on
    Sunday." + ], + [ + "What a disgrace Ron Artest has
    become. And the worst part is,
    the Indiana Pacers guard just
    doesn #39;t get it. Four days
    after fueling one of the
    biggest brawls in the history
    of pro sports, Artest was on
    national" + ], + [ + "Jenson Button has revealed
    dissatisfaction with the way
    his management handled a
    fouled switch to Williams. Not
    only did the move not come
    off, his reputation may have
    been irreparably damaged amid
    news headline" + ], + [ + "Redknapp and his No2 Jim Smith
    resigned from Portsmouth
    yesterday, leaving
    controversial new director
    Velimir Zajec in temporary
    control." + ], + [ + "As the season winds down for
    the Frederick Keys, Manager
    Tom Lawless is starting to
    enjoy the progress his
    pitching staff has made this
    season." + ], + [ + "Britain #39;s Bradley Wiggins
    won the gold medal in men
    #39;s individual pursuit
    Saturday, finishing the
    4,000-meter final in 4:16." + ], + [ + "And when David Akers #39;
    50-yard field goal cleared the
    crossbar in overtime, they did
    just that. They escaped a
    raucous Cleveland Browns
    Stadium relieved but not
    broken, tested but not
    cracked." + ], + [ + "San Antonio, TX (Sports
    Network) - Dean Wilson shot a
    five-under 65 on Friday to
    move into the lead at the
    halfway point of the Texas
    Open." + ], + [ + "Now that Chelsea have added
    Newcastle United to the list
    of clubs that they have given
    what for lately, what price
    Jose Mourinho covering the
    Russian-funded aristocrats of
    west London in glittering
    glory to the tune of four
    trophies?" + ], + [ + "AP - Former Seattle Seahawks
    running back Chris Warren has
    been arrested in Virginia on a
    federal warrant, accused of
    failing to pay #36;137,147 in
    child support for his two
    daughters in Washington state." + ], + [ + "It #39;s official: US Open had
    never gone into the third
    round with only two American
    men, including the defending
    champion, Andy Roddick." + ], + [ + "WASHINGTON, Aug. 17
    (Xinhuanet) -- England coach
    Sven-Goran Eriksson has urged
    the international soccer
    authorities to preserve the
    health of the world superstar
    footballers for major
    tournaments, who expressed his
    will in Slaley of England on
    Tuesday ..." + ], + [ + "Juventus coach Fabio Capello
    has ordered his players not to
    kick the ball out of play when
    an opponent falls to the
    ground apparently hurt because
    he believes some players fake
    injury to stop the match." + ], + [ + "You #39;re angry. You want to
    lash out. The Red Sox are
    doing it to you again. They
    #39;re blowing a playoff
    series, and to the Yankees no
    less." + ], + [ + "TORONTO -- There is no
    mystique to it anymore,
    because after all, the
    Russians have become commoners
    in today's National Hockey
    League, and Finns, Czechs,
    Slovaks, and Swedes also have
    been entrenched in the
    Original 30 long enough to
    turn the ongoing World Cup of
    Hockey into a protracted
    trailer for the NHL season." + ], + [ + "Indianapolis, IN (Sports
    Network) - The Indiana Pacers
    try to win their second
    straight game tonight, as they
    host Kevin Garnett and the
    Minnesota Timberwolves in the
    third of a four-game homestand
    at Conseco Fieldhouse." + ], + [ + "It is a team game, this Ryder
    Cup stuff that will commence
    Friday at Oakland Hills
    Country Club. So what are the
    teams? For the Americans,
    captain Hal Sutton isn't
    saying." + ], + [ + "MANCHESTER United today
    dramatically rejected the
    advances of Malcolm Glazer,
    the US sports boss who is
    mulling an 825m bid for the
    football club." + ], + [ + "As usual, the Big Ten coaches
    were out in full force at
    today #39;s Big Ten
    Teleconference. Both OSU head
    coach Jim Tressel and Iowa
    head coach Kirk Ferentz
    offered some thoughts on the
    upcoming game between OSU" + ], + [ + "New York Knicks #39; Stephon
    Marbury (3) fires over New
    Orleans Hornets #39; Dan
    Dickau (2) during the second
    half in New Orleans Wednesday
    night, Dec. 8, 2004." + ], + [ + "At the very moment when the
    Red Sox desperately need
    someone slightly larger than
    life to rally around, they
    suddenly have the man for the
    job: Thrilling Schilling." + ], + [ + "Dale Earnhardt Jr, right,
    talks with Matt Kenseth, left,
    during a break in practice at
    Lowe #39;s Motor Speedway in
    Concord, NC, Thursday Oct. 14,
    2004 before qualifying for
    Saturday #39;s UAW-GM Quality
    500 NASCAR Nextel Cup race." + ], + [ + "Several starting spots may
    have been usurped or at least
    threatened after relatively
    solid understudy showings
    Sunday, but few players
    welcome the kind of shot
    delivered to Oakland" + ], + [ + "TEMPE, Ariz. -- Neil Rackers
    kicked four field goals and
    the Arizona Cardinals stifled
    rookie Chris Simms and the
    rest of the Tampa Bay offense
    for a 12-7 victory yesterday
    in a matchup of two sputtering
    teams out of playoff
    contention. Coach Jon Gruden's
    team lost its fourth in a row
    to finish 5-11, Tampa Bay's
    worst record since going ..." + ], + [ + "ATHENS France, Britain and the
    United States issued a joint
    challenge Thursday to Germany
    #39;s gold medal in equestrian
    team three-day eventing." + ], + [ + "ATLANTA - Who could have
    imagined Tommy Tuberville in
    this position? Yet there he
    was Friday, standing alongside
    the SEC championship trophy,
    posing for pictures and
    undoubtedly chuckling a bit on
    the inside." + ], + [ + "John Thomson threw shutout
    ball for seven innings
    Wednesday night in carrying
    Atlanta to a 2-0 blanking of
    the New York Mets. New York
    lost for the 21st time in 25
    games on the" + ], + [ + "Vikram Solanki beat the rain
    clouds to register his second
    one-day international century
    as England won the third one-
    day international to wrap up a
    series victory." + ], + [ + "A three-touchdown point spread
    and a recent history of late-
    season collapses had many
    thinking the UCLA football
    team would provide little
    opposition to rival USC #39;s
    march to the BCS-championship
    game at the Orange Bowl." + ], + [ + " PHOENIX (Sports Network) -
    Free agent third baseman Troy
    Glaus is reportedly headed to
    the Arizona Diamondbacks." + ], + [ + "Third-string tailback Chris
    Markey ran for 131 yards to
    lead UCLA to a 34-26 victory
    over Oregon on Saturday.
    Markey, playing because of an
    injury to starter Maurice
    Drew, also caught five passes
    for 84 yards" + ], + [ + "Jay Payton #39;s three-run
    homer led the San Diego Padres
    to a 5-1 win over the San
    Francisco Giants in National
    League play Saturday, despite
    a 701st career blast from
    Barry Bonds." + ], + [ + "The optimists among Rutgers
    fans were delighted, but the
    Scarlet Knights still gave the
    pessimists something to worry
    about." + ], + [ + "Perhaps the sight of Maria
    Sharapova opposite her tonight
    will jog Serena Williams #39;
    memory. Wimbledon. The final.
    You and Maria." + ], + [ + "Five days after making the
    putt that won the Ryder Cup,
    Colin Montgomerie looked set
    to miss the cut at a European
    PGA tour event." + ], + [ + "Karachi - Captain Inzamam ul-
    Haq and coach Bob Woolmer came
    under fire on Thursday for
    choosing to bat first on a
    tricky pitch after Pakistan
    #39;s humiliating defeat in
    the ICC Champions Trophy semi-
    final." + ], + [ + "Scottish champions Celtic saw
    their three-year unbeaten home
    record in Europe broken
    Tuesday as they lost 3-1 to
    Barcelona in the Champions
    League Group F opener." + ], + [ + "AP - Andy Roddick searched out
    Carlos Moya in the throng of
    jumping, screaming Spanish
    tennis players, hoping to
    shake hands." + ], + [ + "ENGLAND captain and Real
    Madrid midfielder David
    Beckham has played down
    speculation that his club are
    moving for England manager
    Sven-Goran Eriksson." + ], + [ + "AFP - National Basketball
    Association players trying to
    win a fourth consecutive
    Olympic gold medal for the
    United States have gotten the
    wake-up call that the \"Dream
    Team\" days are done even if
    supporters have not." + ], + [ + "England will be seeking their
    third clean sweep of the
    summer when the NatWest
    Challenge against India
    concludes at Lord #39;s. After
    beating New Zealand and West
    Indies 3-0 and 4-0 in Tests,
    they have come alive" + ], + [ + "AP - Mark Richt knows he'll
    have to get a little creative
    when he divvies up playing
    time for Georgia's running
    backs next season. Not so on
    Saturday. Thomas Brown is the
    undisputed starter for the
    biggest game of the season." + ], + [ + "Michael Phelps, the six-time
    Olympic champion, issued an
    apology yesterday after being
    arrested and charged with
    drunken driving in the United
    States." + ], + [ + "ATHENS Larry Brown, the US
    coach, leaned back against the
    scorer #39;s table, searching
    for support on a sinking ship.
    His best player, Tim Duncan,
    had just fouled out, and the
    options for an American team
    that" + ], + [ + "Spain have named an unchanged
    team for the Davis Cup final
    against the United States in
    Seville on 3-5 December.
    Carlos Moya, Juan Carlos
    Ferrero, Rafael Nadal and
    Tommy Robredo will take on the
    US in front of 22,000 fans at
    the converted Olympic stadium." + ], + [ + "Last Tuesday night, Harvard
    knocked off rival Boston
    College, which was ranked No.
    1 in the country. Last night,
    the Crimson knocked off
    another local foe, Boston
    University, 2-1, at Walter
    Brown Arena, which marked the
    first time since 1999 that
    Harvard had beaten them both
    in the same season." + ], + [ + "About 500 prospective jurors
    will be in an Eagle, Colorado,
    courtroom Friday, answering an
    82-item questionnaire in
    preparation for the Kobe
    Bryant sexual assault trial." + ], + [ + "BEIJING The NBA has reached
    booming, basketball-crazy
    China _ but the league doesn
    #39;t expect to make any money
    soon. The NBA flew more than
    100 people halfway around the
    world for its first games in
    China, featuring" + ], + [ + " NEW YORK (Reuters) - Curt
    Schilling pitched 6 2/3
    innings and Manny Ramirez hit
    a three-run homer in a seven-
    run fourth frame to lead the
    Boston Red Sox to a 9-3 win
    over the host Anaheim Angels
    in their American League
    Divisional Series opener
    Tuesday." + ], + [ + "AP - The game between the
    Minnesota Twins and the New
    York Yankees on Tuesday night
    was postponed by rain." + ], + [ + "PARIS The verdict is in: The
    world #39;s greatest race car
    driver, the champion of
    champions - all disciplines
    combined - is Heikki
    Kovalainen." + ], + [ + "Pakistan won the toss and
    unsurprisingly chose to bowl
    first as they and West Indies
    did battle at the Rose Bowl
    today for a place in the ICC
    Champions Trophy final against
    hosts England." + ], + [ + "AP - Boston Red Sox center
    fielder Johnny Damon is having
    a recurrence of migraine
    headaches that first bothered
    him after a collision in last
    year's playoffs." + ], + [ + "Felix Cardenas of Colombia won
    the 17th stage of the Spanish
    Vuelta cycling race Wednesday,
    while defending champion
    Roberto Heras held onto the
    overall leader #39;s jersey
    for the sixth day in a row." + ], + [ + "Established star Landon
    Donovan and rising sensation
    Eddie Johnson carried the
    United States into the
    regional qualifying finals for
    the 2006 World Cup in emphatic
    fashion Wednesday night." + ], + [ + "Wizards coach Eddie Jordan
    says the team is making a
    statement that immaturity will
    not be tolerated by suspending
    Kwame Brown one game for not
    taking part in a team huddle
    during a loss to Denver." + ], + [ + "AP - Andy Roddick has yet to
    face a challenge in his U.S.
    Open title defense. He beat
    No. 18 Tommy Robredo of Spain
    6-3, 6-2, 6-4 Tuesday night to
    move into the quarterfinals
    without having lost a set." + ], + [ + "Now that hell froze over in
    Boston, New England braces for
    its rarest season -- a winter
    of content. Red Sox fans are
    adrift from the familiar
    torture of the past." + ], + [ + "THE South Africans have called
    the Wallabies scrum cheats as
    a fresh round of verbal
    warfare opened in the Republic
    last night." + ], + [ + "An overwhelming majority of
    NHL players who expressed
    their opinion in a poll said
    they would not support a
    salary cap even if it meant
    saving a season that was
    supposed to have started Oct.
    13." + ], + [ + "Tony Eury Sr. oversees Dale
    Earnhardt Jr. on the
    racetrack, but Sunday he
    extended his domain to Victory
    Lane. Earnhardt was unbuckling
    to crawl out of the No." + ], + [ + "(Sports Network) - The New
    York Yankees try to move one
    step closer to a division
    title when they conclude their
    critical series with the
    Boston Red Sox at Fenway Park." + ], + [ + "Kurt Busch claimed a stake in
    the points lead in the NASCAR
    Chase for the Nextel Cup
    yesterday, winning the
    Sylvania 300 at New Hampshire
    International Speedway." + ], + [ + "LOUDON, NH - As this
    newfangled stretch drive for
    the Nextel Cup championship
    ensues, Jeff Gordon has to be
    considered the favorite for a
    fifth title." + ], + [ + "By nick giongco. YOU KNOW you
    have reached the status of a
    boxing star when ring
    announcer extraordinaire
    Michael Buffer calls out your
    name in his trademark booming
    voice during a high-profile
    event like yesterday" + ], + [ + "AP - Even with a big lead,
    Eric Gagne wanted to pitch in
    front of his hometown fans one
    last time." + ], + [ + "ATHENS, Greece -- Larry Brown
    was despondent, the head of
    the US selection committee was
    defensive and an irritated
    Allen Iverson was hanging up
    on callers who asked what went
    wrong." + ], + [ + "AP - Courtney Brown refuses to
    surrender to injuries. He's
    already planning another
    comeback." + ], + [ + "It took an off-the-cuff
    reference to a serial
    murderer/cannibal to punctuate
    the Robby Gordon storyline.
    Gordon has been vilified by
    his peers and put on probation" + ], + [ + "By BOBBY ROSS JR. Associated
    Press Writer. play the Atlanta
    Hawks. They will be treated to
    free food and drink and have.
    their pictures taken with
    Mavericks players, dancers and
    team officials." + ], + [ + "ARSENAL #39;S Brazilian World
    Cup winning midfielder
    Gilberto Silva is set to be
    out for at least a month with
    a back injury, the Premiership
    leaders said." + ], + [ + "INDIANAPOLIS -- With a package
    of academic reforms in place,
    the NCAA #39;s next crusade
    will address what its
    president calls a dangerous
    drift toward professionalism
    and sports entertainment." + ], + [ + "AP - It was difficult for
    Southern California's Pete
    Carroll and Oklahoma's Bob
    Stoops to keep from repeating
    each other when the two
    coaches met Thursday." + ], + [ + "Andy Roddick, along with
    Olympic silver medalist Mardy
    Fish and the doubles pair of
    twins Bob and Mike Bryan will
    make up the US team to compete
    with Belarus in the Davis Cup,
    reported CRIENGLISH." + ], + [ + "While the world #39;s best
    athletes fight the noble
    Olympic battle in stadiums and
    pools, their fans storm the
    streets of Athens, turning the
    Greek capital into a huge
    international party every
    night." + ], + [ + "EVERTON showed they would not
    be bullied into selling Wayne
    Rooney last night by rejecting
    a 23.5million bid from
    Newcastle - as Manchester
    United gamble on Goodison
    #39;s resolve to keep the
    striker." + ], + [ + "After months of legal
    wrangling, the case of
    <em>People v. Kobe Bean
    Bryant</em> commences on
    Friday in Eagle, Colo., with
    testimony set to begin next
    month." + ], + [ + "Lyon coach Paul le Guen has
    admitted his side would be
    happy with a draw at Old
    Trafford on Tuesday night. The
    three-times French champions
    have assured themselves of
    qualification for the
    Champions League" + ], + [ + "Reuters - Barry Bonds failed
    to collect a hit in\\his bid to
    join the 700-homer club, but
    he did score a run to\\help the
    San Francisco Giants edge the
    host Milwaukee Brewers\\3-2 in
    National League action on
    Tuesday." + ], + [ + "After waiting an entire summer
    for the snow to fall and
    Beaver Creek to finally open,
    skiers from around the planet
    are coming to check out the
    Birds of Prey World Cup action
    Dec. 1 - 5. Although everyones" + ], + [ + "I confess that I am a complete
    ignoramus when it comes to
    women #39;s beach volleyball.
    In fact, I know only one thing
    about the sport - that it is
    the supreme aesthetic
    experience available on planet
    Earth." + ], + [ + "Manchester United Plc may
    offer US billionaire Malcolm
    Glazer a seat on its board if
    he agrees to drop a takeover
    bid for a year, the Observer
    said, citing an unidentified
    person in the soccer industry." + ], + [ + "For a guy who spent most of
    his first four professional
    seasons on the disabled list,
    Houston Astros reliever Brad
    Lidge has developed into quite
    the ironman these past two
    days." + ], + [ + "Former Bengal Corey Dillon
    found his stride Sunday in his
    second game for the New
    England Patriots. Dillon
    gained 158 yards on 32 carries
    as the Patriots beat the
    Arizona Cardinals, 23-12, for
    their 17th victory in a row." + ], + [ + "If legend is to be believed,
    the end of a victorious war
    was behind Pheidippides #39;
    trek from Marathon to Athens
    2,500 years ago." + ], + [ + " BEIJING (Reuters) - Wimbledon
    champion Maria Sharapova
    demolished fellow Russian
    Tatiana Panova 6-1, 6-1 to
    advance to the quarter-finals
    of the China Open on
    Wednesday." + ], + [ + "Padres general manager Kevin
    Towers called Expos general
    manager Omar Minaya on
    Thursday afternoon and told
    him he needed a shortstop
    because Khalil Greene had
    broken his" + ], + [ + "Motorsport.com. Nine of the
    ten Formula One teams have
    united to propose cost-cutting
    measures for the future, with
    the notable exception of
    Ferrari." + ], + [ + "Steve Francis and Shaquille O
    #39;Neal enjoyed big debuts
    with their new teams. Kobe
    Bryant found out he can #39;t
    carry the Lakers all by
    himself." + ], + [ + "Australia, by winning the
    third Test at Nagpur on
    Friday, also won the four-
    match series 2-0 with one
    match to go. Australia had
    last won a Test series in
    India way back in December
    1969 when Bill Lawry #39;s
    team beat Nawab Pataudi #39;s
    Indian team 3-1." + ], + [ + "Final Score: Connecticut 61,
    New York 51 New York, NY
    (Sports Network) - Nykesha
    Sales scored 15 points to lead
    Connecticut to a 61-51 win
    over New York in Game 1 of
    their best-of-three Eastern
    Conference Finals series at
    Madison Square Garden." + ], + [ + "Charlotte, NC -- LeBron James
    poured in a game-high 19
    points and Jeff McInnis scored
    18 as the Cleveland Cavaliers
    routed the Charlotte Bobcats,
    106-89, at the Charlotte
    Coliseum." + ], + [ + "Lehmann, who was at fault in
    two matches in the tournament
    last season, was blundering
    again with the German set to
    take the rap for both Greek
    goals." + ], + [ + "Jeju Island, South Korea
    (Sports Network) - Grace Park
    and Carin Koch posted matching
    rounds of six-under-par 66 on
    Friday to share the lead after
    the first round of the CJ Nine
    Bridges Classic." + ], + [ + "Benfica and Real Madrid set
    the standard for soccer
    success in Europe in the late
    1950s and early #39;60s. The
    clubs have evolved much
    differently, but both have
    struggled" + ], + [ + "Pakistan are closing in fast
    on Sri Lanka #39;s first
    innings total after impressing
    with both ball and bat on the
    second day of the opening Test
    in Faisalabad." + ], + [ + "Ron Artest has been hit with a
    season long suspension,
    unprecedented for the NBA
    outside doping cases; Stephen
    Jackson banned for 30 games;
    Jermaine O #39;Neal for 25
    games and Anthony Johnson for
    five." + ], + [ + "Three shots behind Grace Park
    with five holes to go, six-
    time LPGA player of the year
    Sorenstam rallied with an
    eagle, birdie and three pars
    to win her fourth Samsung
    World Championship by three
    shots over Park on Sunday." + ], + [ + "NSW Rugby CEO Fraser Neill
    believes Waratah star Mat
    Rogers has learned his lesson
    after he was fined and ordered
    to do community service
    following his controversial
    comments about the club rugby
    competition." + ], + [ + "ATHENS: China, the dominant
    force in world diving for the
    best part of 20 years, won six
    out of eight Olympic titles in
    Athens and prompted
    speculation about a clean
    sweep when they stage the
    Games in Beijing in 2008." + ], + [ + "The Houston Astros won their
    19th straight game at home and
    are one game from winning
    their first playoff series in
    42 years." + ], + [ + "The eighth-seeded American
    fell to sixth-seeded Elena
    Dementieva of Russia, 0-6 6-2
    7-6 (7-5), on Friday - despite
    being up a break on four
    occasions in the third set." + ], + [ + "TUCSON, Arizona (Ticker) --
    No. 20 Arizona State tries to
    post its first three-game
    winning streak over Pac-10
    Conference rival Arizona in 26
    years when they meet Friday." + ], + [ + "AP - Phillip Fulmer kept his
    cool when starting center
    Jason Respert drove off in the
    coach's golf cart at practice." + ], + [ + "GOALS from Wayne Rooney and
    Ruud van Nistelrooy gave
    Manchester United the win at
    Newcastle. Alan Shearer
    briefly levelled matters for
    the Magpies but United managed
    to scrape through." + ], + [ + "AP - Two-time U.S. Open
    doubles champion Max Mirnyi
    will lead the Belarus team
    that faces the United States
    in the Davis Cup semifinals in
    Charleston later this month." + ], + [ + "AP - New York Jets safety Erik
    Coleman got his souvenir
    football from the equipment
    manager and held it tightly." + ], + [ + "Ivan Hlinka coached the Czech
    Republic to the hockey gold
    medal at the 1998 Nagano
    Olympics and became the coach
    of the Pittsburgh Penguins two
    years later." + ], + [ + "AUBURN - Ah, easy street. No
    game this week. Light
    practices. And now Auburn is
    being touted as the No. 3 team
    in the Bowl Championship
    Series standings." + ], + [ + "Portsmouth #39;s Harry
    Redknapp has been named as the
    Barclays manager of the month
    for October. Redknapp #39;s
    side were unbeaten during the
    month and maintained an
    impressive climb to ninth in
    the Premiership." + ], + [ + "Three directors of Manchester
    United have been ousted from
    the board after US tycoon
    Malcolm Glazer, who is
    attempting to buy the club,
    voted against their re-
    election." + ], + [ + "AP - Manny Ramirez singled and
    scored before leaving with a
    bruised knee, and the
    streaking Boston Red Sox beat
    the Detroit Tigers 5-3 Friday
    night for their 10th victory
    in 11 games." + ], + [ + "The news comes fast and
    furious. Pedro Martinez goes
    to Tampa to visit George
    Steinbrenner. Theo Epstein and
    John Henry go to Florida for
    their turn with Pedro. Carl
    Pavano comes to Boston to
    visit Curt Schilling. Jason
    Varitek says he's not a goner.
    Derek Lowe is a goner, but he
    says he wishes it could be
    different. Orlando Cabrera ..." + ], + [ + "FRED Hale Sr, documented as
    the worlds oldest man, has
    died at the age of 113. Hale
    died in his sleep on Friday at
    a hospital in Syracuse, New
    York, while trying to recover
    from a bout of pneumonia, his
    grandson, Fred Hale III said." + ], + [ + "The Oakland Raiders have
    traded Jerry Rice to the
    Seattle Seahawks in a move
    expected to grant the most
    prolific receiver in National
    Football League history his
    wish to get more playing time." + ], + [ + "AP - Florida coach Ron Zook
    was fired Monday but will be
    allowed to finish the season,
    athletic director Jeremy Foley
    told The Gainesville Sun." + ], + [ + "Troy Brown has had to make a
    lot of adjustments while
    playing both sides of the
    football. quot;You always
    want to score when you get the
    ball -- offense or defense" + ], + [ + "Jenson Button will tomorrow
    discover whether he is allowed
    to quit BAR and move to
    Williams for 2005. The
    Englishman has signed
    contracts with both teams but
    prefers a switch to Williams,
    where he began his Formula One
    career in 2000." + ], + [ + "ARSENAL boss Arsene Wenger
    last night suffered a
    Champions League setback as
    Brazilian midfielder Gilberto
    Silva (above) was left facing
    a long-term injury absence." + ], + [ + "American improves to 3-1 on
    the season with a hard-fought
    overtime win, 74-63, against
    Loyala at Bender Arena on
    Friday night." + ], + [ + "Bee Staff Writers. SAN
    FRANCISCO - As Eric Johnson
    drove to the stadium Sunday
    morning, his bruised ribs were
    so sore, he wasn #39;t sure he
    #39;d be able to suit up for
    the game." + ], + [ + "The New England Patriots are
    so single-minded in pursuing
    their third Super Bowl triumph
    in four years that they almost
    have no room for any other
    history." + ], + [ + "Because, while the Eagles are
    certain to stumble at some
    point during the regular
    season, it seems inconceivable
    that they will falter against
    a team with as many offensive
    problems as Baltimore has
    right now." + ], + [ + "AP - J.J. Arrington ran for 84
    of his 121 yards in the second
    half and Aaron Rodgers shook
    off a slow start to throw two
    touchdown passes to help No. 5
    California beat Washington
    42-12 on Saturday." + ], + [ + "SHANGHAI, China The Houston
    Rockets have arrived in
    Shanghai with hometown
    favorite Yao Ming declaring
    himself quot;here on
    business." + ], + [ + "Charleston, SC (Sports
    Network) - Andy Roddick and
    Mardy Fish will play singles
    for the United States in this
    weekend #39;s Davis Cup
    semifinal matchup against
    Belarus." + ], + [ + "RICKY PONTING believes the
    game #39;s watchers have
    fallen for the quot;myth
    quot; that New Zealand know
    how to rattle Australia." + ], + [ + "MILWAUKEE (SportsTicker) -
    Barry Bonds tries to go where
    just two players have gone
    before when the San Francisco
    Giants visit the Milwaukee
    Brewers on Tuesday." + ], + [ + "AP - With Tom Brady as their
    quarterback and a stingy,
    opportunistic defense, it's
    difficult to imagine when the
    New England Patriots might
    lose again. Brady and
    defensive end Richard Seymour
    combined to secure the
    Patriots' record-tying 18th
    straight victory, 31-17 over
    the Buffalo Bills on Sunday." + ], + [ + "Approaching Hurricane Ivan has
    led to postponement of the
    game Thursday night between
    10th-ranked California and
    Southern Mississippi in
    Hattiesburg, Cal #39;s
    athletic director said Monday." + ], + [ + "SAN DIEGO (Ticker) - The San
    Diego Padres lacked speed and
    an experienced bench last
    season, things veteran
    infielder Eric Young is
    capable of providing." + ], + [ + "This is an eye chart,
    reprinted as a public service
    to the New York Mets so they
    may see from what they suffer:
    myopia. Has ever a baseball
    franchise been so shortsighted
    for so long?" + ], + [ + " LONDON (Reuters) - A medical
    product used to treat both
    male hair loss and prostate
    problems has been added to the
    list of banned drugs for
    athletes." + ], + [ + "AP - Curtis Martin and Jerome
    Bettis have the opportunity to
    go over 13,000 career yards
    rushing in the same game when
    Bettis and the Pittsburgh
    Steelers play Martin and the
    New York Jets in a big AFC
    matchup Sunday." + ], + [ + "Michigan Stadium was mostly
    filled with empty seats. The
    only cheers were coming from
    near one end zone -he Iowa
    section. By Carlos Osorio, AP." + ], + [ + "The International Rugby Board
    today confirmed that three
    countries have expressed an
    interest in hosting the 2011
    World Cup. New Zealand, South
    Africa and Japan are leading
    the race to host rugby union
    #39;s global spectacular in
    seven years #39; time." + ], + [ + "MIAMI (Ticker) -- In its first
    season in the Atlantic Coast
    Conference, No. 11 Virginia
    Tech is headed to the BCS.
    Bryan Randall threw two
    touchdown passes and the
    Virginia Tech defense came up
    big all day as the Hokies
    knocked off No." + ], + [ + "(CP) - Somehow, in the span of
    half an hour, the Detroit
    Tigers #39; pitching went from
    brutal to brilliant. Shortly
    after being on the wrong end
    of several records in a 26-5
    thrashing from to the Kansas
    City" + ], + [ + "With the Huskies reeling at
    0-4 - the only member of a
    Bowl Championship Series
    conference left without a win
    -an Jose State suddenly looms
    as the only team left on the
    schedule that UW will be
    favored to beat." + ], + [ + "Darryl Sutter, who coached the
    Calgary Flames to the Stanley
    Cup finals last season, had an
    emergency appendectomy and was
    recovering Friday." + ], + [ + "Athens, Greece (Sports
    Network) - For the second
    straight day a Briton captured
    gold at the Olympic Velodrome.
    Bradley Wiggins won the men
    #39;s individual 4,000-meter
    pursuit Saturday, one day
    after teammate" + ], + [ + "A Steffen Iversen penalty was
    sufficient to secure the
    points for Norway at Hampden
    on Saturday. James McFadden
    was ordered off after 53
    minutes for deliberate
    handball as he punched Claus
    Lundekvam #39;s header off the
    line." + ], + [ + "AP - Ailing St. Louis reliever
    Steve Kline was unavailable
    for Game 3 of the NL
    championship series on
    Saturday, but Cardinals
    manager Tony LaRussa hopes the
    left-hander will pitch later
    this postseason." + ], + [ + "MADRID: A stunning first-half
    free kick from David Beckham
    gave Real Madrid a 1-0 win
    over newly promoted Numancia
    at the Bernabeu last night." + ], + [ + "Favourites Argentina beat
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    leading goalscorer Carlos
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    after 16 minutes, and" + ], + [ + "Shortly after Steve Spurrier
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    the Gators were placed on NCAA
    probation for a year stemming
    from a child-support payment
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    an aviation trade war after
    failing to reach agreement
    during talks Thursday on
    subsidies to Airbus Industrie." + ], + [ + " SAN FRANCISCO (Reuters) -
    Nike Inc. <A HREF=\"http://w
    ww.investor.reuters.com/FullQu
    ote.aspx?ticker=NKE.N target=/
    stocks/quickinfo/fullquote\">
    ;NKE.N</A>, the world's
    largest athletic shoe company,
    on Monday reported a 25
    percent rise in quarterly
    profit, beating analysts'
    estimates, on strong demand
    for high-end running and
    basketball shoes in the
    United States." + ], + [ + "NEW YORK (CBS.MW) - US stocks
    traded mixed Thurday as Merck
    shares lost a quarter of their
    value, dragging blue chips
    lower, but the Nasdaq moved
    higher, buoyed by gains in the
    semiconductor sector." + ], + [ + "Bankrupt ATA Airlines #39;
    withdrawal from Chicago #39;s
    Midway International Airport
    presents a juicy opportunity
    for another airline to beef up
    service in the Midwest" + ], + [ + "The Instinet Group, owner of
    one of the largest electronic
    stock trading systems, is for
    sale, executives briefed on
    the plan say." + ], + [ + "The Auburn Hills-based
    Chrysler Group made a profit
    of \\$269 million in the third
    quarter, even though worldwide
    sales and revenues declined,
    contributing to a \\$1." + ], + [ + "SAN FRANCISCO (CBS.MW) -- UAL
    Corp., parent of United
    Airlines, said Wednesday it
    will overhaul its route
    structure to reduce costs and
    offset rising fuel costs." + ], + [ + "Annual economic growth in
    China has slowed for the third
    quarter in a row, falling to
    9.1 per cent, third-quarter
    data shows. The slowdown shows
    the economy is responding to
    Beijing #39;s efforts to rein
    in break-neck investment and
    lending." + ], + [ + "THE All-India Motor Transport
    Congress (AIMTC) on Saturday
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    strike after finalising an
    agreement with the Government
    on the contentious issue of
    service tax and the various
    demands including tax deducted
    at source (TDS), scrapping" + ], + [ + "AP - The euro-zone economy
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    second quarter of 2004, a
    touch slower than in the first
    three months of the year,
    according to initial estimates
    released Tuesday by the
    European Union." + ], + [ + "A few years ago, casinos
    across the United States were
    closing their poker rooms to
    make space for more popular
    and lucrative slot machines." + ], + [ + "WASHINGTON -- Consumers were
    tightfisted amid soaring
    gasoline costs in August and
    hurricane-related disruptions
    last week sent applications
    for jobless benefits to their
    highest level in seven months." + ], + [ + "Talking kitchens and vanities.
    Musical jump ropes and potty
    seats. Blusterous miniature
    leaf blowers and vacuum
    cleaners -- almost as loud as
    the real things." + ], + [ + "Online merchants in the United
    States have become better at
    weeding out fraudulent credit
    card orders, a new survey
    indicates. But shipping
    overseas remains a risky
    venture. By Joanna Glasner." + ], + [ + "Popping a gadget into a cradle
    to recharge it used to mean
    downtime, but these days
    chargers are doing double
    duty, keeping your portable
    devices playing even when
    they're charging." + ], + [ + " SAN FRANCISCO (Reuters) -
    Texas Instruments Inc. <A H
    REF=\"http://www.investor.reute
    rs.com/FullQuote.aspx?ticker=T
    XN.N target=/stocks/quickinfo/
    fullquote\">TXN.N</A>,
    the largest maker of chips for
    cellular phones, on Monday
    said record demand for its
    handset and television chips
    boosted quarterly profit by
    26 percent, even as it
    struggles with a nagging
    inventory problem." + ], + [ + "LONDON ARM Holdings, a British
    semiconductor designer, said
    on Monday that it would buy
    Artisan Components for \\$913
    million to broaden its product
    range." + ], + [ + "MELBOURNE: Global shopping
    mall owner Westfield Group
    will team up with Australian
    developer Multiplex Group to
    bid 585mil (US\\$1." + ], + [ + "Reuters - Delta Air Lines Inc.
    , which is\\racing to cut costs
    to avoid bankruptcy, on
    Wednesday reported\\a wider
    quarterly loss amid soaring
    fuel prices and weak\\domestic
    airfares." + ], + [ + "Energy utility TXU Corp. on
    Monday more than quadrupled
    its quarterly dividend payment
    and raised its projected
    earnings for 2004 and 2005
    after a companywide
    performance review." + ], + [ + "Northwest Airlines Corp., the
    fourth- largest US airline,
    and its pilots union reached
    tentative agreement on a
    contract that would cut pay
    and benefits, saving the
    company \\$265 million a year." + ], + [ + "Microsoft Corp. MSFT.O and
    cable television provider
    Comcast Corp. CMCSA.O said on
    Monday that they would begin
    deploying set-top boxes
    powered by Microsoft software
    starting next week." + ], + [ + "Philippines mobile phone
    operator Smart Communications
    Inc. is in talks with
    Singapore #39;s Mobile One for
    a possible tie-up,
    BusinessWorld reported Monday." + ], + [ + "Airline warns it may file for
    bankruptcy if too many senior
    pilots take early retirement
    option. Delta Air LInes #39;
    CEO says it faces bankruptcy
    if it can #39;t slow the pace
    of pilots taking early
    retirement." + ], + [ + "Kodak Versamark #39;s parent
    company, Eastman Kodak Co.,
    reported Tuesday it plans to
    eliminate almost 900 jobs this
    year in a restructuring of its
    overseas operations." + ], + [ + "A top official of the US Food
    and Drug Administration said
    Friday that he and his
    colleagues quot;categorically
    reject quot; earlier
    Congressional testimony that
    the agency has failed to
    protect the public against
    dangerous drugs." + ], + [ + "AFP - British retailer Marks
    and Spencer announced a major
    management shake-up as part of
    efforts to revive its
    fortunes, saying trading has
    become tougher ahead of the
    crucial Christmas period." + ], + [ + " ATLANTA (Reuters) - Home
    Depot Inc. <A HREF=\"http://
    www.investor.reuters.com/FullQ
    uote.aspx?ticker=HD.N target=/
    stocks/quickinfo/fullquote\">
    ;HD.N</A>, the top home
    improvement retailer, on
    Tuesday reported a 15 percent
    rise in third-quarter profit,
    topping estimates, aided by
    installed services and sales
    to contractors." + ], + [ + " LONDON (Reuters) - The dollar
    fought back from one-month
    lows against the euro and
    Swiss franc on Wednesday as
    investors viewed its sell-off
    in the wake of the Federal
    Reserve's verdict on interest
    rates as overdone." + ], + [ + "Boston Scientific Corp said on
    Friday it has recalled an ear
    implant the company acquired
    as part of its purchase of
    Advanced Bionics in June." + ], + [ + "MarketWatch.com. Richemont
    sees significant H1 lift,
    unclear on FY (2:53 AM ET)
    LONDON (CBS.MW) -- Swiss
    luxury goods maker
    Richemont(zz:ZZ:001273145:
    news, chart, profile), which
    also is a significant" + ], + [ + "Crude oil climbed more than
    \\$1 in New York on the re-
    election of President George
    W. Bush, who has been filling
    the US Strategic Petroleum
    Reserve." + ], + [ + "Yukos #39; largest oil-
    producing unit regained power
    supplies from Tyumenenergo, a
    Siberia-based electricity
    generator, Friday after the
    subsidiary pledged to pay 440
    million rubles (\\$15 million)
    in debt by Oct. 3." + ], + [ + "US STOCKS vacillated yesterday
    as rising oil prices muted
    Wall Streets excitement over
    strong results from Lehman
    Brothers and Sprints \\$35
    billion acquisition of Nextel
    Communications." + ], + [ + "At the head of the class,
    Rosabeth Moss Kanter is an
    intellectual whirlwind: loud,
    expansive, in constant motion." + ], + [ + "A bitter row between America
    and the European Union over
    alleged subsidies to rival
    aircraft makers Boeing and
    Airbus intensified yesterday." + ], + [ + "Amsterdam (pts) - Dutch
    electronics company Philips
    http://www.philips.com has
    announced today, Friday, that
    it has cut its stake in Atos
    Origin by more than a half." + ], + [ + "TORONTO (CP) - Two-thirds of
    banks around the world have
    reported an increase in the
    volume of suspicious
    activities that they report to
    police, a new report by KPMG
    suggests." + ], + [ + "The Atkins diet frenzy slowed
    growth briefly, but the
    sandwich business is booming,
    with \\$105 billion in sales
    last year." + ], + [ + "Luxury carmaker Jaguar said
    Friday it was stopping
    production at a factory in
    central England, resulting in
    a loss of 1,100 jobs,
    following poor sales in the
    key US market." + ], + [ + "John Gibson said Friday that
    he decided to resign as chief
    executive officer of
    Halliburton Energy Services
    when it became apparent he
    couldn #39;t become the CEO of
    the entire corporation, after
    getting a taste of the No." + ], + [ + "NEW YORK (Reuters) - Outback
    Steakhouse Inc. said Tuesday
    it lost about 130 operating
    days and up to \\$2 million in
    revenue because it had to
    close some restaurants in the
    South due to Hurricane
    Charley." + ], + [ + "State insurance commissioners
    from across the country have
    proposed new rules governing
    insurance brokerage fees,
    winning praise from an
    industry group and criticism
    from" + ], + [ + "SOFTWARE giant Oracle #39;s
    stalled \\$7.7bn (4.2bn) bid to
    take over competitor
    PeopleSoft has received a huge
    boost after a US judge threw
    out an anti-trust lawsuit
    filed by the Department of
    Justice to block the
    acquisition." + ], + [ + "Office Depot Inc. (ODP.N:
    Quote, Profile, Research) on
    Tuesday warned of weaker-than-
    expected profits for the rest
    of the year because of
    disruptions from the string of
    hurricanes" + ], + [ + "THE photo-equipment maker
    Kodak yesterday announced
    plans to axe 600 jobs in the
    UK and close a factory in
    Nottinghamshire, in a move in
    line with the giants global
    restructuring strategy
    unveiled last January." + ], + [ + "China's central bank on
    Thursday raised interest rates
    for the first time in nearly a
    decade, signaling deepening
    unease with the breakneck pace
    of development and an intent
    to reign in a construction
    boom now sowing fears of
    runaway inflation." + ], + [ + "CHICAGO - Delta Air Lines
    (DAL) said Wednesday it plans
    to eliminate between 6,000 and
    6,900 jobs during the next 18
    months, implement a 10 across-
    the-board pay reduction and
    reduce employee benefits." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks were likely to open
    flat on Wednesday, with high
    oil prices and profit warnings
    weighing on the market before
    earnings reports start and key
    jobs data is released this
    week." + ], + [ + "Best known for its popular
    search engine, Google is
    rapidly rolling out new
    products and muscling into
    Microsoft's stronghold: the
    computer desktop. The
    competition means consumers
    get more choices and better
    products." + ], + [ + " MOSCOW (Reuters) - Russia's
    Gazprom said on Tuesday it
    will bid for embattled oil
    firm YUKOS' main unit next
    month, as the Kremlin seeks
    to turn the world's biggest
    gas producer into a major oil
    player." + ], + [ + "Federal Reserve officials
    raised a key short-term
    interest rate Tuesday for the
    fifth time this year, and
    indicated they will gradually
    move rates higher next year to
    keep inflation under control
    as the economy expands." + ], + [ + "Canadians are paying more to
    borrow money for homes, cars
    and other purchases today
    after a quarter-point
    interest-rate increase by the
    Bank of Canada yesterday was
    quickly matched by the
    chartered banks." + ], + [ + "Delta Air Lines is to issue
    millions of new shares without
    shareholder consent as part of
    moves to ensure its survival." + ], + [ + "Genta (GNTA:Nasdaq - news -
    research) is never boring!
    Monday night, the company
    announced that its phase III
    Genasense study in chronic
    lymphocytic leukemia (CLL) met
    its primary endpoint, which
    was tumor shrinkage." + ], + [ + "While the entire airline
    industry #39;s finances are
    under water, ATA Airlines will
    have to hold its breath longer
    than its competitors to keep
    from going belly up." + ], + [ + "One day after ousting its
    chief executive, the nation's
    largest insurance broker said
    it will tell clients exactly
    how much they are paying for
    services and renounce back-
    door payments from carriers." + ], + [ + "LONDON (CBS.MW) -- Outlining
    an expectation for higher oil
    prices and increasing demand,
    Royal Dutch/Shell on Wednesday
    said it #39;s lifting project
    spending to \\$45 billion over
    the next three years." + ], + [ + "Tuesday #39;s meeting could
    hold clues to whether it
    #39;ll be a November or
    December pause in rate hikes.
    By Chris Isidore, CNN/Money
    senior writer." + ], + [ + "The dollar may fall against
    the euro for a third week in
    four on concern near-record
    crude oil prices will temper
    the pace of expansion in the
    US economy, a survey by
    Bloomberg News indicates." + ], + [ + "The battle for the British-
    based Chelsfield plc hotted up
    at the weekend, with reports
    from London that the property
    giant #39;s management was
    working on its own bid to
    thwart the 585 million (\\$A1." + ], + [ + "SAN DIEGO --(Business Wire)--
    Oct. 11, 2004 -- Breakthrough
    PeopleSoft EnterpriseOne 8.11
    Applications Enable
    Manufacturers to Become
    Demand-Driven." + ], + [ + "Reuters - Oil prices rose on
    Friday as tight\\supplies of
    distillate fuel, including
    heating oil, ahead of\\the
    northern hemisphere winter
    spurred buying." + ], + [ + "Nov. 18, 2004 - An FDA
    scientist says the FDA, which
    is charged with protecting
    America #39;s prescription
    drug supply, is incapable of
    doing so." + ], + [ + "The UK's inflation rate fell
    in September, thanks in part
    to a fall in the price of
    airfares, increasing the
    chance that interest rates
    will be kept on hold." + ], + [ + " HONG KONG/SAN FRANCISCO
    (Reuters) - IBM is selling its
    PC-making business to China's
    largest personal computer
    company, Lenovo Group Ltd.,
    for \\$1.25 billion, marking
    the U.S. firm's retreat from
    an industry it helped pioneer
    in 1981." + ], + [ + "Nordstrom reported a strong
    second-quarter profit as it
    continued to select more
    relevant inventory and sell
    more items at full price." + ], + [ + "The Bank of England is set to
    keep interest rates on hold
    following the latest meeting
    of the its Monetary Policy
    Committee." + ], + [ + "The Bush administration upheld
    yesterday the imposition of
    penalty tariffs on shrimp
    imports from China and
    Vietnam, handing a victory to
    beleaguered US shrimp
    producers." + ], + [ + "House prices rose 0.2 percent
    in September compared with the
    month before to stand up 17.8
    percent on a year ago, the
    Nationwide Building Society
    says." + ], + [ + "Nortel Networks says it will
    again delay the release of its
    restated financial results.
    The Canadian telecom vendor
    originally promised to release
    the restated results in
    September." + ], + [ + " CHICAGO (Reuters) - At first
    glance, paying \\$13 or \\$14
    for a hard-wired Internet
    laptop connection in a hotel
    room might seem expensive." + ], + [ + "Reuters - An investigation
    into U.S. insurers\\and brokers
    rattled insurance industry
    stocks for a second day\\on
    Friday as investors, shaken
    further by subpoenas
    delivered\\to the top U.S. life
    insurer, struggled to gauge
    how deep the\\probe might
    reach." + ], + [ + "The Dow Jones Industrial
    Average failed three times
    this year to exceed its
    previous high and fell to
    about 10,000 each time, most
    recently a week ago." + ], + [ + " SINGAPORE (Reuters) - Asian
    share markets staged a broad-
    based retreat on Wednesday,
    led by steelmakers amid
    warnings of price declines,
    but also enveloping technology
    and financial stocks on
    worries that earnings may
    disappoint." + ], + [ + "NEW YORK - CNN has a new boss
    for the second time in 14
    months: former CBS News
    executive Jonathan Klein, who
    will oversee programming and
    editorial direction at the
    second-ranked cable news
    network." + ], + [ + "Cut-price carrier Virgin Blue
    said Tuesday the cost of using
    Australian airports is
    spiraling upward and asked the
    government to review the
    deregulated system of charges." + ], + [ + "Saudi Arabia, Kuwait and the
    United Arab Emirates, which
    account for almost half of
    OPEC #39;s oil output, said
    they #39;re committed to
    boosting capacity to meet
    soaring demand that has driven
    prices to a record." + ], + [ + "The US Commerce Department
    said Thursday personal income
    posted its biggest increase in
    three months in August. The
    government agency also said
    personal spending was
    unchanged after rising
    strongly in July." + ], + [ + " TOKYO (Reuters) - Tokyo's
    Nikkei average opened up 0.54
    percent on Monday with banks
    and exporters leading the way
    as a stronger finish on Wall
    Street and declining oil
    prices soothed worries over
    the global economic outlook." + ], + [ + "Bruce Wasserstein, head of
    Lazard LLC, is asking partners
    to take a one-third pay cut as
    he readies the world #39;s
    largest closely held
    investment bank for a share
    sale, people familiar with the
    situation said." + ], + [ + "The Lemon Bay Manta Rays were
    not going to let a hurricane
    get in the way of football. On
    Friday, they headed to the
    practice field for the first
    time in eight" + ], + [ + "Microsoft Corp. Chairman Bill
    Gates has donated \\$400,000 to
    a campaign in California
    trying to win approval of a
    measure calling for the state
    to sell \\$3 billion in bonds
    to fund stem-cell research." + ], + [ + "Newspaper publisher Pulitzer
    Inc. said Sunday that company
    officials are considering a
    possible sale of the firm to
    boost shareholder value." + ], + [ + "Shares of Merck amp; Co.
    plunged almost 10 percent
    yesterday after a media report
    said that documents show the
    pharmaceutical giant hid or
    denied" + ], + [ + "Reuters - Wall Street was
    expected to dip at\\Thursday's
    opening, but shares of Texas
    Instruments Inc.\\, may climb
    after it gave upbeat earnings
    guidance." + ], + [ + "Late in August, Boeing #39;s
    top sales execs flew to
    Singapore for a crucial sales
    pitch. They were close to
    persuading Singapore Airlines,
    one of the world #39;s leading
    airlines, to buy the American
    company #39;s new jet, the
    mid-sized 7E7." + ], + [ + "SBC Communications and
    BellSouth will acquire
    YellowPages.com with the goal
    of building the site into a
    nationwide online business
    index, the companies said
    Thursday." + ], + [ + "The sounds of tinkling bells
    could be heard above the
    bustle of the Farmers Market
    on the Long Beach Promenade,
    leading shoppers to a row of
    bright red tin kettles dotting
    a pathway Friday." + ], + [ + "LONDON Santander Central
    Hispano of Spain looked
    certain to clinch its bid for
    the British mortgage lender
    Abbey National, after HBOS,
    Britain #39;s biggest home-
    loan company, said Wednesday
    it would not counterbid, and
    after the European Commission
    cleared" + ], + [ + " WASHINGTON (Reuters) - The
    Justice Department is
    investigating possible
    accounting fraud at Fannie
    Mae, bringing greater
    government scrutiny to bear on
    the mortgage finance company,
    already facing a parallel
    inquiry by the SEC, a source
    close to the matter said on
    Thursday." + ], + [ + "Indian industrial group Tata
    agrees to invest \\$2bn in
    Bangladesh, the biggest single
    deal agreed by a firm in the
    south Asian country." + ], + [ + "The steel tubing company
    reports sharply higher
    earnings, but the stock is
    falling." + ], + [ + "Playboy Enterprises, the adult
    entertainment company, has
    announced plans to open a
    private members club in
    Shanghai even though the
    company #39;s flagship men
    #39;s magazine is still banned
    in China." + ], + [ + "TORONTO (CP) - Earnings
    warnings from Celestica and
    Coca-Cola along with a
    slowdown in US industrial
    production sent stock markets
    lower Wednesday." + ], + [ + "Eastman Kodak Co., the world
    #39;s largest maker of
    photographic film, said
    Wednesday it expects sales of
    digital products and services
    to grow at an annual rate of
    36 percent between 2003 and
    2007, above prior growth rate
    estimates of 26 percent
    between 2002" + ], + [ + "AFP - The Iraqi government
    plans to phase out slowly
    subsidies on basic products,
    such as oil and electricity,
    which comprise 50 percent of
    public spending, equal to 15
    billion dollars, the planning
    minister said." + ], + [ + "The federal agency that
    insures pension plans said
    that its deficit, already at
    the highest in its history,
    had doubled in its last fiscal
    year, to \\$23.3 billion." + ], + [ + "A Milan judge on Tuesday opens
    hearings into whether to put
    on trial 32 executives and
    financial institutions over
    the collapse of international
    food group Parmalat in one of
    Europe #39;s biggest fraud
    cases." + ], + [ + "The Bank of England on
    Thursday left its benchmark
    interest rate unchanged, at
    4.75 percent, as policy makers
    assessed whether borrowing
    costs, already the highest in
    the Group of Seven, are
    constraining consumer demand." + ], + [ + "Fashion retailers Austin Reed
    and Ted Baker have reported
    contrasting fortunes on the
    High Street. Austin Reed
    reported interim losses of 2." + ], + [ + " NEW YORK (Reuters) - A
    federal judge on Friday
    approved Citigroup Inc.'s
    \\$2.6 billion settlement with
    WorldCom Inc. investors who
    lost billions when an
    accounting scandal plunged
    the telecommunications company
    into bankruptcy protection." + ], + [ + "An unspecified number of
    cochlear implants to help
    people with severe hearing
    loss are being recalled
    because they may malfunction
    due to ear moisture, the US
    Food and Drug Administration
    announced." + ], + [ + "Profits triple at McDonald's
    Japan after the fast-food
    chain starts selling larger
    burgers." + ], + [ + "Britain #39;s inflation rate
    fell in August further below
    its 2.0 percent government-set
    upper limit target with
    clothing and footwear prices
    actually falling, official
    data showed on Tuesday." + ], + [ + " LONDON (Reuters) - European
    shares shrugged off a spike in
    the euro to a fresh all-time
    high Wednesday, with telecoms
    again leading the way higher
    after interim profits at
    Britain's mm02 beat
    expectations." + ], + [ + "WASHINGTON - Weighed down by
    high energy prices, the US
    economy grew slower than the
    government estimated in the
    April-June quarter, as higher
    oil prices limited consumer
    spending and contributed to a
    record trade deficit." + ], + [ + "CHICAGO United Airlines says
    it will need even more labor
    cuts than anticipated to get
    out of bankruptcy. United told
    a bankruptcy court judge in
    Chicago today that it intends
    to start talks with unions
    next month on a new round of
    cost savings." + ], + [ + "The University of California,
    Berkeley, has signed an
    agreement with the Samoan
    government to isolate, from a
    tree, the gene for a promising
    anti- Aids drug and to share
    any royalties from the sale of
    a gene-derived drug with the
    people of Samoa." + ], + [ + " TOKYO (Reuters) - Tokyo's
    Nikkei average jumped 2.5
    percent by mid-afternoon on
    Monday as semiconductor-
    related stocks such as
    Advantest Corp. mirrored a
    rally by their U.S. peers
    while banks and brokerages
    extended last week's gains." + ], + [ + "General Motors (GM) plans to
    announce a massive
    restructuring Thursday that
    will eliminate as many as
    12,000 jobs in Europe in a
    move to stem the five-year
    flow of red ink from its auto
    operations in the region." + ], + [ + " LONDON (Reuters) - Oil prices
    held firm on Wednesday as
    Hurricane Ivan closed off
    crude output and shut
    refineries in the Gulf of
    Mexico, while OPEC's Gulf
    producers tried to reassure
    traders by recommending an
    output hike." + ], + [ + "State-owned, running a
    monopoly on imports of jet
    fuel to China #39;s fast-
    growing aviation industry and
    a prized member of Singapore
    #39;s Stock Exchange." + ], + [ + "Google has won a trade mark
    dispute, with a District Court
    judge finding that the search
    engines sale of sponsored
    search terms Geico and Geico
    Direct did not breach car
    insurance firm GEICOs rights
    in the trade marked terms." + ], + [ + "Wall Street bounded higher for
    the second straight day
    yesterday as investors reveled
    in sharply falling oil prices
    and the probusiness agenda of
    the second Bush
    administration. The Dow Jones
    industrials gained more than
    177 points for its best day of
    2004, while the Standard amp;
    Poor's 500 closed at its
    highest level since early
    2002." + ], + [ + "Key factors help determine if
    outsourcing benefits or hurts
    Americans." + ], + [ + "The US Trade Representative on
    Monday rejected the European
    Union #39;s assertion that its
    ban on beef from hormone-
    treated cattle is now
    justified by science and that
    US and Canadian retaliatory
    sanctions should be lifted." + ], + [ + "NEW YORK -- Wall Street's
    fourth-quarter rally gave
    stock mutual funds a solid
    performance for 2004, with
    small-cap equity funds and
    real estate funds scoring some
    of the biggest returns. Large-
    cap growth equities and
    technology-focused funds had
    the slimmest gains." + ], + [ + "Big Food Group Plc, the UK
    owner of the Iceland grocery
    chain, said second-quarter
    sales at stores open at least
    a year dropped 3.3 percent,
    the second consecutive
    decline, after competitors cut
    prices." + ], + [ + " WASHINGTON (Reuters) - The
    first case of soybean rust has
    been found on the mainland
    United States and could affect
    U.S. crops for the near
    future, costing farmers
    millions of dollars, the
    Agriculture Department said on
    Wednesday." + ], + [ + "The Supreme Court today
    overturned a five-figure
    damage award to an Alexandria
    man for a local auto dealer
    #39;s alleged loan scam,
    ruling that a Richmond-based
    federal appeals court had
    wrongly" + ], + [ + "Official figures show the
    12-nation eurozone economy
    continues to grow, but there
    are warnings it may slow down
    later in the year." + ], + [ + "In upholding a lower court
    #39;s ruling, the Supreme
    Court rejected arguments that
    the Do Not Call list violates
    telemarketers #39; First
    Amendment rights." + ], + [ + "Infineon Technologies, the
    second-largest chip maker in
    Europe, said Wednesday that it
    planned to invest about \\$1
    billion in a new factory in
    Malaysia to expand its
    automotive chip business and
    be closer to customers in the
    region." + ], + [ + " NEW YORK (Reuters) -
    Washington Post Co. <A HREF
    =\"http://www.investor.reuters.
    com/FullQuote.aspx?ticker=WPO.
    N target=/stocks/quickinfo/ful
    lquote\">WPO.N</A>
    said on Friday that quarterly
    profit jumped, beating
    analysts' forecasts, boosted
    by results at its Kaplan
    education unit and television
    broadcasting operations." + ], + [ + "New orders for US-made durable
    goods increased 0.2pc in
    September, held back by a big
    drop in orders for
    transportation goods, the US
    Commerce Department said
    today." + ], + [ + "Siblings are the first ever to
    be convicted for sending
    boatloads of junk e-mail
    pushing bogus products. Also:
    Microsoft takes MSN music
    download on a Euro trip....
    Nokia begins legal battle
    against European
    counterparts.... and more." + ], + [ + "I always get a kick out of the
    annual list published by
    Forbes singling out the
    richest people in the country.
    It #39;s almost as amusing as
    those on the list bickering
    over their placement." + ], + [ + "Williams-Sonoma Inc., operator
    of home furnishing chains
    including Pottery Barn, said
    third-quarter earnings rose 19
    percent, boosted by store
    openings and catalog sales." + ], + [ + "TOKYO - Mitsubishi Heavy
    Industries said today it #39;s
    in talks to buy a plot of land
    in central Japan #39;s Nagoya
    city from Mitsubishi Motors
    for building aircraft parts." + ], + [ + "Japan #39;s Sumitomo Mitsui
    Financial Group Inc. said
    Tuesday it proposed to UFJ
    Holdings Inc. that the two
    banks merge on an equal basis
    in its latest attempt to woo
    UFJ away from a rival suitor." + ], + [ + "Oil futures prices were little
    changed Thursday as traders
    anxiously watched for
    indications that the supply or
    demand picture would change in
    some way to add pressure to
    the market or take some away." + ], + [ + " CHICAGO (Reuters) - Delta Air
    Lines Inc. <A HREF=\"http://
    www.investor.reuters.com/FullQ
    uote.aspx?ticker=DAL.N target=
    /stocks/quickinfo/fullquote\"&g
    t;DAL.N</A> said on
    Tuesday it will cut wages by
    10 percent and its chief
    executive will go unpaid for
    the rest of the year, but it
    still warned of bankruptcy
    within weeks unless more cuts
    are made." + ], + [ + " WASHINGTON (Reuters) - A
    former Fannie Mae <A HREF=\"
    http://www.investor.reuters.co
    m/FullQuote.aspx?ticker=FNM.N
    target=/stocks/quickinfo/fullq
    uote\">FNM.N</A>
    employee who gave U.S.
    officials information about
    what he saw as accounting
    irregularities will not
    testify as planned before a
    congressional hearing next
    week, a House committee said
    on Friday." + ], + [ + "PARIS, Nov 4 (AFP) - The
    European Aeronautic Defence
    and Space Company reported
    Thursday that its nine-month
    net profit more than doubled,
    thanks largely to sales of
    Airbus aircraft, and raised
    its full-year forecast." + ], + [ + "The number of people claiming
    unemployment benefit last
    month fell by 6,100 to
    830,200, according to the
    Office for National
    Statistics." + ], + [ + "Tyler airlines are gearing up
    for the beginning of holiday
    travel, as officials offer
    tips to help travelers secure
    tickets and pass through
    checkpoints with ease." + ], + [ + "A criminal trial scheduled to
    start Monday involving former
    Enron Corp. executives may
    shine a rare and potentially
    harsh spotlight on the inner
    workings" + ], + [ + "Wal-Mart Stores Inc. #39;s
    Asda, the UK #39;s second
    biggest supermarket chain,
    surpassed Marks amp; Spencer
    Group Plc as Britain #39;s
    largest clothing retailer in
    the last three months,
    according to the Sunday
    Telegraph." + ], + [ + "Oil supply concerns and broker
    downgrades of blue-chip
    companies left stocks mixed
    yesterday, raising doubts that
    Wall Street #39;s year-end
    rally would continue." + ], + [ + "Genentech Inc. said the
    marketing of Rituxan, a cancer
    drug that is the company #39;s
    best-selling product, is the
    subject of a US criminal
    investigation." + ], + [ + " The world's No. 2 soft drink
    company said on Thursday
    quarterly profit rose due to
    tax benefits." + ], + [ + "USATODAY.com - Personal
    finance software programs are
    the computer industry's
    version of veggies: Everyone
    knows they're good for you,
    but it's just hard to get
    anyone excited about them." + ], + [ + " NEW YORK (Reuters) - The
    dollar rebounded on Monday
    after last week's heavy
    selloff, but analysts were
    uncertain if the rally would
    hold after fresh economic data
    suggested the December U.S.
    jobs report due Friday might
    not live up to expectations." + ], + [ + " NEW YORK (Reuters) - U.S.
    stock futures pointed to a
    lower open on Wall Street on
    Thursday, extending the
    previous session's sharp
    fall, with rising energy
    prices feeding investor
    concerns about corporate
    profits and slower growth." + ], + [ + "MILAN General Motors and Fiat
    on Wednesday edged closer to
    initiating a legal battle that
    could pit the two carmakers
    against each other in a New
    York City court room as early
    as next month." + ], + [ + "Two of the Ford Motor Company
    #39;s most senior executives
    retired on Thursday in a sign
    that the company #39;s deep
    financial crisis has abated,
    though serious challenges
    remain." + ], + [ + " LONDON (Reuters) - Wall
    Street was expected to start
    little changed on Friday as
    investors continue to fret
    over the impact of high oil
    prices on earnings, while
    Boeing <A HREF=\"http://www.
    investor.reuters.com/FullQuote
    .aspx?ticker=BA.N target=/stoc
    ks/quickinfo/fullquote\">BA.
    N</A> will be eyed
    after it reiterated its
    earnings forecast." + ], + [ + "Having an always-on, fast net
    connection is changing the way
    Britons use the internet,
    research suggests." + ], + [ + "Crude oil futures prices
    dropped below \\$51 a barrel
    yesterday as supply concerns
    ahead of the Northern
    Hemisphere winter eased after
    an unexpectedly high rise in
    US inventories." + ], + [ + "By Lilly Vitorovich Of DOW
    JONES NEWSWIRES SYDNEY (Dow
    Jones)--Rupert Murdoch has
    seven weeks to convince News
    Corp. (NWS) shareholders a
    move to the US will make the
    media conglomerate more
    attractive to" + ], + [ + "The long-term economic health
    of the United States is
    threatened by \\$53 trillion in
    government debts and
    liabilities that start to come
    due in four years when baby
    boomers begin to retire." + ], + [ + "The Moscow Arbitration Court
    ruled on Monday that the YUKOS
    oil company must pay RUR
    39.113bn (about \\$1.34bn) as
    part of its back tax claim for
    2001." + ], + [ + "NOVEMBER 11, 2004 -- Bankrupt
    US Airways this morning said
    it had reached agreements with
    lenders and lessors to
    continue operating nearly all
    of its mainline and US Airways
    Express fleets." + ], + [ + "The US government asks the
    World Trade Organisation to
    step in to stop EU member
    states from \"subsidising\"
    planemaker Airbus." + ], + [ + "Boston Scientific Corp.
    (BSX.N: Quote, Profile,
    Research) said on Wednesday it
    received US regulatory
    approval for a device to treat
    complications that arise in
    patients with end-stage kidney
    disease who need dialysis." + ], + [ + "With the economy slowly
    turning up, upgrading hardware
    has been on businesses radar
    in the past 12 months as their
    number two priority." + ], + [ + "Toyota Motor Corp. #39;s
    shares fell for a second day,
    after the world #39;s second-
    biggest automaker had an
    unexpected quarterly profit
    drop." + ], + [ + "Britain-based HBOS says it
    will file a complaint to the
    European Commission against
    Spanish bank Santander Central
    Hispano (SCH) in connection
    with SCH #39;s bid to acquire
    British bank Abbey National" + ], + [ + "Verizon Wireless on Thursday
    announced an agreement to
    acquire all the PCS spectrum
    licenses of NextWave Telecom
    Inc. in 23 markets for \\$3
    billion." + ], + [ + " WASHINGTON (Reuters) - The
    PIMCO mutual fund group has
    agreed to pay \\$50 million to
    settle fraud charges involving
    improper rapid dealing in
    mutual fund shares, the U.S.
    Securities and Exchange
    Commission said on Monday." + ], + [ + "The Conference Board reported
    Thursday that the Leading
    Economic Index fell for a
    third consecutive month in
    August, suggesting slower
    economic growth ahead amid
    rising oil prices." + ], + [ + " SAN FRANCISCO (Reuters) -
    Software maker Adobe Systems
    Inc.<A HREF=\"http://www.inv
    estor.reuters.com/FullQuote.as
    px?ticker=ADBE.O target=/stock
    s/quickinfo/fullquote\">ADBE
    .O</A> on Thursday
    posted a quarterly profit that
    rose more than one-third from
    a year ago, but shares fell 3
    percent after the maker of
    Photoshop and Acrobat software
    did not raise forecasts for
    fiscal 2005." + ], + [ + "William Morrison Supermarkets
    has agreed to sell 114 small
    Safeway stores and a
    distribution centre for 260.2
    million pounds. Morrison
    bought these stores as part of
    its 3 billion pound" + ], + [ + "Pepsi pushes a blue version of
    Mountain Dew only at Taco
    Bell. Is this a winning
    strategy?" + ], + [ + "As the election approaches,
    Congress abandons all pretense
    of fiscal responsibility,
    voting tax cuts that would
    drive 10-year deficits past
    \\$3 trillion." + ], + [ + "ServiceMaster profitably
    bundles services and pays a
    healthy 3.5 dividend." + ], + [ + "\\$222.5 million -- in an
    ongoing securities class
    action lawsuit against Enron
    Corp. The settlement,
    announced Friday and" + ], + [ + " NEW YORK (Reuters) -
    Lifestyle guru Martha Stewart
    said on Wednesday she wants
    to start serving her prison
    sentence for lying about a
    suspicious stock sale as soon
    as possible, so she can put
    her \"nightmare\" behind her." + ], + [ + "Apple Computer's iPod remains
    the king of digital music
    players, but robust pretenders
    to the throne have begun to
    emerge in the Windows
    universe. One of them is the
    Zen Touch, from Creative Labs." + ], + [ + "SAN FRANCISCO (CBS.MW) --
    Crude futures closed under
    \\$46 a barrel Wednesday for
    the first time since late
    September and heating-oil and
    unleaded gasoline prices
    dropped more than 6 percent
    following an across-the-board
    climb in US petroleum
    inventories." + ], + [ + "The University of Iowa #39;s
    market for US presidential
    futures, founded 16-years ago,
    has been overtaken by a
    Dublin-based exchange that is
    now 25 times larger." + ], + [ + "President Bush #39;s drive to
    deploy a multibillion-dollar
    shield against ballistic
    missiles was set back on
    Wednesday by what critics
    called a stunning failure of
    its first full flight test in
    two years." + ], + [ + "Air travelers moved one step
    closer to being able to talk
    on cell phones and surf the
    Internet from laptops while in
    flight, thanks to votes by the
    Federal Communications
    Commission yesterday." + ], + [ + "DESPITE the budget deficit,
    continued increases in oil and
    consumer prices, the economy,
    as measured by gross domestic
    product, grew by 6.3 percent
    in the third" + ], + [ + "Consumers who cut it close by
    paying bills from their
    checking accounts a couple of
    days before depositing funds
    will be out of luck under a
    new law that takes effect Oct.
    28." + ], + [ + "Component problems meant
    Brillian's new big screens
    missed the NFL's kickoff
    party." + ], + [ + "A Russian court on Thursday
    rejected an appeal by the
    Yukos oil company seeking to
    overturn a freeze on the
    accounts of the struggling oil
    giant #39;s core subsidiaries." + ], + [ + "Switzerland #39;s struggling
    national airline reported a
    second-quarter profit of 45
    million Swiss francs (\\$35.6
    million) Tuesday, although its
    figures were boosted by a
    legal settlement in France." + ], + [ + "SIPTU has said it is strongly
    opposed to any privatisation
    of Aer Lingus as pressure
    mounts on the Government to
    make a decision on the future
    funding of the airline." + ], + [ + "Molson Inc. Chief Executive
    Officer Daniel O #39;Neill
    said he #39;ll provide
    investors with a positive #39;
    #39; response to their
    concerns over the company
    #39;s plan to let stock-
    option holders vote on its
    planned merger with Adolph
    Coors Co." + ], + [ + " NEW YORK (Reuters) - The
    world's largest gold producer,
    Newmont Mining Corp. <A HRE
    F=\"http://www.investor.reuters
    .com/FullQuote.aspx?ticker=NEM
    .N target=/stocks/quickinfo/fu
    llquote\">NEM.N</A>,
    on Wednesday said higher gold
    prices drove up quarterly
    profit by 12.5 percent, even
    though it sold less of the
    precious metal." + ], + [ + " WASHINGTON (Reuters) - Fannie
    Mae executives and their
    regulator squared off on
    Wednesday, with executives
    denying any accounting
    irregularity and the regulator
    saying the housing finance
    company's management may need
    to go." + ], + [ + "As the first criminal trial
    stemming from the financial
    deals at Enron opened in
    Houston on Monday, it is
    notable as much for who is not
    among the six defendants as
    who is - and for how little
    money was involved compared
    with how much in other Enron" + ], + [ + "LONDON (CBS.MW) -- British
    bank Barclays on Thursday said
    it is in talks to buy a
    majority stake in South
    African bank ABSA. Free!" + ], + [ + "Investors sent stocks sharply
    lower yesterday as oil prices
    continued their climb higher
    and new questions about the
    safety of arthritis drugs
    pressured pharmaceutical
    stocks." + ], + [ + "Reuters - The head of UAL
    Corp.'s United\\Airlines said
    on Thursday the airline's
    restructuring plan\\would lead
    to a significant number of job
    losses, but it was\\not clear
    how many." + ], + [ + "com September 14, 2004, 9:12
    AM PT. With the economy slowly
    turning up, upgrading hardware
    has been on businesses radar
    in the past 12 months as their
    number two priority." + ], + [ + " NEW YORK (Reuters) -
    Children's Place Retail Stores
    Inc. <A HREF=\"http://www.i
    nvestor.reuters.com/FullQuote.
    aspx?ticker=PLCE.O target=/sto
    cks/quickinfo/fullquote\">PL
    CE.O</A> said on
    Wednesday it will buy 313
    retail stores from Walt
    Disney Co., and its stock rose
    more than 14 percent in early
    morning trade." + ], + [ + "ALBANY, N.Y. -- A California-
    based company that brokers
    life, accident, and disability
    policies for leading US
    companies pocketed millions of
    dollars a year in hidden
    payments from insurers and
    from charges on clients'
    unsuspecting workers, New York
    Attorney General Eliot Spitzer
    charged yesterday." + ], + [ + "NORTEL Networks plans to slash
    its workforce by 3500, or ten
    per cent, as it struggles to
    recover from an accounting
    scandal that toppled three top
    executives and led to a
    criminal investigation and
    lawsuits." + ], + [ + "Ebay Inc. (EBAY.O: Quote,
    Profile, Research) said on
    Friday it would buy Rent.com,
    an Internet housing rental
    listing service, for \\$415
    million in a deal that gives
    it access to a new segment of
    the online real estate market." + ], + [ + "Noranda Inc., Canada #39;s
    biggest mining company, began
    exclusive talks on a takeover
    proposal from China Minmetals
    Corp. that would lead to the
    spinoff of Noranda #39;s
    aluminum business to
    shareholders." + ], + [ + "Google warned Thursday that
    increased competition and the
    maturing of the company would
    result in an quot;inevitable
    quot; slowing of its growth." + ], + [ + " TOKYO (Reuters) - The Nikkei
    average rose 0.55 percent by
    midsession on Wednesday as
    some techs including Advantest
    Corp. gained ground after
    Wall Street reacted positively
    to results from Intel Corp.
    released after the U.S. market
    close." + ], + [ + " ATLANTA (Reuters) - Soft
    drink giant Coca-Cola Co.
    <A HREF=\"http://www.investo
    r.reuters.com/FullQuote.aspx?t
    icker=KO.N target=/stocks/quic
    kinfo/fullquote\">KO.N</A
    >, stung by a prolonged
    downturn in North America,
    Germany and other major
    markets, on Thursday lowered
    its key long-term earnings
    and sales targets." + ], + [ + "A Canadian court approved Air
    Canada #39;s (AC.TO: Quote,
    Profile, Research) plan of
    arrangement with creditors on
    Monday, clearing the way for
    the world #39;s 11th largest
    airline to emerge from
    bankruptcy protection at the
    end of next month" + ], + [ + " LONDON (Reuters) - Oil prices
    eased on Monday after rebels
    in Nigeria withdrew a threat
    to target oil operations, but
    lingering concerns over
    stretched supplies ahead of
    winter kept prices close to
    \\$50." + ], + [ + " LONDON (Reuters) - Oil prices
    climbed above \\$42 a barrel on
    Wednesday, rising for the
    third day in a row as cold
    weather gripped the U.S.
    Northeast, the world's biggest
    heating fuel market." + ], + [ + "Travelers headed home for
    Thanksgiving were greeted
    Wednesday with snow-covered
    highways in the Midwest, heavy
    rain and tornadoes in parts of
    the South, and long security
    lines at some of the nation
    #39;s airports." + ], + [ + "The union representing flight
    attendants on Friday said it
    mailed more than 5,000 strike
    authorization ballots to its
    members employed by US Airways
    as both sides continued talks
    that are expected to stretch
    through the weekend." + ], + [ + "LOS ANGELES (CBS.MW) - The US
    Securities and Exchange
    Commission is probing
    transactions between Delphi
    Corp and EDS, which supplies
    the automotive parts and
    components giant with
    technology services, Delphi
    said late Wednesday." + ], + [ + "MONTREAL (CP) - Molson Inc.
    and Adolph Coors Co. are
    sweetening their brewery
    merger plan with a special
    dividend to Molson
    shareholders worth \\$381
    million." + ], + [ + "WELLINGTON: National carrier
    Air New Zealand said yesterday
    the Australian Competition
    Tribunal has approved a
    proposed alliance with Qantas
    Airways Ltd, despite its
    rejection in New Zealand." + ], + [ + "\"Everyone's nervous,\" Acting
    Undersecretary of Defense
    Michael W. Wynne warned in a
    confidential e-mail to Air
    Force Secretary James G. Roche
    on July 8, 2003." + ], + [ + "Reuters - Alpharma Inc. on
    Friday began\\selling a cheaper
    generic version of Pfizer
    Inc.'s #36;3\\billion a year
    epilepsy drug Neurontin
    without waiting for a\\court
    ruling on Pfizer's request to
    block the copycat medicine." + ], + [ + "Opinion: Privacy hysterics
    bring old whine in new bottles
    to the Internet party. The
    desktop search beta from this
    Web search leader doesn #39;t
    do anything you can #39;t do
    already." + ], + [ + "The Nordics fared well because
    of their long-held ideals of
    keeping corruption clamped
    down and respect for
    contracts, rule of law and
    dedication to one-on-one
    business relationships." + ], + [ + "Don't bother with the small
    stuff. Here's what really
    matters to your lender." + ], + [ + "UK interest rates have been
    kept on hold at 4.75 following
    the latest meeting of the Bank
    of England #39;s rate-setting
    committee." + ], + [ + "Resurgent oil prices paused
    for breath as the United
    States prepared to draw on its
    emergency reserves to ease
    supply strains caused by
    Hurricane Ivan." + ], + [ + "President Bush, who credits
    three years of tax relief
    programs with helping
    strengthen the slow economy,
    said Saturday he would sign
    into law the Working Families
    Tax Relief Act to preserve tax
    cuts." + ], + [ + "HEN investors consider the
    bond market these days, the
    low level of interest rates
    should be more cause for worry
    than for gratitude." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks rallied on Monday after
    software maker PeopleSoft Inc.
    <A HREF=\"http://www.investo
    r.reuters.com/FullQuote.aspx?t
    icker=PSFT.O target=/stocks/qu
    ickinfo/fullquote\">PSFT.O&l
    t;/A> accepted a sweetened
    \\$10.3 billion buyout by rival
    Oracle Corp.'s <A HREF=\"htt
    p://www.investor.reuters.com/F
    ullQuote.aspx?ticker=ORCL.O ta
    rget=/stocks/quickinfo/fullquo
    te\">ORCL.O</A> and
    other big deals raised
    expectations of more
    takeovers." + ], + [ + "The New Jersey-based Accoona
    Corporation, an industry
    pioneer in artificial
    intelligence search
    technology, announced on
    Monday the launch of Accoona." + ], + [ + "Goldman Sachs Group Inc. on
    Thursday said fourth-quarter
    profit rose as its fixed-
    income, currency and
    commodities business soared
    while a rebounding stock
    market boosted investment
    banking." + ], + [ + " NEW YORK (Reuters) - The
    dollar rose on Monday in a
    retracement from last week's
    steep losses, but dealers said
    the bias toward a weaker
    greenback remained intact." + ], + [ + "Moscow - Russia plans to
    combine Gazprom, the world
    #39;s biggest natural gas
    producer, with state-owned oil
    producer Rosneft, easing rules
    for trading Gazprom shares and
    creating a company that may
    dominate the country #39;s
    energy industry." + ], + [ + "Diversified manufacturer
    Honeywell International Inc.
    (HON.N: Quote, Profile,
    Research) posted a rise in
    quarterly profit as strong
    demand for aerospace equipment
    and automobile components" + ], + [ + "Reuters - U.S. housing starts
    jumped a\\larger-than-expected
    6.4 percent in October to the
    busiest pace\\since December as
    buyers took advantage of low
    mortgage rates,\\a government
    report showed on Wednesday." + ], + [ + "The Securities and Exchange
    Commission ordered mutual
    funds to stop paying higher
    commissions to brokers who
    promote the companies' funds
    and required portfolio
    managers to reveal investments
    in funds they supervise." + ], + [ + "Sumitomo Mitsui Financial
    Group (SMFG), Japans second
    largest bank, today put
    forward a 3,200 billion (\\$29
    billion) takeover bid for
    United Financial Group (UFJ),
    the countrys fourth biggest
    lender, in an effort to regain
    initiative in its bidding" + ], + [ + " NEW YORK (Reuters) - U.S.
    chain store retail sales
    slipped during the
    Thanksgiving holiday week, as
    consumers took advantage of
    discounted merchandise, a
    retail report said on
    Tuesday." + ], + [ + "By George Chamberlin , Daily
    Transcript Financial
    Correspondent. Concerns about
    oil production leading into
    the winter months sent shivers
    through the stock market
    Wednesday." + ], + [ + "Airbus has withdrawn a filing
    that gave support for
    Microsoft in an antitrust case
    before the European Union
    #39;s Court of First Instance,
    a source close to the
    situation said on Friday." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks rose on Wednesday
    lifted by a merger between
    retailers Kmart and Sears,
    better-than-expected earnings
    from Hewlett-Packard and data
    showing a slight rise in core
    inflation." + ], + [ + "European Commission president
    Romano Prodi has unveiled
    proposals to loosen the
    deficit rules under the EU
    Stability Pact. The loosening
    was drafted by monetary
    affairs commissioner Joaquin
    Almunia, who stood beside the
    president at the announcement." + ], + [ + "Retail sales in Britain saw
    the fastest growth in
    September since January,
    casting doubts on the view
    that the economy is slowing
    down, according to official
    figures released Thursday." + ], + [ + " NEW YORK (Reuters) -
    Interstate Bakeries Corp.
    <A HREF=\"http://www.investo
    r.reuters.com/FullQuote.aspx?t
    icker=IBC.N target=/stocks/qui
    ckinfo/fullquote\">IBC.N<
    /A>, maker of Hostess
    Twinkies and Wonder Bread,
    filed for bankruptcy on
    Wednesday after struggling
    with more than \\$1.3 billion
    in debt and high costs." + ], + [ + "Delta Air Lines (DAL.N: Quote,
    Profile, Research) on Thursday
    said it reached a deal with
    FedEx Express to sell eight
    McDonnell Douglas MD11
    aircraft and four spare
    engines for delivery in 2004." + ], + [ + "Leading OPEC producer Saudi
    Arabia said on Monday in
    Vienna, Austria, that it had
    made a renewed effort to
    deflate record high world oil
    prices by upping crude output
    again." + ], + [ + "The founders of the Pilgrim
    Baxter amp; Associates money-
    management firm agreed
    yesterday to personally fork
    over \\$160 million to settle
    charges they allowed a friend
    to" + ], + [ + "IBM Corp. Tuesday announced
    plans to acquire software
    vendor Systemcorp ALG for an
    undisclosed amount. Systemcorp
    of Montreal makes project
    portfolio management software
    aimed at helping companies
    better manage their IT
    projects." + ], + [ + "Forbes.com - By now you
    probably know that earnings of
    Section 529 college savings
    accounts are free of federal
    tax if used for higher
    education. But taxes are only
    part of the problem. What if
    your investments tank? Just
    ask Laurence and Margo
    Williams of Alexandria, Va. In
    2000 they put #36;45,000 into
    the Virginia Education Savings
    Trust to open accounts for
    daughters Lea, now 5, and
    Anne, now 3. Since then their
    investment has shrunk 5 while
    the average private college
    tuition has climbed 18 to
    #36;18,300." + ], + [ + "Coca-Cola Amatil Ltd.,
    Australia #39;s biggest soft-
    drink maker, offered A\\$500
    million (\\$382 million) in
    cash and stock for fruit
    canner SPC Ardmona Ltd." + ], + [ + "US technology shares tumbled
    on Friday after technology
    bellwether Intel Corp.
    (INTC.O: Quote, Profile,
    Research) slashed its revenue
    forecast, but blue chips were
    only moderately lower as drug
    and industrial stocks made
    solid gains." + ], + [ + " WASHINGTON (Reuters) - Final
    U.S. government tests on an
    animal suspected of having mad
    cow disease were not yet
    complete, the U.S. Agriculture
    Department said, with no
    announcement on the results
    expected on Monday." + ], + [ + "Metro, Germany's biggest
    retailer, turns in weaker-
    than-expected profits as sales
    at its core supermarkets
    division dip lower." + ], + [ + "BOSTON (CBS.MW) -- First
    Command has reached a \\$12
    million settlement with
    federal regulators for making
    misleading statements and
    omitting important information
    when selling mutual funds to
    US military personnel." + ], + [ + "The federal government, banks
    and aircraft lenders are
    putting the clamps on
    airlines, particularly those
    operating under bankruptcy
    protection." + ], + [ + "EURO DISNEY, the financially
    crippled French theme park
    operator, has admitted that
    its annual losses more than
    doubled last financial year as
    it was hit by a surge in
    costs." + ], + [ + "WASHINGTON The idea of a no-
    bid contract for maintaining
    airport security equipment has
    turned into a non-starter for
    the Transportation Security
    Administration." + ], + [ + "Eyetech (EYET:Nasdaq - news -
    research) did not open for
    trading Friday because a Food
    and Drug Administration
    advisory committee is meeting
    to review the small New York-
    based biotech #39;s
    experimental eye disease drug." + ], + [ + "On September 13, 2001, most
    Americans were still reeling
    from the shock of the
    terrorist attacks on New York
    and the Pentagon two days
    before." + ], + [ + "Domestic air travelers could
    be surfing the Web by 2006
    with government-approved
    technology that allows people
    access to high-speed Internet
    connections while they fly." + ], + [ + "GENEVA: Cross-border
    investment is set to bounce in
    2004 after three years of deep
    decline, reflecting a stronger
    world economy and more
    international merger activity,
    the United Nations (UN) said
    overnight." + ], + [ + "Chewing gum giant Wm. Wrigley
    Jr. Co. on Thursday said it
    plans to phase out production
    of its Eclipse breath strips
    at a plant in Phoenix, Arizona
    and shift manufacturing to
    Poznan, Poland." + ], + [ + "com September 16, 2004, 7:58
    AM PT. This fourth priority
    #39;s main focus has been
    improving or obtaining CRM and
    ERP software for the past year
    and a half." + ], + [ + "Tightness in the labour market
    notwithstanding, the prospects
    for hiring in the third
    quarter are down from the
    second quarter, according to
    the new Manpower Employment
    Outlook Survey." + ], + [ + " NEW YORK (Reuters) - The
    dollar rose on Friday, after a
    U.S. report showed consumer
    prices in line with
    expections, reminding
    investors that the Federal
    Reserve was likely to
    continue raising interest
    rates, analysts said." + ], + [ + "Last year some election
    watchers made a bold
    prediction that this
    presidential election would
    set a record: the first half
    billion dollar campaign in
    hard money alone." + ], + [ + "It #39;s one more blow to
    patients who suffer from
    arthritis. Pfizer, the maker
    of Celebrex, says it #39;s
    painkiller poses an increased
    risk of heart attacks to
    patients using the drugs." + ], + [ + "The economic growth rate in
    the July-September period was
    revised slightly downward from
    an already weak preliminary
    report, the government said
    Wednesday." + ], + [ + "A drug company executive who
    spoke out in support of
    Montgomery County's proposal
    to import drugs from Canada
    and similar legislation before
    Congress said that his company
    has launched an investigation
    into his political activities." + ], + [ + "SYDNEY (AFP) - Australia #39;s
    commodity exports are forecast
    to increase by 15 percent to a
    record 95 billion dollars (71
    million US), the government
    #39;s key economic forecaster
    said." + ], + [ + "Google won a major legal
    victory when a federal judge
    ruled that the search engines
    advertising policy does not
    violate federal trademark
    laws." + ], + [ + "AT amp;T Corp. on Thursday
    said it is reducing one fifth
    of its workforce this year and
    will record a non-cash charge
    of approximately \\$11." + ], + [ + "A rift appeared within Canada
    #39;s music industry yesterday
    as prominent artists called on
    the CRTC to embrace satellite
    radio and the industry warned
    of lost revenue and job
    losses." + ], + [ + "NEW DELHI: India and Pakistan
    agreed on Monday to step up
    cooperation in the energy
    sector, which could lead to
    Pakistan importing large
    amounts of diesel fuel from
    its neighbour, according to
    Pakistani Foreign Minister
    Khurshid Mehmood Kasuri." + ], + [ + "TORONTO (CP) - Glamis Gold of
    Reno, Nev., is planning a
    takeover bid for Goldcorp Inc.
    of Toronto - but only if
    Goldcorp drops its
    \\$2.4-billion-Cdn offer for
    another Canadian firm, made in
    early December." + ], + [ + "This week will see the release
    of October new and existing
    home sales, a measure of
    strength in the housing
    industry. But the short
    holiday week will also leave
    investors looking ahead to the
    holiday travel season." + ], + [ + " BETHESDA, Md. (Reuters) - The
    use of some antidepressant
    drugs appears linked to an
    increase in suicidal behavior
    in some children and teen-
    agers, a U.S. advisory panel
    concluded on Tuesday." + ], + [ + " NEW YORK (Reuters) - U.S.
    technology stocks opened lower
    on Thursday after a sales
    warning from Applied Materials
    Inc. <A HREF=\"http://www.i
    nvestor.reuters.com/FullQuote.
    aspx?ticker=AMAT.O target=/sto
    cks/quickinfo/fullquote\">AM
    AT.O</A>, while weekly
    jobless claims data met Wall
    Street's expectations,
    leaving the Dow and S P 500
    market measures little
    changed." + ], + [ + "CHICAGO : Interstate Bakeries
    Corp., the maker of popular,
    old-style snacks Twinkies and
    Hostess Cakes, filed for
    bankruptcy, citing rising
    costs and falling sales." + ], + [ + "Delta Air Lines (DAL:NYSE -
    commentary - research) will
    cut employees and benefits but
    give a bigger-than-expected
    role to Song, its low-cost
    unit, in a widely anticipated
    but still unannounced
    overhaul, TheStreet.com has
    learned." + ], + [ + "Instead of the skinny black
    line, showing a hurricane
    #39;s forecast track,
    forecasters have drafted a
    couple of alternative graphics
    to depict where the storms
    might go -- and they want your
    opinion." + ], + [ + "BERLIN - Volkswagen AG #39;s
    announcement this week that it
    has forged a new partnership
    deal with Malaysian carmaker
    Proton comes as a strong euro
    and Europe #39;s weak economic
    performance triggers a fresh
    wave of German investment in
    Asia." + ], + [ + "Consumers in Dublin pay more
    for basic goods and services
    that people elsewhere in the
    country, according to figures
    released today by the Central
    Statistics Office." + ], + [ + "LIBERTY Media #39;s move last
    week to grab up to 17.1 per
    cent of News Corporation
    voting stock has prompted the
    launch of a defensive
    shareholder rights plan." + ], + [ + "The blue-chip Hang Seng Index
    rose 171.88 points, or 1.22
    percent, to 14,066.91. On
    Friday, the index had slipped
    31.58 points, or 0.2 percent." + ], + [ + "Shares plunge after company
    says its vein graft treatment
    failed to show benefit in
    late-stage test. CHICAGO
    (Reuters) - Biotechnology
    company Corgentech Inc." + ], + [ + "US blue-chip stocks rose
    slightly on Friday as
    government data showed better-
    than-expected demand in August
    for durable goods other than
    transportation equipment, but
    climbing oil prices limited
    gains." + ], + [ + "Business software maker
    PeopleSoft Inc. said Monday
    that it expects third-quarter
    revenue to range between \\$680
    million and \\$695 million,
    above average Wall Street
    estimates of \\$651." + ], + [ + "Reuters - Enron Corp. ,
    desperate to\\meet profit
    targets, \"parked\" unwanted
    power generating barges\\at
    Merrill Lynch in a sham sale
    designed to be reversed,
    a\\prosecutor said on Tuesday
    in the first criminal trial
    of\\former executives at the
    fallen energy company." + ], + [ + " NEW YORK (Reuters) - The
    dollar rebounded on Monday
    after a heavy selloff last
    week, but analysts were
    uncertain if the rally could
    hold as the drumbeat of
    expectation began for to the
    December U.S. jobs report due
    Friday." + ], + [ + "Coles Myer Ltd. Australia
    #39;s biggest retailer,
    increased second-half profit
    by 26 percent after opening
    fuel and convenience stores,
    selling more-profitable
    groceries and cutting costs." + ], + [ + "MOSCOW: Us oil major
    ConocoPhillips is seeking to
    buy up to 25 in Russian oil
    giant Lukoil to add billions
    of barrels of reserves to its
    books, an industry source
    familiar with the matter said
    on Friday." + ], + [ + "US Airways Group (otc: UAIRQ -
    news - people ) on Thursday
    said it #39;ll seek a court
    injunction to prohibit a
    strike by disaffected unions." + ], + [ + "Shares in Unilever fall after
    the Anglo-Dutch consumer goods
    giant issued a surprise
    profits warning." + ], + [ + "SAN FRANCISCO (CBS.MW) - The
    Canadian government will sell
    its 19 percent stake in Petro-
    Canada for \\$2.49 billion,
    according to the final
    prospectus filed with the US
    Securities and Exchange
    Commission Thursday." + ], + [ + " HYDERABAD, India (Reuters) -
    Microsoft Corp. <A HREF=\"ht
    tp://www.investor.reuters.com/
    FullQuote.aspx?ticker=MSFT.O t
    arget=/stocks/quickinfo/fullqu
    ote\">MSFT.O</A> will
    hire several hundred new staff
    at its new Indian campus in
    the next year, its chief
    executive said on Monday, in a
    move aimed at strengthening
    its presence in Asia's fourth-
    biggest economy." + ], + [ + "Alitalia SpA, Italy #39;s
    largest airline, reached an
    agreement with its flight
    attendants #39; unions to cut
    900 jobs, qualifying the
    company for a government
    bailout that will keep it in
    business for another six
    months." + ], + [ + "P amp;Os cutbacks announced
    today are the result of the
    waves of troubles that have
    swamped the ferry industry of
    late. Some would say the
    company has done well to
    weather the storms for as long
    as it has." + ], + [ + "TORONTO (CP) - Russia #39;s
    Severstal has made an offer to
    buy Stelco Inc., in what #39;s
    believed to be one of several
    competing offers emerging for
    the restructuring but
    profitable Hamilton steel
    producer." + ], + [ + "Walt Disney Co. #39;s
    directors nominated Michael
    Ovitz to serve on its board
    for another three years at a
    meeting just weeks before
    forcing him out of his job as" + ], + [ + "The European Union, Japan,
    Brazil and five other
    countries won World Trade
    Organization approval to
    impose tariffs worth more than
    \\$150 million a year on
    imports from the United" + ], + [ + "Industrial conglomerate
    Honeywell International on
    Wednesday said it has filed a
    lawsuit against 34 electronics
    companies including Apple
    Computer and Eastman Kodak,
    claiming patent infringement
    of its liquid crystal display
    technology." + ], + [ + "Australia #39;s Computershare
    has agreed to buy EquiServe of
    the United States for US\\$292
    million (\\$423 million),
    making it the largest US share
    registrar and driving its
    shares up by a third." + ], + [ + "Interest rates on short-term
    Treasury securities were mixed
    in yesterday's auction. The
    Treasury Department sold \\$18
    billion in three-month bills
    at a discount rate of 1.640
    percent, up from 1.635 percent
    last week. An additional \\$16
    billion was sold in six-month
    bills at a rate of 1.840
    percent, down from 1.860
    percent." + ], + [ + "Two top executives of scandal-
    tarred insurance firm Marsh
    Inc. were ousted yesterday,
    the company said, the latest
    casualties of an industry
    probe by New York's attorney
    general." + ], + [ + "USDA #39;s Animal Plant Health
    Inspection Service (APHIS)
    this morning announced it has
    confirmed a detection of
    soybean rust from two test
    plots at Louisiana State
    University near Baton Rouge,
    Louisiana." + ], + [ + "Mexican Cemex, being the third
    largest cement maker in the
    world, agreed to buy its
    British competitor - RMC Group
    - for \\$5.8 billion, as well
    as their debts in order to
    expand their activity on the
    building materials market of
    the USA and Europe." + ], + [ + "The world's largest insurance
    group pays \\$126m in fines as
    part of a settlement with US
    regulators over its dealings
    with two firms." + ], + [ + "The tobacco firm John Player
    amp; Sons has announced plans
    to lay off 90 workers at its
    cigarette factory in Dublin.
    The company said it was
    planning a phased closure of
    the factory between now and
    February as part of a review
    of its global operations." + ], + [ + "AP - Consumers borrowed more
    freely in September,
    especially when it came to
    racking up charges on their
    credit cards, the Federal
    Reserve reported Friday." + ], + [ + "Bold, innovative solutions are
    key to addressing the rapidly
    rising costs of higher
    education and the steady
    reduction in government-
    subsidized help to finance
    such education." + ], + [ + "Just as the PhD crowd emerge
    with different interpretations
    of today's economy, everyday
    Americans battling to balance
    the checkbook hold diverse
    opinions about where things
    stand now and in the future." + ], + [ + "Authorities here are always
    eager to show off their
    accomplishments, so when
    Beijing hosted the World
    Toilet Organization conference
    last week, delegates were
    given a grand tour of the
    city's toilets." + ], + [ + "WASHINGTON Trying to break a
    deadlock on energy policy, a
    diverse group of
    environmentalists, academics
    and former government
    officials were to publish a
    report on Wednesday that
    presents strategies for making
    the United States cleaner,
    more competitive" + ], + [ + "A consortium led by Royal
    Dutch/Shell Group that is
    developing gas reserves off
    Russia #39;s Sakhalin Island
    said Thursday it has struck a
    US\\$6 billion (euro4." + ], + [ + "An audit by international
    observers supported official
    elections results that gave
    President Hugo Chavez a
    victory over a recall vote
    against him, the secretary-
    general of the Organisation of
    American States announced." + ], + [ + "Apple is recalling 28,000
    faulty batteries for its
    15-inch Powerbook G4 laptops." + ], + [ + "The former Chief Executive
    Officer of Computer Associates
    was indicted by a federal
    grand jury in New York
    Wednesday for allegedly
    participating in a massive
    fraud conspiracy and an
    elaborate cover up of a scheme
    that cost investors" + ], + [ + "Honeywell International Inc.,
    the world #39;s largest
    supplier of building controls,
    agreed to buy Novar Plc for
    798 million pounds (\\$1.53
    billion) to expand its
    security, fire and
    ventilation-systems business
    in Europe." + ], + [ + "San Francisco investment bank
    Thomas Weisel Partners on
    Thursday agreed to pay \\$12.5
    million to settle allegations
    that some of the stock
    research the bank published
    during the Internet boom was
    tainted by conflicts of
    interest." + ], + [ + "Forbes.com - Peter Frankling
    tapped an unusual source to
    fund his new business, which
    makes hot-dog-shaped ice cream
    treats known as Cool Dogs: Two
    investors, one a friend and
    the other a professional
    venture capitalist, put in
    more than #36;100,000 each
    from their Individual
    Retirement Accounts. Later
    Franklin added #36;150,000
    from his own IRA." + ], + [ + "A San Diego insurance
    brokerage has been sued by New
    York Attorney General Elliot
    Spitzer for allegedly
    soliciting payoffs in exchange
    for steering business to
    preferred insurance companies." + ], + [ + "The European Union agreed
    Monday to lift penalties that
    have cost American exporters
    \\$300 million, following the
    repeal of a US corporate tax
    break deemed illegal under
    global trade rules." + ], + [ + " LONDON (Reuters) - European
    stock markets scaled
    near-2-1/2 year highs on
    Friday as oil prices held
    below \\$48 a barrel, and the
    euro held off from mounting
    another assault on \\$1.30 but
    hovered near record highs
    against the dollar." + ], + [ + "Reuters - The company behind
    the Atkins Diet\\on Friday
    shrugged off a recent decline
    in interest in low-carb\\diets
    as a seasonal blip, and its
    marketing chief said\\consumers
    would cut out starchy foods
    again after picking up\\pounds
    over the holidays." + ], + [ + "There #39;s something to be
    said for being the quot;first
    mover quot; in an industry
    trend. Those years of extra
    experience in tinkering with a
    new idea can be invaluable in
    helping the first" + ], + [ + " NEW YORK (Reuters) - Shares
    of Chiron Corp. <A HREF=\"ht
    tp://www.investor.reuters.com/
    FullQuote.aspx?ticker=CHIR.O t
    arget=/stocks/quickinfo/fullqu
    ote\">CHIR.O</A> fell
    7 percent before the market
    open on Friday, a day after
    the biopharmaceutical company
    said it is delaying shipment
    of its flu vaccine, Fluvirin,
    because lots containing 4
    million vaccines do not meet
    product sterility standards." + ], + [ + "The nation's top
    telecommunications regulator
    said yesterday he will push --
    before the next president is
    inaugurated -- to protect
    fledgling Internet telephone
    services from getting taxed
    and heavily regulated by the
    50 state governments." + ], + [ + "DALLAS -- Belo Corp. said
    yesterday that it would cut
    250 jobs, more than half of
    them at its flagship
    newspaper, The Dallas Morning
    News, and that an internal
    investigation into circulation
    overstatements" + ], + [ + "The Federal Reserve still has
    some way to go to restore US
    interest rates to more normal
    levels, Philadelphia Federal
    Reserve President Anthony
    Santomero said on Monday." + ], + [ + "Delta Air Lines (DAL.N: Quote,
    Profile, Research) said on
    Wednesday its auditors have
    expressed doubt about the
    airline #39;s financial
    viability." + ], + [ + "Takeover target Ronin Property
    Group said it would respond to
    an offer by Multiplex Group
    for all the securities in the
    company in about three weeks." + ], + [ + "Google Inc. is trying to
    establish an online reading
    room for five major libraries
    by scanning stacks of hard-to-
    find books into its widely
    used Internet search engine." + ], + [ + "HOUSTON--(BUSINESS
    WIRE)--Sept. 1, 2004-- L
    #39;operazione crea una
    centrale globale per l
    #39;analisi strategica el
    #39;approfondimento del
    settore energetico IHS Energy,
    fonte globale leader di
    software, analisi e
    informazioni" + ], + [ + "The US airline industry,
    riddled with excess supply,
    will see a significant drop in
    capacity, or far fewer seats,
    as a result of at least one
    airline liquidating in the
    next year, according to
    AirTran Airways Chief
    Executive Joe Leonard." + ], + [ + "Boeing (nyse: BA - news -
    people ) Chief Executive Harry
    Stonecipher is keeping the
    faith. On Monday, the head of
    the aerospace and military
    contractor insists he #39;s
    confident his firm will
    ultimately win out" + ], + [ + "Australia #39;s biggest
    supplier of fresh milk,
    National Foods, has posted a
    net profit of \\$68.7 million,
    an increase of 14 per cent on
    last financial year." + ], + [ + "Lawyers for customers suing
    Merck amp; Co. want to
    question CEO Raymond Gilmartin
    about what he knew about the
    dangers of Vioxx before the
    company withdrew the drug from
    the market because of health
    hazards." + ], + [ + "New York; September 23, 2004 -
    The Department of Justice
    (DoJ), FBI and US Attorney
    #39;s Office handed down a
    10-count indictment against
    former Computer Associates
    (CA) chairman and CEO Sanjay
    Kumar and Stephen Richards,
    former CA head of worldwide
    sales." + ], + [ + "PACIFIC Hydro shares yesterday
    caught an updraught that sent
    them more than 20 per cent
    higher after the wind farmer
    moved to flush out a bidder." + ], + [ + " NEW YORK (Reuters) - U.S.
    consumer confidence retreated
    in August while Chicago-area
    business activity slowed,
    according to reports on
    Tuesday that added to worries
    the economy's patch of slow
    growth may last beyond the
    summer." + ], + [ + "Unilever has reported a three
    percent rise in third-quarter
    earnings but warned it is
    reviewing its targets up to
    2010, after issuing a shock
    profits warning last month." + ], + [ + " LONDON (Reuters) - Oil prices
    extended recent heavy losses
    on Wednesday ahead of weekly
    U.S. data expected to show
    fuel stocks rising in time
    for peak winter demand." + ], + [ + "Vodafone has increased the
    competition ahead of Christmas
    with plans to launch 10
    handsets before the festive
    season. The Newbury-based
    group said it will begin
    selling the phones in
    November." + ], + [ + "A former assistant treasurer
    at Enron Corp. (ENRNQ.PK:
    Quote, Profile, Research)
    agreed to plead guilty to
    conspiracy to commit
    securities fraud on Tuesday
    and will cooperate with" + ], + [ + "UK house prices fell by 1.1 in
    October, confirming a
    softening of the housing
    market, Halifax has said. The
    UK #39;s biggest mortgage
    lender said prices rose 18." + ], + [ + "More than six newspaper
    companies have received
    letters from the Securities
    and Exchange Commission
    seeking information about
    their circulation practices." + ], + [ + "THE 64,000 dollar -
    correction, make that 500
    million dollar -uestion
    hanging over Shire
    Pharmaceuticals is whether the
    5 per cent jump in the
    companys shares yesterday
    reflects relief that US
    regulators have finally
    approved its drug for" + ], + [ + "Businesses saw inventories
    rise in July and sales picked
    up, the government reported
    Wednesday. The Commerce
    Department said that stocks of
    unsold goods increased 0.9 in
    July, down from a 1.1 rise in
    June." + ], + [ + "Shares of Genta Inc. (GNTA.O:
    Quote, Profile, Research)
    soared nearly 50 percent on
    Monday after the biotechnology
    company presented promising
    data on an experimental
    treatment for blood cancers." + ], + [ + "SBC Communications Inc. plans
    to cut at least 10,000 jobs,
    or 6 percent of its work
    force, by the end of next year
    to compensate for a drop in
    the number of local-telephone
    customers." + ], + [ + " WASHINGTON (Reuters) - Major
    cigarette makers go on trial
    on Tuesday in the U.S.
    government's \\$280 billion
    racketeering case that
    charges the tobacco industry
    with deliberately deceiving
    the public about the risks of
    smoking since the 1950s." + ], + [ + "Federal Reserve policy-makers
    raised the benchmark US
    interest rate a quarter point
    to 2.25 per cent and restated
    a plan to carry out" + ], + [ + " DETROIT (Reuters) - A
    Canadian law firm on Tuesday
    said it had filed a lawsuit
    against Ford Motor Co. <A H
    REF=\"http://www.investor.reute
    rs.com/FullQuote.aspx?ticker=F
    .N target=/stocks/quickinfo/fu
    llquote\">F.N</A> over
    what it claims are defective
    door latches on about 400,000
    of the automaker's popular
    pickup trucks and SUVs." + ], + [ + "AstraZeneca Plc suffered its
    third setback in two months on
    Friday as lung cancer drug
    Iressa failed to help patients
    live longer" + ], + [ + "Bruce Wasserstein, the
    combative chief executive of
    investment bank Lazard, is
    expected to agree this week
    that he will quit the group
    unless he can pull off a
    successful" + ], + [ + "Dr. David J. Graham, the FDA
    drug safety reviewer who
    sounded warnings over five
    drugs he felt could become the
    next Vioxx has turned to a
    Whistleblower protection group
    for legal help." + ], + [ + "The Kmart purchase of Sears,
    Roebuck may be the ultimate
    expression of that old saying
    in real estate: location,
    location, location." + ], + [ + "Sprint Corp. (FON.N: Quote,
    Profile, Research) on Friday
    said it plans to cut up to 700
    jobs as it realigns its
    business to focus on wireless
    and Internet services and
    takes a non-cash network
    impairment charge." + ], + [ + "Says that amount would have
    been earned for the first 9
    months of 2004, before AT
    amp;T purchase. LOS ANGELES,
    (Reuters) - Cingular Wireless
    would have posted a net profit
    of \\$650 million for the first
    nine months" + ], + [ + " CHICAGO (Reuters) - Goodyear
    Tire Rubber Co. <A HREF=\"
    http://www.investor.reuters.co
    m/FullQuote.aspx?ticker=GT.N t
    arget=/stocks/quickinfo/fullqu
    ote\">GT.N</A> said
    on Friday it will cut 340 jobs
    in its engineered products and
    chemical units as part of its
    cost-reduction efforts,
    resulting in a third-quarter
    charge." + ], + [ + "Reuters - Shares of long-
    distance phone\\companies AT T
    Corp. and MCI Inc. have
    plunged\\about 20 percent this
    year, but potential buyers
    seem to be\\holding out for
    clearance sales." + ], + [ + "Russian oil giant Yukos files
    for bankruptcy protection in
    the US in a last ditch effort
    to stop the Kremlin auctioning
    its main production unit." + ], + [ + "British Airways #39; (BA)
    chief executive Rod Eddington
    has admitted that the company
    quot;got it wrong quot; after
    staff shortages led to three
    days of travel chaos for
    passengers." + ], + [ + "The issue of drug advertising
    directly aimed at consumers is
    becoming political." + ], + [ + "AP - China's economic boom is
    still roaring despite efforts
    to cool sizzling growth, with
    the gross domestic product
    climbing 9.5 percent in the
    first three quarters of this
    year, the government reported
    Friday." + ], + [ + "Soaring petroleum prices
    pushed the cost of goods
    imported into the U.S. much
    higher than expected in
    August, the government said
    today." + ], + [ + "Anheuser-Busch teams up with
    Vietnam's largest brewer,
    laying the groundwork for
    future growth in the region." + ], + [ + "CHARLOTTE, NC - Shares of
    Krispy Kreme Doughnuts Inc.
    #39;s fell sharply Monday as a
    79 percent plunge in third-
    quarter earnings and an
    intensifying accounting
    investigation overshadowed the
    pastrymaker #39;s statement
    that the low-carb craze might
    be easing." + ], + [ + "OXNARD - A leak of explosive
    natural gas forced dozens of
    workers to evacuate an
    offshore oil platform for
    hours Thursday but no damage
    or injuries were reported." + ], + [ + "AP - Assets of the nation's
    retail money market mutual
    funds rose by #36;2.85
    billion in the latest week to
    #36;845.69 billion, the
    Investment Company Institute
    said Thursday." + ], + [ + "Peter Mandelson provoked fresh
    Labour in-fighting yesterday
    with an implied attack on
    Gordon Brown #39;s
    quot;exaggerated gloating
    quot; about the health of the
    British economy." + ], + [ + "JC Penney said yesterday that
    Allen I. Questrom, the chief
    executive who has restyled the
    once-beleaguered chain into a
    sleeker and more profitable
    entity, would be succeeded by
    Myron E. Ullman III, another
    longtime retail executive." + ], + [ + " In the cosmetics department
    at Hecht's in downtown
    Washington, construction crews
    have ripped out the
    traditional glass display
    cases, replacing them with a
    system of open shelves stacked
    high with fragrances from
    Chanel, Burberry and Armani,
    now easily within arm's reach
    of the impulse buyer." + ], + [ + " LONDON (Reuters) - Oil prices
    slid from record highs above
    \\$50 a barrel Wednesday as the
    U.S. government reported a
    surprise increase in crude
    stocks and rebels in Nigeria's
    oil-rich delta region agreed
    to a preliminary cease-fire." + ], + [ + "Rocky Shoes and Boots makes an
    accretive acquisition -- and
    gets Dickies and John Deere as
    part of the deal." + ], + [ + "BONN: Deutsche Telekom is
    bidding 2.9 bn for the 26 it
    does not own in T-Online
    International, pulling the
    internet service back into the
    fold four years after selling
    stock to the public." + ], + [ + "Motorola Inc. says it #39;s
    ready to make inroads in the
    cell-phone market after
    posting a third straight
    strong quarter and rolling out
    a series of new handsets in
    time for the holiday selling
    season." + ], + [ + "Costs of employer-sponsored
    health plans are expected to
    climb an average of 8 percent
    in 2005, the first time in
    five years increases have been
    in single digits." + ], + [ + "After again posting record
    earnings for the third
    quarter, Taiwan Semiconductor
    Manufacturing Company (TSMC)
    expects to see its first
    sequential drop in fourth-
    quarter revenues, coupled with
    a sharp drop in capacity
    utilization rates." + ], + [ + "Many people who have never
    bounced a check in their life
    could soon bounce their first
    check if they write checks to
    pay bills a couple of days
    before their paycheck is
    deposited into their checking
    account." + ], + [ + " LUXEMBOURG (Reuters) -
    Microsoft Corp told a judge on
    Thursday that the European
    Commission must be stopped
    from ordering it to give up
    secret technology to
    competitors." + ], + [ + "Dow Jones Industrial Average
    futures declined amid concern
    an upcoming report on
    manufacturing may point to
    slowing economic growth." + ], + [ + "Reuters - U.S. industrial
    output advanced in\\July, as
    American factories operated at
    their highest capacity\\in more
    than three years, a Federal
    Reserve report on
    Tuesday\\showed." + ], + [ + "Sir Martin Sorrell, chief
    executive of WPP, yesterday
    declared he was quot;very
    impressed quot; with Grey
    Global, stoking speculation
    WPP will bid for the US
    advertising company." + ], + [ + "Like introductory credit card
    rates and superior customer
    service, some promises just
    aren #39;t built to last. And
    so it is that Bank of America
    - mere months after its pledge
    to preserve" + ], + [ + "Reuters - Accounting firm KPMG
    will pay #36;10\\million to
    settle charges of improper
    professional conduct\\while
    acting as auditor for Gemstar-
    TV Guide International Inc.\\,
    the U.S. Securities and
    Exchange Commission said
    on\\Wednesday." + ], + [ + "Family matters made public: As
    eager cousins wait for a slice
    of the \\$15 billion cake that
    is the Pritzker Empire,
    Circuit Court Judge David
    Donnersberger has ruled that
    the case will be conducted in
    open court." + ], + [ + "The weekly survey from
    mortgage company Freddie Mac
    had rates on 30-year fixed-
    rate mortgages inching higher
    this week, up to an average
    5.82 percent from last week
    #39;s 5.81 percent." + ], + [ + "The classic power struggle
    between Walt Disney Co. CEO
    Michael Eisner and former
    feared talent agent Michael
    Ovitz makes for high drama in
    the courtroom - and apparently
    on cable." + ], + [ + "LONDON (CBS.MW) -- Elan
    (UK:ELA) (ELN) and partner
    Biogen (BIIB) said the FDA has
    approved new drug Tysabri to
    treat relapsing forms of
    multiple sclerosis." + ], + [ + "Bank of New Zealand has frozen
    all accounts held in the name
    of Access Brokerage, which was
    yesterday placed in
    liquidation after a client
    fund shortfall of around \\$5
    million was discovered." + ], + [ + "ZDNet #39;s survey of IT
    professionals in August kept
    Wired amp; Wireless on top
    for the 18th month in a row.
    Telecommunications equipment
    maker Motorola said Tuesday
    that it would cut 1,000 jobs
    and take related" + ], + [ + "BRITAIN #39;S largest
    financial institutions are
    being urged to take lead roles
    in lawsuits seeking hundreds
    of millions of dollars from
    the scandal-struck US
    insurance industry." + ], + [ + "Bankrupt UAL Corp. (UALAQ.OB:
    Quote, Profile, Research) on
    Thursday reported a narrower
    third-quarter net loss. The
    parent of United Airlines
    posted a loss of \\$274
    million, or" + ], + [ + "The European Union #39;s head
    office issued a bleak economic
    report Tuesday, warning that
    the sharp rise in oil prices
    will quot;take its toll quot;
    on economic growth next year
    while the euro #39;s renewed
    climb could threaten crucial
    exports." + ], + [ + "Though Howard Stern's
    defection from broadcast to
    satellite radio is still 16
    months off, the industry is
    already trying to figure out
    what will fill the crater in
    ad revenue and listenership
    that he is expected to leave
    behind." + ], + [ + " WASHINGTON (Reuters) - The
    United States set final anti-
    dumping duties of up to 112.81
    percent on shrimp imported
    from China and up to 25.76
    percent on shrimp from Vietnam
    to offset unfair pricing, the
    Commerce Department said on
    Tuesday." + ], + [ + "NEW YORK -- When Office Depot
    Inc. stores ran an electronics
    recycling drive last summer
    that accepted everything from
    cellphones to televisions,
    some stores were overwhelmed
    by the amount of e-trash they
    received." + ], + [ + "The International Monetary
    Fund expressed concern Tuesday
    about the impact of the
    troubles besetting oil major
    Yukos on Russia #39;s standing
    as a place to invest." + ], + [ + "LinuxWorld Conference amp;
    Expo will come to Boston for
    the first time in February,
    underscoring the area's
    standing as a hub for the
    open-source software being
    adopted by thousands of
    businesses." + ], + [ + "Interest rates on short-term
    Treasury securities rose in
    yesterday's auction. The
    Treasury Department sold \\$19
    billion in three-month bills
    at a discount rate of 1.710
    percent, up from 1.685 percent
    last week. An additional \\$17
    billion was sold in six-month
    bills at a rate of 1.950
    percent, up from 1.870
    percent." + ], + [ + "Moody #39;s Investors Service
    on Wednesday said it may cut
    its bond ratings on HCA Inc.
    (HCA.N: Quote, Profile,
    Research) deeper into junk,
    citing the hospital operator
    #39;s plan to buy back about
    \\$2." + ], + [ + "Telekom Austria AG, the
    country #39;s biggest phone
    operator, won the right to buy
    Bulgaria #39;s largest mobile
    phone company, MobilTel EAD,
    for 1.6 billion euros (\\$2.1
    billion), an acquisition that
    would add 3 million
    subscribers." + ], + [ + "Shares of ID Biomedical jumped
    after the company reported
    Monday that it signed long-
    term agreements with the three
    largest flu vaccine
    wholesalers in the United
    States in light of the
    shortage of vaccine for the
    current flu season." + ], + [ + "The federal government closed
    its window on the oil industry
    Thursday, saying that it is
    selling its last 19 per cent
    stake in Calgary-based Petro-
    Canada." + ], + [ + " NEW YORK, Nov. 11 -- The 40
    percent share price slide in
    Merck #38; Co. in the five
    weeks after it pulled the
    painkiller Vioxx off the
    market highlighted larger
    problems in the pharmaceutical
    industry that may depress
    performance for years,
    according to academics and
    stock analysts who follow the
    sector." + ], + [ + "Low-fare carrier Southwest
    Airlines Co. said Thursday
    that its third-quarter profit
    rose 12.3 percent to beat Wall
    Street expectations despite
    higher fuel costs." + ], + [ + "Another shock hit the drug
    sector Friday when
    pharmaceutical giant Pfizer
    Inc. announced that it found
    an increased heart risk to
    patients for its blockbuster
    arthritis drug Celebrex." + ], + [ + "The number of US information
    technology workers rose 2
    percent to 10.5 million in the
    first quarter of this year,
    but demand for them is
    dropping, according to a new
    report." + ], + [ + "Insurance firm says its board
    now consists of its new CEO
    Michael Cherkasky and 10
    outside members. NEW YORK
    (Reuters) - Marsh amp;
    McLennan Cos." + ], + [ + "Slot machine maker
    International Game Technology
    (IGT.N: Quote, Profile,
    Research) on Tuesday posted
    better-than-expected quarterly
    earnings, as casinos bought" + ], + [ + "Venezuelan election officials
    say they expect to announce
    Saturday, results of a partial
    audit of last Sunday #39;s
    presidential recall
    referendum." + ], + [ + "Sony Ericsson Mobile
    Communications Ltd., the
    mobile-phone venture owned by
    Sony Corp. and Ericsson AB,
    said third-quarter profit rose
    45 percent on camera phone
    demand and forecast this
    quarter will be its strongest." + ], + [ + "NEW YORK, September 17
    (newratings.com) - Alcatel
    (ALA.NYS) has expanded its
    operations and presence in the
    core North American
    telecommunication market with
    two separate acquisitions for
    about \\$277 million." + ], + [ + "Oil giant Shell swept aside
    nearly 100 years of history
    today when it unveiled plans
    to merge its UK and Dutch
    parent companies. Shell said
    scrapping its twin-board
    structure" + ], + [ + "Caesars Entertainment Inc. on
    Thursday posted a rise in
    third-quarter profit as Las
    Vegas hotels filled up and
    Atlantic City properties
    squeaked out a profit that was
    unexpected by the company." + ], + [ + "One of India #39;s leading
    telecommunications providers
    will use Cisco Systems #39;
    gear to build its new
    Ethernet-based broadband
    network." + ], + [ + "Shares of Beacon Roofing
    Suppler Inc. shot up as much
    as 26 percent in its trading
    debut Thursday, edging out
    bank holding company Valley
    Bancorp as the biggest gainer
    among a handful of new stocks
    that went public this week." + ], + [ + "The first weekend of holiday
    shopping went from red-hot to
    dead white, as a storm that
    delivered freezing, snowy
    weather across Colorado kept
    consumers at home." + ], + [ + " LONDON (Reuters) - The dollar
    fell within half a cent of
    last week's record low against
    the euro on Thursday after
    capital inflows data added to
    worries the United States may
    struggle to fund its current
    account deficit." + ], + [ + "China has pledged to invest
    \\$20 billion in Argentina in
    the next 10 years, La Nacion
    reported Wednesday. The
    announcement came during the
    first day of a two-day visit" + ], + [ + "Toyota Motor Corp., the world
    #39;s biggest carmaker by
    value, will invest 3.8 billion
    yuan (\\$461 million) with its
    partner Guangzhou Automobile
    Group to boost manufacturing
    capacity in" + ], + [ + "BAE Systems has launched a
    search for a senior American
    businessman to become a non-
    executive director. The high-
    profile appointment is
    designed to strengthen the
    board at a time when the
    defence giant is" + ], + [ + "Computer Associates
    International is expected to
    announce that its new chief
    executive will be John
    Swainson, an I.B.M. executive
    with strong technical and
    sales credentials." + ], + [ + "STOCKHOLM (Dow
    Jones)--Expectations for
    Telefon AB LM Ericsson #39;s
    (ERICY) third-quarter
    performance imply that while
    sales of mobile telephony
    equipment are expected to have
    dipped, the company" + ], + [ + "The Anglo-Dutch oil giant
    Shell today sought to draw a
    line under its reserves
    scandal by announcing plans to
    spend \\$15bn (8.4bn) a year to
    replenish reserves and develop
    production in its oil and gas
    business." + ], + [ + "Circulation declined at most
    major US newspapers in the
    last half year, the latest
    blow for an industry already
    rocked by a scandal involving
    circulation misstatements that
    has undermined the confidence
    of investors and advertisers." + ], + [ + "INDIANAPOLIS - ATA Airlines
    has accepted a \\$117 million
    offer from Southwest Airlines
    that would forge close ties
    between two of the largest US
    discount carriers." + ], + [ + "EVERETT Fire investigators
    are still trying to determine
    what caused a two-alarm that
    destroyed a portion of a South
    Everett shopping center this
    morning." + ], + [ + "Sure, the PeopleSoft board
    told shareholders to just say
    no. This battle will go down
    to the wire, and even
    afterward Ellison could
    prevail." + ], + [ + "Investors cheered by falling
    oil prices and an improving
    job picture sent stocks higher
    Tuesday, hoping that the news
    signals a renewal of economic
    strength and a fall rally in
    stocks." + ], + [ + " CHICAGO (Reuters) - Wal-Mart
    Stores Inc. <A HREF=\"http:/
    /www.investor.reuters.com/Full
    Quote.aspx?ticker=WMT.N target
    =/stocks/quickinfo/fullquote\"&
    gt;WMT.N</A>, the
    world's largest retailer, said
    on Saturday it still
    anticipates September U.S.
    sales to be up 2 percent to 4
    percent at stores open at
    least a year." + ], + [ + " TOKYO (Reuters) - Tokyo's
    Nikkei stock average opened
    down 0.15 percent on
    Wednesday as investors took a
    breather from the market's
    recent rises and sold shares
    of gainers such as Sharp
    Corp." + ], + [ + "NEW YORK : World oil prices
    fell, capping a drop of more
    than 14 percent in a two-and-
    a-half-week slide triggered by
    a perception of growing US
    crude oil inventories." + ], + [ + "SUFFOLK -- Virginia Tech
    scientists are preparing to
    protect the state #39;s
    largest crop from a disease
    with strong potential to do
    damage." + ], + [ + " NEW YORK (Reuters) -
    Citigroup Inc. <A HREF=\"htt
    p://www.investor.reuters.com/F
    ullQuote.aspx?ticker=C.N targe
    t=/stocks/quickinfo/fullquote\"
    >C.N</A> the world's
    largest financial services
    company, said on Tuesday it
    will acquire First American
    Bank in the quickly-growing
    Texas market." + ], + [ + " SINGAPORE (Reuters) - U.S.
    oil prices hovered just below
    \\$50 a barrel on Tuesday,
    holding recent gains on a rash
    of crude supply outages and
    fears over thin heating oil
    tanks." + ], + [ + "US retail sales fell 0.3 in
    August as rising energy costs
    and bad weather persuaded
    shoppers to reduce their
    spending." + ], + [ + "GEORGETOWN, Del., Oct. 28 --
    Plaintiffs in a shareholder
    lawsuit over former Walt
    Disney Co. president Michael
    Ovitz's \\$140 million
    severance package attempted
    Thursday to portray Ovitz as a
    dishonest bumbler who botched
    the hiring of a major
    television executive and
    pushed the release of a movie
    that angered the Chinese
    government, damaging Disney's
    business prospects in the
    country." + ], + [ + "US stocks gained ground in
    early trading Thursday after
    tame inflation reports and
    better than expected jobless
    news. Oil prices held steady
    as Hurricane Ivan battered the
    Gulf coast, where oil
    operations have halted." + ], + [ + "PepsiCo. Inc., the world #39;s
    No. 2 soft- drink maker, plans
    to buy General Mills Inc.
    #39;s stake in their European
    joint venture for \\$750
    million in cash, giving it
    complete ownership of Europe
    #39;s largest snack-food
    company." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks opened little changed
    on Friday, after third-
    quarter gross domestic product
    data showed the U.S. economy
    grew at a slower-than-expected
    pace." + ], + [ + " NEW YORK (Reuters) - Limited
    Brands Inc. <A HREF=\"http:/
    /www.investor.reuters.com/Full
    Quote.aspx?ticker=LTD.N target
    =/stocks/quickinfo/fullquote\"&
    gt;LTD.N</A> on
    Thursday reported higher
    quarterly operating profit as
    cost controls and strong
    lingerie sales offset poor
    results at the retailer's
    Express apparel stores." + ], + [ + "General Motors and
    DaimlerChrysler are
    collaborating on development
    of fuel- saving hybrid engines
    in hopes of cashing in on an
    expanding market dominated by
    hybrid leaders Toyota and
    Honda." + ], + [ + " TOKYO (Reuters) - Nintendo
    Co. Ltd. raised its 2004
    shipment target for its DS
    handheld video game device by
    40 percent to 2.8 million
    units on Thursday after many
    stores in Japan and the
    United States sold out in the
    first week of sales." + ], + [ + "Autodesk this week unwrapped
    an updated version of its
    hosted project collaboration
    service targeted at the
    construction and manufacturing
    industries. Autodesk Buzzsaw
    lets multiple, dispersed
    project participants --
    including building owners,
    developers, architects,
    construction teams, and
    facility managers -- share and
    manage data throughout the
    life of a project, according
    to Autodesk officials." + ], + [ + " WASHINGTON (Reuters) - U.S.
    employers hired just 96,000
    workers in September, the
    government said on Friday in a
    weak jobs snapshot, the final
    one ahead of presidential
    elections that also fueled
    speculation about a pause in
    interest-rate rises." + ], + [ + "SAN FRANCISCO (CBS.MW) -- The
    magazine known for evaluating
    cars and electronics is
    setting its sights on finding
    the best value and quality of
    prescription drugs on the
    market." + ], + [ + "A mouse, a house, and your
    tax-planning spouse all factor
    huge in the week of earnings
    that lies ahead." + ], + [ + "DALLAS (CBS.MW) -- Royal
    Dutch/Shell Group will pay a
    \\$120 million penalty to
    settle a Securities and
    Exchange Commission
    investigation of its
    overstatement of nearly 4.5
    billion barrels of proven
    reserves, the federal agency
    said Tuesday." + ], + [ + "PHOENIX America West Airlines
    has backed away from a
    potential bidding war for
    bankrupt ATA Airlines, paving
    the way for AirTran to take
    over ATA operations." + ], + [ + "US stock-index futures
    declined. Dow Jones Industrial
    Average shares including
    General Electric Co. slipped
    in Europe. Citigroup Inc." + ], + [ + "Federal Reserve Chairman Alan
    Greenspan has done it again.
    For at least the fourth time
    this year, he has touched the
    electrified third rail of
    American politics - Social
    Security." + ], + [ + "Description: Scientists say
    the arthritis drug Bextra may
    pose increased risk of
    cardiovascular troubles.
    Bextra is related to Vioxx,
    which was pulled off the
    market in September for the
    same reason." + ], + [ + "The trial of a lawsuit by Walt
    Disney Co. shareholders who
    accuse the board of directors
    of rubberstamping a deal to
    hire Michael Ovitz" + ], + [ + "LOS ANGELES Grocery giant
    Albertsons says it has
    purchased Bristol Farms, which
    operates eleven upscale stores
    in Southern California." + ], + [ + "The figures from a survey
    released today are likely to
    throw more people into the
    ranks of the uninsured,
    analysts said." + ], + [ + "JB Oxford Holdings Inc., a
    Beverly Hills-based discount
    brokerage firm, was sued by
    the Securities and Exchange
    Commission for allegedly
    allowing thousands of improper
    trades in more than 600 mutual
    funds." + ], + [ + "Goldman Sachs reported strong
    fourth quarter and full year
    earnings of \\$1.19bn, up 36
    per cent on the previous
    quarter. Full year profit rose
    52 per cent from the previous
    year to \\$4.55bn." + ], + [ + "Argosy Gaming (AGY:NYSE - news
    - research) jumped in early
    trading Thursday, after the
    company agreed to be acquired
    by Penn National Gaming
    (PENN:Nasdaq - news -
    research) in a \\$1." + ], + [ + "The DuPont Co. has agreed to
    pay up to \\$340 million to
    settle a lawsuit that it
    contaminated water supplies in
    West Virginia and Ohio with a
    chemical used to make Teflon,
    one of its best-known brands." + ], + [ + "Reuters - Jeffrey Greenberg,
    chairman and chief\\executive
    of embattled insurance broker
    Marsh McLennan Cos.\\, is
    expected to step down within
    hours, a newspaper\\reported on
    Friday, citing people close to
    the discussions." + ], + [ + "A state regulatory board
    yesterday handed a five-year
    suspension to a Lawrence
    funeral director accused of
    unprofessional conduct and
    deceptive practices, including
    one case where he refused to
    complete funeral arrangements
    for a client because she had
    purchased a lower-priced
    casket elsewhere." + ], + [ + "The California Public
    Utilities Commission on
    Thursday upheld a \\$12.1
    million fine against Cingular
    Wireless, related to a two-
    year investigation into the
    cellular telephone company
    #39;s business practices." + ], + [ + "Barclays, the British bank
    that left South Africa in 1986
    after apartheid protests, may
    soon resume retail operations
    in the nation." + ], + [ + "Italian-based Parmalat is
    suing its former auditors --
    Grant Thornton International
    and Deloitte Touche Tohmatsu
    -- for billions of dollars in
    damages. Parmalat blames its
    demise on the two companies
    #39; mismanagement of its
    finances." + ], + [ + "update An alliance of
    technology workers on Tuesday
    accused conglomerate Honeywell
    International of planning to
    move thousands of jobs to low-
    cost regions over the next
    five years--a charge that
    Honeywell denies." + ], + [ + " NEW YORK (Reuters) - Northrop
    Grumman Corp. <A HREF=\"http
    ://www.investor.reuters.com/Fu
    llQuote.aspx?ticker=NOC.N targ
    et=/stocks/quickinfo/fullquote
    \">NOC.N</A> reported
    higher third-quarter earnings
    on Wednesday and an 11
    percent increase in sales on
    strength in its mission
    systems, integrated systems,
    ships and space technology
    businesses." + ], + [ + "TORONTO (CP) - Shares of
    Iamgold fell more than 10 per
    cent after its proposed merger
    with Gold Fields Ltd. was
    thrown into doubt Monday as
    South Africa #39;s Harmony
    Gold Mining Company Ltd." + ], + [ + " LONDON (Reuters) - Oil prices
    tumbled again on Monday to an
    8-week low under \\$46 a
    barrel, as growing fuel stocks
    in the United States eased
    fears of a winter supply
    crunch." + ], + [ + "Alcoa Inc., one of the world
    #39;s top producers of
    aluminum, said Monday that it
    received an unsolicited
    quot;mini-tender quot; offer
    from Toronto-based TRC Capital
    Corp." + ], + [ + "The European Commission is to
    warn Greece about publishing
    false information about its
    public finances." + ], + [ + " LONDON (Reuters) - U.S.
    shares were expected to open
    lower on Wednesday after
    crude oil pushed to a fresh
    high overnight, while Web
    search engine Google Inc.
    dented sentiment as it
    slashed the price range on its
    initial public offering." + ], + [ + "The telemarketer at the other
    end of Orlando Castelblanco
    #39;s line promised to reduce
    the consumer #39;s credit card
    debt by at least \\$2,500 and
    get his 20 percent -- and
    growing -- interest rates down
    to single digits." + ], + [ + " NEW YORK (Reuters) - Blue-
    chip stocks fell slightly on
    Monday after No. 1 retailer
    Wal-Mart Stores Inc. <A HRE
    F=\"http://www.investor.reuters
    .com/FullQuote.aspx?ticker=WMT
    .N target=/stocks/quickinfo/fu
    llquote\">WMT.N</A>
    reported lower-than-expected
    Thanksgiving sales, while
    technology shares were lifted
    by a rally in Apple Computer
    Inc. <A HREF=\"http://www.i
    nvestor.reuters.com/FullQuote.
    aspx?ticker=AAPL.O target=/sto
    cks/quickinfo/fullquote\">AA
    PL.O</A>." + ], + [ + "WPP Group Inc., the world
    #39;s second- largest
    marketing and advertising
    company, said it won the
    bidding for Grey Global Group
    Inc." + ], + [ + " NEW YORK (Reuters) - Stocks
    were slightly lower on
    Tuesday, as concerns about
    higher oil prices cutting into
    corporate profits and
    consumer demand weighed on
    sentiment, while retail sales
    posted a larger-than-expected
    decline in August." + ], + [ + " SINGAPORE (Reuters) - Oil
    prices climbed above \\$42 a
    barrel on Wednesday, rising
    for the third day in a row as
    the heavy consuming U.S.
    Northeast feels the first
    chills of winter." + ], + [ + " NEW YORK (Reuters) - Merck
    Co Inc. <A HREF=\"http://www
    .investor.reuters.com/FullQuot
    e.aspx?ticker=MRK.N target=/st
    ocks/quickinfo/fullquote\">M
    RK.N</A> on Thursday
    pulled its arthritis drug
    Vioxx off the market after a
    study showed it doubled the
    risk of heart attack and
    stroke, a move that sent its
    shares plunging and erased
    \\$25 billion from its market
    value." + ], + [ + "The government on Wednesday
    defended its decision to
    radically revise the country
    #39;s deficit figures, ahead
    of a European Commission
    meeting to consider possible
    disciplinary action against
    Greece for submitting faulty
    figures." + ], + [ + "OPEC ministers yesterday
    agreed to increase their
    ceiling for oil production to
    help bring down stubbornly
    high prices in a decision that
    traders and analysts dismissed
    as symbolic because the cartel
    already is pumping more than
    its new target." + ], + [ + "Williams-Sonoma Inc. said
    second- quarter profit rose 55
    percent, boosted by the
    addition of Pottery Barn
    stores and sale of outdoor
    furniture." + ], + [ + " HONG KONG/SAN FRANCISCO
    (Reuters) - China's largest
    personal computer maker,
    Lenovo Group Ltd., said on
    Tuesday it was in acquisition
    talks with a major technology
    company, which a source
    familiar with the situation
    said was IBM." + ], + [ + "Reuters - Oil prices stayed
    close to #36;49 a\\barrel on
    Thursday, supported by a
    forecast for an early
    cold\\snap in the United States
    that could put a strain on a
    thin\\supply cushion of winter
    heating fuel." + ], + [ + "WASHINGTON (CBS.MW) --
    President Bush announced
    Monday that Kellogg chief
    executive Carlos Gutierrez
    would replace Don Evans as
    Commerce secretary, naming the
    first of many expected changes
    to his economic team." + ], + [ + "Iron Mountain moved further
    into the backup and recovery
    space Tuesday with the
    acquisition of Connected Corp.
    for \\$117 million. Connected
    backs up desktop data for more
    than 600 corporations, with
    more than" + ], + [ + "Montgomery County (website -
    news) is a big step closer to
    shopping for prescription
    drugs north of the border. On
    a 7-2 vote, the County Council
    is approving a plan that would
    give county" + ], + [ + "The disclosure this week that
    a Singapore-listed company
    controlled by a Chinese state-
    owned enterprise lost \\$550
    million in derivatives
    transactions" + ], + [ + "VANCOUVER - A Vancouver-based
    firm won #39;t sell 1.2
    million doses of influenza
    vaccine to the United States
    after all, announcing Tuesday
    that it will sell the doses
    within Canada instead." + ], + [ + "consortium led by the Sony
    Corporation of America reached
    a tentative agreement today to
    buy Metro-Goldwyn-Mayer, the
    Hollywood studio famous for
    James Bond and the Pink
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    chain has agreed to pay \\$8.7
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    it segregated black customers,
    subjected them to racial slurs
    and gave black workers
    inferior jobs." + ], + [ + " SINGAPORE (Reuters) -
    Investors bought shares in
    Asian exporters and
    electronics firms such as
    Fujitsu Ltd. on Tuesday,
    buoyed by a favorable outlook
    from U.S. technology
    bellwethers and a slide in oil
    prices." + ], + [ + "Ace Ltd. will stop paying
    brokers for steering business
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    company to make concessions in
    the five days since New York
    Attorney General Eliot Spitzer
    unveiled a probe of the
    insurance industry." + ], + [ + "Wm. Wrigley Jr. Co., the world
    #39;s largest maker of chewing
    gum, agreed to buy candy
    businesses including Altoids
    mints and Life Savers from
    Kraft Foods Inc." + ], + [ + " #39;Reaching a preliminary
    pilot agreement is the single
    most important hurdle they
    have to clear, but certainly
    not the only one." + ], + [ + "TORONTO (CP) - Canada #39;s
    big banks are increasing
    mortgage rates following a
    decision by the Bank of Canada
    to raise its overnight rate by
    one-quarter of a percentage
    point to 2.25 per cent." + ], + [ + "OTTAWA (CP) - The economy
    created another 43,000 jobs
    last month, pushing the
    unemployment rate down to 7.1
    per cent from 7.2 per cent in
    August, Statistics Canada said
    Friday." + ], + [ + " TOKYO (Reuters) - Japan's
    Nikkei average rose 0.39
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    gains in stocks dependent on
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    DaimlerChrysler and General
    Motors will jointly develop
    new hybrid motors to compete
    against Japanese rivals on
    the fuel-saving technology
    that reduces harmful
    emissions, the companies said
    on Monday." + ], + [ + "The bass should be in your
    face. That's what Matt Kelly,
    of Boston's popular punk rock
    band Dropkick Murphys, thinks
    is the mark of a great stereo
    system. And he should know.
    Kelly, 29, is the drummer for
    the band that likes to think
    of itself as a bit of an Irish
    lucky charm for the Red Sox." + ], + [ + "Slumping corporate spending
    and exports caused the economy
    to slow to a crawl in the
    July-September period, with
    real gross domestic product
    expanding just 0.1 percent
    from the previous quarter,
    Cabinet Office data showed
    Friday." + ], + [ + "US President George W. Bush
    signed into law a bill
    replacing an export tax
    subsidy that violated
    international trade rules with
    a \\$145 billion package of new
    corporate tax cuts and a
    buyout for tobacco farmers." + ], + [ + "The Nikkei average was up 0.37
    percent in mid-morning trade
    on Thursday as a recovery in
    the dollar helped auto makers
    among other exporters, but
    trade was slow as investors
    waited for important Japanese
    economic data." + ], + [ + "Jennifer Canada knew she was
    entering a boy's club when she
    enrolled in Southern Methodist
    University's Guildhall school
    of video-game making." + ], + [ + "SYDNEY (Dow Jones)--Colorado
    Group Ltd. (CDO.AU), an
    Australian footwear and
    clothing retailer, said Monday
    it expects net profit for the
    fiscal year ending Jan. 29 to
    be over 30 higher than that of
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    that a tiny nation like
    Antigua and Barbuda could take
    on the United States in an
    international dispute and win?" + ], + [ + "SBC Communications expects to
    cut 10,000 or more jobs by the
    end of next year through
    layoffs and attrition. That
    #39;s about six percent of the
    San Antonio-based company
    #39;s work force." + ], + [ + "Another United Airlines union
    is seeking to oust senior
    management at the troubled
    airline, saying its strategies
    are reckless and incompetent." + ], + [ + "Global oil prices boomed on
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    energy prices will restrain
    economic activity, as traders
    worried about a heating oil
    supply crunch in the American
    winter." + ], + [ + "Custom-designed imported
    furniture was once an
    exclusive realm. Now, it's the
    economical alternative for
    commercial developers and
    designers needing everything
    from seats to beds to desks
    for their projects." + ], + [ + "Short-term interest rate
    futures struggled on Thursday
    after a government report
    showing US core inflation for
    August below market
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    views on Federal Reserve rate
    policy." + ], + [ + "Samsung is now the world #39;s
    second-largest mobile phone
    maker, behind Nokia. According
    to market watcher Gartner, the
    South Korean company has
    finally knocked Motorola into
    third place." + ], + [ + " TOKYO (Reuters) - Tokyo
    stocks climbed to a two week
    high on Friday after Tokyo
    Electron Ltd. and other chip-
    related stocks were boosted
    by a bullish revenue outlook
    from industry leader Intel
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    in quarterly profit, driven by
    robust demand for its
    Photoshop and document-sharing
    software." + ], + [ + "Dow Jones amp; Co., publisher
    of The Wall Street Journal,
    has agreed to buy online
    financial news provider
    MarketWatch Inc. for about
    \\$463 million in a bid to
    boost its revenue from the
    fast-growing Internet
    advertising market." + ], + [ + "Low-fare airline ATA has
    announced plans to lay off
    hundreds of employees and to
    drop most of its flights out
    of Midway Airport in Chicago." + ], + [ + " WASHINGTON (Reuters) -
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    and pay a \\$4.7 million fine
    for taking part in a
    conspiracy to fix the prices
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    price, Russia #39;s No. 1 oil
    company acquired a savvy
    partner, and Putin polished
    Russia #39;s image." + ], + [ + "The maker of Hostess Twinkies,
    a cake bar and a piece of
    Americana children have
    snacked on for almost 75
    years, yesterday raised
    concerns about the company
    #39;s ability to stay in
    business." + ], + [ + "Tenet Healthcare Corp., the
    second- largest US hospital
    chain, said fourth-quarter
    charges may exceed \\$1 billion
    and its loss from continuing
    operations will widen from the
    third quarter #39;s because of
    increased bad debt." + ], + [ + "The London-based brokerage
    Collins Stewart Tullett placed
    55m of new shares yesterday to
    help fund the 69.5m purchase
    of the money and futures
    broker Prebon." + ], + [ + " PARIS (Reuters) - European
    equities flirted with 5-month
    peaks as hopes that economic
    growth was sustainable and a
    small dip in oil prices
    helped lure investors back to
    recent underperformers such
    as technology and insurance
    stocks." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks looked to open higher
    on Friday, as the fourth
    quarter begins on Wall Street
    with oil prices holding below
    \\$50 a barrel." + ], + [ + "LONDON : World oil prices
    stormed above 54 US dollars
    for the first time Tuesday as
    strikes in Nigeria and Norway
    raised worries about possible
    supply shortages during the
    northern hemisphere winter." + ], + [ + "US stocks were little changed
    on Thursday as an upbeat
    earnings report from chip
    maker National Semiconductor
    Corp. (NSM) sparked some
    buying, but higher oil prices
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+ ], + "title": { + "text": "Component 1" + } + } + } + } + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "# get embeddings for all article descriptions\n", + "embeddings = [embedding_from_string(string) for string in article_descriptions]\n", + "# compress the 2048-dimensional embeddings into 2 dimensions using t-SNE\n", + "tsne_components = tsne_components_from_embeddings(embeddings)\n", + "# get the article labels for coloring the chart\n", + "labels = df[\"label\"].tolist()\n", + "\n", + "chart_from_components(\n", + " components=tsne_components,\n", + " labels=labels,\n", + " strings=article_descriptions,\n", + " width=600,\n", + " height=500,\n", + " title=\"t-SNE components of article descriptions\",\n", + ")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "As you can see in the chart above, even the highly compressed embeddings do a good job of clustering article descriptions by category. And it's worth emphasizing: this clustering is done with no knowledge of the labels themselves!\n", + "\n", + "Also, if you look closely at the most egregious outliers, they are often due to mislabeling rather than poor embedding. For example, the majority of the blue World points in the green Sports cluster appear to be Sports stories." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Next, let's recolor the points by whether they are a source article, its nearest neighbors, or other." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": {}, + "outputs": [], + "source": [ + "# create labels for the recommended articles\n", + "def nearest_neighbor_labels(\n", + " list_of_indices: list[int],\n", + " k_nearest_neighbors: int = 5\n", + ") -> list[str]:\n", + " \"\"\"Return a list of labels to color the k nearest neighbors.\"\"\"\n", + " labels = [\"Other\" for _ in list_of_indices]\n", + " source_index = list_of_indices[0]\n", + " labels[source_index] = \"Source\"\n", + " for i in range(k_nearest_neighbors):\n", + " nearest_neighbor_index = list_of_indices[i + 1]\n", + " labels[nearest_neighbor_index] = f\"Nearest neighbor (top {k_nearest_neighbors})\"\n", + " return labels\n", + "\n", + "\n", + "tony_blair_labels = nearest_neighbor_labels(tony_blair_articles, k_nearest_neighbors=5)\n", + "chipset_security_labels = nearest_neighbor_labels(chipset_security_articles, k_nearest_neighbors=5\n", + ")" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": {}, + "outputs": [ + { + "data": { + "application/vnd.plotly.v1+json": { + "config": { + "plotlyServerURL": "https://plot.ly" + }, + "data": [ + { + "customdata": [ + [ + "PC World - Upcoming chip set
    will include built-in security
    features for your PC." + ], + [ + "Newspapers in Greece reflect a
    mixture of exhilaration that
    the Athens Olympics proved
    successful, and relief that
    they passed off without any
    major setback." + ], + [ + "SAN JOSE, Calif. -- Apple
    Computer (Quote, Chart)
    unveiled a batch of new iPods,
    iTunes software and promos
    designed to keep it atop the
    heap of digital music players." + ], + [ + "Any product, any shape, any
    size -- manufactured on your
    desktop! The future is the
    fabricator. By Bruce Sterling
    from Wired magazine." + ], + [ + "KABUL, Sept 22 (AFP): Three US
    soldiers were killed and 14
    wounded in a series of fierce
    clashes with suspected Taliban
    fighters in south and eastern
    Afghanistan this week, the US
    military said Wednesday." + ], + [ + "The EU and US moved closer to
    an aviation trade war after
    failing to reach agreement
    during talks Thursday on
    subsidies to Airbus Industrie." + ], + [ + "AUSTRALIAN journalist John
    Martinkus is lucky to be alive
    after spending 24 hours in the
    hands of Iraqi militants at
    the weekend. Martinkus was in
    Baghdad working for the SBS
    Dateline TV current affairs
    program" + ], + [ + " GAZA (Reuters) - An Israeli
    helicopter fired a missile
    into a town in the southern
    Gaza Strip late on Wednesday,
    witnesses said, hours after a
    Palestinian suicide bomber
    blew herself up in Jerusalem,
    killing two Israeli border
    policemen." + ], + [ + "The Microsoft CEO says one way
    to stem growing piracy of
    Windows and Office in emerging
    markets is to offer low-cost
    computers." + ], + [ + "RIYADH, Saudi Arabia -- Saudi
    police are seeking two young
    men in the killing of a Briton
    in a Riyadh parking lot, the
    Interior Ministry said today,
    and the British ambassador
    called it a terrorist attack." + ], + [ + "Loudon, NH -- As the rain
    began falling late Friday
    afternoon at New Hampshire
    International Speedway, the
    rich in the Nextel Cup garage
    got richer." + ], + [ + "PalmSource surprised the
    mobile vendor community today
    with the announcement that it
    will acquire China MobileSoft
    (CMS), ostensibly to leverage
    that company's expertise in
    building a mobile version of
    the Linux operating system." + ], + [ + "JEFFERSON CITY - An election
    security expert has raised
    questions about Missouri
    Secretary of State Matt Blunt
    #39;s plan to let soldiers at
    remote duty stations or in
    combat areas cast their
    ballots with the help of
    e-mail." + ], + [ + "A gas explosion at a coal mine
    in northern China killed 33
    workers in the 10th deadly
    mine blast reported in three
    months. The explosion occurred
    yesterday at 4:20 pm at Nanlou
    township" + ], + [ + "Reuters - Palestinian leader
    Mahmoud Abbas called\\Israel
    \"the Zionist enemy\" Tuesday,
    unprecedented language for\\the
    relative moderate who is
    expected to succeed Yasser
    Arafat." + ], + [ + "AP - Ottawa Senators right
    wing Marian Hossa is switching
    European teams during the NHL
    lockout." + ], + [ + "A new, optional log-on service
    from America Online that makes
    it harder for scammers to
    access a persons online
    account will not be available
    for Macintosh" + ], + [ + "Nasser al-Qidwa, Palestinian
    representative at the United
    Nations and nephew of late
    leader Yasser Arafat, handed
    Arafat #39;s death report to
    the Palestinian National
    Authority (PNA) on Saturday." + ], + [ + "CAIRO, Egypt - France's
    foreign minister appealed
    Monday for the release of two
    French journalists abducted in
    Baghdad, saying the French
    respect all religions. He did
    not rule out traveling to
    Baghdad..." + ], + [ + "Chelsea terminated Romania
    striker Adrian Mutu #39;s
    contract, citing gross
    misconduct after the player
    failed a doping test for
    cocaine and admitted taking
    the drug, the English soccer
    club said in a statement." + ], + [ + "United Arab Emirates President
    and ruler of Abu Dhabi Sheik
    Zayed bin Sultan al-Nayhan
    died Tuesday, official
    television reports. He was 86." + ], + [ + "PALESTINIAN leader Yasser
    Arafat today issued an urgent
    call for the immediate release
    of two French journalists
    taken hostage in Iraq." + ], + [ + "The al-Qaida terrorist network
    spent less than \\$50,000 on
    each of its major attacks
    except for the Sept. 11, 2001,
    suicide hijackings, and one of
    its hallmarks is using" + ], + [ + " SAN FRANCISCO (Reuters) -
    Nike Inc. <A HREF=\"http://w
    ww.investor.reuters.com/FullQu
    ote.aspx?ticker=NKE.N target=/
    stocks/quickinfo/fullquote\">
    ;NKE.N</A>, the world's
    largest athletic shoe company,
    on Monday reported a 25
    percent rise in quarterly
    profit, beating analysts'
    estimates, on strong demand
    for high-end running and
    basketball shoes in the
    United States." + ], + [ + "A FATHER who scaled the walls
    of a Cardiff court dressed as
    superhero Robin said the
    Buckingham Palace protester
    posed no threat. Fathers 4
    Justice activist Jim Gibson,
    who earlier this year staged
    an eye-catching" + ], + [ + "NEW YORK (CBS.MW) - US stocks
    traded mixed Thurday as Merck
    shares lost a quarter of their
    value, dragging blue chips
    lower, but the Nasdaq moved
    higher, buoyed by gains in the
    semiconductor sector." + ], + [ + "Julia Gillard has reportedly
    bowed out of the race to
    become shadow treasurer,
    taking enormous pressure off
    Opposition Leader Mark Latham." + ], + [ + "It #39;s official. Microsoft
    recently stated
    definitivelyand contrary to
    rumorsthat there will be no
    new versions of Internet
    Explorer for users of older
    versions of Windows." + ], + [ + "The success of Paris #39; bid
    for Olympic Games 2012 would
    bring an exceptional
    development for France for at
    least 6 years, said Jean-
    Francois Lamour, French
    minister for Youth and Sports
    on Tuesday." + ], + [ + "AFP - Maybe it's something to
    do with the fact that the
    playing area is so vast that
    you need a good pair of
    binoculars to see the action
    if it's not taking place right
    in front of the stands." + ], + [ + "Egypt #39;s release of accused
    Israeli spy Azzam Azzam in an
    apparent swap for six Egyptian
    students held on suspicion of
    terrorism is expected to melt
    the ice and perhaps result" + ], + [ + "But fresh antitrust suit is in
    the envelope, says Novell" + ], + [ + "Chips that help a computer's
    main microprocessors perform
    specific types of math
    problems are becoming a big
    business once again.\\" + ], + [ + "GAZA CITY, Gaza Strip: Hamas
    militants killed an Israeli
    soldier and wounded four with
    an explosion in a booby-
    trapped chicken coop on
    Tuesday, in what the Islamic
    group said was an elaborate
    scheme to lure troops to the
    area with the help of a double" + ], + [ + "A rocket carrying two Russian
    cosmonauts and an American
    astronaut to the international
    space station streaked into
    orbit on Thursday, the latest
    flight of a Russian space
    vehicle to fill in for
    grounded US shuttles." + ], + [ + "Bankrupt ATA Airlines #39;
    withdrawal from Chicago #39;s
    Midway International Airport
    presents a juicy opportunity
    for another airline to beef up
    service in the Midwest" + ], + [ + "AP - The 300 men filling out
    forms in the offices of an
    Iranian aid group were offered
    three choices: Train for
    suicide attacks against U.S.
    troops in Iraq, for suicide
    attacks against Israelis or to
    assassinate British author
    Salman Rushdie." + ], + [ + "ATHENS, Greece - Gail Devers,
    the most talented yet star-
    crossed hurdler of her
    generation, was unable to
    complete even one hurdle in
    100-meter event Sunday -
    failing once again to win an
    Olympic hurdling medal.
    Devers, 37, who has three
    world championships in the
    hurdles but has always flopped
    at the Olympics, pulled up
    short and screamed as she slid
    under the first hurdle..." + ], + [ + "OCTOBER 12, 2004
    (COMPUTERWORLD) - Microsoft
    Corp. #39;s monthly patch
    release cycle may be making it
    more predictable for users to
    deploy software updates, but
    there appears to be little
    letup in the number of holes
    being discovered in the
    company #39;s software" + ], + [ + "Wearable tech goes mainstream
    as the Gap introduces jacket
    with built-in radio.\\" + ], + [ + "Madison, WI (Sports Network) -
    Anthony Davis ran for 122
    yards and two touchdowns to
    lead No. 6 Wisconsin over
    Northwestern, 24-12, to
    celebrate Homecoming weekend
    at Camp Randall Stadium." + ], + [ + "LaVar Arrington participated
    in his first practice since
    Oct. 25, when he aggravated a
    knee injury that sidelined him
    for 10 games." + ], + [ + " NEW YORK (Reuters) - Billie
    Jean King cut her final tie
    with the U.S. Fed Cup team
    Tuesday when she retired as
    coach." + ], + [ + "The Instinet Group, owner of
    one of the largest electronic
    stock trading systems, is for
    sale, executives briefed on
    the plan say." + ], + [ + "Funding round of \\$105 million
    brings broadband Internet
    telephony company's total haul
    to \\$208 million." + ], + [ + "The Miami Dolphins arrived for
    their final exhibition game
    last night in New Orleans
    short nine players." + ], + [ + "washingtonpost.com - Anthony
    Casalena was 17 when he got
    his first job as a programmer
    for a start-up called
    HyperOffice, a software
    company that makes e-mail and
    contact management
    applications for the Web.
    Hired as an intern, he became
    a regular programmer after two
    weeks and rewrote the main
    product line." + ], + [ + "The Auburn Hills-based
    Chrysler Group made a profit
    of \\$269 million in the third
    quarter, even though worldwide
    sales and revenues declined,
    contributing to a \\$1." + ], + [ + "SAN FRANCISCO (CBS.MW) -- UAL
    Corp., parent of United
    Airlines, said Wednesday it
    will overhaul its route
    structure to reduce costs and
    offset rising fuel costs." + ], + [ + " NAIROBI (Reuters) - The
    Sudanese government and its
    southern rebel opponents have
    agreed to sign a pledge in the
    Kenyan capital on Friday to
    formally end a brutal 21-year-
    old civil war, with U.N.
    Security Council ambassadors
    as witnesses." + ], + [ + "AP - From LSU at No. 1 to Ohio
    State at No. 10, The AP
    women's basketball poll had no
    changes Monday." + ], + [ + "nother stage victory for race
    leader Petter Solberg, his
    fifth since the start of the
    rally. The Subaru driver is
    not pulling away at a fast
    pace from Gronholm and Loeb
    but the gap is nonetheless
    increasing steadily." + ], + [ + "The fossilized neck bones of a
    230-million-year-old sea
    creature have features
    suggesting that the animal
    #39;s snakelike throat could
    flare open and create suction
    that would pull in prey." + ], + [ + "IBM late on Tuesday announced
    the biggest update of its
    popular WebSphere business
    software in two years, adding
    features such as automatically
    detecting and fixing problems." + ], + [ + "April 1980 -- Players strike
    the last eight days of spring
    training. Ninety-two
    exhibition games are canceled.
    June 1981 -- Players stage
    first midseason strike in
    history." + ], + [ + "AP - Former Guatemalan
    President Alfonso Portillo
    #151; suspected of corruption
    at home #151; is living and
    working part-time in the same
    Mexican city he fled two
    decades ago to avoid arrest on
    murder charges, his close
    associates told The Associated
    Press on Sunday." + ], + [ + "AP - British entrepreneur
    Richard Branson said Monday
    that his company plans to
    launch commercial space
    flights over the next few
    years." + ], + [ + "Annual economic growth in
    China has slowed for the third
    quarter in a row, falling to
    9.1 per cent, third-quarter
    data shows. The slowdown shows
    the economy is responding to
    Beijing #39;s efforts to rein
    in break-neck investment and
    lending." + ], + [ + "washingtonpost.com - BRUSSELS,
    Aug. 26 -- The United States
    will have to wait until next
    year to see its fight with the
    European Union over biotech
    foods resolved, as the World
    Trade Organization agreed to
    an E.U. request to bring
    scientists into the debate,
    officials said Thursday." + ], + [ + "A group of Internet security
    and instant messaging
    providers have teamed up to
    detect and thwart the growing
    threat of IM and peer-to-peer
    viruses and worms, they said
    this week." + ], + [ + "On Sunday, the day after Ohio
    State dropped to 0-3 in the
    Big Ten with a 33-7 loss at
    Iowa, the Columbus Dispatch
    ran a single word above its
    game story on the Buckeyes:
    quot;Embarrassing." + ], + [ + "Insisting that Hurriyat
    Conference is the real
    representative of Kashmiris,
    Pakistan has claimed that
    India is not ready to accept
    ground realities in Kashmir." + ], + [ + "THE All-India Motor Transport
    Congress (AIMTC) on Saturday
    called off its seven-day
    strike after finalising an
    agreement with the Government
    on the contentious issue of
    service tax and the various
    demands including tax deducted
    at source (TDS), scrapping" + ], + [ + "BOSTON - Curt Schilling got
    his 20th win on the eve of
    Boston #39;s big series with
    the New York Yankees. Now he
    wants much more. quot;In a
    couple of weeks, hopefully, it
    will get a lot better, quot;
    he said after becoming" + ], + [ + "Shooed the ghosts of the
    Bambino and the Iron Horse and
    the Yankee Clipper and the
    Mighty Mick, on his 73rd
    birthday, no less, and turned
    Yankee Stadium into a morgue." + ], + [ + "Gold medal-winning Marlon
    Devonish says the men #39;s
    4x100m Olympic relay triumph
    puts British sprinting back on
    the map. Devonish, Darren
    Campbell, Jason Gardener and
    Mark Lewis-Francis edged out
    the American" + ], + [ + "AP - The euro-zone economy
    grew by 0.5 percent in the
    second quarter of 2004, a
    touch slower than in the first
    three months of the year,
    according to initial estimates
    released Tuesday by the
    European Union." + ], + [ + "His first space flight was in
    1965 when he piloted the first
    manned Gemini mission. Later
    he made two trips to the moon
    -- orbiting during a 1969
    flight and then walking on the
    lunar surface during a mission
    in 1972." + ], + [ + "he difficult road conditions
    created a few incidents in the
    first run of the Epynt stage.
    Francois Duval takes his
    second stage victory since the
    start of the rally, nine
    tenths better than Sebastien
    Loeb #39;s performance in
    second position." + ], + [ + "VIENNA -- After two years of
    investigating Iran's atomic
    program, the UN nuclear
    watchdog still cannot rule out
    that Tehran has a secret atom
    bomb project as Washington
    insists, the agency's chief
    said yesterday." + ], + [ + "By RACHEL KONRAD SAN
    FRANCISCO (AP) -- EBay Inc.,
    which has been aggressively
    expanding in Asia, plans to
    increase its stake in South
    Korea's largest online auction
    company. The Internet
    auction giant said Tuesday
    night that it would purchase
    nearly 3 million publicly held
    shares of Internet Auction
    Co..." + ], + [ + "AFP - US Secretary of State
    Colin Powell wrapped up a
    three-nation tour of Asia
    after winning pledges from
    Japan, China and South Korea
    to press North Korea to resume
    stalled talks on its nuclear
    weapons programs." + ], + [ + "Tallahassee, FL (Sports
    Network) - Florida State head
    coach Bobby Bowden has
    suspended senior wide receiver
    Craphonso Thorpe for the
    Seminoles #39; game against
    fellow Atlantic Coast
    Conference member Duke on
    Saturday." + ], + [ + "A few years ago, casinos
    across the United States were
    closing their poker rooms to
    make space for more popular
    and lucrative slot machines." + ], + [ + "CAIRO, Egypt An Egyptian
    company says one of its four
    workers who had been kidnapped
    in Iraq has been freed. It
    says it can #39;t give the
    status of the others being
    held hostage but says it is
    quot;doing its best to secure
    quot; their release." + ], + [ + "WASHINGTON -- Consumers were
    tightfisted amid soaring
    gasoline costs in August and
    hurricane-related disruptions
    last week sent applications
    for jobless benefits to their
    highest level in seven months." + ], + [ + "Talking kitchens and vanities.
    Musical jump ropes and potty
    seats. Blusterous miniature
    leaf blowers and vacuum
    cleaners -- almost as loud as
    the real things." + ], + [ + "Online merchants in the United
    States have become better at
    weeding out fraudulent credit
    card orders, a new survey
    indicates. But shipping
    overseas remains a risky
    venture. By Joanna Glasner." + ], + [ + "Former champion Lleyton Hewitt
    bristled, battled and
    eventually blossomed as he
    took another step towards a
    second US Open title with a
    second-round victory over
    Moroccan Hicham Arazi on
    Friday." + ], + [ + "AP - China's biggest computer
    maker, Lenovo Group, said
    Wednesday it has acquired a
    majority stake in
    International Business
    Machines Corp.'s personal
    computer business for
    #36;1.75 billion, one of the
    biggest Chinese overseas
    acquisitions ever." + ], + [ + "Popping a gadget into a cradle
    to recharge it used to mean
    downtime, but these days
    chargers are doing double
    duty, keeping your portable
    devices playing even when
    they're charging." + ], + [ + "AFP - Hosts India braced
    themselves for a harrowing
    chase on a wearing wicket in
    the first Test after Australia
    declined to enforce the
    follow-on here." + ], + [ + " SAN FRANCISCO (Reuters) -
    Texas Instruments Inc. <A H
    REF=\"http://www.investor.reute
    rs.com/FullQuote.aspx?ticker=T
    XN.N target=/stocks/quickinfo/
    fullquote\">TXN.N</A>,
    the largest maker of chips for
    cellular phones, on Monday
    said record demand for its
    handset and television chips
    boosted quarterly profit by
    26 percent, even as it
    struggles with a nagging
    inventory problem." + ], + [ + "LONDON ARM Holdings, a British
    semiconductor designer, said
    on Monday that it would buy
    Artisan Components for \\$913
    million to broaden its product
    range." + ], + [ + "Big evolutionary insights
    sometimes come in little
    packages. Witness the
    startling discovery, in a cave
    on the eastern Indonesian
    island of Flores, of the
    partial skeleton of a half-
    size Homo species that" + ], + [ + "Prime Minister Paul Martin of
    Canada urged Haitian leaders
    on Sunday to allow the
    political party of the deposed
    president, Jean-Bertrand
    Aristide, to take part in new
    elections." + ], + [ + "Hostage takers holding up to
    240 people at a school in
    southern Russia have refused
    to talk with a top Islamic
    leader and demanded to meet
    with regional leaders instead,
    ITAR-TASS reported on
    Wednesday." + ], + [ + "As the Mets round out their
    search for a new manager, the
    club is giving a last-minute
    nod to its past. Wally
    Backman, an infielder for the
    Mets from 1980-88 who played
    second base on the 1986" + ], + [ + "MELBOURNE: Global shopping
    mall owner Westfield Group
    will team up with Australian
    developer Multiplex Group to
    bid 585mil (US\\$1." + ], + [ + "Three children from a care
    home are missing on the
    Lancashire moors after they
    are separated from a group." + ], + [ + "Luke Skywalker and Darth Vader
    may get all the glory, but a
    new Star Wars video game
    finally gives credit to the
    everyday grunts who couldn
    #39;t summon the Force for
    help." + ], + [ + "AMSTERDAM, Aug 18 (Reuters) -
    Midfielder Edgar Davids #39;s
    leadership qualities and
    never-say-die attitude have
    earned him the captaincy of
    the Netherlands under new
    coach Marco van Basten." + ], + [ + "COUNTY KILKENNY, Ireland (PA)
    -- Hurricane Jeanne has led to
    world No. 1 Vijay Singh
    pulling out of this week #39;s
    WGC-American Express
    Championship at Mount Juliet." + ], + [ + "Green Bay, WI (Sports Network)
    - The Green Bay Packers will
    be without the services of Pro
    Bowl center Mike Flanagan for
    the remainder of the season,
    as he will undergo left knee
    surgery." + ], + [ + "Diabetics should test their
    blood sugar levels more
    regularly to reduce the risk
    of cardiovascular disease, a
    study says." + ], + [ + "COLUMBUS, Ohio -- NCAA
    investigators will return to
    Ohio State University Monday
    to take another look at the
    football program after the
    latest round of allegations
    made by former players,
    according to the Akron Beacon
    Journal." + ], + [ + " LOS ANGELES (Reuters) -
    Federal authorities raided
    three Washington, D.C.-area
    video game stores and arrested
    two people for modifying
    video game consoles to play
    pirated video games, a video
    game industry group said on
    Wednesday." + ], + [ + "Manchester United gave Alex
    Ferguson a 1,000th game
    anniversary present by
    reaching the last 16 of the
    Champions League yesterday,
    while four-time winners Bayern
    Munich romped into the second
    round with a 5-1 beating of
    Maccabi Tel Aviv." + ], + [ + "Iraq's interim Prime Minister
    Ayad Allawi announced that
    proceedings would begin
    against former Baath Party
    leaders." + ], + [ + "Reuters - Delta Air Lines Inc.
    , which is\\racing to cut costs
    to avoid bankruptcy, on
    Wednesday reported\\a wider
    quarterly loss amid soaring
    fuel prices and weak\\domestic
    airfares." + ], + [ + "Energy utility TXU Corp. on
    Monday more than quadrupled
    its quarterly dividend payment
    and raised its projected
    earnings for 2004 and 2005
    after a companywide
    performance review." + ], + [ + "Northwest Airlines Corp., the
    fourth- largest US airline,
    and its pilots union reached
    tentative agreement on a
    contract that would cut pay
    and benefits, saving the
    company \\$265 million a year." + ], + [ + "The last time the Angels
    played the Texas Rangers, they
    dropped two consecutive
    shutouts at home in their most
    agonizing lost weekend of the
    season." + ], + [ + "Microsoft Corp. MSFT.O and
    cable television provider
    Comcast Corp. CMCSA.O said on
    Monday that they would begin
    deploying set-top boxes
    powered by Microsoft software
    starting next week." + ], + [ + "SEATTLE - Chasing a nearly
    forgotten ghost of the game,
    Ichiro Suzuki broke one of
    baseball #39;s oldest records
    Friday night, smoking a single
    up the middle for his 258th
    hit of the year and breaking
    George Sisler #39;s record for
    the most hits in a season" + ], + [ + "Grace Park, her caddie - and
    fans - were poking around in
    the desert brush alongside the
    18th fairway desperately
    looking for her ball." + ], + [ + "Philippines mobile phone
    operator Smart Communications
    Inc. is in talks with
    Singapore #39;s Mobile One for
    a possible tie-up,
    BusinessWorld reported Monday." + ], + [ + "Washington Redskins kicker
    John Hall strained his right
    groin during practice
    Thursday, his third leg injury
    of the season. Hall will be
    held out of practice Friday
    and is questionable for Sunday
    #39;s game against the Chicago
    Bears." + ], + [ + "Airline warns it may file for
    bankruptcy if too many senior
    pilots take early retirement
    option. Delta Air LInes #39;
    CEO says it faces bankruptcy
    if it can #39;t slow the pace
    of pilots taking early
    retirement." + ], + [ + "A toxic batch of home-brewed
    alcohol has killed 31 people
    in several towns in central
    Pakistan, police and hospital
    officials say." + ], + [ + "Thornbugs communicate by
    vibrating the branches they
    live on. Now scientists are
    discovering just what the bugs
    are \"saying.\"" + ], + [ + "The Florida Gators and
    Arkansas Razorbacks meet for
    just the sixth time Saturday.
    The Gators hold a 4-1
    advantage in the previous five
    meetings, including last year
    #39;s 33-28 win." + ], + [ + "Kodak Versamark #39;s parent
    company, Eastman Kodak Co.,
    reported Tuesday it plans to
    eliminate almost 900 jobs this
    year in a restructuring of its
    overseas operations." + ], + [ + "A top official of the US Food
    and Drug Administration said
    Friday that he and his
    colleagues quot;categorically
    reject quot; earlier
    Congressional testimony that
    the agency has failed to
    protect the public against
    dangerous drugs." + ], + [ + " BEIJING (Reuters) - North
    Korea is committed to holding
    six-party talks aimed at
    resolving the crisis over its
    nuclear weapons program, but
    has not indicated when, a top
    British official said on
    Tuesday." + ], + [ + "AP - 1941 #151; Brooklyn
    catcher Mickey Owen dropped a
    third strike on Tommy Henrich
    of what would have been the
    last out of a Dodgers victory
    against the New York Yankees.
    Given the second chance, the
    Yankees scored four runs for a
    7-4 victory and a 3-1 lead in
    the World Series, which they
    ended up winning." + ], + [ + "Federal prosecutors cracked a
    global cartel that had
    illegally fixed prices of
    memory chips in personal
    computers and servers for
    three years." + ], + [ + "AP - Oracle Corp. CEO Larry
    Ellison reiterated his
    determination to prevail in a
    long-running takeover battle
    with rival business software
    maker PeopleSoft Inc.,
    predicting the proposed deal
    will create a more competitive
    company with improved customer
    service." + ], + [ + "Braves shortstop Rafael Furcal
    was arrested on drunken
    driving charges Friday, his
    second D.U.I. arrest in four
    years." + ], + [ + "AFP - British retailer Marks
    and Spencer announced a major
    management shake-up as part of
    efforts to revive its
    fortunes, saying trading has
    become tougher ahead of the
    crucial Christmas period." + ], + [ + " BAGHDAD (Reuters) - Iraq's
    interim government extended
    the closure of Baghdad
    international airport
    indefinitely on Saturday
    under emergency rule imposed
    ahead of this week's U.S.-led
    offensive on Falluja." + ], + [ + " ATHENS (Reuters) - Natalie
    Coughlin's run of bad luck
    finally took a turn for the
    better when she won the gold
    medal in the women's 100
    meters backstroke at the
    Athens Olympics Monday." + ], + [ + " ATLANTA (Reuters) - Home
    Depot Inc. <A HREF=\"http://
    www.investor.reuters.com/FullQ
    uote.aspx?ticker=HD.N target=/
    stocks/quickinfo/fullquote\">
    ;HD.N</A>, the top home
    improvement retailer, on
    Tuesday reported a 15 percent
    rise in third-quarter profit,
    topping estimates, aided by
    installed services and sales
    to contractors." + ], + [ + " LONDON (Reuters) - The dollar
    fought back from one-month
    lows against the euro and
    Swiss franc on Wednesday as
    investors viewed its sell-off
    in the wake of the Federal
    Reserve's verdict on interest
    rates as overdone." + ], + [ + "Rivaling Bush vs. Kerry for
    bitterness, doctors and trial
    lawyers are squaring off this
    fall in an unprecedented four-
    state struggle over limiting
    malpractice awards..." + ], + [ + "Boston Scientific Corp said on
    Friday it has recalled an ear
    implant the company acquired
    as part of its purchase of
    Advanced Bionics in June." + ], + [ + "AP - Pedro Feliz hit a
    tiebreaking grand slam with
    two outs in the eighth inning
    for his fourth hit of the day,
    and the Giants helped their
    playoff chances with a 9-5
    victory over the Los Angeles
    Dodgers on Saturday." + ], + [ + "MarketWatch.com. Richemont
    sees significant H1 lift,
    unclear on FY (2:53 AM ET)
    LONDON (CBS.MW) -- Swiss
    luxury goods maker
    Richemont(zz:ZZ:001273145:
    news, chart, profile), which
    also is a significant" + ], + [ + "Crude oil climbed more than
    \\$1 in New York on the re-
    election of President George
    W. Bush, who has been filling
    the US Strategic Petroleum
    Reserve." + ], + [ + "AP - Hundreds of tribesmen
    gathered Tuesday near the area
    where suspected al-Qaida-
    linked militants are holding
    two Chinese engineers and
    demanding safe passage to
    their reputed leader, a former
    U.S. prisoner from Guantanamo
    Bay, Cuba, officials and
    residents said." + ], + [ + "AP - Scientists warned Tuesday
    that a long-term increase in
    global temperature of 3.5
    degrees could threaten Latin
    American water supplies,
    reduce food yields in Asia and
    result in a rise in extreme
    weather conditions in the
    Caribbean." + ], + [ + "AP - Further proof New York's
    real estate market is
    inflated: The city plans to
    sell space on top of lampposts
    to wireless phone companies
    for #36;21.6 million a year." + ], + [ + "In an alarming development,
    high-precision equipment and
    materials which could be used
    for making nuclear bombs have
    disappeared from some Iraqi
    facilities, the United Nations
    watchdog agency has said." + ], + [ + "Yukos #39; largest oil-
    producing unit regained power
    supplies from Tyumenenergo, a
    Siberia-based electricity
    generator, Friday after the
    subsidiary pledged to pay 440
    million rubles (\\$15 million)
    in debt by Oct. 3." + ], + [ + "The rollout of new servers and
    networking switches in stores
    is part of a five-year
    agreement supporting 7-Eleven
    #39;s Retail Information
    System." + ], + [ + "Top Hollywood studios have
    launched a wave of court cases
    against internet users who
    illegally download film files.
    The Motion Picture Association
    of America, which acts as the
    representative of major film" + ], + [ + "AUSTRALIANS went into a
    television-buying frenzy the
    run-up to the Athens Olympics,
    suggesting that as a nation we
    could easily have scored a
    gold medal for TV purchasing." + ], + [ + "US STOCKS vacillated yesterday
    as rising oil prices muted
    Wall Streets excitement over
    strong results from Lehman
    Brothers and Sprints \\$35
    billion acquisition of Nextel
    Communications." + ], + [ + "The next time you are in your
    bedroom with your PC plus
    Webcam switched on, don #39;t
    think that your privacy is all
    intact. If you have a Webcam
    plugged into an infected
    computer, there is a
    possibility that" + ], + [ + "At the head of the class,
    Rosabeth Moss Kanter is an
    intellectual whirlwind: loud,
    expansive, in constant motion." + ], + [ + "LEVERKUSEN/ROME, Dec 7 (SW) -
    Dynamo Kiev, Bayer Leverkusen,
    and Real Madrid all have a
    good chance of qualifying for
    the Champions League Round of
    16 if they can get the right
    results in Group F on
    Wednesday night." + ], + [ + "Ed Hinkel made a diving,
    fingertip catch for a key
    touchdown and No. 16 Iowa
    stiffened on defense when it
    needed to most to beat Iowa
    State 17-10 Saturday." + ], + [ + "During last Sunday #39;s
    Nextel Cup race, amid the
    ongoing furor over Dale
    Earnhardt Jr. #39;s salty
    language, NBC ran a commercial
    for a show coming on later
    that night called quot;Law
    amp; Order: Criminal Intent." + ], + [ + "AP - After playing in hail,
    fog and chill, top-ranked
    Southern California finishes
    its season in familiar
    comfort. The Trojans (9-0, 6-0
    Pacific-10) have two games at
    home #151; against Arizona on
    Saturday and Notre Dame on
    Nov. 27 #151; before their
    rivalry game at UCLA." + ], + [ + "A US airman dies and two are
    hurt as a helicopter crashes
    due to technical problems in
    western Afghanistan." + ], + [ + "Jacques Chirac has ruled out
    any withdrawal of French
    troops from Ivory Coast,
    despite unrest and anti-French
    attacks, which have forced the
    evacuation of thousands of
    Westerners." + ], + [ + "Japanese Prime Minister
    Junichiro Koizumi reshuffled
    his cabinet yesterday,
    replacing several top
    ministers in an effort to
    boost his popularity,
    consolidate political support
    and quicken the pace of
    reforms in the world #39;s
    second-largest economy." + ], + [ + "The remnants of Hurricane
    Jeanne rained out Monday's
    game between the Mets and the
    Atlanta Braves. It will be
    made up Tuesday as part of a
    doubleheader." + ], + [ + "AP - NASCAR is not expecting
    any immediate changes to its
    top-tier racing series
    following the merger between
    telecommunications giant
    Sprint Corp. and Nextel
    Communications Inc." + ], + [ + "AP - Shawn Fanning's Napster
    software enabled countless
    music fans to swap songs on
    the Internet for free, turning
    him into the recording
    industry's enemy No. 1 in the
    process." + ], + [ + "TBILISI (Reuters) - At least
    two Georgian soldiers were
    killed and five wounded in
    artillery fire with
    separatists in the breakaway
    region of South Ossetia,
    Georgian officials said on
    Wednesday." + ], + [ + "Like wide-open races? You
    #39;ll love the Big 12 North.
    Here #39;s a quick morning
    line of the Big 12 North as it
    opens conference play this
    weekend." + ], + [ + "Reuters - Walt Disney Co.
    is\\increasing investment in
    video games for hand-held
    devices and\\plans to look for
    acquisitions of small game
    publishers and\\developers,
    Disney consumer products
    Chairman Andy Mooney said\\on
    Monday." + ], + [ + "Taquan Dean scored 22 points,
    Francisco Garcia added 19 and
    No. 13 Louisville withstood a
    late rally to beat Florida
    74-70 Saturday." + ], + [ + "BANGKOK - A UN conference last
    week banned commercial trade
    in the rare Irrawaddy dolphin,
    a move environmentalists said
    was needed to save the
    threatened species." + ], + [ + "Laksamana.Net - Two Indonesian
    female migrant workers freed
    by militants in Iraq are
    expected to arrive home within
    a day or two, the Foreign
    Affairs Ministry said
    Wednesday (6/10/04)." + ], + [ + "A bitter row between America
    and the European Union over
    alleged subsidies to rival
    aircraft makers Boeing and
    Airbus intensified yesterday." + ], + [ + "PC World - Updated antivirus
    software for businesses adds
    intrusion prevention features." + ], + [ + "NEW YORK -- This was all about
    Athens, about redemption for
    one and validation for the
    other. Britain's Paula
    Radcliffe, the fastest female
    marathoner in history, failed
    to finish either of her
    Olympic races last summer.
    South Africa's Hendrik Ramaala
    was a five-ringed dropout,
    too, reinforcing his
    reputation as a man who could
    go only half the distance." + ], + [ + "Reuters - Online media company
    Yahoo Inc.\\ late on Monday
    rolled out tests of redesigned
    start\\pages for its popular
    Yahoo.com and My.Yahoo.com
    sites." + ], + [ + "Amsterdam (pts) - Dutch
    electronics company Philips
    http://www.philips.com has
    announced today, Friday, that
    it has cut its stake in Atos
    Origin by more than a half." + ], + [ + "TORONTO (CP) - Two-thirds of
    banks around the world have
    reported an increase in the
    volume of suspicious
    activities that they report to
    police, a new report by KPMG
    suggests." + ], + [ + "The Sun may have captured
    thousands or even millions of
    asteroids from another
    planetary system during an
    encounter more than four
    billion years ago, astronomers
    are reporting." + ], + [ + "LONDON -- Ernie Els has set
    his sights on an improved
    putting display this week at
    the World Golf Championships
    #39; NEC Invitational in
    Akron, Ohio, after the
    disappointment of tying for
    fourth place at the PGA
    Championship last Sunday." + ], + [ + "The Atkins diet frenzy slowed
    growth briefly, but the
    sandwich business is booming,
    with \\$105 billion in sales
    last year." + ], + [ + "Luxury carmaker Jaguar said
    Friday it was stopping
    production at a factory in
    central England, resulting in
    a loss of 1,100 jobs,
    following poor sales in the
    key US market." + ], + [ + "A bus was hijacked today and
    shots were fired at police who
    surrounded it on the outskirts
    of Athens. Police did not know
    how many passengers were
    aboard the bus." + ], + [ + "Thumb through the book - then
    buy a clean copy from Amazon" + ], + [ + "AP - President Bashar Assad
    shuffled his Cabinet on
    Monday, just weeks after the
    United States and the United
    Nations challenged Syria over
    its military presence in
    Lebanon and the security
    situation along its border
    with Iraq." + ], + [ + "Fiji #39;s Vijay Singh
    replaced Tiger Woods as the
    world #39;s No.1 ranked golfer
    today by winning the PGA
    Deutsche Bank Championship." + ], + [ + "LEIPZIG, Germany : Jurgen
    Klinsmann enjoyed his first
    home win as German manager
    with his team defeating ten-
    man Cameroon 3-0 in a friendly
    match." + ], + [ + "AP - Kevin Brown had a chance
    to claim a place in Yankees
    postseason history with his
    start in Game 7 of the AL
    championship series." + ], + [ + "Reuters - High-definition
    television can\\show the sweat
    beading on an athlete's brow,
    but the cost of\\all the
    necessary electronic equipment
    can get a shopper's own\\pulse
    racing." + ], + [ + "HOMESTEAD, Fla. -- Kurt Busch
    got his first big break in
    NASCAR by winning a 1999
    talent audition nicknamed
    quot;The Gong Show. quot; He
    was selected from dozens of
    unknown, young race drivers to
    work for one of the sport
    #39;s most famous team owners,
    Jack Roush." + ], + [ + "AP - President Vladimir Putin
    has signed a bill confirming
    Russia's ratification of the
    Kyoto Protocol, the Kremlin
    said Friday, clearing the way
    for the global climate pact to
    come into force early next
    year." + ], + [ + "John Gibson said Friday that
    he decided to resign as chief
    executive officer of
    Halliburton Energy Services
    when it became apparent he
    couldn #39;t become the CEO of
    the entire corporation, after
    getting a taste of the No." + ], + [ + "MacCentral - Apple Computer
    Inc. on Monday released an
    update for Apple Remote
    Desktop (ARD), the company's
    software solution to assist
    Mac system administrators and
    computer managers with asset
    management, software
    distribution and help desk
    support. ARD 2.1 includes
    several enhancements and bug
    fixes." + ], + [ + "NEW YORK (Reuters) - Outback
    Steakhouse Inc. said Tuesday
    it lost about 130 operating
    days and up to \\$2 million in
    revenue because it had to
    close some restaurants in the
    South due to Hurricane
    Charley." + ], + [ + "State insurance commissioners
    from across the country have
    proposed new rules governing
    insurance brokerage fees,
    winning praise from an
    industry group and criticism
    from" + ], + [ + "AP - The authenticity of newly
    unearthed memos stating that
    George W. Bush failed to meet
    standards of the Texas Air
    National Guard during the
    Vietnam War was questioned
    Thursday by the son of the
    late officer who reportedly
    wrote the memos." + ], + [ + "Zurich, Switzerland (Sports
    Network) - Former world No. 1
    Venus Williams advanced on
    Thursday and will now meet
    Wimbledon champion Maria
    Sharapova in the quarterfinals
    at the \\$1." + ], + [ + "INDIA #39;S cricket chiefs
    began a frenetic search today
    for a broadcaster to show next
    month #39;s home series
    against world champion
    Australia after cancelling a
    controversial \\$US308 million
    (\\$440 million) television
    deal." + ], + [ + "Canadian Press - OAKVILLE,
    Ont. (CP) - The body of a
    missing autistic man was
    pulled from a creek Monday,
    just metres from where a key
    piece of evidence was
    uncovered but originally
    overlooked because searchers
    had the wrong information." + ], + [ + "SOFTWARE giant Oracle #39;s
    stalled \\$7.7bn (4.2bn) bid to
    take over competitor
    PeopleSoft has received a huge
    boost after a US judge threw
    out an anti-trust lawsuit
    filed by the Department of
    Justice to block the
    acquisition." + ], + [ + "The International Olympic
    Committee (IOC) has urged
    Beijing to ensure the city is
    ready to host the 2008 Games
    well in advance, an official
    said on Wednesday." + ], + [ + "AFP - German Chancellor
    Gerhard Schroeder arrived in
    Libya for an official visit
    during which he is to hold
    talks with Libyan leader
    Moamer Kadhafi." + ], + [ + "The fastest-swiveling space
    science observatory ever built
    rocketed into orbit on
    Saturday to scan the universe
    for celestial explosions." + ], + [ + "The government will examine
    claims 100,000 Iraqi civilians
    have been killed since the US-
    led invasion, Jack Straw says." + ], + [ + "Virginia Tech scores 24 points
    off four first-half turnovers
    in a 55-6 wipeout of Maryland
    on Thursday to remain alone
    atop the ACC." + ], + [ + "Copernic Unleashes Desktop
    Search Tool\\\\Copernic
    Technologies Inc. today
    announced Copernic Desktop
    Search(TM) (CDS(TM)), \"The
    Search Engine For Your PC
    (TM).\" Copernic has used the
    experience gained from over 30
    million downloads of its
    Windows-based Web search
    software to develop CDS, a
    desktop search product that
    users are saying is far ..." + ], + [ + "The DVD Forum moved a step
    further toward the advent of
    HD DVD media and drives with
    the approval of key physical
    specifications at a meeting of
    the organisations steering
    committee last week." + ], + [ + "Eton College and Clarence
    House joined forces yesterday
    to deny allegations due to be
    made at an employment tribunal
    today by a former art teacher
    that she improperly helped
    Prince Harry secure an A-level
    pass in art two years ago." + ], + [ + "AFP - Great Britain's chances
    of qualifying for the World
    Group of the Davis Cup were
    evaporating rapidly after
    Austria moved into a 2-1 lead
    following the doubles." + ], + [ + "AP - Martina Navratilova's
    long, illustrious career will
    end without an Olympic medal.
    The 47-year-old Navratilova
    and Lisa Raymond lost 6-4,
    4-6, 6-4 on Thursday night to
    fifth-seeded Shinobu Asagoe
    and Ai Sugiyama of Japan in
    the quarterfinals, one step
    shy of the medal round." + ], + [ + "Often pigeonholed as just a
    seller of televisions and DVD
    players, Royal Philips
    Electronics said third-quarter
    profit surged despite a slide
    into the red by its consumer
    electronics division." + ], + [ + "AP - Google, the Internet
    search engine, has done
    something that law enforcement
    officials and their computer
    tools could not: Identify a
    man who died in an apparent
    hit-and-run accident 11 years
    ago in this small town outside
    Yakima." + ], + [ + "We are all used to IE getting
    a monthly new security bug
    found, but Winamp? In fact,
    this is not the first security
    flaw found in the application." + ], + [ + "The Apache Software Foundation
    and the Debian Project said
    they won't support the Sender
    ID e-mail authentication
    standard in their products." + ], + [ + "USATODAY.com - The newly
    restored THX 1138 arrives on
    DVD today with Star Wars
    creator George Lucas' vision
    of a Brave New World-like
    future." + ], + [ + "Office Depot Inc. (ODP.N:
    Quote, Profile, Research) on
    Tuesday warned of weaker-than-
    expected profits for the rest
    of the year because of
    disruptions from the string of
    hurricanes" + ], + [ + "THE photo-equipment maker
    Kodak yesterday announced
    plans to axe 600 jobs in the
    UK and close a factory in
    Nottinghamshire, in a move in
    line with the giants global
    restructuring strategy
    unveiled last January." + ], + [ + "The chances of scientists
    making any one of five
    discoveries by 2010 have been
    hugely underestimated,
    according to bookmakers." + ], + [ + "Asia-Pacific leaders meet in
    Australia to discuss how to
    keep nuclear weapons out of
    the hands of extremists." + ], + [ + " TALL AFAR, Iraq -- A three-
    foot-high coil of razor wire,
    21-ton armored vehicles and
    American soldiers with black
    M-4 assault rifles stood
    between tens of thousands of
    people and their homes last
    week." + ], + [ + "PSV Eindhoven re-established
    their five-point lead at the
    top of the Dutch Eredivisie
    today with a 2-0 win at
    Vitesse Arnhem. Former
    Sheffield Wednesday striker
    Gerald Sibon put PSV ahead in
    the 56th minute" + ], + [ + "China's central bank on
    Thursday raised interest rates
    for the first time in nearly a
    decade, signaling deepening
    unease with the breakneck pace
    of development and an intent
    to reign in a construction
    boom now sowing fears of
    runaway inflation." + ], + [ + "Deepening its commitment to
    help corporate users create
    SOAs (service-oriented
    architectures) through the use
    of Web services, IBM's Global
    Services unit on Thursday
    announced the formation of an
    SOA Management Practice." + ], + [ + "TODAY AUTO RACING 3 p.m. --
    NASCAR Nextel Cup Sylvania 300
    qualifying at N.H.
    International Speedway,
    Loudon, N.H., TNT PRO BASEBALL
    7 p.m. -- Red Sox at New York
    Yankees, Ch. 38, WEEI (850)
    (on cable systems where Ch. 38
    is not available, the game
    will air on NESN); Chicago
    Cubs at Cincinnati, ESPN 6
    p.m. -- Eastern League finals:
    ..." + ], + [ + "MUMBAI, SEPTEMBER 21: The
    Board of Control for Cricket
    in India (BCCI) today informed
    the Bombay High Court that it
    was cancelling the entire
    tender process regarding
    cricket telecast rights as
    also the conditional deal with
    Zee TV." + ], + [ + "CHICAGO - Delta Air Lines
    (DAL) said Wednesday it plans
    to eliminate between 6,000 and
    6,900 jobs during the next 18
    months, implement a 10 across-
    the-board pay reduction and
    reduce employee benefits." + ], + [ + "LAKE GEORGE, N.Y. - Even
    though he's facing double hip
    replacement surgery, Bill
    Smith is more than happy to
    struggle out the door each
    morning, limp past his brand
    new P.T..." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks were likely to open
    flat on Wednesday, with high
    oil prices and profit warnings
    weighing on the market before
    earnings reports start and key
    jobs data is released this
    week." + ], + [ + "Best known for its popular
    search engine, Google is
    rapidly rolling out new
    products and muscling into
    Microsoft's stronghold: the
    computer desktop. The
    competition means consumers
    get more choices and better
    products." + ], + [ + "Toshiba Corp. #39;s new
    desktop-replacement multimedia
    notebooks, introduced on
    Tuesday, are further evidence
    that US consumers still have
    yet to embrace the mobility
    offered by Intel Corp." + ], + [ + "JEJU, South Korea : Grace Park
    of South Korea won an LPGA
    Tour tournament, firing a
    seven-under-par 65 in the
    final round of the
    1.35-million dollar CJ Nine
    Bridges Classic." + ], + [ + " JERUSALEM (Reuters) - Israeli
    Prime Minister Ariel Sharon
    poured cold water on Tuesday
    on recent international
    efforts to restart stalled
    peace talks with Syria, saying
    there was \"no possibility\" of
    returning to previous
    discussions." + ], + [ + "Dutch smugness was slapped
    hard during the past
    fortnight. The rude awakening
    began with the barbaric
    slaying of controversial
    filmmaker Theo van Gogh on
    November 2. Then followed a
    reciprocal cycle of some" + ], + [ + "AP - The NHL will lock out its
    players Thursday, starting a
    work stoppage that threatens
    to keep the sport off the ice
    for the entire 2004-05 season." + ], + [ + " MOSCOW (Reuters) - Russia's
    Gazprom said on Tuesday it
    will bid for embattled oil
    firm YUKOS' main unit next
    month, as the Kremlin seeks
    to turn the world's biggest
    gas producer into a major oil
    player." + ], + [ + "pee writes quot;A passenger
    on a commuter plane in
    northern Norway attacked both
    pilots and at least one
    passenger with an axe as the
    aircraft was coming in to
    land." + ], + [ + "Aregular Amazon customer,
    Yvette Thompson has found
    shopping online to be mostly
    convenient and trouble-free.
    But last month, after ordering
    two CDs on Amazon.com, the
    Silver Spring reader
    discovered on her bank
    statement that she was double-
    charged for the \\$26.98 order.
    And there was a \\$25 charge
    that was a mystery." + ], + [ + "Prime Minister Ariel Sharon
    pledged Sunday to escalate a
    broad Israeli offensive in
    northern Gaza, saying troops
    will remain until Palestinian
    rocket attacks are halted.
    Israeli officials said the
    offensive -- in which 58
    Palestinians and three
    Israelis have been killed --
    will help clear the way for an
    Israeli withdrawal." + ], + [ + "Federal Reserve officials
    raised a key short-term
    interest rate Tuesday for the
    fifth time this year, and
    indicated they will gradually
    move rates higher next year to
    keep inflation under control
    as the economy expands." + ], + [ + "Canadians are paying more to
    borrow money for homes, cars
    and other purchases today
    after a quarter-point
    interest-rate increase by the
    Bank of Canada yesterday was
    quickly matched by the
    chartered banks." + ], + [ + "NEW YORK - Wall Street
    professionals know to keep
    their expectations in check in
    September, historically the
    worst month of the year for
    stocks. As summertime draws to
    a close, money managers are
    getting back to business,
    cleaning house, and often
    sending the market lower in
    the process..." + ], + [ + "A group linked to al Qaeda
    ally Abu Musab al-Zarqawi said
    it had tried to kill Iraq
    #39;s environment minister on
    Tuesday and warned it would
    not miss next time, according
    to an Internet statement." + ], + [ + "The Israeli military killed
    four Palestinian militants on
    Wednesday as troops in tanks
    and armored vehicles pushed
    into another town in the
    northern Gaza Strip, extending" + ], + [ + "When Paula Radcliffe dropped
    out of the Olympic marathon
    miles from the finish, she
    sobbed uncontrollably.
    Margaret Okayo knew the
    feeling." + ], + [ + "Delta Air Lines is to issue
    millions of new shares without
    shareholder consent as part of
    moves to ensure its survival." + ], + [ + "First baseman Richie Sexson
    agrees to a four-year contract
    with the Seattle Mariners on
    Wednesday." + ], + [ + "KIRKUK, Iraq - A suicide
    attacker detonated a car bomb
    Saturday outside a police
    academy in the northern Iraqi
    city of Kirkuk as hundreds of
    trainees and civilians were
    leaving for the day, killing
    at least 20 people and
    wounding 36, authorities said.
    Separately, U.S and Iraqi
    forces clashed with insurgents
    in another part of northern
    Iraq after launching an
    operation to destroy an
    alleged militant cell in the
    town of Tal Afar, the U.S..." + ], + [ + "Genta (GNTA:Nasdaq - news -
    research) is never boring!
    Monday night, the company
    announced that its phase III
    Genasense study in chronic
    lymphocytic leukemia (CLL) met
    its primary endpoint, which
    was tumor shrinkage." + ], + [ + "Finnish mobile giant Nokia has
    described its new Preminet
    solution, which it launched
    Monday (Oct. 25), as a
    quot;major worldwide
    initiative." + ], + [ + "While the entire airline
    industry #39;s finances are
    under water, ATA Airlines will
    have to hold its breath longer
    than its competitors to keep
    from going belly up." + ], + [ + " SAN FRANCISCO (Reuters) - At
    virtually every turn, Intel
    Corp. executives are heaping
    praise on an emerging long-
    range wireless technology
    known as WiMAX, which can
    blanket entire cities with
    high-speed Internet access." + ], + [ + "One day after ousting its
    chief executive, the nation's
    largest insurance broker said
    it will tell clients exactly
    how much they are paying for
    services and renounce back-
    door payments from carriers." + ], + [ + "LONDON (CBS.MW) -- Outlining
    an expectation for higher oil
    prices and increasing demand,
    Royal Dutch/Shell on Wednesday
    said it #39;s lifting project
    spending to \\$45 billion over
    the next three years." + ], + [ + "Tuesday #39;s meeting could
    hold clues to whether it
    #39;ll be a November or
    December pause in rate hikes.
    By Chris Isidore, CNN/Money
    senior writer." + ], + [ + "Phishing is one of the
    fastest-growing forms of
    personal fraud in the world.
    While consumers are the most
    obvious victims, the damage
    spreads far wider--hurting
    companies #39; finances and
    reputations and potentially" + ], + [ + "Reuters - The U.S. Interior
    Department on\\Friday gave
    final approval to a plan by
    ConocoPhillips and\\partner
    Anadarko Petroleum Corp. to
    develop five tracts around\\the
    oil-rich Alpine field on
    Alaska's North Slope." + ], + [ + "The dollar may fall against
    the euro for a third week in
    four on concern near-record
    crude oil prices will temper
    the pace of expansion in the
    US economy, a survey by
    Bloomberg News indicates." + ], + [ + "The battle for the British-
    based Chelsfield plc hotted up
    at the weekend, with reports
    from London that the property
    giant #39;s management was
    working on its own bid to
    thwart the 585 million (\\$A1." + ], + [ + "Atari has opened the initial
    sign-up phase of the closed
    beta for its Dungeons amp;
    Dragons real-time-strategy
    title, Dragonshard." + ], + [ + "AP - Many states are facing
    legal challenges over possible
    voting problems Nov. 2. A look
    at some of the developments
    Thursday:" + ], + [ + "Israeli troops withdrew from
    the southern Gaza Strip town
    of Khan Yunis on Tuesday
    morning, following a 30-hour
    operation that left 17
    Palestinians dead." + ], + [ + "Notes and quotes from various
    drivers following California
    Speedway #39;s Pop Secret 500.
    Jeff Gordon slipped to second
    in points following an engine
    failure while Jimmie Johnson
    moved back into first." + ], + [ + "PM-designate Omar Karameh
    forms a new 30-member cabinet
    which includes women for the
    first time." + ], + [ + "Bahrain #39;s king pardoned a
    human rights activist who
    convicted of inciting hatred
    of the government and
    sentenced to one year in
    prison Sunday in a case linked
    to criticism of the prime
    minister." + ], + [ + "Big Blue adds features, beefs
    up training efforts in China;
    rival Unisys debuts new line
    and pricing plan." + ], + [ + " MEMPHIS, Tenn. (Sports
    Network) - The Memphis
    Grizzlies signed All-Star
    forward Pau Gasol to a multi-
    year contract on Friday.
    Terms of the deal were not
    announced." + ], + [ + "Leaders from 38 Asian and
    European nations are gathering
    in Vietnam for a summit of the
    Asia-Europe Meeting, know as
    ASEM. One thousand delegates
    are to discuss global trade
    and regional politics during
    the two-day forum." + ], + [ + "A US soldier has pleaded
    guilty to murdering a wounded
    16-year-old Iraqi boy. Staff
    Sergeant Johnny Horne was
    convicted Friday of the
    unpremeditated murder" + ], + [ + "Andre Agassi brushed past
    Jonas Bjorkman 6-3 6-4 at the
    Stockholm Open on Thursday to
    set up a quarter-final meeting
    with Spanish sixth seed
    Fernando Verdasco." + ], + [ + "South Korea's Hynix
    Semiconductor Inc. and Swiss-
    based STMicroelectronics NV
    signed a joint-venture
    agreement on Tuesday to
    construct a memory chip
    manufacturing plant in Wuxi,
    about 100 kilometers west of
    Shanghai, in China." + ], + [ + "SAN DIEGO --(Business Wire)--
    Oct. 11, 2004 -- Breakthrough
    PeopleSoft EnterpriseOne 8.11
    Applications Enable
    Manufacturers to Become
    Demand-Driven." + ], + [ + "Reuters - Oil prices rose on
    Friday as tight\\supplies of
    distillate fuel, including
    heating oil, ahead of\\the
    northern hemisphere winter
    spurred buying." + ], + [ + "Well, Intel did say -
    dismissively of course - that
    wasn #39;t going to try to
    match AMD #39;s little dual-
    core Opteron demo coup of last
    week and show off a dual-core
    Xeon at the Intel Developer
    Forum this week and - as good
    as its word - it didn #39;t." + ], + [ + "Guinea-Bissau #39;s army chief
    of staff and former interim
    president, General Verissimo
    Correia Seabra, was killed
    Wednesday during unrest by
    mutinous soldiers in the
    former Portuguese" + ], + [ + "31 October 2004 -- Exit polls
    show that Prime Minister
    Viktor Yanukovich and
    challenger Viktor Yushchenko
    finished on top in Ukraine
    #39;s presidential election
    today and will face each other
    in a run-off next month." + ], + [ + "Nov. 18, 2004 - An FDA
    scientist says the FDA, which
    is charged with protecting
    America #39;s prescription
    drug supply, is incapable of
    doing so." + ], + [ + "Rock singer Bono pledges to
    spend the rest of his life
    trying to eradicate extreme
    poverty around the world." + ], + [ + "AP - Just when tourists
    thought it was safe to go back
    to the Princess Diana memorial
    fountain, the mud has struck." + ], + [ + "The UK's inflation rate fell
    in September, thanks in part
    to a fall in the price of
    airfares, increasing the
    chance that interest rates
    will be kept on hold." + ], + [ + " HONG KONG/SAN FRANCISCO
    (Reuters) - IBM is selling its
    PC-making business to China's
    largest personal computer
    company, Lenovo Group Ltd.,
    for \\$1.25 billion, marking
    the U.S. firm's retreat from
    an industry it helped pioneer
    in 1981." + ], + [ + "AP - Three times a week, The
    Associated Press picks an
    issue and asks President Bush
    and Democratic presidential
    candidate John Kerry a
    question about it. Today's
    question and their responses:" + ], + [ + " BOSTON (Reuters) - Boston was
    tingling with anticipation on
    Saturday as the Red Sox
    prepared to host Game One of
    the World Series against the
    St. Louis Cardinals and take a
    step toward ridding
    themselves of a hex that has
    hung over the team for eight
    decades." + ], + [ + "FOR the first time since his
    appointment as Newcastle
    United manager, Graeme Souness
    has been required to display
    the strong-arm disciplinary
    qualities that, to Newcastle
    directors, made" + ], + [ + "In an apparent damage control
    exercise, Russian President
    Vladimir Putin on Saturday
    said he favored veto rights
    for India as new permanent
    member of the UN Security
    Council." + ], + [ + "Nordstrom reported a strong
    second-quarter profit as it
    continued to select more
    relevant inventory and sell
    more items at full price." + ], + [ + "WHY IT HAPPENED Tom Coughlin
    won his first game as Giants
    coach and immediately
    announced a fine amnesty for
    all Giants. Just kidding." + ], + [ + "A second-place finish in his
    first tournament since getting
    married was good enough to
    take Tiger Woods from third to
    second in the world rankings." + ], + [ + " COLORADO SPRINGS, Colorado
    (Reuters) - World 400 meters
    champion Jerome Young has been
    given a lifetime ban from
    athletics for a second doping
    offense, the U.S. Anti-Doping
    Agency (USADA) said Wednesday." + ], + [ + "AP - Nigeria's Senate has
    ordered a subsidiary of
    petroleum giant Royal/Dutch
    Shell to pay a Nigerian ethnic
    group #36;1.5 billion for oil
    spills in their homelands, but
    the legislative body can't
    enforce the resolution, an
    official said Wednesday." + ], + [ + "IT #39;S BEEN a heck of an
    interesting two days here in
    Iceland. I #39;ve seen some
    interesting technology, heard
    some inventive speeches and
    met some people with different
    ideas." + ], + [ + "The Bank of England is set to
    keep interest rates on hold
    following the latest meeting
    of the its Monetary Policy
    Committee." + ], + [ + "Australian troops in Baghdad
    came under attack today for
    the first time since the end
    of the Iraq war when a car
    bomb exploded injuring three
    soldiers and damaging an
    Australian armoured convoy." + ], + [ + "The Bush administration upheld
    yesterday the imposition of
    penalty tariffs on shrimp
    imports from China and
    Vietnam, handing a victory to
    beleaguered US shrimp
    producers." + ], + [ + "House prices rose 0.2 percent
    in September compared with the
    month before to stand up 17.8
    percent on a year ago, the
    Nationwide Building Society
    says." + ], + [ + "Reuters - Two clients of
    Germany's Postbank\\(DPBGn.DE)
    fell for an e-mail fraud that
    led them to reveal\\money
    transfer codes to a bogus Web
    site -- the first case of\\this
    scam in German, prosecutors
    said on Thursday." + ], + [ + "US spending on information
    technology goods, services,
    and staff will grow seven
    percent in 2005 and continue
    at a similar pace through
    2008, according to a study
    released Monday by Forrester
    Research." + ], + [ + "LONDON - In two years, Arsenal
    will play their home matches
    in the Emirates stadium. That
    is what their new stadium at
    Ashburton Grove will be called
    after the Premiership
    champions yesterday agreed to
    the" + ], + [ + "KNOXVILLE, Tenn. -- Jason
    Campbell threw for 252 yards
    and two touchdowns, and No. 8
    Auburn proved itself as a
    national title contender by
    overwhelming No. 10 Tennessee,
    34-10, last night." + ], + [ + "Look, Ma, no hands! The U.S.
    space agency's latest
    spacecraft can run an entire
    mission by itself. By Amit
    Asaravala." + ], + [ + "Pakistans decision to refuse
    the International Atomic
    Energy Agency to have direct
    access to Dr AQ Khan is
    correct on both legal and
    political counts." + ], + [ + "MANILA, 4 December 2004 - With
    floods receding, rescuers
    raced to deliver food to
    famished survivors in
    northeastern Philippine
    villages isolated by back-to-
    back storms that left more
    than 650 people dead and
    almost 400 missing." + ], + [ + "Talks on where to build the
    world #39;s first nuclear
    fusion reactor ended without a
    deal on Tuesday but the
    European Union said Japan and
    the United States no longer
    firmly opposed its bid to put
    the plant in France." + ], + [ + "Joining America Online,
    EarthLink and Yahoo against
    spamming, Microsoft Corp.
    today announced the filing of
    three new anti-Spam lawsuits
    under the CAN-SPAM federal law
    as part of its initiative in
    solving the Spam problem for
    Internet users worldwide." + ], + [ + "WASHINGTON -- Another
    Revolution season concluded
    with an overtime elimination.
    Last night, the Revolution
    thrice rallied from deficits
    for a 3-3 tie with D.C. United
    in the Eastern Conference
    final, then lost in the first-
    ever Major League Soccer match
    decided by penalty kicks." + ], + [ + "Dwyane Wade calls himself a
    quot;sidekick, quot; gladly
    accepting the role Kobe Bryant
    never wanted in Los Angeles.
    And not only does second-year
    Heat point" + ], + [ + "A nationwide inspection shows
    Internet users are not as safe
    online as they believe. The
    inspections found most
    consumers have no firewall
    protection, outdated antivirus
    software and dozens of spyware
    programs secretly running on
    their computers." + ], + [ + "World number one golfer Vijay
    Singh of Fiji has captured his
    eighth PGA Tour event of the
    year with a win at the 84
    Lumber Classic in Farmington,
    Pennsylvania." + ], + [ + "The noise was deafening and
    potentially unsettling in the
    minutes before the start of
    the men #39;s Olympic
    200-meter final. The Olympic
    Stadium crowd chanted
    quot;Hellas!" + ], + [ + "CLEVELAND - The White House
    said Vice President Dick
    Cheney faces a \"master
    litigator\" when he debates
    Sen. John Edwards Tuesday
    night, a backhanded compliment
    issued as the Republican
    administration defended itself
    against criticism that it has
    not acknowledged errors in
    waging war in Iraq..." + ], + [ + "Brazilian forward Ronaldinho
    scored a sensational goal for
    Barcelona against Milan,
    making for a last-gasp 2-1.
    The 24-year-old #39;s
    brilliant left-foot hit at the
    Nou Camp Wednesday night sent
    Barcelona atop of Group F." + ], + [ + "Nortel Networks says it will
    again delay the release of its
    restated financial results.
    The Canadian telecom vendor
    originally promised to release
    the restated results in
    September." + ], + [ + " CHICAGO (Reuters) - At first
    glance, paying \\$13 or \\$14
    for a hard-wired Internet
    laptop connection in a hotel
    room might seem expensive." + ], + [ + "SEOUL (Reuters) - The chairman
    of South Korea #39;s ruling
    Uri Party resigned on Thursday
    after saying his father had
    served as a military police
    officer during Japan #39;s
    1910-1945 colonial rule on the
    peninsula." + ], + [ + "ALERE, Uganda -- Kasmiro
    Bongonyinge remembers sitting
    up suddenly in his bed. It was
    just after sunrise on a summer
    morning two years ago, and the
    old man, 87 years old and
    blind, knew something was
    wrong." + ], + [ + "Reuters - An investigation
    into U.S. insurers\\and brokers
    rattled insurance industry
    stocks for a second day\\on
    Friday as investors, shaken
    further by subpoenas
    delivered\\to the top U.S. life
    insurer, struggled to gauge
    how deep the\\probe might
    reach." + ], + [ + "Bee Staff Writer. SANTA CLARA
    - Andre Carter #39;s back
    injury has kept him out of the
    49ers #39; lineup since Week
    1. It #39;s also interfering
    with him rooting on his alma
    mater in person." + ], + [ + "AP - Kellen Winslow Jr. ended
    his second NFL holdout Friday." + ], + [ + "JAKARTA - Official results
    have confirmed former army
    general Susilo Bambang
    Yudhoyono as the winner of
    Indonesia #39;s first direct
    presidential election, while
    incumbent Megawati
    Sukarnoputri urged her nation
    Thursday to wait for the
    official announcement" + ], + [ + "HOUSTON - With only a few
    seconds left in a certain
    victory for Miami, Peyton
    Manning threw a meaningless
    6-yard touchdown pass to
    Marvin Harrison to cut the
    score to 24-15." + ], + [ + "Reuters - A ragged band of
    children\\emerges ghost-like
    from mists in Ethiopia's
    highlands,\\thrusting bunches
    of carrots at a car full of
    foreigners." + ], + [ + "DEADLY SCORER: Manchester
    United #39;s Wayne Rooney
    celebrating his three goals
    against Fenerbahce this week
    at Old Trafford. (AP)." + ], + [ + "AP - A U.N. human rights
    expert criticized the U.S.-led
    coalition forces in
    Afghanistan for violating
    international law by allegedly
    beating Afghans to death and
    forcing some to remove their
    clothes or wear hoods." + ], + [ + "You can feel the confidence
    level, not just with Team
    Canada but with all of Canada.
    There #39;s every expectation,
    from one end of the bench to
    the other, that Martin Brodeur
    is going to hold the fort." + ], + [ + "Heading into the first game of
    a new season, every team has
    question marks. But in 2004,
    the Denver Broncos seemed to
    have more than normal." + ], + [ + " JERUSALEM (Reuters) - Israeli
    Prime Minister Ariel Sharon
    said on Thursday Yasser
    Arafat's death could be a
    turning point for peacemaking
    but he would pursue a
    unilateral plan that would
    strip Palestinians of some
    land they want for a state." + ], + [ + " AL-ASAD AIRBASE, Iraq
    (Reuters) - Defense Secretary
    Donald Rumsfeld swept into an
    airbase in Iraq's western
    desert Sunday to make a
    first-hand evaluation of
    operations to quell a raging
    Iraqi insurgency in his first
    such visit in five months." + ], + [ + "More than three out of four
    (76 percent) consumers are
    experiencing an increase in
    spoofing and phishing
    incidents, and 35 percent
    receive fake e-mails at least
    once a week, according to a
    recent national study." + ], + [ + "The Dow Jones Industrial
    Average failed three times
    this year to exceed its
    previous high and fell to
    about 10,000 each time, most
    recently a week ago." + ], + [ + "AP - Deep in the Atlantic
    Ocean, undersea explorers are
    living a safer life thanks to
    germ-fighting clothing made in
    Kinston." + ], + [ + "Anaheim, Calif. - There is a
    decidedly right lean to the
    three-time champions of the
    American League Central. In a
    span of 26 days, the Minnesota
    Twins lost the left side of
    their infield to free agency." + ], + [ + "Computer Associates Monday
    announced the general
    availability of three
    Unicenter performance
    management products for
    mainframe data management." + ], + [ + "Reuters - The European Union
    approved on\\Wednesday the
    first biotech seeds for
    planting and sale across\\EU
    territory, flying in the face
    of widespread
    consumer\\resistance to
    genetically modified (GMO)
    crops and foods." + ], + [ + "With the NFL trading deadline
    set for 4 p.m. Tuesday,
    Patriots coach Bill Belichick
    said there didn't seem to be
    much happening on the trade
    front around the league." + ], + [ + "WASHINGTON - Democrat John
    Kerry accused President Bush
    on Monday of sending U.S.
    troops to the \"wrong war in
    the wrong place at the wrong
    time\" and said he'd try to
    bring them all home in four
    years..." + ], + [ + " SINGAPORE (Reuters) - Asian
    share markets staged a broad-
    based retreat on Wednesday,
    led by steelmakers amid
    warnings of price declines,
    but also enveloping technology
    and financial stocks on
    worries that earnings may
    disappoint." + ], + [ + "p2pnet.net News:- A Microsoft
    UK quot;WEIGHING THE COST OF
    LINUX VS. WINDOWS? LET #39;S
    REVIEW THE FACTS quot;
    magazine ad has been nailed as
    misleading by Britain #39;s
    Advertising Standards
    Authority (ASA)." + ], + [ + "More lorry drivers are
    bringing supplies to Nepal's
    capital in defiance of an
    indefinite blockade by Maoist
    rebels." + ], + [ + "NEW YORK - CNN has a new boss
    for the second time in 14
    months: former CBS News
    executive Jonathan Klein, who
    will oversee programming and
    editorial direction at the
    second-ranked cable news
    network." + ], + [ + "Cut-price carrier Virgin Blue
    said Tuesday the cost of using
    Australian airports is
    spiraling upward and asked the
    government to review the
    deregulated system of charges." + ], + [ + "The retail sector overall may
    be reporting a sluggish start
    to the season, but holiday
    shoppers are scooping up tech
    goods at a brisk pace -- and
    they're scouring the Web for
    bargains more than ever.
    <FONT face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\" color=\"#666666\">&
    lt;B>-washingtonpost.com<
    ;/B></FONT>" + ], + [ + "AP - David Beckham broke his
    rib moments after scoring
    England's second goal in
    Saturday's 2-0 win over Wales
    in a World Cup qualifying
    game." + ], + [ + "Saudi Arabia, Kuwait and the
    United Arab Emirates, which
    account for almost half of
    OPEC #39;s oil output, said
    they #39;re committed to
    boosting capacity to meet
    soaring demand that has driven
    prices to a record." + ], + [ + "The US Commerce Department
    said Thursday personal income
    posted its biggest increase in
    three months in August. The
    government agency also said
    personal spending was
    unchanged after rising
    strongly in July." + ], + [ + " TOKYO (Reuters) - Tokyo's
    Nikkei average opened up 0.54
    percent on Monday with banks
    and exporters leading the way
    as a stronger finish on Wall
    Street and declining oil
    prices soothed worries over
    the global economic outlook." + ], + [ + " BEIJING (Reuters) - Floods
    and landslides have killed 76
    people in southwest China in
    the past four days and washed
    away homes and roads, knocked
    down power lines and cut off
    at least one city, state
    media said on Monday." + ], + [ + "Nothing changed at the top of
    Serie A as all top teams won
    their games to keep the
    distance between one another
    unaltered. Juventus came back
    from behind against Lazio to
    win thanks to another goal by
    Ibrahimovic" + ], + [ + "The team behind the Beagle 2
    mission has unveiled its
    design for a successor to the
    British Mars lander." + ], + [ + "Survey points to popularity in
    Europe, the Middle East and
    Asia of receivers that skip
    the pay TV and focus on free
    programming." + ], + [ + "RICHMOND, Va. Jeremy Mayfield
    won his first race in over
    four years, taking the
    Chevrolet 400 at Richmond
    International Raceway after
    leader Kurt Busch ran out of
    gas eight laps from the
    finish." + ], + [ + "AP - Victims of the Sept. 11
    attacks were mourned worldwide
    Saturday, but in the Middle
    East, amid sympathy for the
    dead, Arabs said Washington's
    support for Israel and the war
    on terror launched in the
    aftermath of the World Trade
    Center's collapse have only
    fueled anger and violence." + ], + [ + "Linux publisher Red Hat Inc.
    said Tuesday that information-
    technology consulting firm
    Unisys Corp. will begin
    offering a business version of
    the company #39;s open-source
    operating system on its
    servers." + ], + [ + "SEATTLE - Ichiro Suzuki set
    the major league record for
    hits in a season with 258,
    breaking George Sisler's
    84-year-old mark with a pair
    of singles Friday night. The
    Seattle star chopped a leadoff
    single in the first inning,
    then made history with a
    grounder up the middle in the
    third..." + ], + [ + "The intruder who entered
    British Queen Elizabeth II
    #39;s official Scottish
    residence and caused a
    security scare was a reporter
    from the London-based Sunday
    Times newspaper, local media
    reported Friday." + ], + [ + "IBM's p5-575, a specialized
    server geared for high-
    performance computing, has
    eight 1.9GHz Power5
    processors." + ], + [ + "Bruce Wasserstein, head of
    Lazard LLC, is asking partners
    to take a one-third pay cut as
    he readies the world #39;s
    largest closely held
    investment bank for a share
    sale, people familiar with the
    situation said." + ], + [ + "Canadian Press - FREDERICTON
    (CP) - A New Brunswick truck
    driver arrested in Ontario
    this week has been accused by
    police of stealing 50,000 cans
    of Moosehead beer." + ], + [ + "Reuters - British police said
    on Monday they had\\charged a
    man with sending hoax emails
    to relatives of people\\missing
    since the Asian tsunami,
    saying their loved ones
    had\\been confirmed dead." + ], + [ + "The Lemon Bay Manta Rays were
    not going to let a hurricane
    get in the way of football. On
    Friday, they headed to the
    practice field for the first
    time in eight" + ], + [ + "Microsoft Corp. Chairman Bill
    Gates has donated \\$400,000 to
    a campaign in California
    trying to win approval of a
    measure calling for the state
    to sell \\$3 billion in bonds
    to fund stem-cell research." + ], + [ + "AP - Track star Marion Jones
    filed a defamation lawsuit
    Wednesday against the man
    whose company is at the center
    of a federal investigation
    into illegal steroid use among
    some of the nation's top
    athletes." + ], + [ + "LOS ANGELES - On Sept. 1,
    former secretary of
    Agriculture Dan Glickman
    replaced the legendary Jack
    Valenti as president and CEO
    of Hollywood #39;s trade
    group, the Motion Picture
    Association of America." + ], + [ + "England #39;s players hit out
    at cricket #39;s authorities
    tonight and claimed they had
    been used as quot;political
    pawns quot; after the Zimbabwe
    government produced a
    spectacular U-turn to ensure
    the controversial one-day
    series will go ahead." + ], + [ + "Newspaper publisher Pulitzer
    Inc. said Sunday that company
    officials are considering a
    possible sale of the firm to
    boost shareholder value." + ], + [ + "Shares of Merck amp; Co.
    plunged almost 10 percent
    yesterday after a media report
    said that documents show the
    pharmaceutical giant hid or
    denied" + ], + [ + "AP - The Japanese won the
    pregame home run derby. Then
    the game started and the major
    league All-Stars put their
    bats to work. Back-to-back
    home runs by Moises Alou and
    Vernon Wells in the fourth
    inning and by Johnny Estrada
    and Brad Wilkerson in the
    ninth powered the major
    leaguers past the Japanese
    stars 7-3 Sunday for a 3-0
    lead in the eight-game series." + ], + [ + "Reuters - Wall Street was
    expected to dip at\\Thursday's
    opening, but shares of Texas
    Instruments Inc.\\, may climb
    after it gave upbeat earnings
    guidance." + ], + [ + "Chinese authorities detained a
    prominent, U.S.-based Buddhist
    leader in connection with his
    plans to reopen an ancient
    temple complex in the Chinese
    province of Inner Mongolia
    last week and have forced
    dozens of his American
    followers to leave the region,
    local officials said
    Wednesday." + ], + [ + "The director of the National
    Hurricane Center stays calm in
    the midst of a storm, but
    wants everyone in hurricane-
    prone areas to get the message
    from his media advisories:
    Respect the storm's power and
    make proper response plans." + ], + [ + "With Chelsea losing their
    unbeaten record and Manchester
    United failing yet again to
    win, William Hill now make
    Arsenal red-hot 2/5 favourites
    to retain the title." + ], + [ + "Late in August, Boeing #39;s
    top sales execs flew to
    Singapore for a crucial sales
    pitch. They were close to
    persuading Singapore Airlines,
    one of the world #39;s leading
    airlines, to buy the American
    company #39;s new jet, the
    mid-sized 7E7." + ], + [ + "SBC Communications and
    BellSouth will acquire
    YellowPages.com with the goal
    of building the site into a
    nationwide online business
    index, the companies said
    Thursday." + ], + [ + "Theresa special bookcase in Al
    Grohs office completely full
    of game plans from his days in
    the NFL. Green ones are from
    the Jets." + ], + [ + "SAN FRANCISCO Several
    California cities and
    counties, including Los
    Angeles and San Francisco, are
    suing Microsoft for what could
    amount to billions of dollars." + ], + [ + "Newcastle ensured their place
    as top seeds in Friday #39;s
    third round UEFA Cup draw
    after holding Sporting Lisbon
    to a 1-1 draw at St James #39;
    Park." + ], + [ + "Adorned with Turkish and EU
    flags, Turkey #39;s newspapers
    hailed Thursday an official EU
    report recommending the
    country start talks to join
    the bloc, while largely
    ignoring the stringent
    conditions attached to the
    announcement." + ], + [ + "Google plans to release a
    version of its desktop search
    tool for computers that run
    Apple Computer #39;s Mac
    operating system, Google #39;s
    chief executive, Eric Schmidt,
    said Friday." + ], + [ + "AMD : sicurezza e prestazioni
    ottimali con il nuovo
    processore mobile per notebook
    leggeri e sottili; Acer Inc.
    preme sull #39;acceleratore
    con il nuovo notebook a
    marchio Ferrari." + ], + [ + "The sounds of tinkling bells
    could be heard above the
    bustle of the Farmers Market
    on the Long Beach Promenade,
    leading shoppers to a row of
    bright red tin kettles dotting
    a pathway Friday." + ], + [ + "CBC SPORTS ONLINE - Bode
    Miller continued his
    impressive 2004-05 World Cup
    skiing season by winning a
    night slalom race in
    Sestriere, Italy on Monday." + ], + [ + "Firefox use around the world
    climbed 34 percent in the last
    month, according to a report
    published by Web analytics
    company WebSideStory Monday." + ], + [ + "If a plastic card that gives
    you credit for something you
    don't want isn't your idea of
    a great gift, you can put it
    up for sale or swap." + ], + [ + "WASHINGTON Aug. 17, 2004
    Scientists are planning to
    take the pulse of the planet
    and more in an effort to
    improve weather forecasts,
    predict energy needs months in
    advance, anticipate disease
    outbreaks and even tell
    fishermen where the catch will
    be ..." + ], + [ + "Damien Rhodes scored on a
    2-yard run in the second
    overtime, then Syracuse's
    defense stopped Pittsburgh on
    fourth and 1, sending the
    Orange to a 38-31 victory
    yesterday in Syracuse, N.Y." + ], + [ + "AP - Anthony Harris scored 18
    of his career-high 23 points
    in the second half to help
    Miami upset No. 19 Florida
    72-65 Saturday and give first-
    year coach Frank Haith his
    biggest victory." + ], + [ + "LONDON Santander Central
    Hispano of Spain looked
    certain to clinch its bid for
    the British mortgage lender
    Abbey National, after HBOS,
    Britain #39;s biggest home-
    loan company, said Wednesday
    it would not counterbid, and
    after the European Commission
    cleared" + ], + [ + "New communications technology
    could spawn future products.
    Could your purse tell you to
    bring an umbrella?" + ], + [ + " WASHINGTON (Reuters) - The
    Justice Department is
    investigating possible
    accounting fraud at Fannie
    Mae, bringing greater
    government scrutiny to bear on
    the mortgage finance company,
    already facing a parallel
    inquiry by the SEC, a source
    close to the matter said on
    Thursday." + ], + [ + "AP - The five cities looking
    to host the 2012 Summer Games
    submitted bids to the
    International Olympic
    Committee on Monday, entering
    the final stage of a long
    process in hopes of landing
    one of the biggest prizes in
    sports." + ], + [ + "SAP has won a \\$35 million
    contract to install its human
    resources software for the US
    Postal Service. The NetWeaver-
    based system will replace the
    Post Office #39;s current
    25-year-old legacy application" + ], + [ + "The FIA has already cancelled
    todays activities at Suzuka as
    Super Typhoon Ma-On heads
    towards the 5.807km circuit.
    Saturday practice has been
    cancelled altogether while
    pre-qualifying and final
    qualifying" + ], + [ + "Thailand's prime minister
    visits the southern town where
    scores of Muslims died in army
    custody after a rally." + ], + [ + "Indian industrial group Tata
    agrees to invest \\$2bn in
    Bangladesh, the biggest single
    deal agreed by a firm in the
    south Asian country." + ], + [ + "NewsFactor - For years,
    companies large and small have
    been convinced that if they
    want the sophisticated
    functionality of enterprise-
    class software like ERP and
    CRM systems, they must buy
    pre-packaged applications.
    And, to a large extent, that
    remains true." + ], + [ + "Following in the footsteps of
    the RIAA, the MPAA (Motion
    Picture Association of
    America) announced that they
    have began filing lawsuits
    against people who use peer-
    to-peer software to download
    copyrighted movies off the
    Internet." + ], + [ + " GRAND PRAIRIE, Texas
    (Reuters) - Betting on horses
    was banned in Texas until as
    recently as 1987. Times have
    changed rapidly since.
    Saturday, Lone Star Park race
    track hosts the \\$14 million
    Breeders Cup, global racing's
    end-of-season extravaganza." + ], + [ + "MacCentral - At a special
    music event featuring Bono and
    The Edge from rock group U2
    held on Tuesday, Apple took
    the wraps off the iPod Photo,
    a color iPod available in 40GB
    or 60GB storage capacities.
    The company also introduced
    the iPod U2, a special edition
    of Apple's 20GB player clad in
    black, equipped with a red
    Click Wheel and featuring
    engraved U2 band member
    signatures. The iPod Photo is
    available immediately, and
    Apple expects the iPod U2 to
    ship in mid-November." + ], + [ + "Beijing: At least 170 miners
    were trapped underground after
    a gas explosion on Sunday
    ignited a fire in a coalmine
    in north-west China #39;s
    Shaanxi province, reports
    said." + ], + [ + "The steel tubing company
    reports sharply higher
    earnings, but the stock is
    falling." + ], + [ + "It might be a stay of
    execution for Coach P, or it
    might just be a Christmas
    miracle come early. SU #39;s
    upset win over BC has given
    hope to the Orange playing in
    a post season Bowl game." + ], + [ + "PHIL Neville insists
    Manchester United don #39;t
    fear anyone in the Champions
    League last 16 and declared:
    quot;Bring on the Italians." + ], + [ + "Playboy Enterprises, the adult
    entertainment company, has
    announced plans to open a
    private members club in
    Shanghai even though the
    company #39;s flagship men
    #39;s magazine is still banned
    in China." + ], + [ + "Reuters - Oracle Corp is
    likely to win clearance\\from
    the European Commission for
    its hostile #36;7.7
    billion\\takeover of rival
    software firm PeopleSoft Inc.,
    a source close\\to the
    situation said on Friday." + ], + [ + "TORONTO (CP) - Earnings
    warnings from Celestica and
    Coca-Cola along with a
    slowdown in US industrial
    production sent stock markets
    lower Wednesday." + ], + [ + "IBM (Quote, Chart) said it
    would spend a quarter of a
    billion dollars over the next
    year and a half to grow its
    RFID (define) business." + ], + [ + "Eastman Kodak Co., the world
    #39;s largest maker of
    photographic film, said
    Wednesday it expects sales of
    digital products and services
    to grow at an annual rate of
    36 percent between 2003 and
    2007, above prior growth rate
    estimates of 26 percent
    between 2002" + ], + [ + "SAMARRA (Iraq): With renewe d
    wave of skirmishes between the
    Iraqi insurgents and the US-
    led coalition marines, several
    people including top police
    officers were put to death on
    Saturday." + ], + [ + "SPACE.com - NASA released one
    of the best pictures ever made
    of Saturn's moon Titan as the
    Cassini spacecraft begins a
    close-up inspection of the
    satellite today. Cassini is
    making the nearest flyby ever
    of the smog-shrouded moon." + ], + [ + "AFP - The Iraqi government
    plans to phase out slowly
    subsidies on basic products,
    such as oil and electricity,
    which comprise 50 percent of
    public spending, equal to 15
    billion dollars, the planning
    minister said." + ], + [ + "ANNAPOLIS ROYAL, NS - Nova
    Scotia Power officials
    continued to keep the sluice
    gates open at one of the
    utility #39;s hydroelectric
    plants Wednesday in hopes a
    wayward whale would leave the
    area and head for the open
    waters of the Bay of Fundy." + ], + [ + "TORONTO -- Toronto Raptors
    point guard Alvin Williams
    will miss the rest of the
    season after undergoing
    surgery on his right knee
    Monday." + ], + [ + "The federal agency that
    insures pension plans said
    that its deficit, already at
    the highest in its history,
    had doubled in its last fiscal
    year, to \\$23.3 billion." + ], + [ + "AFP - Like most US Latinos,
    members of the extended
    Rodriguez family say they will
    cast their votes for Democrat
    John Kerry in next month's
    presidential polls." + ], + [ + "A Milan judge on Tuesday opens
    hearings into whether to put
    on trial 32 executives and
    financial institutions over
    the collapse of international
    food group Parmalat in one of
    Europe #39;s biggest fraud
    cases." + ], + [ + "AP - Tennessee's two freshmen
    quarterbacks have Volunteers
    fans fantasizing about the
    next four years. Brent
    Schaeffer and Erik Ainge
    surprised many with the nearly
    seamless way they rotated
    throughout a 42-17 victory
    over UNLV on Sunday night." + ], + [ + "In fact, Larry Ellison
    compares himself to the
    warlord, according to
    PeopleSoft's former CEO,
    defending previous remarks he
    made." + ], + [ + "FALLUJAH, Iraq -- Four Iraqi
    fighters huddled in a trench,
    firing rocket-propelled
    grenades at Lieutenant Eric
    Gregory's Bradley Fighting
    Vehicle and the US tanks and
    Humvees that were lumbering
    through tight streets between
    boxlike beige houses." + ], + [ + "MADRID, Aug 18 (Reuters) -
    Portugal captain Luis Figo
    said on Wednesday he was
    taking an indefinite break
    from international football,
    but would not confirm whether
    his decision was final." + ], + [ + "The Bank of England on
    Thursday left its benchmark
    interest rate unchanged, at
    4.75 percent, as policy makers
    assessed whether borrowing
    costs, already the highest in
    the Group of Seven, are
    constraining consumer demand." + ], + [ + "AP - Several thousand
    Christians who packed a
    cathedral compound in the
    Egyptian capital hurled stones
    at riot police Wednesday to
    protest a woman's alleged
    forced conversion to Islam. At
    least 30 people were injured." + ], + [ + "A group of Saudi religious
    scholars have signed an open
    letter urging Iraqis to
    support jihad against US-led
    forces. quot;Fighting the
    occupiers is a duty for all
    those who are able, quot; they
    said in a statement posted on
    the internet at the weekend." + ], + [ + "Fashion retailers Austin Reed
    and Ted Baker have reported
    contrasting fortunes on the
    High Street. Austin Reed
    reported interim losses of 2." + ], + [ + "AP - Shaun Rogers is in the
    backfield as often as some
    running backs. Whether teams
    dare to block Detroit's star
    defensive tackle with one
    player or follow the trend of
    double-teaming him, he often
    rips through offensive lines
    with a rare combination of
    size, speed, strength and
    nimble footwork." + ], + [ + " NEW YORK (Reuters) - A
    federal judge on Friday
    approved Citigroup Inc.'s
    \\$2.6 billion settlement with
    WorldCom Inc. investors who
    lost billions when an
    accounting scandal plunged
    the telecommunications company
    into bankruptcy protection." + ], + [ + "The Lions and Eagles entered
    Sunday #39;s game at Ford
    Field in the same place --
    atop their respective
    divisions -- and with
    identical 2-0 records." + ], + [ + "An unspecified number of
    cochlear implants to help
    people with severe hearing
    loss are being recalled
    because they may malfunction
    due to ear moisture, the US
    Food and Drug Administration
    announced." + ], + [ + "Profits triple at McDonald's
    Japan after the fast-food
    chain starts selling larger
    burgers." + ], + [ + "After Marcos Moreno threw four
    more interceptions in last
    week's 14-13 overtime loss at
    N.C. A T, Bison Coach Ray
    Petty will start Antoine
    Hartfield against Norfolk
    State on Saturday." + ], + [ + "You can empty your pockets of
    change, take off your belt and
    shoes and stick your keys in
    the little tray. But if you've
    had radiation therapy
    recently, you still might set
    off Homeland Security alarms." + ], + [ + "Mountaineers retrieve three
    bodies believed to have been
    buried for 22 years on an
    Indian glacier." + ], + [ + "SEOUL, Oct 19 Asia Pulse -
    LG.Philips LCD Co.
    (KSE:034220), the world #39;s
    second-largest maker of liquid
    crystal display (LCD), said
    Tuesday it has developed the
    world #39;s largest organic
    light emitting diode" + ], + [ + "SOUTH WILLIAMSPORT, Pa. --
    Looking ahead to the US
    championship game almost cost
    Conejo Valley in the
    semifinals of the Little
    League World Series." + ], + [ + "The Cubs didn #39;t need to
    fly anywhere near Florida to
    be in the eye of the storm.
    For a team that is going on
    100 years since last winning a
    championship, the only thing
    they never are at a loss for
    is controversy." + ], + [ + "Security experts warn of
    banner ads with a bad attitude
    --and a link to malicious
    code. Also: Phishers, be gone." + ], + [ + "KETTERING, Ohio Oct. 12, 2004
    - Cincinnati Bengals defensive
    end Justin Smith pleaded not
    guilty to a driving under the
    influence charge." + ], + [ + "com October 15, 2004, 5:11 AM
    PT. Wood paneling and chrome
    made your dad #39;s station
    wagon look like a million
    bucks, and they might also be
    just the ticket for Microsoft
    #39;s fledgling" + ], + [ + "President Thabo Mbeki met with
    Ivory Coast Prime Minister
    Seydou Diarra for three hours
    yesterday as part of talks
    aimed at bringing peace to the
    conflict-wracked Ivory Coast." + ], + [ + "MINNEAPOLIS -- For much of the
    2004 season, Twins pitcher
    Johan Santana didn #39;t just
    beat opposing hitters. Often,
    he overwhelmed and owned them
    in impressive fashion." + ], + [ + "Britain #39;s inflation rate
    fell in August further below
    its 2.0 percent government-set
    upper limit target with
    clothing and footwear prices
    actually falling, official
    data showed on Tuesday." + ], + [ + " KATHMANDU (Reuters) - Nepal's
    Maoist rebels have
    temporarily suspended a
    crippling economic blockade of
    the capital from Wednesday,
    saying the move was in
    response to popular appeals." + ], + [ + "Reuters - An Algerian
    suspected of being a leader\\of
    the Madrid train bombers has
    been identified as one of
    seven\\people who blew
    themselves up in April to
    avoid arrest, Spain's\\Interior
    Ministry said on Friday." + ], + [ + "KABUL: An Afghan man was found
    guilty on Saturday of killing
    four journalists in 2001,
    including two from Reuters,
    and sentenced to death." + ], + [ + "Yasser Arafat, the leader for
    decades of a fight for
    Palestinian independence from
    Israel, has died at a military
    hospital in Paris, according
    to news reports." + ], + [ + " LONDON (Reuters) - European
    shares shrugged off a spike in
    the euro to a fresh all-time
    high Wednesday, with telecoms
    again leading the way higher
    after interim profits at
    Britain's mm02 beat
    expectations." + ], + [ + "WASHINGTON - Weighed down by
    high energy prices, the US
    economy grew slower than the
    government estimated in the
    April-June quarter, as higher
    oil prices limited consumer
    spending and contributed to a
    record trade deficit." + ], + [ + "CHICAGO United Airlines says
    it will need even more labor
    cuts than anticipated to get
    out of bankruptcy. United told
    a bankruptcy court judge in
    Chicago today that it intends
    to start talks with unions
    next month on a new round of
    cost savings." + ], + [ + " JABALYA, Gaza Strip (Reuters)
    - Israel pulled most of its
    forces out of the northern
    Gaza Strip Saturday after a
    four-day incursion it said
    was staged to halt Palestinian
    rocket attacks on southern
    Israeli towns." + ], + [ + "Computer Associates
    International yesterday
    reported a 6 increase in
    revenue during its second
    fiscal quarter, but posted a
    \\$94 million loss after paying
    to settle government
    investigations into the
    company, it said yesterday." + ], + [ + "THE Turkish embassy in Baghdad
    was investigating a television
    report that two Turkish
    hostages had been killed in
    Iraq, but no confirmation was
    available so far, a senior
    Turkish diplomat said today." + ], + [ + "Reuters - Thousands of
    supporters of
    Ukraine's\\opposition leader,
    Viktor Yushchenko, celebrated
    on the streets\\in the early
    hours on Monday after an exit
    poll showed him\\winner of a
    bitterly fought presidential
    election." + ], + [ + "LONDON : The United States
    faced rare criticism over
    human rights from close ally
    Britain, with an official
    British government report
    taking Washington to task over
    concerns about Iraq and the
    Guantanamo Bay jail." + ], + [ + "The University of California,
    Berkeley, has signed an
    agreement with the Samoan
    government to isolate, from a
    tree, the gene for a promising
    anti- Aids drug and to share
    any royalties from the sale of
    a gene-derived drug with the
    people of Samoa." + ], + [ + "PC World - Send your video
    throughout your house--
    wirelessly--with new gateways
    and media adapters." + ], + [ + "At a charity auction in New
    Jersey last weekend, baseball
    memorabilia dealer Warren
    Heller was approached by a man
    with an unusual but topical
    request." + ], + [ + " TOKYO (Reuters) - Tokyo's
    Nikkei average jumped 2.5
    percent by mid-afternoon on
    Monday as semiconductor-
    related stocks such as
    Advantest Corp. mirrored a
    rally by their U.S. peers
    while banks and brokerages
    extended last week's gains." + ], + [ + "INTER Milan coach Roberto
    Mancini believes the club
    #39;s lavish (northern) summer
    signings will enable them to
    mount a serious Serie A
    challenge this season." + ], + [ + "LONDON - A bomb threat that
    mentioned Iraq forced a New
    York-bound Greek airliner to
    make an emergency landing
    Sunday at London's Stansted
    Airport escorted by military
    jets, authorities said. An
    airport spokeswoman said an
    Athens newspaper had received
    a phone call saying there was
    a bomb on board the Olympic
    Airlines plane..." + ], + [ + "Links to this week's topics
    from search engine forums
    across the web: New MSN Search
    Goes LIVE in Beta - Microsoft
    To Launch New Search Engine -
    Google Launches 'Google
    Advertising Professionals' -
    Organic vs Paid Traffic ROI? -
    Making Money With AdWords? -
    Link Building 101" + ], + [ + "AP - Brad Ott shot an 8-under
    64 on Sunday to win the
    Nationwide Tour's Price Cutter
    Charity Championship for his
    first Nationwide victory." + ], + [ + "AP - New York Jets running
    back Curtis Martin passed Eric
    Dickerson and Jerome Bettis on
    the NFL career rushing list
    Sunday against the St. Louis
    Rams, moving to fourth all-
    time." + ], + [ + "Eight conservation groups are
    fighting the US government
    over a plan to poison
    thousands of prairie dogs in
    the grasslands of South
    Dakota, saying wildlife should
    not take a backseat to
    ranching interests." + ], + [ + "ATHENS, Greece - Sheryl
    Swoopes made three big plays
    at the end - two baskets and
    another on defense - to help
    the United States squeeze out
    a 66-62 semifinal victory over
    Russia on Friday. Now, only
    one game stands between the
    U.S..." + ], + [ + "Instead of standing for ante
    meridian and post meridian,
    though, fans will remember the
    time periods of pre-Mia and
    after-Mia. After playing for
    18 years and shattering nearly
    every record" + ], + [ + "General Motors (GM) plans to
    announce a massive
    restructuring Thursday that
    will eliminate as many as
    12,000 jobs in Europe in a
    move to stem the five-year
    flow of red ink from its auto
    operations in the region." + ], + [ + "Scientists are developing a
    device which could improve the
    lives of kidney dialysis
    patients." + ], + [ + "KABUL, Afghanistan The Afghan
    government is blaming drug
    smugglers for yesterday #39;s
    attack on the leading vice
    presidential candidate ." + ], + [ + "Stephon Marbury, concerned
    about his lousy shooting in
    Athens, used an off day to go
    to the gym and work on his
    shot. By finding his range, he
    saved the United States #39;
    hopes for a basketball gold
    medal." + ], + [ + " LONDON (Reuters) - Oil prices
    held firm on Wednesday as
    Hurricane Ivan closed off
    crude output and shut
    refineries in the Gulf of
    Mexico, while OPEC's Gulf
    producers tried to reassure
    traders by recommending an
    output hike." + ], + [ + "State-owned, running a
    monopoly on imports of jet
    fuel to China #39;s fast-
    growing aviation industry and
    a prized member of Singapore
    #39;s Stock Exchange." + ], + [ + "Google has won a trade mark
    dispute, with a District Court
    judge finding that the search
    engines sale of sponsored
    search terms Geico and Geico
    Direct did not breach car
    insurance firm GEICOs rights
    in the trade marked terms." + ], + [ + "Wall Street bounded higher for
    the second straight day
    yesterday as investors reveled
    in sharply falling oil prices
    and the probusiness agenda of
    the second Bush
    administration. The Dow Jones
    industrials gained more than
    177 points for its best day of
    2004, while the Standard amp;
    Poor's 500 closed at its
    highest level since early
    2002." + ], + [ + "Key factors help determine if
    outsourcing benefits or hurts
    Americans." + ], + [ + "The US Trade Representative on
    Monday rejected the European
    Union #39;s assertion that its
    ban on beef from hormone-
    treated cattle is now
    justified by science and that
    US and Canadian retaliatory
    sanctions should be lifted." + ], + [ + "One of the leading figures in
    the Greek Orthodox Church, the
    Patriarch of Alexandria Peter
    VII, has been killed in a
    helicopter crash in the Aegean
    Sea." + ], + [ + "Siding with chip makers,
    Microsoft said it won't charge
    double for its per-processor
    licenses when dual-core chips
    come to market next year." + ], + [ + "NEW YORK -- Wall Street's
    fourth-quarter rally gave
    stock mutual funds a solid
    performance for 2004, with
    small-cap equity funds and
    real estate funds scoring some
    of the biggest returns. Large-
    cap growth equities and
    technology-focused funds had
    the slimmest gains." + ], + [ + "CANBERRA, Australia -- The
    sweat-stained felt hats worn
    by Australian cowboys, as much
    a part of the Outback as
    kangaroos and sun-baked soil,
    may be heading for the history
    books. They fail modern
    industrial safety standards." + ], + [ + "Big Food Group Plc, the UK
    owner of the Iceland grocery
    chain, said second-quarter
    sales at stores open at least
    a year dropped 3.3 percent,
    the second consecutive
    decline, after competitors cut
    prices." + ], + [ + "A London-to-Washington flight
    is diverted after a security
    alert involving the singer
    formerly known as Cat Stevens." + ], + [ + " WASHINGTON (Reuters) - The
    first case of soybean rust has
    been found on the mainland
    United States and could affect
    U.S. crops for the near
    future, costing farmers
    millions of dollars, the
    Agriculture Department said on
    Wednesday." + ], + [ + "IBM and the Spanish government
    have introduced a new
    supercomputer they hope will
    be the most powerful in
    Europe, and one of the 10 most
    powerful in the world." + ], + [ + "The Supreme Court today
    overturned a five-figure
    damage award to an Alexandria
    man for a local auto dealer
    #39;s alleged loan scam,
    ruling that a Richmond-based
    federal appeals court had
    wrongly" + ], + [ + "TheSpaceShipOne team is handed
    the \\$10m cheque and trophy it
    won for claiming the Ansari
    X-Prize." + ], + [ + "Security officials have
    identified six of the
    militants who seized a school
    in southern Russia as being
    from Chechnya, drawing a
    strong connection to the
    Chechen insurgents who have
    been fighting Russian forces
    for years." + ], + [ + "AP - Randy Moss is expected to
    play a meaningful role for the
    Minnesota Vikings this weekend
    against the Giants, even
    without a fully healed right
    hamstring." + ], + [ + "Pros: Fits the recent profile
    (44, past PGA champion, fiery
    Ryder Cup player); the job is
    his if he wants it. Cons:
    Might be too young to be
    willing to burn two years of
    play on tour." + ], + [ + "SEOUL -- North Korea set three
    conditions yesterday to be met
    before it would consider
    returning to six-party talks
    on its nuclear programs." + ], + [ + "Official figures show the
    12-nation eurozone economy
    continues to grow, but there
    are warnings it may slow down
    later in the year." + ], + [ + "Elmer Santos scored in the
    second half, lifting East
    Boston to a 1-0 win over
    Brighton yesterday afternoon
    and giving the Jets an early
    leg up in what is shaping up
    to be a tight Boston City
    League race." + ], + [ + "In upholding a lower court
    #39;s ruling, the Supreme
    Court rejected arguments that
    the Do Not Call list violates
    telemarketers #39; First
    Amendment rights." + ], + [ + "US-backed Iraqi commandos were
    poised Friday to storm rebel
    strongholds in the northern
    city of Mosul, as US military
    commanders said they had
    quot;broken the back quot; of
    the insurgency with their
    assault on the former rebel
    bastion of Fallujah." + ], + [ + "Infineon Technologies, the
    second-largest chip maker in
    Europe, said Wednesday that it
    planned to invest about \\$1
    billion in a new factory in
    Malaysia to expand its
    automotive chip business and
    be closer to customers in the
    region." + ], + [ + "Mozilla's new web browser is
    smart, fast and user-friendly
    while offering a slew of
    advanced, customizable
    functions. By Michelle Delio." + ], + [ + "Saints special teams captain
    Steve Gleason expects to be
    fined by the league after
    being ejected from Sunday's
    game against the Carolina
    Panthers for throwing a punch." + ], + [ + "JERUSALEM (Reuters) - Prime
    Minister Ariel Sharon, facing
    a party mutiny over his plan
    to quit the Gaza Strip, has
    approved 1,000 more Israeli
    settler homes in the West Bank
    in a move that drew a cautious
    response on Tuesday from ..." + ], + [ + "Play has begun in the
    Australian Masters at
    Huntingdale in Melbourne with
    around half the field of 120
    players completing their first
    rounds." + ], + [ + " NEW YORK (Reuters) -
    Washington Post Co. <A HREF
    =\"http://www.investor.reuters.
    com/FullQuote.aspx?ticker=WPO.
    N target=/stocks/quickinfo/ful
    lquote\">WPO.N</A>
    said on Friday that quarterly
    profit jumped, beating
    analysts' forecasts, boosted
    by results at its Kaplan
    education unit and television
    broadcasting operations." + ], + [ + "GHAZNI, Afghanistan, 6 October
    2004 - Wartime security was
    rolled out for Afghanistans
    interim President Hamid Karzai
    as he addressed his first
    election campaign rally
    outside the capital yesterday
    amid spiraling violence." + ], + [ + "LOUISVILLE, Ky. - Louisville
    men #39;s basketball head
    coach Rick Pitino and senior
    forward Ellis Myles met with
    members of the media on Friday
    to preview the Cardinals #39;
    home game against rival
    Kentucky on Satursday." + ], + [ + "AP - Sounds like David
    Letterman is as big a \"Pops\"
    fan as most everyone else." + ], + [ + "originally offered on notebook
    PCs -- to its Opteron 32- and
    64-bit x86 processors for
    server applications. The
    technology will help servers
    to run" + ], + [ + "New orders for US-made durable
    goods increased 0.2pc in
    September, held back by a big
    drop in orders for
    transportation goods, the US
    Commerce Department said
    today." + ], + [ + "Siblings are the first ever to
    be convicted for sending
    boatloads of junk e-mail
    pushing bogus products. Also:
    Microsoft takes MSN music
    download on a Euro trip....
    Nokia begins legal battle
    against European
    counterparts.... and more." + ], + [ + "I always get a kick out of the
    annual list published by
    Forbes singling out the
    richest people in the country.
    It #39;s almost as amusing as
    those on the list bickering
    over their placement." + ], + [ + "MacCentral - After Apple
    unveiled the iMac G5 in Paris
    this week, Vice President of
    Hardware Product Marketing
    Greg Joswiak gave Macworld
    editors a guided tour of the
    desktop's new design. Among
    the topics of conversation:
    the iMac's cooling system, why
    pre-installed Bluetooth
    functionality and FireWire 800
    were left out, and how this
    new model fits in with Apple's
    objectives." + ], + [ + "Williams-Sonoma Inc., operator
    of home furnishing chains
    including Pottery Barn, said
    third-quarter earnings rose 19
    percent, boosted by store
    openings and catalog sales." + ], + [ + "We #39;ve known about
    quot;strained silicon quot;
    for a while--but now there
    #39;s a better way to do it.
    Straining silicon improves
    chip performance." + ], + [ + "This week, Sir Richard Branson
    announced his new company,
    Virgin Galactic, has the
    rights to the first commercial
    flights into space." + ], + [ + "71-inch HDTV comes with a home
    stereo system and components
    painted in 24-karat gold." + ], + [ + "Arsenal was held to a 1-1 tie
    by struggling West Bromwich
    Albion on Saturday, failing to
    pick up a Premier League
    victory when Rob Earnshaw
    scored with 11 minutes left." + ], + [ + "TOKYO - Mitsubishi Heavy
    Industries said today it #39;s
    in talks to buy a plot of land
    in central Japan #39;s Nagoya
    city from Mitsubishi Motors
    for building aircraft parts." + ], + [ + "China has confirmed that it
    found a deadly strain of bird
    flu in pigs as early as two
    years ago. China #39;s
    Agriculture Ministry said two
    cases had been discovered, but
    it did not say exactly where
    the samples had been taken." + ], + [ + "Baseball #39;s executive vice
    president Sandy Alderson
    insisted last month that the
    Cubs, disciplined for an
    assortment of run-ins with
    umpires, would not be targeted
    the rest of the season by
    umpires who might hold a
    grudge." + ], + [ + "As Superman and Batman would
    no doubt reflect during their
    cigarette breaks, the really
    draining thing about being a
    hero was that you have to keep
    riding to the rescue." + ], + [ + "MacCentral - RealNetworks Inc.
    said on Tuesday that it has
    sold more than a million songs
    at its online music store
    since slashing prices last
    week as part of a limited-time
    sale aimed at growing the user
    base of its new digital media
    software." + ], + [ + "With the presidential election
    less than six weeks away,
    activists and security experts
    are ratcheting up concern over
    the use of touch-screen
    machines to cast votes.
    <FONT face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\"\\ color=\"#666666\">
    <B>-washingtonpost.com&l
    t;/B></FONT>" + ], + [ + "NEW YORK, September 14 (New
    Ratings) - Yahoo! Inc
    (YHOO.NAS) has agreed to
    acquire Musicmatch Inc, a
    privately held digital music
    software company, for about
    \\$160 million in cash." + ], + [ + "Japan #39;s Sumitomo Mitsui
    Financial Group Inc. said
    Tuesday it proposed to UFJ
    Holdings Inc. that the two
    banks merge on an equal basis
    in its latest attempt to woo
    UFJ away from a rival suitor." + ], + [ + "Oil futures prices were little
    changed Thursday as traders
    anxiously watched for
    indications that the supply or
    demand picture would change in
    some way to add pressure to
    the market or take some away." + ], + [ + "Gov. Rod Blagojevich plans to
    propose a ban Thursday on the
    sale of violent and sexually
    explicit video games to
    minors, something other states
    have tried with little
    success." + ], + [ + " CHICAGO (Reuters) - Delta Air
    Lines Inc. <A HREF=\"http://
    www.investor.reuters.com/FullQ
    uote.aspx?ticker=DAL.N target=
    /stocks/quickinfo/fullquote\"&g
    t;DAL.N</A> said on
    Tuesday it will cut wages by
    10 percent and its chief
    executive will go unpaid for
    the rest of the year, but it
    still warned of bankruptcy
    within weeks unless more cuts
    are made." + ], + [ + "AP - Ten years after the Irish
    Republican Army's momentous
    cease-fire, negotiations
    resumed Wednesday in hope of
    reviving a Catholic-Protestant
    administration, an elusive
    goal of Northern Ireland's
    hard-fought peace process." + ], + [ + "A cable channel plans to
    resurrect each of the 1,230
    regular-season games listed on
    the league's defunct 2004-2005
    schedule by setting them in
    motion on a video game
    console." + ], + [ + " SANTO DOMINGO, Dominican
    Republic, Sept. 18 -- Tropical
    Storm Jeanne headed for the
    Bahamas on Saturday after an
    assault on the Dominican
    Republic that killed 10
    people, destroyed hundreds of
    houses and forced thousands
    from their homes." + ], + [ + "An explosion tore apart a car
    in Gaza City Monday, killing
    at least one person,
    Palestinian witnesses said.
    They said Israeli warplanes
    were circling overhead at the
    time of the blast, indicating
    a possible missile strike." + ], + [ + " WASHINGTON (Reuters) - A
    former Fannie Mae <A HREF=\"
    http://www.investor.reuters.co
    m/FullQuote.aspx?ticker=FNM.N
    target=/stocks/quickinfo/fullq
    uote\">FNM.N</A>
    employee who gave U.S.
    officials information about
    what he saw as accounting
    irregularities will not
    testify as planned before a
    congressional hearing next
    week, a House committee said
    on Friday." + ], + [ + "Beijing, Oct. 25 (PTI): China
    and the US today agreed to
    work jointly to re-energise
    the six-party talks mechanism
    aimed at dismantling North
    Korea #39;s nuclear programmes
    while Washington urged Beijing
    to resume" + ], + [ + "AFP - Sporadic gunfire and
    shelling took place overnight
    in the disputed Georgian
    region of South Ossetia in
    violation of a fragile
    ceasefire, wounding seven
    Georgian servicemen." + ], + [ + "PARIS, Nov 4 (AFP) - The
    European Aeronautic Defence
    and Space Company reported
    Thursday that its nine-month
    net profit more than doubled,
    thanks largely to sales of
    Airbus aircraft, and raised
    its full-year forecast." + ], + [ + "AP - Eric Hinske and Vernon
    Wells homered, and the Toronto
    Blue Jays completed a three-
    game sweep of the Baltimore
    Orioles with an 8-5 victory
    Sunday." + ], + [ + "SiliconValley.com - When
    \"Halo\" became a smash video
    game hit following Microsoft's
    launch of the Xbox console in
    2001, it was a no-brainer that
    there would be a sequel to the
    science fiction shoot-em-up." + ], + [ + "The number of people claiming
    unemployment benefit last
    month fell by 6,100 to
    830,200, according to the
    Office for National
    Statistics." + ], + [ + " NEW YORK (Reuters) - Todd
    Walker homered, had three hits
    and drove in four runs to lead
    the Chicago Cubs to a 12-5 win
    over the Cincinnati Reds in
    National League play at
    Wrigley Field on Monday." + ], + [ + "PARIS -- The city of Paris
    intends to reduce its
    dependence on software
    suppliers with \"de facto
    monopolies,\" but considers an
    immediate switch of its 17,000
    desktops to open source
    software too costly, it said
    Wednesday." + ], + [ + " FALLUJA, Iraq (Reuters) -
    U.S. forces hit Iraq's rebel
    stronghold of Falluja with the
    fiercest air and ground
    bombardment in months, as
    insurgents struck back on
    Saturday with attacks that
    killed up to 37 people in
    Samarra." + ], + [ + "MIAMI (Sports Network) -
    Shaquille O #39;Neal made his
    home debut, but once again it
    was Dwyane Wade stealing the
    show with 28 points as the
    Miami Heat downed the
    Cleveland Cavaliers, 92-86, in
    front of a record crowd at
    AmericanAirlines Arena." + ], + [ + "AP - The San Diego Chargers
    looked sharp #151; and played
    the same way. Wearing their
    powder-blue throwback jerseys
    and white helmets from the
    1960s, the Chargers did almost
    everything right in beating
    the Jacksonville Jaguars 34-21
    on Sunday." + ], + [ + "The vast majority of consumers
    are unaware that an Apple iPod
    digital music player only
    plays proprietary iTunes
    files, while a smaller
    majority agree that it is
    within RealNetworks #39;
    rights to develop a program
    that will make its music files
    compatible" + ], + [ + "Tyler airlines are gearing up
    for the beginning of holiday
    travel, as officials offer
    tips to help travelers secure
    tickets and pass through
    checkpoints with ease." + ], + [ + " NAJAF, Iraq (Reuters) - The
    fate of a radical Shi'ite
    rebellion in the holy city of
    Najaf was uncertain Friday
    amid disputed reports that
    Iraqi police had gained
    control of the Imam Ali
    Mosque." + ], + [ + " PROVIDENCE, R.I. (Reuters) -
    You change the oil in your car
    every 5,000 miles or so. You
    clean your house every week or
    two. Your PC needs regular
    maintenance as well --
    especially if you're using
    Windows and you spend a lot of
    time on the Internet." + ], + [ + "NERVES - no problem. That
    #39;s the verdict of Jose
    Mourinho today after his
    Chelsea side gave a resolute
    display of character at
    Highbury." + ], + [ + "AP - The latest low point in
    Ron Zook's tenure at Florida
    even has the coach wondering
    what went wrong. Meanwhile,
    Sylvester Croom's first big
    win at Mississippi State has
    given the Bulldogs and their
    fans a reason to believe in
    their first-year leader.
    Jerious Norwood's 37-yard
    touchdown run with 32 seconds
    remaining lifted the Bulldogs
    to a 38-31 upset of the 20th-
    ranked Gators on Saturday." + ], + [ + "A criminal trial scheduled to
    start Monday involving former
    Enron Corp. executives may
    shine a rare and potentially
    harsh spotlight on the inner
    workings" + ], + [ + "The Motley Fool - Here's
    something you don't see every
    day -- the continuing brouhaha
    between Oracle (Nasdaq: ORCL -
    News) and PeopleSoft (Nasdaq:
    PSFT - News) being a notable
    exception. South Africa's
    Harmony Gold Mining Company
    (NYSE: HMY - News) has
    announced a hostile takeover
    bid to acquire fellow South
    African miner Gold Fields
    Limited (NYSE: GFI - News).
    The transaction, if it takes
    place, would be an all-stock
    acquisition, with Harmony
    issuing 1.275 new shares in
    payment for each share of Gold
    Fields. The deal would value
    Gold Fields at more than
    #36;8 billion. ..." + ], + [ + "Someone forgot to inform the
    US Olympic basketball team
    that it was sent to Athens to
    try to win a gold medal, not
    to embarrass its country." + ], + [ + "SPACE.com - NASA's Mars \\rover
    Opportunity nbsp;will back its
    \\way out of a nbsp;crater it
    has spent four months
    exploring after reaching
    terrain nbsp;that appears \\too
    treacherous to tread. nbsp;" + ], + [ + "Sony Corp. announced Tuesday a
    new 20 gigabyte digital music
    player with MP3 support that
    will be available in Great
    Britain and Japan before
    Christmas and elsewhere in
    Europe in early 2005." + ], + [ + "Wal-Mart Stores Inc. #39;s
    Asda, the UK #39;s second
    biggest supermarket chain,
    surpassed Marks amp; Spencer
    Group Plc as Britain #39;s
    largest clothing retailer in
    the last three months,
    according to the Sunday
    Telegraph." + ], + [ + "Ten-man Paris St Germain
    clinched their seventh
    consecutive victory over arch-
    rivals Olympique Marseille
    with a 2-1 triumph in Ligue 1
    on Sunday thanks to a second-
    half winner by substitute
    Edouard Cisse." + ], + [ + "Until this week, only a few
    things about the strange,
    long-ago disappearance of
    Charles Robert Jenkins were
    known beyond a doubt. In the
    bitter cold of Jan. 5, 1965,
    the 24-year-old US Army
    sergeant was leading" + ], + [ + "Roy Oswalt wasn #39;t
    surprised to hear the Astros
    were flying Sunday night
    through the remnants of a
    tropical depression that
    dumped several inches of rain
    in Louisiana and could bring
    showers today in Atlanta." + ], + [ + "AP - This hardly seemed
    possible when Pitt needed
    frantic rallies to overcome
    Division I-AA Furman or Big
    East cellar dweller Temple. Or
    when the Panthers could barely
    move the ball against Ohio
    #151; not Ohio State, but Ohio
    U." + ], + [ + "Everyone is moaning about the
    fallout from last weekend but
    they keep on reporting the
    aftermath. #39;The fall-out
    from the so-called
    quot;Battle of Old Trafford
    quot; continues to settle over
    the nation and the debate" + ], + [ + "Oil supply concerns and broker
    downgrades of blue-chip
    companies left stocks mixed
    yesterday, raising doubts that
    Wall Street #39;s year-end
    rally would continue." + ], + [ + "Genentech Inc. said the
    marketing of Rituxan, a cancer
    drug that is the company #39;s
    best-selling product, is the
    subject of a US criminal
    investigation." + ], + [ + "American Lindsay Davenport
    regained the No. 1 ranking in
    the world for the first time
    since early 2002 by defeating
    Dinara Safina of Russia 6-4,
    6-2 in the second round of the
    Kremlin Cup on Thursday." + ], + [ + " The world's No. 2 soft drink
    company said on Thursday
    quarterly profit rose due to
    tax benefits." + ], + [ + "TOKYO (AP) -- The electronics
    and entertainment giant Sony
    Corp. (SNE) is talking with
    Wal-Mart Stores Inc..." + ], + [ + "After an unprecedented span of
    just five days, SpaceShipOne
    is ready for a return trip to
    space on Monday, its final
    flight to clinch a \\$10
    million prize." + ], + [ + "The United States on Tuesday
    modified slightly a threat of
    sanctions on Sudan #39;s oil
    industry in a revised text of
    its UN resolution on
    atrocities in the country
    #39;s Darfur region." + ], + [ + "Freshman Jeremy Ito kicked
    four field goals and Ryan
    Neill scored on a 31-yard
    interception return to lead
    improving Rutgers to a 19-14
    victory on Saturday over
    visiting Michigan State." + ], + [ + "Hi-tech monitoring of
    livestock at pig farms could
    help improve the animal growth
    process and reduce costs." + ], + [ + "Third-seeded Guillermo Canas
    defeated Guillermo Garcia-
    Lopez of Spain 7-6 (1), 6-3
    Monday on the first day of the
    Shanghai Open on Monday." + ], + [ + "AP - France intensified
    efforts Tuesday to save the
    lives of two journalists held
    hostage in Iraq, and the Arab
    League said the militants'
    deadline for France to revoke
    a ban on Islamic headscarves
    in schools had been extended." + ], + [ + "Cable amp; Wireless plc
    (NYSE: CWP - message board) is
    significantly ramping up its
    investment in local loop
    unbundling (LLU) in the UK,
    and it plans to spend up to 85
    million (\\$152." + ], + [ + "USATODAY.com - Personal
    finance software programs are
    the computer industry's
    version of veggies: Everyone
    knows they're good for you,
    but it's just hard to get
    anyone excited about them." + ], + [ + " NEW YORK (Reuters) - The
    dollar rebounded on Monday
    after last week's heavy
    selloff, but analysts were
    uncertain if the rally would
    hold after fresh economic data
    suggested the December U.S.
    jobs report due Friday might
    not live up to expectations." + ], + [ + "AFP - Microsoft said that it
    had launched a new desktop
    search tool that allows
    personal computer users to
    find documents or messages on
    their PCs." + ], + [ + "At least 12 people die in an
    explosion at a fuel pipeline
    on the outskirts of Nigeria's
    biggest city, Lagos." + ], + [ + "The three largest computer
    makers spearheaded a program
    today designed to standardize
    working conditions for their
    non-US workers." + ], + [ + "Description: Illinois Gov. Rod
    Blagojevich is backing state
    legislation that would ban
    sales or rentals of video
    games with graphic sexual or
    violent content to children
    under 18." + ], + [ + "Volkswagen demanded a two-year
    wage freeze for the
    170,000-strong workforce at
    Europe #39;s biggest car maker
    yesterday, provoking union
    warnings of imminent conflict
    at key pay and conditions
    negotiations." + ], + [ + " NEW YORK (Reuters) - U.S.
    stock futures pointed to a
    lower open on Wall Street on
    Thursday, extending the
    previous session's sharp
    fall, with rising energy
    prices feeding investor
    concerns about corporate
    profits and slower growth." + ], + [ + "But to play as feebly as it
    did for about 35 minutes last
    night in Game 1 of the WNBA
    Finals and lose by only four
    points -- on the road, no less
    -- has to be the best
    confidence builder since Cindy
    St." + ], + [ + "MILAN General Motors and Fiat
    on Wednesday edged closer to
    initiating a legal battle that
    could pit the two carmakers
    against each other in a New
    York City court room as early
    as next month." + ], + [ + "Are you bidding on keywords
    through Overture's Precision
    Match, Google's AdWords or
    another pay-for-placement
    service? If so, you're
    eligible to participate in
    their contextual advertising
    programs." + ], + [ + "Two of the Ford Motor Company
    #39;s most senior executives
    retired on Thursday in a sign
    that the company #39;s deep
    financial crisis has abated,
    though serious challenges
    remain." + ], + [ + "Citing security concerns, the
    U.S. Embassy on Thursday
    banned its employees from
    using the highway linking the
    embassy area to the
    international airport, a
    10-mile stretch of road
    plagued by frequent suicide
    car-bomb attacks." + ], + [ + "Nobel Laureate Wilkins, 87,
    played an important role in
    the discovery of the double
    helix structure of DNA, the
    molecule that carries our
    quot;life code quot;,
    Kazinform refers to BBC News." + ], + [ + "With yesterday #39;s report on
    its athletic department
    violations completed, the
    University of Washington says
    it is pleased to be able to
    move forward." + ], + [ + " LONDON (Reuters) - Wall
    Street was expected to start
    little changed on Friday as
    investors continue to fret
    over the impact of high oil
    prices on earnings, while
    Boeing <A HREF=\"http://www.
    investor.reuters.com/FullQuote
    .aspx?ticker=BA.N target=/stoc
    ks/quickinfo/fullquote\">BA.
    N</A> will be eyed
    after it reiterated its
    earnings forecast." + ], + [ + "AP - Tom Daschle bade his
    fellow Senate Democrats
    farewell Tuesday with a plea
    that they seek common ground
    with Republicans yet continue
    to fight for the less
    fortunate." + ], + [ + "Sammy Sosa was fined \\$87,400
    -- one day's salary -- for
    arriving late to the Cubs'
    regular-season finale at
    Wrigley Field and leaving the
    game early. The slugger's
    agent, Adam Katz , said
    yesterday Sosa most likely
    will file a grievance. Sosa
    arrived 70 minutes before
    Sunday's first pitch, and he
    apparently left 15 minutes
    after the game started without
    ..." + ], + [ + "Having an always-on, fast net
    connection is changing the way
    Britons use the internet,
    research suggests." + ], + [ + "AP - Police defused a bomb in
    a town near Prime Minister
    Silvio Berlusconi's villa on
    the island of Sardinia on
    Wednesday shortly after
    British Prime Minister Tony
    Blair finished a visit there
    with the Italian leader." + ], + [ + "Is the Oklahoma defense a
    notch below its predecessors?
    Is Texas #39; offense a step-
    ahead? Why is Texas Tech
    feeling good about itself
    despite its recent loss?" + ], + [ + "The coffin of Yasser Arafat,
    draped with the Palestinian
    flag, was bound for Ramallah
    in the West Bank Friday,
    following a formal funeral on
    a military compound near
    Cairo." + ], + [ + "US Ambassador to the United
    Nations John Danforth resigned
    on Thursday after serving in
    the post for less than six
    months. Danforth, 68, said in
    a letter released Thursday" + ], + [ + "Crude oil futures prices
    dropped below \\$51 a barrel
    yesterday as supply concerns
    ahead of the Northern
    Hemisphere winter eased after
    an unexpectedly high rise in
    US inventories." + ], + [ + "New York gets 57 combined
    points from its starting
    backcourt of Jamal Crawford
    and Stephon Marbury and tops
    Denver, 107-96." + ], + [ + "ISLAMABAD, Pakistan -- Photos
    were published yesterday in
    newspapers across Pakistan of
    six terror suspects, including
    a senior Al Qaeda operative,
    the government says were
    behind attempts to assassinate
    the nation's president." + ], + [ + "AP - Shawn Marion had a
    season-high 33 points and 15
    rebounds, leading the Phoenix
    Suns on a fourth-quarter
    comeback despite the absence
    of Steve Nash in a 95-86 win
    over the New Orleans Hornets
    on Friday night." + ], + [ + "By Lilly Vitorovich Of DOW
    JONES NEWSWIRES SYDNEY (Dow
    Jones)--Rupert Murdoch has
    seven weeks to convince News
    Corp. (NWS) shareholders a
    move to the US will make the
    media conglomerate more
    attractive to" + ], + [ + "A number of signs point to
    increasing demand for tech
    workers, but not all the
    clouds have been driven away." + ], + [ + "Messina upset defending
    champion AC Milan 2-1
    Wednesday, while Juventus won
    its third straight game to
    stay alone atop the Italian
    league standings." + ], + [ + "Microsoft Corp. (MSFT.O:
    Quote, Profile, Research)
    filed nine new lawsuits
    against spammers who send
    unsolicited e-mail, including
    an e-mail marketing Web
    hosting company, the world
    #39;s largest software maker
    said on Thursday." + ], + [ + "AP - Padraig Harrington
    rallied to a three-stroke
    victory in the German Masters
    on a windy Sunday, closing
    with a 2-under-par 70 and
    giving his game a big boost
    before the Ryder Cup." + ], + [ + " ATHENS (Reuters) - The Athens
    Paralympics canceled
    celebrations at its closing
    ceremony after seven
    schoolchildren traveling to
    watch the event died in a bus
    crash on Monday." + ], + [ + "The rocket plane SpaceShipOne
    is just one flight away from
    claiming the Ansari X-Prize, a
    \\$10m award designed to kick-
    start private space travel." + ], + [ + "Reuters - Three American
    scientists won the\\2004 Nobel
    physics prize on Tuesday for
    showing how tiny
    quark\\particles interact,
    helping to explain everything
    from how a\\coin spins to how
    the universe was built." + ], + [ + "Ironically it was the first
    regular season game for the
    Carolina Panthers that not
    only began the history of the
    franchise, but also saw the
    beginning of a rivalry that
    goes on to this day." + ], + [ + "Baltimore Ravens running back
    Jamal Lewis did not appear at
    his arraignment Friday, but
    his lawyers entered a not
    guilty plea on charges in an
    expanded drug conspiracy
    indictment." + ], + [ + "AP - Sharp Electronics Corp.
    plans to stop selling its
    Linux-based handheld computer
    in the United States, another
    sign of the slowing market for
    personal digital assistants." + ], + [ + "After serving a five-game
    suspension, Milton Bradley
    worked out with the Dodgers as
    they prepared for Tuesday's
    opener against the St. Louis
    Cardinals." + ], + [ + "AP - Prime Time won't be
    playing in prime time this
    time. Deion Sanders was on the
    inactive list and missed a
    chance to strut his stuff on
    \"Monday Night Football.\"" + ], + [ + "Reuters - Glaciers once held
    up by a floating\\ice shelf off
    Antarctica are now sliding off
    into the sea --\\and they are
    going fast, scientists said on
    Tuesday." + ], + [ + "DUBAI : An Islamist group has
    threatened to kill two Italian
    women held hostage in Iraq if
    Rome does not withdraw its
    troops from the war-torn
    country within 24 hours,
    according to an internet
    statement." + ], + [ + "AP - Warning lights flashed
    atop four police cars as the
    caravan wound its way up the
    driveway in a procession fit
    for a presidential candidate.
    At long last, Azy and Indah
    had arrived. They even flew
    through a hurricane to get
    here." + ], + [ + "The man who delivered the
    knockout punch was picked up
    from the Seattle scrap heap
    just after the All-Star Game.
    Before that, John Olerud
    certainly hadn't figured on
    facing Pedro Martinez in
    Yankee Stadium in October." + ], + [ + "\\Female undergraduates work
    harder and are more open-
    minded than males, leading to
    better results, say
    scientists." + ], + [ + "A heavy quake rocked Indonesia
    #39;s Papua province killing
    at least 11 people and
    wounding 75. The quake
    destroyed 150 buildings,
    including churches, mosques
    and schools." + ], + [ + "LONDON : A consortium,
    including former world
    champion Nigel Mansell, claims
    it has agreed terms to ensure
    Silverstone remains one of the
    venues for the 2005 Formula
    One world championship." + ], + [ + " BATON ROUGE, La. (Sports
    Network) - LSU has named Les
    Miles its new head football
    coach, replacing Nick Saban." + ], + [ + "The United Nations annual
    World Robotics Survey predicts
    the use of robots around the
    home will surge seven-fold by
    2007. The boom is expected to
    be seen in robots that can mow
    lawns and vacuum floors, among
    other chores." + ], + [ + "The long-term economic health
    of the United States is
    threatened by \\$53 trillion in
    government debts and
    liabilities that start to come
    due in four years when baby
    boomers begin to retire." + ], + [ + "Reuters - A small group of
    suspected\\gunmen stormed
    Uganda's Water Ministry
    Wednesday and took
    three\\people hostage to
    protest against proposals to
    allow President\\Yoweri
    Museveni for a third
    term.\\Police and soldiers with
    assault rifles cordoned off
    the\\three-story building, just
    328 feet from Uganda's
    parliament\\building in the
    capital Kampala." + ], + [ + "The Moscow Arbitration Court
    ruled on Monday that the YUKOS
    oil company must pay RUR
    39.113bn (about \\$1.34bn) as
    part of its back tax claim for
    2001." + ], + [ + "NOVEMBER 11, 2004 -- Bankrupt
    US Airways this morning said
    it had reached agreements with
    lenders and lessors to
    continue operating nearly all
    of its mainline and US Airways
    Express fleets." + ], + [ + "Venezuela suggested Friday
    that exiles living in Florida
    may have masterminded the
    assassination of a prosecutor
    investigating a short-lived
    coup against leftist President
    Hugo Chvez" + ], + [ + "Want to dive deep -- really
    deep -- into the technical
    literature about search
    engines? Here's a road map to
    some of the best web
    information retrieval
    resources available online." + ], + [ + "Reuters - Ancel Keys, a
    pioneer in public health\\best
    known for identifying the
    connection between
    a\\cholesterol-rich diet and
    heart disease, has died." + ], + [ + "The US government asks the
    World Trade Organisation to
    step in to stop EU member
    states from \"subsidising\"
    planemaker Airbus." + ], + [ + "Reuters - T-Mobile USA, the
    U.S. wireless unit\\of Deutsche
    Telekom AG (DTEGn.DE), does
    not expect to offer\\broadband
    mobile data services for at
    least the next two years,\\its
    chief executive said on
    Thursday." + ], + [ + "Verizon Communications is
    stepping further into video as
    a way to compete against cable
    companies." + ], + [ + "Facing a popular outcry at
    home and stern warnings from
    Europe, the Turkish government
    discreetly stepped back
    Tuesday from a plan to
    introduce a motion into a
    crucial penal reform bill to
    make adultery a crime
    punishable by prison." + ], + [ + "Boston Scientific Corp.
    (BSX.N: Quote, Profile,
    Research) said on Wednesday it
    received US regulatory
    approval for a device to treat
    complications that arise in
    patients with end-stage kidney
    disease who need dialysis." + ], + [ + "North-west Norfolk MP Henry
    Bellingham has called for the
    release of an old college
    friend accused of plotting a
    coup in Equatorial Guinea." + ], + [ + "With the economy slowly
    turning up, upgrading hardware
    has been on businesses radar
    in the past 12 months as their
    number two priority." + ], + [ + "AP - The Chicago Blackhawks
    re-signed goaltender Michael
    Leighton to a one-year
    contract Wednesday." + ], + [ + "Oracle Corp. plans to release
    the latest version of its CRM
    (customer relationship
    management) applications
    within the next two months, as
    part of an ongoing update of
    its E-Business Suite." + ], + [ + "Toyota Motor Corp. #39;s
    shares fell for a second day,
    after the world #39;s second-
    biggest automaker had an
    unexpected quarterly profit
    drop." + ], + [ + "AFP - Want to buy a castle?
    Head for the former East
    Germany." + ], + [ + "Hosted CRM service provider
    Salesforce.com took another
    step forward last week in its
    strategy to build an online
    ecosystem of vendors that
    offer software as a service." + ], + [ + "Britain-based HBOS says it
    will file a complaint to the
    European Commission against
    Spanish bank Santander Central
    Hispano (SCH) in connection
    with SCH #39;s bid to acquire
    British bank Abbey National" + ], + [ + "AFP - Steven Gerrard has moved
    to allay Liverpool fans' fears
    that he could be out until
    Christmas after breaking a
    metatarsal bone in his left
    foot." + ], + [ + "Verizon Wireless on Thursday
    announced an agreement to
    acquire all the PCS spectrum
    licenses of NextWave Telecom
    Inc. in 23 markets for \\$3
    billion." + ], + [ + "washingtonpost.com -
    Technology giants IBM and
    Hewlett-Packard are injecting
    hundreds of millions of
    dollars into radio-frequency
    identification technology,
    which aims to advance the
    tracking of items from ho-hum
    bar codes to smart tags packed
    with data." + ], + [ + "ATHENS -- She won her first
    Olympic gold medal in kayaking
    when she was 18, the youngest
    paddler to do so in Games
    history. Yesterday, at 42,
    Germany #39;s golden girl
    Birgit Fischer won her eighth
    Olympic gold in the four-woman
    500-metre kayak race." + ], + [ + "England boss Sven Goran
    Eriksson has defended
    goalkeeper David James after
    last night #39;s 2-2 draw in
    Austria. James allowed Andreas
    Ivanschitz #39;s shot to slip
    through his fingers to
    complete Austria comeback from
    two goals down." + ], + [ + "MINSK - Legislative elections
    in Belarus held at the same
    time as a referendum on
    whether President Alexander
    Lukashenko should be allowed
    to seek a third term fell
    significantly short of
    democratic standards, foreign
    observers said here Monday." + ], + [ + "An Olympic sailor is charged
    with the manslaughter of a
    Briton who died after being
    hit by a car in Athens." + ], + [ + "The Norfolk Broads are on
    their way to getting a clean
    bill of ecological health
    after a century of stagnation." + ], + [ + "AP - Secretary of State Colin
    Powell on Friday praised the
    peace deal that ended fighting
    in Iraq's holy city of Najaf
    and said the presence of U.S.
    forces in the area helped make
    it possible." + ], + [ + "The quot;future quot; is
    getting a chance to revive the
    presently struggling New York
    Giants. Two other teams also
    decided it was time for a
    change at quarterback, but the
    Buffalo Bills are not one of
    them." + ], + [ + "For the second time this year,
    an alliance of major Internet
    providers - including Atlanta-
    based EarthLink -iled a
    coordinated group of lawsuits
    aimed at stemming the flood of
    online junk mail." + ], + [ + " WASHINGTON (Reuters) - The
    PIMCO mutual fund group has
    agreed to pay \\$50 million to
    settle fraud charges involving
    improper rapid dealing in
    mutual fund shares, the U.S.
    Securities and Exchange
    Commission said on Monday." + ], + [ + "Via Technologies has released
    a version of the open-source
    Xine media player that is
    designed to take advantage of
    hardware digital video
    acceleration capabilities in
    two of the company #39;s PC
    chipsets, the CN400 and
    CLE266." + ], + [ + "The Conference Board reported
    Thursday that the Leading
    Economic Index fell for a
    third consecutive month in
    August, suggesting slower
    economic growth ahead amid
    rising oil prices." + ], + [ + " SAN FRANCISCO (Reuters) -
    Software maker Adobe Systems
    Inc.<A HREF=\"http://www.inv
    estor.reuters.com/FullQuote.as
    px?ticker=ADBE.O target=/stock
    s/quickinfo/fullquote\">ADBE
    .O</A> on Thursday
    posted a quarterly profit that
    rose more than one-third from
    a year ago, but shares fell 3
    percent after the maker of
    Photoshop and Acrobat software
    did not raise forecasts for
    fiscal 2005." + ], + [ + "William Morrison Supermarkets
    has agreed to sell 114 small
    Safeway stores and a
    distribution centre for 260.2
    million pounds. Morrison
    bought these stores as part of
    its 3 billion pound" + ], + [ + "SCO Group has a plan to keep
    itself fit enough to continue
    its legal battles against
    Linux and to develop its Unix-
    on-Intel operating systems." + ], + [ + "Flushing Meadows, NY (Sports
    Network) - The men #39;s
    semifinals at the 2004 US Open
    will be staged on Saturday,
    with three of the tournament
    #39;s top-five seeds ready for
    action at the USTA National
    Tennis Center." + ], + [ + "Pepsi pushes a blue version of
    Mountain Dew only at Taco
    Bell. Is this a winning
    strategy?" + ], + [ + "New software helps corporate
    travel managers track down
    business travelers who
    overspend. But it also poses a
    dilemma for honest travelers
    who are only trying to save
    money." + ], + [ + "NATO Secretary-General Jaap de
    Hoop Scheffer has called a
    meeting of NATO states and
    Russia on Tuesday to discuss
    the siege of a school by
    Chechen separatists in which
    more than 335 people died, a
    NATO spokesman said." + ], + [ + "26 August 2004 -- Iraq #39;s
    top Shi #39;ite cleric, Grand
    Ayatollah Ali al-Sistani,
    arrived in the city of Al-
    Najaf today in a bid to end a
    weeks-long conflict between US
    forces and militiamen loyal to
    Shi #39;ite cleric Muqtada al-
    Sadr." + ], + [ + "AFP - Senior executives at
    business software group
    PeopleSoft unanimously
    recommended that its
    shareholders reject a 8.8
    billion dollar takeover bid
    from Oracle Corp, PeopleSoft
    said in a statement Wednesday." + ], + [ + "Reuters - Neolithic people in
    China may have\\been the first
    in the world to make wine,
    according to\\scientists who
    have found the earliest
    evidence of winemaking\\from
    pottery shards dating from
    7,000 BC in northern China." + ], + [ + "Given nearly a week to examine
    the security issues raised by
    the now-infamous brawl between
    players and fans in Auburn
    Hills, Mich., Nov. 19, the
    Celtics returned to the
    FleetCenter last night with
    two losses and few concerns
    about their on-court safety." + ], + [ + " TOKYO (Reuters) - Electronics
    conglomerate Sony Corp.
    unveiled eight new flat-screen
    televisions on Thursday in a
    product push it hopes will
    help it secure a leading 35
    percent of the domestic
    market in the key month of
    December." + ], + [ + "As the election approaches,
    Congress abandons all pretense
    of fiscal responsibility,
    voting tax cuts that would
    drive 10-year deficits past
    \\$3 trillion." + ], + [ + "PARIS : French trade unions
    called on workers at France
    Telecom to stage a 24-hour
    strike September 7 to protest
    government plans to privatize
    the public telecommunications
    operator, union sources said." + ], + [ + "ServiceMaster profitably
    bundles services and pays a
    healthy 3.5 dividend." + ], + [ + "The Indonesian tourism
    industry has so far not been
    affected by last week #39;s
    bombing outside the Australian
    embassy in Jakarta and
    officials said they do not
    expect a significant drop in
    visitor numbers as a result of
    the attack." + ], + [ + "\\$222.5 million -- in an
    ongoing securities class
    action lawsuit against Enron
    Corp. The settlement,
    announced Friday and" + ], + [ + "Arsenals Thierry Henry today
    missed out on the European
    Footballer of the Year award
    as Andriy Shevchenko took the
    honour. AC Milan frontman
    Shevchenko held off
    competition from Barcelona
    pair Deco and" + ], + [ + "Donald Halsted, one target of
    a class-action suit alleging
    financial improprieties at
    bankrupt Polaroid, officially
    becomes CFO." + ], + [ + "Neil Mellor #39;s sensational
    late winner for Liverpool
    against Arsenal on Sunday has
    earned the back-up striker the
    chance to salvage a career
    that had appeared to be
    drifting irrevocably towards
    the lower divisions." + ], + [ + "ABOUT 70,000 people were
    forced to evacuate Real Madrid
    #39;s Santiago Bernabeu
    stadium minutes before the end
    of a Primera Liga match
    yesterday after a bomb threat
    in the name of ETA Basque
    separatist guerillas." + ], + [ + "The team learned on Monday
    that full-back Jon Ritchie
    will miss the rest of the
    season with a torn anterior
    cruciate ligament in his left
    knee." + ], + [ + " NEW YORK (Reuters) -
    Lifestyle guru Martha Stewart
    said on Wednesday she wants
    to start serving her prison
    sentence for lying about a
    suspicious stock sale as soon
    as possible, so she can put
    her \"nightmare\" behind her." + ], + [ + "Apple Computer's iPod remains
    the king of digital music
    players, but robust pretenders
    to the throne have begun to
    emerge in the Windows
    universe. One of them is the
    Zen Touch, from Creative Labs." + ], + [ + "The 7710 model features a
    touch screen, pen input, a
    digital camera, an Internet
    browser, a radio, video
    playback and streaming and
    recording capabilities, the
    company said." + ], + [ + "SAN FRANCISCO (CBS.MW) --
    Crude futures closed under
    \\$46 a barrel Wednesday for
    the first time since late
    September and heating-oil and
    unleaded gasoline prices
    dropped more than 6 percent
    following an across-the-board
    climb in US petroleum
    inventories." + ], + [ + "The University of Iowa #39;s
    market for US presidential
    futures, founded 16-years ago,
    has been overtaken by a
    Dublin-based exchange that is
    now 25 times larger." + ], + [ + "Venus Williams barely kept
    alive her hopes of qualifying
    for next week #39;s WTA Tour
    Championships. Williams,
    seeded fifth, survived a
    third-set tiebreaker to
    outlast Yuilana Fedak of the
    Ukraine, 6-4 2-6 7-6" + ], + [ + " SYDNEY (Reuters) - Arnold
    Palmer has taken a swing at
    America's top players,
    criticizing their increasing
    reluctance to travel abroad
    to play in tournaments." + ], + [ + "MARK Thatcher will have to
    wait until at least next April
    to face trial on allegations
    he helped bankroll a coup
    attempt in oil-rich Equatorial
    Guinea." + ], + [ + "A top Red Hat executive has
    attacked the open-source
    credentials of its sometime
    business partner Sun
    Microsystems. In a Web log
    posting Thursday, Michael
    Tiemann, Red Hat #39;s vice
    president of open-source
    affairs" + ], + [ + "President Bush #39;s drive to
    deploy a multibillion-dollar
    shield against ballistic
    missiles was set back on
    Wednesday by what critics
    called a stunning failure of
    its first full flight test in
    two years." + ], + [ + "Although he was well-beaten by
    Retief Goosen in Sunday #39;s
    final round of The Tour
    Championship in Atlanta, there
    has been some compensation for
    the former world number one,
    Tiger Woods." + ], + [ + "WAYNE Rooney and Henrik
    Larsson are among the players
    nominated for FIFAs
    prestigious World Player of
    the Year award. Rooney is one
    of four Manchester United
    players on a list which is
    heavily influenced by the
    Premiership." + ], + [ + "It didn #39;t look good when
    it happened on the field, and
    it looked worse in the
    clubhouse. Angels second
    baseman Adam Kennedy left the
    Angels #39; 5-2 win over the
    Seattle Mariners" + ], + [ + "Air travelers moved one step
    closer to being able to talk
    on cell phones and surf the
    Internet from laptops while in
    flight, thanks to votes by the
    Federal Communications
    Commission yesterday." + ], + [ + "MySQL developers turn to an
    unlikely source for database
    tool: Microsoft. Also: SGI
    visualizes Linux, and the
    return of Java veteran Kim
    Polese." + ], + [ + "DESPITE the budget deficit,
    continued increases in oil and
    consumer prices, the economy,
    as measured by gross domestic
    product, grew by 6.3 percent
    in the third" + ], + [ + "NEW YORK - A drop in oil
    prices and upbeat outlooks
    from Wal-Mart and Lowe's
    helped send stocks sharply
    higher Monday on Wall Street,
    with the swing exaggerated by
    thin late summer trading. The
    Dow Jones industrials surged
    nearly 130 points..." + ], + [ + "Freshman Darius Walker ran for
    115 yards and scored two
    touchdowns, helping revive an
    Irish offense that had managed
    just one touchdown in the
    season's first six quarters." + ], + [ + "Consumers who cut it close by
    paying bills from their
    checking accounts a couple of
    days before depositing funds
    will be out of luck under a
    new law that takes effect Oct.
    28." + ], + [ + "Dell Inc. said its profit
    surged 25 percent in the third
    quarter as the world's largest
    personal computer maker posted
    record sales due to rising
    technology spending in the
    corporate and government
    sectors in the United States
    and abroad." + ], + [ + "AP - NBC is adding a 5-second
    delay to its NASCAR telecasts
    after Dale Earnhardt Jr. used
    a vulgarity during a postrace
    TV interview last weekend." + ], + [ + " BOSTON (Sports Network) - The
    New York Yankees will start
    Orlando \"El Duque\" Hernandez
    in Game 4 of the American
    League Championship Series on
    Saturday against the Boston
    Red Sox." + ], + [ + "The future of Sven-Goran
    Eriksson as England coach is
    the subject of intense
    discussion after the draw in
    Austria. Has the Swede lost
    the confidence of the nation
    or does he remain the best man
    for the job?" + ], + [ + "Component problems meant
    Brillian's new big screens
    missed the NFL's kickoff
    party." + ], + [ + "Spain begin their third final
    in five seasons at the Olympic
    stadium hoping to secure their
    second title since their first
    in Barcelona against Australia
    in 2000." + ], + [ + "Second-seeded David Nalbandian
    of Argentina lost at the Japan
    Open on Thursday, beaten by
    Gilles Muller of Luxembourg
    7-6 (4), 3-6, 6-4 in the third
    round." + ], + [ + "Thursday #39;s unexpected
    resignation of Memphis
    Grizzlies coach Hubie Brown
    left a lot of questions
    unanswered. In his unique way
    of putting things, the
    71-year-old Brown seemed to
    indicate he was burned out and
    had some health concerns." + ], + [ + "RUDI Voller had quit as coach
    of Roma after a 3-1 defeat
    away to Bologna, the Serie A
    club said today. Under the
    former Germany coach, Roma had
    taken just four league points
    from a possible 12." + ], + [ + "A Russian court on Thursday
    rejected an appeal by the
    Yukos oil company seeking to
    overturn a freeze on the
    accounts of the struggling oil
    giant #39;s core subsidiaries." + ], + [ + "ONE by one, the players #39;
    faces had flashed up on the
    giant Ibrox screens offering
    season #39;s greetings to the
    Rangers fans. But the main
    presents were reserved for
    Auxerre." + ], + [ + "Switzerland #39;s struggling
    national airline reported a
    second-quarter profit of 45
    million Swiss francs (\\$35.6
    million) Tuesday, although its
    figures were boosted by a
    legal settlement in France." + ], + [ + "ROSTOV-ON-DON, Russia --
    Hundreds of protesters
    ransacked and occupied the
    regional administration
    building in a southern Russian
    province Tuesday, demanding
    the resignation of the region
    #39;s president, whose former
    son-in-law has been linked to
    a multiple" + ], + [ + "SIPTU has said it is strongly
    opposed to any privatisation
    of Aer Lingus as pressure
    mounts on the Government to
    make a decision on the future
    funding of the airline." + ], + [ + "Reuters - SBC Communications
    said on Monday it\\would offer
    a television set-top box that
    can handle music,\\photos and
    Internet downloads, part of
    SBC's efforts to expand\\into
    home entertainment." + ], + [ + "Molson Inc. Chief Executive
    Officer Daniel O #39;Neill
    said he #39;ll provide
    investors with a positive #39;
    #39; response to their
    concerns over the company
    #39;s plan to let stock-
    option holders vote on its
    planned merger with Adolph
    Coors Co." + ], + [ + "South Korea #39;s Grace Park
    shot a seven-under-par 65 to
    triumph at the CJ Nine Bridges
    Classic on Sunday. Park #39;s
    victory made up her final-
    round collapse at the Samsung
    World Championship two weeks
    ago." + ], + [ + " WASHINGTON (Reuters) - Hopes
    -- and worries -- that U.S.
    regulators will soon end the
    ban on using wireless phones
    during U.S. commercial flights
    are likely at least a year or
    two early, government
    officials and analysts say." + ], + [ + "AFP - Iraqi Foreign Minister
    Hoshyar Zebari arrived
    unexpectedly in the holy city
    of Mecca Wednesday where he
    met Crown Prince Abdullah bin
    Abdul Aziz, the official SPA
    news agency reported." + ], + [ + "Titans QB Steve McNair was
    released from a Nashville
    hospital after a two-night
    stay for treatment of a
    bruised sternum. McNair was
    injured during the fourth
    quarter of the Titans #39;
    15-12 loss to Jacksonville on
    Sunday." + ], + [ + "Keith Miller, Australia #39;s
    most prolific all-rounder in
    Test cricket, died today at a
    nursing home, Cricket
    Australia said. He was 84." + ], + [ + "Haitian police and UN troops
    moved into a slum neighborhood
    on Sunday and cleared street
    barricades that paralyzed a
    part of the capital." + ], + [ + "TORONTO Former Toronto pitcher
    John Cerutti (seh-ROO
    #39;-tee) was found dead in
    his hotel room today,
    according to the team. He was
    44." + ], + [ + "withdrawal of troops and
    settlers from occupied Gaza
    next year. Militants seek to
    claim any pullout as a
    victory. quot;Islamic Jihad
    will not be broken by this
    martyrdom, quot; said Khaled
    al-Batsh, a senior political
    leader in Gaza." + ], + [ + " NEW YORK (Reuters) - The
    world's largest gold producer,
    Newmont Mining Corp. <A HRE
    F=\"http://www.investor.reuters
    .com/FullQuote.aspx?ticker=NEM
    .N target=/stocks/quickinfo/fu
    llquote\">NEM.N</A>,
    on Wednesday said higher gold
    prices drove up quarterly
    profit by 12.5 percent, even
    though it sold less of the
    precious metal." + ], + [ + "The U.S. military has found
    nearly 20 houses where
    intelligence officers believe
    hostages were tortured or
    killed in this city, including
    the house with the cage that
    held a British contractor who
    was beheaded last month." + ], + [ + "AFP - Opponents of the Lao
    government may be plotting
    bomb attacks in Vientiane and
    other areas of Laos timed to
    coincide with a summit of
    Southeast Asian leaders the
    country is hosting next month,
    the United States said." + ], + [ + "After a year of pilots and
    trials, Siebel Systems jumped
    with both feet into the SMB
    market Tuesday, announcing a
    new approach to offer Siebel
    Professional CRM applications
    to SMBs (small and midsize
    businesses) -- companies with
    revenues up to about \\$500
    million." + ], + [ + "AP - Russia agreed Thursday to
    send warships to help NATO
    naval patrols that monitor
    suspicious vessels in the
    Mediterranean, part of a push
    for closer counterterrorism
    cooperation between Moscow and
    the western alliance." + ], + [ + "Intel won't release a 4-GHz
    version of its flagship
    Pentium 4 product, having
    decided instead to realign its
    engineers around the company's
    new design priorities, an
    Intel spokesman said today.
    \\\\" + ], + [ + "AP - A Soyuz spacecraft
    carrying two Russians and an
    American rocketed closer
    Friday to its docking with the
    international space station,
    where the current three-man
    crew made final departure
    preparations." + ], + [ + "Defense: Ken Lucas. His
    biggest play was his first
    one. The fourth-year
    cornerback intercepted a Ken
    Dorsey pass that kissed off
    the hands of wide receiver
    Rashaun Woods and returned it
    25 yards to set up the
    Seahawks #39; first score." + ], + [ + "Scientists have manipulated
    carbon atoms to create a
    material that could be used to
    create light-based, versus
    electronic, switches. The
    material could lead to a
    supercharged Internet based
    entirely on light, scientists
    say." + ], + [ + "A military plane crashed in
    the mountains near Caracas,
    killing all 16 persons on
    board, including two high-
    ranking military officers,
    officials said." + ], + [ + "The powerful St. Louis trio of
    Albert Pujols, Scott Rolen and
    Jim Edmonds is 4 for 23 with
    one RBI in the series and with
    runners on base, they are 1
    for 13." + ], + [ + "A voice recording said to be
    that of suspected Al Qaeda
    commander Abu Mussab al-
    Zarqawi, claims Iraq #39;s
    Prime Minister Iyad Allawi is
    the militant network #39;s
    number one target." + ], + [ + "BEIJING -- More than a year
    after becoming China's
    president, Hu Jintao was
    handed the full reins of power
    yesterday when his
    predecessor, Jiang Zemin, gave
    up the nation's most powerful
    military post." + ], + [ + "LOS ANGELES (Reuters)
    Qualcomm has dropped an \\$18
    million claim for monetary
    damages from rival Texas
    Instruments for publicly
    discussing terms of a
    licensing pact, a TI
    spokeswoman confirmed Tuesday." + ], + [ + "Hewlett-Packard is the latest
    IT vendor to try blogging. But
    analysts wonder if the weblog
    trend is the 21st century
    equivalent of CB radios, which
    made a big splash in the 1970s
    before fading." + ], + [ + " WASHINGTON (Reuters) - Fannie
    Mae executives and their
    regulator squared off on
    Wednesday, with executives
    denying any accounting
    irregularity and the regulator
    saying the housing finance
    company's management may need
    to go." + ], + [ + "The scientists behind Dolly
    the sheep apply for a license
    to clone human embryos. They
    want to take stem cells from
    the embryos to study Lou
    Gehrig's disease." + ], + [ + "As the first criminal trial
    stemming from the financial
    deals at Enron opened in
    Houston on Monday, it is
    notable as much for who is not
    among the six defendants as
    who is - and for how little
    money was involved compared
    with how much in other Enron" + ], + [ + "LONDON (CBS.MW) -- British
    bank Barclays on Thursday said
    it is in talks to buy a
    majority stake in South
    African bank ABSA. Free!" + ], + [ + "The Jets signed 33-year-old
    cornerback Terrell Buckley,
    who was released by New
    England on Sunday, after
    putting nickel back Ray
    Mickens on season-ending
    injured reserve yesterday with
    a torn ACL in his left knee." + ], + [ + "Some of the silly tunes
    Japanese pay to download to
    use as the ring tone for their
    mobile phones sure have their
    knockers, but it #39;s for
    precisely that reason that a
    well-known counselor is raking
    it in at the moment, according
    to Shukan Gendai (10/2)." + ], + [ + "WEST INDIES thrilling victory
    yesterday in the International
    Cricket Council Champions
    Trophy meant the world to the
    five million people of the
    Caribbean." + ], + [ + "AP - Greenpeace activists
    scaled the walls of Ford Motor
    Co.'s Norwegian headquarters
    Tuesday to protest plans to
    destroy hundreds of non-
    polluting electric cars." + ], + [ + "Investors sent stocks sharply
    lower yesterday as oil prices
    continued their climb higher
    and new questions about the
    safety of arthritis drugs
    pressured pharmaceutical
    stocks." + ], + [ + "Scotland manager Berti Vogts
    insists he is expecting
    nothing but victory against
    Moldova on Wednesday. The game
    certainly is a must-win affair
    if the Scots are to have any
    chance of qualifying for the
    2006 World Cup finals." + ], + [ + "IBM announced yesterday that
    it will invest US\\$250 million
    (S\\$425 million) over the next
    five years and employ 1,000
    people in a new business unit
    to support products and
    services related to sensor
    networks." + ], + [ + "AFP - The chances of Rupert
    Murdoch's News Corp relocating
    from Australia to the United
    States have increased after
    one of its biggest
    institutional investors has
    chosen to abstain from a vote
    next week on the move." + ], + [ + "AFP - An Indian minister said
    a school text-book used in the
    violence-prone western state
    of Gujarat portrayed Adolf
    Hitler as a role model." + ], + [ + "Reuters - The head of UAL
    Corp.'s United\\Airlines said
    on Thursday the airline's
    restructuring plan\\would lead
    to a significant number of job
    losses, but it was\\not clear
    how many." + ], + [ + "DOVER, N.H. (AP) -- Democrat
    John Kerry is seizing on the
    Bush administration's failure
    to secure hundreds of tons of
    explosives now missing in
    Iraq." + ], + [ + "AP - Microsoft Corp. goes into
    round two Friday of its battle
    to get the European Union's
    sweeping antitrust ruling
    lifted having told a judge
    that it had been prepared
    during settlement talks to
    share more software code with
    its rivals than the EU
    ultimately demanded." + ], + [ + " INDIANAPOLIS (Reuters) -
    Jenny Thompson will take the
    spotlight from injured U.S.
    team mate Michael Phelps at
    the world short course
    championships Saturday as she
    brings down the curtain on a
    spectacular swimming career." + ], + [ + "Martin Brodeur made 27 saves,
    and Brad Richards, Kris Draper
    and Joe Sakic scored goals to
    help Canada beat Russia 3-1
    last night in the World Cup of
    Hockey, giving the Canadians a
    3-0 record in round-robin
    play." + ], + [ + "AP - Sears, Roebuck and Co.,
    which has successfully sold
    its tools and appliances on
    the Web, is counting on having
    the same magic with bedspreads
    and sweaters, thanks in part
    to expertise gained by its
    purchase of Lands' End Inc." + ], + [ + "com September 14, 2004, 9:12
    AM PT. With the economy slowly
    turning up, upgrading hardware
    has been on businesses radar
    in the past 12 months as their
    number two priority." + ], + [ + " NEW YORK (Reuters) -
    Children's Place Retail Stores
    Inc. <A HREF=\"http://www.i
    nvestor.reuters.com/FullQuote.
    aspx?ticker=PLCE.O target=/sto
    cks/quickinfo/fullquote\">PL
    CE.O</A> said on
    Wednesday it will buy 313
    retail stores from Walt
    Disney Co., and its stock rose
    more than 14 percent in early
    morning trade." + ], + [ + "ALBANY, N.Y. -- A California-
    based company that brokers
    life, accident, and disability
    policies for leading US
    companies pocketed millions of
    dollars a year in hidden
    payments from insurers and
    from charges on clients'
    unsuspecting workers, New York
    Attorney General Eliot Spitzer
    charged yesterday." + ], + [ + "NORTEL Networks plans to slash
    its workforce by 3500, or ten
    per cent, as it struggles to
    recover from an accounting
    scandal that toppled three top
    executives and led to a
    criminal investigation and
    lawsuits." + ], + [ + "Ebay Inc. (EBAY.O: Quote,
    Profile, Research) said on
    Friday it would buy Rent.com,
    an Internet housing rental
    listing service, for \\$415
    million in a deal that gives
    it access to a new segment of
    the online real estate market." + ], + [ + "Austin police are working with
    overseas officials to bring
    charges against an English man
    for sexual assault of a child,
    a second-degree felony." + ], + [ + "United Nations officials
    report security breaches in
    internally displaced people
    and refugee camps in Sudan
    #39;s embattled Darfur region
    and neighboring Chad." + ], + [ + "Noranda Inc., Canada #39;s
    biggest mining company, began
    exclusive talks on a takeover
    proposal from China Minmetals
    Corp. that would lead to the
    spinoff of Noranda #39;s
    aluminum business to
    shareholders." + ], + [ + "San Francisco
    developer/publisher lands
    coveted Paramount sci-fi
    license, \\$6.5 million in
    funding on same day. Although
    it is less than two years old,
    Perpetual Entertainment has
    acquired one of the most
    coveted sci-fi licenses on the
    market." + ], + [ + "ST. LOUIS -- Mike Martz #39;s
    week of anger was no empty
    display. He saw the defending
    NFC West champions slipping
    and thought taking potshots at
    his players might be his best
    shot at turning things around." + ], + [ + "Google warned Thursday that
    increased competition and the
    maturing of the company would
    result in an quot;inevitable
    quot; slowing of its growth." + ], + [ + "By KELLY WIESE JEFFERSON
    CITY, Mo. (AP) -- Missouri
    will allow members of the
    military stationed overseas to
    return absentee ballots via
    e-mail, raising concerns from
    Internet security experts
    about fraud and ballot
    secrecy..." + ], + [ + "Avis Europe PLC has dumped a
    new ERP system based on
    software from PeopleSoft Inc.
    before it was even rolled out,
    citing substantial delays and
    higher-than-expected costs." + ], + [ + " TOKYO (Reuters) - The Nikkei
    average rose 0.55 percent by
    midsession on Wednesday as
    some techs including Advantest
    Corp. gained ground after
    Wall Street reacted positively
    to results from Intel Corp.
    released after the U.S. market
    close." + ], + [ + "Yahoo #39;s (Quote, Chart)
    public embrace of the RSS
    content syndication format
    took a major leap forward with
    the release of a revamped My
    Yahoo portal seeking to
    introduce the technology to
    mainstream consumers." + ], + [ + "KINGSTON, Jamaica - Hurricane
    Ivan's deadly winds and
    monstrous waves bore down on
    Jamaica on Friday, threatening
    a direct hit on its densely
    populated capital after
    ravaging Grenada and killing
    at least 33 people. The
    Jamaican government ordered
    the evacuation of half a
    million people from coastal
    areas, where rains on Ivan's
    outer edges were already
    flooding roads..." + ], + [ + "North Korea has denounced as
    quot;wicked terrorists quot;
    the South Korean officials who
    orchestrated last month #39;s
    airlift to Seoul of 468 North
    Korean defectors." + ], + [ + "The Black Watch regiment has
    returned to its base in Basra
    in southern Iraq after a
    month-long mission standing in
    for US troops in a more
    violent part of the country,
    the Ministry of Defence says." + ], + [ + "Romanian soccer star Adrian
    Mutu as he arrives at the
    British Football Association
    in London, ahead of his
    disciplinary hearing, Thursday
    Nov. 4, 2004." + ], + [ + "Australia completed an
    emphatic Test series sweep
    over New Zealand with a
    213-run win Tuesday, prompting
    a caution from Black Caps
    skipper Stephen Fleming for
    other cricket captains around
    the globe." + ], + [ + "AP - A senior Congolese
    official said Tuesday his
    nation had been invaded by
    neighboring Rwanda, and U.N.
    officials said they were
    investigating claims of
    Rwandan forces clashing with
    militias in the east." + ], + [ + "Liverpool, England (Sports
    Network) - Former English
    international and Liverpool
    great Emlyn Hughes passed away
    Tuesday from a brain tumor." + ], + [ + " ATLANTA (Reuters) - Soft
    drink giant Coca-Cola Co.
    <A HREF=\"http://www.investo
    r.reuters.com/FullQuote.aspx?t
    icker=KO.N target=/stocks/quic
    kinfo/fullquote\">KO.N</A
    >, stung by a prolonged
    downturn in North America,
    Germany and other major
    markets, on Thursday lowered
    its key long-term earnings
    and sales targets." + ], + [ + "JACKSONVILLE, Fla. -- They
    were singing in the Colts #39;
    locker room today, singing
    like a bunch of wounded
    songbirds. Never mind that
    Marcus Pollard, Dallas Clark
    and Ben Hartsock won #39;t be
    recording a remake of Kenny
    Chesney #39;s song,
    quot;Young, quot; any time
    soon." + ], + [ + "TOKYO (AP) -- Japanese
    electronics and entertainment
    giant Sony Corp. (SNE) plans
    to begin selling a camcorder
    designed for consumers that
    takes video at digital high-
    definition quality and is
    being priced at about
    \\$3,600..." + ], + [ + "As the close-knit NASCAR
    community mourns the loss of
    team owner Rick Hendrick #39;s
    son, brother, twin nieces and
    six others in a plane crash
    Sunday, perhaps no one outside
    of the immediate family
    grieves more deeply than
    Darrell Waltrip." + ], + [ + "AP - Purdue quarterback Kyle
    Orton has no trouble
    remembering how he felt after
    last year's game at Michigan." + ], + [ + "UNITED NATIONS - The United
    Nations #39; nuclear agency
    says it is concerned about the
    disappearance of equipment and
    materials from Iraq that could
    be used to make nuclear
    weapons." + ], + [ + " BRUSSELS (Reuters) - The EU's
    historic deal with Turkey to
    open entry talks with the vast
    Muslim country was hailed by
    supporters as a bridge builder
    between Europe and the Islamic
    world." + ], + [ + "Iraqi President Ghazi al-
    Yawar, who was due in Paris on
    Sunday to start a European
    tour, has postponed his visit
    to France due to the ongoing
    hostage drama involving two
    French journalists, Arab
    diplomats said Friday." + ], + [ + " SAO PAULO, Brazil (Reuters) -
    President Luiz Inacio Lula da
    Silva's Workers' Party (PT)
    won the mayoralty of six state
    capitals in Sunday's municipal
    vote but was forced into a
    run-off to defend its hold on
    the race's biggest prize, the
    city of Sao Paulo." + ], + [ + "ATHENS, Greece - They are
    America's newest golden girls
    - powerful and just a shade
    from perfection. The U.S..." + ], + [ + "AMMAN, Sept. 15. - The owner
    of a Jordanian truck company
    announced today that he had
    ordered its Iraq operations
    stopped in a bid to save the
    life of a driver held hostage
    by a militant group." + ], + [ + "AP - U.S. State Department
    officials learned that seven
    American children had been
    abandoned at a Nigerian
    orphanage but waited more than
    a week to check on the youths,
    who were suffering from
    malnutrition, malaria and
    typhoid, a newspaper reported
    Saturday." + ], + [ + "\\Angry mobs in Ivory Coast's
    main city, Abidjan, marched on
    the airport, hours after it
    came under French control." + ], + [ + "Several workers are believed
    to have been killed and others
    injured after a contruction
    site collapsed at Dubai
    airport. The workers were
    trapped under rubble at the
    site of a \\$4." + ], + [ + "Talks between Sudan #39;s
    government and two rebel
    groups to resolve the nearly
    two-year battle resume Friday.
    By Abraham McLaughlin Staff
    writer of The Christian
    Science Monitor." + ], + [ + "In a meeting of the cinematic
    with the scientific, Hollywood
    helicopter stunt pilots will
    try to snatch a returning NASA
    space probe out of the air
    before it hits the ground." + ], + [ + "Legend has it (incorrectly, it
    seems) that infamous bank
    robber Willie Sutton, when
    asked why banks were his
    favorite target, responded,
    quot;Because that #39;s where
    the money is." + ], + [ + "Brown is a second year player
    from Memphis and has spent the
    2004 season on the Steelers
    #39; practice squad. He played
    in two games last year." + ], + [ + "A Canadian court approved Air
    Canada #39;s (AC.TO: Quote,
    Profile, Research) plan of
    arrangement with creditors on
    Monday, clearing the way for
    the world #39;s 11th largest
    airline to emerge from
    bankruptcy protection at the
    end of next month" + ], + [ + "Pfizer, GlaxoSmithKline and
    Purdue Pharma are the first
    drugmakers willing to take the
    plunge and use radio frequency
    identification technology to
    protect their US drug supply
    chains from counterfeiters." + ], + [ + "Barret Jackman, the last of
    the Blues left to sign before
    the league #39;s probable
    lockout on Wednesday,
    finalized a deal Monday that
    is rare in the current
    economic climate but fitting
    for him." + ], + [ + " LONDON (Reuters) - Oil prices
    eased on Monday after rebels
    in Nigeria withdrew a threat
    to target oil operations, but
    lingering concerns over
    stretched supplies ahead of
    winter kept prices close to
    \\$50." + ], + [ + "AP - In the tumult of the
    visitors' clubhouse at Yankee
    Stadium, champagne pouring all
    around him, Theo Epstein held
    a beer. \"I came in and there
    was no champagne left,\" he
    said this week. \"I said, 'I'll
    have champagne if we win it
    all.'\" Get ready to pour a
    glass of bubbly for Epstein.
    No I.D. necessary." + ], + [ + "Search any fee-based digital
    music service for the best-
    loved musical artists of the
    20th century and most of the
    expected names show up." + ], + [ + "Barcelona held on from an
    early Deco goal to edge game
    local rivals Espanyol 1-0 and
    carve out a five point
    tabletop cushion. Earlier,
    Ronaldo rescued a point for
    Real Madrid, who continued
    their middling form with a 1-1
    draw at Real Betis." + ], + [ + "MONTREAL (CP) - The Expos may
    be history, but their demise
    has heated up the market for
    team memorabilia. Vintage
    1970s and 1980s shirts are
    already sold out, but
    everything from caps, beer
    glasses and key-chains to
    dolls of mascot Youppi!" + ], + [ + "Stansted airport is the
    designated emergency landing
    ground for planes in British
    airspace hit by in-flight
    security alerts. Emergency
    services at Stansted have
    successfully dealt" + ], + [ + "The massive military operation
    to retake Fallujah has been
    quot;accomplished quot;, a
    senior Iraqi official said.
    Fierce fighting continued in
    the war-torn city where
    pockets of resistance were
    still holding out against US
    forces." + ], + [ + "There are some signs of
    progress in resolving the
    Nigerian conflict that is
    riling global oil markets. The
    leader of militia fighters
    threatening to widen a battle
    for control of Nigeria #39;s
    oil-rich south has" + ], + [ + "A strong earthquake hit Taiwan
    on Monday, shaking buildings
    in the capital Taipei for
    several seconds. No casualties
    were reported." + ], + [ + "America Online Inc. is
    packaging new features to
    combat viruses, spam and
    spyware in response to growing
    online security threats.
    Subscribers will be able to
    get the free tools" + ], + [ + "A 76th minute goal from
    European Footballer of the
    Year Pavel Nedved gave
    Juventus a 1-0 win over Bayern
    Munich on Tuesday handing the
    Italians clear control at the
    top of Champions League Group
    C." + ], + [ + " LONDON (Reuters) - Oil prices
    climbed above \\$42 a barrel on
    Wednesday, rising for the
    third day in a row as cold
    weather gripped the U.S.
    Northeast, the world's biggest
    heating fuel market." + ], + [ + "A policeman ran amok at a
    security camp in Indian-
    controlled Kashmir after an
    argument and shot dead seven
    colleagues before he was
    gunned down, police said on
    Sunday." + ], + [ + "ANN ARBOR, Mich. -- Some NHL
    players who took part in a
    charity hockey game at the
    University of Michigan on
    Thursday were hopeful the news
    that the NHL and the players
    association will resume talks
    next week" + ], + [ + "New York police have developed
    a pre-emptive strike policy,
    cutting off demonstrations
    before they grow large." + ], + [ + "CAPE CANAVERAL-- NASA aims to
    launch its first post-Columbia
    shuttle mission during a
    shortened nine-day window
    March, and failure to do so
    likely would delay a planned
    return to flight until at
    least May." + ], + [ + "Travelers headed home for
    Thanksgiving were greeted
    Wednesday with snow-covered
    highways in the Midwest, heavy
    rain and tornadoes in parts of
    the South, and long security
    lines at some of the nation
    #39;s airports." + ], + [ + "BOULDER, Colo. -- Vernand
    Morency ran for 165 yards and
    two touchdowns and Donovan
    Woods threw for three more
    scores, lifting No. 22
    Oklahoma State to a 42-14
    victory over Colorado
    yesterday." + ], + [ + "The Chinese city of Beijing
    has cancelled an order for
    Microsoft software, apparently
    bowing to protectionist
    sentiment. The deal has come
    under fire in China, which is
    trying to build a domestic
    software industry." + ], + [ + "Apple says it will deliver its
    iTunes music service to more
    European countries next month.
    Corroborating several reports
    in recent months, Reuters is
    reporting today that Apple
    Computer is planning the next" + ], + [ + "Reuters - Motorola Inc., the
    world's\\second-largest mobile
    phone maker, said on Tuesday
    it expects\\to sustain strong
    sales growth in the second
    half of 2004\\thanks to new
    handsets with innovative
    designs and features." + ], + [ + "PULLMAN - Last week, in
    studying USC game film, Cougar
    coaches thought they found a
    chink in the national
    champions armor. And not just
    any chink - one with the
    potential, right from the get
    go, to" + ], + [ + "The union representing flight
    attendants on Friday said it
    mailed more than 5,000 strike
    authorization ballots to its
    members employed by US Airways
    as both sides continued talks
    that are expected to stretch
    through the weekend." + ], + [ + "AP - Matt Leinart was quite a
    baseball prospect growing up,
    showing so much promise as a
    left-handed pitcher that
    scouts took notice before high
    school." + ], + [ + "PRAGUE, Czech Republic --
    Eugene Cernan, the last man to
    walk on the moon during the
    final Apollo landing, said
    Thursday he doesn't expect
    space tourism to become
    reality in the near future,
    despite a strong demand.
    Cernan, now 70, who was
    commander of NASA's Apollo 17
    mission and set foot on the
    lunar surface in December 1972
    during his third space flight,
    acknowledged that \"there are
    many people interested in
    space tourism.\" But the
    former astronaut said he
    believed \"we are a long way
    away from the day when we can
    send a bus of tourists to the
    moon.\" He spoke to reporters
    before being awarded a medal
    by the Czech Academy of
    Sciences for his contribution
    to science..." + ], + [ + "Never shy about entering a
    market late, Microsoft Corp.
    is planning to open the
    virtual doors of its long-
    planned Internet music store
    next week. <FONT
    face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\"\\ color=\"#666666\">
    <B>-Leslie
    Walker</B></FONT>" + ], + [ + "AP - On his first birthday
    Thursday, giant panda cub Mei
    Sheng delighted visitors by
    playing for the first time in
    snow delivered to him at the
    San Diego Zoo. He also sat on
    his ice cake, wrestled with
    his mom, got his coat
    incredibly dirty, and didn't
    read any of the more than 700
    birthday wishes sent him via
    e-mail from as far away as
    Ireland and Argentina." + ], + [ + "AP - Researchers put a
    satellite tracking device on a
    15-foot shark that appeared to
    be lost in shallow water off
    Cape Cod, the first time a
    great white has been tagged
    that way in the Atlantic." + ], + [ + "LSU will stick with a two-
    quarterback rotation Saturday
    at Auburn, according to Tigers
    coach Nick Saban, who seemed
    to have some fun telling the
    media what he will and won
    #39;t discuss Monday." + ], + [ + "Bulgaria has started its first
    co-mission with the EU in
    Bosnia and Herzegovina, along
    with some 30 countries,
    including Canada and Turkey." + ], + [ + "The Windows Future Storage
    (WinFS) technology that got
    cut out of Windows
    quot;Longhorn quot; is in
    serious trouble, and not just
    the hot water a feature might
    encounter for missing its
    intended production vehicle." + ], + [ + "Seattle -- - Not so long ago,
    the 49ers were inflicting on
    other teams the kind of pain
    and embarrassment they felt in
    their 34-0 loss to the
    Seahawks on Sunday." + ], + [ + "AP - The pileup of events in
    the city next week, including
    the Republican National
    Convention, will add to the
    security challenge for the New
    York Police Department, but
    commissioner Ray Kelly says,
    \"With a big, experienced
    police force, we can do it.\"" + ], + [ + "washingtonpost.com - Let the
    games begin. Not the Olympics
    again, but the all-out battle
    between video game giants Sony
    Corp. and Nintendo Co. Ltd.
    The two Japanese companies are
    rolling out new gaming
    consoles, but Nintendo has
    beaten Sony to the punch by
    announcing an earlier launch
    date for its new hand-held
    game player." + ], + [ + "London - Manchester City held
    fierce crosstown rivals
    Manchester United to a 0-0
    draw on Sunday, keeping the
    Red Devils eleven points
    behind leaders Chelsea." + ], + [ + "LONDON, Dec 11 (IranMania) -
    Iraqi Vice-President Ibrahim
    al-Jaafari refused to believe
    in remarks published Friday
    that Iran was attempting to
    influence Iraqi polls with the
    aim of creating a
    quot;crescent quot; dominated
    by Shiites in the region." + ], + [ + "LOS ANGELES (CBS.MW) - The US
    Securities and Exchange
    Commission is probing
    transactions between Delphi
    Corp and EDS, which supplies
    the automotive parts and
    components giant with
    technology services, Delphi
    said late Wednesday." + ], + [ + "MONTREAL (CP) - Molson Inc.
    and Adolph Coors Co. are
    sweetening their brewery
    merger plan with a special
    dividend to Molson
    shareholders worth \\$381
    million." + ], + [ + "AP - Echoing what NASA
    officials said a day earlier,
    a Russian space official on
    Friday said the two-man crew
    on the international space
    station could be forced to
    return to Earth if a planned
    resupply flight cannot reach
    them with food supplies later
    this month." + ], + [ + "InfoWorld - SANTA CLARA,
    CALIF. -- Accommodating large
    patch sets in Linux is
    expected to mean forking off
    of the 2.7 version of the
    platform to accommodate these
    changes, according to Andrew
    Morton, lead maintainer of the
    Linux kernel for Open Source
    Development Labs (OSDL)." + ], + [ + "AMSTERDAM The mobile phone
    giants Vodafone and Nokia
    teamed up on Thursday to
    simplify cellphone software
    written with the Java computer
    language." + ], + [ + "WELLINGTON: National carrier
    Air New Zealand said yesterday
    the Australian Competition
    Tribunal has approved a
    proposed alliance with Qantas
    Airways Ltd, despite its
    rejection in New Zealand." + ], + [ + "The late Princess Dianas
    former bodyguard, Ken Wharfe,
    dismisses her suspicions that
    one of her lovers was bumped
    off. Princess Diana had an
    affair with Barry Mannakee, a
    policeman who was assigned to
    protect her." + ], + [ + "Long considered beyond the
    reach of mainland mores, the
    Florida city is trying to
    limit blatant displays of
    sexual behavior." + ], + [ + "The overall Linux market is
    far larger than previous
    estimates show, a new study
    says. In an analysis of the
    Linux market released late
    Tuesday, market research firm
    IDC estimated that the Linux
    market -- including" + ], + [ + "By PAUL ELIAS SAN FRANCISCO
    (AP) -- Several California
    cities and counties, including
    San Francisco and Los Angeles,
    sued Microsoft Corp. (MSFT) on
    Friday, accusing the software
    giant of illegally charging
    inflated prices for its
    products because of monopoly
    control of the personal
    computer operating system
    market..." + ], + [ + "New Ole Miss head coach Ed
    Orgeron, speaking for the
    first time since his hiring,
    made clear the goal of his
    football program. quot;The
    goal of this program will be
    to go to the Sugar Bowl, quot;
    Orgeron said." + ], + [ + "\"Everyone's nervous,\" Acting
    Undersecretary of Defense
    Michael W. Wynne warned in a
    confidential e-mail to Air
    Force Secretary James G. Roche
    on July 8, 2003." + ], + [ + "Reuters - Alpharma Inc. on
    Friday began\\selling a cheaper
    generic version of Pfizer
    Inc.'s #36;3\\billion a year
    epilepsy drug Neurontin
    without waiting for a\\court
    ruling on Pfizer's request to
    block the copycat medicine." + ], + [ + "Public opinion of the database
    giant sinks to 12-year low, a
    new report indicates." + ], + [ + "Opinion: Privacy hysterics
    bring old whine in new bottles
    to the Internet party. The
    desktop search beta from this
    Web search leader doesn #39;t
    do anything you can #39;t do
    already." + ], + [ + "It is much too easy to call
    Pedro Martinez the selfish
    one, to say he is walking out
    on the Red Sox, his baseball
    family, for the extra year of
    the Mets #39; crazy money." + ], + [ + "It is impossible for young
    tennis players today to know
    what it was like to be Althea
    Gibson and not to be able to
    quot;walk in the front door,
    quot; Garrison said." + ], + [ + "Senator John Kerry said today
    that the war in Iraq was a
    \"profound diversion\" from the
    war on terror and Osama bin
    Laden." + ], + [ + "For spammers, it #39;s been a
    summer of love. Two newly
    issued reports tracking the
    circulation of unsolicited
    e-mails say pornographic spam
    dominated this summer, nearly
    all of it originating from
    Internet addresses in North
    America." + ], + [ + "The Nordics fared well because
    of their long-held ideals of
    keeping corruption clamped
    down and respect for
    contracts, rule of law and
    dedication to one-on-one
    business relationships." + ], + [ + "Microsoft portrayed its
    Longhorn decision as a
    necessary winnowing to hit the
    2006 timetable. The
    announcement on Friday,
    Microsoft executives insisted,
    did not point to a setback in
    software" + ], + [ + "Here #39;s an obvious word of
    advice to Florida athletic
    director Jeremy Foley as he
    kicks off another search for
    the Gators football coach: Get
    Steve Spurrier on board." + ], + [ + "Don't bother with the small
    stuff. Here's what really
    matters to your lender." + ], + [ + "A problem in the Service Pack
    2 update for Windows XP may
    keep owners of AMD-based
    computers from using the long-
    awaited security package,
    according to Microsoft." + ], + [ + "Five years ago, running a
    telephone company was an
    immensely profitable
    proposition. Since then, those
    profits have inexorably
    declined, and now that decline
    has taken another gut-
    wrenching dip." + ], + [ + "NEW YORK - The litigious
    Recording Industry Association
    of America (RIAA) is involved
    in another legal dispute with
    a P-to-P (peer-to-peer)
    technology maker, but this
    time, the RIAA is on defense.
    Altnet Inc. filed a lawsuit
    Wednesday accusing the RIAA
    and several of its partners of
    infringing an Altnet patent
    covering technology for
    identifying requested files on
    a P-to-P network." + ], + [ + "A group claiming to have
    captured two Indonesian women
    in Iraq has said it will
    release them if Jakarta frees
    Muslim cleric Abu Bakar Bashir
    being held for alleged
    terrorist links." + ], + [ + "Amid the stormy gloom in
    Gotham, the rain-idled Yankees
    last night had plenty of time
    to gather in front of their
    televisions and watch the Red
    Sox Express roar toward them.
    The national telecast might
    have been enough to send a
    jittery Boss Steinbrenner
    searching his Bartlett's
    Familiar Quotations for some
    quot;Little Engine That Could
    quot; metaphor." + ], + [ + "FULHAM fans would have been
    singing the late Elvis #39;
    hit #39;The wonder of you
    #39; to their player Elvis
    Hammond. If not for Frank
    Lampard spoiling the party,
    with his dedication to his
    late grandfather." + ], + [ + "Indonesian police said
    yesterday that DNA tests had
    identified a suicide bomber
    involved in a deadly attack
    this month on the Australian
    embassy in Jakarta." + ], + [ + "NEW YORK - Wal-Mart Stores
    Inc.'s warning of
    disappointing sales sent
    stocks fluctuating Monday as
    investors' concerns about a
    slowing economy offset their
    relief over a drop in oil
    prices. October contracts
    for a barrel of light crude
    were quoted at \\$46.48, down
    24 cents, on the New York
    Mercantile Exchange..." + ], + [ + "Iraq #39;s top Shi #39;ite
    cleric made a sudden return to
    the country on Wednesday and
    said he had a plan to end an
    uprising in the quot;burning
    city quot; of Najaf, where
    fighting is creeping ever
    closer to its holiest shrine." + ], + [ + "For two weeks before MTV
    debuted U2 #39;s video for the
    new single quot;Vertigo,
    quot; fans had a chance to see
    the band perform the song on
    TV -- in an iPod commercial." + ], + [ + "A bird #39;s eye view of the
    circuit at Shanghai shows what
    an event Sunday #39;s Chinese
    Grand Prix will be. The course
    is arguably one of the best
    there is, and so it should be
    considering the amount of
    money that has been spent on
    it." + ], + [ + "KABUL, Afghanistan Aug. 22,
    2004 - US soldiers sprayed a
    pickup truck with bullets
    after it failed to stop at a
    roadblock in central
    Afghanistan, killing two women
    and a man and critically
    wounding two other" + ], + [ + "Oct. 26, 2004 - The US-
    European spacecraft Cassini-
    Huygens on Tuesday made a
    historic flyby of Titan,
    Saturn #39;s largest moon,
    passing so low as to almost
    touch the fringes of its
    atmosphere." + ], + [ + "Reuters - A volcano in central
    Japan sent smoke and\\ash high
    into the sky and spat out
    molten rock as it erupted\\for
    a fourth straight day on
    Friday, but experts said the
    peak\\appeared to be quieting
    slightly." + ], + [ + "Shares of Google Inc. made
    their market debut on Thursday
    and quickly traded up 19
    percent at \\$101.28. The Web
    search company #39;s initial
    public offering priced at \\$85" + ], + [ + "SYDNEY -- Prime Minister John
    Howard of Australia, a key US
    ally and supporter of the Iraq
    war, celebrated his election
    win over opposition Labor
    after voters enjoying the
    fruits of a strong economy
    gave him another term." + ], + [ + "Reuters - Global warming is
    melting\\Ecuador's cherished
    mountain glaciers and could
    cause several\\of them to
    disappear over the next two
    decades, Ecuadorean and\\French
    scientists said on Wednesday." + ], + [ + "AP - Sirius Satellite Radio
    signed a deal to air the men's
    NCAA basketball tournament
    through 2007, the latest move
    made in an attempt to draw
    customers through sports
    programming." + ], + [ + "Rather than tell you, Dan
    Kranzler chooses instead to
    show you how he turned Mforma
    into a worldwide publisher of
    video games, ringtones and
    other hot downloads for mobile
    phones." + ], + [ + "UK interest rates have been
    kept on hold at 4.75 following
    the latest meeting of the Bank
    of England #39;s rate-setting
    committee." + ], + [ + "BAGHDAD, Iraq - Two rockets
    hit a downtown Baghdad hotel
    housing foreigners and
    journalists Thursday, and
    gunfire erupted in the
    neighborhood across the Tigris
    River from the U.S. Embassy
    compound..." + ], + [ + "The Prevention of Terrorism
    Act 2002 (Pota) polarised the
    country, not just by the
    manner in which it was pushed
    through by the NDA government
    through a joint session of
    Parliament but by the shabby
    and often biased manner in
    which it was enforced." + ], + [ + "Not being part of a culture
    with a highly developed
    language, could limit your
    thoughts, at least as far as
    numbers are concerned, reveals
    a new study conducted by a
    psychologist at the Columbia
    University in New York." + ], + [ + "CAMBRIDGE, Mass. A native of
    Red Oak, Iowa, who was a
    pioneer in astronomy who
    proposed the quot;dirty
    snowball quot; theory for the
    substance of comets, has died." + ], + [ + "Resurgent oil prices paused
    for breath as the United
    States prepared to draw on its
    emergency reserves to ease
    supply strains caused by
    Hurricane Ivan." + ], + [ + "(Sports Network) - The
    inconsistent San Diego Padres
    will try for consecutive wins
    for the first time since
    August 28-29 tonight, when
    they begin a huge four-game
    set against the Los Angeles
    Dodgers at Dodger Stadium." + ], + [ + "A Portuguese-sounding version
    of the virus has appeared in
    the wild. Be wary of mail from
    Manaus." + ], + [ + " NEW YORK (Reuters) - Top seed
    Roger Federer survived a
    stirring comeback from twice
    champion Andre Agassi to reach
    the semifinals of the U.S.
    Open for the first time on
    Thursday, squeezing through
    6-3, 2-6, 7-5, 3-6, 6-3." + ], + [ + "President Bush, who credits
    three years of tax relief
    programs with helping
    strengthen the slow economy,
    said Saturday he would sign
    into law the Working Families
    Tax Relief Act to preserve tax
    cuts." + ], + [ + "HEN investors consider the
    bond market these days, the
    low level of interest rates
    should be more cause for worry
    than for gratitude." + ], + [ + "Reuters - Hunters soon may be
    able to sit at\\their computers
    and blast away at animals on a
    Texas ranch via\\the Internet,
    a prospect that has state
    wildlife officials up\\in arms." + ], + [ + "The Bedminster-based company
    yesterday said it was pushing
    into 21 new markets with the
    service, AT amp;T CallVantage,
    and extending an introductory
    rate offer until Sept. 30. In
    addition, the company is
    offering in-home installation
    of up to five ..." + ], + [ + "Samsung Electronics Co., Ltd.
    has developed a new LCD
    (liquid crystal display)
    technology that builds a touch
    screen into the display, a
    development that could lead to
    thinner and cheaper display
    panels for mobile phones, the
    company said Tuesday." + ], + [ + "The US military says marines
    in Fallujah shot and killed an
    insurgent who engaged them as
    he was faking being dead, a
    week after footage of a marine
    killing an apparently unarmed
    and wounded Iraqi caused a
    stir in the region." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks rallied on Monday after
    software maker PeopleSoft Inc.
    <A HREF=\"http://www.investo
    r.reuters.com/FullQuote.aspx?t
    icker=PSFT.O target=/stocks/qu
    ickinfo/fullquote\">PSFT.O&l
    t;/A> accepted a sweetened
    \\$10.3 billion buyout by rival
    Oracle Corp.'s <A HREF=\"htt
    p://www.investor.reuters.com/F
    ullQuote.aspx?ticker=ORCL.O ta
    rget=/stocks/quickinfo/fullquo
    te\">ORCL.O</A> and
    other big deals raised
    expectations of more
    takeovers." + ], + [ + "SAN FRANCISCO - What Babe Ruth
    was to the first half of the
    20th century and Hank Aaron
    was to the second, Barry Bonds
    has become for the home run
    generation." + ], + [ + "Description: NPR #39;s Alex
    Chadwick talks to Colin Brown,
    deputy political editor for
    the United Kingdom #39;s
    Independent newspaper,
    currently covering the British
    Labour Party Conference." + ], + [ + "Birgit Fischer settled for
    silver, leaving the 42-year-
    old Olympian with two medals
    in two days against decidedly
    younger competition." + ], + [ + "The New Jersey-based Accoona
    Corporation, an industry
    pioneer in artificial
    intelligence search
    technology, announced on
    Monday the launch of Accoona." + ], + [ + "Hamas vowed revenge yesterday
    after an Israeli airstrike in
    Gaza killed one of its senior
    commanders - the latest
    assassination to have weakened
    the militant group." + ], + [ + "There was no mystery ... no
    secret strategy ... no baited
    trap that snapped shut and
    changed the course of history
    #39;s most lucrative non-
    heavyweight fight." + ], + [ + "Notre Dame accepted an
    invitation Sunday to play in
    the Insight Bowl in Phoenix
    against a Pac-10 team on Dec.
    28. The Irish (6-5) accepted
    the bid a day after losing to
    Southern California" + ], + [ + "Greg Anderson has so dominated
    Pro Stock this season that his
    championship quest has evolved
    into a pursuit of NHRA
    history. By Bob Hesser, Racers
    Edge Photography." + ], + [ + "Goldman Sachs Group Inc. on
    Thursday said fourth-quarter
    profit rose as its fixed-
    income, currency and
    commodities business soared
    while a rebounding stock
    market boosted investment
    banking." + ], + [ + " NEW YORK (Reuters) - The
    dollar rose on Monday in a
    retracement from last week's
    steep losses, but dealers said
    the bias toward a weaker
    greenback remained intact." + ], + [ + "Michael Powell, chairman of
    the FCC, said Wednesday he was
    disappointed with ABC for
    airing a sexually suggestive
    opening to \"Monday Night
    Football.\"" + ], + [ + "The message against illegally
    copying CDs for uses such as
    in file-sharing over the
    Internet has widely sunk in,
    said the company in it #39;s
    recent announcement to drop
    the Copy-Control program." + ], + [ + "Former Washington football
    coach Rick Neuheisel looked
    forward to a return to
    coaching Wednesday after being
    cleared by the NCAA of
    wrongdoing related to his
    gambling on basketball games." + ], + [ + "Reuters - California will
    become hotter and\\drier by the
    end of the century, menacing
    the valuable wine and\\dairy
    industries, even if dramatic
    steps are taken to curb\\global
    warming, researchers said on
    Monday." + ], + [ + "A senior member of the
    Palestinian resistance group
    Hamas has been released from
    an Israeli prison after
    completing a two-year
    sentence." + ], + [ + "IBM said Monday that it won a
    500 million (AUD\\$1.25
    billion), seven-year services
    contract to help move UK bank
    Lloyds TBS from its
    traditional voice
    infrastructure to a converged
    voice and data network." + ], + [ + "MPs have announced a new
    inquiry into family courts and
    whether parents are treated
    fairly over issues such as
    custody or contact with their
    children." + ], + [ + "Canadian Press - MELBOURNE,
    Australia (AP) - A 36-year-old
    businesswoman was believed to
    be the first woman to walk
    around Australia on Friday
    after striding into her
    hometown of Melbourne to
    complete her 16,700-kilometre
    trek in 365 days." + ], + [ + "Most remaining Pakistani
    prisoners held at the US
    Guantanamo Bay prison camp are
    freed, officials say." + ], + [ + "Space shuttle astronauts will
    fly next year without the
    ability to repair in orbit the
    type of damage that destroyed
    the Columbia vehicle in
    February 2003." + ], + [ + "Moscow - Russia plans to
    combine Gazprom, the world
    #39;s biggest natural gas
    producer, with state-owned oil
    producer Rosneft, easing rules
    for trading Gazprom shares and
    creating a company that may
    dominate the country #39;s
    energy industry." + ], + [ + "AP - Rutgers basketball player
    Shalicia Hurns was suspended
    from the team after pleading
    guilty to punching and tying
    up her roommate during a
    dispute over painkilling
    drugs." + ], + [ + "By cutting WinFS from Longhorn
    and indefinitely delaying the
    storage system, Microsoft
    Corp. has also again delayed
    the Microsoft Business
    Framework (MBF), a new Windows
    programming layer that is
    closely tied to WinFS." + ], + [ + "French police are
    investigating an arson-caused
    fire at a Jewish Social Center
    that might have killed dozens
    without the quick response of
    firefighters." + ], + [ + "Rodney King, whose videotaped
    beating led to riots in Los
    Angeles in 1992, is out of
    jail now and talking frankly
    for the first time about the
    riots, himself and the
    American way of life." + ], + [ + "AFP - Radical Islamic cleric
    Abu Hamza al-Masri was set to
    learn Thursday whether he
    would be charged under
    Britain's anti-terrorism law,
    thus delaying his possible
    extradition to the United
    States to face terrorism-
    related charges." + ], + [ + "Diversified manufacturer
    Honeywell International Inc.
    (HON.N: Quote, Profile,
    Research) posted a rise in
    quarterly profit as strong
    demand for aerospace equipment
    and automobile components" + ], + [ + "This was the event Michael
    Phelps didn't really need to
    compete in if his goal was to
    win eight golds. He probably
    would have had a better chance
    somewhere else." + ], + [ + "Samsung's new SPH-V5400 mobile
    phone sports a built-in
    1-inch, 1.5-gigabyte hard disk
    that can store about 15 times
    more data than conventional
    handsets, Samsung said." + ], + [ + "Reuters - U.S. housing starts
    jumped a\\larger-than-expected
    6.4 percent in October to the
    busiest pace\\since December as
    buyers took advantage of low
    mortgage rates,\\a government
    report showed on Wednesday." + ], + [ + "Gabe Kapler became the first
    player to leave the World
    Series champion Boston Red
    Sox, agreeing to a one-year
    contract with the Yomiuri
    Giants in Tokyo." + ], + [ + "Louisen Louis, 30, walked
    Monday in the middle of a
    street that resembled a small
    river with brown rivulets and
    waves. He wore sandals and had
    a cut on one of his big toes." + ], + [ + "BALI, Indonesia - Svetlana
    Kuznetsova, fresh off her
    championship at the US Open,
    defeated Australian qualifier
    Samantha Stosur 6-4, 6-4
    Thursday to reach the
    quarterfinals of the Wismilak
    International." + ], + [ + "The Securities and Exchange
    Commission ordered mutual
    funds to stop paying higher
    commissions to brokers who
    promote the companies' funds
    and required portfolio
    managers to reveal investments
    in funds they supervise." + ], + [ + "A car bomb exploded outside
    the main hospital in Chechny
    #39;s capital, Grozny, on
    Sunday, injuring 17 people in
    an attack apparently targeting
    members of a Chechen security
    force bringing in wounded from
    an earlier explosion" + ], + [ + "AP - Just like the old days in
    Dallas, Emmitt Smith made life
    miserable for the New York
    Giants on Sunday." + ], + [ + "Sumitomo Mitsui Financial
    Group (SMFG), Japans second
    largest bank, today put
    forward a 3,200 billion (\\$29
    billion) takeover bid for
    United Financial Group (UFJ),
    the countrys fourth biggest
    lender, in an effort to regain
    initiative in its bidding" + ], + [ + "AP - Gay marriage is emerging
    as a big enough issue in
    several states to influence
    races both for Congress and
    the presidency." + ], + [ + "TAMPA, Fla. - Chris Simms
    first NFL start lasted 19
    plays, and it might be a while
    before he plays again for the
    Tampa Bay Buccaneers." + ], + [ + "Vendor says it #39;s
    developing standards-based
    servers in various form
    factors for the telecom
    market. By Darrell Dunn.
    Hewlett-Packard on Thursday
    unveiled plans to create a
    portfolio of products and
    services" + ], + [ + "Jarno Trulli made the most of
    the conditions in qualifying
    to claim pole ahead of Michael
    Schumacher, while Fernando
    finished third." + ], + [ + "More than 30 aid workers have
    been airlifted to safety from
    a town in Sudan #39;s troubled
    Darfur region after fighting
    broke out and their base was
    bombed, a British charity
    says." + ], + [ + " NEW YORK (Reuters) - U.S.
    chain store retail sales
    slipped during the
    Thanksgiving holiday week, as
    consumers took advantage of
    discounted merchandise, a
    retail report said on
    Tuesday." + ], + [ + "Ziff Davis - The company this
    week will unveil more programs
    and technologies designed to
    ease users of its high-end
    servers onto its Integrity
    line, which uses Intel's
    64-bit Itanium processor." + ], + [ + "The Mac maker says it will
    replace about 28,000 batteries
    in one model of PowerBook G4
    and tells people to stop using
    the notebook." + ], + [ + "It #39;s the mildest of mild
    winters down here in the south
    of Italy and, last weekend at
    Bcoli, a pretty suburb by the
    seaside west of Naples, the
    customers of Pizzeria quot;Da
    Enrico quot; were making the
    most of it." + ], + [ + "By George Chamberlin , Daily
    Transcript Financial
    Correspondent. Concerns about
    oil production leading into
    the winter months sent shivers
    through the stock market
    Wednesday." + ], + [ + "Airbus has withdrawn a filing
    that gave support for
    Microsoft in an antitrust case
    before the European Union
    #39;s Court of First Instance,
    a source close to the
    situation said on Friday." + ], + [ + "WASHINGTON - A spotty job
    market and stagnant paychecks
    cloud this Labor Day holiday
    for many workers, highlighting
    the importance of pocketbook
    issues in the presidential
    election. \"Working harder
    and enjoying it less,\" said
    economist Ken Mayland,
    president of ClearView
    Economics, summing up the
    state of working America..." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks rose on Wednesday
    lifted by a merger between
    retailers Kmart and Sears,
    better-than-expected earnings
    from Hewlett-Packard and data
    showing a slight rise in core
    inflation." + ], + [ + "AP - Authorities are
    investigating whether bettors
    at New York's top thoroughbred
    tracks were properly informed
    when jockeys came in
    overweight at races, a source
    familiar with the probe told
    The Associated Press." + ], + [ + "European Commission president
    Romano Prodi has unveiled
    proposals to loosen the
    deficit rules under the EU
    Stability Pact. The loosening
    was drafted by monetary
    affairs commissioner Joaquin
    Almunia, who stood beside the
    president at the announcement." + ], + [ + "Canadian Press - MONTREAL (CP)
    - A 19-year-old man charged in
    a firebombing at a Jewish
    elementary school pleaded
    guilty Thursday to arson." + ], + [ + "Retail sales in Britain saw
    the fastest growth in
    September since January,
    casting doubts on the view
    that the economy is slowing
    down, according to official
    figures released Thursday." + ], + [ + " NEW YORK (Reuters) -
    Interstate Bakeries Corp.
    <A HREF=\"http://www.investo
    r.reuters.com/FullQuote.aspx?t
    icker=IBC.N target=/stocks/qui
    ckinfo/fullquote\">IBC.N<
    /A>, maker of Hostess
    Twinkies and Wonder Bread,
    filed for bankruptcy on
    Wednesday after struggling
    with more than \\$1.3 billion
    in debt and high costs." + ], + [ + "Delta Air Lines (DAL.N: Quote,
    Profile, Research) on Thursday
    said it reached a deal with
    FedEx Express to sell eight
    McDonnell Douglas MD11
    aircraft and four spare
    engines for delivery in 2004." + ], + [ + "Michael Owen scored his first
    goal for Real Madrid in a 1-0
    home victory over Dynamo Kiev
    in the Champions League. The
    England striker toe-poked home
    Ronaldo #39;s cross in the
    35th minute to join the
    Russians" + ], + [ + " quot;Resuming uranium
    enrichment is not in our
    agenda. We are still committed
    to the suspension, quot;
    Foreign Ministry spokesman
    Hamid Reza." + ], + [ + "Leading OPEC producer Saudi
    Arabia said on Monday in
    Vienna, Austria, that it had
    made a renewed effort to
    deflate record high world oil
    prices by upping crude output
    again." + ], + [ + "The U.S. military presence in
    Iraq will grow to 150,000
    troops by next month, the
    highest level since the
    invasion last year." + ], + [ + "UPDATE, SUN 9PM: More than a
    million people have left their
    homes in Cuba, as Hurricane
    Ivan approaches. The ferocious
    storm is headed that way,
    after ripping through the
    Cayman Islands, tearing off
    roofs, flooding homes and
    causing general havoc." + ], + [ + "Jason Giambi has returned to
    the New York Yankees'
    clubhouse but is still
    clueless as to when he will be
    able to play again." + ], + [ + "Seventy-five National Hockey
    League players met with union
    leaders yesterday to get an
    update on a lockout that shows
    no sign of ending." + ], + [ + "Howard, at 6-4 overall and 3-3
    in the Mid-Eastern Athletic
    Conference, can clinch a
    winning record in the MEAC
    with a win over Delaware State
    on Saturday." + ], + [ + "Millions of casual US anglers
    are having are larger than
    appreciated impact on sea fish
    stocks, scientists claim." + ], + [ + "The founders of the Pilgrim
    Baxter amp; Associates money-
    management firm agreed
    yesterday to personally fork
    over \\$160 million to settle
    charges they allowed a friend
    to" + ], + [ + " ATHENS (Reuters) - Top-ranked
    Argentina booked their berth
    in the women's hockey semi-
    finals at the Athens Olympics
    on Friday but defending
    champions Australia now face
    an obstacle course to qualify
    for the medal matches." + ], + [ + "IBM Corp. Tuesday announced
    plans to acquire software
    vendor Systemcorp ALG for an
    undisclosed amount. Systemcorp
    of Montreal makes project
    portfolio management software
    aimed at helping companies
    better manage their IT
    projects." + ], + [ + "The Notre Dame message boards
    are no longer discussing
    whether Tyrone Willingham
    should be fired. Theyre
    already arguing about whether
    the next coach should be Barry
    Alvarez or Steve Spurrier." + ], + [ + "Forbes.com - By now you
    probably know that earnings of
    Section 529 college savings
    accounts are free of federal
    tax if used for higher
    education. But taxes are only
    part of the problem. What if
    your investments tank? Just
    ask Laurence and Margo
    Williams of Alexandria, Va. In
    2000 they put #36;45,000 into
    the Virginia Education Savings
    Trust to open accounts for
    daughters Lea, now 5, and
    Anne, now 3. Since then their
    investment has shrunk 5 while
    the average private college
    tuition has climbed 18 to
    #36;18,300." + ], + [ + "German Chancellor Gerhard
    Schroeder said Sunday that
    there was quot;no problem
    quot; with Germany #39;s
    support to the start of
    negotiations on Turkey #39;s
    entrance into EU." + ], + [ + "Coca-Cola Amatil Ltd.,
    Australia #39;s biggest soft-
    drink maker, offered A\\$500
    million (\\$382 million) in
    cash and stock for fruit
    canner SPC Ardmona Ltd." + ], + [ + "US technology shares tumbled
    on Friday after technology
    bellwether Intel Corp.
    (INTC.O: Quote, Profile,
    Research) slashed its revenue
    forecast, but blue chips were
    only moderately lower as drug
    and industrial stocks made
    solid gains." + ], + [ + " WASHINGTON (Reuters) - Final
    U.S. government tests on an
    animal suspected of having mad
    cow disease were not yet
    complete, the U.S. Agriculture
    Department said, with no
    announcement on the results
    expected on Monday." + ], + [ + "MANILA Fernando Poe Jr., the
    popular actor who challenged
    President Gloria Macapagal
    Arroyo in the presidential
    elections this year, died
    early Tuesday." + ], + [ + " THOMASTOWN, Ireland (Reuters)
    - World number three Ernie
    Els overcame difficult weather
    conditions to fire a sparkling
    eight-under-par 64 and move
    two shots clear after two
    rounds of the WGC-American
    Express Championship Friday." + ], + [ + "Lamar Odom supplemented 20
    points with 13 rebounds and
    Kobe Bryant added 19 points to
    overcome a big night from Yao
    Ming as the Los Angeles Lakers
    ground out an 84-79 win over
    the Rockets in Houston
    Saturday." + ], + [ + "AMSTERDAM, NETHERLANDS - A
    Dutch filmmaker who outraged
    members of the Muslim
    community by making a film
    critical of the mistreatment
    of women in Islamic society
    was gunned down and stabbed to
    death Tuesday on an Amsterdam
    street." + ], + [ + "Microsoft Xbox Live traffic on
    service provider networks
    quadrupled following the
    November 9th launch of Halo-II
    -- which set entertainment
    industry records by selling
    2.4-million units in the US
    and Canada on the first day of
    availability, driving cash" + ], + [ + "Lawyers in a California class
    action suit against Microsoft
    will get less than half the
    payout they had hoped for. A
    judge in San Francisco ruled
    that the attorneys will
    collect only \\$112." + ], + [ + "Google Browser May Become
    Reality\\\\There has been much
    fanfare in the Mozilla fan
    camps about the possibility of
    Google using Mozilla browser
    technology to produce a
    GBrowser - the Google Browser.
    Over the past two weeks, the
    news and speculation has
    escalated to the point where
    even Google itself is ..." + ], + [ + "Metro, Germany's biggest
    retailer, turns in weaker-
    than-expected profits as sales
    at its core supermarkets
    division dip lower." + ], + [ + "It rained Sunday, of course,
    and but another soppy, sloppy
    gray day at Westside Tennis
    Club did nothing to deter
    Roger Federer from his
    appointed rounds." + ], + [ + "Hewlett-Packard has joined
    with Brocade to integrate
    Brocade #39;s storage area
    network switching technology
    into HP Bladesystem servers to
    reduce the amount of fabric
    infrastructure needed in a
    datacentre." + ], + [ + "Zimbabwe #39;s most persecuted
    white MP began a year of hard
    labour last night after
    parliament voted to jail him
    for shoving the Justice
    Minister during a debate over
    land seizures." + ], + [ + "BOSTON (CBS.MW) -- First
    Command has reached a \\$12
    million settlement with
    federal regulators for making
    misleading statements and
    omitting important information
    when selling mutual funds to
    US military personnel." + ], + [ + "A smashing blow is being dealt
    to thousands of future
    pensioners by a law that has
    just been brought into force
    by the Federal Government." + ], + [ + "The US Senate Commerce
    Committee on Wednesday
    approved a measure that would
    provide up to \\$1 billion to
    ensure consumers can still
    watch television when
    broadcasters switch to new,
    crisp digital signals." + ], + [ + "Amelie Mauresmo was handed a
    place in the Advanta
    Championships final after
    Maria Sharapova withdrew from
    their semi-final because of
    injury." + ], + [ + "AP - An Israeli helicopter
    fired two missiles in Gaza
    City after nightfall
    Wednesday, one at a building
    in the Zeitoun neighborhood,
    witnesses said, setting a
    fire." + ], + [ + "Reuters - Internet stocks
    are\\as volatile as ever, with
    growth-starved investors
    flocking to\\the sector in the
    hope they've bought shares in
    the next online\\blue chip." + ], + [ + "The federal government, banks
    and aircraft lenders are
    putting the clamps on
    airlines, particularly those
    operating under bankruptcy
    protection." + ], + [ + "EURO DISNEY, the financially
    crippled French theme park
    operator, has admitted that
    its annual losses more than
    doubled last financial year as
    it was hit by a surge in
    costs." + ], + [ + " EAST RUTHERFORD, New Jersey
    (Sports Network) - Retired NBA
    center and seven-time All-
    Star Alonzo Mourning is going
    to give his playing career
    one more shot." + ], + [ + "NASA has released an inventory
    of the scientific devices to
    be put on board the Mars
    Science Laboratory rover
    scheduled to land on the
    surface of Mars in 2009, NASAs
    news release reads." + ], + [ + "The U.S. Congress needs to
    invest more in the U.S.
    education system and do more
    to encourage broadband
    adoption, the chief executive
    of Cisco said Wednesday.<p&
    gt;ADVERTISEMENT</p><
    p><img src=\"http://ad.do
    ubleclick.net/ad/idg.us.ifw.ge
    neral/sbcspotrssfeed;sz=1x1;or
    d=200301151450?\" width=\"1\"
    height=\"1\"
    border=\"0\"/><a href=\"htt
    p://ad.doubleclick.net/clk;922
    8975;9651165;a?http://www.info
    world.com/spotlights/sbc/main.
    html?lpid0103035400730000idlp\"
    >SBC Case Study: Crate Ba
    rrel</a><br/>What
    sold them on improving their
    network? A system that could
    cut management costs from the
    get-go. Find out
    more.</p>" + ], + [ + "The Philippines put the toll
    at more than 1,000 dead or
    missing in four storms in two
    weeks but, even with a break
    in the weather on Saturday" + ], + [ + "Reuters - Four explosions were
    reported at petrol\\stations in
    the Madrid area on Friday,
    Spanish radio stations\\said,
    following a phone warning in
    the name of the armed
    Basque\\separatist group ETA to
    a Basque newspaper." + ], + [ + "Thirty-two countries and
    regions will participate the
    Fifth China International
    Aviation and Aerospace
    Exhibition, opening Nov. 1 in
    Zhuhai, a city in south China
    #39;s Guangdong Province." + ], + [ + "Jordan have confirmed that
    Timo Glock will replace
    Giorgio Pantano for this
    weekend #39;s Chinese GP as
    the team has terminated its
    contract with Pantano." + ], + [ + "WEST PALM BEACH, Fla. -
    Hurricane Jeanne got stronger,
    bigger and faster as it
    battered the Bahamas and bore
    down on Florida Saturday,
    sending huge waves crashing
    onto beaches and forcing
    thousands into shelters just
    weeks after Frances ravaged
    this area..." + ], + [ + "p2pnet.net News:- Virgin
    Electronics has joined the mp3
    race with a \\$250, five gig
    player which also handles
    Microsoft #39;s WMA format." + ], + [ + "WASHINGTON The idea of a no-
    bid contract for maintaining
    airport security equipment has
    turned into a non-starter for
    the Transportation Security
    Administration." + ], + [ + "Eyetech (EYET:Nasdaq - news -
    research) did not open for
    trading Friday because a Food
    and Drug Administration
    advisory committee is meeting
    to review the small New York-
    based biotech #39;s
    experimental eye disease drug." + ], + [ + "The continuing heartache of
    Wake Forest #39;s ACC football
    season was best described by
    fifth-ranked Florida State
    coach Bobby Bowden, after his
    Seminoles had edged the
    Deacons 20-17 Saturday at
    Groves Stadium." + ], + [ + "On September 13, 2001, most
    Americans were still reeling
    from the shock of the
    terrorist attacks on New York
    and the Pentagon two days
    before." + ], + [ + "Microsoft has suspended the
    beta testing of the next
    version of its MSN Messenger
    client because of a potential
    security problem, a company
    spokeswoman said Wednesday." + ], + [ + "AP - Business software maker
    Oracle Corp. attacked the
    credibility and motives of
    PeopleSoft Inc.'s board of
    directors Monday, hoping to
    rally investor support as the
    17-month takeover battle
    between the bitter business
    software rivals nears a
    climactic showdown." + ], + [ + "NEW YORK - Elena Dementieva
    shook off a subpar serve that
    produced 15 double-faults, an
    aching left thigh and an upset
    stomach to advance to the
    semifinals at the U.S. Open
    with a 4-6, 6-4, 7-6 (1)
    victory Tuesday over Amelie
    Mauresmo..." + ], + [ + "THE glory days have returned
    to White Hart Lane. When Spurs
    new first-team coach Martin
    Jol promised a return to the
    traditions of the 1960s,
    nobody could have believed he
    was so determined to act so
    quickly and so literally." + ], + [ + "A new worm has been discovered
    in the wild that #39;s not
    just settling for invading
    users #39; PCs--it wants to
    invade their homes too." + ], + [ + "Domestic air travelers could
    be surfing the Web by 2006
    with government-approved
    technology that allows people
    access to high-speed Internet
    connections while they fly." + ], + [ + "GENEVA: Cross-border
    investment is set to bounce in
    2004 after three years of deep
    decline, reflecting a stronger
    world economy and more
    international merger activity,
    the United Nations (UN) said
    overnight." + ], + [ + "Researchers have for the first
    time established the existence
    of odd-parity superconductors,
    materials that can carry
    electric current without any
    resistance." + ], + [ + "Chewing gum giant Wm. Wrigley
    Jr. Co. on Thursday said it
    plans to phase out production
    of its Eclipse breath strips
    at a plant in Phoenix, Arizona
    and shift manufacturing to
    Poznan, Poland." + ], + [ + "Prime Minister Dr Manmohan
    Singh inaugurated a research
    centre in the Capital on
    Thursday to mark 400 years of
    compilation of Sikh holy book
    the Guru Granth Sahib." + ], + [ + "com September 16, 2004, 7:58
    AM PT. This fourth priority
    #39;s main focus has been
    improving or obtaining CRM and
    ERP software for the past year
    and a half." + ], + [ + "BRUSSELS, Belgium (AP) --
    European antitrust regulators
    said Monday they have extended
    their review of a deal between
    Microsoft Corp. (MSFT) and
    Time Warner Inc..." + ], + [ + "AP - When Paula Radcliffe
    dropped out of the Olympic
    marathon miles from the
    finish, she sobbed
    uncontrollably. Margaret Okayo
    knew the feeling. Okayo pulled
    out of the marathon at the
    15th mile with a left leg
    injury, and she cried, too.
    When she watched Radcliffe
    quit, Okayo thought, \"Let's
    cry together.\"" + ], + [ + "Tightness in the labour market
    notwithstanding, the prospects
    for hiring in the third
    quarter are down from the
    second quarter, according to
    the new Manpower Employment
    Outlook Survey." + ], + [ + "Fans who can't get enough of
    \"The Apprentice\" can visit a
    new companion Web site each
    week and watch an extra 40
    minutes of video not broadcast
    on the Thursday
    show.<br><FONT
    face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\"\\ color=\"#666666\">
    <B>-Leslie
    Walker</b></font>" + ], + [ + " ATLANTA (Sports Network) -
    The Atlanta Hawks signed free
    agent Kevin Willis on
    Wednesday, nearly a decade
    after the veteran big man
    ended an 11- year stint with
    the team." + ], + [ + "An adult Web site publisher is
    suing Google, saying the
    search engine company made it
    easier for users to see the
    site #39;s copyrighted nude
    photographs without paying or
    gaining access through the
    proper channels." + ], + [ + "When his right-front tire went
    flying off early in the Ford
    400, the final race of the
    NASCAR Nextel Cup Series
    season, Kurt Busch, it seemed,
    was destined to spend his
    offseason" + ], + [ + "A Washington-based public
    opinion firm has released the
    results of an election day
    survey of Nevada voters
    showing 81 support for the
    issuance of paper receipts
    when votes are cast
    electronically." + ], + [ + "NAPSTER creator SHAWN FANNING
    has revealed his plans for a
    new licensed file-sharing
    service with an almost
    unlimited selection of tracks." + ], + [ + " NEW YORK (Reuters) - The
    dollar rose on Friday, after a
    U.S. report showed consumer
    prices in line with
    expections, reminding
    investors that the Federal
    Reserve was likely to
    continue raising interest
    rates, analysts said." + ], + [ + "Brandon Backe and Woody
    Williams pitched well last
    night even though neither
    earned a win. But their
    outings will show up in the
    record books." + ], + [ + "President George W. Bush
    pledged Friday to spend some
    of the political capital from
    his re-election trying to
    secure a lasting Middle East
    peace, and he envisioned the
    establishment" + ], + [ + "The Football Association today
    decided not to charge David
    Beckham with bringing the game
    into disrepute. The FA made
    the surprise announcement
    after their compliance unit
    ruled" + ], + [ + "Last year some election
    watchers made a bold
    prediction that this
    presidential election would
    set a record: the first half
    billion dollar campaign in
    hard money alone." + ], + [ + "It #39;s one more blow to
    patients who suffer from
    arthritis. Pfizer, the maker
    of Celebrex, says it #39;s
    painkiller poses an increased
    risk of heart attacks to
    patients using the drugs." + ], + [ + "NEW DELHI - A bomb exploded
    during an Independence Day
    parade in India's remote
    northeast on Sunday, killing
    at least 15 people, officials
    said, just an hour after Prime
    Minister Manmohan Singh
    pledged to fight terrorism.
    The outlawed United Liberation
    Front of Asom was suspected of
    being behind the attack in
    Assam state and a second one
    later in the area, said Assam
    Inspector General of Police
    Khagen Sharma..." + ], + [ + "Two separate studies by U.S.
    researchers find that super
    drug-resistant strains of
    tuberculosis are at the
    tipping point of a global
    epidemic, and only small
    changes could help them spread
    quickly." + ], + [ + " CRANS-SUR-SIERRE, Switzerland
    (Reuters) - World number
    three Ernie Els says he feels
    a failure after narrowly
    missing out on three of the
    year's four major
    championships." + ], + [ + "A UN envoy to Sudan will visit
    Darfur tomorrow to check on
    the government #39;s claim
    that some 70,000 people
    displaced by conflict there
    have voluntarily returned to
    their homes, a spokesman said." + ], + [ + "Dell cut prices on some
    servers and PCs by as much as
    22 percent because it #39;s
    paying less for parts. The
    company will pass the savings
    on components such as memory
    and liquid crystal displays" + ], + [ + "AP - Most of the presidential
    election provisional ballots
    rejected so far in Ohio came
    from people who were not even
    registered to vote, election
    officials said after spending
    nearly two weeks poring over
    thousands of disputed votes." + ], + [ + "Striker Bonaventure Kalou
    netted twice to send AJ
    Auxerre through to the first
    knockout round of the UEFA Cup
    at the expense of Rangers on
    Wednesday." + ], + [ + "AP - Rival inmates fought each
    other with knives and sticks
    Wednesday at a San Salvador
    prison, leaving at least 31
    people dead and two dozen
    injured, officials said." + ], + [ + "WASHINGTON: The European-
    American Cassini-Huygens space
    probe has detected traces of
    ice flowing on the surface of
    Saturn #39;s largest moon,
    Titan, suggesting the
    existence of an ice volcano,
    NASA said Tuesday." + ], + [ + "The economic growth rate in
    the July-September period was
    revised slightly downward from
    an already weak preliminary
    report, the government said
    Wednesday." + ], + [ + "All ISS systems continue to
    function nominally, except
    those noted previously or
    below. Day 7 of joint
    Exp.9/Exp.10 operations and
    last full day before 8S
    undocking." + ], + [ + "BAGHDAD - Two Egyptian
    employees of a mobile phone
    company were seized when
    gunmen stormed into their
    Baghdad office, the latest in
    a series of kidnappings in the
    country." + ], + [ + "BRISBANE, Australia - The body
    of a whale resembling a giant
    dolphin that washed up on an
    eastern Australian beach has
    intrigued local scientists,
    who agreed Wednesday that it
    is rare but are not sure just
    how rare." + ], + [ + " quot;Magic can happen. quot;
    Sirius Satellite Radio
    (nasdaq: SIRI - news - people
    ) may have signed Howard Stern
    and the men #39;s NCAA
    basketball tournaments, but XM
    Satellite Radio (nasdaq: XMSR
    - news - people ) has its
    sights on your cell phone." + ], + [ + "Trick-or-treaters can expect
    an early Halloween treat on
    Wednesday night, when a total
    lunar eclipse makes the moon
    look like a glowing pumpkin." + ], + [ + "THIS weekend sees the
    quot;other quot; showdown
    between New York and New
    England as the Jets and
    Patriots clash in a battle of
    the unbeaten teams." + ], + [ + "Sifting through millions of
    documents to locate a valuable
    few is tedious enough, but
    what happens when those files
    are scattered across different
    repositories?" + ], + [ + "President Bush aims to
    highlight American drug-
    fighting aid in Colombia and
    boost a conservative Latin
    American leader with a stop in
    the Andean nation where
    thousands of security forces
    are deployed to safeguard his
    brief stay." + ], + [ + "Dubai - Former Palestinian
    security minister Mohammed
    Dahlan said on Monday that a
    quot;gang of mercenaries quot;
    known to the Palestinian
    police were behind the
    shooting that resulted in two
    deaths in a mourning tent for
    Yasser Arafat in Gaza." + ], + [ + "A drug company executive who
    spoke out in support of
    Montgomery County's proposal
    to import drugs from Canada
    and similar legislation before
    Congress said that his company
    has launched an investigation
    into his political activities." + ], + [ + "Nicolas Anelka is fit for
    Manchester City #39;s
    Premiership encounter against
    Tottenham at Eastlands, but
    the 13million striker will
    have to be content with a
    place on the bench." + ], + [ + " PITTSBURGH (Reuters) - Ben
    Roethlisberger passed for 183
    yards and two touchdowns,
    Hines Ward scored twice and
    the Pittsburgh Steelers
    rolled to a convincing 27-3
    victory over Philadelphia on
    Sunday for their second
    straight win against an
    undefeated opponent." + ], + [ + " ATHENS (Reuters) - The U.S.
    men's basketball team got
    their first comfortable win
    at the Olympic basketball
    tournament Monday, routing
    winless Angola 89-53 in their
    final preliminary round game." + ], + [ + "A Frenchman working for Thales
    SA, Europe #39;s biggest maker
    of military electronics, was
    shot dead while driving home
    at night in the Saudi Arabian
    city of Jeddah." + ], + [ + "Often, the older a pitcher
    becomes, the less effective he
    is on the mound. Roger Clemens
    apparently didn #39;t get that
    memo. On Tuesday, the 42-year-
    old Clemens won an
    unprecedented" + ], + [ + "NASA #39;s Mars rovers have
    uncovered more tantalizing
    evidence of a watery past on
    the Red Planet, scientists
    said Wednesday. And the
    rovers, Spirit and
    Opportunity, are continuing to
    do their jobs months after
    they were expected to ..." + ], + [ + "SYDNEY (AFP) - Australia #39;s
    commodity exports are forecast
    to increase by 15 percent to a
    record 95 billion dollars (71
    million US), the government
    #39;s key economic forecaster
    said." + ], + [ + "Google won a major legal
    victory when a federal judge
    ruled that the search engines
    advertising policy does not
    violate federal trademark
    laws." + ], + [ + "Intel Chief Technology Officer
    Pat Gelsinger said on
    Thursday, Sept. 9, that the
    Internet needed to be upgraded
    in order to deal with problems
    that will become real issues
    soon." + ], + [ + "China will take tough measures
    this winter to improve the
    country #39;s coal mine safety
    and prevent accidents. State
    Councilor Hua Jianmin said
    Thursday the industry should
    take" + ], + [ + "AT amp;T Corp. on Thursday
    said it is reducing one fifth
    of its workforce this year and
    will record a non-cash charge
    of approximately \\$11." + ], + [ + "BAGHDAD (Iraq): As the
    intensity of skirmishes
    swelled on the soils of Iraq,
    dozens of people were put to
    death with toxic shots by the
    US helicopter gunship, which
    targeted the civilians,
    milling around a burning
    American vehicle in a Baghdad
    street on" + ], + [ + " LONDON (Reuters) - Television
    junkies of the world, get
    ready for \"Friends,\" \"Big
    Brother\" and \"The Simpsons\" to
    phone home." + ], + [ + "A rift appeared within Canada
    #39;s music industry yesterday
    as prominent artists called on
    the CRTC to embrace satellite
    radio and the industry warned
    of lost revenue and job
    losses." + ], + [ + "Reuters - A key Iranian
    nuclear facility which
    the\\U.N.'s nuclear watchdog
    has urged Tehran to shut down
    is\\nearing completion, a
    senior Iranian nuclear
    official said on\\Sunday." + ], + [ + "Spain's Football Federation
    launches an investigation into
    racist comments made by
    national coach Luis Aragones." + ], + [ + "Bricks and plaster blew inward
    from the wall, as the windows
    all shattered and I fell to
    the floorwhether from the
    shock wave, or just fright, it
    wasn #39;t clear." + ], + [ + "Surfersvillage Global Surf
    News, 13 September 2004: - -
    Hurricane Ivan, one of the
    most powerful storms to ever
    hit the Caribbean, killed at
    least 16 people in Jamaica,
    where it wrecked houses and
    washed away roads on Saturday,
    but appears to have spared" + ], + [ + "I #39;M FEELING a little bit
    better about the hundreds of
    junk e-mails I get every day
    now that I #39;ve read that
    someone else has much bigger
    e-mail troubles." + ], + [ + "NEW DELHI: India and Pakistan
    agreed on Monday to step up
    cooperation in the energy
    sector, which could lead to
    Pakistan importing large
    amounts of diesel fuel from
    its neighbour, according to
    Pakistani Foreign Minister
    Khurshid Mehmood Kasuri." + ], + [ + "LONDON, England -- A US
    scientist is reported to have
    observed a surprising jump in
    the amount of carbon dioxide,
    the main greenhouse gas." + ], + [ + "Microsoft's antispam Sender ID
    technology continues to get
    the cold shoulder. Now AOL
    adds its voice to a growing
    chorus of businesses and
    organizations shunning the
    proprietary e-mail
    authentication system." + ], + [ + "PSV Eindhoven faces Arsenal at
    Highbury tomorrow night on the
    back of a free-scoring start
    to the season. Despite losing
    Mateja Kezman to Chelsea in
    the summer, the Dutch side has
    scored 12 goals in the first" + ], + [ + "Through the World Community
    Grid, your computer could help
    address the world's health and
    social problems." + ], + [ + "Zimbabwe #39;s ruling Zanu-PF
    old guard has emerged on top
    after a bitter power struggle
    in the deeply divided party
    during its five-yearly
    congress, which ended
    yesterday." + ], + [ + "PLAYER OF THE GAME: Playing
    with a broken nose, Seattle
    point guard Sue Bird set a
    WNBA playoff record for
    assists with 14, also pumping
    in 10 points as the Storm
    claimed the Western Conference
    title last night." + ], + [ + "Reuters - Thousands of
    demonstrators pressing
    to\\install Ukraine's
    opposition leader as president
    after a\\disputed election
    launched fresh street rallies
    in the capital\\for the third
    day Wednesday." + ], + [ + "Michael Jackson wishes he had
    fought previous child
    molestation claims instead of
    trying to \"buy peace\", his
    lawyer says." + ], + [ + "North Korea says it will not
    abandon its weapons programme
    after the South admitted
    nuclear activities." + ], + [ + "While there is growing
    attention to ongoing genocide
    in Darfur, this has not
    translated into either a
    meaningful international
    response or an accurate
    rendering of the scale and
    evident course of the
    catastrophe." + ], + [ + "A planned component for
    Microsoft #39;s next version
    of Windows is causing
    consternation among antivirus
    experts, who say that the new
    module, a scripting platform
    called Microsoft Shell, could
    give birth to a whole new
    generation of viruses and
    remotely" + ], + [ + "THE prosecution on terrorism
    charges of extremist Islamic
    cleric and accused Jemaah
    Islamiah leader Abu Bakar
    Bashir will rely heavily on
    the potentially tainted
    testimony of at least two
    convicted Bali bombers, his
    lawyers have said." + ], + [ + "SAN JOSE, California Yahoo
    will likely have a tough time
    getting American courts to
    intervene in a dispute over
    the sale of Nazi memorabilia
    in France after a US appeals
    court ruling." + ], + [ + "TORONTO (CP) - Glamis Gold of
    Reno, Nev., is planning a
    takeover bid for Goldcorp Inc.
    of Toronto - but only if
    Goldcorp drops its
    \\$2.4-billion-Cdn offer for
    another Canadian firm, made in
    early December." + ], + [ + "Clashes between US troops and
    Sadr militiamen escalated
    Thursday, as the US surrounded
    Najaf for possible siege." + ], + [ + "eBay Style director Constance
    White joins Post fashion
    editor Robin Givhan and host
    Janet Bennett to discuss how
    to find trends and bargains
    and pull together a wardrobe
    online." + ], + [ + "This week will see the release
    of October new and existing
    home sales, a measure of
    strength in the housing
    industry. But the short
    holiday week will also leave
    investors looking ahead to the
    holiday travel season." + ], + [ + "Frankfurt - World Cup winners
    Brazil were on Monday drawn to
    meet European champions
    Greece, Gold Cup winners
    Mexico and Asian champions
    Japan at the 2005
    Confederations Cup." + ], + [ + "Third baseman Vinny Castilla
    said he fits fine with the
    Colorado youth movement, even
    though he #39;ll turn 38 next
    season and the Rockies are
    coming off the second-worst" + ], + [ + "With a sudden shudder, the
    ground collapsed and the pipe
    pushed upward, buckling into a
    humped shape as Cornell
    University scientists produced
    the first simulated earthquake" + ], + [ + "(Sports Network) - Two of the
    top teams in the American
    League tangle in a possible
    American League Division
    Series preview tonight, as the
    West-leading Oakland Athletics
    host the wild card-leading
    Boston Red Sox for the first
    of a three-game set at the" + ], + [ + "over half the children in the
    world - suffer extreme
    deprivation because of war,
    HIV/AIDS or poverty, according
    to a report released yesterday
    by the United Nations Children
    #39;s Fund." + ], + [ + "Microsoft (Quote, Chart) has
    fired another salvo in its
    ongoing spam battle, this time
    against porn peddlers who don
    #39;t keep their smut inside
    the digital equivalent of a
    quot;Brown Paper Wrapper." + ], + [ + " BETHESDA, Md. (Reuters) - The
    use of some antidepressant
    drugs appears linked to an
    increase in suicidal behavior
    in some children and teen-
    agers, a U.S. advisory panel
    concluded on Tuesday." + ], + [ + " SEATTLE (Reuters) - The next
    version of the Windows
    operating system, Microsoft
    Corp.'s <A HREF=\"http://www
    .reuters.co.uk/financeQuoteLoo
    kup.jhtml?ticker=MSFT.O
    qtype=sym infotype=info
    qcat=news\">MSFT.O</A>
    flagship product, will ship
    in 2006, the world's largest
    software maker said on
    Friday." + ], + [ + "Reuters - Philippine rescue
    teams\\evacuated thousands of
    people from the worst flooding
    in the\\central Luzon region
    since the 1970s as hungry
    victims hunted\\rats and birds
    for food." + ], + [ + "Inverness Caledonian Thistle
    appointed Craig Brewster as
    its new manager-player
    Thursday although he #39;s
    unable to play for the team
    until January." + ], + [ + "The Afghan president expresses
    deep concern after a bomb
    attack which left at least
    seven people dead." + ], + [ + " NEW YORK (Reuters) - U.S.
    technology stocks opened lower
    on Thursday after a sales
    warning from Applied Materials
    Inc. <A HREF=\"http://www.i
    nvestor.reuters.com/FullQuote.
    aspx?ticker=AMAT.O target=/sto
    cks/quickinfo/fullquote\">AM
    AT.O</A>, while weekly
    jobless claims data met Wall
    Street's expectations,
    leaving the Dow and S P 500
    market measures little
    changed." + ], + [ + "ATHENS, Greece -- Alan Shearer
    converted an 87th-minute
    penalty to give Newcastle a
    1-0 win over Panionios in
    their UEFA Cup Group D match." + ], + [ + "Fossil remains of the oldest
    and smallest known ancestor of
    Tyrannosaurus rex, the world
    #39;s favorite ferocious
    dinosaur, have been discovered
    in China with evidence that
    its body was cloaked in downy
    quot;protofeathers." + ], + [ + "Derek Jeter turned a season
    that started with a terrible
    slump into one of the best in
    his accomplished 10-year
    career. quot;I don #39;t
    think there is any question,
    quot; the New York Yankees
    manager said." + ], + [ + "Gardez (Afghanistan), Sept. 16
    (Reuters): Afghan President
    Hamid Karzai escaped an
    assassination bid today when a
    rocket was fired at his US
    military helicopter as it was
    landing in the southeastern
    town of Gardez." + ], + [ + "The Jets came up with four
    turnovers by Dolphins
    quarterback Jay Fiedler in the
    second half, including an
    interception returned 66 yards
    for a touchdown." + ], + [ + "China's Guo Jingjing easily
    won the women's 3-meter
    springboard last night, and Wu
    Minxia made it a 1-2 finish
    for the world's diving
    superpower, taking the silver." + ], + [ + "GREEN BAY, Wisconsin (Ticker)
    -- Brett Favre will be hoping
    his 200th consecutive start
    turns out better than his last
    two have against the St." + ], + [ + "People fishing for sport are
    doing far more damage to US
    marine fish stocks than anyone
    thought, accounting for nearly
    a quarter of the" + ], + [ + "When an NFL team opens with a
    prolonged winning streak,
    former Miami Dolphins coach
    Don Shula and his players from
    the 17-0 team of 1972 root
    unabashedly for the next
    opponent." + ], + [ + "MIANNE Bagger, the transsexual
    golfer who prompted a change
    in the rules to allow her to
    compete on the professional
    circuit, made history
    yesterday by qualifying to
    play full-time on the Ladies
    European Tour." + ], + [ + "Great Britain #39;s gold medal
    tally now stands at five after
    Leslie Law was handed the
    individual three day eventing
    title - in a courtroom." + ], + [ + "This particular index is
    produced by the University of
    Michigan Business School, in
    partnership with the American
    Society for Quality and CFI
    Group, and is supported in
    part by ForeSee Results" + ], + [ + "CHICAGO : Interstate Bakeries
    Corp., the maker of popular,
    old-style snacks Twinkies and
    Hostess Cakes, filed for
    bankruptcy, citing rising
    costs and falling sales." + ], + [ + "Delta Air Lines (DAL:NYSE -
    commentary - research) will
    cut employees and benefits but
    give a bigger-than-expected
    role to Song, its low-cost
    unit, in a widely anticipated
    but still unannounced
    overhaul, TheStreet.com has
    learned." + ], + [ + "PC World - Symantec, McAfee
    hope raising virus-definition
    fees will move users to\\
    suites." + ], + [ + "By byron kho. A consortium of
    movie and record companies
    joined forces on Friday to
    request that the US Supreme
    Court take another look at
    peer-to-peer file-sharing
    programs." + ], + [ + "DUBLIN -- Prime Minister
    Bertie Ahern urged Irish
    Republican Army commanders
    yesterday to meet what he
    acknowledged was ''a heavy
    burden quot;: disarming and
    disbanding their organization
    in support of Northern
    Ireland's 1998 peace accord." + ], + [ + "Mayor Tom Menino must be
    proud. His Boston Red Sox just
    won their first World Series
    in 86 years and his Hyde Park
    Blue Stars yesterday clinched
    their first Super Bowl berth
    in 32 years, defeating
    O'Bryant, 14-0. Who would have
    thought?" + ], + [ + "While reproductive planning
    and women #39;s equality have
    improved substantially over
    the past decade, says a United
    Nations report, world
    population will increase from
    6.4 billion today to 8.9
    billion by 2050, with the 50
    poorest countries tripling in" + ], + [ + "Instead of the skinny black
    line, showing a hurricane
    #39;s forecast track,
    forecasters have drafted a
    couple of alternative graphics
    to depict where the storms
    might go -- and they want your
    opinion." + ], + [ + "South Korea have appealed to
    sport #39;s supreme legal body
    in an attempt to award Yang
    Tae-young the Olympic
    gymnastics all-round gold
    medal after a scoring error
    robbed him of the title in
    Athens." + ], + [ + "BERLIN - Volkswagen AG #39;s
    announcement this week that it
    has forged a new partnership
    deal with Malaysian carmaker
    Proton comes as a strong euro
    and Europe #39;s weak economic
    performance triggers a fresh
    wave of German investment in
    Asia." + ], + [ + "AP - Johan Santana had an
    early lead and was well on his
    way to his 10th straight win
    when the rain started to fall." + ], + [ + "ATHENS-In one of the biggest
    shocks in Olympic judo
    history, defending champion
    Kosei Inoue was defeated by
    Dutchman Elco van der Geest in
    the men #39;s 100-kilogram
    category Thursday." + ], + [ + "Consumers in Dublin pay more
    for basic goods and services
    that people elsewhere in the
    country, according to figures
    released today by the Central
    Statistics Office." + ], + [ + "LIBERTY Media #39;s move last
    week to grab up to 17.1 per
    cent of News Corporation
    voting stock has prompted the
    launch of a defensive
    shareholder rights plan." + ], + [ + "NBC is adding a 5-second delay
    to its Nascar telecasts after
    Dale Earnhardt Jr. used a
    vulgarity during a postrace
    interview last weekend." + ], + [ + "LONDON - Wild capuchin monkeys
    can understand cause and
    effect well enough to use
    rocks to dig for food,
    scientists have found.
    Capuchin monkeys often use
    tools and solve problems in
    captivity and sometimes" + ], + [ + "San Francisco Giants
    outfielder Barry Bonds, who
    became the third player in
    Major League Baseball history
    to hit 700 career home runs,
    won the National League Most
    Valuable Player Award" + ], + [ + "The blue-chip Hang Seng Index
    rose 171.88 points, or 1.22
    percent, to 14,066.91. On
    Friday, the index had slipped
    31.58 points, or 0.2 percent." + ], + [ + "BAR's Anthony Davidson and
    Jenson Button set the pace at
    the first Chinese Grand Prix." + ], + [ + "Shares plunge after company
    says its vein graft treatment
    failed to show benefit in
    late-stage test. CHICAGO
    (Reuters) - Biotechnology
    company Corgentech Inc." + ], + [ + "WASHINGTON - Contradicting the
    main argument for a war that
    has cost more than 1,000
    American lives, the top U.S.
    arms inspector reported
    Wednesday that he found no
    evidence that Iraq produced
    any weapons of mass
    destruction after 1991..." + ], + [ + "The key to hidden treasure
    lies in your handheld GPS
    unit. GPS-based \"geocaching\"
    is a high-tech sport being
    played by thousands of people
    across the globe." + ], + [ + "AFP - Style mavens will be
    scanning the catwalks in Paris
    this week for next spring's
    must-have handbag, as a
    sweeping exhibition at the
    French capital's fashion and
    textile museum reveals the bag
    in all its forms." + ], + [ + "Canadian Press - SAINT-
    QUENTIN, N.B. (CP) - A major
    highway in northern New
    Brunswick remained closed to
    almost all traffic Monday, as
    local residents protested
    planned health care cuts." + ], + [ + "The U.S. information tech
    sector lost 403,300 jobs
    between March 2001 and April
    2004, and the market for tech
    workers remains bleak,
    according to a new report." + ], + [ + " NAJAF, Iraq (Reuters) - A
    radical Iraqi cleric leading a
    Shi'ite uprising agreed on
    Wednesday to disarm his
    militia and leave one of the
    country's holiest Islamic
    shrines after warnings of an
    onslaught by government
    forces." + ], + [ + "Saudi security forces have
    killed a wanted militant near
    the scene of a deadly shootout
    Thursday. Officials say the
    militant was killed in a
    gunbattle Friday in the
    northern town of Buraida,
    hours after one" + ], + [ + "Portsmouth chairman Milan
    Mandaric said on Tuesday that
    Harry Redknapp, who resigned
    as manager last week, was
    innocent of any wrong-doing
    over agent or transfer fees." + ], + [ + "This record is for all the
    little guys, for all the
    players who have to leg out
    every hit instead of taking a
    relaxing trot around the
    bases, for all the batters
    whose muscles aren #39;t" + ], + [ + "Two South Africans acquitted
    by a Zimbabwean court of
    charges related to the alleged
    coup plot in Equatorial Guinea
    are to be questioned today by
    the South African authorities." + ], + [ + "Charlie Hodgson #39;s record-
    equalling performance against
    South Africa was praised by
    coach Andy Robinson after the
    Sale flyhalf scored 27 points
    in England #39;s 32-16 victory
    here at Twickenham on
    Saturday." + ], + [ + "com September 30, 2004, 11:11
    AM PT. SanDisk announced
    Thursday increased capacities
    for several different flash
    memory cards. The Sunnyvale,
    Calif." + ], + [ + "MOSCOW (CP) - Russia mourned
    89 victims of a double air
    disaster today as debate
    intensified over whether the
    two passenger liners could
    have plunged almost
    simultaneously from the sky by
    accident." + ], + [ + "US blue-chip stocks rose
    slightly on Friday as
    government data showed better-
    than-expected demand in August
    for durable goods other than
    transportation equipment, but
    climbing oil prices limited
    gains." + ], + [ + "BASEBALL Atlanta (NL):
    Optioned P Roman Colon to
    Greenville (Southern);
    recalled OF Dewayne Wise from
    Richmond (IL). Boston (AL):
    Purchased C Sandy Martinez
    from Cleveland (AL) and
    assigned him to Pawtucket
    (IL). Cleveland (AL): Recalled
    OF Ryan Ludwick from Buffalo
    (IL). Chicago (NL): Acquired
    OF Ben Grieve from Milwaukee
    (NL) for player to be named
    and cash; acquired C Mike ..." + ], + [ + "Australia #39;s prime minister
    says a body found in Fallujah
    is likely that of kidnapped
    aid worker Margaret Hassan.
    John Howard told Parliament a
    videotape of an Iraqi
    terrorist group executing a
    Western woman appears to have
    been genuine." + ], + [ + "roundup Plus: Tech firms rally
    against copyright bill...Apple
    .Mac customers suffer e-mail
    glitches...Alvarion expands
    wireless broadband in China." + ], + [ + "BRONX, New York (Ticker) --
    Kelvim Escobar was the latest
    Anaheim Angels #39; pitcher to
    subdue the New York Yankees.
    Escobar pitched seven strong
    innings and Bengie Molina tied
    a career-high with four hits,
    including" + ], + [ + "Business software maker
    PeopleSoft Inc. said Monday
    that it expects third-quarter
    revenue to range between \\$680
    million and \\$695 million,
    above average Wall Street
    estimates of \\$651." + ], + [ + "Sep 08 - Vijay Singh revelled
    in his status as the new world
    number one after winning the
    Deutsche Bank Championship by
    three shots in Boston on
    Monday." + ], + [ + "Reuters - Enron Corp. ,
    desperate to\\meet profit
    targets, \"parked\" unwanted
    power generating barges\\at
    Merrill Lynch in a sham sale
    designed to be reversed,
    a\\prosecutor said on Tuesday
    in the first criminal trial
    of\\former executives at the
    fallen energy company." + ], + [ + " NEW YORK (Reuters) - The
    dollar rebounded on Monday
    after a heavy selloff last
    week, but analysts were
    uncertain if the rally could
    hold as the drumbeat of
    expectation began for to the
    December U.S. jobs report due
    Friday." + ], + [ + "AP - Their first debate less
    than a week away, President
    Bush and Democrat John Kerry
    kept their public schedules
    clear on Saturday and began to
    focus on their prime-time
    showdown." + ], + [ + "Many people in golf are asking
    that today. He certainly wasn
    #39;t A-list and he wasn #39;t
    Larry Nelson either. But you
    couldn #39;t find a more solid
    guy to lead the United States
    into Ireland for the 2006
    Ryder Cup Matches." + ], + [ + "Coles Myer Ltd. Australia
    #39;s biggest retailer,
    increased second-half profit
    by 26 percent after opening
    fuel and convenience stores,
    selling more-profitable
    groceries and cutting costs." + ], + [ + "MOSCOW: Us oil major
    ConocoPhillips is seeking to
    buy up to 25 in Russian oil
    giant Lukoil to add billions
    of barrels of reserves to its
    books, an industry source
    familiar with the matter said
    on Friday." + ], + [ + "Australian Stuart Appleby, who
    was the joint second-round
    leader, returned a two-over 74
    to drop to third at three-
    under while American Chris
    DiMarco moved into fourth with
    a round of 69." + ], + [ + "PARIS Getting to the bottom of
    what killed Yassar Arafat
    could shape up to be an ugly
    family tug-of-war. Arafat
    #39;s half-brother and nephew
    want copies of Arafat #39;s
    medical records from the
    suburban Paris hospital" + ], + [ + "Red Hat is acquiring security
    and authentication tools from
    Netscape Security Solutions to
    bolster its software arsenal.
    Red Hat #39;s CEO and chairman
    Matthew Szulik spoke about the
    future strategy of the Linux
    supplier." + ], + [ + "With a doubleheader sweep of
    the Minnesota Twins, the New
    York Yankees moved to the
    verge of clinching their
    seventh straight AL East
    title." + ], + [ + "Global Web portal Yahoo! Inc.
    Wednesday night made available
    a beta version of a new search
    service for videos. Called
    Yahoo! Video Search, the
    search engine crawls the Web
    for different types of media
    files" + ], + [ + "Interactive posters at 25
    underground stations are
    helping Londoners travel
    safely over Christmas." + ], + [ + "Athens, Greece (Sports
    Network) - The first official
    track event took place this
    morning and Italy #39;s Ivano
    Brugnetti won the men #39;s
    20km walk at the Summer
    Olympics in Athens." + ], + [ + " THE HAGUE (Reuters) - Former
    Yugoslav President Slobodan
    Milosevic condemned his war
    crimes trial as a \"pure farce\"
    on Wednesday in a defiant
    finish to his opening defense
    statement against charges of
    ethnic cleansing in the
    Balkans." + ], + [ + "US Airways Group (otc: UAIRQ -
    news - people ) on Thursday
    said it #39;ll seek a court
    injunction to prohibit a
    strike by disaffected unions." + ], + [ + "Shares in Unilever fall after
    the Anglo-Dutch consumer goods
    giant issued a surprise
    profits warning." + ], + [ + "SAN FRANCISCO (CBS.MW) - The
    Canadian government will sell
    its 19 percent stake in Petro-
    Canada for \\$2.49 billion,
    according to the final
    prospectus filed with the US
    Securities and Exchange
    Commission Thursday." + ], + [ + "Champions Arsenal opened a
    five-point lead at the top of
    the Premier League after a 4-0
    thrashing of Charlton Athletic
    at Highbury Saturday." + ], + [ + "The Redskins and Browns have
    traded field goals and are
    tied, 3-3, in the first
    quarter in Cleveland." + ], + [ + " HYDERABAD, India (Reuters) -
    Microsoft Corp. <A HREF=\"ht
    tp://www.investor.reuters.com/
    FullQuote.aspx?ticker=MSFT.O t
    arget=/stocks/quickinfo/fullqu
    ote\">MSFT.O</A> will
    hire several hundred new staff
    at its new Indian campus in
    the next year, its chief
    executive said on Monday, in a
    move aimed at strengthening
    its presence in Asia's fourth-
    biggest economy." + ], + [ + "According to Swiss
    authorities, history was made
    Sunday when 2723 people in
    four communities in canton
    Geneva, Switzerland, voted
    online in a national federal
    referendum." + ], + [ + " GUWAHATI, India (Reuters) -
    People braved a steady drizzle
    to come out to vote in a
    remote northeast Indian state
    on Thursday, as troops
    guarded polling stations in an
    election being held under the
    shadow of violence." + ], + [ + "AFP - Three of the nine
    Canadian sailors injured when
    their newly-delivered,
    British-built submarine caught
    fire in the North Atlantic
    were airlifted Wednesday to
    hospital in northwest Ireland,
    officials said." + ], + [ + "Alitalia SpA, Italy #39;s
    largest airline, reached an
    agreement with its flight
    attendants #39; unions to cut
    900 jobs, qualifying the
    company for a government
    bailout that will keep it in
    business for another six
    months." + ], + [ + "BAGHDAD, Iraq - A series of
    strong explosions shook
    central Baghdad near dawn
    Sunday, and columns of thick
    black smoke rose from the
    Green Zone where U.S. and
    Iraqi government offices are
    located..." + ], + [ + "Forget September call-ups. The
    Red Sox may tap their minor
    league system for an extra
    player or two when the rules
    allow them to expand their
    25-man roster Wednesday, but
    any help from the farm is
    likely to pale against the
    abundance of talent they gain
    from the return of numerous
    players, including Trot Nixon
    , from the disabled list." + ], + [ + "P amp;Os cutbacks announced
    today are the result of the
    waves of troubles that have
    swamped the ferry industry of
    late. Some would say the
    company has done well to
    weather the storms for as long
    as it has." + ], + [ + "Sven-Goran Eriksson may gamble
    by playing goalkeeper Paul
    Robinson and striker Jermain
    Defoe in Poland." + ], + [ + "Foreign Secretary Jack Straw
    has flown to Khartoum on a
    mission to pile the pressure
    on the Sudanese government to
    tackle the humanitarian
    catastrophe in Darfur." + ], + [ + " Nextel was the big story in
    telecommunications yesterday,
    thanks to the Reston company's
    mega-merger with Sprint, but
    the future of wireless may be
    percolating in dozens of
    Washington area start-ups." + ], + [ + "Reuters - A senior U.S.
    official said on
    Wednesday\\deals should not be
    done with hostage-takers ahead
    of the\\latest deadline set by
    Afghan Islamic militants who
    have\\threatened to kill three
    kidnapped U.N. workers." + ], + [ + "I have been anticipating this
    day like a child waits for
    Christmas. Today, PalmOne
    introduces the Treo 650, the
    answer to my quot;what smart
    phone will I buy?" + ], + [ + "THOUSAND OAKS -- Anonymity is
    only a problem if you want it
    to be, and it is obvious Vijay
    Singh doesn #39;t want it to
    be. Let others chase fame." + ], + [ + "Wikipedia has surprised Web
    watchers by growing fast and
    maturing into one of the most
    popular reference sites." + ], + [ + "It only takes 20 minutes on
    the Internet for an
    unprotected computer running
    Microsoft Windows to be taken
    over by a hacker. Any personal
    or financial information
    stored" + ], + [ + "TORONTO (CP) - Russia #39;s
    Severstal has made an offer to
    buy Stelco Inc., in what #39;s
    believed to be one of several
    competing offers emerging for
    the restructuring but
    profitable Hamilton steel
    producer." + ], + [ + "Prices for flash memory cards
    -- the little modules used by
    digital cameras, handheld
    organizers, MP3 players and
    cell phones to store pictures,
    music and other data -- are
    headed down -- way down. Past
    trends suggest that prices
    will drop 35 percent a year,
    but industry analysts think
    that rate will be more like 40
    or 50 percent this year and
    next, due to more
    manufacturers entering the
    market." + ], + [ + "Walt Disney Co. #39;s
    directors nominated Michael
    Ovitz to serve on its board
    for another three years at a
    meeting just weeks before
    forcing him out of his job as" + ], + [ + "The European Union, Japan,
    Brazil and five other
    countries won World Trade
    Organization approval to
    impose tariffs worth more than
    \\$150 million a year on
    imports from the United" + ], + [ + "Industrial conglomerate
    Honeywell International on
    Wednesday said it has filed a
    lawsuit against 34 electronics
    companies including Apple
    Computer and Eastman Kodak,
    claiming patent infringement
    of its liquid crystal display
    technology." + ], + [ + "Sinn Fein leader Gerry Adams
    has put the pressure for the
    success or failure of the
    Northern Ireland assembly
    talks firmly on the shoulders
    of Ian Paisley." + ], + [ + "Australia #39;s Computershare
    has agreed to buy EquiServe of
    the United States for US\\$292
    million (\\$423 million),
    making it the largest US share
    registrar and driving its
    shares up by a third." + ], + [ + "David Coulthard #39;s season-
    long search for a Formula One
    drive next year is almost
    over. Negotiations between Red
    Bull Racing and Coulthard, who
    tested for the Austrian team
    for the first time" + ], + [ + "Interest rates on short-term
    Treasury securities were mixed
    in yesterday's auction. The
    Treasury Department sold \\$18
    billion in three-month bills
    at a discount rate of 1.640
    percent, up from 1.635 percent
    last week. An additional \\$16
    billion was sold in six-month
    bills at a rate of 1.840
    percent, down from 1.860
    percent." + ], + [ + "Two top executives of scandal-
    tarred insurance firm Marsh
    Inc. were ousted yesterday,
    the company said, the latest
    casualties of an industry
    probe by New York's attorney
    general." + ], + [ + "AP - Southern California
    tailback LenDale White
    remembers Justin Holland from
    high school. The Colorado
    State quarterback made quite
    an impression." + ], + [ + "TIM HENMAN last night admitted
    all of his energy has been
    drained away as he bowed out
    of the Madrid Masters. The top
    seed, who had a blood test on
    Wednesday to get to the bottom
    of his fatigue, went down" + ], + [ + "USDA #39;s Animal Plant Health
    Inspection Service (APHIS)
    this morning announced it has
    confirmed a detection of
    soybean rust from two test
    plots at Louisiana State
    University near Baton Rouge,
    Louisiana." + ], + [ + " JAKARTA (Reuters) - President
    Megawati Sukarnoputri urged
    Indonesians on Thursday to
    accept the results of the
    country's first direct
    election of a leader, but
    stopped short of conceding
    defeat." + ], + [ + "ISLAMABAD: Pakistan early
    Monday test-fired its
    indigenously developed short-
    range nuclear-capable Ghaznavi
    missile, the Inter Services
    Public Relations (ISPR) said
    in a statement." + ], + [ + "While the US software giant
    Microsoft has achieved almost
    sweeping victories in
    government procurement
    projects in several Chinese
    provinces and municipalities,
    the process" + ], + [ + "Mexican Cemex, being the third
    largest cement maker in the
    world, agreed to buy its
    British competitor - RMC Group
    - for \\$5.8 billion, as well
    as their debts in order to
    expand their activity on the
    building materials market of
    the USA and Europe." + ], + [ + "Microsoft Corp. has delayed
    automated distribution of a
    major security upgrade to its
    Windows XP Professional
    operating system, citing a
    desire to give companies more
    time to test it." + ], + [ + "The trial of a man accused of
    murdering York backpacker
    Caroline Stuttle begins in
    Australia." + ], + [ + "Gateway computers will be more
    widely available at Office
    Depot, in the PC maker #39;s
    latest move to broaden
    distribution at retail stores
    since acquiring rival
    eMachines this year." + ], + [ + "ATHENS -- US sailors needed a
    big day to bring home gold and
    bronze medals from the sailing
    finale here yesterday. But
    rolling the dice on windshifts
    and starting tactics backfired
    both in Star and Tornado
    classes, and the Americans had
    to settle for a single silver
    medal." + ], + [ + "Intel Corp. is refreshing its
    64-bit Itanium 2 processor
    line with six new chips based
    on the Madison core. The new
    processors represent the last
    single-core Itanium chips that
    the Santa Clara, Calif." + ], + [ + "The world's largest insurance
    group pays \\$126m in fines as
    part of a settlement with US
    regulators over its dealings
    with two firms." + ], + [ + "BRUSSELS: The EU sought
    Wednesday to keep pressure on
    Turkey over its bid to start
    talks on joining the bloc, as
    last-minute haggling seemed
    set to go down to the wire at
    a summit poised to give a
    green light to Ankara." + ], + [ + "AP - J. Cofer Black, the State
    Department official in charge
    of counterterrorism, is
    leaving government in the next
    few weeks." + ], + [ + "For the first time, broadband
    connections are reaching more
    than half (51 percent) of the
    American online population at
    home, according to measurement
    taken in July by
    Nielsen/NetRatings, a
    Milpitas-based Internet
    audience measurement and
    research ..." + ], + [ + "AP - Cavaliers forward Luke
    Jackson was activated
    Wednesday after missing five
    games because of tendinitis in
    his right knee. Cleveland also
    placed forward Sasha Pavlovic
    on the injured list." + ], + [ + "The tobacco firm John Player
    amp; Sons has announced plans
    to lay off 90 workers at its
    cigarette factory in Dublin.
    The company said it was
    planning a phased closure of
    the factory between now and
    February as part of a review
    of its global operations." + ], + [ + "AP - Consumers borrowed more
    freely in September,
    especially when it came to
    racking up charges on their
    credit cards, the Federal
    Reserve reported Friday." + ], + [ + "AFP - The United States
    presented a draft UN
    resolution that steps up the
    pressure on Sudan over the
    crisis in Darfur, including
    possible international
    sanctions against its oil
    sector." + ], + [ + "AFP - At least 33 people were
    killed and dozens others
    wounded when two bombs ripped
    through a congregation of
    Sunni Muslims in Pakistan's
    central city of Multan, police
    said." + ], + [ + "Bold, innovative solutions are
    key to addressing the rapidly
    rising costs of higher
    education and the steady
    reduction in government-
    subsidized help to finance
    such education." + ], + [ + "Just as the PhD crowd emerge
    with different interpretations
    of today's economy, everyday
    Americans battling to balance
    the checkbook hold diverse
    opinions about where things
    stand now and in the future." + ], + [ + "The Brisbane Lions #39;
    football manager stepped out
    of the changerooms just before
    six o #39;clock last night and
    handed one of the milling
    supporters a six-pack of beer." + ], + [ + "Authorities here are always
    eager to show off their
    accomplishments, so when
    Beijing hosted the World
    Toilet Organization conference
    last week, delegates were
    given a grand tour of the
    city's toilets." + ], + [ + "Cavaliers owner Gordon Gund is
    in quot;serious quot;
    negotiations to sell the NBA
    franchise, which has enjoyed a
    dramatic financial turnaround
    since the arrival of star
    LeBron James." + ], + [ + "WASHINGTON Trying to break a
    deadlock on energy policy, a
    diverse group of
    environmentalists, academics
    and former government
    officials were to publish a
    report on Wednesday that
    presents strategies for making
    the United States cleaner,
    more competitive" + ], + [ + "After two days of gloom, China
    was back on the winning rails
    on Thursday with Liu Chunhong
    winning a weightlifting title
    on her record-shattering binge
    and its shuttlers contributing
    two golds in the cliff-hanging
    finals." + ], + [ + "One question that arises
    whenever a player is linked to
    steroids is, \"What would he
    have done without them?\"
    Baseball history whispers an
    answer." + ], + [ + "AFP - A series of torchlight
    rallies and vigils were held
    after darkness fell on this
    central Indian city as victims
    and activists jointly
    commemorated a night of horror
    20 years ago when lethal gas
    leaked from a pesticide plant
    and killed thousands." + ], + [ + "Consider the New World of
    Information - stuff that,
    unlike the paper days of the
    past, doesn't always
    physically exist. You've got
    notes, scrawlings and
    snippets, Web graphics, photos
    and sounds. Stuff needs to be
    cut, pasted, highlighted,
    annotated, crossed out,
    dragged away. And, as Ross
    Perot used to say (or maybe it
    was Dana Carvey impersonating
    him), don't forget the graphs
    and charts." + ], + [ + "The second round of the
    Canadian Open golf tournament
    continues Saturday Glenn Abbey
    Golf Club in Oakville,
    Ontario, after play was
    suspended late Friday due to
    darkness." + ], + [ + "A consortium led by Royal
    Dutch/Shell Group that is
    developing gas reserves off
    Russia #39;s Sakhalin Island
    said Thursday it has struck a
    US\\$6 billion (euro4." + ], + [ + "Major Hollywood studios on
    Tuesday announced scores of
    lawsuits against computer
    server operators worldwide,
    including eDonkey, BitTorrent
    and DirectConnect networks,
    for allowing trading of
    pirated movies." + ], + [ + "A massive plan to attract the
    2012 Summer Olympics to New
    York, touting the city's
    diversity, financial and media
    power, was revealed Wednesday." + ], + [ + "A Zimbabwe court Friday
    convicted a British man
    accused of leading a coup plot
    against the government of oil-
    rich Equatorial Guinea on
    weapons charges, but acquitted
    most of the 69 other men held
    with him." + ], + [ + "But will Wi-Fi, high-
    definition broadcasts, mobile
    messaging and other
    enhancements improve the game,
    or wreck it?\\<br />
    Photos of tech-friendly parks\\" + ], + [ + "An audit by international
    observers supported official
    elections results that gave
    President Hugo Chavez a
    victory over a recall vote
    against him, the secretary-
    general of the Organisation of
    American States announced." + ], + [ + "Canadian Press - TORONTO (CP)
    - The fatal stabbing of a
    young man trying to eject
    unwanted party guests from his
    family home, the third such
    knifing in just weeks, has
    police worried about a
    potentially fatal holiday
    recipe: teens, alcohol and
    knives." + ], + [ + "NICK Heidfeld #39;s test with
    Williams has been brought
    forward after BAR blocked
    plans for Anthony Davidson to
    drive its Formula One rival
    #39;s car." + ], + [ + "MOSCOW - A female suicide
    bomber set off a shrapnel-
    filled explosive device
    outside a busy Moscow subway
    station on Tuesday night,
    officials said, killing 10
    people and injuring more than
    50." + ], + [ + "Grace Park closed with an
    eagle and two birdies for a
    7-under-par 65 and a two-
    stroke lead after three rounds
    of the Wachovia LPGA Classic
    on Saturday." + ], + [ + "ABIDJAN (AFP) - Two Ivory
    Coast military aircraft
    carried out a second raid on
    Bouake, the stronghold of the
    former rebel New Forces (FN)
    in the divided west African
    country, a French military
    source told AFP." + ], + [ + "Carlos Beltran drives in five
    runs to carry the Astros to a
    12-3 rout of the Braves in
    Game 5 of their first-round NL
    playoff series." + ], + [ + "On-demand viewing isn't just
    for TiVo owners anymore.
    Television over internet
    protocol, or TVIP, offers
    custom programming over
    standard copper wires." + ], + [ + "Apple is recalling 28,000
    faulty batteries for its
    15-inch Powerbook G4 laptops." + ], + [ + "Since Lennox Lewis #39;s
    retirement, the heavyweight
    division has been knocked for
    having more quantity than
    quality. Eight heavyweights on
    Saturday night #39;s card at
    Madison Square Garden hope to
    change that perception, at
    least for one night." + ], + [ + "PalmSource #39;s European
    developer conference is going
    on now in Germany, and this
    company is using this
    opportunity to show off Palm
    OS Cobalt 6.1, the latest
    version of its operating
    system." + ], + [ + "The former Chief Executive
    Officer of Computer Associates
    was indicted by a federal
    grand jury in New York
    Wednesday for allegedly
    participating in a massive
    fraud conspiracy and an
    elaborate cover up of a scheme
    that cost investors" + ], + [ + "Speaking to members of the
    Massachusetts Software
    Council, Microsoft CEO Steve
    Ballmer touted a bright future
    for technology but warned his
    listeners to think twice
    before adopting open-source
    products like Linux." + ], + [ + "MIAMI - The Trillian instant
    messaging (IM) application
    will feature several
    enhancements in its upcoming
    version 3.0, including new
    video and audio chat
    capabilities, enhanced IM
    session logs and integration
    with the Wikipedia online
    encyclopedia, according to
    information posted Friday on
    the product developer's Web
    site." + ], + [ + "Honeywell International Inc.,
    the world #39;s largest
    supplier of building controls,
    agreed to buy Novar Plc for
    798 million pounds (\\$1.53
    billion) to expand its
    security, fire and
    ventilation-systems business
    in Europe." + ], + [ + "San Francisco investment bank
    Thomas Weisel Partners on
    Thursday agreed to pay \\$12.5
    million to settle allegations
    that some of the stock
    research the bank published
    during the Internet boom was
    tainted by conflicts of
    interest." + ], + [ + "AFP - A state of civil
    emergency in the rebellion-hit
    Indonesian province of Aceh
    has been formally extended by
    six month, as the country's
    president pledged to end
    violence there without foreign
    help." + ], + [ + "Forbes.com - Peter Frankling
    tapped an unusual source to
    fund his new business, which
    makes hot-dog-shaped ice cream
    treats known as Cool Dogs: Two
    investors, one a friend and
    the other a professional
    venture capitalist, put in
    more than #36;100,000 each
    from their Individual
    Retirement Accounts. Later
    Franklin added #36;150,000
    from his own IRA." + ], + [ + "Reuters - Online DVD rental
    service Netflix Inc.\\and TiVo
    Inc., maker of a digital video
    recorder, on Thursday\\said
    they have agreed to develop a
    joint entertainment\\offering,
    driving shares of both
    companies higher." + ], + [ + "A San Diego insurance
    brokerage has been sued by New
    York Attorney General Elliot
    Spitzer for allegedly
    soliciting payoffs in exchange
    for steering business to
    preferred insurance companies." + ], + [ + "The European Union agreed
    Monday to lift penalties that
    have cost American exporters
    \\$300 million, following the
    repeal of a US corporate tax
    break deemed illegal under
    global trade rules." + ], + [ + "US Secretary of State Colin
    Powell on Monday said he had
    spoken to both Indian Foreign
    Minister K Natwar Singh and
    his Pakistani counterpart
    Khurshid Mahmud Kasuri late
    last week before the two met
    in New Delhi this week for
    talks." + ], + [ + "NEW YORK - Victor Diaz hit a
    tying, three-run homer with
    two outs in the ninth inning,
    and Craig Brazell's first
    major league home run in the
    11th gave the New York Mets a
    stunning 4-3 victory over the
    Chicago Cubs on Saturday.
    The Cubs had much on the
    line..." + ], + [ + "AFP - At least 54 people have
    died and more than a million
    have fled their homes as
    torrential rains lashed parts
    of India and Bangladesh,
    officials said." + ], + [ + "LOS ANGELES - California has
    adopted the world's first
    rules to reduce greenhouse
    emissions for autos, taking
    what supporters see as a
    dramatic step toward cleaning
    up the environment but also
    ensuring higher costs for
    drivers. The rules may lead
    to sweeping changes in
    vehicles nationwide,
    especially if other states opt
    to follow California's
    example..." + ], + [ + " LONDON (Reuters) - European
    stock markets scaled
    near-2-1/2 year highs on
    Friday as oil prices held
    below \\$48 a barrel, and the
    euro held off from mounting
    another assault on \\$1.30 but
    hovered near record highs
    against the dollar." + ], + [ + "Tim Duncan had 17 points and
    10 rebounds, helping the San
    Antonio Spurs to a 99-81
    victory over the New York
    Kicks. This was the Spurs
    fourth straight win this
    season." + ], + [ + "Nagpur: India suffered a
    double blow even before the
    first ball was bowled in the
    crucial third cricket Test
    against Australia on Tuesday
    when captain Sourav Ganguly
    and off spinner Harbhajan
    Singh were ruled out of the
    match." + ], + [ + "AFP - Republican and
    Democratic leaders each
    declared victory after the
    first head-to-head sparring
    match between President George
    W. Bush and Democratic
    presidential hopeful John
    Kerry." + ], + [ + "THIS YULE is all about console
    supply and there #39;s
    precious little units around,
    it has emerged. Nintendo has
    announced that it is going to
    ship another 400,000 units of
    its DS console to the United
    States to meet the shortfall
    there." + ], + [ + "Annual global semiconductor
    sales growth will probably
    fall by half in 2005 and
    memory chip sales could
    collapse as a supply glut saps
    prices, world-leading memory
    chip maker Samsung Electronics
    said on Monday." + ], + [ + "NEW YORK - Traditional phone
    systems may be going the way
    of the Pony Express. Voice-
    over-Internet Protocol,
    technology that allows users
    to make and receive phone
    calls using the Internet, is
    giving the old circuit-
    switched system a run for its
    money." + ], + [ + "AP - Former New York Yankees
    hitting coach Rick Down was
    hired for the same job by the
    Mets on Friday, reuniting him
    with new manager Willie
    Randolph." + ], + [ + "Last night in New York, the UN
    secretary-general was given a
    standing ovation - a robust
    response to a series of
    attacks in past weeks." + ], + [ + "FILDERSTADT (Germany) - Amelie
    Mauresmo and Lindsay Davenport
    took their battle for the No.
    1 ranking and Porsche Grand
    Prix title into the semi-
    finals with straight-sets
    victories on Friday." + ], + [ + "Reuters - The company behind
    the Atkins Diet\\on Friday
    shrugged off a recent decline
    in interest in low-carb\\diets
    as a seasonal blip, and its
    marketing chief said\\consumers
    would cut out starchy foods
    again after picking up\\pounds
    over the holidays." + ], + [ + "There #39;s something to be
    said for being the quot;first
    mover quot; in an industry
    trend. Those years of extra
    experience in tinkering with a
    new idea can be invaluable in
    helping the first" + ], + [ + "JOHANNESBURG -- Meeting in
    Nigeria four years ago,
    African leaders set a goal
    that 60 percent of children
    and pregnant women in malaria-
    affected areas around the
    continent would be sleeping
    under bed nets by the end of
    2005." + ], + [ + "AP - The first U.S. cases of
    the fungus soybean rust, which
    hinders plant growth and
    drastically cuts crop
    production, were found at two
    research sites in Louisiana,
    officials said Wednesday." + ], + [ + "Carter returned, but it was
    running back Curtis Martin and
    the offensive line that put
    the Jets ahead. Martin rushed
    for all but 10 yards of a
    45-yard drive that stalled at
    the Cardinals 10." + ], + [ + " quot;There #39;s no way
    anyone would hire them to
    fight viruses, quot; said
    Sophos security analyst Gregg
    Mastoras. quot;For one, no
    security firm could maintain
    its reputation by employing
    hackers." + ], + [ + "Symantec has revoked its
    decision to blacklist a
    program that allows Web
    surfers in China to browse
    government-blocked Web sites.
    The move follows reports that
    the firm labelled the Freegate
    program, which" + ], + [ + " NEW YORK (Reuters) - Shares
    of Chiron Corp. <A HREF=\"ht
    tp://www.investor.reuters.com/
    FullQuote.aspx?ticker=CHIR.O t
    arget=/stocks/quickinfo/fullqu
    ote\">CHIR.O</A> fell
    7 percent before the market
    open on Friday, a day after
    the biopharmaceutical company
    said it is delaying shipment
    of its flu vaccine, Fluvirin,
    because lots containing 4
    million vaccines do not meet
    product sterility standards." + ], + [ + "The nation's top
    telecommunications regulator
    said yesterday he will push --
    before the next president is
    inaugurated -- to protect
    fledgling Internet telephone
    services from getting taxed
    and heavily regulated by the
    50 state governments." + ], + [ + "Microsoft has signed a pact to
    work with the United Nations
    Educational, Scientific and
    Cultural Organization (UNESCO)
    to increase computer use,
    Internet access and teacher
    training in developing
    countries." + ], + [ + "DENVER (Ticker) -- Jake
    Plummer more than made up for
    a lack of a running game.
    Plummer passed for 294 yards
    and two touchdowns as the
    Denver Broncos posted a 23-13
    victory over the San Diego
    Chargers in a battle of AFC
    West Division rivals." + ], + [ + "DALLAS -- Belo Corp. said
    yesterday that it would cut
    250 jobs, more than half of
    them at its flagship
    newspaper, The Dallas Morning
    News, and that an internal
    investigation into circulation
    overstatements" + ], + [ + "AP - Duke Bainum outspent Mufi
    Hannemann in Honolulu's most
    expensive mayoral race, but
    apparently failed to garner
    enough votes in Saturday's
    primary to claim the office
    outright." + ], + [ + "roundup Plus: Microsoft tests
    Windows Marketplace...Nortel
    delays financials
    again...Microsoft updates
    SharePoint." + ], + [ + "The Federal Reserve still has
    some way to go to restore US
    interest rates to more normal
    levels, Philadelphia Federal
    Reserve President Anthony
    Santomero said on Monday." + ], + [ + "It took all of about five
    minutes of an introductory
    press conference Wednesday at
    Heritage Hall for USC
    basketball to gain something
    it never really had before." + ], + [ + "Delta Air Lines (DAL.N: Quote,
    Profile, Research) said on
    Wednesday its auditors have
    expressed doubt about the
    airline #39;s financial
    viability." + ], + [ + "POLITICIANS and aid agencies
    yesterday stressed the
    importance of the media in
    keeping the spotlight on the
    appalling human rights abuses
    taking place in the Darfur
    region of Sudan." + ], + [ + "AP - The Boston Red Sox looked
    at the out-of-town scoreboard
    and could hardly believe what
    they saw. The New York Yankees
    were trailing big at home
    against the Cleveland Indians
    in what would be the worst
    loss in the 101-year history
    of the storied franchise." + ], + [ + "The Red Sox will either
    complete an amazing comeback
    as the first team to rebound
    from a 3-0 deficit in
    postseason history, or the
    Yankees will stop them." + ], + [ + "\\Children who have a poor diet
    are more likely to become
    aggressive and anti-social, US
    researchers believe." + ], + [ + "OPEN SOURCE champion Microsoft
    is expanding its programme to
    give government organisations
    some of its source code. In a
    communique from the lair of
    the Vole, in Redmond,
    spinsters have said that
    Microsoft" + ], + [ + "The Red Sox have reached
    agreement with free agent
    pitcher Matt Clement yesterday
    on a three-year deal that will
    pay him around \\$25 million,
    his agent confirmed yesterday." + ], + [ + "Takeover target Ronin Property
    Group said it would respond to
    an offer by Multiplex Group
    for all the securities in the
    company in about three weeks." + ], + [ + "Canadian Press - OTTAWA (CP) -
    Contrary to Immigration
    Department claims, there is no
    shortage of native-borne
    exotic dancers in Canada, says
    a University of Toronto law
    professor who has studied the
    strip club business." + ], + [ + "HEN Manny Ramirez and David
    Ortiz hit consecutive home
    runs Sunday night in Chicago
    to put the Red Sox ahead,
    there was dancing in the
    streets in Boston." + ], + [ + "Google Inc. is trying to
    establish an online reading
    room for five major libraries
    by scanning stacks of hard-to-
    find books into its widely
    used Internet search engine." + ], + [ + "HOUSTON--(BUSINESS
    WIRE)--Sept. 1, 2004-- L
    #39;operazione crea una
    centrale globale per l
    #39;analisi strategica el
    #39;approfondimento del
    settore energetico IHS Energy,
    fonte globale leader di
    software, analisi e
    informazioni" + ], + [ + "The European Union presidency
    yesterday expressed optimism
    that a deal could be struck
    over Turkey #39;s refusal to
    recognize Cyprus in the lead-
    up to next weekend #39;s EU
    summit, which will decide
    whether to give Ankara a date
    for the start of accession
    talks." + ], + [ + "WASHINGTON Can you always tell
    when somebody #39;s lying? If
    so, you might be a wizard of
    the fib. A California
    psychology professor says
    there #39;s a tiny subculture
    of people that can pick out a
    lie nearly every time they
    hear one." + ], + [ + " KHARTOUM (Reuters) - Sudan on
    Saturday questioned U.N.
    estimates that up to 70,000
    people have died from hunger
    and disease in its remote
    Darfur region since a
    rebellion began 20 months
    ago." + ], + [ + "Type design was once the
    province of skilled artisans.
    With the help of new computer
    programs, neophytes have
    flooded the Internet with
    their creations." + ], + [ + "RCN Inc., co-owner of
    Starpower Communications LLC,
    the Washington area
    television, telephone and
    Internet provider, filed a
    plan of reorganization
    yesterday that it said puts
    the company" + ], + [ + "MIAMI -- Bryan Randall grabbed
    a set of Mardi Gras beads and
    waved them aloft, while his
    teammates exalted in the
    prospect of a trip to New
    Orleans." + ], + [ + "<strong>Letters</stro
    ng> Reports of demise
    premature" + ], + [ + "TORONTO (CP) - With an injured
    Vince Carter on the bench, the
    Toronto Raptors dropped their
    sixth straight game Friday,
    101-87 to the Denver Nuggets." + ], + [ + "The US airline industry,
    riddled with excess supply,
    will see a significant drop in
    capacity, or far fewer seats,
    as a result of at least one
    airline liquidating in the
    next year, according to
    AirTran Airways Chief
    Executive Joe Leonard." + ], + [ + "Boeing (nyse: BA - news -
    people ) Chief Executive Harry
    Stonecipher is keeping the
    faith. On Monday, the head of
    the aerospace and military
    contractor insists he #39;s
    confident his firm will
    ultimately win out" + ], + [ + "While not quite a return to
    glory, Monday represents the
    Redskins' return to the
    national consciousness." + ], + [ + "Australia #39;s biggest
    supplier of fresh milk,
    National Foods, has posted a
    net profit of \\$68.7 million,
    an increase of 14 per cent on
    last financial year." + ], + [ + "Lawyers for customers suing
    Merck amp; Co. want to
    question CEO Raymond Gilmartin
    about what he knew about the
    dangers of Vioxx before the
    company withdrew the drug from
    the market because of health
    hazards." + ], + [ + "Vijay Singh has won the US PGA
    Tour player of the year award
    for the first time, ending
    Tiger Woods #39;s five-year
    hold on the honour." + ], + [ + "New York; September 23, 2004 -
    The Department of Justice
    (DoJ), FBI and US Attorney
    #39;s Office handed down a
    10-count indictment against
    former Computer Associates
    (CA) chairman and CEO Sanjay
    Kumar and Stephen Richards,
    former CA head of worldwide
    sales." + ], + [ + "AFP - At least four Georgian
    soldiers were killed and five
    wounded in overnight clashes
    in Georgia's separatist, pro-
    Russian region of South
    Ossetia, Georgian officers
    near the frontline with
    Ossetian forces said early
    Thursday." + ], + [ + "Intel, the world #39;s largest
    chip maker, scrapped a plan
    Thursday to enter the digital
    television chip business,
    marking a major retreat from
    its push into consumer
    electronics." + ], + [ + "PACIFIC Hydro shares yesterday
    caught an updraught that sent
    them more than 20 per cent
    higher after the wind farmer
    moved to flush out a bidder." + ], + [ + "The European Commission is
    expected later this week to
    recommend EU membership talks
    with Turkey. Meanwhile, German
    Chancellor Gerhard Schroeder
    and Turkish Prime Minister
    Tayyip Erdogan are
    anticipating a quot;positive
    report." + ], + [ + "The US is the originator of
    over 42 of the worlds
    unsolicited commercial e-mail,
    making it the worst offender
    in a league table of the top
    12 spam producing countries
    published yesterday by anti-
    virus firm Sophos." + ], + [ + " NEW YORK (Reuters) - U.S.
    consumer confidence retreated
    in August while Chicago-area
    business activity slowed,
    according to reports on
    Tuesday that added to worries
    the economy's patch of slow
    growth may last beyond the
    summer." + ], + [ + "Intel is drawing the curtain
    on some of its future research
    projects to continue making
    transistors smaller, faster,
    and less power-hungry out as
    far as 2020." + ], + [ + "England got strikes from
    sparkling debut starter
    Jermain Defoe and Michael Owen
    to defeat Poland in a Uefa
    World Cup qualifier in
    Chorzow." + ], + [ + "The Canadian government
    signalled its intention
    yesterday to reintroduce
    legislation to decriminalise
    the possession of small
    amounts of marijuana." + ], + [ + "A screensaver targeting spam-
    related websites appears to
    have been too successful." + ], + [ + "Titleholder Ernie Els moved
    within sight of a record sixth
    World Match Play title on
    Saturday by solving a putting
    problem to overcome injured
    Irishman Padraig Harrington 5
    and 4." + ], + [ + "If the Washington Nationals
    never win a pennant, they have
    no reason to ever doubt that
    DC loves them. Yesterday, the
    District City Council
    tentatively approved a tab for
    a publicly financed ballpark
    that could amount to as much
    as \\$630 million." + ], + [ + "Plus: Experts fear Check 21
    could lead to massive bank
    fraud." + ], + [ + "By SIOBHAN McDONOUGH
    WASHINGTON (AP) -- Fewer
    American youths are using
    marijuana, LSD and Ecstasy,
    but more are abusing
    prescription drugs, says a
    government report released
    Thursday. The 2003 National
    Survey on Drug Use and Health
    also found youths and young
    adults are more aware of the
    risks of using pot once a
    month or more frequently..." + ], + [ + " ABIDJAN (Reuters) - Ivory
    Coast warplanes killed nine
    French soldiers on Saturday in
    a bombing raid during the
    fiercest clashes with rebels
    for 18 months and France hit
    back by destroying most of
    the West African country's
    small airforce." + ], + [ + "Unilever has reported a three
    percent rise in third-quarter
    earnings but warned it is
    reviewing its targets up to
    2010, after issuing a shock
    profits warning last month." + ], + [ + "A TNO engineer prepares to
    start capturing images for the
    world's biggest digital photo." + ], + [ + "Defensive back Brandon
    Johnson, who had two
    interceptions for Tennessee at
    Mississippi, was suspended
    indefinitely Monday for
    violation of team rules." + ], + [ + "Points leader Kurt Busch spun
    out and ran out of fuel, and
    his misfortune was one of the
    reasons crew chief Jimmy
    Fennig elected not to pit with
    20 laps to go." + ], + [ + " LONDON (Reuters) - Oil prices
    extended recent heavy losses
    on Wednesday ahead of weekly
    U.S. data expected to show
    fuel stocks rising in time
    for peak winter demand." + ], + [ + "(CP) - The NHL all-star game
    hasn #39;t been cancelled
    after all. It #39;s just been
    moved to Russia. The agent for
    New York Rangers winger
    Jaromir Jagr confirmed Monday
    that the Czech star had joined
    Omsk Avangard" + ], + [ + "PalmOne is aiming to sharpen
    up its image with the launch
    of the Treo 650 on Monday. As
    previously reported, the smart
    phone update has a higher-
    resolution screen and a faster
    processor than the previous
    top-of-the-line model, the
    Treo 600." + ], + [ + "GAZA CITY, Gaza Strip --
    Islamic militant groups behind
    many suicide bombings
    dismissed yesterday a call
    from Mahmoud Abbas, the
    interim Palestinian leader, to
    halt attacks in the run-up to
    a Jan. 9 election to replace
    Yasser Arafat." + ], + [ + "Secretary of State Colin
    Powell is wrapping up an East
    Asia trip focused on prodding
    North Korea to resume talks
    aimed at ending its nuclear-
    weapons program." + ], + [ + "HERE in the land of myth, that
    familiar god of sports --
    karma -- threw a bolt of
    lightning into the Olympic
    stadium yesterday. Marion
    Jones lunged desperately with
    her baton in the 4 x 100m
    relay final, but couldn #39;t
    reach her target." + ], + [ + "kinrowan writes quot;MIT,
    inventor of Kerberos, has
    announced a pair of
    vulnerabities in the software
    that will allow an attacker to
    either execute a DOS attack or
    execute code on the machine." + ], + [ + "The risk of intestinal damage
    from common painkillers may be
    higher than thought, research
    suggests." + ], + [ + "AN earthquake measuring 7.3 on
    the Richter Scale hit western
    Japan this morning, just hours
    after another strong quake
    rocked the area." + ], + [ + "Vodafone has increased the
    competition ahead of Christmas
    with plans to launch 10
    handsets before the festive
    season. The Newbury-based
    group said it will begin
    selling the phones in
    November." + ], + [ + "Reuters - Former Pink Floyd
    mainman Roger\\Waters released
    two new songs, both inspired
    by the U.S.-led\\invasion of
    Iraq, via online download
    outlets Tuesday." + ], + [ + "A former assistant treasurer
    at Enron Corp. (ENRNQ.PK:
    Quote, Profile, Research)
    agreed to plead guilty to
    conspiracy to commit
    securities fraud on Tuesday
    and will cooperate with" + ], + [ + "Britain #39;s Prince Harry,
    struggling to shed a growing
    quot;wild child quot; image,
    won #39;t apologize to a
    photographer he scuffled with
    outside an exclusive London
    nightclub, a royal spokesman
    said on Saturday." + ], + [ + "UK house prices fell by 1.1 in
    October, confirming a
    softening of the housing
    market, Halifax has said. The
    UK #39;s biggest mortgage
    lender said prices rose 18." + ], + [ + "Pakistan #39;s interim Prime
    Minister Chaudhry Shaujaat
    Hussain has announced his
    resignation, paving the way
    for his successor Shauket
    Aziz." + ], + [ + "A previously unknown group
    calling itself Jamaat Ansar
    al-Jihad al-Islamiya says it
    set fire to a Jewish soup
    kitchen in Paris, according to
    an Internet statement." + ], + [ + "More than six newspaper
    companies have received
    letters from the Securities
    and Exchange Commission
    seeking information about
    their circulation practices." + ], + [ + "THE 64,000 dollar -
    correction, make that 500
    million dollar -uestion
    hanging over Shire
    Pharmaceuticals is whether the
    5 per cent jump in the
    companys shares yesterday
    reflects relief that US
    regulators have finally
    approved its drug for" + ], + [ + "The deadliest attack on
    Americans in Iraq since May
    came as Iraqi officials
    announced that Saddam
    Hussein's deputy had not been
    captured on Sunday." + ], + [ + "AP - Kenny Rogers lost at the
    Coliseum for the first time in
    more than 10 years, with Bobby
    Crosby's three-run double in
    the fifth inning leading the
    Athletics to a 5-4 win over
    the Texas Rangers on Thursday." + ], + [ + "A fundraising concert will be
    held in London in memory of
    the hundreds of victims of the
    Beslan school siege." + ], + [ + "Dambulla, Sri Lanka - Kumar
    Sangakkara and Avishka
    Gunawardene slammed impressive
    half-centuries to help an
    under-strength Sri Lanka crush
    South Africa by seven wickets
    in the fourth one-day
    international here on
    Saturday." + ], + [ + "Fresh off being the worst team
    in baseball, the Arizona
    Diamondbacks set a new record
    this week: fastest team to
    both hire and fire a manager." + ], + [ + "Nebraska head coach Bill
    Callahan runs off the field at
    halftime of the game against
    Baylor in Lincoln, Neb.,
    Saturday, Oct. 16, 2004." + ], + [ + "NASA has been working on a
    test flight project for the
    last few years involving
    hypersonic flight. Hypersonic
    flight is fight at speeds
    greater than Mach 5, or five
    times the speed of sound." + ], + [ + "AFP - The landmark trial of a
    Rwandan Roman Catholic priest
    accused of supervising the
    massacre of 2,000 of his Tutsi
    parishioners during the
    central African country's 1994
    genocide opens at a UN court
    in Tanzania." + ], + [ + "The Irish government has
    stepped up its efforts to free
    the British hostage in Iraq,
    Ken Bigley, whose mother is
    from Ireland, by talking to
    diplomats from Iran and
    Jordan." + ], + [ + "AP - Republican Sen. Lincoln
    Chafee, who flirted with
    changing political parties in
    the wake of President Bush's
    re-election victory, says he
    will stay in the GOP." + ], + [ + "AP - Microsoft Corp. announced
    Wednesday that it would offer
    a low-cost, localized version
    of its Windows XP operating
    system in India to tap the
    large market potential in this
    country of 1 billion people,
    most of whom do not speak
    English." + ], + [ + "Businesses saw inventories
    rise in July and sales picked
    up, the government reported
    Wednesday. The Commerce
    Department said that stocks of
    unsold goods increased 0.9 in
    July, down from a 1.1 rise in
    June." + ], + [ + " EAST RUTHERFORD, N.J. (Sports
    Network) - The Toronto
    Raptors have traded All-Star
    swingman Vince Carter to the
    New Jersey Nets in exchange
    for center Alonzo Mourning,
    forward Eric Williams,
    center/forward Aaron Williams
    and two first- round draft
    picks." + ], + [ + "AP - Utah State University has
    secured a #36;40 million
    contract with NASA to build an
    orbiting telescope that will
    examine galaxies and try to
    find new stars." + ], + [ + "South Korean President Roh
    Moo-hyun pays a surprise visit
    to troops in Iraq, after his
    government decided to extend
    their mandate." + ], + [ + "As the European Union
    approaches a contentious
    decision - whether to let
    Turkey join the club - the
    Continent #39;s rulers seem to
    have left their citizens
    behind." + ], + [ + "What riot? quot;An Argentine
    friend of mine was a little
    derisive of the Pacers-Pistons
    eruption, quot; says reader
    Mike Gaynes. quot;He snorted,
    #39;Is that what Americans
    call a riot?" + ], + [ + "All season, Chris Barnicle
    seemed prepared for just about
    everything, but the Newton
    North senior was not ready for
    the hot weather he encountered
    yesterday in San Diego at the
    Footlocker Cross-Country
    National Championships. Racing
    in humid conditions with
    temperatures in the 70s, the
    Massachusetts Division 1 state
    champion finished sixth in 15
    minutes 34 seconds in the
    5-kilometer race. ..." + ], + [ + "Shares of Genta Inc. (GNTA.O:
    Quote, Profile, Research)
    soared nearly 50 percent on
    Monday after the biotechnology
    company presented promising
    data on an experimental
    treatment for blood cancers." + ], + [ + "I spend anywhere from three to
    eight hours every week
    sweating along with a motley
    crew of local misfits,
    shelving, sorting, and hauling
    ton after ton of written
    matter in a rowhouse basement
    in Baltimore. We have no heat
    nor air conditioning, but
    still, every week, we come and
    work. Volunteer night is
    Wednesday, but many of us also
    work on the weekends, when
    we're open to the public.
    There are times when we're
    freezing and we have to wear
    coats and gloves inside,
    making handling books somewhat
    tricky; other times, we're all
    soaked with sweat, since it's
    90 degrees out and the
    basement is thick with bodies.
    One learns to forget about
    personal space when working at
    The Book Thing, since you can
    scarcely breathe without
    bumping into someone, and we
    are all so accustomed to
    having to scrape by each other
    that most of us no longer
    bother to say \"excuse me\"
    unless some particularly
    dramatic brushing occurs." + ], + [ + " BAGHDAD (Reuters) - A
    deadline set by militants who
    have threatened to kill two
    Americans and a Briton seized
    in Iraq was due to expire
    Monday, and more than two
    dozen other hostages were
    also facing death unless rebel
    demands were met." + ], + [ + "Alarmed by software glitches,
    security threats and computer
    crashes with ATM-like voting
    machines, officials from
    Washington, D.C., to
    California are considering an
    alternative from an unlikely
    place: Nevada." + ], + [ + "Some Venezuelan television
    channels began altering their
    programs Thursday, citing
    fears of penalties under a new
    law restricting violence and
    sexual content over the
    airwaves." + ], + [ + "SBC Communications Inc. plans
    to cut at least 10,000 jobs,
    or 6 percent of its work
    force, by the end of next year
    to compensate for a drop in
    the number of local-telephone
    customers." + ], + [ + "afrol News, 4 October -
    Hundred years of conservation
    efforts have lifted the
    southern black rhino
    population from about hundred
    to 11,000 animals." + ], + [ + "THE death of Philippine movie
    star and defeated presidential
    candidate Fernando Poe has
    drawn some political backlash,
    with some people seeking to
    use his sudden demise as a
    platform to attack President
    Gloria Arroyo." + ], + [ + " WASHINGTON (Reuters) - Major
    cigarette makers go on trial
    on Tuesday in the U.S.
    government's \\$280 billion
    racketeering case that
    charges the tobacco industry
    with deliberately deceiving
    the public about the risks of
    smoking since the 1950s." + ], + [ + "More Americans are quitting
    their jobs and taking the risk
    of starting a business despite
    a still-lackluster job market." + ], + [ + "AP - Coach Tyrone Willingham
    was fired by Notre Dame on
    Tuesday after three seasons in
    which he failed to return one
    of the nation's most storied
    football programs to
    prominence." + ], + [ + "COLLEGE PARK, Md. -- Joel
    Statham completed 18 of 25
    passes for 268 yards and two
    touchdowns in No. 23
    Maryland's 45-22 victory over
    Temple last night, the
    Terrapins' 12th straight win
    at Byrd Stadium." + ], + [ + "Manchester United boss Sir
    Alex Ferguson wants the FA to
    punish Arsenal good guy Dennis
    Bergkamp for taking a swing at
    Alan Smith last Sunday." + ], + [ + "VIENTIANE, Laos China moved
    yet another step closer in
    cementing its economic and
    diplomatic relationships with
    Southeast Asia today when
    Prime Minister Wen Jiabao
    signed a trade accord at a
    regional summit that calls for
    zero tariffs on a wide range
    of" + ], + [ + "SPACE.com - With nbsp;food
    stores nbsp;running low, the
    two \\astronauts living aboard
    the International Space
    Station (ISS) are cutting back
    \\their meal intake and
    awaiting a critical cargo
    nbsp;delivery expected to
    arrive \\on Dec. 25." + ], + [ + "British judges in London
    Tuesday ordered radical Muslim
    imam Abu Hamza to stand trial
    for soliciting murder and
    inciting racial hatred." + ], + [ + "Federal Reserve policy-makers
    raised the benchmark US
    interest rate a quarter point
    to 2.25 per cent and restated
    a plan to carry out" + ], + [ + " DETROIT (Reuters) - A
    Canadian law firm on Tuesday
    said it had filed a lawsuit
    against Ford Motor Co. <A H
    REF=\"http://www.investor.reute
    rs.com/FullQuote.aspx?ticker=F
    .N target=/stocks/quickinfo/fu
    llquote\">F.N</A> over
    what it claims are defective
    door latches on about 400,000
    of the automaker's popular
    pickup trucks and SUVs." + ], + [ + "Published reports say Barry
    Bonds has testified that he
    used a clear substance and a
    cream given to him by a
    trainer who was indicted in a
    steroid-distribution ring." + ], + [ + "SARASOTA, Fla. - The
    devastation brought on by
    Hurricane Charley has been
    especially painful for an
    elderly population that is
    among the largest in the
    nation..." + ], + [ + " ATHENS (Reuters) - Christos
    Angourakis added his name to
    Greece's list of Paralympic
    medal winners when he claimed
    a bronze in the T53 shot put
    competition Thursday." + ], + [ + " quot;He is charged for having
    a part in the Bali incident,
    quot; state prosecutor Andi
    Herman told Reuters on
    Saturday. bombing attack at
    the US-run JW." + ], + [ + "Jay Fiedler threw for one
    touchdown, Sage Rosenfels
    threw for another and the
    Miami Dolphins got a victory
    in a game they did not want to
    play, beating the New Orleans
    Saints 20-19 Friday night." + ], + [ + " NEW YORK (Reuters) - Terrell
    Owens scored three touchdowns
    and the Philadelphia Eagles
    amassed 35 first-half points
    on the way to a 49-21
    drubbing of the Dallas Cowboys
    in Irving, Texas, Monday." + ], + [ + "AstraZeneca Plc suffered its
    third setback in two months on
    Friday as lung cancer drug
    Iressa failed to help patients
    live longer" + ], + [ + "Virgin Electronics hopes its
    slim Virgin Player, which
    debuts today and is smaller
    than a deck of cards, will
    rise as a lead competitor to
    Apple's iPod." + ], + [ + "Researchers train a monkey to
    feed itself by guiding a
    mechanical arm with its mind.
    It could be a big step forward
    for prosthetics. By David
    Cohn." + ], + [ + "Bruce Wasserstein, the
    combative chief executive of
    investment bank Lazard, is
    expected to agree this week
    that he will quit the group
    unless he can pull off a
    successful" + ], + [ + "A late strike by Salomon Kalou
    sealed a 2-1 win for Feyenoord
    over NEC Nijmegen, while
    second placed AZ Alkmaar
    defeated ADO Den Haag 2-0 in
    the Dutch first division on
    Sunday." + ], + [ + "The United Nations called on
    Monday for an immediate
    ceasefire in eastern Congo as
    fighting between rival army
    factions flared for a third
    day." + ], + [ + "What a disgrace Ron Artest has
    become. And the worst part is,
    the Indiana Pacers guard just
    doesn #39;t get it. Four days
    after fueling one of the
    biggest brawls in the history
    of pro sports, Artest was on
    national" + ], + [ + "Allowing dozens of casinos to
    be built in the UK would bring
    investment and thousands of
    jobs, Tony Blair says." + ], + [ + "The shock here was not just
    from the awful fact itself,
    that two vibrant young Italian
    women were kidnapped in Iraq,
    dragged from their office by
    attackers who, it seems, knew
    their names." + ], + [ + "Tehran/Vianna, Sept. 19 (NNN):
    Iran on Sunday rejected the
    International Atomic Energy
    Agency (IAEA) call to suspend
    of all its nuclear activities,
    saying that it will not agree
    to halt uranium enrichment." + ], + [ + "A 15-year-old Argentine
    student opened fire at his
    classmates on Tuesday in a
    middle school in the south of
    the Buenos Aires province,
    leaving at least four dead and
    five others wounded, police
    said." + ], + [ + "Dr. David J. Graham, the FDA
    drug safety reviewer who
    sounded warnings over five
    drugs he felt could become the
    next Vioxx has turned to a
    Whistleblower protection group
    for legal help." + ], + [ + "AP - Scientists in 17
    countries will scout waterways
    to locate and study the
    world's largest freshwater
    fish species, many of which
    are declining in numbers,
    hoping to learn how to better
    protect them, researchers
    announced Thursday." + ], + [ + "AP - Google Inc.'s plans to
    move ahead with its initial
    public stock offering ran into
    a roadblock when the
    Securities and Exchange
    Commission didn't approve the
    Internet search giant's
    regulatory paperwork as
    requested." + ], + [ + "Jenson Button has revealed
    dissatisfaction with the way
    his management handled a
    fouled switch to Williams. Not
    only did the move not come
    off, his reputation may have
    been irreparably damaged amid
    news headline" + ], + [ + "The Kmart purchase of Sears,
    Roebuck may be the ultimate
    expression of that old saying
    in real estate: location,
    location, location." + ], + [ + "Citing security risks, a state
    university is urging students
    to drop Internet Explorer in
    favor of alternative Web
    browsers such as Firefox and
    Safari." + ], + [ + "Redknapp and his No2 Jim Smith
    resigned from Portsmouth
    yesterday, leaving
    controversial new director
    Velimir Zajec in temporary
    control." + ], + [ + "Despite being ranked eleventh
    in the world in broadband
    penetration, the United States
    is rolling out high-speed
    services on a quot;reasonable
    and timely basis to all
    Americans, quot; according to
    a new report narrowly approved
    today by the Federal
    Communications" + ], + [ + "Sprint Corp. (FON.N: Quote,
    Profile, Research) on Friday
    said it plans to cut up to 700
    jobs as it realigns its
    business to focus on wireless
    and Internet services and
    takes a non-cash network
    impairment charge." + ], + [ + "As the season winds down for
    the Frederick Keys, Manager
    Tom Lawless is starting to
    enjoy the progress his
    pitching staff has made this
    season." + ], + [ + "Britain #39;s Bradley Wiggins
    won the gold medal in men
    #39;s individual pursuit
    Saturday, finishing the
    4,000-meter final in 4:16." + ], + [ + "And when David Akers #39;
    50-yard field goal cleared the
    crossbar in overtime, they did
    just that. They escaped a
    raucous Cleveland Browns
    Stadium relieved but not
    broken, tested but not
    cracked." + ], + [ + "NEW YORK - A cable pay-per-
    view company has decided not
    to show a three-hour election
    eve special with filmmaker
    Michael Moore that included a
    showing of his documentary
    \"Fahrenheit 9/11,\" which is
    sharply critical of President
    Bush. The company, iN
    DEMAND, said Friday that its
    decision is due to \"legitimate
    business and legal concerns.\"
    A spokesman would not
    elaborate..." + ], + [ + "Democracy candidates picked up
    at least one more seat in
    parliament, according to exit
    polls." + ], + [ + "The IOC wants suspended
    Olympic member Ivan Slavkov to
    be thrown out of the
    organisation." + ], + [ + " BANGKOK (Reuters) - The
    ouster of Myanmar's prime
    minister, architect of a
    tentative \"roadmap to
    democracy,\" has dashed faint
    hopes for an end to military
    rule and leaves Southeast
    Asia's policy of constructive
    engagement in tatters." + ], + [ + "San Antonio, TX (Sports
    Network) - Dean Wilson shot a
    five-under 65 on Friday to
    move into the lead at the
    halfway point of the Texas
    Open." + ], + [ + "Now that Chelsea have added
    Newcastle United to the list
    of clubs that they have given
    what for lately, what price
    Jose Mourinho covering the
    Russian-funded aristocrats of
    west London in glittering
    glory to the tune of four
    trophies?" + ], + [ + "AP - Former Seattle Seahawks
    running back Chris Warren has
    been arrested in Virginia on a
    federal warrant, accused of
    failing to pay #36;137,147 in
    child support for his two
    daughters in Washington state." + ], + [ + "The anguish of hostage Kenneth
    Bigley in Iraq hangs over
    Prime Minister Tony Blair
    today as he faces the twin
    test of a local election and a
    debate by his Labour Party
    about the divisive war." + ], + [ + "Says that amount would have
    been earned for the first 9
    months of 2004, before AT
    amp;T purchase. LOS ANGELES,
    (Reuters) - Cingular Wireless
    would have posted a net profit
    of \\$650 million for the first
    nine months" + ], + [ + "NewsFactor - Taking an
    innovative approach to the
    marketing of high-performance
    \\computing, Sun Microsystems
    (Nasdaq: SUNW) is offering its
    N1 Grid program in a pay-for-
    use pricing model that mimics
    the way such commodities as
    electricity and wireless phone
    plans are sold." + ], + [ + " CHICAGO (Reuters) - Goodyear
    Tire Rubber Co. <A HREF=\"
    http://www.investor.reuters.co
    m/FullQuote.aspx?ticker=GT.N t
    arget=/stocks/quickinfo/fullqu
    ote\">GT.N</A> said
    on Friday it will cut 340 jobs
    in its engineered products and
    chemical units as part of its
    cost-reduction efforts,
    resulting in a third-quarter
    charge." + ], + [ + " quot;There were 16 people
    travelling aboard. ... It
    crashed into a mountain, quot;
    Col. Antonio Rivero, head of
    the Civil Protection service,
    told." + ], + [ + "Reuters - Shares of long-
    distance phone\\companies AT T
    Corp. and MCI Inc. have
    plunged\\about 20 percent this
    year, but potential buyers
    seem to be\\holding out for
    clearance sales." + ], + [ + "Reuters - Madonna and m-Qube
    have\\made it possible for the
    star's North American fans to
    download\\polyphonic ring tones
    and other licensed mobile
    content from\\her official Web
    site, across most major
    carriers and without\\the need
    for a credit card." + ], + [ + "President Bush is reveling in
    winning the popular vote and
    feels he can no longer be
    considered a one-term accident
    of history." + ], + [ + "Russian oil giant Yukos files
    for bankruptcy protection in
    the US in a last ditch effort
    to stop the Kremlin auctioning
    its main production unit." + ], + [ + "British Airways #39; (BA)
    chief executive Rod Eddington
    has admitted that the company
    quot;got it wrong quot; after
    staff shortages led to three
    days of travel chaos for
    passengers." + ], + [ + "It #39;s official: US Open had
    never gone into the third
    round with only two American
    men, including the defending
    champion, Andy Roddick." + ], + [ + "Canadian Press - TORONTO (CP)
    - Thousands of Royal Bank
    clerks are being asked to
    display rainbow stickers at
    their desks and cubicles to
    promote a safe work
    environment for gays,
    lesbians, and bisexuals." + ], + [ + "The chipmaker is back on a
    buying spree, having scooped
    up five other companies this
    year." + ], + [ + "The issue of drug advertising
    directly aimed at consumers is
    becoming political." + ], + [ + "WASHINGTON, Aug. 17
    (Xinhuanet) -- England coach
    Sven-Goran Eriksson has urged
    the international soccer
    authorities to preserve the
    health of the world superstar
    footballers for major
    tournaments, who expressed his
    will in Slaley of England on
    Tuesday ..." + ], + [ + " BAGHDAD (Reuters) - A car
    bomb killed two American
    soldiers and wounded eight
    when it exploded in Baghdad on
    Saturday, the U.S. military
    said in a statement." + ], + [ + "Juventus coach Fabio Capello
    has ordered his players not to
    kick the ball out of play when
    an opponent falls to the
    ground apparently hurt because
    he believes some players fake
    injury to stop the match." + ], + [ + "AP - China's economic boom is
    still roaring despite efforts
    to cool sizzling growth, with
    the gross domestic product
    climbing 9.5 percent in the
    first three quarters of this
    year, the government reported
    Friday." + ], + [ + "Soaring petroleum prices
    pushed the cost of goods
    imported into the U.S. much
    higher than expected in
    August, the government said
    today." + ], + [ + "Anheuser-Busch teams up with
    Vietnam's largest brewer,
    laying the groundwork for
    future growth in the region." + ], + [ + "You #39;re angry. You want to
    lash out. The Red Sox are
    doing it to you again. They
    #39;re blowing a playoff
    series, and to the Yankees no
    less." + ], + [ + "TORONTO -- There is no
    mystique to it anymore,
    because after all, the
    Russians have become commoners
    in today's National Hockey
    League, and Finns, Czechs,
    Slovaks, and Swedes also have
    been entrenched in the
    Original 30 long enough to
    turn the ongoing World Cup of
    Hockey into a protracted
    trailer for the NHL season." + ], + [ + "Sudanese authorities have
    moved hundreds of pro-
    government fighters from the
    crisis-torn Darfur region to
    other parts of the country to
    keep them out of sight of
    foreign military observers
    demanding the militia #39;s
    disarmament, a rebel leader
    charged" + ], + [ + "CHARLOTTE, NC - Shares of
    Krispy Kreme Doughnuts Inc.
    #39;s fell sharply Monday as a
    79 percent plunge in third-
    quarter earnings and an
    intensifying accounting
    investigation overshadowed the
    pastrymaker #39;s statement
    that the low-carb craze might
    be easing." + ], + [ + "The company hopes to lure
    software partners by promising
    to save them from
    infrastructure headaches." + ], + [ + "BAGHDAD, Iraq - A car bomb
    Tuesday ripped through a busy
    market near a Baghdad police
    headquarters where Iraqis were
    waiting to apply for jobs on
    the force, and gunmen opened
    fire on a van carrying police
    home from work in Baqouba,
    killing at least 59 people
    total and wounding at least
    114. The attacks were the
    latest attempts by militants
    to wreck the building of a
    strong Iraqi security force, a
    keystone of the U.S..." + ], + [ + "The Israeli prime minister
    said today that he wanted to
    begin withdrawing settlers
    from Gaza next May or June." + ], + [ + "Indianapolis, IN (Sports
    Network) - The Indiana Pacers
    try to win their second
    straight game tonight, as they
    host Kevin Garnett and the
    Minnesota Timberwolves in the
    third of a four-game homestand
    at Conseco Fieldhouse." + ], + [ + "OXNARD - A leak of explosive
    natural gas forced dozens of
    workers to evacuate an
    offshore oil platform for
    hours Thursday but no damage
    or injuries were reported." + ], + [ + "AP - Assets of the nation's
    retail money market mutual
    funds rose by #36;2.85
    billion in the latest week to
    #36;845.69 billion, the
    Investment Company Institute
    said Thursday." + ], + [ + "Peter Mandelson provoked fresh
    Labour in-fighting yesterday
    with an implied attack on
    Gordon Brown #39;s
    quot;exaggerated gloating
    quot; about the health of the
    British economy." + ], + [ + "Queen Elizabeth II stopped
    short of apologizing for the
    Allies #39; bombing of Dresden
    during her first state visit
    to Germany in 12 years and
    instead acknowledged quot;the
    appalling suffering of war on
    both sides." + ], + [ + "JC Penney said yesterday that
    Allen I. Questrom, the chief
    executive who has restyled the
    once-beleaguered chain into a
    sleeker and more profitable
    entity, would be succeeded by
    Myron E. Ullman III, another
    longtime retail executive." + ], + [ + " In the cosmetics department
    at Hecht's in downtown
    Washington, construction crews
    have ripped out the
    traditional glass display
    cases, replacing them with a
    system of open shelves stacked
    high with fragrances from
    Chanel, Burberry and Armani,
    now easily within arm's reach
    of the impulse buyer." + ], + [ + " LONDON (Reuters) - Oil prices
    slid from record highs above
    \\$50 a barrel Wednesday as the
    U.S. government reported a
    surprise increase in crude
    stocks and rebels in Nigeria's
    oil-rich delta region agreed
    to a preliminary cease-fire." + ], + [ + "Rocky Shoes and Boots makes an
    accretive acquisition -- and
    gets Dickies and John Deere as
    part of the deal." + ], + [ + "AP - Fugitive Taliban leader
    Mullah Mohammed Omar has
    fallen out with some of his
    lieutenants, who blame him for
    the rebels' failure to disrupt
    the landmark Afghan
    presidential election, the
    U.S. military said Wednesday." + ], + [ + "HAVANA -- Cuban President
    Fidel Castro's advancing age
    -- and ultimately his
    mortality -- were brought home
    yesterday, a day after he
    fractured a knee and arm when
    he tripped and fell at a
    public event." + ], + [ + " BRASILIA, Brazil (Reuters) -
    The United States and Brazil
    predicted on Tuesday Latin
    America's largest country
    would resolve a dispute with
    the U.N. nuclear watchdog over
    inspections of a uranium
    enrichment plant." + ], + [ + "Call it the Maximus factor.
    Archaeologists working at the
    site of an old Roman temple
    near Guy #39;s hospital in
    London have uncovered a pot of
    cosmetic cream dating back to
    AD2." + ], + [ + "It is a team game, this Ryder
    Cup stuff that will commence
    Friday at Oakland Hills
    Country Club. So what are the
    teams? For the Americans,
    captain Hal Sutton isn't
    saying." + ], + [ + "Two bombs exploded today near
    a tea shop and wounded 20
    people in southern Thailand,
    police said, as violence
    continued unabated in the
    Muslim-majority region where
    residents are seething over
    the deaths of 78 detainees
    while in military custody." + ], + [ + "BONN: Deutsche Telekom is
    bidding 2.9 bn for the 26 it
    does not own in T-Online
    International, pulling the
    internet service back into the
    fold four years after selling
    stock to the public." + ], + [ + "Motorola Inc. says it #39;s
    ready to make inroads in the
    cell-phone market after
    posting a third straight
    strong quarter and rolling out
    a series of new handsets in
    time for the holiday selling
    season." + ], + [ + "Prime Minister Junichiro
    Koizumi, back in Tokyo after
    an 11-day diplomatic mission
    to the Americas, hunkered down
    with senior ruling party
    officials on Friday to focus
    on a major reshuffle of
    cabinet and top party posts." + ], + [ + "Costs of employer-sponsored
    health plans are expected to
    climb an average of 8 percent
    in 2005, the first time in
    five years increases have been
    in single digits." + ], + [ + "NEW YORK - A sluggish gross
    domestic product reading was
    nonetheless better than
    expected, prompting investors
    to send stocks slightly higher
    Friday on hopes that the
    economic slowdown would not be
    as bad as first thought.
    The 2.8 percent GDP growth in
    the second quarter, a revision
    from the 3 percent preliminary
    figure reported in July, is a
    far cry from the 4.5 percent
    growth in the first quarter..." + ], + [ + "After again posting record
    earnings for the third
    quarter, Taiwan Semiconductor
    Manufacturing Company (TSMC)
    expects to see its first
    sequential drop in fourth-
    quarter revenues, coupled with
    a sharp drop in capacity
    utilization rates." + ], + [ + " SEOUL (Reuters) - A huge
    explosion in North Korea last
    week may have been due to a
    combination of demolition work
    for a power plant and
    atmospheric clouds, South
    Korea's spy agency said on
    Wednesday." + ], + [ + "The deal comes as Cisco pushes
    to develop a market for CRS-1,
    a new line of routers aimed at
    telephone, wireless and cable
    companies." + ], + [ + "Many people who have never
    bounced a check in their life
    could soon bounce their first
    check if they write checks to
    pay bills a couple of days
    before their paycheck is
    deposited into their checking
    account." + ], + [ + " LUXEMBOURG (Reuters) -
    Microsoft Corp told a judge on
    Thursday that the European
    Commission must be stopped
    from ordering it to give up
    secret technology to
    competitors." + ], + [ + "SEPTEMBER 14, 2004 (IDG NEWS
    SERVICE) - Sun Microsystems
    Inc. and Microsoft Corp. next
    month plan to provide more
    details on the work they are
    doing to make their products
    interoperable, a Sun executive
    said yesterday." + ], + [ + "MANCHESTER United today
    dramatically rejected the
    advances of Malcolm Glazer,
    the US sports boss who is
    mulling an 825m bid for the
    football club." + ], + [ + "Dow Jones Industrial Average
    futures declined amid concern
    an upcoming report on
    manufacturing may point to
    slowing economic growth." + ], + [ + "AP - Astronomy buffs and
    amateur stargazers turned out
    to watch a total lunar eclipse
    Wednesday night #151; the
    last one Earth will get for
    nearly two and a half years." + ], + [ + "Reuters - U.S. industrial
    output advanced in\\July, as
    American factories operated at
    their highest capacity\\in more
    than three years, a Federal
    Reserve report on
    Tuesday\\showed." + ], + [ + "As usual, the Big Ten coaches
    were out in full force at
    today #39;s Big Ten
    Teleconference. Both OSU head
    coach Jim Tressel and Iowa
    head coach Kirk Ferentz
    offered some thoughts on the
    upcoming game between OSU" + ], + [ + "Sir Martin Sorrell, chief
    executive of WPP, yesterday
    declared he was quot;very
    impressed quot; with Grey
    Global, stoking speculation
    WPP will bid for the US
    advertising company." + ], + [ + "The compact disc has at least
    another five years as the most
    popular music format before
    online downloads chip away at
    its dominance, a new study
    said on Tuesday." + ], + [ + "New York Knicks #39; Stephon
    Marbury (3) fires over New
    Orleans Hornets #39; Dan
    Dickau (2) during the second
    half in New Orleans Wednesday
    night, Dec. 8, 2004." + ], + [ + "AP - Sudan's U.N. ambassador
    challenged the United States
    to send troops to the Darfur
    region if it really believes a
    genocide is taking place as
    the U.S. Congress and
    President Bush's
    administration have
    determined." + ], + [ + "At the very moment when the
    Red Sox desperately need
    someone slightly larger than
    life to rally around, they
    suddenly have the man for the
    job: Thrilling Schilling." + ], + [ + "Does Your Site Need a Custom
    Search Engine
    Toolbar?\\\\Today's surfers
    aren't always too comfortable
    installing software on their
    computers. Especially free
    software that they don't
    necessarily understand. With
    all the horror stories of
    viruses, spyware, and adware
    that make the front page these
    days, it's no wonder. So is
    there ..." + ], + [ + "Dale Earnhardt Jr, right,
    talks with Matt Kenseth, left,
    during a break in practice at
    Lowe #39;s Motor Speedway in
    Concord, NC, Thursday Oct. 14,
    2004 before qualifying for
    Saturday #39;s UAW-GM Quality
    500 NASCAR Nextel Cup race." + ], + [ + "Several starting spots may
    have been usurped or at least
    threatened after relatively
    solid understudy showings
    Sunday, but few players
    welcome the kind of shot
    delivered to Oakland" + ], + [ + "TEMPE, Ariz. -- Neil Rackers
    kicked four field goals and
    the Arizona Cardinals stifled
    rookie Chris Simms and the
    rest of the Tampa Bay offense
    for a 12-7 victory yesterday
    in a matchup of two sputtering
    teams out of playoff
    contention. Coach Jon Gruden's
    team lost its fourth in a row
    to finish 5-11, Tampa Bay's
    worst record since going ..." + ], + [ + "AFP - A junior British
    minister said that people
    living in camps after fleeing
    their villages because of
    conflict in Sudan's Darfur
    region lived in fear of
    leaving their temporary homes
    despite a greater presence now
    of aid workers." + ], + [ + "US Deputy Secretary of State
    Richard Armitage (L) shakes
    hands with Pakistani Foreign
    Minister Khurshid Kasuri prior
    to their meeting in Islamabad,
    09 November 2004." + ], + [ + "Spammers aren't ducking
    antispam laws by operating
    offshore, they're just
    ignoring it." + ], + [ + "AP - Biologists plan to use
    large nets and traps this week
    in Chicago's Burnham Harbor to
    search for the northern
    snakehead #151; a type of
    fish known for its voracious
    appetite and ability to wreak
    havoc on freshwater
    ecosystems." + ], + [ + "BAE Systems PLC and Northrop
    Grumman Corp. won \\$45 million
    contracts yesterday to develop
    prototypes of anti-missile
    technology that could protect
    commercial aircraft from
    shoulder-fired missiles." + ], + [ + "AP - Democrat John Edwards
    kept up a long-distance debate
    over his \"two Americas\"
    campaign theme with Vice
    President Dick Cheney on
    Tuesday, saying it was no
    illusion to thousands of laid-
    off workers in Ohio." + ], + [ + "Like introductory credit card
    rates and superior customer
    service, some promises just
    aren #39;t built to last. And
    so it is that Bank of America
    - mere months after its pledge
    to preserve" + ], + [ + "A group of 76 Eritreans on a
    repatriation flight from Libya
    Friday forced their plane to
    change course and land in the
    Sudanese capital, Khartoum,
    where they" + ], + [ + "Reuters - Accounting firm KPMG
    will pay #36;10\\million to
    settle charges of improper
    professional conduct\\while
    acting as auditor for Gemstar-
    TV Guide International Inc.\\,
    the U.S. Securities and
    Exchange Commission said
    on\\Wednesday." + ], + [ + "Family matters made public: As
    eager cousins wait for a slice
    of the \\$15 billion cake that
    is the Pritzker Empire,
    Circuit Court Judge David
    Donnersberger has ruled that
    the case will be conducted in
    open court." + ], + [ + "Users of Microsoft #39;s
    Hotmail, most of whom are
    accustomed to getting regular
    sales pitches for premium
    e-mail accounts, got a
    pleasant surprise in their
    inboxes this week: extra
    storage for free." + ], + [ + "AP - They say that opposites
    attract, and in the case of
    Sen. John Kerry and Teresa
    Heinz Kerry, that may be true
    #151; at least in their public
    personas." + ], + [ + "The weekly survey from
    mortgage company Freddie Mac
    had rates on 30-year fixed-
    rate mortgages inching higher
    this week, up to an average
    5.82 percent from last week
    #39;s 5.81 percent." + ], + [ + "The classic power struggle
    between Walt Disney Co. CEO
    Michael Eisner and former
    feared talent agent Michael
    Ovitz makes for high drama in
    the courtroom - and apparently
    on cable." + ], + [ + "ATHENS France, Britain and the
    United States issued a joint
    challenge Thursday to Germany
    #39;s gold medal in equestrian
    team three-day eventing." + ], + [ + "LONDON (CBS.MW) -- Elan
    (UK:ELA) (ELN) and partner
    Biogen (BIIB) said the FDA has
    approved new drug Tysabri to
    treat relapsing forms of
    multiple sclerosis." + ], + [ + "IT services provider
    Electronic Data Systems
    yesterday reported a net loss
    of \\$153 million for the third
    quarter, with earnings hit in
    part by an asset impairment
    charge of \\$375 million
    connected with EDS's N/MCI
    project." + ], + [ + "ATLANTA - Who could have
    imagined Tommy Tuberville in
    this position? Yet there he
    was Friday, standing alongside
    the SEC championship trophy,
    posing for pictures and
    undoubtedly chuckling a bit on
    the inside." + ], + [ + "AP - Impoverished North Korea
    might resort to selling
    weapons-grade plutonium to
    terrorists for much-needed
    cash, and that would be
    \"disastrous for the world,\"
    the top U.S. military
    commander in South Korea said
    Friday." + ], + [ + "International Business
    Machines Corp. said Wednesday
    it had agreed to settle most
    of the issues in a suit over
    changes in its pension plans
    on terms that allow the
    company to continue to appeal
    a key question while capping
    its liability at \\$1.7
    billion. <BR><FONT
    face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\"\\ color=\"#666666\">
    <B>-The Washington
    Post</B></FONT>" + ], + [ + "Dubai: Osama bin Laden on
    Thursday called on his
    fighters to strike Gulf oil
    supplies and warned Saudi
    leaders they risked a popular
    uprising in an audio message
    said to be by the Western
    world #39;s most wanted terror
    mastermind." + ], + [ + "Bank of New Zealand has frozen
    all accounts held in the name
    of Access Brokerage, which was
    yesterday placed in
    liquidation after a client
    fund shortfall of around \\$5
    million was discovered." + ], + [ + "Edward Kozel, Cisco's former
    chief technology officer,
    joins the board of Linux
    seller Red Hat." + ], + [ + "Reuters - International
    Business Machines\\Corp. late
    on Wednesday rolled out a new
    version of its\\database
    software aimed at users of
    Linux and Unix
    operating\\systems that it
    hopes will help the company
    take away market\\share from
    market leader Oracle Corp. ." + ], + [ + "NASA #39;s Mars Odyssey
    mission, originally scheduled
    to end on Tuesday, has been
    granted a stay of execution
    until at least September 2006,
    reveal NASA scientists." + ], + [ + "ZDNet #39;s survey of IT
    professionals in August kept
    Wired amp; Wireless on top
    for the 18th month in a row.
    Telecommunications equipment
    maker Motorola said Tuesday
    that it would cut 1,000 jobs
    and take related" + ], + [ + "John Thomson threw shutout
    ball for seven innings
    Wednesday night in carrying
    Atlanta to a 2-0 blanking of
    the New York Mets. New York
    lost for the 21st time in 25
    games on the" + ], + [ + "BEIJING -- Chinese authorities
    have arrested or reprimanded
    more than 750 officials in
    recent months in connection
    with billions of dollars in
    financial irregularities
    ranging from unpaid taxes to
    embezzlement, according to a
    report made public yesterday." + ], + [ + "Canadian Press - GAUHATI,
    India (AP) - Residents of
    northeastern India were
    bracing for more violence
    Friday, a day after bombs
    ripped through two buses and a
    grenade was hurled into a
    crowded market in attacks that
    killed four people and wounded
    54." + ], + [ + "Vikram Solanki beat the rain
    clouds to register his second
    one-day international century
    as England won the third one-
    day international to wrap up a
    series victory." + ], + [ + "Some 28 people are charged
    with trying to overthrow
    Sudan's President Bashir,
    reports say." + ], + [ + "Hong Kong #39;s Beijing-backed
    chief executive yesterday
    ruled out any early moves to
    pass a controversial national
    security law which last year
    sparked a street protest by
    half a million people." + ], + [ + "BRITAIN #39;S largest
    financial institutions are
    being urged to take lead roles
    in lawsuits seeking hundreds
    of millions of dollars from
    the scandal-struck US
    insurance industry." + ], + [ + "Bankrupt UAL Corp. (UALAQ.OB:
    Quote, Profile, Research) on
    Thursday reported a narrower
    third-quarter net loss. The
    parent of United Airlines
    posted a loss of \\$274
    million, or" + ], + [ + "A three-touchdown point spread
    and a recent history of late-
    season collapses had many
    thinking the UCLA football
    team would provide little
    opposition to rival USC #39;s
    march to the BCS-championship
    game at the Orange Bowl." + ], + [ + " PHOENIX (Sports Network) -
    Free agent third baseman Troy
    Glaus is reportedly headed to
    the Arizona Diamondbacks." + ], + [ + "The European Union #39;s head
    office issued a bleak economic
    report Tuesday, warning that
    the sharp rise in oil prices
    will quot;take its toll quot;
    on economic growth next year
    while the euro #39;s renewed
    climb could threaten crucial
    exports." + ], + [ + "Third-string tailback Chris
    Markey ran for 131 yards to
    lead UCLA to a 34-26 victory
    over Oregon on Saturday.
    Markey, playing because of an
    injury to starter Maurice
    Drew, also caught five passes
    for 84 yards" + ], + [ + "Though Howard Stern's
    defection from broadcast to
    satellite radio is still 16
    months off, the industry is
    already trying to figure out
    what will fill the crater in
    ad revenue and listenership
    that he is expected to leave
    behind." + ], + [ + " WASHINGTON (Reuters) - The
    United States set final anti-
    dumping duties of up to 112.81
    percent on shrimp imported
    from China and up to 25.76
    percent on shrimp from Vietnam
    to offset unfair pricing, the
    Commerce Department said on
    Tuesday." + ], + [ + "Jay Payton #39;s three-run
    homer led the San Diego Padres
    to a 5-1 win over the San
    Francisco Giants in National
    League play Saturday, despite
    a 701st career blast from
    Barry Bonds." + ], + [ + "The optimists among Rutgers
    fans were delighted, but the
    Scarlet Knights still gave the
    pessimists something to worry
    about." + ], + [ + "Beer consumption has doubled
    over the past five years,
    prompting legislators to
    implement new rules." + ], + [ + "BusinessWeek Online - The
    jubilation that swept East
    Germany after the fall of the
    Berlin Wall in 1989 long ago
    gave way to the sober reality
    of globalization and market
    forces. Now a decade of
    resentment seems to be boiling
    over. In Eastern cities such
    as Leipzig or Chemnitz,
    thousands have taken to the
    streets since July to protest
    cuts in unemployment benefits,
    the main source of livelihood
    for 1.6 million East Germans.
    Discontent among
    reunification's losers fueled
    big gains by the far left and
    far right in Brandenburg and
    Saxony state elections Sept.
    19. ..." + ], + [ + "NEW YORK -- When Office Depot
    Inc. stores ran an electronics
    recycling drive last summer
    that accepted everything from
    cellphones to televisions,
    some stores were overwhelmed
    by the amount of e-trash they
    received." + ], + [ + "In a discovery sure to set off
    a firestorm of debate over
    human migration to the western
    hemisphere, archaeologists in
    South Carolina say they have
    uncovered evidence that people
    lived in eastern North America
    at least 50,000 years ago -
    far earlier than" + ], + [ + "AP - Former president Bill
    Clinton on Monday helped
    launch a new Internet search
    company backed by the Chinese
    government which says its
    technology uses artificial
    intelligence to produce better
    results than Google Inc." + ], + [ + "The International Monetary
    Fund expressed concern Tuesday
    about the impact of the
    troubles besetting oil major
    Yukos on Russia #39;s standing
    as a place to invest." + ], + [ + "Perhaps the sight of Maria
    Sharapova opposite her tonight
    will jog Serena Williams #39;
    memory. Wimbledon. The final.
    You and Maria." + ], + [ + "At least 79 people were killed
    and 74 were missing after some
    of the worst rainstorms in
    recent years triggered
    landslides and flash floods in
    southwest China, disaster
    relief officials said
    yesterday." + ], + [ + "LinuxWorld Conference amp;
    Expo will come to Boston for
    the first time in February,
    underscoring the area's
    standing as a hub for the
    open-source software being
    adopted by thousands of
    businesses." + ], + [ + "BANGKOK Thai military aircraft
    dropped 100 million paper
    birds over southern Thailand
    on Sunday in a gesture
    intended to promote peace in
    mainly Muslim provinces, where
    more than 500 people have died
    this year in attacks by
    separatist militants and" + ], + [ + "Insurgents threatened on
    Saturday to cut the throats of
    two Americans and a Briton
    seized in Baghdad, and
    launched a suicide car bomb
    attack on Iraqi security
    forces in Kirkuk that killed
    at least 23 people." + ], + [ + "Five days after making the
    putt that won the Ryder Cup,
    Colin Montgomerie looked set
    to miss the cut at a European
    PGA tour event." + ], + [ + "Sun Microsystems plans to
    release its Sun Studio 10
    development tool in the fourth
    quarter of this year,
    featuring support for 64-bit
    applications running on AMD
    Opteron and Nocona processors,
    Sun officials said on Tuesday." + ], + [ + "Karachi - Captain Inzamam ul-
    Haq and coach Bob Woolmer came
    under fire on Thursday for
    choosing to bat first on a
    tricky pitch after Pakistan
    #39;s humiliating defeat in
    the ICC Champions Trophy semi-
    final." + ], + [ + "Scottish champions Celtic saw
    their three-year unbeaten home
    record in Europe broken
    Tuesday as they lost 3-1 to
    Barcelona in the Champions
    League Group F opener." + ], + [ + "Interest rates on short-term
    Treasury securities rose in
    yesterday's auction. The
    Treasury Department sold \\$19
    billion in three-month bills
    at a discount rate of 1.710
    percent, up from 1.685 percent
    last week. An additional \\$17
    billion was sold in six-month
    bills at a rate of 1.950
    percent, up from 1.870
    percent." + ], + [ + "Most IT Managers won #39;t
    question the importance of
    security, but this priority
    has been sliding between the
    third and fourth most
    important focus for companies." + ], + [ + "AP - Andy Roddick searched out
    Carlos Moya in the throng of
    jumping, screaming Spanish
    tennis players, hoping to
    shake hands." + ], + [ + "AP - The Federal Trade
    Commission on Thursday filed
    the first case in the country
    against software companies
    accused of infecting computers
    with intrusive \"spyware\" and
    then trying to sell people the
    solution." + ], + [ + "While shares of Apple have
    climbed more than 10 percent
    this week, reaching 52-week
    highs, two research firms told
    investors Friday they continue
    to remain highly bullish about
    the stock." + ], + [ + "Moody #39;s Investors Service
    on Wednesday said it may cut
    its bond ratings on HCA Inc.
    (HCA.N: Quote, Profile,
    Research) deeper into junk,
    citing the hospital operator
    #39;s plan to buy back about
    \\$2." + ], + [ + "States are now barred from
    imposing telecommunications
    regulations on Net phone
    providers." + ], + [ + "Telekom Austria AG, the
    country #39;s biggest phone
    operator, won the right to buy
    Bulgaria #39;s largest mobile
    phone company, MobilTel EAD,
    for 1.6 billion euros (\\$2.1
    billion), an acquisition that
    would add 3 million
    subscribers." + ], + [ + "Shares of ID Biomedical jumped
    after the company reported
    Monday that it signed long-
    term agreements with the three
    largest flu vaccine
    wholesalers in the United
    States in light of the
    shortage of vaccine for the
    current flu season." + ], + [ + "ENGLAND captain and Real
    Madrid midfielder David
    Beckham has played down
    speculation that his club are
    moving for England manager
    Sven-Goran Eriksson." + ], + [ + "The federal government closed
    its window on the oil industry
    Thursday, saying that it is
    selling its last 19 per cent
    stake in Calgary-based Petro-
    Canada." + ], + [ + "Strong sales of new mobile
    phone models boosts profits at
    Carphone Warehouse but the
    retailer's shares fall on
    concerns at a decline in
    profits from pre-paid phones." + ], + [ + " NEW YORK (Reuters) - IBM and
    top scientific research
    organizations are joining
    forces in a humanitarian
    effort to tap the unused
    power of millions of computers
    and help solve complex social
    problems." + ], + [ + " NEW YORK, Nov. 11 -- The 40
    percent share price slide in
    Merck #38; Co. in the five
    weeks after it pulled the
    painkiller Vioxx off the
    market highlighted larger
    problems in the pharmaceutical
    industry that may depress
    performance for years,
    according to academics and
    stock analysts who follow the
    sector." + ], + [ + "Low-fare carrier Southwest
    Airlines Co. said Thursday
    that its third-quarter profit
    rose 12.3 percent to beat Wall
    Street expectations despite
    higher fuel costs." + ], + [ + "Another shock hit the drug
    sector Friday when
    pharmaceutical giant Pfizer
    Inc. announced that it found
    an increased heart risk to
    patients for its blockbuster
    arthritis drug Celebrex." + ], + [ + "AFP - National Basketball
    Association players trying to
    win a fourth consecutive
    Olympic gold medal for the
    United States have gotten the
    wake-up call that the \"Dream
    Team\" days are done even if
    supporters have not." + ], + [ + " BEIJING (Reuters) - Secretary
    of State Colin Powell will
    urge China Monday to exert its
    influence over North Korea to
    resume six-party negotiations
    on scrapping its suspected
    nuclear weapons program." + ], + [ + "Reuters - An Israeli missile
    strike killed at least\\two
    Hamas gunmen in Gaza City
    Monday, a day after a
    top\\commander of the Islamic
    militant group was killed in a
    similar\\attack, Palestinian
    witnesses and security sources
    said." + ], + [ + "England will be seeking their
    third clean sweep of the
    summer when the NatWest
    Challenge against India
    concludes at Lord #39;s. After
    beating New Zealand and West
    Indies 3-0 and 4-0 in Tests,
    they have come alive" + ], + [ + "AP - The Pentagon has restored
    access to a Web site that
    assists soldiers and other
    Americans living overseas in
    voting, after receiving
    complaints that its security
    measures were preventing
    legitimate voters from using
    it." + ], + [ + "The number of US information
    technology workers rose 2
    percent to 10.5 million in the
    first quarter of this year,
    but demand for them is
    dropping, according to a new
    report." + ], + [ + "Montreal - The British-built
    Canadian submarine HMCS
    Chicoutimi, crippled since a
    fire at sea that killed one of
    its crew, is under tow and is
    expected in Faslane, Scotland,
    early next week, Canadian
    naval commander Tyrone Pile
    said on Friday." + ], + [ + "AP - Grizzly and black bears
    killed a majority of elk
    calves in northern Yellowstone
    National Park for the second
    year in a row, preliminary
    study results show." + ], + [ + "Thirty-five Pakistanis freed
    from the US Guantanamo Bay
    prison camp arrived home on
    Saturday and were taken
    straight to prison for further
    interrogation, the interior
    minister said." + ], + [ + "AP - Mark Richt knows he'll
    have to get a little creative
    when he divvies up playing
    time for Georgia's running
    backs next season. Not so on
    Saturday. Thomas Brown is the
    undisputed starter for the
    biggest game of the season." + ], + [ + "Witnesses in the trial of a US
    soldier charged with abusing
    prisoners at Abu Ghraib have
    told the court that the CIA
    sometimes directed abuse and
    orders were received from
    military command to toughen
    interrogations." + ], + [ + "Insurance firm says its board
    now consists of its new CEO
    Michael Cherkasky and 10
    outside members. NEW YORK
    (Reuters) - Marsh amp;
    McLennan Cos." + ], + [ + "Michael Phelps, the six-time
    Olympic champion, issued an
    apology yesterday after being
    arrested and charged with
    drunken driving in the United
    States." + ], + [ + "PC World - The one-time World
    Class Product of the Year PDA
    gets a much-needed upgrade." + ], + [ + "AP - Italian and Lebanese
    authorities have arrested 10
    suspected terrorists who
    planned to blow up the Italian
    Embassy in Beirut, an Italian
    news agency and the Defense
    Ministry said Tuesday." + ], + [ + "CORAL GABLES, Fla. -- John F.
    Kerry and President Bush
    clashed sharply over the war
    in Iraq last night during the
    first debate of the
    presidential campaign season,
    with the senator from
    Massachusetts accusing" + ], + [ + "GONAIVES, Haiti -- While
    desperately hungry flood
    victims wander the streets of
    Gonaives searching for help,
    tons of food aid is stacking
    up in a warehouse guarded by
    United Nations peacekeepers." + ], + [ + "As part of its much-touted new
    MSN Music offering, Microsoft
    Corp. (MSFT) is testing a Web-
    based radio service that
    mimics nearly 1,000 local
    radio stations, allowing users
    to hear a version of their
    favorite radio station with
    far fewer interruptions." + ], + [ + "AFP - The resignation of
    disgraced Fiji Vice-President
    Ratu Jope Seniloli failed to
    quell anger among opposition
    leaders and the military over
    his surprise release from
    prison after serving just
    three months of a four-year
    jail term for his role in a
    failed 2000 coup." + ], + [ + "ATHENS Larry Brown, the US
    coach, leaned back against the
    scorer #39;s table, searching
    for support on a sinking ship.
    His best player, Tim Duncan,
    had just fouled out, and the
    options for an American team
    that" + ], + [ + "Sourav Ganguly files an appeal
    against a two-match ban
    imposed for time-wasting." + ], + [ + "Ziff Davis - A quick
    resolution to the Mambo open-
    source copyright dispute seems
    unlikely now that one of the
    parties has rejected an offer
    for mediation." + ], + [ + "Greek police surrounded a bus
    full of passengers seized by
    armed hijackers along a
    highway from an Athens suburb
    Wednesday, police said." + ], + [ + "Spain have named an unchanged
    team for the Davis Cup final
    against the United States in
    Seville on 3-5 December.
    Carlos Moya, Juan Carlos
    Ferrero, Rafael Nadal and
    Tommy Robredo will take on the
    US in front of 22,000 fans at
    the converted Olympic stadium." + ], + [ + "BAGHDAD: A suicide car bomber
    attacked a police convoy in
    Baghdad yesterday as guerillas
    kept pressure on Iraq #39;s
    security forces despite a
    bloody rout of insurgents in
    Fallujah." + ], + [ + "Slot machine maker
    International Game Technology
    (IGT.N: Quote, Profile,
    Research) on Tuesday posted
    better-than-expected quarterly
    earnings, as casinos bought" + ], + [ + "Fixtures and fittings from
    Damien Hirst's restaurant
    Pharmacy sell for 11.1, 8m
    more than expected." + ], + [ + "Last Tuesday night, Harvard
    knocked off rival Boston
    College, which was ranked No.
    1 in the country. Last night,
    the Crimson knocked off
    another local foe, Boston
    University, 2-1, at Walter
    Brown Arena, which marked the
    first time since 1999 that
    Harvard had beaten them both
    in the same season." + ], + [ + "Venezuelan election officials
    say they expect to announce
    Saturday, results of a partial
    audit of last Sunday #39;s
    presidential recall
    referendum." + ], + [ + "About 500 prospective jurors
    will be in an Eagle, Colorado,
    courtroom Friday, answering an
    82-item questionnaire in
    preparation for the Kobe
    Bryant sexual assault trial." + ], + [ + "Students take note - endless
    journeys to the library could
    become a thing of the past
    thanks to a new multimillion-
    pound scheme to make classic
    texts available at the click
    of a mouse." + ], + [ + "Moscow - The next crew of the
    International Space Station
    (ISS) is to contribute to the
    Russian search for a vaccine
    against Aids, Russian
    cosmonaut Salijan Sharipovthe
    said on Thursday." + ], + [ + "Locked away in fossils is
    evidence of a sudden solar
    cooling. Kate Ravilious meets
    the experts who say it could
    explain a 3,000-year-old mass
    migration - and today #39;s
    global warming." + ], + [ + "By ANICK JESDANUN NEW YORK
    (AP) -- Taran Rampersad didn't
    complain when he failed to
    find anything on his hometown
    in the online encyclopedia
    Wikipedia. Instead, he simply
    wrote his own entry for San
    Fernando, Trinidad and
    Tobago..." + ], + [ + "Five years after catastrophic
    floods and mudslides killed
    thousands along Venezuela's
    Caribbean coast, survivors in
    this town still see the signs
    of destruction - shattered
    concrete walls and tall weeds
    growing atop streets covered
    in dried mud." + ], + [ + "How do you top a battle
    between Marines and an alien
    religious cult fighting to the
    death on a giant corona in
    outer space? The next logical
    step is to take that battle to
    Earth, which is exactly what
    Microsoft" + ], + [ + "Sony Ericsson Mobile
    Communications Ltd., the
    mobile-phone venture owned by
    Sony Corp. and Ericsson AB,
    said third-quarter profit rose
    45 percent on camera phone
    demand and forecast this
    quarter will be its strongest." + ], + [ + "NEW YORK, September 17
    (newratings.com) - Alcatel
    (ALA.NYS) has expanded its
    operations and presence in the
    core North American
    telecommunication market with
    two separate acquisitions for
    about \\$277 million." + ], + [ + "Four U.S. agencies yesterday
    announced a coordinated attack
    to stem the global trade in
    counterfeit merchandise and
    pirated music and movies, an
    underground industry that law-
    enforcement officials estimate
    to be worth \\$500 billion each
    year." + ], + [ + "BEIJING The NBA has reached
    booming, basketball-crazy
    China _ but the league doesn
    #39;t expect to make any money
    soon. The NBA flew more than
    100 people halfway around the
    world for its first games in
    China, featuring" + ], + [ + "The UN nuclear watchdog
    confirmed Monday that nearly
    400 tons of powerful
    explosives that could be used
    in conventional or nuclear
    missiles disappeared from an
    unguarded military
    installation in Iraq." + ], + [ + " NEW YORK (Reuters) - Curt
    Schilling pitched 6 2/3
    innings and Manny Ramirez hit
    a three-run homer in a seven-
    run fourth frame to lead the
    Boston Red Sox to a 9-3 win
    over the host Anaheim Angels
    in their American League
    Divisional Series opener
    Tuesday." + ], + [ + "AP - The game between the
    Minnesota Twins and the New
    York Yankees on Tuesday night
    was postponed by rain." + ], + [ + "Oil giant Shell swept aside
    nearly 100 years of history
    today when it unveiled plans
    to merge its UK and Dutch
    parent companies. Shell said
    scrapping its twin-board
    structure" + ], + [ + "Caesars Entertainment Inc. on
    Thursday posted a rise in
    third-quarter profit as Las
    Vegas hotels filled up and
    Atlantic City properties
    squeaked out a profit that was
    unexpected by the company." + ], + [ + "Half of all U.S. Secret
    Service agents are dedicated
    to protecting President
    Washington #151;and all the
    other Presidents on U.S.
    currency #151;from
    counterfeiters." + ], + [ + "One of India #39;s leading
    telecommunications providers
    will use Cisco Systems #39;
    gear to build its new
    Ethernet-based broadband
    network." + ], + [ + "Militants in Iraq have killed
    the second of two US civilians
    they were holding hostage,
    according to a statement on an
    Islamist website." + ], + [ + "PARIS The verdict is in: The
    world #39;s greatest race car
    driver, the champion of
    champions - all disciplines
    combined - is Heikki
    Kovalainen." + ], + [ + "Pakistan won the toss and
    unsurprisingly chose to bowl
    first as they and West Indies
    did battle at the Rose Bowl
    today for a place in the ICC
    Champions Trophy final against
    hosts England." + ], + [ + "Shares of Beacon Roofing
    Suppler Inc. shot up as much
    as 26 percent in its trading
    debut Thursday, edging out
    bank holding company Valley
    Bancorp as the biggest gainer
    among a handful of new stocks
    that went public this week." + ], + [ + "More than 4,000 American and
    Iraqi soldiers mounted a
    military assault on the
    insurgent-held city of
    Samarra." + ], + [ + "Maxime Faget conceived and
    proposed the development of
    the one-man spacecraft used in
    Project Mercury, which put the
    first American astronauts into
    suborbital flight, then
    orbital flight" + ], + [ + "The first weekend of holiday
    shopping went from red-hot to
    dead white, as a storm that
    delivered freezing, snowy
    weather across Colorado kept
    consumers at home." + ], + [ + " MEXICO CITY (Reuters) -
    Mexican President Vicente Fox
    said on Monday he hoped
    President Bush's re-election
    meant a bilateral accord on
    migration would be reached
    before his own term runs out
    at the end of 2006." + ], + [ + " LONDON (Reuters) - The dollar
    fell within half a cent of
    last week's record low against
    the euro on Thursday after
    capital inflows data added to
    worries the United States may
    struggle to fund its current
    account deficit." + ], + [ + " BRUSSELS (Reuters) - French
    President Jacques Chirac said
    on Friday he had not refused
    to meet Iraqi interim Prime
    Minister Iyad Allawi after
    reports he was snubbing the
    Iraqi leader by leaving an EU
    meeting in Brussels early." + ], + [ + "China has pledged to invest
    \\$20 billion in Argentina in
    the next 10 years, La Nacion
    reported Wednesday. The
    announcement came during the
    first day of a two-day visit" + ], + [ + "NEW YORK - Former President
    Bill Clinton was in good
    spirits Saturday, walking
    around his hospital room in
    street clothes and buoyed by
    thousands of get-well messages
    as he awaited heart bypass
    surgery early this coming
    week, people close to the
    family said. Clinton was
    expected to undergo surgery as
    early as Monday but probably
    Tuesday, said Democratic Party
    Chairman Terry McAuliffe, who
    said the former president was
    \"upbeat\" when he spoke to him
    by phone Friday..." + ], + [ + "Toyota Motor Corp., the world
    #39;s biggest carmaker by
    value, will invest 3.8 billion
    yuan (\\$461 million) with its
    partner Guangzhou Automobile
    Group to boost manufacturing
    capacity in" + ], + [ + "Poland will cut its troops in
    Iraq early next year and won
    #39;t stay in the country
    quot;an hour longer quot; than
    needed, the country #39;s
    prime minister said Friday." + ], + [ + "AP - Boston Red Sox center
    fielder Johnny Damon is having
    a recurrence of migraine
    headaches that first bothered
    him after a collision in last
    year's playoffs." + ], + [ + "The patch fixes a flaw in the
    e-mail server software that
    could be used to get access to
    in-boxes and information." + ], + [ + "Felix Cardenas of Colombia won
    the 17th stage of the Spanish
    Vuelta cycling race Wednesday,
    while defending champion
    Roberto Heras held onto the
    overall leader #39;s jersey
    for the sixth day in a row." + ], + [ + "AP - Being the biggest dog may
    pay off at feeding time, but
    species that grow too large
    may be more vulnerable to
    extinction, new research
    suggests." + ], + [ + "A car bomb exploded at a US
    military convoy in the
    northern Iraqi city of Mosul,
    causing several casualties,
    the army and Iraqi officials
    said." + ], + [ + "Newest Efficeon processor also
    offers higher frequency using
    less power." + ], + [ + "BAE Systems has launched a
    search for a senior American
    businessman to become a non-
    executive director. The high-
    profile appointment is
    designed to strengthen the
    board at a time when the
    defence giant is" + ], + [ + "AP - In what it calls a first
    in international air travel,
    Finnair says it will let its
    frequent fliers check in using
    text messages on mobile
    phones." + ], + [ + "Computer Associates
    International is expected to
    announce that its new chief
    executive will be John
    Swainson, an I.B.M. executive
    with strong technical and
    sales credentials." + ], + [ + "Established star Landon
    Donovan and rising sensation
    Eddie Johnson carried the
    United States into the
    regional qualifying finals for
    the 2006 World Cup in emphatic
    fashion Wednesday night." + ], + [ + "STOCKHOLM (Dow
    Jones)--Expectations for
    Telefon AB LM Ericsson #39;s
    (ERICY) third-quarter
    performance imply that while
    sales of mobile telephony
    equipment are expected to have
    dipped, the company" + ], + [ + "ATHENS, Greece - Michael
    Phelps took care of qualifying
    for the Olympic 200-meter
    freestyle semifinals Sunday,
    and then found out he had been
    added to the American team for
    the evening's 400 freestyle
    relay final. Phelps' rivals
    Ian Thorpe and Pieter van den
    Hoogenband and teammate Klete
    Keller were faster than the
    teenager in the 200 free
    preliminaries..." + ], + [ + " JERUSALEM (Reuters) - Israeli
    Prime Minister Ariel Sharon
    intends to present a timetable
    for withdrawal from Gaza to
    lawmakers from his Likud Party
    Tuesday despite a mutiny in
    the right-wing bloc over the
    plan." + ], + [ + "Search Engine for Programming
    Code\\\\An article at Newsforge
    pointed me to Koders (
    http://www.koders.com ) a
    search engine for finding
    programming code. Nifty.\\\\The
    front page allows you to
    specify keywords, sixteen
    languages (from ASP to VB.NET)
    and sixteen licenses (from AFL
    to ZPL -- fortunately there's
    an information page to ..." + ], + [ + " #147;Apple once again was the
    star of the show at the annual
    MacUser awards, #148; reports
    MacUser, #147;taking away six
    Maxine statues including
    Product of the Year for the
    iTunes Music Store. #148;
    Other Apple award winners:
    Power Mac G5, AirPort Express,
    20-inch Apple Cinema Display,
    GarageBand and DVD Studio Pro
    3. Nov 22" + ], + [ + " KABUL (Reuters) - Three U.N.
    workers held by militants in
    Afghanistan were in their
    third week of captivity on
    Friday after calls from both
    sides for the crisis to be
    resolved ahead of this
    weekend's Muslim festival of
    Eid al-Fitr." + ], + [ + "AP - A sweeping wildlife
    preserve in southwestern
    Arizona is among the nation's
    10 most endangered refuges,
    due in large part to illegal
    drug and immigrant traffic and
    Border Patrol operations, a
    conservation group said
    Friday." + ], + [ + "ROME, Oct 29 (AFP) - French
    President Jacques Chirac urged
    the head of the incoming
    European Commission Friday to
    take quot;the appropriate
    decisions quot; to resolve a
    row over his EU executive team
    which has left the EU in
    limbo." + ], + [ + "The Anglo-Dutch oil giant
    Shell today sought to draw a
    line under its reserves
    scandal by announcing plans to
    spend \\$15bn (8.4bn) a year to
    replenish reserves and develop
    production in its oil and gas
    business." + ], + [ + "SEPTEMBER 20, 2004
    (COMPUTERWORLD) - At
    PeopleSoft Inc. #39;s Connect
    2004 conference in San
    Francisco this week, the
    software vendor is expected to
    face questions from users
    about its ability to fend off
    Oracle Corp." + ], + [ + "Russia's parliament will
    launch an inquiry into a
    school siege that killed over
    300 hostages, President
    Vladimir Putin said on Friday,
    but analysts doubt it will
    satisfy those who blame the
    carnage on security services." + ], + [ + "Tony Blair talks to business
    leaders about new proposals
    for a major shake-up of the
    English exam system." + ], + [ + "KUALA LUMPUR, Malaysia A
    Malaysian woman has claimed a
    new world record after living
    with over six-thousand
    scorpions for 36 days
    straight." + ], + [ + "PBS's Charlie Rose quizzes Sun
    co-founder Bill Joy,
    Monster.com chief Jeff Taylor,
    and venture capitalist John
    Doerr." + ], + [ + "Circulation declined at most
    major US newspapers in the
    last half year, the latest
    blow for an industry already
    rocked by a scandal involving
    circulation misstatements that
    has undermined the confidence
    of investors and advertisers." + ], + [ + "Skype Technologies is teaming
    with Siemens to offer cordless
    phone users the ability to
    make Internet telephony calls,
    in addition to traditional
    calls, from their handsets." + ], + [ + "AP - After years of
    resistance, the U.S. trucking
    industry says it will not try
    to impede or delay a new
    federal rule aimed at cutting
    diesel pollution." + ], + [ + "INDIANAPOLIS - ATA Airlines
    has accepted a \\$117 million
    offer from Southwest Airlines
    that would forge close ties
    between two of the largest US
    discount carriers." + ], + [ + "could take drastic steps if
    the talks did not proceed as
    Tehran wants. Mehr news agency
    quoted him as saying on
    Wednesday. that could be used" + ], + [ + "Wizards coach Eddie Jordan
    says the team is making a
    statement that immaturity will
    not be tolerated by suspending
    Kwame Brown one game for not
    taking part in a team huddle
    during a loss to Denver." + ], + [ + "EVERETT Fire investigators
    are still trying to determine
    what caused a two-alarm that
    destroyed a portion of a South
    Everett shopping center this
    morning." + ], + [ + "Umesh Patel, a 36-year old
    software engineer from
    California, debated until the
    last minute." + ], + [ + "Sure, the PeopleSoft board
    told shareholders to just say
    no. This battle will go down
    to the wire, and even
    afterward Ellison could
    prevail." + ], + [ + "Investors cheered by falling
    oil prices and an improving
    job picture sent stocks higher
    Tuesday, hoping that the news
    signals a renewal of economic
    strength and a fall rally in
    stocks." + ], + [ + "AP - Andy Roddick has yet to
    face a challenge in his U.S.
    Open title defense. He beat
    No. 18 Tommy Robredo of Spain
    6-3, 6-2, 6-4 Tuesday night to
    move into the quarterfinals
    without having lost a set." + ], + [ + "Now that hell froze over in
    Boston, New England braces for
    its rarest season -- a winter
    of content. Red Sox fans are
    adrift from the familiar
    torture of the past." + ], + [ + " CHICAGO (Reuters) - Wal-Mart
    Stores Inc. <A HREF=\"http:/
    /www.investor.reuters.com/Full
    Quote.aspx?ticker=WMT.N target
    =/stocks/quickinfo/fullquote\"&
    gt;WMT.N</A>, the
    world's largest retailer, said
    on Saturday it still
    anticipates September U.S.
    sales to be up 2 percent to 4
    percent at stores open at
    least a year." + ], + [ + " TOKYO (Reuters) - Tokyo's
    Nikkei stock average opened
    down 0.15 percent on
    Wednesday as investors took a
    breather from the market's
    recent rises and sold shares
    of gainers such as Sharp
    Corp." + ], + [ + "AP - From now until the start
    of winter, male tarantulas are
    roaming around, searching for
    female mates, an ideal time to
    find out where the spiders
    flourish in Arkansas." + ], + [ + "Israeli authorities have
    launched an investigation into
    death threats against Israeli
    Prime Minister Ariel Sharon
    and other officials supporting
    his disengagement plan from
    Gaza and parts of the West
    Bank, Jerusalem police said
    Tuesday." + ], + [ + "NewsFactor - While a U.S.
    District Court continues to
    weigh the legality of
    \\Oracle's (Nasdaq: ORCL)
    attempted takeover of
    \\PeopleSoft (Nasdaq: PSFT),
    Oracle has taken the necessary
    steps to ensure the offer does
    not die on the vine with
    PeopleSoft's shareholders." + ], + [ + "NEW YORK : World oil prices
    fell, capping a drop of more
    than 14 percent in a two-and-
    a-half-week slide triggered by
    a perception of growing US
    crude oil inventories." + ], + [ + "SUFFOLK -- Virginia Tech
    scientists are preparing to
    protect the state #39;s
    largest crop from a disease
    with strong potential to do
    damage." + ], + [ + "It was a carefully scripted
    moment when Russian President
    Vladimir Putin began quoting
    Taras Shevchenko, this country
    #39;s 19th-century bard,
    during a live television" + ], + [ + "China says it welcomes Russia
    #39;s ratification of the
    Kyoto Protocol that aims to
    stem global warming by
    reducing greenhouse-gas
    emissions." + ], + [ + "Analysts said the smartphone
    enhancements hold the
    potential to bring PalmSource
    into an expanding market that
    still has room despite early
    inroads by Symbian and
    Microsoft." + ], + [ + "The Metropolitan
    Transportation Authority is
    proposing a tax increase to
    raise \\$900 million a year to
    help pay for a five-year
    rebuilding program." + ], + [ + "AFP - The Canadian armed
    forces chief of staff on was
    elected to take over as head
    of NATO's Military Committee,
    the alliance's highest
    military authority, military
    and diplomatic sources said." + ], + [ + "AFP - Two proposed resolutions
    condemning widespread rights
    abuses in Sudan and Zimbabwe
    failed to pass a UN committee,
    mired in debate between
    African and western nations." + ], + [ + " NEW YORK (Reuters) -
    Citigroup Inc. <A HREF=\"htt
    p://www.investor.reuters.com/F
    ullQuote.aspx?ticker=C.N targe
    t=/stocks/quickinfo/fullquote\"
    >C.N</A> the world's
    largest financial services
    company, said on Tuesday it
    will acquire First American
    Bank in the quickly-growing
    Texas market." + ], + [ + " SINGAPORE (Reuters) - U.S.
    oil prices hovered just below
    \\$50 a barrel on Tuesday,
    holding recent gains on a rash
    of crude supply outages and
    fears over thin heating oil
    tanks." + ], + [ + "THE South Africans have called
    the Wallabies scrum cheats as
    a fresh round of verbal
    warfare opened in the Republic
    last night." + ], + [ + "The return of noted reformer
    Nabil Amr to Palestinian
    politics comes at a crucial
    juncture." + ], + [ + "An overwhelming majority of
    NHL players who expressed
    their opinion in a poll said
    they would not support a
    salary cap even if it meant
    saving a season that was
    supposed to have started Oct.
    13." + ], + [ + "Tony Eury Sr. oversees Dale
    Earnhardt Jr. on the
    racetrack, but Sunday he
    extended his domain to Victory
    Lane. Earnhardt was unbuckling
    to crawl out of the No." + ], + [ + "AP - After two debates, voters
    have seen President Bush look
    peevish and heard him pass the
    buck. They've watched Sen.
    John Kerry deny he's a flip-
    flopper and then argue that
    Saddam Hussein was a threat
    #151; and wasn't. It's no
    wonder so few minds have
    changed." + ], + [ + "(Sports Network) - The New
    York Yankees try to move one
    step closer to a division
    title when they conclude their
    critical series with the
    Boston Red Sox at Fenway Park." + ], + [ + "US retail sales fell 0.3 in
    August as rising energy costs
    and bad weather persuaded
    shoppers to reduce their
    spending." + ], + [ + "The United States says the
    Lebanese parliament #39;s
    decision Friday to extend the
    term of pro-Syrian President
    Emile Lahoud made a
    quot;crude mockery of
    democratic principles." + ], + [ + "Kurt Busch claimed a stake in
    the points lead in the NASCAR
    Chase for the Nextel Cup
    yesterday, winning the
    Sylvania 300 at New Hampshire
    International Speedway." + ], + [ + "TOKYO (AFP) - Japan #39;s
    Fujitsu and Cisco Systems of
    the US said they have agreed
    to form a strategic alliance
    focusing on routers and
    switches that will enable
    businesses to build advanced
    Internet Protocol (IP)
    networks." + ], + [ + "GEORGETOWN, Del., Oct. 28 --
    Plaintiffs in a shareholder
    lawsuit over former Walt
    Disney Co. president Michael
    Ovitz's \\$140 million
    severance package attempted
    Thursday to portray Ovitz as a
    dishonest bumbler who botched
    the hiring of a major
    television executive and
    pushed the release of a movie
    that angered the Chinese
    government, damaging Disney's
    business prospects in the
    country." + ], + [ + "US stocks gained ground in
    early trading Thursday after
    tame inflation reports and
    better than expected jobless
    news. Oil prices held steady
    as Hurricane Ivan battered the
    Gulf coast, where oil
    operations have halted." + ], + [ + "It #39;s a new internet
    browser. The first full
    version, Firefox 1.0, was
    launched earlier this month.
    In the sense it #39;s a
    browser, yes, but the
    differences are larger than
    the similarities." + ], + [ + "LOUDON, NH - As this
    newfangled stretch drive for
    the Nextel Cup championship
    ensues, Jeff Gordon has to be
    considered the favorite for a
    fifth title." + ], + [ + "PepsiCo. Inc., the world #39;s
    No. 2 soft- drink maker, plans
    to buy General Mills Inc.
    #39;s stake in their European
    joint venture for \\$750
    million in cash, giving it
    complete ownership of Europe
    #39;s largest snack-food
    company." + ], + [ + "Microsoft will accelerate SP2
    distribution to meet goal of
    100 million downloads in two
    months." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks opened little changed
    on Friday, after third-
    quarter gross domestic product
    data showed the U.S. economy
    grew at a slower-than-expected
    pace." + ], + [ + "International Business
    Machines Corp., taken to court
    by workers over changes it
    made to its traditional
    pension plan, has decided to
    stop offering any such plan to
    new employees." + ], + [ + "NEW YORK - With four of its
    executives pleading guilty to
    price-fixing charges today,
    Infineon will have a hard time
    arguing that it didn #39;t fix
    prices in its ongoing
    litigation with Rambus." + ], + [ + "By nick giongco. YOU KNOW you
    have reached the status of a
    boxing star when ring
    announcer extraordinaire
    Michael Buffer calls out your
    name in his trademark booming
    voice during a high-profile
    event like yesterday" + ], + [ + "Canadian Press - OTTAWA (CP) -
    The prime minister says he's
    been assured by President
    George W. Bush that the U.S.
    missile defence plan does not
    necessarily involve the
    weaponization of space." + ], + [ + "AP - Even with a big lead,
    Eric Gagne wanted to pitch in
    front of his hometown fans one
    last time." + ], + [ + " NEW YORK (Reuters) - Limited
    Brands Inc. <A HREF=\"http:/
    /www.investor.reuters.com/Full
    Quote.aspx?ticker=LTD.N target
    =/stocks/quickinfo/fullquote\"&
    gt;LTD.N</A> on
    Thursday reported higher
    quarterly operating profit as
    cost controls and strong
    lingerie sales offset poor
    results at the retailer's
    Express apparel stores." + ], + [ + "General Motors and
    DaimlerChrysler are
    collaborating on development
    of fuel- saving hybrid engines
    in hopes of cashing in on an
    expanding market dominated by
    hybrid leaders Toyota and
    Honda." + ], + [ + "ATHENS, Greece -- Larry Brown
    was despondent, the head of
    the US selection committee was
    defensive and an irritated
    Allen Iverson was hanging up
    on callers who asked what went
    wrong." + ], + [ + "Fifty-six miners are dead and
    another 92 are still stranded
    underground after a gas blast
    at the Daping coal mine in
    Central China #39;s Henan
    Province, safety officials
    said Thursday." + ], + [ + "BEIRUT, Lebanon - Three
    Lebanese men and their Iraqi
    driver have been kidnapped in
    Iraq, the Lebanese Foreign
    Ministry said Sunday, as
    Iraq's prime minister said his
    government was working for the
    release of two Americans and a
    Briton also being held
    hostage. Gunmen snatched
    the three Lebanese, who worked
    for a travel agency with a
    branch in Baghdad, as they
    drove on the highway between
    the capital and Fallujah on
    Friday night, a ministry
    official said..." + ], + [ + "PORTLAND, Ore. - It's been
    almost a year since singer-
    songwriter Elliott Smith
    committed suicide, and fans
    and friends will be looking
    for answers as the posthumous
    \"From a Basement on the Hill\"
    is released..." + ], + [ + "AP - Courtney Brown refuses to
    surrender to injuries. He's
    already planning another
    comeback." + ], + [ + " GAZA (Reuters) - Israel
    expanded its military
    offensive in northern Gaza,
    launching two air strikes
    early on Monday that killed
    at least three Palestinians
    and wounded two, including a
    senior Hamas leader, witnesses
    and medics said." + ], + [ + " TOKYO (Reuters) - Nintendo
    Co. Ltd. raised its 2004
    shipment target for its DS
    handheld video game device by
    40 percent to 2.8 million
    units on Thursday after many
    stores in Japan and the
    United States sold out in the
    first week of sales." + ], + [ + "It took an off-the-cuff
    reference to a serial
    murderer/cannibal to punctuate
    the Robby Gordon storyline.
    Gordon has been vilified by
    his peers and put on probation" + ], + [ + "By BOBBY ROSS JR. Associated
    Press Writer. play the Atlanta
    Hawks. They will be treated to
    free food and drink and have.
    their pictures taken with
    Mavericks players, dancers and
    team officials." + ], + [ + "ARSENAL #39;S Brazilian World
    Cup winning midfielder
    Gilberto Silva is set to be
    out for at least a month with
    a back injury, the Premiership
    leaders said." + ], + [ + "INDIANAPOLIS -- With a package
    of academic reforms in place,
    the NCAA #39;s next crusade
    will address what its
    president calls a dangerous
    drift toward professionalism
    and sports entertainment." + ], + [ + "Autodesk this week unwrapped
    an updated version of its
    hosted project collaboration
    service targeted at the
    construction and manufacturing
    industries. Autodesk Buzzsaw
    lets multiple, dispersed
    project participants --
    including building owners,
    developers, architects,
    construction teams, and
    facility managers -- share and
    manage data throughout the
    life of a project, according
    to Autodesk officials." + ], + [ + "AP - It was difficult for
    Southern California's Pete
    Carroll and Oklahoma's Bob
    Stoops to keep from repeating
    each other when the two
    coaches met Thursday." + ], + [ + "Reuters - Sudan's government
    resumed\\talks with rebels in
    the oil-producing south on
    Thursday while\\the United
    Nations set up a panel to
    investigate charges
    of\\genocide in the west of
    Africa's largest country." + ], + [ + "Andy Roddick, along with
    Olympic silver medalist Mardy
    Fish and the doubles pair of
    twins Bob and Mike Bryan will
    make up the US team to compete
    with Belarus in the Davis Cup,
    reported CRIENGLISH." + ], + [ + "NEW YORK - Optimism that the
    embattled technology sector
    was due for a recovery sent
    stocks modestly higher Monday
    despite a new revenue warning
    from semiconductor company
    Broadcom Inc. While
    Broadcom, which makes chips
    for television set-top boxes
    and other electronics, said
    high inventories resulted in
    delayed shipments, investors
    were encouraged as it said
    future quarters looked
    brighter..." + ], + [ + "A report indicates that many
    giants of the industry have
    been able to capture lasting
    feelings of customer loyalty." + ], + [ + "It is the news that Internet
    users do not want to hear: the
    worldwide web is in danger of
    collapsing around us. Patrick
    Gelsinger, the chief
    technology officer for
    computer chip maker Intel,
    told a conference" + ], + [ + " WASHINGTON (Reuters) - U.S.
    employers hired just 96,000
    workers in September, the
    government said on Friday in a
    weak jobs snapshot, the final
    one ahead of presidential
    elections that also fueled
    speculation about a pause in
    interest-rate rises." + ], + [ + "Cisco Systems CEO John
    Chambers said today that his
    company plans to offer twice
    as many new products this year
    as ever before." + ], + [ + "While the world #39;s best
    athletes fight the noble
    Olympic battle in stadiums and
    pools, their fans storm the
    streets of Athens, turning the
    Greek capital into a huge
    international party every
    night." + ], + [ + "Carlos Barrios Orta squeezed
    himself into his rubber diving
    suit, pulled on an 18-pound
    helmet that made him look like
    an astronaut, then lowered
    himself into the sewer. He
    disappeared into the filthy
    water, which looked like some
    cauldron of rancid beef stew,
    until the only sign of him was
    air bubbles breaking the
    surface." + ], + [ + "The mutilated body of a woman
    of about sixty years,
    amputated of arms and legs and
    with the sliced throat, has
    been discovered this morning
    in a street of the south of
    Faluya by the Marines,
    according to a statement by
    AFP photographer." + ], + [ + "Every day, somewhere in the
    universe, there #39;s an
    explosion that puts the power
    of the Sun in the shade. Steve
    Connor investigates the riddle
    of gamma-ray bursts." + ], + [ + "Ten months after NASA #39;s
    twin rovers landed on Mars,
    scientists reported this week
    that both robotic vehicles are
    still navigating their rock-
    studded landscapes with all
    instruments operating" + ], + [ + "EVERTON showed they would not
    be bullied into selling Wayne
    Rooney last night by rejecting
    a 23.5million bid from
    Newcastle - as Manchester
    United gamble on Goodison
    #39;s resolve to keep the
    striker." + ], + [ + "SAN FRANCISCO (CBS.MW) -- The
    magazine known for evaluating
    cars and electronics is
    setting its sights on finding
    the best value and quality of
    prescription drugs on the
    market." + ], + [ + "After months of legal
    wrangling, the case of
    <em>People v. Kobe Bean
    Bryant</em> commences on
    Friday in Eagle, Colo., with
    testimony set to begin next
    month." + ], + [ + "(SH) - In Afghanistan, Hamid
    Karzai defeated a raft of
    candidates to win his historic
    election. In Iraq, more than
    200 political parties have
    registered for next month
    #39;s elections." + ], + [ + "Lyon coach Paul le Guen has
    admitted his side would be
    happy with a draw at Old
    Trafford on Tuesday night. The
    three-times French champions
    have assured themselves of
    qualification for the
    Champions League" + ], + [ + "British scientists say they
    have found a new, greener way
    to power cars and homes using
    sunflower oil, a commodity
    more commonly used for cooking
    fries." + ], + [ + "A mouse, a house, and your
    tax-planning spouse all factor
    huge in the week of earnings
    that lies ahead." + ], + [ + "The United States #39; top
    computer-security official has
    resigned after a little more
    than a year on the job, the US
    Department of Homeland
    Security said on Friday." + ], + [ + "Reuters - Barry Bonds failed
    to collect a hit in\\his bid to
    join the 700-homer club, but
    he did score a run to\\help the
    San Francisco Giants edge the
    host Milwaukee Brewers\\3-2 in
    National League action on
    Tuesday." + ], + [ + " SAN FRANCISCO (Reuters) -
    Intel Corp. <A HREF=\"http:/
    /www.reuters.co.uk/financeQuot
    eLookup.jhtml?ticker=INTC.O
    qtype=sym infotype=info
    qcat=news\">INTC.O</A>
    on Thursday outlined its
    vision of the Internet of the
    future, one in which millions
    of computer servers would
    analyze and direct network
    traffic to make the Web safer
    and more efficient." + ], + [ + "Margaret Hassan, who works for
    charity Care International,
    was taken hostage while on her
    way to work in Baghdad. Here
    are the main events since her
    kidnapping." + ], + [ + "Check out new gadgets as
    holiday season approaches." + ], + [ + "DALLAS (CBS.MW) -- Royal
    Dutch/Shell Group will pay a
    \\$120 million penalty to
    settle a Securities and
    Exchange Commission
    investigation of its
    overstatement of nearly 4.5
    billion barrels of proven
    reserves, the federal agency
    said Tuesday." + ], + [ + "After waiting an entire summer
    for the snow to fall and
    Beaver Creek to finally open,
    skiers from around the planet
    are coming to check out the
    Birds of Prey World Cup action
    Dec. 1 - 5. Although everyones" + ], + [ + "TOKYO, Sep 08, 2004 (AFX-UK
    via COMTEX) -- Sony Corp will
    launch its popular projection
    televisions with large liquid
    crystal display (LCD) screens
    in China early next year, a
    company spokeswoman said." + ], + [ + "I confess that I am a complete
    ignoramus when it comes to
    women #39;s beach volleyball.
    In fact, I know only one thing
    about the sport - that it is
    the supreme aesthetic
    experience available on planet
    Earth." + ], + [ + "Manchester United Plc may
    offer US billionaire Malcolm
    Glazer a seat on its board if
    he agrees to drop a takeover
    bid for a year, the Observer
    said, citing an unidentified
    person in the soccer industry." + ], + [ + "PHOENIX America West Airlines
    has backed away from a
    potential bidding war for
    bankrupt ATA Airlines, paving
    the way for AirTran to take
    over ATA operations." + ], + [ + "US stock-index futures
    declined. Dow Jones Industrial
    Average shares including
    General Electric Co. slipped
    in Europe. Citigroup Inc." + ], + [ + "For a guy who spent most of
    his first four professional
    seasons on the disabled list,
    Houston Astros reliever Brad
    Lidge has developed into quite
    the ironman these past two
    days." + ], + [ + "AFP - Outgoing EU competition
    commissioner Mario Monti wants
    to reach a decision on
    Oracle's proposed takeover of
    business software rival
    PeopleSoft by the end of next
    month, an official said." + ], + [ + "Former Bengal Corey Dillon
    found his stride Sunday in his
    second game for the New
    England Patriots. Dillon
    gained 158 yards on 32 carries
    as the Patriots beat the
    Arizona Cardinals, 23-12, for
    their 17th victory in a row." + ], + [ + "If legend is to be believed,
    the end of a victorious war
    was behind Pheidippides #39;
    trek from Marathon to Athens
    2,500 years ago." + ], + [ + " WASHINGTON (Reuters) -
    Satellite companies would be
    able to retransmit
    broadcasters' television
    signals for another five
    years but would have to offer
    those signals on a single
    dish, under legislation
    approved by Congress on
    Saturday." + ], + [ + " BEIJING (Reuters) - Wimbledon
    champion Maria Sharapova
    demolished fellow Russian
    Tatiana Panova 6-1, 6-1 to
    advance to the quarter-finals
    of the China Open on
    Wednesday." + ], + [ + "Sven Jaschan, who may face
    five years in prison for
    spreading the Netsky and
    Sasser worms, is now working
    in IT security. Photo: AFP." + ], + [ + "Padres general manager Kevin
    Towers called Expos general
    manager Omar Minaya on
    Thursday afternoon and told
    him he needed a shortstop
    because Khalil Greene had
    broken his" + ], + [ + "Motorsport.com. Nine of the
    ten Formula One teams have
    united to propose cost-cutting
    measures for the future, with
    the notable exception of
    Ferrari." + ], + [ + "Reuters - The United States
    modified its\\call for U.N.
    sanctions against Sudan on
    Tuesday but still kept\\up the
    threat of punitive measures if
    Khartoum did not
    stop\\atrocities in its Darfur
    region." + ], + [ + "Human rights and environmental
    activists have hailed the
    award of the 2004 Nobel Peace
    Prize to Wangari Maathai of
    Kenya as fitting recognition
    of the growing role of civil
    society" + ], + [ + "Federal Reserve Chairman Alan
    Greenspan has done it again.
    For at least the fourth time
    this year, he has touched the
    electrified third rail of
    American politics - Social
    Security." + ], + [ + "An Al Qaeda-linked militant
    group beheaded an American
    hostage in Iraq and threatened
    last night to kill another two
    Westerners in 24 hours unless
    women prisoners were freed
    from Iraqi jails." + ], + [ + "TechWeb - An Indian Institute
    of Technology professor--and
    open-source evangelist--
    discusses the role of Linux
    and open source in India." + ], + [ + "Here are some of the latest
    health and medical news
    developments, compiled by
    editors of HealthDay: -----
    Contaminated Fish in Many U.S.
    Lakes and Rivers Fish
    that may be contaminated with
    dioxin, mercury, PCBs and
    pesticides are swimming in
    more than one-third of the
    United States' lakes and
    nearly one-quarter of its
    rivers, according to a list of
    advisories released by the
    Environmental Protection
    Agency..." + ], + [ + "Argentina, Denmark, Greece,
    Japan and Tanzania on Friday
    won coveted two-year terms on
    the UN Security Council at a
    time when pressure is mounting
    to expand the powerful
    15-nation body." + ], + [ + "Description: Scientists say
    the arthritis drug Bextra may
    pose increased risk of
    cardiovascular troubles.
    Bextra is related to Vioxx,
    which was pulled off the
    market in September for the
    same reason." + ], + [ + "Steve Francis and Shaquille O
    #39;Neal enjoyed big debuts
    with their new teams. Kobe
    Bryant found out he can #39;t
    carry the Lakers all by
    himself." + ], + [ + "The trial of a lawsuit by Walt
    Disney Co. shareholders who
    accuse the board of directors
    of rubberstamping a deal to
    hire Michael Ovitz" + ], + [ + "Australia, by winning the
    third Test at Nagpur on
    Friday, also won the four-
    match series 2-0 with one
    match to go. Australia had
    last won a Test series in
    India way back in December
    1969 when Bill Lawry #39;s
    team beat Nawab Pataudi #39;s
    Indian team 3-1." + ], + [ + "LOS ANGELES Grocery giant
    Albertsons says it has
    purchased Bristol Farms, which
    operates eleven upscale stores
    in Southern California." + ], + [ + "ISLAMABAD: Pakistan and India
    agreed on Friday to re-open
    the Khokhropar-Munabao railway
    link, which was severed nearly
    40 years ago." + ], + [ + "Final Score: Connecticut 61,
    New York 51 New York, NY
    (Sports Network) - Nykesha
    Sales scored 15 points to lead
    Connecticut to a 61-51 win
    over New York in Game 1 of
    their best-of-three Eastern
    Conference Finals series at
    Madison Square Garden." + ], + [ + "NEW YORK - Stocks moved higher
    Friday as a stronger than
    expected retail sales report
    showed that higher oil prices
    aren't scaring consumers away
    from spending. Federal Reserve
    Chairman Alan Greenspan's
    positive comments on oil
    prices also encouraged
    investors..." + ], + [ + "The figures from a survey
    released today are likely to
    throw more people into the
    ranks of the uninsured,
    analysts said." + ], + [ + "Bill Gates is giving his big
    speech right now at Microsofts
    big Digital Entertainment
    Anywhere event in Los Angeles.
    Well have a proper report for
    you soon, but in the meantime
    we figured wed" + ], + [ + "Spinal and non-spinal
    fractures are reduced by
    almost a third in women age 80
    or older who take a drug
    called strontium ranelate,
    European investigators
    announced" + ], + [ + "JB Oxford Holdings Inc., a
    Beverly Hills-based discount
    brokerage firm, was sued by
    the Securities and Exchange
    Commission for allegedly
    allowing thousands of improper
    trades in more than 600 mutual
    funds." + ], + [ + "BERLIN: Apple Computer Inc is
    planning the next wave of
    expansion for its popular
    iTunes online music store with
    a multi-country European
    launch in October, the
    services chief architect said
    on Wednesday." + ], + [ + "Charlotte, NC -- LeBron James
    poured in a game-high 19
    points and Jeff McInnis scored
    18 as the Cleveland Cavaliers
    routed the Charlotte Bobcats,
    106-89, at the Charlotte
    Coliseum." + ], + [ + "Goldman Sachs reported strong
    fourth quarter and full year
    earnings of \\$1.19bn, up 36
    per cent on the previous
    quarter. Full year profit rose
    52 per cent from the previous
    year to \\$4.55bn." + ], + [ + "Saddam Hussein lives in an
    air-conditioned 10-by-13 foot
    cell on the grounds of one of
    his former palaces, tending
    plants and proclaiming himself
    Iraq's lawful ruler." + ], + [ + "Lehmann, who was at fault in
    two matches in the tournament
    last season, was blundering
    again with the German set to
    take the rap for both Greek
    goals." + ], + [ + "Calls to 13 other countries
    will be blocked to thwart
    auto-dialer software." + ], + [ + "AP - Brazilian U.N.
    peacekeepers will remain in
    Haiti until presidential
    elections are held in that
    Caribbean nation sometime next
    year, President Luiz Inacio
    Lula da Silva said Monday." + ], + [ + "com September 27, 2004, 5:00
    AM PT. This fourth priority
    #39;s main focus has been
    improving or obtaining CRM and
    ERP software for the past year
    and a half." + ], + [ + "Jeju Island, South Korea
    (Sports Network) - Grace Park
    and Carin Koch posted matching
    rounds of six-under-par 66 on
    Friday to share the lead after
    the first round of the CJ Nine
    Bridges Classic." + ], + [ + "SPACE.com - The Zero Gravity
    Corporation \\ has been given
    the thumbs up by the Federal
    Aviation Administration (FAA)
    to \\ conduct quot;weightless
    flights quot; for the general
    public, providing the \\
    sensation of floating in
    space." + ], + [ + "A 32-year-old woman in Belgium
    has become the first woman
    ever to give birth after
    having ovarian tissue removed,
    frozen and then implanted back
    in her body." + ], + [ + "Argosy Gaming (AGY:NYSE - news
    - research) jumped in early
    trading Thursday, after the
    company agreed to be acquired
    by Penn National Gaming
    (PENN:Nasdaq - news -
    research) in a \\$1." + ], + [ + "No sweat, says the new CEO,
    who's imported from IBM. He
    also knows this will be the
    challenge of a lifetime." + ], + [ + "Iraq is quot;working 24 hours
    a day to ... stop the
    terrorists, quot; interim
    Iraqi Prime Minister Ayad
    Allawi said Monday. Iraqis are
    pushing ahead with reforms and
    improvements, Allawi told" + ], + [ + "AP - Their discoveries may be
    hard for many to comprehend,
    but this year's Nobel Prize
    winners still have to explain
    what they did and how they did
    it." + ], + [ + "The DuPont Co. has agreed to
    pay up to \\$340 million to
    settle a lawsuit that it
    contaminated water supplies in
    West Virginia and Ohio with a
    chemical used to make Teflon,
    one of its best-known brands." + ], + [ + "Benfica and Real Madrid set
    the standard for soccer
    success in Europe in the late
    1950s and early #39;60s. The
    clubs have evolved much
    differently, but both have
    struggled" + ], + [ + "AP - Musicians, composers and
    authors were among the more
    than two dozen people
    Wednesday honored with
    National Medal of Arts and
    National Humanities awards at
    the White House." + ], + [ + "More than 15 million homes in
    the UK will be able to get on-
    demand movies by 2008, say
    analysts." + ], + [ + "Wynton Marsalis, the trumpet-
    playing star and artistic
    director of Jazz at Lincoln
    Center, has become an entity
    above and around the daily
    jazz world." + ], + [ + "A well-researched study by the
    National Academy of Sciences
    makes a persuasive case for
    refurbishing the Hubble Space
    Telescope. NASA, which is
    reluctant to take this
    mission, should rethink its
    position." + ], + [ + "BUENOS AIRES: Pakistan has
    ratified the Kyoto Protocol on
    Climatic Change, Environment
    Minister Malik Khan said on
    Thursday. He said that
    Islamabad has notified UN
    authorities of ratification,
    which formally comes into
    effect in February 2005." + ], + [ + "Reuters - Jeffrey Greenberg,
    chairman and chief\\executive
    of embattled insurance broker
    Marsh McLennan Cos.\\, is
    expected to step down within
    hours, a newspaper\\reported on
    Friday, citing people close to
    the discussions." + ], + [ + "Staff Sgt. Johnny Horne, Jr.,
    and Staff Sgt. Cardenas Alban
    have been charged with murder
    in the death of an Iraqi, the
    1st Cavalry Division announced
    Monday." + ], + [ + "LONDON -- In a deal that
    appears to buck the growing
    trend among governments to
    adopt open-source
    alternatives, the U.K.'s
    Office of Government Commerce
    (OGC) is negotiating a renewal
    of a three-year agreement with
    Microsoft Corp." + ], + [ + "WASHINGTON - A Pennsylvania
    law requiring Internet service
    providers (ISPs) to block Web
    sites the state's prosecuting
    attorneys deem to be child
    pornography has been reversed
    by a U.S. federal court, with
    the judge ruling the law
    violated free speech rights." + ], + [ + "The Microsoft Corp. chairman
    receives four million e-mails
    a day, but practically an
    entire department at the
    company he founded is
    dedicated to ensuring that
    nothing unwanted gets into his
    inbox, the company #39;s chief
    executive said Thursday." + ], + [ + "The 7100t has a mobile phone,
    e-mail, instant messaging, Web
    browsing and functions as an
    organiser. The device looks
    like a mobile phone and has
    the features of the other
    BlackBerry models, with a
    large screen" + ], + [ + "President Vladimir Putin makes
    a speech as he hosts Russia
    #39;s Olympic athletes at a
    Kremlin banquet in Moscow,
    Thursday, Nov. 4, 2004." + ], + [ + "Apple Computer has unveiled
    two new versions of its hugely
    successful iPod: the iPod
    Photo and the U2 iPod. Apple
    also has expanded" + ], + [ + "Pakistan are closing in fast
    on Sri Lanka #39;s first
    innings total after impressing
    with both ball and bat on the
    second day of the opening Test
    in Faisalabad." + ], + [ + "A state regulatory board
    yesterday handed a five-year
    suspension to a Lawrence
    funeral director accused of
    unprofessional conduct and
    deceptive practices, including
    one case where he refused to
    complete funeral arrangements
    for a client because she had
    purchased a lower-priced
    casket elsewhere." + ], + [ + "AP - This year's hurricanes
    spread citrus canker to at
    least 11,000 trees in
    Charlotte County, one of the
    largest outbreaks of the
    fruit-damaging infection to
    ever affect Florida's citrus
    industry, state officials
    said." + ], + [ + "AFP - Hundreds of Buddhists in
    southern Russia marched
    through snow to see and hear
    the Dalai Lama as he continued
    a long-awaited visit to the
    country in spite of Chinese
    protests." + ], + [ + "British officials were on
    diplomatic tenterhooks as they
    awaited the arrival on
    Thursday of French President
    Jacques Chirac for a two-day
    state visit to Britain." + ], + [ + " SYDNEY (Reuters) - A group of
    women on Pitcairn Island in
    the South Pacific are standing
    by their men, who face
    underage sex charges, saying
    having sex at age 12 is a
    tradition dating back to 18th
    century mutineers who settled
    on the island." + ], + [ + "Two aircraft are flying out
    from the UK on Sunday to
    deliver vital aid and supplies
    to Haiti, which has been
    devastated by tropical storm
    Jeanne." + ], + [ + "A scare triggered by a
    vibrating sex toy shut down a
    major Australian regional
    airport for almost an hour
    Monday, police said. The
    vibrating object was
    discovered Monday morning" + ], + [ + "IBM moved back into the iSCSI
    (Internet SCSI) market Friday
    with a new array priced at
    US\\$3,000 and aimed at the
    small and midsize business
    market." + ], + [ + "Ron Artest has been hit with a
    season long suspension,
    unprecedented for the NBA
    outside doping cases; Stephen
    Jackson banned for 30 games;
    Jermaine O #39;Neal for 25
    games and Anthony Johnson for
    five." + ], + [ + "The California Public
    Utilities Commission on
    Thursday upheld a \\$12.1
    million fine against Cingular
    Wireless, related to a two-
    year investigation into the
    cellular telephone company
    #39;s business practices." + ], + [ + "Barclays, the British bank
    that left South Africa in 1986
    after apartheid protests, may
    soon resume retail operations
    in the nation." + ], + [ + "AP - An Indiana congressional
    candidate abruptly walked off
    the set of a debate because
    she got stage fright." + ], + [ + "Italian-based Parmalat is
    suing its former auditors --
    Grant Thornton International
    and Deloitte Touche Tohmatsu
    -- for billions of dollars in
    damages. Parmalat blames its
    demise on the two companies
    #39; mismanagement of its
    finances." + ], + [ + "Reuters - Having reached out
    to Kashmiris during a two-day
    visit to the region, the prime
    minister heads this weekend to
    the country's volatile
    northeast, where public anger
    is high over alleged abuses by
    Indian soldiers." + ], + [ + "SAN JUAN IXTAYOPAN, Mexico --
    A pair of wooden crosses
    outside the elementary school
    are all that mark the macabre
    site where, just weeks ago, an
    angry mob captured two federal
    police officers, beat them
    unconscious, and set them on
    fire." + ], + [ + "Three shots behind Grace Park
    with five holes to go, six-
    time LPGA player of the year
    Sorenstam rallied with an
    eagle, birdie and three pars
    to win her fourth Samsung
    World Championship by three
    shots over Park on Sunday." + ], + [ + "update An alliance of
    technology workers on Tuesday
    accused conglomerate Honeywell
    International of planning to
    move thousands of jobs to low-
    cost regions over the next
    five years--a charge that
    Honeywell denies." + ], + [ + "NSW Rugby CEO Fraser Neill
    believes Waratah star Mat
    Rogers has learned his lesson
    after he was fined and ordered
    to do community service
    following his controversial
    comments about the club rugby
    competition." + ], + [ + "American Technology Research
    analyst Shaw Wu has initiated
    coverage of Apple Computer
    (AAPL) with a #39;buy #39;
    recommendation, and a 12-month
    target of US\\$78 per share." + ], + [ + "washingtonpost.com - Cell
    phone shoppers looking for new
    deals and features didn't find
    them yesterday, a day after
    Sprint Corp. and Nextel
    Communications Inc. announced
    a merger that the phone
    companies promised would shake
    up the wireless business." + ], + [ + "ATHENS: China, the dominant
    force in world diving for the
    best part of 20 years, won six
    out of eight Olympic titles in
    Athens and prompted
    speculation about a clean
    sweep when they stage the
    Games in Beijing in 2008." + ], + [ + " NEW YORK (Reuters) - Northrop
    Grumman Corp. <A HREF=\"http
    ://www.investor.reuters.com/Fu
    llQuote.aspx?ticker=NOC.N targ
    et=/stocks/quickinfo/fullquote
    \">NOC.N</A> reported
    higher third-quarter earnings
    on Wednesday and an 11
    percent increase in sales on
    strength in its mission
    systems, integrated systems,
    ships and space technology
    businesses." + ], + [ + "JAPANESE GIANT Sharp has
    pulled the plug on its Linux-
    based PDAs in the United
    States as no-one seems to want
    them. The company said that it
    will continue to sell them in
    Japan where they sell like hot
    cakes" + ], + [ + "DUBLIN : An Olympic Airlines
    plane diverted to Ireland
    following a bomb alert, the
    second faced by the Greek
    carrier in four days, resumed
    its journey to New York after
    no device was found on board,
    airport officials said." + ], + [ + "New NavOne offers handy PDA
    and Pocket PC connectivity,
    but fails to impress on
    everything else." + ], + [ + "TORONTO (CP) - Shares of
    Iamgold fell more than 10 per
    cent after its proposed merger
    with Gold Fields Ltd. was
    thrown into doubt Monday as
    South Africa #39;s Harmony
    Gold Mining Company Ltd." + ], + [ + "The Marvel deal calls for
    Mforma to work with the comic
    book giant to develop a
    variety of mobile applications
    based on Marvel content." + ], + [ + " LONDON (Reuters) - Oil prices
    tumbled again on Monday to an
    8-week low under \\$46 a
    barrel, as growing fuel stocks
    in the United States eased
    fears of a winter supply
    crunch." + ], + [ + " WASHINGTON (Reuters) - The
    United States said on Friday
    it is preparing a new U.N.
    resolution on Darfur and that
    Secretary of State Colin
    Powell might address next week
    whether the violence in
    western Sudan constitutes
    genocide." + ], + [ + "The Houston Astros won their
    19th straight game at home and
    are one game from winning
    their first playoff series in
    42 years." + ], + [ + "washingtonpost.com - The
    Portable Media Center -- a
    new, Microsoft-conceived
    handheld device that presents
    video and photos as well as
    music -- would be a decent
    idea if there weren't such a
    thing as lampposts. Or street
    signs. Or trees. Or other
    cars." + ], + [ + "Alcoa Inc., one of the world
    #39;s top producers of
    aluminum, said Monday that it
    received an unsolicited
    quot;mini-tender quot; offer
    from Toronto-based TRC Capital
    Corp." + ], + [ + "The European Commission is to
    warn Greece about publishing
    false information about its
    public finances." + ], + [ + "The Air Force Reserve #39;s
    Hurricane Hunters, those
    fearless crews who dive into
    the eyewalls of hurricanes to
    relay critical data on
    tropical systems, were chased
    from their base on the
    Mississippi Gulf Coast by
    Hurricane Ivan." + ], + [ + " LONDON (Reuters) - U.S.
    shares were expected to open
    lower on Wednesday after
    crude oil pushed to a fresh
    high overnight, while Web
    search engine Google Inc.
    dented sentiment as it
    slashed the price range on its
    initial public offering." + ], + [ + "The eighth-seeded American
    fell to sixth-seeded Elena
    Dementieva of Russia, 0-6 6-2
    7-6 (7-5), on Friday - despite
    being up a break on four
    occasions in the third set." + ], + [ + "TUCSON, Arizona (Ticker) --
    No. 20 Arizona State tries to
    post its first three-game
    winning streak over Pac-10
    Conference rival Arizona in 26
    years when they meet Friday." + ], + [ + "NAJAF, Iraq : Iraq #39;s top
    Shiite Muslim clerics, back in
    control of Najaf #39;s Imam
    Ali shrine after a four-month
    militia occupation, met amid
    the ruins of the city as life
    spluttered back to normality." + ], + [ + "Embargo or not, Fidel Castro's
    socialist paradise has quietly
    become a pharmaceutical
    powerhouse. (They're still
    working on the capitalism
    thing.) By Douglas Starr from
    Wired magazine." + ], + [ + "AP - Phillip Fulmer kept his
    cool when starting center
    Jason Respert drove off in the
    coach's golf cart at practice." + ], + [ + "GOALS from Wayne Rooney and
    Ruud van Nistelrooy gave
    Manchester United the win at
    Newcastle. Alan Shearer
    briefly levelled matters for
    the Magpies but United managed
    to scrape through." + ], + [ + "The telemarketer at the other
    end of Orlando Castelblanco
    #39;s line promised to reduce
    the consumer #39;s credit card
    debt by at least \\$2,500 and
    get his 20 percent -- and
    growing -- interest rates down
    to single digits." + ], + [ + " NEW YORK (Reuters) - Blue-
    chip stocks fell slightly on
    Monday after No. 1 retailer
    Wal-Mart Stores Inc. <A HRE
    F=\"http://www.investor.reuters
    .com/FullQuote.aspx?ticker=WMT
    .N target=/stocks/quickinfo/fu
    llquote\">WMT.N</A>
    reported lower-than-expected
    Thanksgiving sales, while
    technology shares were lifted
    by a rally in Apple Computer
    Inc. <A HREF=\"http://www.i
    nvestor.reuters.com/FullQuote.
    aspx?ticker=AAPL.O target=/sto
    cks/quickinfo/fullquote\">AA
    PL.O</A>." + ], + [ + "Reuters - House of
    Representatives
    Majority\\Leader Tom DeLay,
    admonished twice in six days
    by his chamber's\\ethics
    committee, withstood calls on
    Thursday by rival\\Democrats
    and citizen groups that he
    step aside." + ], + [ + "WPP Group Inc., the world
    #39;s second- largest
    marketing and advertising
    company, said it won the
    bidding for Grey Global Group
    Inc." + ], + [ + "AFP - Australia has accounted
    for all its nationals known to
    be working in Iraq following a
    claim by a radical Islamic
    group to have kidnapped two
    Australians, Foreign Affairs
    Minister Alexander Downer
    said." + ], + [ + "A new worm can spy on users by
    hijacking their Web cameras, a
    security firm warned Monday.
    The Rbot.gr worm -- the latest
    in a long line of similar
    worms; one security firm
    estimates that more than 4,000
    variations" + ], + [ + " NEW YORK (Reuters) - Stocks
    were slightly lower on
    Tuesday, as concerns about
    higher oil prices cutting into
    corporate profits and
    consumer demand weighed on
    sentiment, while retail sales
    posted a larger-than-expected
    decline in August." + ], + [ + "The separation of PalmOne and
    PalmSource will be complete
    with Eric Benhamou's
    resignation as the latter's
    chairman." + ], + [ + "AP - Two-time U.S. Open
    doubles champion Max Mirnyi
    will lead the Belarus team
    that faces the United States
    in the Davis Cup semifinals in
    Charleston later this month." + ], + [ + "The hurricane appeared to be
    strengthening and forecasters
    warned of an increased threat
    of serious flooding and wind
    damage." + ], + [ + "AP - New York Jets safety Erik
    Coleman got his souvenir
    football from the equipment
    manager and held it tightly." + ], + [ + " SINGAPORE (Reuters) - Oil
    prices climbed above \\$42 a
    barrel on Wednesday, rising
    for the third day in a row as
    the heavy consuming U.S.
    Northeast feels the first
    chills of winter." + ], + [ + "\\\\\"(CNN) -- A longtime
    associate of al Qaeda leader
    Osama bin Laden surrendered
    to\\Saudi Arabian officials
    Tuesday, a Saudi Interior
    Ministry official said.\"\\\\\"But
    it is unclear what role, if
    any, Khaled al-Harbi may have
    had in any terror\\attacks
    because no public charges have
    been filed against him.\"\\\\\"The
    Saudi government -- in a
    statement released by its
    embassy in Washington
    --\\called al-Harbi's surrender
    \"the latest direct result\" of
    its limited, one-month\\offer
    of leniency to terror
    suspects.\"\\\\This is great! I
    hope this really starts to pay
    off. Creative solutions
    to\\terrorism that don't
    involve violence. \\\\How
    refreshing! \\\\Are you paying
    attention Bush
    administration?\\\\" + ], + [ + " NEW YORK (Reuters) - Merck
    Co Inc. <A HREF=\"http://www
    .investor.reuters.com/FullQuot
    e.aspx?ticker=MRK.N target=/st
    ocks/quickinfo/fullquote\">M
    RK.N</A> on Thursday
    pulled its arthritis drug
    Vioxx off the market after a
    study showed it doubled the
    risk of heart attack and
    stroke, a move that sent its
    shares plunging and erased
    \\$25 billion from its market
    value." + ], + [ + "AT T Corp. is cutting 7,400
    more jobs and slashing the
    book value of its assets by
    \\$11.4 billion, drastic moves
    prompted by the company's plan
    to retreat from the
    traditional consumer telephone
    business following a lost
    court battle." + ], + [ + "The government on Wednesday
    defended its decision to
    radically revise the country
    #39;s deficit figures, ahead
    of a European Commission
    meeting to consider possible
    disciplinary action against
    Greece for submitting faulty
    figures." + ], + [ + "Ivan Hlinka coached the Czech
    Republic to the hockey gold
    medal at the 1998 Nagano
    Olympics and became the coach
    of the Pittsburgh Penguins two
    years later." + ], + [ + "Four homeless men were
    bludgeoned to death and six
    were in critical condition on
    Friday following early morning
    attacks by unknown assailants
    in downtown streets of Sao
    Paulo, a police spokesman
    said." + ], + [ + "OPEC ministers yesterday
    agreed to increase their
    ceiling for oil production to
    help bring down stubbornly
    high prices in a decision that
    traders and analysts dismissed
    as symbolic because the cartel
    already is pumping more than
    its new target." + ], + [ + "Williams-Sonoma Inc. said
    second- quarter profit rose 55
    percent, boosted by the
    addition of Pottery Barn
    stores and sale of outdoor
    furniture." + ], + [ + "Celerons form the basis of
    Intel #39;s entry-level
    platform which includes
    integrated/value motherboards
    as well. The Celeron 335D is
    the fastest - until the
    Celeron 340D - 2.93GHz -
    becomes mainstream - of the" + ], + [ + "The entertainment industry
    asks the Supreme Court to
    reverse the Grokster decision,
    which held that peer-to-peer
    networks are not liable for
    copyright abuses of their
    users. By Michael Grebb." + ], + [ + " HONG KONG/SAN FRANCISCO
    (Reuters) - China's largest
    personal computer maker,
    Lenovo Group Ltd., said on
    Tuesday it was in acquisition
    talks with a major technology
    company, which a source
    familiar with the situation
    said was IBM." + ], + [ + "Phone companies are not doing
    enough to warn customers about
    internet \"rogue-dialling\"
    scams, watchdog Icstis warns." + ], + [ + "Reuters - Oil prices stayed
    close to #36;49 a\\barrel on
    Thursday, supported by a
    forecast for an early
    cold\\snap in the United States
    that could put a strain on a
    thin\\supply cushion of winter
    heating fuel." + ], + [ + "AUBURN - Ah, easy street. No
    game this week. Light
    practices. And now Auburn is
    being touted as the No. 3 team
    in the Bowl Championship
    Series standings." + ], + [ + "Portsmouth #39;s Harry
    Redknapp has been named as the
    Barclays manager of the month
    for October. Redknapp #39;s
    side were unbeaten during the
    month and maintained an
    impressive climb to ninth in
    the Premiership." + ], + [ + "India's main opposition party
    takes action against senior
    party member Uma Bharti after
    a public row." + ], + [ + "Hewlett-Packard will shell out
    \\$16.1 billion for chips in
    2005, but Dell's wallet is
    wide open, too." + ], + [ + "WASHINGTON (CBS.MW) --
    President Bush announced
    Monday that Kellogg chief
    executive Carlos Gutierrez
    would replace Don Evans as
    Commerce secretary, naming the
    first of many expected changes
    to his economic team." + ], + [ + "Iron Mountain moved further
    into the backup and recovery
    space Tuesday with the
    acquisition of Connected Corp.
    for \\$117 million. Connected
    backs up desktop data for more
    than 600 corporations, with
    more than" + ], + [ + "Three directors of Manchester
    United have been ousted from
    the board after US tycoon
    Malcolm Glazer, who is
    attempting to buy the club,
    voted against their re-
    election." + ], + [ + "The Russian military yesterday
    extended its offer of a \\$10
    million reward for information
    leading to the capture of two
    separatist leaders who, the
    Kremlin claims, were behind
    the Beslan massacre." + ], + [ + "AP - Manny Ramirez singled and
    scored before leaving with a
    bruised knee, and the
    streaking Boston Red Sox beat
    the Detroit Tigers 5-3 Friday
    night for their 10th victory
    in 11 games." + ], + [ + "A remote attacker could take
    complete control over
    computers running many
    versions of Microsoft software
    by inserting malicious code in
    a JPEG image that executes
    through an unchecked buffer" + ], + [ + "Montgomery County (website -
    news) is a big step closer to
    shopping for prescription
    drugs north of the border. On
    a 7-2 vote, the County Council
    is approving a plan that would
    give county" + ], + [ + "Israels Shin Bet security
    service has tightened
    protection of the prime
    minister, MPs and parliament
    ahead of next weeks crucial
    vote on a Gaza withdrawal." + ], + [ + "The news comes fast and
    furious. Pedro Martinez goes
    to Tampa to visit George
    Steinbrenner. Theo Epstein and
    John Henry go to Florida for
    their turn with Pedro. Carl
    Pavano comes to Boston to
    visit Curt Schilling. Jason
    Varitek says he's not a goner.
    Derek Lowe is a goner, but he
    says he wishes it could be
    different. Orlando Cabrera ..." + ], + [ + "The disclosure this week that
    a Singapore-listed company
    controlled by a Chinese state-
    owned enterprise lost \\$550
    million in derivatives
    transactions" + ], + [ + "Reuters - Iraq's interim
    defense minister
    accused\\neighbors Iran and
    Syria on Wednesday of aiding
    al Qaeda\\Islamist Abu Musab
    al-Zarqawi and former agents
    of Saddam\\Hussein to promote a
    \"terrorist\" insurgency in
    Iraq." + ], + [ + "AP - The Senate race in
    Kentucky stayed at fever pitch
    on Thursday as Democratic
    challenger Daniel Mongiardo
    stressed his opposition to gay
    marriage while accusing
    Republican incumbent Jim
    Bunning of fueling personal
    attacks that seemed to suggest
    Mongiardo is gay." + ], + [ + "The lawsuit claims the
    companies use a patented
    Honeywell technology for
    brightening images and
    reducing interference on
    displays." + ], + [ + "The co-president of Oracle
    testified that her company was
    serious about its takeover
    offer for PeopleSoft and was
    not trying to scare off its
    customers." + ], + [ + "VANCOUVER - A Vancouver-based
    firm won #39;t sell 1.2
    million doses of influenza
    vaccine to the United States
    after all, announcing Tuesday
    that it will sell the doses
    within Canada instead." + ], + [ + "An extremely rare Hawaiian
    bird dies in captivity,
    possibly marking the
    extinction of its entire
    species only 31 years after it
    was first discovered." + ], + [ + "Does Geico's trademark lawsuit
    against Google have merit? How
    will the case be argued?
    What's the likely outcome of
    the trial? A mock court of
    trademark experts weighs in
    with their verdict." + ], + [ + "Treo 650 boasts a high-res
    display, an improved keyboard
    and camera, a removable
    battery, and more. PalmOne
    this week is announcing the
    Treo 650, a hybrid PDA/cell-
    phone device that addresses
    many of the shortcomings" + ], + [ + "FRED Hale Sr, documented as
    the worlds oldest man, has
    died at the age of 113. Hale
    died in his sleep on Friday at
    a hospital in Syracuse, New
    York, while trying to recover
    from a bout of pneumonia, his
    grandson, Fred Hale III said." + ], + [ + "The Oakland Raiders have
    traded Jerry Rice to the
    Seattle Seahawks in a move
    expected to grant the most
    prolific receiver in National
    Football League history his
    wish to get more playing time." + ], + [ + "consortium led by the Sony
    Corporation of America reached
    a tentative agreement today to
    buy Metro-Goldwyn-Mayer, the
    Hollywood studio famous for
    James Bond and the Pink
    Panther, for" + ], + [ + "International Business
    Machines Corp.'s possible exit
    from the personal computer
    business would be the latest
    move in what amounts to a long
    goodbye from a field it
    pioneered and revolutionized." + ], + [ + "Leipzig Game Convention in
    Germany, the stage for price-
    slash revelations. Sony has
    announced that it #39;s
    slashing the cost of PS2 in
    the UK and Europe to 104.99
    GBP." + ], + [ + "AP - Florida coach Ron Zook
    was fired Monday but will be
    allowed to finish the season,
    athletic director Jeremy Foley
    told The Gainesville Sun." + ], + [ + "The country-cooking restaurant
    chain has agreed to pay \\$8.7
    million over allegations that
    it segregated black customers,
    subjected them to racial slurs
    and gave black workers
    inferior jobs." + ], + [ + "Troy Brown has had to make a
    lot of adjustments while
    playing both sides of the
    football. quot;You always
    want to score when you get the
    ball -- offense or defense" + ], + [ + " PORT LOUIS, Aug. 17
    (Xinhuanet) -- Southern
    African countries Tuesday
    pledged better trade and
    investment relations with
    China as well as India in the
    final communique released at
    the end of their two-day
    summit." + ], + [ + "In yet another devastating
    body blow to the company,
    Intel (Nasdaq: INTC) announced
    it would be canceling its
    4-GHz Pentium chip. The
    semiconductor bellwether said
    it was switching" + ], + [ + "Jenson Button will tomorrow
    discover whether he is allowed
    to quit BAR and move to
    Williams for 2005. The
    Englishman has signed
    contracts with both teams but
    prefers a switch to Williams,
    where he began his Formula One
    career in 2000." + ], + [ + "Seagate #39;s native SATA
    interface technology with
    Native Command Queuing (NCQ)
    allows the Barracuda 7200.8 to
    match the performance of
    10,000-rpm SATA drives without
    sacrificing capacity" + ], + [ + "ARSENAL boss Arsene Wenger
    last night suffered a
    Champions League setback as
    Brazilian midfielder Gilberto
    Silva (above) was left facing
    a long-term injury absence." + ], + [ + "BAGHDAD - A militant group has
    released a video saying it
    kidnapped a missing journalist
    in Iraq and would kill him
    unless US forces left Najaf
    within 48 hours." + ], + [ + "18 August 2004 -- There has
    been renewed fighting in the
    Iraqi city of Al-Najaf between
    US and Iraqi troops and Shi
    #39;a militiamen loyal to
    radical cleric Muqtada al-
    Sadr." + ], + [ + "You #39;re probably already
    familiar with one of the most
    common questions we hear at
    iPodlounge: quot;how can I
    load my iPod up with free
    music?" + ], + [ + " SINGAPORE (Reuters) -
    Investors bought shares in
    Asian exporters and
    electronics firms such as
    Fujitsu Ltd. on Tuesday,
    buoyed by a favorable outlook
    from U.S. technology
    bellwethers and a slide in oil
    prices." + ], + [ + "Ace Ltd. will stop paying
    brokers for steering business
    its way, becoming the third
    company to make concessions in
    the five days since New York
    Attorney General Eliot Spitzer
    unveiled a probe of the
    insurance industry." + ], + [ + "Vice chairman of the office of
    the chief executive officer in
    Novell Inc, Chris Stone is
    leaving the company to pursue
    other opportunities in life." + ], + [ + "Wm. Wrigley Jr. Co., the world
    #39;s largest maker of chewing
    gum, agreed to buy candy
    businesses including Altoids
    mints and Life Savers from
    Kraft Foods Inc." + ], + [ + "American improves to 3-1 on
    the season with a hard-fought
    overtime win, 74-63, against
    Loyala at Bender Arena on
    Friday night." + ], + [ + "Australia tighten their grip
    on the third Test and the
    series after dominating India
    on day two in Nagpur." + ], + [ + " #39;Reaching a preliminary
    pilot agreement is the single
    most important hurdle they
    have to clear, but certainly
    not the only one." + ], + [ + "Bee Staff Writers. SAN
    FRANCISCO - As Eric Johnson
    drove to the stadium Sunday
    morning, his bruised ribs were
    so sore, he wasn #39;t sure he
    #39;d be able to suit up for
    the game." + ], + [ + "The New England Patriots are
    so single-minded in pursuing
    their third Super Bowl triumph
    in four years that they almost
    have no room for any other
    history." + ], + [ + "TORONTO (CP) - Canada #39;s
    big banks are increasing
    mortgage rates following a
    decision by the Bank of Canada
    to raise its overnight rate by
    one-quarter of a percentage
    point to 2.25 per cent." + ], + [ + " SEOUL (Reuters) - North
    Korea's two-year-old nuclear
    crisis has taxed the world's
    patience, the chief United
    Nations nuclear regulator
    said on Wednesday, urging
    communist Pyongyang to return
    to its disarmament treaty
    obligations." + ], + [ + "washingtonpost.com - Microsoft
    is going to Tinseltown today
    to announce plans for its
    revamped Windows XP Media
    Center, part of an aggressive
    push to get ahead in the
    digital entertainment race." + ], + [ + "GROZNY, Russia - The Russian
    government's choice for
    president of war-ravaged
    Chechnya appeared to be the
    victor Sunday in an election
    tainted by charges of fraud
    and shadowed by last week's
    terrorist destruction of two
    airliners. Little more than
    two hours after polls closed,
    acting Chechen president
    Sergei Abramov said
    preliminary results showed
    Maj..." + ], + [ + "Because, while the Eagles are
    certain to stumble at some
    point during the regular
    season, it seems inconceivable
    that they will falter against
    a team with as many offensive
    problems as Baltimore has
    right now." + ], + [ + "AP - J.J. Arrington ran for 84
    of his 121 yards in the second
    half and Aaron Rodgers shook
    off a slow start to throw two
    touchdown passes to help No. 5
    California beat Washington
    42-12 on Saturday." + ], + [ + "BAGHDAD, Sept 5 (AFP) - Izzat
    Ibrahim al-Duri, Saddam
    Hussein #39;s deputy whose
    capture was announced Sunday,
    is 62 and riddled with cancer,
    but was public enemy number
    two in Iraq for the world
    #39;s most powerful military." + ], + [ + "AP - An explosion targeted the
    Baghdad governor's convoy as
    he was traveling through the
    capital Tuesday, killing two
    people but leaving him
    uninjured, the Interior
    Ministry said." + ], + [ + "GENEVA: Rescuers have found
    the bodies of five Swiss
    firemen who died after the
    ceiling of an underground car
    park collapsed during a fire,
    a police spokesman said last
    night." + ], + [ + "John Kerry has held 10 \"front
    porch visit\" events an actual
    front porch is optional where
    perhaps 100 people ask
    questions in a low-key
    campaigning style." + ], + [ + "AP - Most of the turkeys
    gracing the nation's dinner
    tables Thursday have been
    selectively bred for their
    white meat for so many
    generations that simply
    walking can be a problem for
    many of the big-breasted birds
    and sex is no longer possible." + ], + [ + "OTTAWA (CP) - The economy
    created another 43,000 jobs
    last month, pushing the
    unemployment rate down to 7.1
    per cent from 7.2 per cent in
    August, Statistics Canada said
    Friday." + ], + [ + "The right-win opposition
    Conservative Party and Liberal
    Center Union won 43 seats in
    the 141-member Lithuanian
    parliament, after more than 99
    percent of the votes were
    counted" + ], + [ + " TOKYO (Reuters) - Japan's
    Nikkei average rose 0.39
    percent by midsession on
    Friday, bolstered by solid
    gains in stocks dependent on
    domestic business such as Kao
    Corp. <A HREF=\"http://www.i
    nvestor.reuters.com/FullQuote.
    aspx?ticker=4452.T target=/sto
    cks/quickinfo/fullquote\">44
    52.T</A>." + ], + [ + " FRANKFURT (Reuters) -
    DaimlerChrysler and General
    Motors will jointly develop
    new hybrid motors to compete
    against Japanese rivals on
    the fuel-saving technology
    that reduces harmful
    emissions, the companies said
    on Monday." + ], + [ + " SEATTLE (Reuters) - Microsoft
    Corp. <A HREF=\"http://www.r
    euters.co.uk/financeQuoteLooku
    p.jhtml?ticker=MSFT.O
    qtype=sym infotype=info
    qcat=news\">MSFT.O</A>
    is making a renewed push this
    week to get its software into
    living rooms with the launch
    of a new version of its
    Windows XP Media Center, a
    personal computer designed for
    viewing movies, listening to
    music and scrolling through
    digital pictures." + ], + [ + "KHARTOUM, Aug 18 (Reuters) -
    The United Nations said on
    Wednesday it was very
    concerned by Sudan #39;s lack
    of practical progress in
    bringing security to Darfur,
    where more than a million
    people have fled their homes
    for fear of militia ..." + ], + [ + "SHANGHAI, China The Houston
    Rockets have arrived in
    Shanghai with hometown
    favorite Yao Ming declaring
    himself quot;here on
    business." + ], + [ + "Charleston, SC (Sports
    Network) - Andy Roddick and
    Mardy Fish will play singles
    for the United States in this
    weekend #39;s Davis Cup
    semifinal matchup against
    Belarus." + ], + [ + "AFP - With less than two
    months until the November 2
    election, President George W.
    Bush is working to shore up
    support among his staunchest
    supporters even as he courts
    undecided voters." + ], + [ + "The bass should be in your
    face. That's what Matt Kelly,
    of Boston's popular punk rock
    band Dropkick Murphys, thinks
    is the mark of a great stereo
    system. And he should know.
    Kelly, 29, is the drummer for
    the band that likes to think
    of itself as a bit of an Irish
    lucky charm for the Red Sox." + ], + [ + "Chile's government says it
    will build a prison for
    officers convicted of human
    rights abuses in the Pinochet
    era." + ], + [ + "Slumping corporate spending
    and exports caused the economy
    to slow to a crawl in the
    July-September period, with
    real gross domestic product
    expanding just 0.1 percent
    from the previous quarter,
    Cabinet Office data showed
    Friday." + ], + [ + "US President George W. Bush
    signed into law a bill
    replacing an export tax
    subsidy that violated
    international trade rules with
    a \\$145 billion package of new
    corporate tax cuts and a
    buyout for tobacco farmers." + ], + [ + "The Nikkei average was up 0.37
    percent in mid-morning trade
    on Thursday as a recovery in
    the dollar helped auto makers
    among other exporters, but
    trade was slow as investors
    waited for important Japanese
    economic data." + ], + [ + "Jennifer Canada knew she was
    entering a boy's club when she
    enrolled in Southern Methodist
    University's Guildhall school
    of video-game making." + ], + [ + "RICKY PONTING believes the
    game #39;s watchers have
    fallen for the quot;myth
    quot; that New Zealand know
    how to rattle Australia." + ], + [ + "MILWAUKEE (SportsTicker) -
    Barry Bonds tries to go where
    just two players have gone
    before when the San Francisco
    Giants visit the Milwaukee
    Brewers on Tuesday." + ], + [ + "Palestinian leader Mahmoud
    Abbas reiterated calls for his
    people to drop their weapons
    in the struggle for a state. a
    clear change of strategy for
    peace with Israel after Yasser
    Arafat #39;s death." + ], + [ + "The new software is designed
    to simplify the process of
    knitting together back-office
    business applications." + ], + [ + "SYDNEY (Dow Jones)--Colorado
    Group Ltd. (CDO.AU), an
    Australian footwear and
    clothing retailer, said Monday
    it expects net profit for the
    fiscal year ending Jan. 29 to
    be over 30 higher than that of
    a year earlier." + ], + [ + "NEW YORK - What are the odds
    that a tiny nation like
    Antigua and Barbuda could take
    on the United States in an
    international dispute and win?" + ], + [ + "AP - With Tom Brady as their
    quarterback and a stingy,
    opportunistic defense, it's
    difficult to imagine when the
    New England Patriots might
    lose again. Brady and
    defensive end Richard Seymour
    combined to secure the
    Patriots' record-tying 18th
    straight victory, 31-17 over
    the Buffalo Bills on Sunday." + ], + [ + "FRANKFURT, GERMANY -- The
    German subsidiaries of
    Hewlett-Packard Co. (HP) and
    Novell Inc. are teaming to
    offer Linux-based products to
    the country's huge public
    sector." + ], + [ + "SBC Communications expects to
    cut 10,000 or more jobs by the
    end of next year through
    layoffs and attrition. That
    #39;s about six percent of the
    San Antonio-based company
    #39;s work force." + ], + [ + " BAGHDAD (Reuters) - Iraq's
    U.S.-backed government said on
    Tuesday that \"major neglect\"
    by its American-led military
    allies led to a massacre of 49
    army recruits at the weekend." + ], + [ + "SiliconValley.com - \"I'm
    back,\" declared Apple
    Computer's Steve Jobs on
    Thursday morning in his first
    public appearance before
    reporters since cancer surgery
    in late July." + ], + [ + "BEIJING -- Police have
    detained a man accused of
    slashing as many as nine boys
    to death as they slept in
    their high school dormitory in
    central China, state media
    reported today." + ], + [ + "Health India: London, Nov 4 :
    Cosmetic face cream used by
    fashionable Roman women was
    discovered at an ongoing
    archaeological dig in London,
    in a metal container, complete
    with the lid and contents." + ], + [ + "Israeli Prime Minister Ariel
    Sharon #39;s Likud party
    agreed on Thursday to a
    possible alliance with
    opposition Labour in a vote
    that averted a snap election
    and strengthened his Gaza
    withdrawal plan." + ], + [ + "Another United Airlines union
    is seeking to oust senior
    management at the troubled
    airline, saying its strategies
    are reckless and incompetent." + ], + [ + "Approaching Hurricane Ivan has
    led to postponement of the
    game Thursday night between
    10th-ranked California and
    Southern Mississippi in
    Hattiesburg, Cal #39;s
    athletic director said Monday." + ], + [ + "Global oil prices boomed on
    Wednesday, spreading fear that
    energy prices will restrain
    economic activity, as traders
    worried about a heating oil
    supply crunch in the American
    winter." + ], + [ + "Custom-designed imported
    furniture was once an
    exclusive realm. Now, it's the
    economical alternative for
    commercial developers and
    designers needing everything
    from seats to beds to desks
    for their projects." + ], + [ + "SAN DIEGO (Ticker) - The San
    Diego Padres lacked speed and
    an experienced bench last
    season, things veteran
    infielder Eric Young is
    capable of providing." + ], + [ + "This is an eye chart,
    reprinted as a public service
    to the New York Mets so they
    may see from what they suffer:
    myopia. Has ever a baseball
    franchise been so shortsighted
    for so long?" + ], + [ + "Short-term interest rate
    futures struggled on Thursday
    after a government report
    showing US core inflation for
    August below market
    expectations failed to alter
    views on Federal Reserve rate
    policy." + ], + [ + "MacCentral - Microsoft's
    Macintosh Business Unit on
    Tuesday issued a patch for
    Virtual PC 7 that fixes a
    problem that occurred when
    running the software on Power
    Mac G5 computers with more
    than 2GB of RAM installed.
    Previously, Virtual PC 7 would
    not run on those computers,
    causing a fatal error that
    crashed the application.
    Microsoft also noted that
    Virtual PC 7.0.1 also offers
    stability improvements,
    although it wasn't more
    specific than that -- some
    users have reported problems
    with using USB devices and
    other issues." + ], + [ + "Samsung is now the world #39;s
    second-largest mobile phone
    maker, behind Nokia. According
    to market watcher Gartner, the
    South Korean company has
    finally knocked Motorola into
    third place." + ], + [ + "Don't bother buying Star Wars:
    Battlefront if you're looking
    for a first-class shooter --
    there are far better games out
    there. But if you're a Star
    Wars freak and need a fix,
    this title will suffice. Lore
    Sjberg reviews Battlefront." + ], + [ + "While the American forces are
    preparing to launch a large-
    scale attack against Falluja
    and al-Ramadi, one of the
    chieftains of Falluja said
    that contacts are still
    continuous yesterday between
    members representing the city
    #39;s delegation and the
    interim" + ], + [ + "UN Security Council
    ambassadors were still
    quibbling over how best to
    pressure Sudan and rebels to
    end two different wars in the
    country even as they left for
    Kenya on Tuesday for a meeting
    on the crisis." + ], + [ + "HENDALA, Sri Lanka -- Day
    after day, locked in a cement
    room somewhere in Iraq, the
    hooded men beat him. They told
    him he would be beheaded.
    ''Ameriqi! quot; they shouted,
    even though he comes from this
    poor Sri Lankan fishing
    village." + ], + [ + "THE kidnappers of British aid
    worker Margaret Hassan last
    night threatened to turn her
    over to the group which
    beheaded Ken Bigley if the
    British Government refuses to
    pull its troops out of Iraq." + ], + [ + " TOKYO (Reuters) - Tokyo
    stocks climbed to a two week
    high on Friday after Tokyo
    Electron Ltd. and other chip-
    related stocks were boosted
    by a bullish revenue outlook
    from industry leader Intel
    Corp." + ], + [ + "More than by brain size or
    tool-making ability, the human
    species was set apart from its
    ancestors by the ability to
    jog mile after lung-stabbing
    mile with greater endurance
    than any other primate,
    according to research
    published today in the journal" + ], + [ + " LONDON (Reuters) - A medical
    product used to treat both
    male hair loss and prostate
    problems has been added to the
    list of banned drugs for
    athletes." + ], + [ + "AP - Curtis Martin and Jerome
    Bettis have the opportunity to
    go over 13,000 career yards
    rushing in the same game when
    Bettis and the Pittsburgh
    Steelers play Martin and the
    New York Jets in a big AFC
    matchup Sunday." + ], + [ + "<p></p><p>
    BOGOTA, Colombia (Reuters) -
    Ten Colombian police
    officerswere killed on Tuesday
    in an ambush by the National
    LiberationArmy, or ELN, in the
    worst attack by the Marxist
    group inyears, authorities
    told Reuters.</p>" + ], + [ + "Woodland Hills-based Brilliant
    Digital Entertainment and its
    subsidiary Altnet announced
    yesterday that they have filed
    a patent infringement suit
    against the Recording Industry
    Association of America (RIAA)." + ], + [ + "AFP - US President George W.
    Bush called his Philippines
    counterpart Gloria Arroyo and
    said their countries should
    keep strong ties, a spokesman
    said after a spat over
    Arroyo's handling of an Iraq
    kidnapping." + ], + [ + "A scathing judgment from the
    UK #39;s highest court
    condemning the UK government
    #39;s indefinite detention of
    foreign terror suspects as a
    threat to the life of the
    nation, left anti-terrorist
    laws in the UK in tatters on
    Thursday." + ], + [ + "Sony Ericsson and Cingular
    provide Z500a phones and
    service for military families
    to stay connected on today
    #39;s quot;Dr. Phil Show
    quot;." + ], + [ + "Reuters - Australia's
    conservative Prime
    Minister\\John Howard, handed
    the most powerful mandate in a
    generation,\\got down to work
    on Monday with reform of
    telecommunications,\\labor and
    media laws high on his agenda." + ], + [ + " TOKYO (Reuters) - Typhoon
    Megi killed one person as it
    slammed ashore in northern
    Japan on Friday, bringing the
    death toll to at least 13 and
    cutting power to thousands of
    homes before heading out into
    the Pacific." + ], + [ + "Cairo: Egyptian President
    Hosni Mubarak held telephone
    talks with Palestinian leader
    Yasser Arafat about Israel
    #39;s plan to pull troops and
    the 8,000 Jewish settlers out
    of the Gaza Strip next year,
    the Egyptian news agency Mena
    said." + ], + [ + " NEW YORK (Reuters) - Adobe
    Systems Inc. <A HREF=\"http:
    //www.investor.reuters.com/Ful
    lQuote.aspx?ticker=ADBE.O targ
    et=/stocks/quickinfo/fullquote
    \">ADBE.O</A> on
    Monday reported a sharp rise
    in quarterly profit, driven by
    robust demand for its
    Photoshop and document-sharing
    software." + ], + [ + "Dow Jones amp; Co., publisher
    of The Wall Street Journal,
    has agreed to buy online
    financial news provider
    MarketWatch Inc. for about
    \\$463 million in a bid to
    boost its revenue from the
    fast-growing Internet
    advertising market." + ], + [ + "The first American military
    intelligence soldier to be
    court-martialed over the Abu
    Ghraib abuse scandal was
    sentenced Saturday to eight
    months in jail, a reduction in
    rank and a bad-conduct
    discharge." + ], + [ + " TOKYO (Reuters) - Japan will
    seek an explanation at weekend
    talks with North Korea on
    activity indicating Pyongyang
    may be preparing a missile
    test, although Tokyo does not
    think a launch is imminent,
    Japan's top government
    spokesman said." + ], + [ + "Michigan Stadium was mostly
    filled with empty seats. The
    only cheers were coming from
    near one end zone -he Iowa
    section. By Carlos Osorio, AP." + ], + [ + "The International Rugby Board
    today confirmed that three
    countries have expressed an
    interest in hosting the 2011
    World Cup. New Zealand, South
    Africa and Japan are leading
    the race to host rugby union
    #39;s global spectacular in
    seven years #39; time." + ], + [ + "Officials at EarthLink #39;s
    (Quote, Chart) R amp;D
    facility have quietly released
    a proof-of-concept file-
    sharing application based on
    the Session Initiated Protocol
    (define)." + ], + [ + "Low-fare airline ATA has
    announced plans to lay off
    hundreds of employees and to
    drop most of its flights out
    of Midway Airport in Chicago." + ], + [ + " BEIJING (Reuters) - Iran will
    never be prepared to
    dismantle its nuclear program
    entirely but remains committed
    to the non-proliferation
    treaty (NPT), its chief
    delegate to the International
    Atomic Energy Agency said on
    Wednesday." + ], + [ + " WASHINGTON (Reuters) -
    Germany's Bayer AG <A HREF=
    \"http://www.investor.reuters.c
    om/FullQuote.aspx?ticker=BAYG.
    DE target=/stocks/quickinfo/fu
    llquote\">BAYG.DE</A>
    has agreed to plead guilty
    and pay a \\$4.7 million fine
    for taking part in a
    conspiracy to fix the prices
    of synthetic rubber, the U.S.
    Justice Department said on
    Wednesday." + ], + [ + "MIAMI (Ticker) -- In its first
    season in the Atlantic Coast
    Conference, No. 11 Virginia
    Tech is headed to the BCS.
    Bryan Randall threw two
    touchdown passes and the
    Virginia Tech defense came up
    big all day as the Hokies
    knocked off No." + ], + [ + "(CP) - Somehow, in the span of
    half an hour, the Detroit
    Tigers #39; pitching went from
    brutal to brilliant. Shortly
    after being on the wrong end
    of several records in a 26-5
    thrashing from to the Kansas
    City" + ], + [ + "<p></p><p>
    SANTIAGO, Chile (Reuters) -
    President Bush on
    Saturdayreached into a throng
    of squabbling bodyguards and
    pulled aSecret Service agent
    away from Chilean security
    officers afterthey stopped the
    U.S. agent from accompanying
    the president ata
    dinner.</p>" + ], + [ + " quot;It #39;s your mail,
    quot; the Google Web site
    said. quot;You should be able
    to choose how and where you
    read it. You can even switch
    to other e-mail services
    without having to worry about
    losing access to your
    messages." + ], + [ + "The US oil giant got a good
    price, Russia #39;s No. 1 oil
    company acquired a savvy
    partner, and Putin polished
    Russia #39;s image." + ], + [ + "With the Huskies reeling at
    0-4 - the only member of a
    Bowl Championship Series
    conference left without a win
    -an Jose State suddenly looms
    as the only team left on the
    schedule that UW will be
    favored to beat." + ], + [ + "Darryl Sutter, who coached the
    Calgary Flames to the Stanley
    Cup finals last season, had an
    emergency appendectomy and was
    recovering Friday." + ], + [ + "The maker of Hostess Twinkies,
    a cake bar and a piece of
    Americana children have
    snacked on for almost 75
    years, yesterday raised
    concerns about the company
    #39;s ability to stay in
    business." + ], + [ + "Iranian deputy foreign
    minister Gholamali Khoshrou
    denied Tuesday that his
    country #39;s top leaders were
    at odds over whether nuclear
    weapons were un-Islamic,
    insisting that it will
    quot;never quot; make the
    bomb." + ], + [ + " quot;Israel mercenaries
    assisting the Ivory Coast army
    operated unmanned aircraft
    that aided aerial bombings of
    a French base in the country,
    quot; claimed" + ], + [ + "Athens, Greece (Sports
    Network) - For the second
    straight day a Briton captured
    gold at the Olympic Velodrome.
    Bradley Wiggins won the men
    #39;s individual 4,000-meter
    pursuit Saturday, one day
    after teammate" + ], + [ + "AFP - SAP, the world's leading
    maker of business software,
    may need an extra year to
    achieve its medium-term profit
    target of an operating margin
    of 30 percent, its chief
    financial officer said." + ], + [ + "Tenet Healthcare Corp., the
    second- largest US hospital
    chain, said fourth-quarter
    charges may exceed \\$1 billion
    and its loss from continuing
    operations will widen from the
    third quarter #39;s because of
    increased bad debt." + ], + [ + "AFP - The airline Swiss said
    it had managed to cut its
    first-half net loss by about
    90 percent but warned that
    spiralling fuel costs were
    hampering a turnaround despite
    increasing passenger travel." + ], + [ + "Vulnerable youngsters expelled
    from schools in England are
    being let down by the system,
    say inspectors." + ], + [ + "Yasser Arafat was undergoing
    tests for possible leukaemia
    at a military hospital outside
    Paris last night after being
    airlifted from his Ramallah
    headquarters to an anxious
    farewell from Palestinian
    well-wishers." + ], + [ + "The world #39;s only captive
    great white shark made history
    this week when she ate several
    salmon fillets, marking the
    first time that a white shark
    in captivity" + ], + [ + "Nepal #39;s main opposition
    party urged the government on
    Monday to call a unilateral
    ceasefire with Maoist rebels
    and seek peace talks to end a
    road blockade that has cut the
    capital off from the rest of
    the country." + ], + [ + "Communications aggregator
    iPass said Monday that it is
    adding in-flight Internet
    access to its access
    portfolio. Specifically, iPass
    said it will add access
    offered by Connexion by Boeing
    to its list of access
    providers." + ], + [ + "The London-based brokerage
    Collins Stewart Tullett placed
    55m of new shares yesterday to
    help fund the 69.5m purchase
    of the money and futures
    broker Prebon." + ], + [ + "BOSTON - The New York Yankees
    and Boston were tied 4-4 after
    13 innings Monday night with
    the Red Sox trying to stay
    alive in the AL championship
    series. Boston tied the
    game with two runs in the
    eighth inning on David Ortiz's
    solo homer, a walk to Kevin
    Millar, a single by Trot Nixon
    and a sacrifice fly by Jason
    Varitek..." + ], + [ + "A Steffen Iversen penalty was
    sufficient to secure the
    points for Norway at Hampden
    on Saturday. James McFadden
    was ordered off after 53
    minutes for deliberate
    handball as he punched Claus
    Lundekvam #39;s header off the
    line." + ], + [ + "Reuters - The country may be
    more\\or less evenly divided
    along partisan lines when it
    comes to\\the presidential
    race, but the Republican Party
    prevailed in\\the Nielsen
    polling of this summer's
    nominating conventions." + ], + [ + " PARIS (Reuters) - European
    equities flirted with 5-month
    peaks as hopes that economic
    growth was sustainable and a
    small dip in oil prices
    helped lure investors back to
    recent underperformers such
    as technology and insurance
    stocks." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks looked to open higher
    on Friday, as the fourth
    quarter begins on Wall Street
    with oil prices holding below
    \\$50 a barrel." + ], + [ + "LONDON : World oil prices
    stormed above 54 US dollars
    for the first time Tuesday as
    strikes in Nigeria and Norway
    raised worries about possible
    supply shortages during the
    northern hemisphere winter." + ], + [ + "AP - Ailing St. Louis reliever
    Steve Kline was unavailable
    for Game 3 of the NL
    championship series on
    Saturday, but Cardinals
    manager Tony LaRussa hopes the
    left-hander will pitch later
    this postseason." + ], + [ + "Company launches free test
    version of service that
    fosters popular Internet
    activity." + ], + [ + "Outdated computer systems are
    hampering the work of
    inspectors, says the UN
    nuclear agency." + ], + [ + " In Vice President Cheney's
    final push before next
    Tuesday's election, talk of
    nuclear annihilation and
    escalating war rhetoric have
    blended with balloon drops,
    confetti cannons and the other
    trappings of modern
    campaigning with such ferocity
    that it is sometimes tough to
    tell just who the enemy is." + ], + [ + "MADRID: A stunning first-half
    free kick from David Beckham
    gave Real Madrid a 1-0 win
    over newly promoted Numancia
    at the Bernabeu last night." + ], + [ + "MacCentral - You Software Inc.
    announced on Tuesday the
    availability of You Control:
    iTunes, a free\\download that
    places iTunes controls in the
    Mac OS X menu bar.
    Without\\leaving the current
    application, you can pause,
    play, rewind or skip songs,\\as
    well as control iTunes' volume
    and even browse your entire
    music library\\by album, artist
    or genre. Each time a new song
    plays, You Control:
    iTunes\\also pops up a window
    that displays the artist and
    song name and the
    album\\artwork, if it's in the
    library. System requirements
    call for Mac OS X\\v10.2.6 and
    10MB free hard drive space.
    ..." + ], + [ + "Favourites Argentina beat
    Italy 3-0 this morning to
    claim their place in the final
    of the men #39;s Olympic
    football tournament. Goals by
    leading goalscorer Carlos
    Tevez, with a smart volley
    after 16 minutes, and" + ], + [ + "Shortly after Steve Spurrier
    arrived at Florida in 1990,
    the Gators were placed on NCAA
    probation for a year stemming
    from a child-support payment
    former coach Galen Hall made
    for a player." + ], + [ + "The US Secret Service Thursday
    announced arrests in eight
    states and six foreign
    countries of 28 suspected
    cybercrime gangsters on
    charges of identity theft,
    computer fraud, credit-card
    fraud, and conspiracy." + ], + [ + "US stocks were little changed
    on Thursday as an upbeat
    earnings report from chip
    maker National Semiconductor
    Corp. (NSM) sparked some
    buying, but higher oil prices
    limited gains." + ] + ], + "hovertemplate": "label=Other
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"white", + "zerolinewidth": 2 + } + } + }, + "title": { + "text": "Nearest neighbors of the Tony Blair article" + }, + "width": 600, + "xaxis": { + "anchor": "y", + "domain": [ + 0, + 1 + ], + "title": { + "text": "Component 0" + } + }, + "yaxis": { + "anchor": "x", + "domain": [ + 0, + 1 + ], + "title": { + "text": "Component 1" + } + } + } + } + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "# a 2D chart of nearest neighbors of the Tony Blair article\n", + "chart_from_components(\n", + " components=tsne_components,\n", + " labels=tony_blair_labels,\n", + " strings=article_descriptions,\n", + " width=600,\n", + " height=500,\n", + " title=\"Nearest neighbors of the Tony Blair article\",\n", + " category_orders={\"label\": [\"Other\", \"Nearest neighbor (top 5)\", \"Source\"]},\n", + ")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Looking at the 2D chart above, we can see that the articles about Tony Blair are somewhat close together inside of the World news cluster. Interestingly, although the 5 nearest neighbors (red) were closest in high dimensional space, they are not the closest points in this compressed 2D space. Compressing the embeddings from 2048 dimensions to 2 dimensions discards much of their information, and the nearest neighbors in the 2D space don't seem to be as relevant as those in the full embedding space." + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": {}, + "outputs": [ + { + "data": { + "application/vnd.plotly.v1+json": { + "config": { + "plotlyServerURL": "https://plot.ly" + }, + "data": [ + { + "customdata": [ + [ + "BRITAIN: BLAIR WARNS OF
    CLIMATE THREAT Prime Minister
    Tony Blair urged the
    international community to
    consider global warming a dire
    threat and agree on a plan of
    action to curb the
    quot;alarming quot; growth of
    greenhouse gases." + ], + [ + "Newspapers in Greece reflect a
    mixture of exhilaration that
    the Athens Olympics proved
    successful, and relief that
    they passed off without any
    major setback." + ], + [ + "SAN JOSE, Calif. -- Apple
    Computer (Quote, Chart)
    unveiled a batch of new iPods,
    iTunes software and promos
    designed to keep it atop the
    heap of digital music players." + ], + [ + "Any product, any shape, any
    size -- manufactured on your
    desktop! The future is the
    fabricator. By Bruce Sterling
    from Wired magazine." + ], + [ + "KABUL, Sept 22 (AFP): Three US
    soldiers were killed and 14
    wounded in a series of fierce
    clashes with suspected Taliban
    fighters in south and eastern
    Afghanistan this week, the US
    military said Wednesday." + ], + [ + "The EU and US moved closer to
    an aviation trade war after
    failing to reach agreement
    during talks Thursday on
    subsidies to Airbus Industrie." + ], + [ + "AUSTRALIAN journalist John
    Martinkus is lucky to be alive
    after spending 24 hours in the
    hands of Iraqi militants at
    the weekend. Martinkus was in
    Baghdad working for the SBS
    Dateline TV current affairs
    program" + ], + [ + " GAZA (Reuters) - An Israeli
    helicopter fired a missile
    into a town in the southern
    Gaza Strip late on Wednesday,
    witnesses said, hours after a
    Palestinian suicide bomber
    blew herself up in Jerusalem,
    killing two Israeli border
    policemen." + ], + [ + "The Microsoft CEO says one way
    to stem growing piracy of
    Windows and Office in emerging
    markets is to offer low-cost
    computers." + ], + [ + "RIYADH, Saudi Arabia -- Saudi
    police are seeking two young
    men in the killing of a Briton
    in a Riyadh parking lot, the
    Interior Ministry said today,
    and the British ambassador
    called it a terrorist attack." + ], + [ + "Loudon, NH -- As the rain
    began falling late Friday
    afternoon at New Hampshire
    International Speedway, the
    rich in the Nextel Cup garage
    got richer." + ], + [ + "PalmSource surprised the
    mobile vendor community today
    with the announcement that it
    will acquire China MobileSoft
    (CMS), ostensibly to leverage
    that company's expertise in
    building a mobile version of
    the Linux operating system." + ], + [ + "JEFFERSON CITY - An election
    security expert has raised
    questions about Missouri
    Secretary of State Matt Blunt
    #39;s plan to let soldiers at
    remote duty stations or in
    combat areas cast their
    ballots with the help of
    e-mail." + ], + [ + "A gas explosion at a coal mine
    in northern China killed 33
    workers in the 10th deadly
    mine blast reported in three
    months. The explosion occurred
    yesterday at 4:20 pm at Nanlou
    township" + ], + [ + "Reuters - Palestinian leader
    Mahmoud Abbas called\\Israel
    \"the Zionist enemy\" Tuesday,
    unprecedented language for\\the
    relative moderate who is
    expected to succeed Yasser
    Arafat." + ], + [ + "AP - Ottawa Senators right
    wing Marian Hossa is switching
    European teams during the NHL
    lockout." + ], + [ + "A new, optional log-on service
    from America Online that makes
    it harder for scammers to
    access a persons online
    account will not be available
    for Macintosh" + ], + [ + "Nasser al-Qidwa, Palestinian
    representative at the United
    Nations and nephew of late
    leader Yasser Arafat, handed
    Arafat #39;s death report to
    the Palestinian National
    Authority (PNA) on Saturday." + ], + [ + "CAIRO, Egypt - France's
    foreign minister appealed
    Monday for the release of two
    French journalists abducted in
    Baghdad, saying the French
    respect all religions. He did
    not rule out traveling to
    Baghdad..." + ], + [ + "Chelsea terminated Romania
    striker Adrian Mutu #39;s
    contract, citing gross
    misconduct after the player
    failed a doping test for
    cocaine and admitted taking
    the drug, the English soccer
    club said in a statement." + ], + [ + "United Arab Emirates President
    and ruler of Abu Dhabi Sheik
    Zayed bin Sultan al-Nayhan
    died Tuesday, official
    television reports. He was 86." + ], + [ + "PALESTINIAN leader Yasser
    Arafat today issued an urgent
    call for the immediate release
    of two French journalists
    taken hostage in Iraq." + ], + [ + "The al-Qaida terrorist network
    spent less than \\$50,000 on
    each of its major attacks
    except for the Sept. 11, 2001,
    suicide hijackings, and one of
    its hallmarks is using" + ], + [ + " SAN FRANCISCO (Reuters) -
    Nike Inc. <A HREF=\"http://w
    ww.investor.reuters.com/FullQu
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    stocks/quickinfo/fullquote\">
    ;NKE.N</A>, the world's
    largest athletic shoe company,
    on Monday reported a 25
    percent rise in quarterly
    profit, beating analysts'
    estimates, on strong demand
    for high-end running and
    basketball shoes in the
    United States." + ], + [ + "A FATHER who scaled the walls
    of a Cardiff court dressed as
    superhero Robin said the
    Buckingham Palace protester
    posed no threat. Fathers 4
    Justice activist Jim Gibson,
    who earlier this year staged
    an eye-catching" + ], + [ + "NEW YORK (CBS.MW) - US stocks
    traded mixed Thurday as Merck
    shares lost a quarter of their
    value, dragging blue chips
    lower, but the Nasdaq moved
    higher, buoyed by gains in the
    semiconductor sector." + ], + [ + "Julia Gillard has reportedly
    bowed out of the race to
    become shadow treasurer,
    taking enormous pressure off
    Opposition Leader Mark Latham." + ], + [ + "It #39;s official. Microsoft
    recently stated
    definitivelyand contrary to
    rumorsthat there will be no
    new versions of Internet
    Explorer for users of older
    versions of Windows." + ], + [ + "The success of Paris #39; bid
    for Olympic Games 2012 would
    bring an exceptional
    development for France for at
    least 6 years, said Jean-
    Francois Lamour, French
    minister for Youth and Sports
    on Tuesday." + ], + [ + "AFP - Maybe it's something to
    do with the fact that the
    playing area is so vast that
    you need a good pair of
    binoculars to see the action
    if it's not taking place right
    in front of the stands." + ], + [ + "Egypt #39;s release of accused
    Israeli spy Azzam Azzam in an
    apparent swap for six Egyptian
    students held on suspicion of
    terrorism is expected to melt
    the ice and perhaps result" + ], + [ + "But fresh antitrust suit is in
    the envelope, says Novell" + ], + [ + "Chips that help a computer's
    main microprocessors perform
    specific types of math
    problems are becoming a big
    business once again.\\" + ], + [ + "GAZA CITY, Gaza Strip: Hamas
    militants killed an Israeli
    soldier and wounded four with
    an explosion in a booby-
    trapped chicken coop on
    Tuesday, in what the Islamic
    group said was an elaborate
    scheme to lure troops to the
    area with the help of a double" + ], + [ + "A rocket carrying two Russian
    cosmonauts and an American
    astronaut to the international
    space station streaked into
    orbit on Thursday, the latest
    flight of a Russian space
    vehicle to fill in for
    grounded US shuttles." + ], + [ + "Bankrupt ATA Airlines #39;
    withdrawal from Chicago #39;s
    Midway International Airport
    presents a juicy opportunity
    for another airline to beef up
    service in the Midwest" + ], + [ + "AP - The 300 men filling out
    forms in the offices of an
    Iranian aid group were offered
    three choices: Train for
    suicide attacks against U.S.
    troops in Iraq, for suicide
    attacks against Israelis or to
    assassinate British author
    Salman Rushdie." + ], + [ + "ATHENS, Greece - Gail Devers,
    the most talented yet star-
    crossed hurdler of her
    generation, was unable to
    complete even one hurdle in
    100-meter event Sunday -
    failing once again to win an
    Olympic hurdling medal.
    Devers, 37, who has three
    world championships in the
    hurdles but has always flopped
    at the Olympics, pulled up
    short and screamed as she slid
    under the first hurdle..." + ], + [ + "OCTOBER 12, 2004
    (COMPUTERWORLD) - Microsoft
    Corp. #39;s monthly patch
    release cycle may be making it
    more predictable for users to
    deploy software updates, but
    there appears to be little
    letup in the number of holes
    being discovered in the
    company #39;s software" + ], + [ + "Wearable tech goes mainstream
    as the Gap introduces jacket
    with built-in radio.\\" + ], + [ + "Madison, WI (Sports Network) -
    Anthony Davis ran for 122
    yards and two touchdowns to
    lead No. 6 Wisconsin over
    Northwestern, 24-12, to
    celebrate Homecoming weekend
    at Camp Randall Stadium." + ], + [ + "LaVar Arrington participated
    in his first practice since
    Oct. 25, when he aggravated a
    knee injury that sidelined him
    for 10 games." + ], + [ + " NEW YORK (Reuters) - Billie
    Jean King cut her final tie
    with the U.S. Fed Cup team
    Tuesday when she retired as
    coach." + ], + [ + "The Instinet Group, owner of
    one of the largest electronic
    stock trading systems, is for
    sale, executives briefed on
    the plan say." + ], + [ + "Funding round of \\$105 million
    brings broadband Internet
    telephony company's total haul
    to \\$208 million." + ], + [ + "The Miami Dolphins arrived for
    their final exhibition game
    last night in New Orleans
    short nine players." + ], + [ + "washingtonpost.com - Anthony
    Casalena was 17 when he got
    his first job as a programmer
    for a start-up called
    HyperOffice, a software
    company that makes e-mail and
    contact management
    applications for the Web.
    Hired as an intern, he became
    a regular programmer after two
    weeks and rewrote the main
    product line." + ], + [ + "The Auburn Hills-based
    Chrysler Group made a profit
    of \\$269 million in the third
    quarter, even though worldwide
    sales and revenues declined,
    contributing to a \\$1." + ], + [ + "SAN FRANCISCO (CBS.MW) -- UAL
    Corp., parent of United
    Airlines, said Wednesday it
    will overhaul its route
    structure to reduce costs and
    offset rising fuel costs." + ], + [ + " NAIROBI (Reuters) - The
    Sudanese government and its
    southern rebel opponents have
    agreed to sign a pledge in the
    Kenyan capital on Friday to
    formally end a brutal 21-year-
    old civil war, with U.N.
    Security Council ambassadors
    as witnesses." + ], + [ + "AP - From LSU at No. 1 to Ohio
    State at No. 10, The AP
    women's basketball poll had no
    changes Monday." + ], + [ + "nother stage victory for race
    leader Petter Solberg, his
    fifth since the start of the
    rally. The Subaru driver is
    not pulling away at a fast
    pace from Gronholm and Loeb
    but the gap is nonetheless
    increasing steadily." + ], + [ + "The fossilized neck bones of a
    230-million-year-old sea
    creature have features
    suggesting that the animal
    #39;s snakelike throat could
    flare open and create suction
    that would pull in prey." + ], + [ + "IBM late on Tuesday announced
    the biggest update of its
    popular WebSphere business
    software in two years, adding
    features such as automatically
    detecting and fixing problems." + ], + [ + "April 1980 -- Players strike
    the last eight days of spring
    training. Ninety-two
    exhibition games are canceled.
    June 1981 -- Players stage
    first midseason strike in
    history." + ], + [ + "AP - Former Guatemalan
    President Alfonso Portillo
    #151; suspected of corruption
    at home #151; is living and
    working part-time in the same
    Mexican city he fled two
    decades ago to avoid arrest on
    murder charges, his close
    associates told The Associated
    Press on Sunday." + ], + [ + "AP - British entrepreneur
    Richard Branson said Monday
    that his company plans to
    launch commercial space
    flights over the next few
    years." + ], + [ + "Annual economic growth in
    China has slowed for the third
    quarter in a row, falling to
    9.1 per cent, third-quarter
    data shows. The slowdown shows
    the economy is responding to
    Beijing #39;s efforts to rein
    in break-neck investment and
    lending." + ], + [ + "washingtonpost.com - BRUSSELS,
    Aug. 26 -- The United States
    will have to wait until next
    year to see its fight with the
    European Union over biotech
    foods resolved, as the World
    Trade Organization agreed to
    an E.U. request to bring
    scientists into the debate,
    officials said Thursday." + ], + [ + "A group of Internet security
    and instant messaging
    providers have teamed up to
    detect and thwart the growing
    threat of IM and peer-to-peer
    viruses and worms, they said
    this week." + ], + [ + "On Sunday, the day after Ohio
    State dropped to 0-3 in the
    Big Ten with a 33-7 loss at
    Iowa, the Columbus Dispatch
    ran a single word above its
    game story on the Buckeyes:
    quot;Embarrassing." + ], + [ + "Insisting that Hurriyat
    Conference is the real
    representative of Kashmiris,
    Pakistan has claimed that
    India is not ready to accept
    ground realities in Kashmir." + ], + [ + "THE All-India Motor Transport
    Congress (AIMTC) on Saturday
    called off its seven-day
    strike after finalising an
    agreement with the Government
    on the contentious issue of
    service tax and the various
    demands including tax deducted
    at source (TDS), scrapping" + ], + [ + "BOSTON - Curt Schilling got
    his 20th win on the eve of
    Boston #39;s big series with
    the New York Yankees. Now he
    wants much more. quot;In a
    couple of weeks, hopefully, it
    will get a lot better, quot;
    he said after becoming" + ], + [ + "Shooed the ghosts of the
    Bambino and the Iron Horse and
    the Yankee Clipper and the
    Mighty Mick, on his 73rd
    birthday, no less, and turned
    Yankee Stadium into a morgue." + ], + [ + "Gold medal-winning Marlon
    Devonish says the men #39;s
    4x100m Olympic relay triumph
    puts British sprinting back on
    the map. Devonish, Darren
    Campbell, Jason Gardener and
    Mark Lewis-Francis edged out
    the American" + ], + [ + "AP - The euro-zone economy
    grew by 0.5 percent in the
    second quarter of 2004, a
    touch slower than in the first
    three months of the year,
    according to initial estimates
    released Tuesday by the
    European Union." + ], + [ + "His first space flight was in
    1965 when he piloted the first
    manned Gemini mission. Later
    he made two trips to the moon
    -- orbiting during a 1969
    flight and then walking on the
    lunar surface during a mission
    in 1972." + ], + [ + "he difficult road conditions
    created a few incidents in the
    first run of the Epynt stage.
    Francois Duval takes his
    second stage victory since the
    start of the rally, nine
    tenths better than Sebastien
    Loeb #39;s performance in
    second position." + ], + [ + "VIENNA -- After two years of
    investigating Iran's atomic
    program, the UN nuclear
    watchdog still cannot rule out
    that Tehran has a secret atom
    bomb project as Washington
    insists, the agency's chief
    said yesterday." + ], + [ + "By RACHEL KONRAD SAN
    FRANCISCO (AP) -- EBay Inc.,
    which has been aggressively
    expanding in Asia, plans to
    increase its stake in South
    Korea's largest online auction
    company. The Internet
    auction giant said Tuesday
    night that it would purchase
    nearly 3 million publicly held
    shares of Internet Auction
    Co..." + ], + [ + "AFP - US Secretary of State
    Colin Powell wrapped up a
    three-nation tour of Asia
    after winning pledges from
    Japan, China and South Korea
    to press North Korea to resume
    stalled talks on its nuclear
    weapons programs." + ], + [ + "Tallahassee, FL (Sports
    Network) - Florida State head
    coach Bobby Bowden has
    suspended senior wide receiver
    Craphonso Thorpe for the
    Seminoles #39; game against
    fellow Atlantic Coast
    Conference member Duke on
    Saturday." + ], + [ + "A few years ago, casinos
    across the United States were
    closing their poker rooms to
    make space for more popular
    and lucrative slot machines." + ], + [ + "CAIRO, Egypt An Egyptian
    company says one of its four
    workers who had been kidnapped
    in Iraq has been freed. It
    says it can #39;t give the
    status of the others being
    held hostage but says it is
    quot;doing its best to secure
    quot; their release." + ], + [ + "WASHINGTON -- Consumers were
    tightfisted amid soaring
    gasoline costs in August and
    hurricane-related disruptions
    last week sent applications
    for jobless benefits to their
    highest level in seven months." + ], + [ + "Talking kitchens and vanities.
    Musical jump ropes and potty
    seats. Blusterous miniature
    leaf blowers and vacuum
    cleaners -- almost as loud as
    the real things." + ], + [ + "Online merchants in the United
    States have become better at
    weeding out fraudulent credit
    card orders, a new survey
    indicates. But shipping
    overseas remains a risky
    venture. By Joanna Glasner." + ], + [ + "Former champion Lleyton Hewitt
    bristled, battled and
    eventually blossomed as he
    took another step towards a
    second US Open title with a
    second-round victory over
    Moroccan Hicham Arazi on
    Friday." + ], + [ + "AP - China's biggest computer
    maker, Lenovo Group, said
    Wednesday it has acquired a
    majority stake in
    International Business
    Machines Corp.'s personal
    computer business for
    #36;1.75 billion, one of the
    biggest Chinese overseas
    acquisitions ever." + ], + [ + "Popping a gadget into a cradle
    to recharge it used to mean
    downtime, but these days
    chargers are doing double
    duty, keeping your portable
    devices playing even when
    they're charging." + ], + [ + "AFP - Hosts India braced
    themselves for a harrowing
    chase on a wearing wicket in
    the first Test after Australia
    declined to enforce the
    follow-on here." + ], + [ + " SAN FRANCISCO (Reuters) -
    Texas Instruments Inc. <A H
    REF=\"http://www.investor.reute
    rs.com/FullQuote.aspx?ticker=T
    XN.N target=/stocks/quickinfo/
    fullquote\">TXN.N</A>,
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    cellular phones, on Monday
    said record demand for its
    handset and television chips
    boosted quarterly profit by
    26 percent, even as it
    struggles with a nagging
    inventory problem." + ], + [ + "LONDON ARM Holdings, a British
    semiconductor designer, said
    on Monday that it would buy
    Artisan Components for \\$913
    million to broaden its product
    range." + ], + [ + "Big evolutionary insights
    sometimes come in little
    packages. Witness the
    startling discovery, in a cave
    on the eastern Indonesian
    island of Flores, of the
    partial skeleton of a half-
    size Homo species that" + ], + [ + "Prime Minister Paul Martin of
    Canada urged Haitian leaders
    on Sunday to allow the
    political party of the deposed
    president, Jean-Bertrand
    Aristide, to take part in new
    elections." + ], + [ + "Hostage takers holding up to
    240 people at a school in
    southern Russia have refused
    to talk with a top Islamic
    leader and demanded to meet
    with regional leaders instead,
    ITAR-TASS reported on
    Wednesday." + ], + [ + "As the Mets round out their
    search for a new manager, the
    club is giving a last-minute
    nod to its past. Wally
    Backman, an infielder for the
    Mets from 1980-88 who played
    second base on the 1986" + ], + [ + "MELBOURNE: Global shopping
    mall owner Westfield Group
    will team up with Australian
    developer Multiplex Group to
    bid 585mil (US\\$1." + ], + [ + "Three children from a care
    home are missing on the
    Lancashire moors after they
    are separated from a group." + ], + [ + "Luke Skywalker and Darth Vader
    may get all the glory, but a
    new Star Wars video game
    finally gives credit to the
    everyday grunts who couldn
    #39;t summon the Force for
    help." + ], + [ + "AMSTERDAM, Aug 18 (Reuters) -
    Midfielder Edgar Davids #39;s
    leadership qualities and
    never-say-die attitude have
    earned him the captaincy of
    the Netherlands under new
    coach Marco van Basten." + ], + [ + "COUNTY KILKENNY, Ireland (PA)
    -- Hurricane Jeanne has led to
    world No. 1 Vijay Singh
    pulling out of this week #39;s
    WGC-American Express
    Championship at Mount Juliet." + ], + [ + "Green Bay, WI (Sports Network)
    - The Green Bay Packers will
    be without the services of Pro
    Bowl center Mike Flanagan for
    the remainder of the season,
    as he will undergo left knee
    surgery." + ], + [ + "Diabetics should test their
    blood sugar levels more
    regularly to reduce the risk
    of cardiovascular disease, a
    study says." + ], + [ + "COLUMBUS, Ohio -- NCAA
    investigators will return to
    Ohio State University Monday
    to take another look at the
    football program after the
    latest round of allegations
    made by former players,
    according to the Akron Beacon
    Journal." + ], + [ + " LOS ANGELES (Reuters) -
    Federal authorities raided
    three Washington, D.C.-area
    video game stores and arrested
    two people for modifying
    video game consoles to play
    pirated video games, a video
    game industry group said on
    Wednesday." + ], + [ + "Manchester United gave Alex
    Ferguson a 1,000th game
    anniversary present by
    reaching the last 16 of the
    Champions League yesterday,
    while four-time winners Bayern
    Munich romped into the second
    round with a 5-1 beating of
    Maccabi Tel Aviv." + ], + [ + "Iraq's interim Prime Minister
    Ayad Allawi announced that
    proceedings would begin
    against former Baath Party
    leaders." + ], + [ + "Reuters - Delta Air Lines Inc.
    , which is\\racing to cut costs
    to avoid bankruptcy, on
    Wednesday reported\\a wider
    quarterly loss amid soaring
    fuel prices and weak\\domestic
    airfares." + ], + [ + "Energy utility TXU Corp. on
    Monday more than quadrupled
    its quarterly dividend payment
    and raised its projected
    earnings for 2004 and 2005
    after a companywide
    performance review." + ], + [ + "Northwest Airlines Corp., the
    fourth- largest US airline,
    and its pilots union reached
    tentative agreement on a
    contract that would cut pay
    and benefits, saving the
    company \\$265 million a year." + ], + [ + "The last time the Angels
    played the Texas Rangers, they
    dropped two consecutive
    shutouts at home in their most
    agonizing lost weekend of the
    season." + ], + [ + "Microsoft Corp. MSFT.O and
    cable television provider
    Comcast Corp. CMCSA.O said on
    Monday that they would begin
    deploying set-top boxes
    powered by Microsoft software
    starting next week." + ], + [ + "SEATTLE - Chasing a nearly
    forgotten ghost of the game,
    Ichiro Suzuki broke one of
    baseball #39;s oldest records
    Friday night, smoking a single
    up the middle for his 258th
    hit of the year and breaking
    George Sisler #39;s record for
    the most hits in a season" + ], + [ + "Grace Park, her caddie - and
    fans - were poking around in
    the desert brush alongside the
    18th fairway desperately
    looking for her ball." + ], + [ + "Philippines mobile phone
    operator Smart Communications
    Inc. is in talks with
    Singapore #39;s Mobile One for
    a possible tie-up,
    BusinessWorld reported Monday." + ], + [ + "Washington Redskins kicker
    John Hall strained his right
    groin during practice
    Thursday, his third leg injury
    of the season. Hall will be
    held out of practice Friday
    and is questionable for Sunday
    #39;s game against the Chicago
    Bears." + ], + [ + "Airline warns it may file for
    bankruptcy if too many senior
    pilots take early retirement
    option. Delta Air LInes #39;
    CEO says it faces bankruptcy
    if it can #39;t slow the pace
    of pilots taking early
    retirement." + ], + [ + "A toxic batch of home-brewed
    alcohol has killed 31 people
    in several towns in central
    Pakistan, police and hospital
    officials say." + ], + [ + "Thornbugs communicate by
    vibrating the branches they
    live on. Now scientists are
    discovering just what the bugs
    are \"saying.\"" + ], + [ + "The Florida Gators and
    Arkansas Razorbacks meet for
    just the sixth time Saturday.
    The Gators hold a 4-1
    advantage in the previous five
    meetings, including last year
    #39;s 33-28 win." + ], + [ + "Kodak Versamark #39;s parent
    company, Eastman Kodak Co.,
    reported Tuesday it plans to
    eliminate almost 900 jobs this
    year in a restructuring of its
    overseas operations." + ], + [ + "A top official of the US Food
    and Drug Administration said
    Friday that he and his
    colleagues quot;categorically
    reject quot; earlier
    Congressional testimony that
    the agency has failed to
    protect the public against
    dangerous drugs." + ], + [ + " BEIJING (Reuters) - North
    Korea is committed to holding
    six-party talks aimed at
    resolving the crisis over its
    nuclear weapons program, but
    has not indicated when, a top
    British official said on
    Tuesday." + ], + [ + "AP - 1941 #151; Brooklyn
    catcher Mickey Owen dropped a
    third strike on Tommy Henrich
    of what would have been the
    last out of a Dodgers victory
    against the New York Yankees.
    Given the second chance, the
    Yankees scored four runs for a
    7-4 victory and a 3-1 lead in
    the World Series, which they
    ended up winning." + ], + [ + "Federal prosecutors cracked a
    global cartel that had
    illegally fixed prices of
    memory chips in personal
    computers and servers for
    three years." + ], + [ + "AP - Oracle Corp. CEO Larry
    Ellison reiterated his
    determination to prevail in a
    long-running takeover battle
    with rival business software
    maker PeopleSoft Inc.,
    predicting the proposed deal
    will create a more competitive
    company with improved customer
    service." + ], + [ + "Braves shortstop Rafael Furcal
    was arrested on drunken
    driving charges Friday, his
    second D.U.I. arrest in four
    years." + ], + [ + "AFP - British retailer Marks
    and Spencer announced a major
    management shake-up as part of
    efforts to revive its
    fortunes, saying trading has
    become tougher ahead of the
    crucial Christmas period." + ], + [ + " BAGHDAD (Reuters) - Iraq's
    interim government extended
    the closure of Baghdad
    international airport
    indefinitely on Saturday
    under emergency rule imposed
    ahead of this week's U.S.-led
    offensive on Falluja." + ], + [ + " ATHENS (Reuters) - Natalie
    Coughlin's run of bad luck
    finally took a turn for the
    better when she won the gold
    medal in the women's 100
    meters backstroke at the
    Athens Olympics Monday." + ], + [ + " ATLANTA (Reuters) - Home
    Depot Inc. <A HREF=\"http://
    www.investor.reuters.com/FullQ
    uote.aspx?ticker=HD.N target=/
    stocks/quickinfo/fullquote\">
    ;HD.N</A>, the top home
    improvement retailer, on
    Tuesday reported a 15 percent
    rise in third-quarter profit,
    topping estimates, aided by
    installed services and sales
    to contractors." + ], + [ + " LONDON (Reuters) - The dollar
    fought back from one-month
    lows against the euro and
    Swiss franc on Wednesday as
    investors viewed its sell-off
    in the wake of the Federal
    Reserve's verdict on interest
    rates as overdone." + ], + [ + "Rivaling Bush vs. Kerry for
    bitterness, doctors and trial
    lawyers are squaring off this
    fall in an unprecedented four-
    state struggle over limiting
    malpractice awards..." + ], + [ + "Boston Scientific Corp said on
    Friday it has recalled an ear
    implant the company acquired
    as part of its purchase of
    Advanced Bionics in June." + ], + [ + "AP - Pedro Feliz hit a
    tiebreaking grand slam with
    two outs in the eighth inning
    for his fourth hit of the day,
    and the Giants helped their
    playoff chances with a 9-5
    victory over the Los Angeles
    Dodgers on Saturday." + ], + [ + "MarketWatch.com. Richemont
    sees significant H1 lift,
    unclear on FY (2:53 AM ET)
    LONDON (CBS.MW) -- Swiss
    luxury goods maker
    Richemont(zz:ZZ:001273145:
    news, chart, profile), which
    also is a significant" + ], + [ + "Crude oil climbed more than
    \\$1 in New York on the re-
    election of President George
    W. Bush, who has been filling
    the US Strategic Petroleum
    Reserve." + ], + [ + "AP - Hundreds of tribesmen
    gathered Tuesday near the area
    where suspected al-Qaida-
    linked militants are holding
    two Chinese engineers and
    demanding safe passage to
    their reputed leader, a former
    U.S. prisoner from Guantanamo
    Bay, Cuba, officials and
    residents said." + ], + [ + "AP - Scientists warned Tuesday
    that a long-term increase in
    global temperature of 3.5
    degrees could threaten Latin
    American water supplies,
    reduce food yields in Asia and
    result in a rise in extreme
    weather conditions in the
    Caribbean." + ], + [ + "AP - Further proof New York's
    real estate market is
    inflated: The city plans to
    sell space on top of lampposts
    to wireless phone companies
    for #36;21.6 million a year." + ], + [ + "In an alarming development,
    high-precision equipment and
    materials which could be used
    for making nuclear bombs have
    disappeared from some Iraqi
    facilities, the United Nations
    watchdog agency has said." + ], + [ + "Yukos #39; largest oil-
    producing unit regained power
    supplies from Tyumenenergo, a
    Siberia-based electricity
    generator, Friday after the
    subsidiary pledged to pay 440
    million rubles (\\$15 million)
    in debt by Oct. 3." + ], + [ + "The rollout of new servers and
    networking switches in stores
    is part of a five-year
    agreement supporting 7-Eleven
    #39;s Retail Information
    System." + ], + [ + "Top Hollywood studios have
    launched a wave of court cases
    against internet users who
    illegally download film files.
    The Motion Picture Association
    of America, which acts as the
    representative of major film" + ], + [ + "AUSTRALIANS went into a
    television-buying frenzy the
    run-up to the Athens Olympics,
    suggesting that as a nation we
    could easily have scored a
    gold medal for TV purchasing." + ], + [ + "US STOCKS vacillated yesterday
    as rising oil prices muted
    Wall Streets excitement over
    strong results from Lehman
    Brothers and Sprints \\$35
    billion acquisition of Nextel
    Communications." + ], + [ + "The next time you are in your
    bedroom with your PC plus
    Webcam switched on, don #39;t
    think that your privacy is all
    intact. If you have a Webcam
    plugged into an infected
    computer, there is a
    possibility that" + ], + [ + "At the head of the class,
    Rosabeth Moss Kanter is an
    intellectual whirlwind: loud,
    expansive, in constant motion." + ], + [ + "LEVERKUSEN/ROME, Dec 7 (SW) -
    Dynamo Kiev, Bayer Leverkusen,
    and Real Madrid all have a
    good chance of qualifying for
    the Champions League Round of
    16 if they can get the right
    results in Group F on
    Wednesday night." + ], + [ + "Ed Hinkel made a diving,
    fingertip catch for a key
    touchdown and No. 16 Iowa
    stiffened on defense when it
    needed to most to beat Iowa
    State 17-10 Saturday." + ], + [ + "During last Sunday #39;s
    Nextel Cup race, amid the
    ongoing furor over Dale
    Earnhardt Jr. #39;s salty
    language, NBC ran a commercial
    for a show coming on later
    that night called quot;Law
    amp; Order: Criminal Intent." + ], + [ + "AP - After playing in hail,
    fog and chill, top-ranked
    Southern California finishes
    its season in familiar
    comfort. The Trojans (9-0, 6-0
    Pacific-10) have two games at
    home #151; against Arizona on
    Saturday and Notre Dame on
    Nov. 27 #151; before their
    rivalry game at UCLA." + ], + [ + "A US airman dies and two are
    hurt as a helicopter crashes
    due to technical problems in
    western Afghanistan." + ], + [ + "Jacques Chirac has ruled out
    any withdrawal of French
    troops from Ivory Coast,
    despite unrest and anti-French
    attacks, which have forced the
    evacuation of thousands of
    Westerners." + ], + [ + "Japanese Prime Minister
    Junichiro Koizumi reshuffled
    his cabinet yesterday,
    replacing several top
    ministers in an effort to
    boost his popularity,
    consolidate political support
    and quicken the pace of
    reforms in the world #39;s
    second-largest economy." + ], + [ + "The remnants of Hurricane
    Jeanne rained out Monday's
    game between the Mets and the
    Atlanta Braves. It will be
    made up Tuesday as part of a
    doubleheader." + ], + [ + "AP - NASCAR is not expecting
    any immediate changes to its
    top-tier racing series
    following the merger between
    telecommunications giant
    Sprint Corp. and Nextel
    Communications Inc." + ], + [ + "AP - Shawn Fanning's Napster
    software enabled countless
    music fans to swap songs on
    the Internet for free, turning
    him into the recording
    industry's enemy No. 1 in the
    process." + ], + [ + "TBILISI (Reuters) - At least
    two Georgian soldiers were
    killed and five wounded in
    artillery fire with
    separatists in the breakaway
    region of South Ossetia,
    Georgian officials said on
    Wednesday." + ], + [ + "Like wide-open races? You
    #39;ll love the Big 12 North.
    Here #39;s a quick morning
    line of the Big 12 North as it
    opens conference play this
    weekend." + ], + [ + "Reuters - Walt Disney Co.
    is\\increasing investment in
    video games for hand-held
    devices and\\plans to look for
    acquisitions of small game
    publishers and\\developers,
    Disney consumer products
    Chairman Andy Mooney said\\on
    Monday." + ], + [ + "Taquan Dean scored 22 points,
    Francisco Garcia added 19 and
    No. 13 Louisville withstood a
    late rally to beat Florida
    74-70 Saturday." + ], + [ + "BANGKOK - A UN conference last
    week banned commercial trade
    in the rare Irrawaddy dolphin,
    a move environmentalists said
    was needed to save the
    threatened species." + ], + [ + "Laksamana.Net - Two Indonesian
    female migrant workers freed
    by militants in Iraq are
    expected to arrive home within
    a day or two, the Foreign
    Affairs Ministry said
    Wednesday (6/10/04)." + ], + [ + "A bitter row between America
    and the European Union over
    alleged subsidies to rival
    aircraft makers Boeing and
    Airbus intensified yesterday." + ], + [ + "NEW YORK -- This was all about
    Athens, about redemption for
    one and validation for the
    other. Britain's Paula
    Radcliffe, the fastest female
    marathoner in history, failed
    to finish either of her
    Olympic races last summer.
    South Africa's Hendrik Ramaala
    was a five-ringed dropout,
    too, reinforcing his
    reputation as a man who could
    go only half the distance." + ], + [ + "Reuters - Online media company
    Yahoo Inc.\\ late on Monday
    rolled out tests of redesigned
    start\\pages for its popular
    Yahoo.com and My.Yahoo.com
    sites." + ], + [ + "Amsterdam (pts) - Dutch
    electronics company Philips
    http://www.philips.com has
    announced today, Friday, that
    it has cut its stake in Atos
    Origin by more than a half." + ], + [ + "TORONTO (CP) - Two-thirds of
    banks around the world have
    reported an increase in the
    volume of suspicious
    activities that they report to
    police, a new report by KPMG
    suggests." + ], + [ + "The Sun may have captured
    thousands or even millions of
    asteroids from another
    planetary system during an
    encounter more than four
    billion years ago, astronomers
    are reporting." + ], + [ + "LONDON -- Ernie Els has set
    his sights on an improved
    putting display this week at
    the World Golf Championships
    #39; NEC Invitational in
    Akron, Ohio, after the
    disappointment of tying for
    fourth place at the PGA
    Championship last Sunday." + ], + [ + "The Atkins diet frenzy slowed
    growth briefly, but the
    sandwich business is booming,
    with \\$105 billion in sales
    last year." + ], + [ + "Luxury carmaker Jaguar said
    Friday it was stopping
    production at a factory in
    central England, resulting in
    a loss of 1,100 jobs,
    following poor sales in the
    key US market." + ], + [ + "A bus was hijacked today and
    shots were fired at police who
    surrounded it on the outskirts
    of Athens. Police did not know
    how many passengers were
    aboard the bus." + ], + [ + "Thumb through the book - then
    buy a clean copy from Amazon" + ], + [ + "AP - President Bashar Assad
    shuffled his Cabinet on
    Monday, just weeks after the
    United States and the United
    Nations challenged Syria over
    its military presence in
    Lebanon and the security
    situation along its border
    with Iraq." + ], + [ + "Fiji #39;s Vijay Singh
    replaced Tiger Woods as the
    world #39;s No.1 ranked golfer
    today by winning the PGA
    Deutsche Bank Championship." + ], + [ + "LEIPZIG, Germany : Jurgen
    Klinsmann enjoyed his first
    home win as German manager
    with his team defeating ten-
    man Cameroon 3-0 in a friendly
    match." + ], + [ + "AP - Kevin Brown had a chance
    to claim a place in Yankees
    postseason history with his
    start in Game 7 of the AL
    championship series." + ], + [ + "Reuters - High-definition
    television can\\show the sweat
    beading on an athlete's brow,
    but the cost of\\all the
    necessary electronic equipment
    can get a shopper's own\\pulse
    racing." + ], + [ + "HOMESTEAD, Fla. -- Kurt Busch
    got his first big break in
    NASCAR by winning a 1999
    talent audition nicknamed
    quot;The Gong Show. quot; He
    was selected from dozens of
    unknown, young race drivers to
    work for one of the sport
    #39;s most famous team owners,
    Jack Roush." + ], + [ + "AP - President Vladimir Putin
    has signed a bill confirming
    Russia's ratification of the
    Kyoto Protocol, the Kremlin
    said Friday, clearing the way
    for the global climate pact to
    come into force early next
    year." + ], + [ + "John Gibson said Friday that
    he decided to resign as chief
    executive officer of
    Halliburton Energy Services
    when it became apparent he
    couldn #39;t become the CEO of
    the entire corporation, after
    getting a taste of the No." + ], + [ + "MacCentral - Apple Computer
    Inc. on Monday released an
    update for Apple Remote
    Desktop (ARD), the company's
    software solution to assist
    Mac system administrators and
    computer managers with asset
    management, software
    distribution and help desk
    support. ARD 2.1 includes
    several enhancements and bug
    fixes." + ], + [ + "NEW YORK (Reuters) - Outback
    Steakhouse Inc. said Tuesday
    it lost about 130 operating
    days and up to \\$2 million in
    revenue because it had to
    close some restaurants in the
    South due to Hurricane
    Charley." + ], + [ + "State insurance commissioners
    from across the country have
    proposed new rules governing
    insurance brokerage fees,
    winning praise from an
    industry group and criticism
    from" + ], + [ + "AP - The authenticity of newly
    unearthed memos stating that
    George W. Bush failed to meet
    standards of the Texas Air
    National Guard during the
    Vietnam War was questioned
    Thursday by the son of the
    late officer who reportedly
    wrote the memos." + ], + [ + "Zurich, Switzerland (Sports
    Network) - Former world No. 1
    Venus Williams advanced on
    Thursday and will now meet
    Wimbledon champion Maria
    Sharapova in the quarterfinals
    at the \\$1." + ], + [ + "INDIA #39;S cricket chiefs
    began a frenetic search today
    for a broadcaster to show next
    month #39;s home series
    against world champion
    Australia after cancelling a
    controversial \\$US308 million
    (\\$440 million) television
    deal." + ], + [ + "Canadian Press - OAKVILLE,
    Ont. (CP) - The body of a
    missing autistic man was
    pulled from a creek Monday,
    just metres from where a key
    piece of evidence was
    uncovered but originally
    overlooked because searchers
    had the wrong information." + ], + [ + "SOFTWARE giant Oracle #39;s
    stalled \\$7.7bn (4.2bn) bid to
    take over competitor
    PeopleSoft has received a huge
    boost after a US judge threw
    out an anti-trust lawsuit
    filed by the Department of
    Justice to block the
    acquisition." + ], + [ + "The International Olympic
    Committee (IOC) has urged
    Beijing to ensure the city is
    ready to host the 2008 Games
    well in advance, an official
    said on Wednesday." + ], + [ + "AFP - German Chancellor
    Gerhard Schroeder arrived in
    Libya for an official visit
    during which he is to hold
    talks with Libyan leader
    Moamer Kadhafi." + ], + [ + "The fastest-swiveling space
    science observatory ever built
    rocketed into orbit on
    Saturday to scan the universe
    for celestial explosions." + ], + [ + "The government will examine
    claims 100,000 Iraqi civilians
    have been killed since the US-
    led invasion, Jack Straw says." + ], + [ + "Virginia Tech scores 24 points
    off four first-half turnovers
    in a 55-6 wipeout of Maryland
    on Thursday to remain alone
    atop the ACC." + ], + [ + "Copernic Unleashes Desktop
    Search Tool\\\\Copernic
    Technologies Inc. today
    announced Copernic Desktop
    Search(TM) (CDS(TM)), \"The
    Search Engine For Your PC
    (TM).\" Copernic has used the
    experience gained from over 30
    million downloads of its
    Windows-based Web search
    software to develop CDS, a
    desktop search product that
    users are saying is far ..." + ], + [ + "The DVD Forum moved a step
    further toward the advent of
    HD DVD media and drives with
    the approval of key physical
    specifications at a meeting of
    the organisations steering
    committee last week." + ], + [ + "Eton College and Clarence
    House joined forces yesterday
    to deny allegations due to be
    made at an employment tribunal
    today by a former art teacher
    that she improperly helped
    Prince Harry secure an A-level
    pass in art two years ago." + ], + [ + "AFP - Great Britain's chances
    of qualifying for the World
    Group of the Davis Cup were
    evaporating rapidly after
    Austria moved into a 2-1 lead
    following the doubles." + ], + [ + "AP - Martina Navratilova's
    long, illustrious career will
    end without an Olympic medal.
    The 47-year-old Navratilova
    and Lisa Raymond lost 6-4,
    4-6, 6-4 on Thursday night to
    fifth-seeded Shinobu Asagoe
    and Ai Sugiyama of Japan in
    the quarterfinals, one step
    shy of the medal round." + ], + [ + "Often pigeonholed as just a
    seller of televisions and DVD
    players, Royal Philips
    Electronics said third-quarter
    profit surged despite a slide
    into the red by its consumer
    electronics division." + ], + [ + "AP - Google, the Internet
    search engine, has done
    something that law enforcement
    officials and their computer
    tools could not: Identify a
    man who died in an apparent
    hit-and-run accident 11 years
    ago in this small town outside
    Yakima." + ], + [ + "We are all used to IE getting
    a monthly new security bug
    found, but Winamp? In fact,
    this is not the first security
    flaw found in the application." + ], + [ + "The Apache Software Foundation
    and the Debian Project said
    they won't support the Sender
    ID e-mail authentication
    standard in their products." + ], + [ + "USATODAY.com - The newly
    restored THX 1138 arrives on
    DVD today with Star Wars
    creator George Lucas' vision
    of a Brave New World-like
    future." + ], + [ + "Office Depot Inc. (ODP.N:
    Quote, Profile, Research) on
    Tuesday warned of weaker-than-
    expected profits for the rest
    of the year because of
    disruptions from the string of
    hurricanes" + ], + [ + "THE photo-equipment maker
    Kodak yesterday announced
    plans to axe 600 jobs in the
    UK and close a factory in
    Nottinghamshire, in a move in
    line with the giants global
    restructuring strategy
    unveiled last January." + ], + [ + "The chances of scientists
    making any one of five
    discoveries by 2010 have been
    hugely underestimated,
    according to bookmakers." + ], + [ + "Asia-Pacific leaders meet in
    Australia to discuss how to
    keep nuclear weapons out of
    the hands of extremists." + ], + [ + " TALL AFAR, Iraq -- A three-
    foot-high coil of razor wire,
    21-ton armored vehicles and
    American soldiers with black
    M-4 assault rifles stood
    between tens of thousands of
    people and their homes last
    week." + ], + [ + "PSV Eindhoven re-established
    their five-point lead at the
    top of the Dutch Eredivisie
    today with a 2-0 win at
    Vitesse Arnhem. Former
    Sheffield Wednesday striker
    Gerald Sibon put PSV ahead in
    the 56th minute" + ], + [ + "China's central bank on
    Thursday raised interest rates
    for the first time in nearly a
    decade, signaling deepening
    unease with the breakneck pace
    of development and an intent
    to reign in a construction
    boom now sowing fears of
    runaway inflation." + ], + [ + "Deepening its commitment to
    help corporate users create
    SOAs (service-oriented
    architectures) through the use
    of Web services, IBM's Global
    Services unit on Thursday
    announced the formation of an
    SOA Management Practice." + ], + [ + "TODAY AUTO RACING 3 p.m. --
    NASCAR Nextel Cup Sylvania 300
    qualifying at N.H.
    International Speedway,
    Loudon, N.H., TNT PRO BASEBALL
    7 p.m. -- Red Sox at New York
    Yankees, Ch. 38, WEEI (850)
    (on cable systems where Ch. 38
    is not available, the game
    will air on NESN); Chicago
    Cubs at Cincinnati, ESPN 6
    p.m. -- Eastern League finals:
    ..." + ], + [ + "MUMBAI, SEPTEMBER 21: The
    Board of Control for Cricket
    in India (BCCI) today informed
    the Bombay High Court that it
    was cancelling the entire
    tender process regarding
    cricket telecast rights as
    also the conditional deal with
    Zee TV." + ], + [ + "CHICAGO - Delta Air Lines
    (DAL) said Wednesday it plans
    to eliminate between 6,000 and
    6,900 jobs during the next 18
    months, implement a 10 across-
    the-board pay reduction and
    reduce employee benefits." + ], + [ + "LAKE GEORGE, N.Y. - Even
    though he's facing double hip
    replacement surgery, Bill
    Smith is more than happy to
    struggle out the door each
    morning, limp past his brand
    new P.T..." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks were likely to open
    flat on Wednesday, with high
    oil prices and profit warnings
    weighing on the market before
    earnings reports start and key
    jobs data is released this
    week." + ], + [ + "Best known for its popular
    search engine, Google is
    rapidly rolling out new
    products and muscling into
    Microsoft's stronghold: the
    computer desktop. The
    competition means consumers
    get more choices and better
    products." + ], + [ + "Toshiba Corp. #39;s new
    desktop-replacement multimedia
    notebooks, introduced on
    Tuesday, are further evidence
    that US consumers still have
    yet to embrace the mobility
    offered by Intel Corp." + ], + [ + "JEJU, South Korea : Grace Park
    of South Korea won an LPGA
    Tour tournament, firing a
    seven-under-par 65 in the
    final round of the
    1.35-million dollar CJ Nine
    Bridges Classic." + ], + [ + " JERUSALEM (Reuters) - Israeli
    Prime Minister Ariel Sharon
    poured cold water on Tuesday
    on recent international
    efforts to restart stalled
    peace talks with Syria, saying
    there was \"no possibility\" of
    returning to previous
    discussions." + ], + [ + "Dutch smugness was slapped
    hard during the past
    fortnight. The rude awakening
    began with the barbaric
    slaying of controversial
    filmmaker Theo van Gogh on
    November 2. Then followed a
    reciprocal cycle of some" + ], + [ + "AP - The NHL will lock out its
    players Thursday, starting a
    work stoppage that threatens
    to keep the sport off the ice
    for the entire 2004-05 season." + ], + [ + " MOSCOW (Reuters) - Russia's
    Gazprom said on Tuesday it
    will bid for embattled oil
    firm YUKOS' main unit next
    month, as the Kremlin seeks
    to turn the world's biggest
    gas producer into a major oil
    player." + ], + [ + "pee writes quot;A passenger
    on a commuter plane in
    northern Norway attacked both
    pilots and at least one
    passenger with an axe as the
    aircraft was coming in to
    land." + ], + [ + "Aregular Amazon customer,
    Yvette Thompson has found
    shopping online to be mostly
    convenient and trouble-free.
    But last month, after ordering
    two CDs on Amazon.com, the
    Silver Spring reader
    discovered on her bank
    statement that she was double-
    charged for the \\$26.98 order.
    And there was a \\$25 charge
    that was a mystery." + ], + [ + "Prime Minister Ariel Sharon
    pledged Sunday to escalate a
    broad Israeli offensive in
    northern Gaza, saying troops
    will remain until Palestinian
    rocket attacks are halted.
    Israeli officials said the
    offensive -- in which 58
    Palestinians and three
    Israelis have been killed --
    will help clear the way for an
    Israeli withdrawal." + ], + [ + "Federal Reserve officials
    raised a key short-term
    interest rate Tuesday for the
    fifth time this year, and
    indicated they will gradually
    move rates higher next year to
    keep inflation under control
    as the economy expands." + ], + [ + "Canadians are paying more to
    borrow money for homes, cars
    and other purchases today
    after a quarter-point
    interest-rate increase by the
    Bank of Canada yesterday was
    quickly matched by the
    chartered banks." + ], + [ + "NEW YORK - Wall Street
    professionals know to keep
    their expectations in check in
    September, historically the
    worst month of the year for
    stocks. As summertime draws to
    a close, money managers are
    getting back to business,
    cleaning house, and often
    sending the market lower in
    the process..." + ], + [ + "A group linked to al Qaeda
    ally Abu Musab al-Zarqawi said
    it had tried to kill Iraq
    #39;s environment minister on
    Tuesday and warned it would
    not miss next time, according
    to an Internet statement." + ], + [ + "The Israeli military killed
    four Palestinian militants on
    Wednesday as troops in tanks
    and armored vehicles pushed
    into another town in the
    northern Gaza Strip, extending" + ], + [ + "When Paula Radcliffe dropped
    out of the Olympic marathon
    miles from the finish, she
    sobbed uncontrollably.
    Margaret Okayo knew the
    feeling." + ], + [ + "Delta Air Lines is to issue
    millions of new shares without
    shareholder consent as part of
    moves to ensure its survival." + ], + [ + "First baseman Richie Sexson
    agrees to a four-year contract
    with the Seattle Mariners on
    Wednesday." + ], + [ + "KIRKUK, Iraq - A suicide
    attacker detonated a car bomb
    Saturday outside a police
    academy in the northern Iraqi
    city of Kirkuk as hundreds of
    trainees and civilians were
    leaving for the day, killing
    at least 20 people and
    wounding 36, authorities said.
    Separately, U.S and Iraqi
    forces clashed with insurgents
    in another part of northern
    Iraq after launching an
    operation to destroy an
    alleged militant cell in the
    town of Tal Afar, the U.S..." + ], + [ + "Genta (GNTA:Nasdaq - news -
    research) is never boring!
    Monday night, the company
    announced that its phase III
    Genasense study in chronic
    lymphocytic leukemia (CLL) met
    its primary endpoint, which
    was tumor shrinkage." + ], + [ + "Finnish mobile giant Nokia has
    described its new Preminet
    solution, which it launched
    Monday (Oct. 25), as a
    quot;major worldwide
    initiative." + ], + [ + "While the entire airline
    industry #39;s finances are
    under water, ATA Airlines will
    have to hold its breath longer
    than its competitors to keep
    from going belly up." + ], + [ + " SAN FRANCISCO (Reuters) - At
    virtually every turn, Intel
    Corp. executives are heaping
    praise on an emerging long-
    range wireless technology
    known as WiMAX, which can
    blanket entire cities with
    high-speed Internet access." + ], + [ + "One day after ousting its
    chief executive, the nation's
    largest insurance broker said
    it will tell clients exactly
    how much they are paying for
    services and renounce back-
    door payments from carriers." + ], + [ + "LONDON (CBS.MW) -- Outlining
    an expectation for higher oil
    prices and increasing demand,
    Royal Dutch/Shell on Wednesday
    said it #39;s lifting project
    spending to \\$45 billion over
    the next three years." + ], + [ + "Tuesday #39;s meeting could
    hold clues to whether it
    #39;ll be a November or
    December pause in rate hikes.
    By Chris Isidore, CNN/Money
    senior writer." + ], + [ + "Phishing is one of the
    fastest-growing forms of
    personal fraud in the world.
    While consumers are the most
    obvious victims, the damage
    spreads far wider--hurting
    companies #39; finances and
    reputations and potentially" + ], + [ + "Reuters - The U.S. Interior
    Department on\\Friday gave
    final approval to a plan by
    ConocoPhillips and\\partner
    Anadarko Petroleum Corp. to
    develop five tracts around\\the
    oil-rich Alpine field on
    Alaska's North Slope." + ], + [ + "The dollar may fall against
    the euro for a third week in
    four on concern near-record
    crude oil prices will temper
    the pace of expansion in the
    US economy, a survey by
    Bloomberg News indicates." + ], + [ + "The battle for the British-
    based Chelsfield plc hotted up
    at the weekend, with reports
    from London that the property
    giant #39;s management was
    working on its own bid to
    thwart the 585 million (\\$A1." + ], + [ + "Atari has opened the initial
    sign-up phase of the closed
    beta for its Dungeons amp;
    Dragons real-time-strategy
    title, Dragonshard." + ], + [ + "AP - Many states are facing
    legal challenges over possible
    voting problems Nov. 2. A look
    at some of the developments
    Thursday:" + ], + [ + "Israeli troops withdrew from
    the southern Gaza Strip town
    of Khan Yunis on Tuesday
    morning, following a 30-hour
    operation that left 17
    Palestinians dead." + ], + [ + "Notes and quotes from various
    drivers following California
    Speedway #39;s Pop Secret 500.
    Jeff Gordon slipped to second
    in points following an engine
    failure while Jimmie Johnson
    moved back into first." + ], + [ + "PM-designate Omar Karameh
    forms a new 30-member cabinet
    which includes women for the
    first time." + ], + [ + "Bahrain #39;s king pardoned a
    human rights activist who
    convicted of inciting hatred
    of the government and
    sentenced to one year in
    prison Sunday in a case linked
    to criticism of the prime
    minister." + ], + [ + "Big Blue adds features, beefs
    up training efforts in China;
    rival Unisys debuts new line
    and pricing plan." + ], + [ + " MEMPHIS, Tenn. (Sports
    Network) - The Memphis
    Grizzlies signed All-Star
    forward Pau Gasol to a multi-
    year contract on Friday.
    Terms of the deal were not
    announced." + ], + [ + "Leaders from 38 Asian and
    European nations are gathering
    in Vietnam for a summit of the
    Asia-Europe Meeting, know as
    ASEM. One thousand delegates
    are to discuss global trade
    and regional politics during
    the two-day forum." + ], + [ + "A US soldier has pleaded
    guilty to murdering a wounded
    16-year-old Iraqi boy. Staff
    Sergeant Johnny Horne was
    convicted Friday of the
    unpremeditated murder" + ], + [ + "Andre Agassi brushed past
    Jonas Bjorkman 6-3 6-4 at the
    Stockholm Open on Thursday to
    set up a quarter-final meeting
    with Spanish sixth seed
    Fernando Verdasco." + ], + [ + "South Korea's Hynix
    Semiconductor Inc. and Swiss-
    based STMicroelectronics NV
    signed a joint-venture
    agreement on Tuesday to
    construct a memory chip
    manufacturing plant in Wuxi,
    about 100 kilometers west of
    Shanghai, in China." + ], + [ + "SAN DIEGO --(Business Wire)--
    Oct. 11, 2004 -- Breakthrough
    PeopleSoft EnterpriseOne 8.11
    Applications Enable
    Manufacturers to Become
    Demand-Driven." + ], + [ + "Reuters - Oil prices rose on
    Friday as tight\\supplies of
    distillate fuel, including
    heating oil, ahead of\\the
    northern hemisphere winter
    spurred buying." + ], + [ + "Well, Intel did say -
    dismissively of course - that
    wasn #39;t going to try to
    match AMD #39;s little dual-
    core Opteron demo coup of last
    week and show off a dual-core
    Xeon at the Intel Developer
    Forum this week and - as good
    as its word - it didn #39;t." + ], + [ + "Guinea-Bissau #39;s army chief
    of staff and former interim
    president, General Verissimo
    Correia Seabra, was killed
    Wednesday during unrest by
    mutinous soldiers in the
    former Portuguese" + ], + [ + "31 October 2004 -- Exit polls
    show that Prime Minister
    Viktor Yanukovich and
    challenger Viktor Yushchenko
    finished on top in Ukraine
    #39;s presidential election
    today and will face each other
    in a run-off next month." + ], + [ + "Nov. 18, 2004 - An FDA
    scientist says the FDA, which
    is charged with protecting
    America #39;s prescription
    drug supply, is incapable of
    doing so." + ], + [ + "Rock singer Bono pledges to
    spend the rest of his life
    trying to eradicate extreme
    poverty around the world." + ], + [ + "AP - Just when tourists
    thought it was safe to go back
    to the Princess Diana memorial
    fountain, the mud has struck." + ], + [ + "The UK's inflation rate fell
    in September, thanks in part
    to a fall in the price of
    airfares, increasing the
    chance that interest rates
    will be kept on hold." + ], + [ + " HONG KONG/SAN FRANCISCO
    (Reuters) - IBM is selling its
    PC-making business to China's
    largest personal computer
    company, Lenovo Group Ltd.,
    for \\$1.25 billion, marking
    the U.S. firm's retreat from
    an industry it helped pioneer
    in 1981." + ], + [ + "AP - Three times a week, The
    Associated Press picks an
    issue and asks President Bush
    and Democratic presidential
    candidate John Kerry a
    question about it. Today's
    question and their responses:" + ], + [ + " BOSTON (Reuters) - Boston was
    tingling with anticipation on
    Saturday as the Red Sox
    prepared to host Game One of
    the World Series against the
    St. Louis Cardinals and take a
    step toward ridding
    themselves of a hex that has
    hung over the team for eight
    decades." + ], + [ + "FOR the first time since his
    appointment as Newcastle
    United manager, Graeme Souness
    has been required to display
    the strong-arm disciplinary
    qualities that, to Newcastle
    directors, made" + ], + [ + "In an apparent damage control
    exercise, Russian President
    Vladimir Putin on Saturday
    said he favored veto rights
    for India as new permanent
    member of the UN Security
    Council." + ], + [ + "Nordstrom reported a strong
    second-quarter profit as it
    continued to select more
    relevant inventory and sell
    more items at full price." + ], + [ + "WHY IT HAPPENED Tom Coughlin
    won his first game as Giants
    coach and immediately
    announced a fine amnesty for
    all Giants. Just kidding." + ], + [ + "A second-place finish in his
    first tournament since getting
    married was good enough to
    take Tiger Woods from third to
    second in the world rankings." + ], + [ + " COLORADO SPRINGS, Colorado
    (Reuters) - World 400 meters
    champion Jerome Young has been
    given a lifetime ban from
    athletics for a second doping
    offense, the U.S. Anti-Doping
    Agency (USADA) said Wednesday." + ], + [ + "AP - Nigeria's Senate has
    ordered a subsidiary of
    petroleum giant Royal/Dutch
    Shell to pay a Nigerian ethnic
    group #36;1.5 billion for oil
    spills in their homelands, but
    the legislative body can't
    enforce the resolution, an
    official said Wednesday." + ], + [ + "IT #39;S BEEN a heck of an
    interesting two days here in
    Iceland. I #39;ve seen some
    interesting technology, heard
    some inventive speeches and
    met some people with different
    ideas." + ], + [ + "The Bank of England is set to
    keep interest rates on hold
    following the latest meeting
    of the its Monetary Policy
    Committee." + ], + [ + "Australian troops in Baghdad
    came under attack today for
    the first time since the end
    of the Iraq war when a car
    bomb exploded injuring three
    soldiers and damaging an
    Australian armoured convoy." + ], + [ + "The Bush administration upheld
    yesterday the imposition of
    penalty tariffs on shrimp
    imports from China and
    Vietnam, handing a victory to
    beleaguered US shrimp
    producers." + ], + [ + "House prices rose 0.2 percent
    in September compared with the
    month before to stand up 17.8
    percent on a year ago, the
    Nationwide Building Society
    says." + ], + [ + "Reuters - Two clients of
    Germany's Postbank\\(DPBGn.DE)
    fell for an e-mail fraud that
    led them to reveal\\money
    transfer codes to a bogus Web
    site -- the first case of\\this
    scam in German, prosecutors
    said on Thursday." + ], + [ + "US spending on information
    technology goods, services,
    and staff will grow seven
    percent in 2005 and continue
    at a similar pace through
    2008, according to a study
    released Monday by Forrester
    Research." + ], + [ + "LONDON - In two years, Arsenal
    will play their home matches
    in the Emirates stadium. That
    is what their new stadium at
    Ashburton Grove will be called
    after the Premiership
    champions yesterday agreed to
    the" + ], + [ + "KNOXVILLE, Tenn. -- Jason
    Campbell threw for 252 yards
    and two touchdowns, and No. 8
    Auburn proved itself as a
    national title contender by
    overwhelming No. 10 Tennessee,
    34-10, last night." + ], + [ + "Look, Ma, no hands! The U.S.
    space agency's latest
    spacecraft can run an entire
    mission by itself. By Amit
    Asaravala." + ], + [ + "Pakistans decision to refuse
    the International Atomic
    Energy Agency to have direct
    access to Dr AQ Khan is
    correct on both legal and
    political counts." + ], + [ + "MANILA, 4 December 2004 - With
    floods receding, rescuers
    raced to deliver food to
    famished survivors in
    northeastern Philippine
    villages isolated by back-to-
    back storms that left more
    than 650 people dead and
    almost 400 missing." + ], + [ + "Talks on where to build the
    world #39;s first nuclear
    fusion reactor ended without a
    deal on Tuesday but the
    European Union said Japan and
    the United States no longer
    firmly opposed its bid to put
    the plant in France." + ], + [ + "Joining America Online,
    EarthLink and Yahoo against
    spamming, Microsoft Corp.
    today announced the filing of
    three new anti-Spam lawsuits
    under the CAN-SPAM federal law
    as part of its initiative in
    solving the Spam problem for
    Internet users worldwide." + ], + [ + "WASHINGTON -- Another
    Revolution season concluded
    with an overtime elimination.
    Last night, the Revolution
    thrice rallied from deficits
    for a 3-3 tie with D.C. United
    in the Eastern Conference
    final, then lost in the first-
    ever Major League Soccer match
    decided by penalty kicks." + ], + [ + "Dwyane Wade calls himself a
    quot;sidekick, quot; gladly
    accepting the role Kobe Bryant
    never wanted in Los Angeles.
    And not only does second-year
    Heat point" + ], + [ + "A nationwide inspection shows
    Internet users are not as safe
    online as they believe. The
    inspections found most
    consumers have no firewall
    protection, outdated antivirus
    software and dozens of spyware
    programs secretly running on
    their computers." + ], + [ + "World number one golfer Vijay
    Singh of Fiji has captured his
    eighth PGA Tour event of the
    year with a win at the 84
    Lumber Classic in Farmington,
    Pennsylvania." + ], + [ + "The noise was deafening and
    potentially unsettling in the
    minutes before the start of
    the men #39;s Olympic
    200-meter final. The Olympic
    Stadium crowd chanted
    quot;Hellas!" + ], + [ + "CLEVELAND - The White House
    said Vice President Dick
    Cheney faces a \"master
    litigator\" when he debates
    Sen. John Edwards Tuesday
    night, a backhanded compliment
    issued as the Republican
    administration defended itself
    against criticism that it has
    not acknowledged errors in
    waging war in Iraq..." + ], + [ + "Brazilian forward Ronaldinho
    scored a sensational goal for
    Barcelona against Milan,
    making for a last-gasp 2-1.
    The 24-year-old #39;s
    brilliant left-foot hit at the
    Nou Camp Wednesday night sent
    Barcelona atop of Group F." + ], + [ + "Nortel Networks says it will
    again delay the release of its
    restated financial results.
    The Canadian telecom vendor
    originally promised to release
    the restated results in
    September." + ], + [ + " CHICAGO (Reuters) - At first
    glance, paying \\$13 or \\$14
    for a hard-wired Internet
    laptop connection in a hotel
    room might seem expensive." + ], + [ + "SEOUL (Reuters) - The chairman
    of South Korea #39;s ruling
    Uri Party resigned on Thursday
    after saying his father had
    served as a military police
    officer during Japan #39;s
    1910-1945 colonial rule on the
    peninsula." + ], + [ + "ALERE, Uganda -- Kasmiro
    Bongonyinge remembers sitting
    up suddenly in his bed. It was
    just after sunrise on a summer
    morning two years ago, and the
    old man, 87 years old and
    blind, knew something was
    wrong." + ], + [ + "Reuters - An investigation
    into U.S. insurers\\and brokers
    rattled insurance industry
    stocks for a second day\\on
    Friday as investors, shaken
    further by subpoenas
    delivered\\to the top U.S. life
    insurer, struggled to gauge
    how deep the\\probe might
    reach." + ], + [ + "Bee Staff Writer. SANTA CLARA
    - Andre Carter #39;s back
    injury has kept him out of the
    49ers #39; lineup since Week
    1. It #39;s also interfering
    with him rooting on his alma
    mater in person." + ], + [ + "AP - Kellen Winslow Jr. ended
    his second NFL holdout Friday." + ], + [ + "JAKARTA - Official results
    have confirmed former army
    general Susilo Bambang
    Yudhoyono as the winner of
    Indonesia #39;s first direct
    presidential election, while
    incumbent Megawati
    Sukarnoputri urged her nation
    Thursday to wait for the
    official announcement" + ], + [ + "HOUSTON - With only a few
    seconds left in a certain
    victory for Miami, Peyton
    Manning threw a meaningless
    6-yard touchdown pass to
    Marvin Harrison to cut the
    score to 24-15." + ], + [ + "Reuters - A ragged band of
    children\\emerges ghost-like
    from mists in Ethiopia's
    highlands,\\thrusting bunches
    of carrots at a car full of
    foreigners." + ], + [ + "DEADLY SCORER: Manchester
    United #39;s Wayne Rooney
    celebrating his three goals
    against Fenerbahce this week
    at Old Trafford. (AP)." + ], + [ + "AP - A U.N. human rights
    expert criticized the U.S.-led
    coalition forces in
    Afghanistan for violating
    international law by allegedly
    beating Afghans to death and
    forcing some to remove their
    clothes or wear hoods." + ], + [ + "You can feel the confidence
    level, not just with Team
    Canada but with all of Canada.
    There #39;s every expectation,
    from one end of the bench to
    the other, that Martin Brodeur
    is going to hold the fort." + ], + [ + "Heading into the first game of
    a new season, every team has
    question marks. But in 2004,
    the Denver Broncos seemed to
    have more than normal." + ], + [ + " JERUSALEM (Reuters) - Israeli
    Prime Minister Ariel Sharon
    said on Thursday Yasser
    Arafat's death could be a
    turning point for peacemaking
    but he would pursue a
    unilateral plan that would
    strip Palestinians of some
    land they want for a state." + ], + [ + " AL-ASAD AIRBASE, Iraq
    (Reuters) - Defense Secretary
    Donald Rumsfeld swept into an
    airbase in Iraq's western
    desert Sunday to make a
    first-hand evaluation of
    operations to quell a raging
    Iraqi insurgency in his first
    such visit in five months." + ], + [ + "More than three out of four
    (76 percent) consumers are
    experiencing an increase in
    spoofing and phishing
    incidents, and 35 percent
    receive fake e-mails at least
    once a week, according to a
    recent national study." + ], + [ + "The Dow Jones Industrial
    Average failed three times
    this year to exceed its
    previous high and fell to
    about 10,000 each time, most
    recently a week ago." + ], + [ + "AP - Deep in the Atlantic
    Ocean, undersea explorers are
    living a safer life thanks to
    germ-fighting clothing made in
    Kinston." + ], + [ + "Anaheim, Calif. - There is a
    decidedly right lean to the
    three-time champions of the
    American League Central. In a
    span of 26 days, the Minnesota
    Twins lost the left side of
    their infield to free agency." + ], + [ + "Computer Associates Monday
    announced the general
    availability of three
    Unicenter performance
    management products for
    mainframe data management." + ], + [ + "Reuters - The European Union
    approved on\\Wednesday the
    first biotech seeds for
    planting and sale across\\EU
    territory, flying in the face
    of widespread
    consumer\\resistance to
    genetically modified (GMO)
    crops and foods." + ], + [ + "With the NFL trading deadline
    set for 4 p.m. Tuesday,
    Patriots coach Bill Belichick
    said there didn't seem to be
    much happening on the trade
    front around the league." + ], + [ + "WASHINGTON - Democrat John
    Kerry accused President Bush
    on Monday of sending U.S.
    troops to the \"wrong war in
    the wrong place at the wrong
    time\" and said he'd try to
    bring them all home in four
    years..." + ], + [ + " SINGAPORE (Reuters) - Asian
    share markets staged a broad-
    based retreat on Wednesday,
    led by steelmakers amid
    warnings of price declines,
    but also enveloping technology
    and financial stocks on
    worries that earnings may
    disappoint." + ], + [ + "p2pnet.net News:- A Microsoft
    UK quot;WEIGHING THE COST OF
    LINUX VS. WINDOWS? LET #39;S
    REVIEW THE FACTS quot;
    magazine ad has been nailed as
    misleading by Britain #39;s
    Advertising Standards
    Authority (ASA)." + ], + [ + "More lorry drivers are
    bringing supplies to Nepal's
    capital in defiance of an
    indefinite blockade by Maoist
    rebels." + ], + [ + "NEW YORK - CNN has a new boss
    for the second time in 14
    months: former CBS News
    executive Jonathan Klein, who
    will oversee programming and
    editorial direction at the
    second-ranked cable news
    network." + ], + [ + "Cut-price carrier Virgin Blue
    said Tuesday the cost of using
    Australian airports is
    spiraling upward and asked the
    government to review the
    deregulated system of charges." + ], + [ + "The retail sector overall may
    be reporting a sluggish start
    to the season, but holiday
    shoppers are scooping up tech
    goods at a brisk pace -- and
    they're scouring the Web for
    bargains more than ever.
    <FONT face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\" color=\"#666666\">&
    lt;B>-washingtonpost.com<
    ;/B></FONT>" + ], + [ + "AP - David Beckham broke his
    rib moments after scoring
    England's second goal in
    Saturday's 2-0 win over Wales
    in a World Cup qualifying
    game." + ], + [ + "Saudi Arabia, Kuwait and the
    United Arab Emirates, which
    account for almost half of
    OPEC #39;s oil output, said
    they #39;re committed to
    boosting capacity to meet
    soaring demand that has driven
    prices to a record." + ], + [ + "The US Commerce Department
    said Thursday personal income
    posted its biggest increase in
    three months in August. The
    government agency also said
    personal spending was
    unchanged after rising
    strongly in July." + ], + [ + " TOKYO (Reuters) - Tokyo's
    Nikkei average opened up 0.54
    percent on Monday with banks
    and exporters leading the way
    as a stronger finish on Wall
    Street and declining oil
    prices soothed worries over
    the global economic outlook." + ], + [ + " BEIJING (Reuters) - Floods
    and landslides have killed 76
    people in southwest China in
    the past four days and washed
    away homes and roads, knocked
    down power lines and cut off
    at least one city, state
    media said on Monday." + ], + [ + "Nothing changed at the top of
    Serie A as all top teams won
    their games to keep the
    distance between one another
    unaltered. Juventus came back
    from behind against Lazio to
    win thanks to another goal by
    Ibrahimovic" + ], + [ + "The team behind the Beagle 2
    mission has unveiled its
    design for a successor to the
    British Mars lander." + ], + [ + "Survey points to popularity in
    Europe, the Middle East and
    Asia of receivers that skip
    the pay TV and focus on free
    programming." + ], + [ + "RICHMOND, Va. Jeremy Mayfield
    won his first race in over
    four years, taking the
    Chevrolet 400 at Richmond
    International Raceway after
    leader Kurt Busch ran out of
    gas eight laps from the
    finish." + ], + [ + "AP - Victims of the Sept. 11
    attacks were mourned worldwide
    Saturday, but in the Middle
    East, amid sympathy for the
    dead, Arabs said Washington's
    support for Israel and the war
    on terror launched in the
    aftermath of the World Trade
    Center's collapse have only
    fueled anger and violence." + ], + [ + "Linux publisher Red Hat Inc.
    said Tuesday that information-
    technology consulting firm
    Unisys Corp. will begin
    offering a business version of
    the company #39;s open-source
    operating system on its
    servers." + ], + [ + "SEATTLE - Ichiro Suzuki set
    the major league record for
    hits in a season with 258,
    breaking George Sisler's
    84-year-old mark with a pair
    of singles Friday night. The
    Seattle star chopped a leadoff
    single in the first inning,
    then made history with a
    grounder up the middle in the
    third..." + ], + [ + "The intruder who entered
    British Queen Elizabeth II
    #39;s official Scottish
    residence and caused a
    security scare was a reporter
    from the London-based Sunday
    Times newspaper, local media
    reported Friday." + ], + [ + "IBM's p5-575, a specialized
    server geared for high-
    performance computing, has
    eight 1.9GHz Power5
    processors." + ], + [ + "Bruce Wasserstein, head of
    Lazard LLC, is asking partners
    to take a one-third pay cut as
    he readies the world #39;s
    largest closely held
    investment bank for a share
    sale, people familiar with the
    situation said." + ], + [ + "Canadian Press - FREDERICTON
    (CP) - A New Brunswick truck
    driver arrested in Ontario
    this week has been accused by
    police of stealing 50,000 cans
    of Moosehead beer." + ], + [ + "Reuters - British police said
    on Monday they had\\charged a
    man with sending hoax emails
    to relatives of people\\missing
    since the Asian tsunami,
    saying their loved ones
    had\\been confirmed dead." + ], + [ + "The Lemon Bay Manta Rays were
    not going to let a hurricane
    get in the way of football. On
    Friday, they headed to the
    practice field for the first
    time in eight" + ], + [ + "Microsoft Corp. Chairman Bill
    Gates has donated \\$400,000 to
    a campaign in California
    trying to win approval of a
    measure calling for the state
    to sell \\$3 billion in bonds
    to fund stem-cell research." + ], + [ + "AP - Track star Marion Jones
    filed a defamation lawsuit
    Wednesday against the man
    whose company is at the center
    of a federal investigation
    into illegal steroid use among
    some of the nation's top
    athletes." + ], + [ + "LOS ANGELES - On Sept. 1,
    former secretary of
    Agriculture Dan Glickman
    replaced the legendary Jack
    Valenti as president and CEO
    of Hollywood #39;s trade
    group, the Motion Picture
    Association of America." + ], + [ + "England #39;s players hit out
    at cricket #39;s authorities
    tonight and claimed they had
    been used as quot;political
    pawns quot; after the Zimbabwe
    government produced a
    spectacular U-turn to ensure
    the controversial one-day
    series will go ahead." + ], + [ + "Newspaper publisher Pulitzer
    Inc. said Sunday that company
    officials are considering a
    possible sale of the firm to
    boost shareholder value." + ], + [ + "Shares of Merck amp; Co.
    plunged almost 10 percent
    yesterday after a media report
    said that documents show the
    pharmaceutical giant hid or
    denied" + ], + [ + "AP - The Japanese won the
    pregame home run derby. Then
    the game started and the major
    league All-Stars put their
    bats to work. Back-to-back
    home runs by Moises Alou and
    Vernon Wells in the fourth
    inning and by Johnny Estrada
    and Brad Wilkerson in the
    ninth powered the major
    leaguers past the Japanese
    stars 7-3 Sunday for a 3-0
    lead in the eight-game series." + ], + [ + "Reuters - Wall Street was
    expected to dip at\\Thursday's
    opening, but shares of Texas
    Instruments Inc.\\, may climb
    after it gave upbeat earnings
    guidance." + ], + [ + "Chinese authorities detained a
    prominent, U.S.-based Buddhist
    leader in connection with his
    plans to reopen an ancient
    temple complex in the Chinese
    province of Inner Mongolia
    last week and have forced
    dozens of his American
    followers to leave the region,
    local officials said
    Wednesday." + ], + [ + "The director of the National
    Hurricane Center stays calm in
    the midst of a storm, but
    wants everyone in hurricane-
    prone areas to get the message
    from his media advisories:
    Respect the storm's power and
    make proper response plans." + ], + [ + "With Chelsea losing their
    unbeaten record and Manchester
    United failing yet again to
    win, William Hill now make
    Arsenal red-hot 2/5 favourites
    to retain the title." + ], + [ + "Late in August, Boeing #39;s
    top sales execs flew to
    Singapore for a crucial sales
    pitch. They were close to
    persuading Singapore Airlines,
    one of the world #39;s leading
    airlines, to buy the American
    company #39;s new jet, the
    mid-sized 7E7." + ], + [ + "SBC Communications and
    BellSouth will acquire
    YellowPages.com with the goal
    of building the site into a
    nationwide online business
    index, the companies said
    Thursday." + ], + [ + "Theresa special bookcase in Al
    Grohs office completely full
    of game plans from his days in
    the NFL. Green ones are from
    the Jets." + ], + [ + "SAN FRANCISCO Several
    California cities and
    counties, including Los
    Angeles and San Francisco, are
    suing Microsoft for what could
    amount to billions of dollars." + ], + [ + "Newcastle ensured their place
    as top seeds in Friday #39;s
    third round UEFA Cup draw
    after holding Sporting Lisbon
    to a 1-1 draw at St James #39;
    Park." + ], + [ + "Adorned with Turkish and EU
    flags, Turkey #39;s newspapers
    hailed Thursday an official EU
    report recommending the
    country start talks to join
    the bloc, while largely
    ignoring the stringent
    conditions attached to the
    announcement." + ], + [ + "Google plans to release a
    version of its desktop search
    tool for computers that run
    Apple Computer #39;s Mac
    operating system, Google #39;s
    chief executive, Eric Schmidt,
    said Friday." + ], + [ + "AMD : sicurezza e prestazioni
    ottimali con il nuovo
    processore mobile per notebook
    leggeri e sottili; Acer Inc.
    preme sull #39;acceleratore
    con il nuovo notebook a
    marchio Ferrari." + ], + [ + "The sounds of tinkling bells
    could be heard above the
    bustle of the Farmers Market
    on the Long Beach Promenade,
    leading shoppers to a row of
    bright red tin kettles dotting
    a pathway Friday." + ], + [ + "CBC SPORTS ONLINE - Bode
    Miller continued his
    impressive 2004-05 World Cup
    skiing season by winning a
    night slalom race in
    Sestriere, Italy on Monday." + ], + [ + "Firefox use around the world
    climbed 34 percent in the last
    month, according to a report
    published by Web analytics
    company WebSideStory Monday." + ], + [ + "If a plastic card that gives
    you credit for something you
    don't want isn't your idea of
    a great gift, you can put it
    up for sale or swap." + ], + [ + "WASHINGTON Aug. 17, 2004
    Scientists are planning to
    take the pulse of the planet
    and more in an effort to
    improve weather forecasts,
    predict energy needs months in
    advance, anticipate disease
    outbreaks and even tell
    fishermen where the catch will
    be ..." + ], + [ + "Damien Rhodes scored on a
    2-yard run in the second
    overtime, then Syracuse's
    defense stopped Pittsburgh on
    fourth and 1, sending the
    Orange to a 38-31 victory
    yesterday in Syracuse, N.Y." + ], + [ + "AP - Anthony Harris scored 18
    of his career-high 23 points
    in the second half to help
    Miami upset No. 19 Florida
    72-65 Saturday and give first-
    year coach Frank Haith his
    biggest victory." + ], + [ + "LONDON Santander Central
    Hispano of Spain looked
    certain to clinch its bid for
    the British mortgage lender
    Abbey National, after HBOS,
    Britain #39;s biggest home-
    loan company, said Wednesday
    it would not counterbid, and
    after the European Commission
    cleared" + ], + [ + "New communications technology
    could spawn future products.
    Could your purse tell you to
    bring an umbrella?" + ], + [ + " WASHINGTON (Reuters) - The
    Justice Department is
    investigating possible
    accounting fraud at Fannie
    Mae, bringing greater
    government scrutiny to bear on
    the mortgage finance company,
    already facing a parallel
    inquiry by the SEC, a source
    close to the matter said on
    Thursday." + ], + [ + "AP - The five cities looking
    to host the 2012 Summer Games
    submitted bids to the
    International Olympic
    Committee on Monday, entering
    the final stage of a long
    process in hopes of landing
    one of the biggest prizes in
    sports." + ], + [ + "SAP has won a \\$35 million
    contract to install its human
    resources software for the US
    Postal Service. The NetWeaver-
    based system will replace the
    Post Office #39;s current
    25-year-old legacy application" + ], + [ + "The FIA has already cancelled
    todays activities at Suzuka as
    Super Typhoon Ma-On heads
    towards the 5.807km circuit.
    Saturday practice has been
    cancelled altogether while
    pre-qualifying and final
    qualifying" + ], + [ + "Thailand's prime minister
    visits the southern town where
    scores of Muslims died in army
    custody after a rally." + ], + [ + "Indian industrial group Tata
    agrees to invest \\$2bn in
    Bangladesh, the biggest single
    deal agreed by a firm in the
    south Asian country." + ], + [ + "NewsFactor - For years,
    companies large and small have
    been convinced that if they
    want the sophisticated
    functionality of enterprise-
    class software like ERP and
    CRM systems, they must buy
    pre-packaged applications.
    And, to a large extent, that
    remains true." + ], + [ + "Following in the footsteps of
    the RIAA, the MPAA (Motion
    Picture Association of
    America) announced that they
    have began filing lawsuits
    against people who use peer-
    to-peer software to download
    copyrighted movies off the
    Internet." + ], + [ + " GRAND PRAIRIE, Texas
    (Reuters) - Betting on horses
    was banned in Texas until as
    recently as 1987. Times have
    changed rapidly since.
    Saturday, Lone Star Park race
    track hosts the \\$14 million
    Breeders Cup, global racing's
    end-of-season extravaganza." + ], + [ + "MacCentral - At a special
    music event featuring Bono and
    The Edge from rock group U2
    held on Tuesday, Apple took
    the wraps off the iPod Photo,
    a color iPod available in 40GB
    or 60GB storage capacities.
    The company also introduced
    the iPod U2, a special edition
    of Apple's 20GB player clad in
    black, equipped with a red
    Click Wheel and featuring
    engraved U2 band member
    signatures. The iPod Photo is
    available immediately, and
    Apple expects the iPod U2 to
    ship in mid-November." + ], + [ + "Beijing: At least 170 miners
    were trapped underground after
    a gas explosion on Sunday
    ignited a fire in a coalmine
    in north-west China #39;s
    Shaanxi province, reports
    said." + ], + [ + "The steel tubing company
    reports sharply higher
    earnings, but the stock is
    falling." + ], + [ + "It might be a stay of
    execution for Coach P, or it
    might just be a Christmas
    miracle come early. SU #39;s
    upset win over BC has given
    hope to the Orange playing in
    a post season Bowl game." + ], + [ + "PHIL Neville insists
    Manchester United don #39;t
    fear anyone in the Champions
    League last 16 and declared:
    quot;Bring on the Italians." + ], + [ + "Playboy Enterprises, the adult
    entertainment company, has
    announced plans to open a
    private members club in
    Shanghai even though the
    company #39;s flagship men
    #39;s magazine is still banned
    in China." + ], + [ + "Reuters - Oracle Corp is
    likely to win clearance\\from
    the European Commission for
    its hostile #36;7.7
    billion\\takeover of rival
    software firm PeopleSoft Inc.,
    a source close\\to the
    situation said on Friday." + ], + [ + "TORONTO (CP) - Earnings
    warnings from Celestica and
    Coca-Cola along with a
    slowdown in US industrial
    production sent stock markets
    lower Wednesday." + ], + [ + "IBM (Quote, Chart) said it
    would spend a quarter of a
    billion dollars over the next
    year and a half to grow its
    RFID (define) business." + ], + [ + "Eastman Kodak Co., the world
    #39;s largest maker of
    photographic film, said
    Wednesday it expects sales of
    digital products and services
    to grow at an annual rate of
    36 percent between 2003 and
    2007, above prior growth rate
    estimates of 26 percent
    between 2002" + ], + [ + "SAMARRA (Iraq): With renewe d
    wave of skirmishes between the
    Iraqi insurgents and the US-
    led coalition marines, several
    people including top police
    officers were put to death on
    Saturday." + ], + [ + "SPACE.com - NASA released one
    of the best pictures ever made
    of Saturn's moon Titan as the
    Cassini spacecraft begins a
    close-up inspection of the
    satellite today. Cassini is
    making the nearest flyby ever
    of the smog-shrouded moon." + ], + [ + "AFP - The Iraqi government
    plans to phase out slowly
    subsidies on basic products,
    such as oil and electricity,
    which comprise 50 percent of
    public spending, equal to 15
    billion dollars, the planning
    minister said." + ], + [ + "ANNAPOLIS ROYAL, NS - Nova
    Scotia Power officials
    continued to keep the sluice
    gates open at one of the
    utility #39;s hydroelectric
    plants Wednesday in hopes a
    wayward whale would leave the
    area and head for the open
    waters of the Bay of Fundy." + ], + [ + "TORONTO -- Toronto Raptors
    point guard Alvin Williams
    will miss the rest of the
    season after undergoing
    surgery on his right knee
    Monday." + ], + [ + "The federal agency that
    insures pension plans said
    that its deficit, already at
    the highest in its history,
    had doubled in its last fiscal
    year, to \\$23.3 billion." + ], + [ + "AFP - Like most US Latinos,
    members of the extended
    Rodriguez family say they will
    cast their votes for Democrat
    John Kerry in next month's
    presidential polls." + ], + [ + "A Milan judge on Tuesday opens
    hearings into whether to put
    on trial 32 executives and
    financial institutions over
    the collapse of international
    food group Parmalat in one of
    Europe #39;s biggest fraud
    cases." + ], + [ + "AP - Tennessee's two freshmen
    quarterbacks have Volunteers
    fans fantasizing about the
    next four years. Brent
    Schaeffer and Erik Ainge
    surprised many with the nearly
    seamless way they rotated
    throughout a 42-17 victory
    over UNLV on Sunday night." + ], + [ + "In fact, Larry Ellison
    compares himself to the
    warlord, according to
    PeopleSoft's former CEO,
    defending previous remarks he
    made." + ], + [ + "FALLUJAH, Iraq -- Four Iraqi
    fighters huddled in a trench,
    firing rocket-propelled
    grenades at Lieutenant Eric
    Gregory's Bradley Fighting
    Vehicle and the US tanks and
    Humvees that were lumbering
    through tight streets between
    boxlike beige houses." + ], + [ + "MADRID, Aug 18 (Reuters) -
    Portugal captain Luis Figo
    said on Wednesday he was
    taking an indefinite break
    from international football,
    but would not confirm whether
    his decision was final." + ], + [ + "The Bank of England on
    Thursday left its benchmark
    interest rate unchanged, at
    4.75 percent, as policy makers
    assessed whether borrowing
    costs, already the highest in
    the Group of Seven, are
    constraining consumer demand." + ], + [ + "AP - Several thousand
    Christians who packed a
    cathedral compound in the
    Egyptian capital hurled stones
    at riot police Wednesday to
    protest a woman's alleged
    forced conversion to Islam. At
    least 30 people were injured." + ], + [ + "A group of Saudi religious
    scholars have signed an open
    letter urging Iraqis to
    support jihad against US-led
    forces. quot;Fighting the
    occupiers is a duty for all
    those who are able, quot; they
    said in a statement posted on
    the internet at the weekend." + ], + [ + "Fashion retailers Austin Reed
    and Ted Baker have reported
    contrasting fortunes on the
    High Street. Austin Reed
    reported interim losses of 2." + ], + [ + "AP - Shaun Rogers is in the
    backfield as often as some
    running backs. Whether teams
    dare to block Detroit's star
    defensive tackle with one
    player or follow the trend of
    double-teaming him, he often
    rips through offensive lines
    with a rare combination of
    size, speed, strength and
    nimble footwork." + ], + [ + " NEW YORK (Reuters) - A
    federal judge on Friday
    approved Citigroup Inc.'s
    \\$2.6 billion settlement with
    WorldCom Inc. investors who
    lost billions when an
    accounting scandal plunged
    the telecommunications company
    into bankruptcy protection." + ], + [ + "The Lions and Eagles entered
    Sunday #39;s game at Ford
    Field in the same place --
    atop their respective
    divisions -- and with
    identical 2-0 records." + ], + [ + "An unspecified number of
    cochlear implants to help
    people with severe hearing
    loss are being recalled
    because they may malfunction
    due to ear moisture, the US
    Food and Drug Administration
    announced." + ], + [ + "Profits triple at McDonald's
    Japan after the fast-food
    chain starts selling larger
    burgers." + ], + [ + "After Marcos Moreno threw four
    more interceptions in last
    week's 14-13 overtime loss at
    N.C. A T, Bison Coach Ray
    Petty will start Antoine
    Hartfield against Norfolk
    State on Saturday." + ], + [ + "You can empty your pockets of
    change, take off your belt and
    shoes and stick your keys in
    the little tray. But if you've
    had radiation therapy
    recently, you still might set
    off Homeland Security alarms." + ], + [ + "Mountaineers retrieve three
    bodies believed to have been
    buried for 22 years on an
    Indian glacier." + ], + [ + "SEOUL, Oct 19 Asia Pulse -
    LG.Philips LCD Co.
    (KSE:034220), the world #39;s
    second-largest maker of liquid
    crystal display (LCD), said
    Tuesday it has developed the
    world #39;s largest organic
    light emitting diode" + ], + [ + "SOUTH WILLIAMSPORT, Pa. --
    Looking ahead to the US
    championship game almost cost
    Conejo Valley in the
    semifinals of the Little
    League World Series." + ], + [ + "The Cubs didn #39;t need to
    fly anywhere near Florida to
    be in the eye of the storm.
    For a team that is going on
    100 years since last winning a
    championship, the only thing
    they never are at a loss for
    is controversy." + ], + [ + "Security experts warn of
    banner ads with a bad attitude
    --and a link to malicious
    code. Also: Phishers, be gone." + ], + [ + "KETTERING, Ohio Oct. 12, 2004
    - Cincinnati Bengals defensive
    end Justin Smith pleaded not
    guilty to a driving under the
    influence charge." + ], + [ + "com October 15, 2004, 5:11 AM
    PT. Wood paneling and chrome
    made your dad #39;s station
    wagon look like a million
    bucks, and they might also be
    just the ticket for Microsoft
    #39;s fledgling" + ], + [ + "President Thabo Mbeki met with
    Ivory Coast Prime Minister
    Seydou Diarra for three hours
    yesterday as part of talks
    aimed at bringing peace to the
    conflict-wracked Ivory Coast." + ], + [ + "MINNEAPOLIS -- For much of the
    2004 season, Twins pitcher
    Johan Santana didn #39;t just
    beat opposing hitters. Often,
    he overwhelmed and owned them
    in impressive fashion." + ], + [ + "Britain #39;s inflation rate
    fell in August further below
    its 2.0 percent government-set
    upper limit target with
    clothing and footwear prices
    actually falling, official
    data showed on Tuesday." + ], + [ + " KATHMANDU (Reuters) - Nepal's
    Maoist rebels have
    temporarily suspended a
    crippling economic blockade of
    the capital from Wednesday,
    saying the move was in
    response to popular appeals." + ], + [ + "Reuters - An Algerian
    suspected of being a leader\\of
    the Madrid train bombers has
    been identified as one of
    seven\\people who blew
    themselves up in April to
    avoid arrest, Spain's\\Interior
    Ministry said on Friday." + ], + [ + "KABUL: An Afghan man was found
    guilty on Saturday of killing
    four journalists in 2001,
    including two from Reuters,
    and sentenced to death." + ], + [ + "Yasser Arafat, the leader for
    decades of a fight for
    Palestinian independence from
    Israel, has died at a military
    hospital in Paris, according
    to news reports." + ], + [ + " LONDON (Reuters) - European
    shares shrugged off a spike in
    the euro to a fresh all-time
    high Wednesday, with telecoms
    again leading the way higher
    after interim profits at
    Britain's mm02 beat
    expectations." + ], + [ + "WASHINGTON - Weighed down by
    high energy prices, the US
    economy grew slower than the
    government estimated in the
    April-June quarter, as higher
    oil prices limited consumer
    spending and contributed to a
    record trade deficit." + ], + [ + "CHICAGO United Airlines says
    it will need even more labor
    cuts than anticipated to get
    out of bankruptcy. United told
    a bankruptcy court judge in
    Chicago today that it intends
    to start talks with unions
    next month on a new round of
    cost savings." + ], + [ + " JABALYA, Gaza Strip (Reuters)
    - Israel pulled most of its
    forces out of the northern
    Gaza Strip Saturday after a
    four-day incursion it said
    was staged to halt Palestinian
    rocket attacks on southern
    Israeli towns." + ], + [ + "Computer Associates
    International yesterday
    reported a 6 increase in
    revenue during its second
    fiscal quarter, but posted a
    \\$94 million loss after paying
    to settle government
    investigations into the
    company, it said yesterday." + ], + [ + "THE Turkish embassy in Baghdad
    was investigating a television
    report that two Turkish
    hostages had been killed in
    Iraq, but no confirmation was
    available so far, a senior
    Turkish diplomat said today." + ], + [ + "Reuters - Thousands of
    supporters of
    Ukraine's\\opposition leader,
    Viktor Yushchenko, celebrated
    on the streets\\in the early
    hours on Monday after an exit
    poll showed him\\winner of a
    bitterly fought presidential
    election." + ], + [ + "LONDON : The United States
    faced rare criticism over
    human rights from close ally
    Britain, with an official
    British government report
    taking Washington to task over
    concerns about Iraq and the
    Guantanamo Bay jail." + ], + [ + "The University of California,
    Berkeley, has signed an
    agreement with the Samoan
    government to isolate, from a
    tree, the gene for a promising
    anti- Aids drug and to share
    any royalties from the sale of
    a gene-derived drug with the
    people of Samoa." + ], + [ + "At a charity auction in New
    Jersey last weekend, baseball
    memorabilia dealer Warren
    Heller was approached by a man
    with an unusual but topical
    request." + ], + [ + " TOKYO (Reuters) - Tokyo's
    Nikkei average jumped 2.5
    percent by mid-afternoon on
    Monday as semiconductor-
    related stocks such as
    Advantest Corp. mirrored a
    rally by their U.S. peers
    while banks and brokerages
    extended last week's gains." + ], + [ + "INTER Milan coach Roberto
    Mancini believes the club
    #39;s lavish (northern) summer
    signings will enable them to
    mount a serious Serie A
    challenge this season." + ], + [ + "LONDON - A bomb threat that
    mentioned Iraq forced a New
    York-bound Greek airliner to
    make an emergency landing
    Sunday at London's Stansted
    Airport escorted by military
    jets, authorities said. An
    airport spokeswoman said an
    Athens newspaper had received
    a phone call saying there was
    a bomb on board the Olympic
    Airlines plane..." + ], + [ + "Links to this week's topics
    from search engine forums
    across the web: New MSN Search
    Goes LIVE in Beta - Microsoft
    To Launch New Search Engine -
    Google Launches 'Google
    Advertising Professionals' -
    Organic vs Paid Traffic ROI? -
    Making Money With AdWords? -
    Link Building 101" + ], + [ + "AP - Brad Ott shot an 8-under
    64 on Sunday to win the
    Nationwide Tour's Price Cutter
    Charity Championship for his
    first Nationwide victory." + ], + [ + "AP - New York Jets running
    back Curtis Martin passed Eric
    Dickerson and Jerome Bettis on
    the NFL career rushing list
    Sunday against the St. Louis
    Rams, moving to fourth all-
    time." + ], + [ + "Eight conservation groups are
    fighting the US government
    over a plan to poison
    thousands of prairie dogs in
    the grasslands of South
    Dakota, saying wildlife should
    not take a backseat to
    ranching interests." + ], + [ + "ATHENS, Greece - Sheryl
    Swoopes made three big plays
    at the end - two baskets and
    another on defense - to help
    the United States squeeze out
    a 66-62 semifinal victory over
    Russia on Friday. Now, only
    one game stands between the
    U.S..." + ], + [ + "Instead of standing for ante
    meridian and post meridian,
    though, fans will remember the
    time periods of pre-Mia and
    after-Mia. After playing for
    18 years and shattering nearly
    every record" + ], + [ + "General Motors (GM) plans to
    announce a massive
    restructuring Thursday that
    will eliminate as many as
    12,000 jobs in Europe in a
    move to stem the five-year
    flow of red ink from its auto
    operations in the region." + ], + [ + "Scientists are developing a
    device which could improve the
    lives of kidney dialysis
    patients." + ], + [ + "KABUL, Afghanistan The Afghan
    government is blaming drug
    smugglers for yesterday #39;s
    attack on the leading vice
    presidential candidate ." + ], + [ + "Stephon Marbury, concerned
    about his lousy shooting in
    Athens, used an off day to go
    to the gym and work on his
    shot. By finding his range, he
    saved the United States #39;
    hopes for a basketball gold
    medal." + ], + [ + " LONDON (Reuters) - Oil prices
    held firm on Wednesday as
    Hurricane Ivan closed off
    crude output and shut
    refineries in the Gulf of
    Mexico, while OPEC's Gulf
    producers tried to reassure
    traders by recommending an
    output hike." + ], + [ + "State-owned, running a
    monopoly on imports of jet
    fuel to China #39;s fast-
    growing aviation industry and
    a prized member of Singapore
    #39;s Stock Exchange." + ], + [ + "Google has won a trade mark
    dispute, with a District Court
    judge finding that the search
    engines sale of sponsored
    search terms Geico and Geico
    Direct did not breach car
    insurance firm GEICOs rights
    in the trade marked terms." + ], + [ + "Wall Street bounded higher for
    the second straight day
    yesterday as investors reveled
    in sharply falling oil prices
    and the probusiness agenda of
    the second Bush
    administration. The Dow Jones
    industrials gained more than
    177 points for its best day of
    2004, while the Standard amp;
    Poor's 500 closed at its
    highest level since early
    2002." + ], + [ + "Key factors help determine if
    outsourcing benefits or hurts
    Americans." + ], + [ + "The US Trade Representative on
    Monday rejected the European
    Union #39;s assertion that its
    ban on beef from hormone-
    treated cattle is now
    justified by science and that
    US and Canadian retaliatory
    sanctions should be lifted." + ], + [ + "One of the leading figures in
    the Greek Orthodox Church, the
    Patriarch of Alexandria Peter
    VII, has been killed in a
    helicopter crash in the Aegean
    Sea." + ], + [ + "Siding with chip makers,
    Microsoft said it won't charge
    double for its per-processor
    licenses when dual-core chips
    come to market next year." + ], + [ + "NEW YORK -- Wall Street's
    fourth-quarter rally gave
    stock mutual funds a solid
    performance for 2004, with
    small-cap equity funds and
    real estate funds scoring some
    of the biggest returns. Large-
    cap growth equities and
    technology-focused funds had
    the slimmest gains." + ], + [ + "CANBERRA, Australia -- The
    sweat-stained felt hats worn
    by Australian cowboys, as much
    a part of the Outback as
    kangaroos and sun-baked soil,
    may be heading for the history
    books. They fail modern
    industrial safety standards." + ], + [ + "Big Food Group Plc, the UK
    owner of the Iceland grocery
    chain, said second-quarter
    sales at stores open at least
    a year dropped 3.3 percent,
    the second consecutive
    decline, after competitors cut
    prices." + ], + [ + "A London-to-Washington flight
    is diverted after a security
    alert involving the singer
    formerly known as Cat Stevens." + ], + [ + " WASHINGTON (Reuters) - The
    first case of soybean rust has
    been found on the mainland
    United States and could affect
    U.S. crops for the near
    future, costing farmers
    millions of dollars, the
    Agriculture Department said on
    Wednesday." + ], + [ + "IBM and the Spanish government
    have introduced a new
    supercomputer they hope will
    be the most powerful in
    Europe, and one of the 10 most
    powerful in the world." + ], + [ + "The Supreme Court today
    overturned a five-figure
    damage award to an Alexandria
    man for a local auto dealer
    #39;s alleged loan scam,
    ruling that a Richmond-based
    federal appeals court had
    wrongly" + ], + [ + "AP - President Bush declared
    Friday that charges of voter
    fraud have cast doubt on the
    Ukrainian election, and warned
    that any European-negotiated
    pact on Iran's nuclear program
    must ensure the world can
    verify Tehran's compliance." + ], + [ + "TheSpaceShipOne team is handed
    the \\$10m cheque and trophy it
    won for claiming the Ansari
    X-Prize." + ], + [ + "Security officials have
    identified six of the
    militants who seized a school
    in southern Russia as being
    from Chechnya, drawing a
    strong connection to the
    Chechen insurgents who have
    been fighting Russian forces
    for years." + ], + [ + "AP - Randy Moss is expected to
    play a meaningful role for the
    Minnesota Vikings this weekend
    against the Giants, even
    without a fully healed right
    hamstring." + ], + [ + "Pros: Fits the recent profile
    (44, past PGA champion, fiery
    Ryder Cup player); the job is
    his if he wants it. Cons:
    Might be too young to be
    willing to burn two years of
    play on tour." + ], + [ + "SEOUL -- North Korea set three
    conditions yesterday to be met
    before it would consider
    returning to six-party talks
    on its nuclear programs." + ], + [ + "Official figures show the
    12-nation eurozone economy
    continues to grow, but there
    are warnings it may slow down
    later in the year." + ], + [ + "Elmer Santos scored in the
    second half, lifting East
    Boston to a 1-0 win over
    Brighton yesterday afternoon
    and giving the Jets an early
    leg up in what is shaping up
    to be a tight Boston City
    League race." + ], + [ + "In upholding a lower court
    #39;s ruling, the Supreme
    Court rejected arguments that
    the Do Not Call list violates
    telemarketers #39; First
    Amendment rights." + ], + [ + "US-backed Iraqi commandos were
    poised Friday to storm rebel
    strongholds in the northern
    city of Mosul, as US military
    commanders said they had
    quot;broken the back quot; of
    the insurgency with their
    assault on the former rebel
    bastion of Fallujah." + ], + [ + "Infineon Technologies, the
    second-largest chip maker in
    Europe, said Wednesday that it
    planned to invest about \\$1
    billion in a new factory in
    Malaysia to expand its
    automotive chip business and
    be closer to customers in the
    region." + ], + [ + "Mozilla's new web browser is
    smart, fast and user-friendly
    while offering a slew of
    advanced, customizable
    functions. By Michelle Delio." + ], + [ + "Saints special teams captain
    Steve Gleason expects to be
    fined by the league after
    being ejected from Sunday's
    game against the Carolina
    Panthers for throwing a punch." + ], + [ + "JERUSALEM (Reuters) - Prime
    Minister Ariel Sharon, facing
    a party mutiny over his plan
    to quit the Gaza Strip, has
    approved 1,000 more Israeli
    settler homes in the West Bank
    in a move that drew a cautious
    response on Tuesday from ..." + ], + [ + "Play has begun in the
    Australian Masters at
    Huntingdale in Melbourne with
    around half the field of 120
    players completing their first
    rounds." + ], + [ + " NEW YORK (Reuters) -
    Washington Post Co. <A HREF
    =\"http://www.investor.reuters.
    com/FullQuote.aspx?ticker=WPO.
    N target=/stocks/quickinfo/ful
    lquote\">WPO.N</A>
    said on Friday that quarterly
    profit jumped, beating
    analysts' forecasts, boosted
    by results at its Kaplan
    education unit and television
    broadcasting operations." + ], + [ + "GHAZNI, Afghanistan, 6 October
    2004 - Wartime security was
    rolled out for Afghanistans
    interim President Hamid Karzai
    as he addressed his first
    election campaign rally
    outside the capital yesterday
    amid spiraling violence." + ], + [ + "LOUISVILLE, Ky. - Louisville
    men #39;s basketball head
    coach Rick Pitino and senior
    forward Ellis Myles met with
    members of the media on Friday
    to preview the Cardinals #39;
    home game against rival
    Kentucky on Satursday." + ], + [ + "AP - Sounds like David
    Letterman is as big a \"Pops\"
    fan as most everyone else." + ], + [ + "originally offered on notebook
    PCs -- to its Opteron 32- and
    64-bit x86 processors for
    server applications. The
    technology will help servers
    to run" + ], + [ + "New orders for US-made durable
    goods increased 0.2pc in
    September, held back by a big
    drop in orders for
    transportation goods, the US
    Commerce Department said
    today." + ], + [ + "Siblings are the first ever to
    be convicted for sending
    boatloads of junk e-mail
    pushing bogus products. Also:
    Microsoft takes MSN music
    download on a Euro trip....
    Nokia begins legal battle
    against European
    counterparts.... and more." + ], + [ + "I always get a kick out of the
    annual list published by
    Forbes singling out the
    richest people in the country.
    It #39;s almost as amusing as
    those on the list bickering
    over their placement." + ], + [ + "MacCentral - After Apple
    unveiled the iMac G5 in Paris
    this week, Vice President of
    Hardware Product Marketing
    Greg Joswiak gave Macworld
    editors a guided tour of the
    desktop's new design. Among
    the topics of conversation:
    the iMac's cooling system, why
    pre-installed Bluetooth
    functionality and FireWire 800
    were left out, and how this
    new model fits in with Apple's
    objectives." + ], + [ + "Williams-Sonoma Inc., operator
    of home furnishing chains
    including Pottery Barn, said
    third-quarter earnings rose 19
    percent, boosted by store
    openings and catalog sales." + ], + [ + "We #39;ve known about
    quot;strained silicon quot;
    for a while--but now there
    #39;s a better way to do it.
    Straining silicon improves
    chip performance." + ], + [ + "This week, Sir Richard Branson
    announced his new company,
    Virgin Galactic, has the
    rights to the first commercial
    flights into space." + ], + [ + "71-inch HDTV comes with a home
    stereo system and components
    painted in 24-karat gold." + ], + [ + "Arsenal was held to a 1-1 tie
    by struggling West Bromwich
    Albion on Saturday, failing to
    pick up a Premier League
    victory when Rob Earnshaw
    scored with 11 minutes left." + ], + [ + "TOKYO - Mitsubishi Heavy
    Industries said today it #39;s
    in talks to buy a plot of land
    in central Japan #39;s Nagoya
    city from Mitsubishi Motors
    for building aircraft parts." + ], + [ + "China has confirmed that it
    found a deadly strain of bird
    flu in pigs as early as two
    years ago. China #39;s
    Agriculture Ministry said two
    cases had been discovered, but
    it did not say exactly where
    the samples had been taken." + ], + [ + "Baseball #39;s executive vice
    president Sandy Alderson
    insisted last month that the
    Cubs, disciplined for an
    assortment of run-ins with
    umpires, would not be targeted
    the rest of the season by
    umpires who might hold a
    grudge." + ], + [ + "As Superman and Batman would
    no doubt reflect during their
    cigarette breaks, the really
    draining thing about being a
    hero was that you have to keep
    riding to the rescue." + ], + [ + "MacCentral - RealNetworks Inc.
    said on Tuesday that it has
    sold more than a million songs
    at its online music store
    since slashing prices last
    week as part of a limited-time
    sale aimed at growing the user
    base of its new digital media
    software." + ], + [ + "With the presidential election
    less than six weeks away,
    activists and security experts
    are ratcheting up concern over
    the use of touch-screen
    machines to cast votes.
    <FONT face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\"\\ color=\"#666666\">
    <B>-washingtonpost.com&l
    t;/B></FONT>" + ], + [ + "NEW YORK, September 14 (New
    Ratings) - Yahoo! Inc
    (YHOO.NAS) has agreed to
    acquire Musicmatch Inc, a
    privately held digital music
    software company, for about
    \\$160 million in cash." + ], + [ + "Japan #39;s Sumitomo Mitsui
    Financial Group Inc. said
    Tuesday it proposed to UFJ
    Holdings Inc. that the two
    banks merge on an equal basis
    in its latest attempt to woo
    UFJ away from a rival suitor." + ], + [ + "Oil futures prices were little
    changed Thursday as traders
    anxiously watched for
    indications that the supply or
    demand picture would change in
    some way to add pressure to
    the market or take some away." + ], + [ + "Gov. Rod Blagojevich plans to
    propose a ban Thursday on the
    sale of violent and sexually
    explicit video games to
    minors, something other states
    have tried with little
    success." + ], + [ + " CHICAGO (Reuters) - Delta Air
    Lines Inc. <A HREF=\"http://
    www.investor.reuters.com/FullQ
    uote.aspx?ticker=DAL.N target=
    /stocks/quickinfo/fullquote\"&g
    t;DAL.N</A> said on
    Tuesday it will cut wages by
    10 percent and its chief
    executive will go unpaid for
    the rest of the year, but it
    still warned of bankruptcy
    within weeks unless more cuts
    are made." + ], + [ + "AP - Ten years after the Irish
    Republican Army's momentous
    cease-fire, negotiations
    resumed Wednesday in hope of
    reviving a Catholic-Protestant
    administration, an elusive
    goal of Northern Ireland's
    hard-fought peace process." + ], + [ + "A cable channel plans to
    resurrect each of the 1,230
    regular-season games listed on
    the league's defunct 2004-2005
    schedule by setting them in
    motion on a video game
    console." + ], + [ + " SANTO DOMINGO, Dominican
    Republic, Sept. 18 -- Tropical
    Storm Jeanne headed for the
    Bahamas on Saturday after an
    assault on the Dominican
    Republic that killed 10
    people, destroyed hundreds of
    houses and forced thousands
    from their homes." + ], + [ + "An explosion tore apart a car
    in Gaza City Monday, killing
    at least one person,
    Palestinian witnesses said.
    They said Israeli warplanes
    were circling overhead at the
    time of the blast, indicating
    a possible missile strike." + ], + [ + " WASHINGTON (Reuters) - A
    former Fannie Mae <A HREF=\"
    http://www.investor.reuters.co
    m/FullQuote.aspx?ticker=FNM.N
    target=/stocks/quickinfo/fullq
    uote\">FNM.N</A>
    employee who gave U.S.
    officials information about
    what he saw as accounting
    irregularities will not
    testify as planned before a
    congressional hearing next
    week, a House committee said
    on Friday." + ], + [ + "Beijing, Oct. 25 (PTI): China
    and the US today agreed to
    work jointly to re-energise
    the six-party talks mechanism
    aimed at dismantling North
    Korea #39;s nuclear programmes
    while Washington urged Beijing
    to resume" + ], + [ + "AFP - Sporadic gunfire and
    shelling took place overnight
    in the disputed Georgian
    region of South Ossetia in
    violation of a fragile
    ceasefire, wounding seven
    Georgian servicemen." + ], + [ + "PARIS, Nov 4 (AFP) - The
    European Aeronautic Defence
    and Space Company reported
    Thursday that its nine-month
    net profit more than doubled,
    thanks largely to sales of
    Airbus aircraft, and raised
    its full-year forecast." + ], + [ + "AP - Eric Hinske and Vernon
    Wells homered, and the Toronto
    Blue Jays completed a three-
    game sweep of the Baltimore
    Orioles with an 8-5 victory
    Sunday." + ], + [ + "SiliconValley.com - When
    \"Halo\" became a smash video
    game hit following Microsoft's
    launch of the Xbox console in
    2001, it was a no-brainer that
    there would be a sequel to the
    science fiction shoot-em-up." + ], + [ + "The number of people claiming
    unemployment benefit last
    month fell by 6,100 to
    830,200, according to the
    Office for National
    Statistics." + ], + [ + " NEW YORK (Reuters) - Todd
    Walker homered, had three hits
    and drove in four runs to lead
    the Chicago Cubs to a 12-5 win
    over the Cincinnati Reds in
    National League play at
    Wrigley Field on Monday." + ], + [ + "PARIS -- The city of Paris
    intends to reduce its
    dependence on software
    suppliers with \"de facto
    monopolies,\" but considers an
    immediate switch of its 17,000
    desktops to open source
    software too costly, it said
    Wednesday." + ], + [ + " FALLUJA, Iraq (Reuters) -
    U.S. forces hit Iraq's rebel
    stronghold of Falluja with the
    fiercest air and ground
    bombardment in months, as
    insurgents struck back on
    Saturday with attacks that
    killed up to 37 people in
    Samarra." + ], + [ + "MIAMI (Sports Network) -
    Shaquille O #39;Neal made his
    home debut, but once again it
    was Dwyane Wade stealing the
    show with 28 points as the
    Miami Heat downed the
    Cleveland Cavaliers, 92-86, in
    front of a record crowd at
    AmericanAirlines Arena." + ], + [ + "AP - The San Diego Chargers
    looked sharp #151; and played
    the same way. Wearing their
    powder-blue throwback jerseys
    and white helmets from the
    1960s, the Chargers did almost
    everything right in beating
    the Jacksonville Jaguars 34-21
    on Sunday." + ], + [ + "The vast majority of consumers
    are unaware that an Apple iPod
    digital music player only
    plays proprietary iTunes
    files, while a smaller
    majority agree that it is
    within RealNetworks #39;
    rights to develop a program
    that will make its music files
    compatible" + ], + [ + "Tyler airlines are gearing up
    for the beginning of holiday
    travel, as officials offer
    tips to help travelers secure
    tickets and pass through
    checkpoints with ease." + ], + [ + " NAJAF, Iraq (Reuters) - The
    fate of a radical Shi'ite
    rebellion in the holy city of
    Najaf was uncertain Friday
    amid disputed reports that
    Iraqi police had gained
    control of the Imam Ali
    Mosque." + ], + [ + " PROVIDENCE, R.I. (Reuters) -
    You change the oil in your car
    every 5,000 miles or so. You
    clean your house every week or
    two. Your PC needs regular
    maintenance as well --
    especially if you're using
    Windows and you spend a lot of
    time on the Internet." + ], + [ + "NERVES - no problem. That
    #39;s the verdict of Jose
    Mourinho today after his
    Chelsea side gave a resolute
    display of character at
    Highbury." + ], + [ + "AP - The latest low point in
    Ron Zook's tenure at Florida
    even has the coach wondering
    what went wrong. Meanwhile,
    Sylvester Croom's first big
    win at Mississippi State has
    given the Bulldogs and their
    fans a reason to believe in
    their first-year leader.
    Jerious Norwood's 37-yard
    touchdown run with 32 seconds
    remaining lifted the Bulldogs
    to a 38-31 upset of the 20th-
    ranked Gators on Saturday." + ], + [ + "A criminal trial scheduled to
    start Monday involving former
    Enron Corp. executives may
    shine a rare and potentially
    harsh spotlight on the inner
    workings" + ], + [ + "The Motley Fool - Here's
    something you don't see every
    day -- the continuing brouhaha
    between Oracle (Nasdaq: ORCL -
    News) and PeopleSoft (Nasdaq:
    PSFT - News) being a notable
    exception. South Africa's
    Harmony Gold Mining Company
    (NYSE: HMY - News) has
    announced a hostile takeover
    bid to acquire fellow South
    African miner Gold Fields
    Limited (NYSE: GFI - News).
    The transaction, if it takes
    place, would be an all-stock
    acquisition, with Harmony
    issuing 1.275 new shares in
    payment for each share of Gold
    Fields. The deal would value
    Gold Fields at more than
    #36;8 billion. ..." + ], + [ + "Someone forgot to inform the
    US Olympic basketball team
    that it was sent to Athens to
    try to win a gold medal, not
    to embarrass its country." + ], + [ + "SPACE.com - NASA's Mars \\rover
    Opportunity nbsp;will back its
    \\way out of a nbsp;crater it
    has spent four months
    exploring after reaching
    terrain nbsp;that appears \\too
    treacherous to tread. nbsp;" + ], + [ + "Sony Corp. announced Tuesday a
    new 20 gigabyte digital music
    player with MP3 support that
    will be available in Great
    Britain and Japan before
    Christmas and elsewhere in
    Europe in early 2005." + ], + [ + "Wal-Mart Stores Inc. #39;s
    Asda, the UK #39;s second
    biggest supermarket chain,
    surpassed Marks amp; Spencer
    Group Plc as Britain #39;s
    largest clothing retailer in
    the last three months,
    according to the Sunday
    Telegraph." + ], + [ + "Ten-man Paris St Germain
    clinched their seventh
    consecutive victory over arch-
    rivals Olympique Marseille
    with a 2-1 triumph in Ligue 1
    on Sunday thanks to a second-
    half winner by substitute
    Edouard Cisse." + ], + [ + "Until this week, only a few
    things about the strange,
    long-ago disappearance of
    Charles Robert Jenkins were
    known beyond a doubt. In the
    bitter cold of Jan. 5, 1965,
    the 24-year-old US Army
    sergeant was leading" + ], + [ + "Roy Oswalt wasn #39;t
    surprised to hear the Astros
    were flying Sunday night
    through the remnants of a
    tropical depression that
    dumped several inches of rain
    in Louisiana and could bring
    showers today in Atlanta." + ], + [ + "AP - This hardly seemed
    possible when Pitt needed
    frantic rallies to overcome
    Division I-AA Furman or Big
    East cellar dweller Temple. Or
    when the Panthers could barely
    move the ball against Ohio
    #151; not Ohio State, but Ohio
    U." + ], + [ + "Everyone is moaning about the
    fallout from last weekend but
    they keep on reporting the
    aftermath. #39;The fall-out
    from the so-called
    quot;Battle of Old Trafford
    quot; continues to settle over
    the nation and the debate" + ], + [ + "Oil supply concerns and broker
    downgrades of blue-chip
    companies left stocks mixed
    yesterday, raising doubts that
    Wall Street #39;s year-end
    rally would continue." + ], + [ + "Genentech Inc. said the
    marketing of Rituxan, a cancer
    drug that is the company #39;s
    best-selling product, is the
    subject of a US criminal
    investigation." + ], + [ + "American Lindsay Davenport
    regained the No. 1 ranking in
    the world for the first time
    since early 2002 by defeating
    Dinara Safina of Russia 6-4,
    6-2 in the second round of the
    Kremlin Cup on Thursday." + ], + [ + " The world's No. 2 soft drink
    company said on Thursday
    quarterly profit rose due to
    tax benefits." + ], + [ + "TOKYO (AP) -- The electronics
    and entertainment giant Sony
    Corp. (SNE) is talking with
    Wal-Mart Stores Inc..." + ], + [ + "After an unprecedented span of
    just five days, SpaceShipOne
    is ready for a return trip to
    space on Monday, its final
    flight to clinch a \\$10
    million prize." + ], + [ + "The United States on Tuesday
    modified slightly a threat of
    sanctions on Sudan #39;s oil
    industry in a revised text of
    its UN resolution on
    atrocities in the country
    #39;s Darfur region." + ], + [ + "Freshman Jeremy Ito kicked
    four field goals and Ryan
    Neill scored on a 31-yard
    interception return to lead
    improving Rutgers to a 19-14
    victory on Saturday over
    visiting Michigan State." + ], + [ + "Hi-tech monitoring of
    livestock at pig farms could
    help improve the animal growth
    process and reduce costs." + ], + [ + "Third-seeded Guillermo Canas
    defeated Guillermo Garcia-
    Lopez of Spain 7-6 (1), 6-3
    Monday on the first day of the
    Shanghai Open on Monday." + ], + [ + "AP - France intensified
    efforts Tuesday to save the
    lives of two journalists held
    hostage in Iraq, and the Arab
    League said the militants'
    deadline for France to revoke
    a ban on Islamic headscarves
    in schools had been extended." + ], + [ + "Cable amp; Wireless plc
    (NYSE: CWP - message board) is
    significantly ramping up its
    investment in local loop
    unbundling (LLU) in the UK,
    and it plans to spend up to 85
    million (\\$152." + ], + [ + "USATODAY.com - Personal
    finance software programs are
    the computer industry's
    version of veggies: Everyone
    knows they're good for you,
    but it's just hard to get
    anyone excited about them." + ], + [ + " NEW YORK (Reuters) - The
    dollar rebounded on Monday
    after last week's heavy
    selloff, but analysts were
    uncertain if the rally would
    hold after fresh economic data
    suggested the December U.S.
    jobs report due Friday might
    not live up to expectations." + ], + [ + "AFP - Microsoft said that it
    had launched a new desktop
    search tool that allows
    personal computer users to
    find documents or messages on
    their PCs." + ], + [ + "At least 12 people die in an
    explosion at a fuel pipeline
    on the outskirts of Nigeria's
    biggest city, Lagos." + ], + [ + "The three largest computer
    makers spearheaded a program
    today designed to standardize
    working conditions for their
    non-US workers." + ], + [ + "Description: Illinois Gov. Rod
    Blagojevich is backing state
    legislation that would ban
    sales or rentals of video
    games with graphic sexual or
    violent content to children
    under 18." + ], + [ + "Volkswagen demanded a two-year
    wage freeze for the
    170,000-strong workforce at
    Europe #39;s biggest car maker
    yesterday, provoking union
    warnings of imminent conflict
    at key pay and conditions
    negotiations." + ], + [ + " NEW YORK (Reuters) - U.S.
    stock futures pointed to a
    lower open on Wall Street on
    Thursday, extending the
    previous session's sharp
    fall, with rising energy
    prices feeding investor
    concerns about corporate
    profits and slower growth." + ], + [ + "But to play as feebly as it
    did for about 35 minutes last
    night in Game 1 of the WNBA
    Finals and lose by only four
    points -- on the road, no less
    -- has to be the best
    confidence builder since Cindy
    St." + ], + [ + "MILAN General Motors and Fiat
    on Wednesday edged closer to
    initiating a legal battle that
    could pit the two carmakers
    against each other in a New
    York City court room as early
    as next month." + ], + [ + "Are you bidding on keywords
    through Overture's Precision
    Match, Google's AdWords or
    another pay-for-placement
    service? If so, you're
    eligible to participate in
    their contextual advertising
    programs." + ], + [ + "Two of the Ford Motor Company
    #39;s most senior executives
    retired on Thursday in a sign
    that the company #39;s deep
    financial crisis has abated,
    though serious challenges
    remain." + ], + [ + "Citing security concerns, the
    U.S. Embassy on Thursday
    banned its employees from
    using the highway linking the
    embassy area to the
    international airport, a
    10-mile stretch of road
    plagued by frequent suicide
    car-bomb attacks." + ], + [ + "Nobel Laureate Wilkins, 87,
    played an important role in
    the discovery of the double
    helix structure of DNA, the
    molecule that carries our
    quot;life code quot;,
    Kazinform refers to BBC News." + ], + [ + "With yesterday #39;s report on
    its athletic department
    violations completed, the
    University of Washington says
    it is pleased to be able to
    move forward." + ], + [ + " LONDON (Reuters) - Wall
    Street was expected to start
    little changed on Friday as
    investors continue to fret
    over the impact of high oil
    prices on earnings, while
    Boeing <A HREF=\"http://www.
    investor.reuters.com/FullQuote
    .aspx?ticker=BA.N target=/stoc
    ks/quickinfo/fullquote\">BA.
    N</A> will be eyed
    after it reiterated its
    earnings forecast." + ], + [ + "AP - Tom Daschle bade his
    fellow Senate Democrats
    farewell Tuesday with a plea
    that they seek common ground
    with Republicans yet continue
    to fight for the less
    fortunate." + ], + [ + "Sammy Sosa was fined \\$87,400
    -- one day's salary -- for
    arriving late to the Cubs'
    regular-season finale at
    Wrigley Field and leaving the
    game early. The slugger's
    agent, Adam Katz , said
    yesterday Sosa most likely
    will file a grievance. Sosa
    arrived 70 minutes before
    Sunday's first pitch, and he
    apparently left 15 minutes
    after the game started without
    ..." + ], + [ + "Having an always-on, fast net
    connection is changing the way
    Britons use the internet,
    research suggests." + ], + [ + "AP - Police defused a bomb in
    a town near Prime Minister
    Silvio Berlusconi's villa on
    the island of Sardinia on
    Wednesday shortly after
    British Prime Minister Tony
    Blair finished a visit there
    with the Italian leader." + ], + [ + "Is the Oklahoma defense a
    notch below its predecessors?
    Is Texas #39; offense a step-
    ahead? Why is Texas Tech
    feeling good about itself
    despite its recent loss?" + ], + [ + "The coffin of Yasser Arafat,
    draped with the Palestinian
    flag, was bound for Ramallah
    in the West Bank Friday,
    following a formal funeral on
    a military compound near
    Cairo." + ], + [ + "US Ambassador to the United
    Nations John Danforth resigned
    on Thursday after serving in
    the post for less than six
    months. Danforth, 68, said in
    a letter released Thursday" + ], + [ + "Crude oil futures prices
    dropped below \\$51 a barrel
    yesterday as supply concerns
    ahead of the Northern
    Hemisphere winter eased after
    an unexpectedly high rise in
    US inventories." + ], + [ + "New York gets 57 combined
    points from its starting
    backcourt of Jamal Crawford
    and Stephon Marbury and tops
    Denver, 107-96." + ], + [ + "ISLAMABAD, Pakistan -- Photos
    were published yesterday in
    newspapers across Pakistan of
    six terror suspects, including
    a senior Al Qaeda operative,
    the government says were
    behind attempts to assassinate
    the nation's president." + ], + [ + "AP - Shawn Marion had a
    season-high 33 points and 15
    rebounds, leading the Phoenix
    Suns on a fourth-quarter
    comeback despite the absence
    of Steve Nash in a 95-86 win
    over the New Orleans Hornets
    on Friday night." + ], + [ + "By Lilly Vitorovich Of DOW
    JONES NEWSWIRES SYDNEY (Dow
    Jones)--Rupert Murdoch has
    seven weeks to convince News
    Corp. (NWS) shareholders a
    move to the US will make the
    media conglomerate more
    attractive to" + ], + [ + "A number of signs point to
    increasing demand for tech
    workers, but not all the
    clouds have been driven away." + ], + [ + "Messina upset defending
    champion AC Milan 2-1
    Wednesday, while Juventus won
    its third straight game to
    stay alone atop the Italian
    league standings." + ], + [ + "Microsoft Corp. (MSFT.O:
    Quote, Profile, Research)
    filed nine new lawsuits
    against spammers who send
    unsolicited e-mail, including
    an e-mail marketing Web
    hosting company, the world
    #39;s largest software maker
    said on Thursday." + ], + [ + "AP - Padraig Harrington
    rallied to a three-stroke
    victory in the German Masters
    on a windy Sunday, closing
    with a 2-under-par 70 and
    giving his game a big boost
    before the Ryder Cup." + ], + [ + " ATHENS (Reuters) - The Athens
    Paralympics canceled
    celebrations at its closing
    ceremony after seven
    schoolchildren traveling to
    watch the event died in a bus
    crash on Monday." + ], + [ + "The rocket plane SpaceShipOne
    is just one flight away from
    claiming the Ansari X-Prize, a
    \\$10m award designed to kick-
    start private space travel." + ], + [ + "Reuters - Three American
    scientists won the\\2004 Nobel
    physics prize on Tuesday for
    showing how tiny
    quark\\particles interact,
    helping to explain everything
    from how a\\coin spins to how
    the universe was built." + ], + [ + "Ironically it was the first
    regular season game for the
    Carolina Panthers that not
    only began the history of the
    franchise, but also saw the
    beginning of a rivalry that
    goes on to this day." + ], + [ + "Baltimore Ravens running back
    Jamal Lewis did not appear at
    his arraignment Friday, but
    his lawyers entered a not
    guilty plea on charges in an
    expanded drug conspiracy
    indictment." + ], + [ + "AP - Sharp Electronics Corp.
    plans to stop selling its
    Linux-based handheld computer
    in the United States, another
    sign of the slowing market for
    personal digital assistants." + ], + [ + "After serving a five-game
    suspension, Milton Bradley
    worked out with the Dodgers as
    they prepared for Tuesday's
    opener against the St. Louis
    Cardinals." + ], + [ + "AP - Prime Time won't be
    playing in prime time this
    time. Deion Sanders was on the
    inactive list and missed a
    chance to strut his stuff on
    \"Monday Night Football.\"" + ], + [ + "Reuters - Glaciers once held
    up by a floating\\ice shelf off
    Antarctica are now sliding off
    into the sea --\\and they are
    going fast, scientists said on
    Tuesday." + ], + [ + "DUBAI : An Islamist group has
    threatened to kill two Italian
    women held hostage in Iraq if
    Rome does not withdraw its
    troops from the war-torn
    country within 24 hours,
    according to an internet
    statement." + ], + [ + "AP - Warning lights flashed
    atop four police cars as the
    caravan wound its way up the
    driveway in a procession fit
    for a presidential candidate.
    At long last, Azy and Indah
    had arrived. They even flew
    through a hurricane to get
    here." + ], + [ + "The man who delivered the
    knockout punch was picked up
    from the Seattle scrap heap
    just after the All-Star Game.
    Before that, John Olerud
    certainly hadn't figured on
    facing Pedro Martinez in
    Yankee Stadium in October." + ], + [ + "\\Female undergraduates work
    harder and are more open-
    minded than males, leading to
    better results, say
    scientists." + ], + [ + "A heavy quake rocked Indonesia
    #39;s Papua province killing
    at least 11 people and
    wounding 75. The quake
    destroyed 150 buildings,
    including churches, mosques
    and schools." + ], + [ + "LONDON : A consortium,
    including former world
    champion Nigel Mansell, claims
    it has agreed terms to ensure
    Silverstone remains one of the
    venues for the 2005 Formula
    One world championship." + ], + [ + " BATON ROUGE, La. (Sports
    Network) - LSU has named Les
    Miles its new head football
    coach, replacing Nick Saban." + ], + [ + "The United Nations annual
    World Robotics Survey predicts
    the use of robots around the
    home will surge seven-fold by
    2007. The boom is expected to
    be seen in robots that can mow
    lawns and vacuum floors, among
    other chores." + ], + [ + "The long-term economic health
    of the United States is
    threatened by \\$53 trillion in
    government debts and
    liabilities that start to come
    due in four years when baby
    boomers begin to retire." + ], + [ + "Reuters - A small group of
    suspected\\gunmen stormed
    Uganda's Water Ministry
    Wednesday and took
    three\\people hostage to
    protest against proposals to
    allow President\\Yoweri
    Museveni for a third
    term.\\Police and soldiers with
    assault rifles cordoned off
    the\\three-story building, just
    328 feet from Uganda's
    parliament\\building in the
    capital Kampala." + ], + [ + "The Moscow Arbitration Court
    ruled on Monday that the YUKOS
    oil company must pay RUR
    39.113bn (about \\$1.34bn) as
    part of its back tax claim for
    2001." + ], + [ + "NOVEMBER 11, 2004 -- Bankrupt
    US Airways this morning said
    it had reached agreements with
    lenders and lessors to
    continue operating nearly all
    of its mainline and US Airways
    Express fleets." + ], + [ + "Venezuela suggested Friday
    that exiles living in Florida
    may have masterminded the
    assassination of a prosecutor
    investigating a short-lived
    coup against leftist President
    Hugo Chvez" + ], + [ + "Want to dive deep -- really
    deep -- into the technical
    literature about search
    engines? Here's a road map to
    some of the best web
    information retrieval
    resources available online." + ], + [ + "Reuters - Ancel Keys, a
    pioneer in public health\\best
    known for identifying the
    connection between
    a\\cholesterol-rich diet and
    heart disease, has died." + ], + [ + "The US government asks the
    World Trade Organisation to
    step in to stop EU member
    states from \"subsidising\"
    planemaker Airbus." + ], + [ + "Reuters - T-Mobile USA, the
    U.S. wireless unit\\of Deutsche
    Telekom AG (DTEGn.DE), does
    not expect to offer\\broadband
    mobile data services for at
    least the next two years,\\its
    chief executive said on
    Thursday." + ], + [ + "Verizon Communications is
    stepping further into video as
    a way to compete against cable
    companies." + ], + [ + "Facing a popular outcry at
    home and stern warnings from
    Europe, the Turkish government
    discreetly stepped back
    Tuesday from a plan to
    introduce a motion into a
    crucial penal reform bill to
    make adultery a crime
    punishable by prison." + ], + [ + "Boston Scientific Corp.
    (BSX.N: Quote, Profile,
    Research) said on Wednesday it
    received US regulatory
    approval for a device to treat
    complications that arise in
    patients with end-stage kidney
    disease who need dialysis." + ], + [ + "North-west Norfolk MP Henry
    Bellingham has called for the
    release of an old college
    friend accused of plotting a
    coup in Equatorial Guinea." + ], + [ + "With the economy slowly
    turning up, upgrading hardware
    has been on businesses radar
    in the past 12 months as their
    number two priority." + ], + [ + "AP - The Chicago Blackhawks
    re-signed goaltender Michael
    Leighton to a one-year
    contract Wednesday." + ], + [ + "Oracle Corp. plans to release
    the latest version of its CRM
    (customer relationship
    management) applications
    within the next two months, as
    part of an ongoing update of
    its E-Business Suite." + ], + [ + "Toyota Motor Corp. #39;s
    shares fell for a second day,
    after the world #39;s second-
    biggest automaker had an
    unexpected quarterly profit
    drop." + ], + [ + "AFP - Want to buy a castle?
    Head for the former East
    Germany." + ], + [ + "Hosted CRM service provider
    Salesforce.com took another
    step forward last week in its
    strategy to build an online
    ecosystem of vendors that
    offer software as a service." + ], + [ + "Britain-based HBOS says it
    will file a complaint to the
    European Commission against
    Spanish bank Santander Central
    Hispano (SCH) in connection
    with SCH #39;s bid to acquire
    British bank Abbey National" + ], + [ + "AFP - Steven Gerrard has moved
    to allay Liverpool fans' fears
    that he could be out until
    Christmas after breaking a
    metatarsal bone in his left
    foot." + ], + [ + "Verizon Wireless on Thursday
    announced an agreement to
    acquire all the PCS spectrum
    licenses of NextWave Telecom
    Inc. in 23 markets for \\$3
    billion." + ], + [ + "washingtonpost.com -
    Technology giants IBM and
    Hewlett-Packard are injecting
    hundreds of millions of
    dollars into radio-frequency
    identification technology,
    which aims to advance the
    tracking of items from ho-hum
    bar codes to smart tags packed
    with data." + ], + [ + "ATHENS -- She won her first
    Olympic gold medal in kayaking
    when she was 18, the youngest
    paddler to do so in Games
    history. Yesterday, at 42,
    Germany #39;s golden girl
    Birgit Fischer won her eighth
    Olympic gold in the four-woman
    500-metre kayak race." + ], + [ + "England boss Sven Goran
    Eriksson has defended
    goalkeeper David James after
    last night #39;s 2-2 draw in
    Austria. James allowed Andreas
    Ivanschitz #39;s shot to slip
    through his fingers to
    complete Austria comeback from
    two goals down." + ], + [ + "MINSK - Legislative elections
    in Belarus held at the same
    time as a referendum on
    whether President Alexander
    Lukashenko should be allowed
    to seek a third term fell
    significantly short of
    democratic standards, foreign
    observers said here Monday." + ], + [ + "An Olympic sailor is charged
    with the manslaughter of a
    Briton who died after being
    hit by a car in Athens." + ], + [ + "The Norfolk Broads are on
    their way to getting a clean
    bill of ecological health
    after a century of stagnation." + ], + [ + "AP - Secretary of State Colin
    Powell on Friday praised the
    peace deal that ended fighting
    in Iraq's holy city of Najaf
    and said the presence of U.S.
    forces in the area helped make
    it possible." + ], + [ + "The quot;future quot; is
    getting a chance to revive the
    presently struggling New York
    Giants. Two other teams also
    decided it was time for a
    change at quarterback, but the
    Buffalo Bills are not one of
    them." + ], + [ + "For the second time this year,
    an alliance of major Internet
    providers - including Atlanta-
    based EarthLink -iled a
    coordinated group of lawsuits
    aimed at stemming the flood of
    online junk mail." + ], + [ + " WASHINGTON (Reuters) - The
    PIMCO mutual fund group has
    agreed to pay \\$50 million to
    settle fraud charges involving
    improper rapid dealing in
    mutual fund shares, the U.S.
    Securities and Exchange
    Commission said on Monday." + ], + [ + "Via Technologies has released
    a version of the open-source
    Xine media player that is
    designed to take advantage of
    hardware digital video
    acceleration capabilities in
    two of the company #39;s PC
    chipsets, the CN400 and
    CLE266." + ], + [ + "The Conference Board reported
    Thursday that the Leading
    Economic Index fell for a
    third consecutive month in
    August, suggesting slower
    economic growth ahead amid
    rising oil prices." + ], + [ + " SAN FRANCISCO (Reuters) -
    Software maker Adobe Systems
    Inc.<A HREF=\"http://www.inv
    estor.reuters.com/FullQuote.as
    px?ticker=ADBE.O target=/stock
    s/quickinfo/fullquote\">ADBE
    .O</A> on Thursday
    posted a quarterly profit that
    rose more than one-third from
    a year ago, but shares fell 3
    percent after the maker of
    Photoshop and Acrobat software
    did not raise forecasts for
    fiscal 2005." + ], + [ + "William Morrison Supermarkets
    has agreed to sell 114 small
    Safeway stores and a
    distribution centre for 260.2
    million pounds. Morrison
    bought these stores as part of
    its 3 billion pound" + ], + [ + "SCO Group has a plan to keep
    itself fit enough to continue
    its legal battles against
    Linux and to develop its Unix-
    on-Intel operating systems." + ], + [ + "Flushing Meadows, NY (Sports
    Network) - The men #39;s
    semifinals at the 2004 US Open
    will be staged on Saturday,
    with three of the tournament
    #39;s top-five seeds ready for
    action at the USTA National
    Tennis Center." + ], + [ + "Pepsi pushes a blue version of
    Mountain Dew only at Taco
    Bell. Is this a winning
    strategy?" + ], + [ + "New software helps corporate
    travel managers track down
    business travelers who
    overspend. But it also poses a
    dilemma for honest travelers
    who are only trying to save
    money." + ], + [ + "NATO Secretary-General Jaap de
    Hoop Scheffer has called a
    meeting of NATO states and
    Russia on Tuesday to discuss
    the siege of a school by
    Chechen separatists in which
    more than 335 people died, a
    NATO spokesman said." + ], + [ + "26 August 2004 -- Iraq #39;s
    top Shi #39;ite cleric, Grand
    Ayatollah Ali al-Sistani,
    arrived in the city of Al-
    Najaf today in a bid to end a
    weeks-long conflict between US
    forces and militiamen loyal to
    Shi #39;ite cleric Muqtada al-
    Sadr." + ], + [ + "AFP - Senior executives at
    business software group
    PeopleSoft unanimously
    recommended that its
    shareholders reject a 8.8
    billion dollar takeover bid
    from Oracle Corp, PeopleSoft
    said in a statement Wednesday." + ], + [ + "Reuters - Neolithic people in
    China may have\\been the first
    in the world to make wine,
    according to\\scientists who
    have found the earliest
    evidence of winemaking\\from
    pottery shards dating from
    7,000 BC in northern China." + ], + [ + "Given nearly a week to examine
    the security issues raised by
    the now-infamous brawl between
    players and fans in Auburn
    Hills, Mich., Nov. 19, the
    Celtics returned to the
    FleetCenter last night with
    two losses and few concerns
    about their on-court safety." + ], + [ + " TOKYO (Reuters) - Electronics
    conglomerate Sony Corp.
    unveiled eight new flat-screen
    televisions on Thursday in a
    product push it hopes will
    help it secure a leading 35
    percent of the domestic
    market in the key month of
    December." + ], + [ + "As the election approaches,
    Congress abandons all pretense
    of fiscal responsibility,
    voting tax cuts that would
    drive 10-year deficits past
    \\$3 trillion." + ], + [ + "PARIS : French trade unions
    called on workers at France
    Telecom to stage a 24-hour
    strike September 7 to protest
    government plans to privatize
    the public telecommunications
    operator, union sources said." + ], + [ + "ServiceMaster profitably
    bundles services and pays a
    healthy 3.5 dividend." + ], + [ + "The Indonesian tourism
    industry has so far not been
    affected by last week #39;s
    bombing outside the Australian
    embassy in Jakarta and
    officials said they do not
    expect a significant drop in
    visitor numbers as a result of
    the attack." + ], + [ + "\\$222.5 million -- in an
    ongoing securities class
    action lawsuit against Enron
    Corp. The settlement,
    announced Friday and" + ], + [ + "Arsenals Thierry Henry today
    missed out on the European
    Footballer of the Year award
    as Andriy Shevchenko took the
    honour. AC Milan frontman
    Shevchenko held off
    competition from Barcelona
    pair Deco and" + ], + [ + "Donald Halsted, one target of
    a class-action suit alleging
    financial improprieties at
    bankrupt Polaroid, officially
    becomes CFO." + ], + [ + "Neil Mellor #39;s sensational
    late winner for Liverpool
    against Arsenal on Sunday has
    earned the back-up striker the
    chance to salvage a career
    that had appeared to be
    drifting irrevocably towards
    the lower divisions." + ], + [ + "ABOUT 70,000 people were
    forced to evacuate Real Madrid
    #39;s Santiago Bernabeu
    stadium minutes before the end
    of a Primera Liga match
    yesterday after a bomb threat
    in the name of ETA Basque
    separatist guerillas." + ], + [ + "The team learned on Monday
    that full-back Jon Ritchie
    will miss the rest of the
    season with a torn anterior
    cruciate ligament in his left
    knee." + ], + [ + " NEW YORK (Reuters) -
    Lifestyle guru Martha Stewart
    said on Wednesday she wants
    to start serving her prison
    sentence for lying about a
    suspicious stock sale as soon
    as possible, so she can put
    her \"nightmare\" behind her." + ], + [ + "Apple Computer's iPod remains
    the king of digital music
    players, but robust pretenders
    to the throne have begun to
    emerge in the Windows
    universe. One of them is the
    Zen Touch, from Creative Labs." + ], + [ + "The 7710 model features a
    touch screen, pen input, a
    digital camera, an Internet
    browser, a radio, video
    playback and streaming and
    recording capabilities, the
    company said." + ], + [ + "SAN FRANCISCO (CBS.MW) --
    Crude futures closed under
    \\$46 a barrel Wednesday for
    the first time since late
    September and heating-oil and
    unleaded gasoline prices
    dropped more than 6 percent
    following an across-the-board
    climb in US petroleum
    inventories." + ], + [ + "The University of Iowa #39;s
    market for US presidential
    futures, founded 16-years ago,
    has been overtaken by a
    Dublin-based exchange that is
    now 25 times larger." + ], + [ + "Venus Williams barely kept
    alive her hopes of qualifying
    for next week #39;s WTA Tour
    Championships. Williams,
    seeded fifth, survived a
    third-set tiebreaker to
    outlast Yuilana Fedak of the
    Ukraine, 6-4 2-6 7-6" + ], + [ + " SYDNEY (Reuters) - Arnold
    Palmer has taken a swing at
    America's top players,
    criticizing their increasing
    reluctance to travel abroad
    to play in tournaments." + ], + [ + "MARK Thatcher will have to
    wait until at least next April
    to face trial on allegations
    he helped bankroll a coup
    attempt in oil-rich Equatorial
    Guinea." + ], + [ + "A top Red Hat executive has
    attacked the open-source
    credentials of its sometime
    business partner Sun
    Microsystems. In a Web log
    posting Thursday, Michael
    Tiemann, Red Hat #39;s vice
    president of open-source
    affairs" + ], + [ + "President Bush #39;s drive to
    deploy a multibillion-dollar
    shield against ballistic
    missiles was set back on
    Wednesday by what critics
    called a stunning failure of
    its first full flight test in
    two years." + ], + [ + "Although he was well-beaten by
    Retief Goosen in Sunday #39;s
    final round of The Tour
    Championship in Atlanta, there
    has been some compensation for
    the former world number one,
    Tiger Woods." + ], + [ + "WAYNE Rooney and Henrik
    Larsson are among the players
    nominated for FIFAs
    prestigious World Player of
    the Year award. Rooney is one
    of four Manchester United
    players on a list which is
    heavily influenced by the
    Premiership." + ], + [ + "It didn #39;t look good when
    it happened on the field, and
    it looked worse in the
    clubhouse. Angels second
    baseman Adam Kennedy left the
    Angels #39; 5-2 win over the
    Seattle Mariners" + ], + [ + "Air travelers moved one step
    closer to being able to talk
    on cell phones and surf the
    Internet from laptops while in
    flight, thanks to votes by the
    Federal Communications
    Commission yesterday." + ], + [ + "MySQL developers turn to an
    unlikely source for database
    tool: Microsoft. Also: SGI
    visualizes Linux, and the
    return of Java veteran Kim
    Polese." + ], + [ + "DESPITE the budget deficit,
    continued increases in oil and
    consumer prices, the economy,
    as measured by gross domestic
    product, grew by 6.3 percent
    in the third" + ], + [ + "NEW YORK - A drop in oil
    prices and upbeat outlooks
    from Wal-Mart and Lowe's
    helped send stocks sharply
    higher Monday on Wall Street,
    with the swing exaggerated by
    thin late summer trading. The
    Dow Jones industrials surged
    nearly 130 points..." + ], + [ + "Freshman Darius Walker ran for
    115 yards and scored two
    touchdowns, helping revive an
    Irish offense that had managed
    just one touchdown in the
    season's first six quarters." + ], + [ + "Consumers who cut it close by
    paying bills from their
    checking accounts a couple of
    days before depositing funds
    will be out of luck under a
    new law that takes effect Oct.
    28." + ], + [ + "Dell Inc. said its profit
    surged 25 percent in the third
    quarter as the world's largest
    personal computer maker posted
    record sales due to rising
    technology spending in the
    corporate and government
    sectors in the United States
    and abroad." + ], + [ + "AP - NBC is adding a 5-second
    delay to its NASCAR telecasts
    after Dale Earnhardt Jr. used
    a vulgarity during a postrace
    TV interview last weekend." + ], + [ + " BOSTON (Sports Network) - The
    New York Yankees will start
    Orlando \"El Duque\" Hernandez
    in Game 4 of the American
    League Championship Series on
    Saturday against the Boston
    Red Sox." + ], + [ + "The future of Sven-Goran
    Eriksson as England coach is
    the subject of intense
    discussion after the draw in
    Austria. Has the Swede lost
    the confidence of the nation
    or does he remain the best man
    for the job?" + ], + [ + "Component problems meant
    Brillian's new big screens
    missed the NFL's kickoff
    party." + ], + [ + "Spain begin their third final
    in five seasons at the Olympic
    stadium hoping to secure their
    second title since their first
    in Barcelona against Australia
    in 2000." + ], + [ + "Second-seeded David Nalbandian
    of Argentina lost at the Japan
    Open on Thursday, beaten by
    Gilles Muller of Luxembourg
    7-6 (4), 3-6, 6-4 in the third
    round." + ], + [ + "Thursday #39;s unexpected
    resignation of Memphis
    Grizzlies coach Hubie Brown
    left a lot of questions
    unanswered. In his unique way
    of putting things, the
    71-year-old Brown seemed to
    indicate he was burned out and
    had some health concerns." + ], + [ + "RUDI Voller had quit as coach
    of Roma after a 3-1 defeat
    away to Bologna, the Serie A
    club said today. Under the
    former Germany coach, Roma had
    taken just four league points
    from a possible 12." + ], + [ + "A Russian court on Thursday
    rejected an appeal by the
    Yukos oil company seeking to
    overturn a freeze on the
    accounts of the struggling oil
    giant #39;s core subsidiaries." + ], + [ + "ONE by one, the players #39;
    faces had flashed up on the
    giant Ibrox screens offering
    season #39;s greetings to the
    Rangers fans. But the main
    presents were reserved for
    Auxerre." + ], + [ + "Switzerland #39;s struggling
    national airline reported a
    second-quarter profit of 45
    million Swiss francs (\\$35.6
    million) Tuesday, although its
    figures were boosted by a
    legal settlement in France." + ], + [ + "ROSTOV-ON-DON, Russia --
    Hundreds of protesters
    ransacked and occupied the
    regional administration
    building in a southern Russian
    province Tuesday, demanding
    the resignation of the region
    #39;s president, whose former
    son-in-law has been linked to
    a multiple" + ], + [ + "SIPTU has said it is strongly
    opposed to any privatisation
    of Aer Lingus as pressure
    mounts on the Government to
    make a decision on the future
    funding of the airline." + ], + [ + "Reuters - SBC Communications
    said on Monday it\\would offer
    a television set-top box that
    can handle music,\\photos and
    Internet downloads, part of
    SBC's efforts to expand\\into
    home entertainment." + ], + [ + "Molson Inc. Chief Executive
    Officer Daniel O #39;Neill
    said he #39;ll provide
    investors with a positive #39;
    #39; response to their
    concerns over the company
    #39;s plan to let stock-
    option holders vote on its
    planned merger with Adolph
    Coors Co." + ], + [ + "South Korea #39;s Grace Park
    shot a seven-under-par 65 to
    triumph at the CJ Nine Bridges
    Classic on Sunday. Park #39;s
    victory made up her final-
    round collapse at the Samsung
    World Championship two weeks
    ago." + ], + [ + " WASHINGTON (Reuters) - Hopes
    -- and worries -- that U.S.
    regulators will soon end the
    ban on using wireless phones
    during U.S. commercial flights
    are likely at least a year or
    two early, government
    officials and analysts say." + ], + [ + "AFP - Iraqi Foreign Minister
    Hoshyar Zebari arrived
    unexpectedly in the holy city
    of Mecca Wednesday where he
    met Crown Prince Abdullah bin
    Abdul Aziz, the official SPA
    news agency reported." + ], + [ + "Titans QB Steve McNair was
    released from a Nashville
    hospital after a two-night
    stay for treatment of a
    bruised sternum. McNair was
    injured during the fourth
    quarter of the Titans #39;
    15-12 loss to Jacksonville on
    Sunday." + ], + [ + "Keith Miller, Australia #39;s
    most prolific all-rounder in
    Test cricket, died today at a
    nursing home, Cricket
    Australia said. He was 84." + ], + [ + "Haitian police and UN troops
    moved into a slum neighborhood
    on Sunday and cleared street
    barricades that paralyzed a
    part of the capital." + ], + [ + "TORONTO Former Toronto pitcher
    John Cerutti (seh-ROO
    #39;-tee) was found dead in
    his hotel room today,
    according to the team. He was
    44." + ], + [ + "withdrawal of troops and
    settlers from occupied Gaza
    next year. Militants seek to
    claim any pullout as a
    victory. quot;Islamic Jihad
    will not be broken by this
    martyrdom, quot; said Khaled
    al-Batsh, a senior political
    leader in Gaza." + ], + [ + " NEW YORK (Reuters) - The
    world's largest gold producer,
    Newmont Mining Corp. <A HRE
    F=\"http://www.investor.reuters
    .com/FullQuote.aspx?ticker=NEM
    .N target=/stocks/quickinfo/fu
    llquote\">NEM.N</A>,
    on Wednesday said higher gold
    prices drove up quarterly
    profit by 12.5 percent, even
    though it sold less of the
    precious metal." + ], + [ + "The U.S. military has found
    nearly 20 houses where
    intelligence officers believe
    hostages were tortured or
    killed in this city, including
    the house with the cage that
    held a British contractor who
    was beheaded last month." + ], + [ + "AFP - Opponents of the Lao
    government may be plotting
    bomb attacks in Vientiane and
    other areas of Laos timed to
    coincide with a summit of
    Southeast Asian leaders the
    country is hosting next month,
    the United States said." + ], + [ + "After a year of pilots and
    trials, Siebel Systems jumped
    with both feet into the SMB
    market Tuesday, announcing a
    new approach to offer Siebel
    Professional CRM applications
    to SMBs (small and midsize
    businesses) -- companies with
    revenues up to about \\$500
    million." + ], + [ + "AP - Russia agreed Thursday to
    send warships to help NATO
    naval patrols that monitor
    suspicious vessels in the
    Mediterranean, part of a push
    for closer counterterrorism
    cooperation between Moscow and
    the western alliance." + ], + [ + "Intel won't release a 4-GHz
    version of its flagship
    Pentium 4 product, having
    decided instead to realign its
    engineers around the company's
    new design priorities, an
    Intel spokesman said today.
    \\\\" + ], + [ + "AP - A Soyuz spacecraft
    carrying two Russians and an
    American rocketed closer
    Friday to its docking with the
    international space station,
    where the current three-man
    crew made final departure
    preparations." + ], + [ + "Defense: Ken Lucas. His
    biggest play was his first
    one. The fourth-year
    cornerback intercepted a Ken
    Dorsey pass that kissed off
    the hands of wide receiver
    Rashaun Woods and returned it
    25 yards to set up the
    Seahawks #39; first score." + ], + [ + "Scientists have manipulated
    carbon atoms to create a
    material that could be used to
    create light-based, versus
    electronic, switches. The
    material could lead to a
    supercharged Internet based
    entirely on light, scientists
    say." + ], + [ + "A military plane crashed in
    the mountains near Caracas,
    killing all 16 persons on
    board, including two high-
    ranking military officers,
    officials said." + ], + [ + "The powerful St. Louis trio of
    Albert Pujols, Scott Rolen and
    Jim Edmonds is 4 for 23 with
    one RBI in the series and with
    runners on base, they are 1
    for 13." + ], + [ + "A voice recording said to be
    that of suspected Al Qaeda
    commander Abu Mussab al-
    Zarqawi, claims Iraq #39;s
    Prime Minister Iyad Allawi is
    the militant network #39;s
    number one target." + ], + [ + "BEIJING -- More than a year
    after becoming China's
    president, Hu Jintao was
    handed the full reins of power
    yesterday when his
    predecessor, Jiang Zemin, gave
    up the nation's most powerful
    military post." + ], + [ + "LOS ANGELES (Reuters)
    Qualcomm has dropped an \\$18
    million claim for monetary
    damages from rival Texas
    Instruments for publicly
    discussing terms of a
    licensing pact, a TI
    spokeswoman confirmed Tuesday." + ], + [ + "Hewlett-Packard is the latest
    IT vendor to try blogging. But
    analysts wonder if the weblog
    trend is the 21st century
    equivalent of CB radios, which
    made a big splash in the 1970s
    before fading." + ], + [ + " WASHINGTON (Reuters) - Fannie
    Mae executives and their
    regulator squared off on
    Wednesday, with executives
    denying any accounting
    irregularity and the regulator
    saying the housing finance
    company's management may need
    to go." + ], + [ + "The scientists behind Dolly
    the sheep apply for a license
    to clone human embryos. They
    want to take stem cells from
    the embryos to study Lou
    Gehrig's disease." + ], + [ + "As the first criminal trial
    stemming from the financial
    deals at Enron opened in
    Houston on Monday, it is
    notable as much for who is not
    among the six defendants as
    who is - and for how little
    money was involved compared
    with how much in other Enron" + ], + [ + "LONDON (CBS.MW) -- British
    bank Barclays on Thursday said
    it is in talks to buy a
    majority stake in South
    African bank ABSA. Free!" + ], + [ + "The Jets signed 33-year-old
    cornerback Terrell Buckley,
    who was released by New
    England on Sunday, after
    putting nickel back Ray
    Mickens on season-ending
    injured reserve yesterday with
    a torn ACL in his left knee." + ], + [ + "Some of the silly tunes
    Japanese pay to download to
    use as the ring tone for their
    mobile phones sure have their
    knockers, but it #39;s for
    precisely that reason that a
    well-known counselor is raking
    it in at the moment, according
    to Shukan Gendai (10/2)." + ], + [ + "WEST INDIES thrilling victory
    yesterday in the International
    Cricket Council Champions
    Trophy meant the world to the
    five million people of the
    Caribbean." + ], + [ + "AP - Greenpeace activists
    scaled the walls of Ford Motor
    Co.'s Norwegian headquarters
    Tuesday to protest plans to
    destroy hundreds of non-
    polluting electric cars." + ], + [ + "Investors sent stocks sharply
    lower yesterday as oil prices
    continued their climb higher
    and new questions about the
    safety of arthritis drugs
    pressured pharmaceutical
    stocks." + ], + [ + "Scotland manager Berti Vogts
    insists he is expecting
    nothing but victory against
    Moldova on Wednesday. The game
    certainly is a must-win affair
    if the Scots are to have any
    chance of qualifying for the
    2006 World Cup finals." + ], + [ + "IBM announced yesterday that
    it will invest US\\$250 million
    (S\\$425 million) over the next
    five years and employ 1,000
    people in a new business unit
    to support products and
    services related to sensor
    networks." + ], + [ + "AFP - The chances of Rupert
    Murdoch's News Corp relocating
    from Australia to the United
    States have increased after
    one of its biggest
    institutional investors has
    chosen to abstain from a vote
    next week on the move." + ], + [ + "AFP - An Indian minister said
    a school text-book used in the
    violence-prone western state
    of Gujarat portrayed Adolf
    Hitler as a role model." + ], + [ + "Reuters - The head of UAL
    Corp.'s United\\Airlines said
    on Thursday the airline's
    restructuring plan\\would lead
    to a significant number of job
    losses, but it was\\not clear
    how many." + ], + [ + "DOVER, N.H. (AP) -- Democrat
    John Kerry is seizing on the
    Bush administration's failure
    to secure hundreds of tons of
    explosives now missing in
    Iraq." + ], + [ + "AP - Microsoft Corp. goes into
    round two Friday of its battle
    to get the European Union's
    sweeping antitrust ruling
    lifted having told a judge
    that it had been prepared
    during settlement talks to
    share more software code with
    its rivals than the EU
    ultimately demanded." + ], + [ + " INDIANAPOLIS (Reuters) -
    Jenny Thompson will take the
    spotlight from injured U.S.
    team mate Michael Phelps at
    the world short course
    championships Saturday as she
    brings down the curtain on a
    spectacular swimming career." + ], + [ + "Martin Brodeur made 27 saves,
    and Brad Richards, Kris Draper
    and Joe Sakic scored goals to
    help Canada beat Russia 3-1
    last night in the World Cup of
    Hockey, giving the Canadians a
    3-0 record in round-robin
    play." + ], + [ + "AP - Sears, Roebuck and Co.,
    which has successfully sold
    its tools and appliances on
    the Web, is counting on having
    the same magic with bedspreads
    and sweaters, thanks in part
    to expertise gained by its
    purchase of Lands' End Inc." + ], + [ + "com September 14, 2004, 9:12
    AM PT. With the economy slowly
    turning up, upgrading hardware
    has been on businesses radar
    in the past 12 months as their
    number two priority." + ], + [ + " NEW YORK (Reuters) -
    Children's Place Retail Stores
    Inc. <A HREF=\"http://www.i
    nvestor.reuters.com/FullQuote.
    aspx?ticker=PLCE.O target=/sto
    cks/quickinfo/fullquote\">PL
    CE.O</A> said on
    Wednesday it will buy 313
    retail stores from Walt
    Disney Co., and its stock rose
    more than 14 percent in early
    morning trade." + ], + [ + "ALBANY, N.Y. -- A California-
    based company that brokers
    life, accident, and disability
    policies for leading US
    companies pocketed millions of
    dollars a year in hidden
    payments from insurers and
    from charges on clients'
    unsuspecting workers, New York
    Attorney General Eliot Spitzer
    charged yesterday." + ], + [ + "NORTEL Networks plans to slash
    its workforce by 3500, or ten
    per cent, as it struggles to
    recover from an accounting
    scandal that toppled three top
    executives and led to a
    criminal investigation and
    lawsuits." + ], + [ + "Ebay Inc. (EBAY.O: Quote,
    Profile, Research) said on
    Friday it would buy Rent.com,
    an Internet housing rental
    listing service, for \\$415
    million in a deal that gives
    it access to a new segment of
    the online real estate market." + ], + [ + "Austin police are working with
    overseas officials to bring
    charges against an English man
    for sexual assault of a child,
    a second-degree felony." + ], + [ + "United Nations officials
    report security breaches in
    internally displaced people
    and refugee camps in Sudan
    #39;s embattled Darfur region
    and neighboring Chad." + ], + [ + "Noranda Inc., Canada #39;s
    biggest mining company, began
    exclusive talks on a takeover
    proposal from China Minmetals
    Corp. that would lead to the
    spinoff of Noranda #39;s
    aluminum business to
    shareholders." + ], + [ + "San Francisco
    developer/publisher lands
    coveted Paramount sci-fi
    license, \\$6.5 million in
    funding on same day. Although
    it is less than two years old,
    Perpetual Entertainment has
    acquired one of the most
    coveted sci-fi licenses on the
    market." + ], + [ + "ST. LOUIS -- Mike Martz #39;s
    week of anger was no empty
    display. He saw the defending
    NFC West champions slipping
    and thought taking potshots at
    his players might be his best
    shot at turning things around." + ], + [ + "Google warned Thursday that
    increased competition and the
    maturing of the company would
    result in an quot;inevitable
    quot; slowing of its growth." + ], + [ + "By KELLY WIESE JEFFERSON
    CITY, Mo. (AP) -- Missouri
    will allow members of the
    military stationed overseas to
    return absentee ballots via
    e-mail, raising concerns from
    Internet security experts
    about fraud and ballot
    secrecy..." + ], + [ + "Avis Europe PLC has dumped a
    new ERP system based on
    software from PeopleSoft Inc.
    before it was even rolled out,
    citing substantial delays and
    higher-than-expected costs." + ], + [ + " TOKYO (Reuters) - The Nikkei
    average rose 0.55 percent by
    midsession on Wednesday as
    some techs including Advantest
    Corp. gained ground after
    Wall Street reacted positively
    to results from Intel Corp.
    released after the U.S. market
    close." + ], + [ + "Yahoo #39;s (Quote, Chart)
    public embrace of the RSS
    content syndication format
    took a major leap forward with
    the release of a revamped My
    Yahoo portal seeking to
    introduce the technology to
    mainstream consumers." + ], + [ + "KINGSTON, Jamaica - Hurricane
    Ivan's deadly winds and
    monstrous waves bore down on
    Jamaica on Friday, threatening
    a direct hit on its densely
    populated capital after
    ravaging Grenada and killing
    at least 33 people. The
    Jamaican government ordered
    the evacuation of half a
    million people from coastal
    areas, where rains on Ivan's
    outer edges were already
    flooding roads..." + ], + [ + "North Korea has denounced as
    quot;wicked terrorists quot;
    the South Korean officials who
    orchestrated last month #39;s
    airlift to Seoul of 468 North
    Korean defectors." + ], + [ + "The Black Watch regiment has
    returned to its base in Basra
    in southern Iraq after a
    month-long mission standing in
    for US troops in a more
    violent part of the country,
    the Ministry of Defence says." + ], + [ + "Romanian soccer star Adrian
    Mutu as he arrives at the
    British Football Association
    in London, ahead of his
    disciplinary hearing, Thursday
    Nov. 4, 2004." + ], + [ + "Australia completed an
    emphatic Test series sweep
    over New Zealand with a
    213-run win Tuesday, prompting
    a caution from Black Caps
    skipper Stephen Fleming for
    other cricket captains around
    the globe." + ], + [ + "AP - A senior Congolese
    official said Tuesday his
    nation had been invaded by
    neighboring Rwanda, and U.N.
    officials said they were
    investigating claims of
    Rwandan forces clashing with
    militias in the east." + ], + [ + "Liverpool, England (Sports
    Network) - Former English
    international and Liverpool
    great Emlyn Hughes passed away
    Tuesday from a brain tumor." + ], + [ + " ATLANTA (Reuters) - Soft
    drink giant Coca-Cola Co.
    <A HREF=\"http://www.investo
    r.reuters.com/FullQuote.aspx?t
    icker=KO.N target=/stocks/quic
    kinfo/fullquote\">KO.N</A
    >, stung by a prolonged
    downturn in North America,
    Germany and other major
    markets, on Thursday lowered
    its key long-term earnings
    and sales targets." + ], + [ + "JACKSONVILLE, Fla. -- They
    were singing in the Colts #39;
    locker room today, singing
    like a bunch of wounded
    songbirds. Never mind that
    Marcus Pollard, Dallas Clark
    and Ben Hartsock won #39;t be
    recording a remake of Kenny
    Chesney #39;s song,
    quot;Young, quot; any time
    soon." + ], + [ + "TOKYO (AP) -- Japanese
    electronics and entertainment
    giant Sony Corp. (SNE) plans
    to begin selling a camcorder
    designed for consumers that
    takes video at digital high-
    definition quality and is
    being priced at about
    \\$3,600..." + ], + [ + "As the close-knit NASCAR
    community mourns the loss of
    team owner Rick Hendrick #39;s
    son, brother, twin nieces and
    six others in a plane crash
    Sunday, perhaps no one outside
    of the immediate family
    grieves more deeply than
    Darrell Waltrip." + ], + [ + "AP - Purdue quarterback Kyle
    Orton has no trouble
    remembering how he felt after
    last year's game at Michigan." + ], + [ + "UNITED NATIONS - The United
    Nations #39; nuclear agency
    says it is concerned about the
    disappearance of equipment and
    materials from Iraq that could
    be used to make nuclear
    weapons." + ], + [ + " BRUSSELS (Reuters) - The EU's
    historic deal with Turkey to
    open entry talks with the vast
    Muslim country was hailed by
    supporters as a bridge builder
    between Europe and the Islamic
    world." + ], + [ + "Iraqi President Ghazi al-
    Yawar, who was due in Paris on
    Sunday to start a European
    tour, has postponed his visit
    to France due to the ongoing
    hostage drama involving two
    French journalists, Arab
    diplomats said Friday." + ], + [ + " SAO PAULO, Brazil (Reuters) -
    President Luiz Inacio Lula da
    Silva's Workers' Party (PT)
    won the mayoralty of six state
    capitals in Sunday's municipal
    vote but was forced into a
    run-off to defend its hold on
    the race's biggest prize, the
    city of Sao Paulo." + ], + [ + "ATHENS, Greece - They are
    America's newest golden girls
    - powerful and just a shade
    from perfection. The U.S..." + ], + [ + "AMMAN, Sept. 15. - The owner
    of a Jordanian truck company
    announced today that he had
    ordered its Iraq operations
    stopped in a bid to save the
    life of a driver held hostage
    by a militant group." + ], + [ + "Israel is prepared to back a
    Middle East conference
    convened by Tony Blair early
    next year despite having
    expressed fears that the
    British plans were over-
    ambitious and designed" + ], + [ + "AP - U.S. State Department
    officials learned that seven
    American children had been
    abandoned at a Nigerian
    orphanage but waited more than
    a week to check on the youths,
    who were suffering from
    malnutrition, malaria and
    typhoid, a newspaper reported
    Saturday." + ], + [ + "\\Angry mobs in Ivory Coast's
    main city, Abidjan, marched on
    the airport, hours after it
    came under French control." + ], + [ + "Several workers are believed
    to have been killed and others
    injured after a contruction
    site collapsed at Dubai
    airport. The workers were
    trapped under rubble at the
    site of a \\$4." + ], + [ + "Talks between Sudan #39;s
    government and two rebel
    groups to resolve the nearly
    two-year battle resume Friday.
    By Abraham McLaughlin Staff
    writer of The Christian
    Science Monitor." + ], + [ + "In a meeting of the cinematic
    with the scientific, Hollywood
    helicopter stunt pilots will
    try to snatch a returning NASA
    space probe out of the air
    before it hits the ground." + ], + [ + "Legend has it (incorrectly, it
    seems) that infamous bank
    robber Willie Sutton, when
    asked why banks were his
    favorite target, responded,
    quot;Because that #39;s where
    the money is." + ], + [ + "Brown is a second year player
    from Memphis and has spent the
    2004 season on the Steelers
    #39; practice squad. He played
    in two games last year." + ], + [ + "A Canadian court approved Air
    Canada #39;s (AC.TO: Quote,
    Profile, Research) plan of
    arrangement with creditors on
    Monday, clearing the way for
    the world #39;s 11th largest
    airline to emerge from
    bankruptcy protection at the
    end of next month" + ], + [ + "Pfizer, GlaxoSmithKline and
    Purdue Pharma are the first
    drugmakers willing to take the
    plunge and use radio frequency
    identification technology to
    protect their US drug supply
    chains from counterfeiters." + ], + [ + "Barret Jackman, the last of
    the Blues left to sign before
    the league #39;s probable
    lockout on Wednesday,
    finalized a deal Monday that
    is rare in the current
    economic climate but fitting
    for him." + ], + [ + " LONDON (Reuters) - Oil prices
    eased on Monday after rebels
    in Nigeria withdrew a threat
    to target oil operations, but
    lingering concerns over
    stretched supplies ahead of
    winter kept prices close to
    \\$50." + ], + [ + "AP - In the tumult of the
    visitors' clubhouse at Yankee
    Stadium, champagne pouring all
    around him, Theo Epstein held
    a beer. \"I came in and there
    was no champagne left,\" he
    said this week. \"I said, 'I'll
    have champagne if we win it
    all.'\" Get ready to pour a
    glass of bubbly for Epstein.
    No I.D. necessary." + ], + [ + "Search any fee-based digital
    music service for the best-
    loved musical artists of the
    20th century and most of the
    expected names show up." + ], + [ + "Barcelona held on from an
    early Deco goal to edge game
    local rivals Espanyol 1-0 and
    carve out a five point
    tabletop cushion. Earlier,
    Ronaldo rescued a point for
    Real Madrid, who continued
    their middling form with a 1-1
    draw at Real Betis." + ], + [ + "MONTREAL (CP) - The Expos may
    be history, but their demise
    has heated up the market for
    team memorabilia. Vintage
    1970s and 1980s shirts are
    already sold out, but
    everything from caps, beer
    glasses and key-chains to
    dolls of mascot Youppi!" + ], + [ + "Stansted airport is the
    designated emergency landing
    ground for planes in British
    airspace hit by in-flight
    security alerts. Emergency
    services at Stansted have
    successfully dealt" + ], + [ + "The massive military operation
    to retake Fallujah has been
    quot;accomplished quot;, a
    senior Iraqi official said.
    Fierce fighting continued in
    the war-torn city where
    pockets of resistance were
    still holding out against US
    forces." + ], + [ + "There are some signs of
    progress in resolving the
    Nigerian conflict that is
    riling global oil markets. The
    leader of militia fighters
    threatening to widen a battle
    for control of Nigeria #39;s
    oil-rich south has" + ], + [ + "A strong earthquake hit Taiwan
    on Monday, shaking buildings
    in the capital Taipei for
    several seconds. No casualties
    were reported." + ], + [ + "America Online Inc. is
    packaging new features to
    combat viruses, spam and
    spyware in response to growing
    online security threats.
    Subscribers will be able to
    get the free tools" + ], + [ + "A 76th minute goal from
    European Footballer of the
    Year Pavel Nedved gave
    Juventus a 1-0 win over Bayern
    Munich on Tuesday handing the
    Italians clear control at the
    top of Champions League Group
    C." + ], + [ + " LONDON (Reuters) - Oil prices
    climbed above \\$42 a barrel on
    Wednesday, rising for the
    third day in a row as cold
    weather gripped the U.S.
    Northeast, the world's biggest
    heating fuel market." + ], + [ + "A policeman ran amok at a
    security camp in Indian-
    controlled Kashmir after an
    argument and shot dead seven
    colleagues before he was
    gunned down, police said on
    Sunday." + ], + [ + "ANN ARBOR, Mich. -- Some NHL
    players who took part in a
    charity hockey game at the
    University of Michigan on
    Thursday were hopeful the news
    that the NHL and the players
    association will resume talks
    next week" + ], + [ + "New York police have developed
    a pre-emptive strike policy,
    cutting off demonstrations
    before they grow large." + ], + [ + "CAPE CANAVERAL-- NASA aims to
    launch its first post-Columbia
    shuttle mission during a
    shortened nine-day window
    March, and failure to do so
    likely would delay a planned
    return to flight until at
    least May." + ], + [ + "Travelers headed home for
    Thanksgiving were greeted
    Wednesday with snow-covered
    highways in the Midwest, heavy
    rain and tornadoes in parts of
    the South, and long security
    lines at some of the nation
    #39;s airports." + ], + [ + "BOULDER, Colo. -- Vernand
    Morency ran for 165 yards and
    two touchdowns and Donovan
    Woods threw for three more
    scores, lifting No. 22
    Oklahoma State to a 42-14
    victory over Colorado
    yesterday." + ], + [ + "The Chinese city of Beijing
    has cancelled an order for
    Microsoft software, apparently
    bowing to protectionist
    sentiment. The deal has come
    under fire in China, which is
    trying to build a domestic
    software industry." + ], + [ + "Apple says it will deliver its
    iTunes music service to more
    European countries next month.
    Corroborating several reports
    in recent months, Reuters is
    reporting today that Apple
    Computer is planning the next" + ], + [ + "Reuters - Motorola Inc., the
    world's\\second-largest mobile
    phone maker, said on Tuesday
    it expects\\to sustain strong
    sales growth in the second
    half of 2004\\thanks to new
    handsets with innovative
    designs and features." + ], + [ + "PULLMAN - Last week, in
    studying USC game film, Cougar
    coaches thought they found a
    chink in the national
    champions armor. And not just
    any chink - one with the
    potential, right from the get
    go, to" + ], + [ + "The union representing flight
    attendants on Friday said it
    mailed more than 5,000 strike
    authorization ballots to its
    members employed by US Airways
    as both sides continued talks
    that are expected to stretch
    through the weekend." + ], + [ + "AP - Matt Leinart was quite a
    baseball prospect growing up,
    showing so much promise as a
    left-handed pitcher that
    scouts took notice before high
    school." + ], + [ + "PRAGUE, Czech Republic --
    Eugene Cernan, the last man to
    walk on the moon during the
    final Apollo landing, said
    Thursday he doesn't expect
    space tourism to become
    reality in the near future,
    despite a strong demand.
    Cernan, now 70, who was
    commander of NASA's Apollo 17
    mission and set foot on the
    lunar surface in December 1972
    during his third space flight,
    acknowledged that \"there are
    many people interested in
    space tourism.\" But the
    former astronaut said he
    believed \"we are a long way
    away from the day when we can
    send a bus of tourists to the
    moon.\" He spoke to reporters
    before being awarded a medal
    by the Czech Academy of
    Sciences for his contribution
    to science..." + ], + [ + "Never shy about entering a
    market late, Microsoft Corp.
    is planning to open the
    virtual doors of its long-
    planned Internet music store
    next week. <FONT
    face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\"\\ color=\"#666666\">
    <B>-Leslie
    Walker</B></FONT>" + ], + [ + "AP - On his first birthday
    Thursday, giant panda cub Mei
    Sheng delighted visitors by
    playing for the first time in
    snow delivered to him at the
    San Diego Zoo. He also sat on
    his ice cake, wrestled with
    his mom, got his coat
    incredibly dirty, and didn't
    read any of the more than 700
    birthday wishes sent him via
    e-mail from as far away as
    Ireland and Argentina." + ], + [ + "AP - Researchers put a
    satellite tracking device on a
    15-foot shark that appeared to
    be lost in shallow water off
    Cape Cod, the first time a
    great white has been tagged
    that way in the Atlantic." + ], + [ + "LSU will stick with a two-
    quarterback rotation Saturday
    at Auburn, according to Tigers
    coach Nick Saban, who seemed
    to have some fun telling the
    media what he will and won
    #39;t discuss Monday." + ], + [ + "Bulgaria has started its first
    co-mission with the EU in
    Bosnia and Herzegovina, along
    with some 30 countries,
    including Canada and Turkey." + ], + [ + "The Windows Future Storage
    (WinFS) technology that got
    cut out of Windows
    quot;Longhorn quot; is in
    serious trouble, and not just
    the hot water a feature might
    encounter for missing its
    intended production vehicle." + ], + [ + "Seattle -- - Not so long ago,
    the 49ers were inflicting on
    other teams the kind of pain
    and embarrassment they felt in
    their 34-0 loss to the
    Seahawks on Sunday." + ], + [ + "AP - The pileup of events in
    the city next week, including
    the Republican National
    Convention, will add to the
    security challenge for the New
    York Police Department, but
    commissioner Ray Kelly says,
    \"With a big, experienced
    police force, we can do it.\"" + ], + [ + "washingtonpost.com - Let the
    games begin. Not the Olympics
    again, but the all-out battle
    between video game giants Sony
    Corp. and Nintendo Co. Ltd.
    The two Japanese companies are
    rolling out new gaming
    consoles, but Nintendo has
    beaten Sony to the punch by
    announcing an earlier launch
    date for its new hand-held
    game player." + ], + [ + "London - Manchester City held
    fierce crosstown rivals
    Manchester United to a 0-0
    draw on Sunday, keeping the
    Red Devils eleven points
    behind leaders Chelsea." + ], + [ + "LONDON, Dec 11 (IranMania) -
    Iraqi Vice-President Ibrahim
    al-Jaafari refused to believe
    in remarks published Friday
    that Iran was attempting to
    influence Iraqi polls with the
    aim of creating a
    quot;crescent quot; dominated
    by Shiites in the region." + ], + [ + "LOS ANGELES (CBS.MW) - The US
    Securities and Exchange
    Commission is probing
    transactions between Delphi
    Corp and EDS, which supplies
    the automotive parts and
    components giant with
    technology services, Delphi
    said late Wednesday." + ], + [ + "MONTREAL (CP) - Molson Inc.
    and Adolph Coors Co. are
    sweetening their brewery
    merger plan with a special
    dividend to Molson
    shareholders worth \\$381
    million." + ], + [ + "AP - Echoing what NASA
    officials said a day earlier,
    a Russian space official on
    Friday said the two-man crew
    on the international space
    station could be forced to
    return to Earth if a planned
    resupply flight cannot reach
    them with food supplies later
    this month." + ], + [ + "InfoWorld - SANTA CLARA,
    CALIF. -- Accommodating large
    patch sets in Linux is
    expected to mean forking off
    of the 2.7 version of the
    platform to accommodate these
    changes, according to Andrew
    Morton, lead maintainer of the
    Linux kernel for Open Source
    Development Labs (OSDL)." + ], + [ + "AMSTERDAM The mobile phone
    giants Vodafone and Nokia
    teamed up on Thursday to
    simplify cellphone software
    written with the Java computer
    language." + ], + [ + "WELLINGTON: National carrier
    Air New Zealand said yesterday
    the Australian Competition
    Tribunal has approved a
    proposed alliance with Qantas
    Airways Ltd, despite its
    rejection in New Zealand." + ], + [ + "The late Princess Dianas
    former bodyguard, Ken Wharfe,
    dismisses her suspicions that
    one of her lovers was bumped
    off. Princess Diana had an
    affair with Barry Mannakee, a
    policeman who was assigned to
    protect her." + ], + [ + "Long considered beyond the
    reach of mainland mores, the
    Florida city is trying to
    limit blatant displays of
    sexual behavior." + ], + [ + "The overall Linux market is
    far larger than previous
    estimates show, a new study
    says. In an analysis of the
    Linux market released late
    Tuesday, market research firm
    IDC estimated that the Linux
    market -- including" + ], + [ + "By PAUL ELIAS SAN FRANCISCO
    (AP) -- Several California
    cities and counties, including
    San Francisco and Los Angeles,
    sued Microsoft Corp. (MSFT) on
    Friday, accusing the software
    giant of illegally charging
    inflated prices for its
    products because of monopoly
    control of the personal
    computer operating system
    market..." + ], + [ + "New Ole Miss head coach Ed
    Orgeron, speaking for the
    first time since his hiring,
    made clear the goal of his
    football program. quot;The
    goal of this program will be
    to go to the Sugar Bowl, quot;
    Orgeron said." + ], + [ + "\"Everyone's nervous,\" Acting
    Undersecretary of Defense
    Michael W. Wynne warned in a
    confidential e-mail to Air
    Force Secretary James G. Roche
    on July 8, 2003." + ], + [ + "Reuters - Alpharma Inc. on
    Friday began\\selling a cheaper
    generic version of Pfizer
    Inc.'s #36;3\\billion a year
    epilepsy drug Neurontin
    without waiting for a\\court
    ruling on Pfizer's request to
    block the copycat medicine." + ], + [ + "Public opinion of the database
    giant sinks to 12-year low, a
    new report indicates." + ], + [ + "Opinion: Privacy hysterics
    bring old whine in new bottles
    to the Internet party. The
    desktop search beta from this
    Web search leader doesn #39;t
    do anything you can #39;t do
    already." + ], + [ + "It is much too easy to call
    Pedro Martinez the selfish
    one, to say he is walking out
    on the Red Sox, his baseball
    family, for the extra year of
    the Mets #39; crazy money." + ], + [ + "It is impossible for young
    tennis players today to know
    what it was like to be Althea
    Gibson and not to be able to
    quot;walk in the front door,
    quot; Garrison said." + ], + [ + "Senator John Kerry said today
    that the war in Iraq was a
    \"profound diversion\" from the
    war on terror and Osama bin
    Laden." + ], + [ + "For spammers, it #39;s been a
    summer of love. Two newly
    issued reports tracking the
    circulation of unsolicited
    e-mails say pornographic spam
    dominated this summer, nearly
    all of it originating from
    Internet addresses in North
    America." + ], + [ + "The Nordics fared well because
    of their long-held ideals of
    keeping corruption clamped
    down and respect for
    contracts, rule of law and
    dedication to one-on-one
    business relationships." + ], + [ + "Microsoft portrayed its
    Longhorn decision as a
    necessary winnowing to hit the
    2006 timetable. The
    announcement on Friday,
    Microsoft executives insisted,
    did not point to a setback in
    software" + ], + [ + "Here #39;s an obvious word of
    advice to Florida athletic
    director Jeremy Foley as he
    kicks off another search for
    the Gators football coach: Get
    Steve Spurrier on board." + ], + [ + "Don't bother with the small
    stuff. Here's what really
    matters to your lender." + ], + [ + "A problem in the Service Pack
    2 update for Windows XP may
    keep owners of AMD-based
    computers from using the long-
    awaited security package,
    according to Microsoft." + ], + [ + "Five years ago, running a
    telephone company was an
    immensely profitable
    proposition. Since then, those
    profits have inexorably
    declined, and now that decline
    has taken another gut-
    wrenching dip." + ], + [ + "NEW YORK - The litigious
    Recording Industry Association
    of America (RIAA) is involved
    in another legal dispute with
    a P-to-P (peer-to-peer)
    technology maker, but this
    time, the RIAA is on defense.
    Altnet Inc. filed a lawsuit
    Wednesday accusing the RIAA
    and several of its partners of
    infringing an Altnet patent
    covering technology for
    identifying requested files on
    a P-to-P network." + ], + [ + "A group claiming to have
    captured two Indonesian women
    in Iraq has said it will
    release them if Jakarta frees
    Muslim cleric Abu Bakar Bashir
    being held for alleged
    terrorist links." + ], + [ + "Amid the stormy gloom in
    Gotham, the rain-idled Yankees
    last night had plenty of time
    to gather in front of their
    televisions and watch the Red
    Sox Express roar toward them.
    The national telecast might
    have been enough to send a
    jittery Boss Steinbrenner
    searching his Bartlett's
    Familiar Quotations for some
    quot;Little Engine That Could
    quot; metaphor." + ], + [ + "FULHAM fans would have been
    singing the late Elvis #39;
    hit #39;The wonder of you
    #39; to their player Elvis
    Hammond. If not for Frank
    Lampard spoiling the party,
    with his dedication to his
    late grandfather." + ], + [ + "Indonesian police said
    yesterday that DNA tests had
    identified a suicide bomber
    involved in a deadly attack
    this month on the Australian
    embassy in Jakarta." + ], + [ + "NEW YORK - Wal-Mart Stores
    Inc.'s warning of
    disappointing sales sent
    stocks fluctuating Monday as
    investors' concerns about a
    slowing economy offset their
    relief over a drop in oil
    prices. October contracts
    for a barrel of light crude
    were quoted at \\$46.48, down
    24 cents, on the New York
    Mercantile Exchange..." + ], + [ + "Iraq #39;s top Shi #39;ite
    cleric made a sudden return to
    the country on Wednesday and
    said he had a plan to end an
    uprising in the quot;burning
    city quot; of Najaf, where
    fighting is creeping ever
    closer to its holiest shrine." + ], + [ + "For two weeks before MTV
    debuted U2 #39;s video for the
    new single quot;Vertigo,
    quot; fans had a chance to see
    the band perform the song on
    TV -- in an iPod commercial." + ], + [ + "A bird #39;s eye view of the
    circuit at Shanghai shows what
    an event Sunday #39;s Chinese
    Grand Prix will be. The course
    is arguably one of the best
    there is, and so it should be
    considering the amount of
    money that has been spent on
    it." + ], + [ + "KABUL, Afghanistan Aug. 22,
    2004 - US soldiers sprayed a
    pickup truck with bullets
    after it failed to stop at a
    roadblock in central
    Afghanistan, killing two women
    and a man and critically
    wounding two other" + ], + [ + "Oct. 26, 2004 - The US-
    European spacecraft Cassini-
    Huygens on Tuesday made a
    historic flyby of Titan,
    Saturn #39;s largest moon,
    passing so low as to almost
    touch the fringes of its
    atmosphere." + ], + [ + "Reuters - A volcano in central
    Japan sent smoke and\\ash high
    into the sky and spat out
    molten rock as it erupted\\for
    a fourth straight day on
    Friday, but experts said the
    peak\\appeared to be quieting
    slightly." + ], + [ + "Shares of Google Inc. made
    their market debut on Thursday
    and quickly traded up 19
    percent at \\$101.28. The Web
    search company #39;s initial
    public offering priced at \\$85" + ], + [ + "SYDNEY -- Prime Minister John
    Howard of Australia, a key US
    ally and supporter of the Iraq
    war, celebrated his election
    win over opposition Labor
    after voters enjoying the
    fruits of a strong economy
    gave him another term." + ], + [ + "Reuters - Global warming is
    melting\\Ecuador's cherished
    mountain glaciers and could
    cause several\\of them to
    disappear over the next two
    decades, Ecuadorean and\\French
    scientists said on Wednesday." + ], + [ + "AP - Sirius Satellite Radio
    signed a deal to air the men's
    NCAA basketball tournament
    through 2007, the latest move
    made in an attempt to draw
    customers through sports
    programming." + ], + [ + "Rather than tell you, Dan
    Kranzler chooses instead to
    show you how he turned Mforma
    into a worldwide publisher of
    video games, ringtones and
    other hot downloads for mobile
    phones." + ], + [ + "UK interest rates have been
    kept on hold at 4.75 following
    the latest meeting of the Bank
    of England #39;s rate-setting
    committee." + ], + [ + "BAGHDAD, Iraq - Two rockets
    hit a downtown Baghdad hotel
    housing foreigners and
    journalists Thursday, and
    gunfire erupted in the
    neighborhood across the Tigris
    River from the U.S. Embassy
    compound..." + ], + [ + "The Prevention of Terrorism
    Act 2002 (Pota) polarised the
    country, not just by the
    manner in which it was pushed
    through by the NDA government
    through a joint session of
    Parliament but by the shabby
    and often biased manner in
    which it was enforced." + ], + [ + "Not being part of a culture
    with a highly developed
    language, could limit your
    thoughts, at least as far as
    numbers are concerned, reveals
    a new study conducted by a
    psychologist at the Columbia
    University in New York." + ], + [ + "CAMBRIDGE, Mass. A native of
    Red Oak, Iowa, who was a
    pioneer in astronomy who
    proposed the quot;dirty
    snowball quot; theory for the
    substance of comets, has died." + ], + [ + "Resurgent oil prices paused
    for breath as the United
    States prepared to draw on its
    emergency reserves to ease
    supply strains caused by
    Hurricane Ivan." + ], + [ + "(Sports Network) - The
    inconsistent San Diego Padres
    will try for consecutive wins
    for the first time since
    August 28-29 tonight, when
    they begin a huge four-game
    set against the Los Angeles
    Dodgers at Dodger Stadium." + ], + [ + "A Portuguese-sounding version
    of the virus has appeared in
    the wild. Be wary of mail from
    Manaus." + ], + [ + " NEW YORK (Reuters) - Top seed
    Roger Federer survived a
    stirring comeback from twice
    champion Andre Agassi to reach
    the semifinals of the U.S.
    Open for the first time on
    Thursday, squeezing through
    6-3, 2-6, 7-5, 3-6, 6-3." + ], + [ + "President Bush, who credits
    three years of tax relief
    programs with helping
    strengthen the slow economy,
    said Saturday he would sign
    into law the Working Families
    Tax Relief Act to preserve tax
    cuts." + ], + [ + "HEN investors consider the
    bond market these days, the
    low level of interest rates
    should be more cause for worry
    than for gratitude." + ], + [ + "Reuters - Hunters soon may be
    able to sit at\\their computers
    and blast away at animals on a
    Texas ranch via\\the Internet,
    a prospect that has state
    wildlife officials up\\in arms." + ], + [ + "The Bedminster-based company
    yesterday said it was pushing
    into 21 new markets with the
    service, AT amp;T CallVantage,
    and extending an introductory
    rate offer until Sept. 30. In
    addition, the company is
    offering in-home installation
    of up to five ..." + ], + [ + "Samsung Electronics Co., Ltd.
    has developed a new LCD
    (liquid crystal display)
    technology that builds a touch
    screen into the display, a
    development that could lead to
    thinner and cheaper display
    panels for mobile phones, the
    company said Tuesday." + ], + [ + "The US military says marines
    in Fallujah shot and killed an
    insurgent who engaged them as
    he was faking being dead, a
    week after footage of a marine
    killing an apparently unarmed
    and wounded Iraqi caused a
    stir in the region." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks rallied on Monday after
    software maker PeopleSoft Inc.
    <A HREF=\"http://www.investo
    r.reuters.com/FullQuote.aspx?t
    icker=PSFT.O target=/stocks/qu
    ickinfo/fullquote\">PSFT.O&l
    t;/A> accepted a sweetened
    \\$10.3 billion buyout by rival
    Oracle Corp.'s <A HREF=\"htt
    p://www.investor.reuters.com/F
    ullQuote.aspx?ticker=ORCL.O ta
    rget=/stocks/quickinfo/fullquo
    te\">ORCL.O</A> and
    other big deals raised
    expectations of more
    takeovers." + ], + [ + "SAN FRANCISCO - What Babe Ruth
    was to the first half of the
    20th century and Hank Aaron
    was to the second, Barry Bonds
    has become for the home run
    generation." + ], + [ + "Description: NPR #39;s Alex
    Chadwick talks to Colin Brown,
    deputy political editor for
    the United Kingdom #39;s
    Independent newspaper,
    currently covering the British
    Labour Party Conference." + ], + [ + "Birgit Fischer settled for
    silver, leaving the 42-year-
    old Olympian with two medals
    in two days against decidedly
    younger competition." + ], + [ + "The New Jersey-based Accoona
    Corporation, an industry
    pioneer in artificial
    intelligence search
    technology, announced on
    Monday the launch of Accoona." + ], + [ + "Hamas vowed revenge yesterday
    after an Israeli airstrike in
    Gaza killed one of its senior
    commanders - the latest
    assassination to have weakened
    the militant group." + ], + [ + "There was no mystery ... no
    secret strategy ... no baited
    trap that snapped shut and
    changed the course of history
    #39;s most lucrative non-
    heavyweight fight." + ], + [ + "Notre Dame accepted an
    invitation Sunday to play in
    the Insight Bowl in Phoenix
    against a Pac-10 team on Dec.
    28. The Irish (6-5) accepted
    the bid a day after losing to
    Southern California" + ], + [ + "Greg Anderson has so dominated
    Pro Stock this season that his
    championship quest has evolved
    into a pursuit of NHRA
    history. By Bob Hesser, Racers
    Edge Photography." + ], + [ + "Goldman Sachs Group Inc. on
    Thursday said fourth-quarter
    profit rose as its fixed-
    income, currency and
    commodities business soared
    while a rebounding stock
    market boosted investment
    banking." + ], + [ + " NEW YORK (Reuters) - The
    dollar rose on Monday in a
    retracement from last week's
    steep losses, but dealers said
    the bias toward a weaker
    greenback remained intact." + ], + [ + "Michael Powell, chairman of
    the FCC, said Wednesday he was
    disappointed with ABC for
    airing a sexually suggestive
    opening to \"Monday Night
    Football.\"" + ], + [ + "The message against illegally
    copying CDs for uses such as
    in file-sharing over the
    Internet has widely sunk in,
    said the company in it #39;s
    recent announcement to drop
    the Copy-Control program." + ], + [ + "Former Washington football
    coach Rick Neuheisel looked
    forward to a return to
    coaching Wednesday after being
    cleared by the NCAA of
    wrongdoing related to his
    gambling on basketball games." + ], + [ + "Reuters - California will
    become hotter and\\drier by the
    end of the century, menacing
    the valuable wine and\\dairy
    industries, even if dramatic
    steps are taken to curb\\global
    warming, researchers said on
    Monday." + ], + [ + "A senior member of the
    Palestinian resistance group
    Hamas has been released from
    an Israeli prison after
    completing a two-year
    sentence." + ], + [ + "IBM said Monday that it won a
    500 million (AUD\\$1.25
    billion), seven-year services
    contract to help move UK bank
    Lloyds TBS from its
    traditional voice
    infrastructure to a converged
    voice and data network." + ], + [ + "MPs have announced a new
    inquiry into family courts and
    whether parents are treated
    fairly over issues such as
    custody or contact with their
    children." + ], + [ + "Canadian Press - MELBOURNE,
    Australia (AP) - A 36-year-old
    businesswoman was believed to
    be the first woman to walk
    around Australia on Friday
    after striding into her
    hometown of Melbourne to
    complete her 16,700-kilometre
    trek in 365 days." + ], + [ + "Most remaining Pakistani
    prisoners held at the US
    Guantanamo Bay prison camp are
    freed, officials say." + ], + [ + "Space shuttle astronauts will
    fly next year without the
    ability to repair in orbit the
    type of damage that destroyed
    the Columbia vehicle in
    February 2003." + ], + [ + "Moscow - Russia plans to
    combine Gazprom, the world
    #39;s biggest natural gas
    producer, with state-owned oil
    producer Rosneft, easing rules
    for trading Gazprom shares and
    creating a company that may
    dominate the country #39;s
    energy industry." + ], + [ + "AP - Rutgers basketball player
    Shalicia Hurns was suspended
    from the team after pleading
    guilty to punching and tying
    up her roommate during a
    dispute over painkilling
    drugs." + ], + [ + "By cutting WinFS from Longhorn
    and indefinitely delaying the
    storage system, Microsoft
    Corp. has also again delayed
    the Microsoft Business
    Framework (MBF), a new Windows
    programming layer that is
    closely tied to WinFS." + ], + [ + "French police are
    investigating an arson-caused
    fire at a Jewish Social Center
    that might have killed dozens
    without the quick response of
    firefighters." + ], + [ + "Rodney King, whose videotaped
    beating led to riots in Los
    Angeles in 1992, is out of
    jail now and talking frankly
    for the first time about the
    riots, himself and the
    American way of life." + ], + [ + "AFP - Radical Islamic cleric
    Abu Hamza al-Masri was set to
    learn Thursday whether he
    would be charged under
    Britain's anti-terrorism law,
    thus delaying his possible
    extradition to the United
    States to face terrorism-
    related charges." + ], + [ + "Diversified manufacturer
    Honeywell International Inc.
    (HON.N: Quote, Profile,
    Research) posted a rise in
    quarterly profit as strong
    demand for aerospace equipment
    and automobile components" + ], + [ + "This was the event Michael
    Phelps didn't really need to
    compete in if his goal was to
    win eight golds. He probably
    would have had a better chance
    somewhere else." + ], + [ + "Samsung's new SPH-V5400 mobile
    phone sports a built-in
    1-inch, 1.5-gigabyte hard disk
    that can store about 15 times
    more data than conventional
    handsets, Samsung said." + ], + [ + "Reuters - U.S. housing starts
    jumped a\\larger-than-expected
    6.4 percent in October to the
    busiest pace\\since December as
    buyers took advantage of low
    mortgage rates,\\a government
    report showed on Wednesday." + ], + [ + "Gabe Kapler became the first
    player to leave the World
    Series champion Boston Red
    Sox, agreeing to a one-year
    contract with the Yomiuri
    Giants in Tokyo." + ], + [ + "Louisen Louis, 30, walked
    Monday in the middle of a
    street that resembled a small
    river with brown rivulets and
    waves. He wore sandals and had
    a cut on one of his big toes." + ], + [ + "BALI, Indonesia - Svetlana
    Kuznetsova, fresh off her
    championship at the US Open,
    defeated Australian qualifier
    Samantha Stosur 6-4, 6-4
    Thursday to reach the
    quarterfinals of the Wismilak
    International." + ], + [ + "The Securities and Exchange
    Commission ordered mutual
    funds to stop paying higher
    commissions to brokers who
    promote the companies' funds
    and required portfolio
    managers to reveal investments
    in funds they supervise." + ], + [ + "A car bomb exploded outside
    the main hospital in Chechny
    #39;s capital, Grozny, on
    Sunday, injuring 17 people in
    an attack apparently targeting
    members of a Chechen security
    force bringing in wounded from
    an earlier explosion" + ], + [ + "AP - Just like the old days in
    Dallas, Emmitt Smith made life
    miserable for the New York
    Giants on Sunday." + ], + [ + "Sumitomo Mitsui Financial
    Group (SMFG), Japans second
    largest bank, today put
    forward a 3,200 billion (\\$29
    billion) takeover bid for
    United Financial Group (UFJ),
    the countrys fourth biggest
    lender, in an effort to regain
    initiative in its bidding" + ], + [ + "AP - Gay marriage is emerging
    as a big enough issue in
    several states to influence
    races both for Congress and
    the presidency." + ], + [ + "TAMPA, Fla. - Chris Simms
    first NFL start lasted 19
    plays, and it might be a while
    before he plays again for the
    Tampa Bay Buccaneers." + ], + [ + "Vendor says it #39;s
    developing standards-based
    servers in various form
    factors for the telecom
    market. By Darrell Dunn.
    Hewlett-Packard on Thursday
    unveiled plans to create a
    portfolio of products and
    services" + ], + [ + "Jarno Trulli made the most of
    the conditions in qualifying
    to claim pole ahead of Michael
    Schumacher, while Fernando
    finished third." + ], + [ + "More than 30 aid workers have
    been airlifted to safety from
    a town in Sudan #39;s troubled
    Darfur region after fighting
    broke out and their base was
    bombed, a British charity
    says." + ], + [ + " NEW YORK (Reuters) - U.S.
    chain store retail sales
    slipped during the
    Thanksgiving holiday week, as
    consumers took advantage of
    discounted merchandise, a
    retail report said on
    Tuesday." + ], + [ + "The Mac maker says it will
    replace about 28,000 batteries
    in one model of PowerBook G4
    and tells people to stop using
    the notebook." + ], + [ + "It #39;s the mildest of mild
    winters down here in the south
    of Italy and, last weekend at
    Bcoli, a pretty suburb by the
    seaside west of Naples, the
    customers of Pizzeria quot;Da
    Enrico quot; were making the
    most of it." + ], + [ + "By George Chamberlin , Daily
    Transcript Financial
    Correspondent. Concerns about
    oil production leading into
    the winter months sent shivers
    through the stock market
    Wednesday." + ], + [ + "Airbus has withdrawn a filing
    that gave support for
    Microsoft in an antitrust case
    before the European Union
    #39;s Court of First Instance,
    a source close to the
    situation said on Friday." + ], + [ + "WASHINGTON - A spotty job
    market and stagnant paychecks
    cloud this Labor Day holiday
    for many workers, highlighting
    the importance of pocketbook
    issues in the presidential
    election. \"Working harder
    and enjoying it less,\" said
    economist Ken Mayland,
    president of ClearView
    Economics, summing up the
    state of working America..." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks rose on Wednesday
    lifted by a merger between
    retailers Kmart and Sears,
    better-than-expected earnings
    from Hewlett-Packard and data
    showing a slight rise in core
    inflation." + ], + [ + "AP - Authorities are
    investigating whether bettors
    at New York's top thoroughbred
    tracks were properly informed
    when jockeys came in
    overweight at races, a source
    familiar with the probe told
    The Associated Press." + ], + [ + "European Commission president
    Romano Prodi has unveiled
    proposals to loosen the
    deficit rules under the EU
    Stability Pact. The loosening
    was drafted by monetary
    affairs commissioner Joaquin
    Almunia, who stood beside the
    president at the announcement." + ], + [ + "Canadian Press - MONTREAL (CP)
    - A 19-year-old man charged in
    a firebombing at a Jewish
    elementary school pleaded
    guilty Thursday to arson." + ], + [ + "Retail sales in Britain saw
    the fastest growth in
    September since January,
    casting doubts on the view
    that the economy is slowing
    down, according to official
    figures released Thursday." + ], + [ + " NEW YORK (Reuters) -
    Interstate Bakeries Corp.
    <A HREF=\"http://www.investo
    r.reuters.com/FullQuote.aspx?t
    icker=IBC.N target=/stocks/qui
    ckinfo/fullquote\">IBC.N<
    /A>, maker of Hostess
    Twinkies and Wonder Bread,
    filed for bankruptcy on
    Wednesday after struggling
    with more than \\$1.3 billion
    in debt and high costs." + ], + [ + "Delta Air Lines (DAL.N: Quote,
    Profile, Research) on Thursday
    said it reached a deal with
    FedEx Express to sell eight
    McDonnell Douglas MD11
    aircraft and four spare
    engines for delivery in 2004." + ], + [ + "Michael Owen scored his first
    goal for Real Madrid in a 1-0
    home victory over Dynamo Kiev
    in the Champions League. The
    England striker toe-poked home
    Ronaldo #39;s cross in the
    35th minute to join the
    Russians" + ], + [ + " quot;Resuming uranium
    enrichment is not in our
    agenda. We are still committed
    to the suspension, quot;
    Foreign Ministry spokesman
    Hamid Reza." + ], + [ + "Leading OPEC producer Saudi
    Arabia said on Monday in
    Vienna, Austria, that it had
    made a renewed effort to
    deflate record high world oil
    prices by upping crude output
    again." + ], + [ + "The U.S. military presence in
    Iraq will grow to 150,000
    troops by next month, the
    highest level since the
    invasion last year." + ], + [ + "UPDATE, SUN 9PM: More than a
    million people have left their
    homes in Cuba, as Hurricane
    Ivan approaches. The ferocious
    storm is headed that way,
    after ripping through the
    Cayman Islands, tearing off
    roofs, flooding homes and
    causing general havoc." + ], + [ + "Jason Giambi has returned to
    the New York Yankees'
    clubhouse but is still
    clueless as to when he will be
    able to play again." + ], + [ + "Seventy-five National Hockey
    League players met with union
    leaders yesterday to get an
    update on a lockout that shows
    no sign of ending." + ], + [ + "Howard, at 6-4 overall and 3-3
    in the Mid-Eastern Athletic
    Conference, can clinch a
    winning record in the MEAC
    with a win over Delaware State
    on Saturday." + ], + [ + "Millions of casual US anglers
    are having are larger than
    appreciated impact on sea fish
    stocks, scientists claim." + ], + [ + "The founders of the Pilgrim
    Baxter amp; Associates money-
    management firm agreed
    yesterday to personally fork
    over \\$160 million to settle
    charges they allowed a friend
    to" + ], + [ + " ATHENS (Reuters) - Top-ranked
    Argentina booked their berth
    in the women's hockey semi-
    finals at the Athens Olympics
    on Friday but defending
    champions Australia now face
    an obstacle course to qualify
    for the medal matches." + ], + [ + "IBM Corp. Tuesday announced
    plans to acquire software
    vendor Systemcorp ALG for an
    undisclosed amount. Systemcorp
    of Montreal makes project
    portfolio management software
    aimed at helping companies
    better manage their IT
    projects." + ], + [ + "The Notre Dame message boards
    are no longer discussing
    whether Tyrone Willingham
    should be fired. Theyre
    already arguing about whether
    the next coach should be Barry
    Alvarez or Steve Spurrier." + ], + [ + "Forbes.com - By now you
    probably know that earnings of
    Section 529 college savings
    accounts are free of federal
    tax if used for higher
    education. But taxes are only
    part of the problem. What if
    your investments tank? Just
    ask Laurence and Margo
    Williams of Alexandria, Va. In
    2000 they put #36;45,000 into
    the Virginia Education Savings
    Trust to open accounts for
    daughters Lea, now 5, and
    Anne, now 3. Since then their
    investment has shrunk 5 while
    the average private college
    tuition has climbed 18 to
    #36;18,300." + ], + [ + "German Chancellor Gerhard
    Schroeder said Sunday that
    there was quot;no problem
    quot; with Germany #39;s
    support to the start of
    negotiations on Turkey #39;s
    entrance into EU." + ], + [ + "Coca-Cola Amatil Ltd.,
    Australia #39;s biggest soft-
    drink maker, offered A\\$500
    million (\\$382 million) in
    cash and stock for fruit
    canner SPC Ardmona Ltd." + ], + [ + "US technology shares tumbled
    on Friday after technology
    bellwether Intel Corp.
    (INTC.O: Quote, Profile,
    Research) slashed its revenue
    forecast, but blue chips were
    only moderately lower as drug
    and industrial stocks made
    solid gains." + ], + [ + " WASHINGTON (Reuters) - Final
    U.S. government tests on an
    animal suspected of having mad
    cow disease were not yet
    complete, the U.S. Agriculture
    Department said, with no
    announcement on the results
    expected on Monday." + ], + [ + "MANILA Fernando Poe Jr., the
    popular actor who challenged
    President Gloria Macapagal
    Arroyo in the presidential
    elections this year, died
    early Tuesday." + ], + [ + " THOMASTOWN, Ireland (Reuters)
    - World number three Ernie
    Els overcame difficult weather
    conditions to fire a sparkling
    eight-under-par 64 and move
    two shots clear after two
    rounds of the WGC-American
    Express Championship Friday." + ], + [ + "Lamar Odom supplemented 20
    points with 13 rebounds and
    Kobe Bryant added 19 points to
    overcome a big night from Yao
    Ming as the Los Angeles Lakers
    ground out an 84-79 win over
    the Rockets in Houston
    Saturday." + ], + [ + "AMSTERDAM, NETHERLANDS - A
    Dutch filmmaker who outraged
    members of the Muslim
    community by making a film
    critical of the mistreatment
    of women in Islamic society
    was gunned down and stabbed to
    death Tuesday on an Amsterdam
    street." + ], + [ + "Microsoft Xbox Live traffic on
    service provider networks
    quadrupled following the
    November 9th launch of Halo-II
    -- which set entertainment
    industry records by selling
    2.4-million units in the US
    and Canada on the first day of
    availability, driving cash" + ], + [ + "Lawyers in a California class
    action suit against Microsoft
    will get less than half the
    payout they had hoped for. A
    judge in San Francisco ruled
    that the attorneys will
    collect only \\$112." + ], + [ + "Google Browser May Become
    Reality\\\\There has been much
    fanfare in the Mozilla fan
    camps about the possibility of
    Google using Mozilla browser
    technology to produce a
    GBrowser - the Google Browser.
    Over the past two weeks, the
    news and speculation has
    escalated to the point where
    even Google itself is ..." + ], + [ + "Metro, Germany's biggest
    retailer, turns in weaker-
    than-expected profits as sales
    at its core supermarkets
    division dip lower." + ], + [ + "It rained Sunday, of course,
    and but another soppy, sloppy
    gray day at Westside Tennis
    Club did nothing to deter
    Roger Federer from his
    appointed rounds." + ], + [ + "Hewlett-Packard has joined
    with Brocade to integrate
    Brocade #39;s storage area
    network switching technology
    into HP Bladesystem servers to
    reduce the amount of fabric
    infrastructure needed in a
    datacentre." + ], + [ + "Zimbabwe #39;s most persecuted
    white MP began a year of hard
    labour last night after
    parliament voted to jail him
    for shoving the Justice
    Minister during a debate over
    land seizures." + ], + [ + "BOSTON (CBS.MW) -- First
    Command has reached a \\$12
    million settlement with
    federal regulators for making
    misleading statements and
    omitting important information
    when selling mutual funds to
    US military personnel." + ], + [ + "A smashing blow is being dealt
    to thousands of future
    pensioners by a law that has
    just been brought into force
    by the Federal Government." + ], + [ + "The US Senate Commerce
    Committee on Wednesday
    approved a measure that would
    provide up to \\$1 billion to
    ensure consumers can still
    watch television when
    broadcasters switch to new,
    crisp digital signals." + ], + [ + "Amelie Mauresmo was handed a
    place in the Advanta
    Championships final after
    Maria Sharapova withdrew from
    their semi-final because of
    injury." + ], + [ + "AP - An Israeli helicopter
    fired two missiles in Gaza
    City after nightfall
    Wednesday, one at a building
    in the Zeitoun neighborhood,
    witnesses said, setting a
    fire." + ], + [ + "Reuters - Internet stocks
    are\\as volatile as ever, with
    growth-starved investors
    flocking to\\the sector in the
    hope they've bought shares in
    the next online\\blue chip." + ], + [ + "The federal government, banks
    and aircraft lenders are
    putting the clamps on
    airlines, particularly those
    operating under bankruptcy
    protection." + ], + [ + "EURO DISNEY, the financially
    crippled French theme park
    operator, has admitted that
    its annual losses more than
    doubled last financial year as
    it was hit by a surge in
    costs." + ], + [ + " EAST RUTHERFORD, New Jersey
    (Sports Network) - Retired NBA
    center and seven-time All-
    Star Alonzo Mourning is going
    to give his playing career
    one more shot." + ], + [ + "NASA has released an inventory
    of the scientific devices to
    be put on board the Mars
    Science Laboratory rover
    scheduled to land on the
    surface of Mars in 2009, NASAs
    news release reads." + ], + [ + "The U.S. Congress needs to
    invest more in the U.S.
    education system and do more
    to encourage broadband
    adoption, the chief executive
    of Cisco said Wednesday.<p&
    gt;ADVERTISEMENT</p><
    p><img src=\"http://ad.do
    ubleclick.net/ad/idg.us.ifw.ge
    neral/sbcspotrssfeed;sz=1x1;or
    d=200301151450?\" width=\"1\"
    height=\"1\"
    border=\"0\"/><a href=\"htt
    p://ad.doubleclick.net/clk;922
    8975;9651165;a?http://www.info
    world.com/spotlights/sbc/main.
    html?lpid0103035400730000idlp\"
    >SBC Case Study: Crate Ba
    rrel</a><br/>What
    sold them on improving their
    network? A system that could
    cut management costs from the
    get-go. Find out
    more.</p>" + ], + [ + "The Philippines put the toll
    at more than 1,000 dead or
    missing in four storms in two
    weeks but, even with a break
    in the weather on Saturday" + ], + [ + "Reuters - Four explosions were
    reported at petrol\\stations in
    the Madrid area on Friday,
    Spanish radio stations\\said,
    following a phone warning in
    the name of the armed
    Basque\\separatist group ETA to
    a Basque newspaper." + ], + [ + "Thirty-two countries and
    regions will participate the
    Fifth China International
    Aviation and Aerospace
    Exhibition, opening Nov. 1 in
    Zhuhai, a city in south China
    #39;s Guangdong Province." + ], + [ + "Jordan have confirmed that
    Timo Glock will replace
    Giorgio Pantano for this
    weekend #39;s Chinese GP as
    the team has terminated its
    contract with Pantano." + ], + [ + "WEST PALM BEACH, Fla. -
    Hurricane Jeanne got stronger,
    bigger and faster as it
    battered the Bahamas and bore
    down on Florida Saturday,
    sending huge waves crashing
    onto beaches and forcing
    thousands into shelters just
    weeks after Frances ravaged
    this area..." + ], + [ + "p2pnet.net News:- Virgin
    Electronics has joined the mp3
    race with a \\$250, five gig
    player which also handles
    Microsoft #39;s WMA format." + ], + [ + "WASHINGTON The idea of a no-
    bid contract for maintaining
    airport security equipment has
    turned into a non-starter for
    the Transportation Security
    Administration." + ], + [ + "Eyetech (EYET:Nasdaq - news -
    research) did not open for
    trading Friday because a Food
    and Drug Administration
    advisory committee is meeting
    to review the small New York-
    based biotech #39;s
    experimental eye disease drug." + ], + [ + "The continuing heartache of
    Wake Forest #39;s ACC football
    season was best described by
    fifth-ranked Florida State
    coach Bobby Bowden, after his
    Seminoles had edged the
    Deacons 20-17 Saturday at
    Groves Stadium." + ], + [ + "On September 13, 2001, most
    Americans were still reeling
    from the shock of the
    terrorist attacks on New York
    and the Pentagon two days
    before." + ], + [ + "Microsoft has suspended the
    beta testing of the next
    version of its MSN Messenger
    client because of a potential
    security problem, a company
    spokeswoman said Wednesday." + ], + [ + "AP - Business software maker
    Oracle Corp. attacked the
    credibility and motives of
    PeopleSoft Inc.'s board of
    directors Monday, hoping to
    rally investor support as the
    17-month takeover battle
    between the bitter business
    software rivals nears a
    climactic showdown." + ], + [ + "NEW YORK - Elena Dementieva
    shook off a subpar serve that
    produced 15 double-faults, an
    aching left thigh and an upset
    stomach to advance to the
    semifinals at the U.S. Open
    with a 4-6, 6-4, 7-6 (1)
    victory Tuesday over Amelie
    Mauresmo..." + ], + [ + "THE glory days have returned
    to White Hart Lane. When Spurs
    new first-team coach Martin
    Jol promised a return to the
    traditions of the 1960s,
    nobody could have believed he
    was so determined to act so
    quickly and so literally." + ], + [ + "A new worm has been discovered
    in the wild that #39;s not
    just settling for invading
    users #39; PCs--it wants to
    invade their homes too." + ], + [ + "Domestic air travelers could
    be surfing the Web by 2006
    with government-approved
    technology that allows people
    access to high-speed Internet
    connections while they fly." + ], + [ + "GENEVA: Cross-border
    investment is set to bounce in
    2004 after three years of deep
    decline, reflecting a stronger
    world economy and more
    international merger activity,
    the United Nations (UN) said
    overnight." + ], + [ + "Researchers have for the first
    time established the existence
    of odd-parity superconductors,
    materials that can carry
    electric current without any
    resistance." + ], + [ + "Chewing gum giant Wm. Wrigley
    Jr. Co. on Thursday said it
    plans to phase out production
    of its Eclipse breath strips
    at a plant in Phoenix, Arizona
    and shift manufacturing to
    Poznan, Poland." + ], + [ + "Prime Minister Dr Manmohan
    Singh inaugurated a research
    centre in the Capital on
    Thursday to mark 400 years of
    compilation of Sikh holy book
    the Guru Granth Sahib." + ], + [ + "com September 16, 2004, 7:58
    AM PT. This fourth priority
    #39;s main focus has been
    improving or obtaining CRM and
    ERP software for the past year
    and a half." + ], + [ + "BRUSSELS, Belgium (AP) --
    European antitrust regulators
    said Monday they have extended
    their review of a deal between
    Microsoft Corp. (MSFT) and
    Time Warner Inc..." + ], + [ + "AP - When Paula Radcliffe
    dropped out of the Olympic
    marathon miles from the
    finish, she sobbed
    uncontrollably. Margaret Okayo
    knew the feeling. Okayo pulled
    out of the marathon at the
    15th mile with a left leg
    injury, and she cried, too.
    When she watched Radcliffe
    quit, Okayo thought, \"Let's
    cry together.\"" + ], + [ + "Tightness in the labour market
    notwithstanding, the prospects
    for hiring in the third
    quarter are down from the
    second quarter, according to
    the new Manpower Employment
    Outlook Survey." + ], + [ + "THE re-election of British
    Prime Minister Tony Blair
    would be seen as an
    endorsement of the military
    action in Iraq, Prime Minister
    John Howard said today." + ], + [ + "Fans who can't get enough of
    \"The Apprentice\" can visit a
    new companion Web site each
    week and watch an extra 40
    minutes of video not broadcast
    on the Thursday
    show.<br><FONT
    face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\"\\ color=\"#666666\">
    <B>-Leslie
    Walker</b></font>" + ], + [ + " ATLANTA (Sports Network) -
    The Atlanta Hawks signed free
    agent Kevin Willis on
    Wednesday, nearly a decade
    after the veteran big man
    ended an 11- year stint with
    the team." + ], + [ + "An adult Web site publisher is
    suing Google, saying the
    search engine company made it
    easier for users to see the
    site #39;s copyrighted nude
    photographs without paying or
    gaining access through the
    proper channels." + ], + [ + "When his right-front tire went
    flying off early in the Ford
    400, the final race of the
    NASCAR Nextel Cup Series
    season, Kurt Busch, it seemed,
    was destined to spend his
    offseason" + ], + [ + "A Washington-based public
    opinion firm has released the
    results of an election day
    survey of Nevada voters
    showing 81 support for the
    issuance of paper receipts
    when votes are cast
    electronically." + ], + [ + "NAPSTER creator SHAWN FANNING
    has revealed his plans for a
    new licensed file-sharing
    service with an almost
    unlimited selection of tracks." + ], + [ + " NEW YORK (Reuters) - The
    dollar rose on Friday, after a
    U.S. report showed consumer
    prices in line with
    expections, reminding
    investors that the Federal
    Reserve was likely to
    continue raising interest
    rates, analysts said." + ], + [ + "Brandon Backe and Woody
    Williams pitched well last
    night even though neither
    earned a win. But their
    outings will show up in the
    record books." + ], + [ + "President George W. Bush
    pledged Friday to spend some
    of the political capital from
    his re-election trying to
    secure a lasting Middle East
    peace, and he envisioned the
    establishment" + ], + [ + "The Football Association today
    decided not to charge David
    Beckham with bringing the game
    into disrepute. The FA made
    the surprise announcement
    after their compliance unit
    ruled" + ], + [ + "Last year some election
    watchers made a bold
    prediction that this
    presidential election would
    set a record: the first half
    billion dollar campaign in
    hard money alone." + ], + [ + "It #39;s one more blow to
    patients who suffer from
    arthritis. Pfizer, the maker
    of Celebrex, says it #39;s
    painkiller poses an increased
    risk of heart attacks to
    patients using the drugs." + ], + [ + "NEW DELHI - A bomb exploded
    during an Independence Day
    parade in India's remote
    northeast on Sunday, killing
    at least 15 people, officials
    said, just an hour after Prime
    Minister Manmohan Singh
    pledged to fight terrorism.
    The outlawed United Liberation
    Front of Asom was suspected of
    being behind the attack in
    Assam state and a second one
    later in the area, said Assam
    Inspector General of Police
    Khagen Sharma..." + ], + [ + "Two separate studies by U.S.
    researchers find that super
    drug-resistant strains of
    tuberculosis are at the
    tipping point of a global
    epidemic, and only small
    changes could help them spread
    quickly." + ], + [ + " CRANS-SUR-SIERRE, Switzerland
    (Reuters) - World number
    three Ernie Els says he feels
    a failure after narrowly
    missing out on three of the
    year's four major
    championships." + ], + [ + "A UN envoy to Sudan will visit
    Darfur tomorrow to check on
    the government #39;s claim
    that some 70,000 people
    displaced by conflict there
    have voluntarily returned to
    their homes, a spokesman said." + ], + [ + "Dell cut prices on some
    servers and PCs by as much as
    22 percent because it #39;s
    paying less for parts. The
    company will pass the savings
    on components such as memory
    and liquid crystal displays" + ], + [ + "AP - Most of the presidential
    election provisional ballots
    rejected so far in Ohio came
    from people who were not even
    registered to vote, election
    officials said after spending
    nearly two weeks poring over
    thousands of disputed votes." + ], + [ + "Striker Bonaventure Kalou
    netted twice to send AJ
    Auxerre through to the first
    knockout round of the UEFA Cup
    at the expense of Rangers on
    Wednesday." + ], + [ + "AP - Rival inmates fought each
    other with knives and sticks
    Wednesday at a San Salvador
    prison, leaving at least 31
    people dead and two dozen
    injured, officials said." + ], + [ + "WASHINGTON: The European-
    American Cassini-Huygens space
    probe has detected traces of
    ice flowing on the surface of
    Saturn #39;s largest moon,
    Titan, suggesting the
    existence of an ice volcano,
    NASA said Tuesday." + ], + [ + "The economic growth rate in
    the July-September period was
    revised slightly downward from
    an already weak preliminary
    report, the government said
    Wednesday." + ], + [ + "All ISS systems continue to
    function nominally, except
    those noted previously or
    below. Day 7 of joint
    Exp.9/Exp.10 operations and
    last full day before 8S
    undocking." + ], + [ + "BAGHDAD - Two Egyptian
    employees of a mobile phone
    company were seized when
    gunmen stormed into their
    Baghdad office, the latest in
    a series of kidnappings in the
    country." + ], + [ + "BRISBANE, Australia - The body
    of a whale resembling a giant
    dolphin that washed up on an
    eastern Australian beach has
    intrigued local scientists,
    who agreed Wednesday that it
    is rare but are not sure just
    how rare." + ], + [ + " quot;Magic can happen. quot;
    Sirius Satellite Radio
    (nasdaq: SIRI - news - people
    ) may have signed Howard Stern
    and the men #39;s NCAA
    basketball tournaments, but XM
    Satellite Radio (nasdaq: XMSR
    - news - people ) has its
    sights on your cell phone." + ], + [ + "Trick-or-treaters can expect
    an early Halloween treat on
    Wednesday night, when a total
    lunar eclipse makes the moon
    look like a glowing pumpkin." + ], + [ + "THIS weekend sees the
    quot;other quot; showdown
    between New York and New
    England as the Jets and
    Patriots clash in a battle of
    the unbeaten teams." + ], + [ + "Sifting through millions of
    documents to locate a valuable
    few is tedious enough, but
    what happens when those files
    are scattered across different
    repositories?" + ], + [ + "President Bush aims to
    highlight American drug-
    fighting aid in Colombia and
    boost a conservative Latin
    American leader with a stop in
    the Andean nation where
    thousands of security forces
    are deployed to safeguard his
    brief stay." + ], + [ + "Dubai - Former Palestinian
    security minister Mohammed
    Dahlan said on Monday that a
    quot;gang of mercenaries quot;
    known to the Palestinian
    police were behind the
    shooting that resulted in two
    deaths in a mourning tent for
    Yasser Arafat in Gaza." + ], + [ + "A drug company executive who
    spoke out in support of
    Montgomery County's proposal
    to import drugs from Canada
    and similar legislation before
    Congress said that his company
    has launched an investigation
    into his political activities." + ], + [ + "Nicolas Anelka is fit for
    Manchester City #39;s
    Premiership encounter against
    Tottenham at Eastlands, but
    the 13million striker will
    have to be content with a
    place on the bench." + ], + [ + " PITTSBURGH (Reuters) - Ben
    Roethlisberger passed for 183
    yards and two touchdowns,
    Hines Ward scored twice and
    the Pittsburgh Steelers
    rolled to a convincing 27-3
    victory over Philadelphia on
    Sunday for their second
    straight win against an
    undefeated opponent." + ], + [ + " ATHENS (Reuters) - The U.S.
    men's basketball team got
    their first comfortable win
    at the Olympic basketball
    tournament Monday, routing
    winless Angola 89-53 in their
    final preliminary round game." + ], + [ + "A Frenchman working for Thales
    SA, Europe #39;s biggest maker
    of military electronics, was
    shot dead while driving home
    at night in the Saudi Arabian
    city of Jeddah." + ], + [ + "Often, the older a pitcher
    becomes, the less effective he
    is on the mound. Roger Clemens
    apparently didn #39;t get that
    memo. On Tuesday, the 42-year-
    old Clemens won an
    unprecedented" + ], + [ + "NASA #39;s Mars rovers have
    uncovered more tantalizing
    evidence of a watery past on
    the Red Planet, scientists
    said Wednesday. And the
    rovers, Spirit and
    Opportunity, are continuing to
    do their jobs months after
    they were expected to ..." + ], + [ + "SYDNEY (AFP) - Australia #39;s
    commodity exports are forecast
    to increase by 15 percent to a
    record 95 billion dollars (71
    million US), the government
    #39;s key economic forecaster
    said." + ], + [ + "Google won a major legal
    victory when a federal judge
    ruled that the search engines
    advertising policy does not
    violate federal trademark
    laws." + ], + [ + "Intel Chief Technology Officer
    Pat Gelsinger said on
    Thursday, Sept. 9, that the
    Internet needed to be upgraded
    in order to deal with problems
    that will become real issues
    soon." + ], + [ + "China will take tough measures
    this winter to improve the
    country #39;s coal mine safety
    and prevent accidents. State
    Councilor Hua Jianmin said
    Thursday the industry should
    take" + ], + [ + "AT amp;T Corp. on Thursday
    said it is reducing one fifth
    of its workforce this year and
    will record a non-cash charge
    of approximately \\$11." + ], + [ + "BAGHDAD (Iraq): As the
    intensity of skirmishes
    swelled on the soils of Iraq,
    dozens of people were put to
    death with toxic shots by the
    US helicopter gunship, which
    targeted the civilians,
    milling around a burning
    American vehicle in a Baghdad
    street on" + ], + [ + " LONDON (Reuters) - Television
    junkies of the world, get
    ready for \"Friends,\" \"Big
    Brother\" and \"The Simpsons\" to
    phone home." + ], + [ + "A rift appeared within Canada
    #39;s music industry yesterday
    as prominent artists called on
    the CRTC to embrace satellite
    radio and the industry warned
    of lost revenue and job
    losses." + ], + [ + "Reuters - A key Iranian
    nuclear facility which
    the\\U.N.'s nuclear watchdog
    has urged Tehran to shut down
    is\\nearing completion, a
    senior Iranian nuclear
    official said on\\Sunday." + ], + [ + "Spain's Football Federation
    launches an investigation into
    racist comments made by
    national coach Luis Aragones." + ], + [ + "Bricks and plaster blew inward
    from the wall, as the windows
    all shattered and I fell to
    the floorwhether from the
    shock wave, or just fright, it
    wasn #39;t clear." + ], + [ + "Surfersvillage Global Surf
    News, 13 September 2004: - -
    Hurricane Ivan, one of the
    most powerful storms to ever
    hit the Caribbean, killed at
    least 16 people in Jamaica,
    where it wrecked houses and
    washed away roads on Saturday,
    but appears to have spared" + ], + [ + "I #39;M FEELING a little bit
    better about the hundreds of
    junk e-mails I get every day
    now that I #39;ve read that
    someone else has much bigger
    e-mail troubles." + ], + [ + "NEW DELHI: India and Pakistan
    agreed on Monday to step up
    cooperation in the energy
    sector, which could lead to
    Pakistan importing large
    amounts of diesel fuel from
    its neighbour, according to
    Pakistani Foreign Minister
    Khurshid Mehmood Kasuri." + ], + [ + "LONDON, England -- A US
    scientist is reported to have
    observed a surprising jump in
    the amount of carbon dioxide,
    the main greenhouse gas." + ], + [ + "Microsoft's antispam Sender ID
    technology continues to get
    the cold shoulder. Now AOL
    adds its voice to a growing
    chorus of businesses and
    organizations shunning the
    proprietary e-mail
    authentication system." + ], + [ + "PSV Eindhoven faces Arsenal at
    Highbury tomorrow night on the
    back of a free-scoring start
    to the season. Despite losing
    Mateja Kezman to Chelsea in
    the summer, the Dutch side has
    scored 12 goals in the first" + ], + [ + "Through the World Community
    Grid, your computer could help
    address the world's health and
    social problems." + ], + [ + "Zimbabwe #39;s ruling Zanu-PF
    old guard has emerged on top
    after a bitter power struggle
    in the deeply divided party
    during its five-yearly
    congress, which ended
    yesterday." + ], + [ + "PLAYER OF THE GAME: Playing
    with a broken nose, Seattle
    point guard Sue Bird set a
    WNBA playoff record for
    assists with 14, also pumping
    in 10 points as the Storm
    claimed the Western Conference
    title last night." + ], + [ + "Reuters - Thousands of
    demonstrators pressing
    to\\install Ukraine's
    opposition leader as president
    after a\\disputed election
    launched fresh street rallies
    in the capital\\for the third
    day Wednesday." + ], + [ + "Michael Jackson wishes he had
    fought previous child
    molestation claims instead of
    trying to \"buy peace\", his
    lawyer says." + ], + [ + "North Korea says it will not
    abandon its weapons programme
    after the South admitted
    nuclear activities." + ], + [ + "While there is growing
    attention to ongoing genocide
    in Darfur, this has not
    translated into either a
    meaningful international
    response or an accurate
    rendering of the scale and
    evident course of the
    catastrophe." + ], + [ + "A planned component for
    Microsoft #39;s next version
    of Windows is causing
    consternation among antivirus
    experts, who say that the new
    module, a scripting platform
    called Microsoft Shell, could
    give birth to a whole new
    generation of viruses and
    remotely" + ], + [ + "THE prosecution on terrorism
    charges of extremist Islamic
    cleric and accused Jemaah
    Islamiah leader Abu Bakar
    Bashir will rely heavily on
    the potentially tainted
    testimony of at least two
    convicted Bali bombers, his
    lawyers have said." + ], + [ + "SAN JOSE, California Yahoo
    will likely have a tough time
    getting American courts to
    intervene in a dispute over
    the sale of Nazi memorabilia
    in France after a US appeals
    court ruling." + ], + [ + "TORONTO (CP) - Glamis Gold of
    Reno, Nev., is planning a
    takeover bid for Goldcorp Inc.
    of Toronto - but only if
    Goldcorp drops its
    \\$2.4-billion-Cdn offer for
    another Canadian firm, made in
    early December." + ], + [ + "Clashes between US troops and
    Sadr militiamen escalated
    Thursday, as the US surrounded
    Najaf for possible siege." + ], + [ + "eBay Style director Constance
    White joins Post fashion
    editor Robin Givhan and host
    Janet Bennett to discuss how
    to find trends and bargains
    and pull together a wardrobe
    online." + ], + [ + "This week will see the release
    of October new and existing
    home sales, a measure of
    strength in the housing
    industry. But the short
    holiday week will also leave
    investors looking ahead to the
    holiday travel season." + ], + [ + "Frankfurt - World Cup winners
    Brazil were on Monday drawn to
    meet European champions
    Greece, Gold Cup winners
    Mexico and Asian champions
    Japan at the 2005
    Confederations Cup." + ], + [ + "Third baseman Vinny Castilla
    said he fits fine with the
    Colorado youth movement, even
    though he #39;ll turn 38 next
    season and the Rockies are
    coming off the second-worst" + ], + [ + "With a sudden shudder, the
    ground collapsed and the pipe
    pushed upward, buckling into a
    humped shape as Cornell
    University scientists produced
    the first simulated earthquake" + ], + [ + "AFP - A battle group of
    British troops rolled out of
    southern Iraq on a US-
    requested mission to deadlier
    areas near Baghdad, in a major
    political gamble for British
    Prime Minister Tony Blair." + ], + [ + "(Sports Network) - Two of the
    top teams in the American
    League tangle in a possible
    American League Division
    Series preview tonight, as the
    West-leading Oakland Athletics
    host the wild card-leading
    Boston Red Sox for the first
    of a three-game set at the" + ], + [ + "over half the children in the
    world - suffer extreme
    deprivation because of war,
    HIV/AIDS or poverty, according
    to a report released yesterday
    by the United Nations Children
    #39;s Fund." + ], + [ + "Microsoft (Quote, Chart) has
    fired another salvo in its
    ongoing spam battle, this time
    against porn peddlers who don
    #39;t keep their smut inside
    the digital equivalent of a
    quot;Brown Paper Wrapper." + ], + [ + " BETHESDA, Md. (Reuters) - The
    use of some antidepressant
    drugs appears linked to an
    increase in suicidal behavior
    in some children and teen-
    agers, a U.S. advisory panel
    concluded on Tuesday." + ], + [ + " SEATTLE (Reuters) - The next
    version of the Windows
    operating system, Microsoft
    Corp.'s <A HREF=\"http://www
    .reuters.co.uk/financeQuoteLoo
    kup.jhtml?ticker=MSFT.O
    qtype=sym infotype=info
    qcat=news\">MSFT.O</A>
    flagship product, will ship
    in 2006, the world's largest
    software maker said on
    Friday." + ], + [ + "Reuters - Philippine rescue
    teams\\evacuated thousands of
    people from the worst flooding
    in the\\central Luzon region
    since the 1970s as hungry
    victims hunted\\rats and birds
    for food." + ], + [ + "Inverness Caledonian Thistle
    appointed Craig Brewster as
    its new manager-player
    Thursday although he #39;s
    unable to play for the team
    until January." + ], + [ + "The Afghan president expresses
    deep concern after a bomb
    attack which left at least
    seven people dead." + ], + [ + " NEW YORK (Reuters) - U.S.
    technology stocks opened lower
    on Thursday after a sales
    warning from Applied Materials
    Inc. <A HREF=\"http://www.i
    nvestor.reuters.com/FullQuote.
    aspx?ticker=AMAT.O target=/sto
    cks/quickinfo/fullquote\">AM
    AT.O</A>, while weekly
    jobless claims data met Wall
    Street's expectations,
    leaving the Dow and S P 500
    market measures little
    changed." + ], + [ + "ATHENS, Greece -- Alan Shearer
    converted an 87th-minute
    penalty to give Newcastle a
    1-0 win over Panionios in
    their UEFA Cup Group D match." + ], + [ + "Fossil remains of the oldest
    and smallest known ancestor of
    Tyrannosaurus rex, the world
    #39;s favorite ferocious
    dinosaur, have been discovered
    in China with evidence that
    its body was cloaked in downy
    quot;protofeathers." + ], + [ + "Derek Jeter turned a season
    that started with a terrible
    slump into one of the best in
    his accomplished 10-year
    career. quot;I don #39;t
    think there is any question,
    quot; the New York Yankees
    manager said." + ], + [ + "Gardez (Afghanistan), Sept. 16
    (Reuters): Afghan President
    Hamid Karzai escaped an
    assassination bid today when a
    rocket was fired at his US
    military helicopter as it was
    landing in the southeastern
    town of Gardez." + ], + [ + "The Jets came up with four
    turnovers by Dolphins
    quarterback Jay Fiedler in the
    second half, including an
    interception returned 66 yards
    for a touchdown." + ], + [ + "China's Guo Jingjing easily
    won the women's 3-meter
    springboard last night, and Wu
    Minxia made it a 1-2 finish
    for the world's diving
    superpower, taking the silver." + ], + [ + "GREEN BAY, Wisconsin (Ticker)
    -- Brett Favre will be hoping
    his 200th consecutive start
    turns out better than his last
    two have against the St." + ], + [ + "People fishing for sport are
    doing far more damage to US
    marine fish stocks than anyone
    thought, accounting for nearly
    a quarter of the" + ], + [ + "When an NFL team opens with a
    prolonged winning streak,
    former Miami Dolphins coach
    Don Shula and his players from
    the 17-0 team of 1972 root
    unabashedly for the next
    opponent." + ], + [ + "MIANNE Bagger, the transsexual
    golfer who prompted a change
    in the rules to allow her to
    compete on the professional
    circuit, made history
    yesterday by qualifying to
    play full-time on the Ladies
    European Tour." + ], + [ + "Great Britain #39;s gold medal
    tally now stands at five after
    Leslie Law was handed the
    individual three day eventing
    title - in a courtroom." + ], + [ + "This particular index is
    produced by the University of
    Michigan Business School, in
    partnership with the American
    Society for Quality and CFI
    Group, and is supported in
    part by ForeSee Results" + ], + [ + "CHICAGO : Interstate Bakeries
    Corp., the maker of popular,
    old-style snacks Twinkies and
    Hostess Cakes, filed for
    bankruptcy, citing rising
    costs and falling sales." + ], + [ + "Delta Air Lines (DAL:NYSE -
    commentary - research) will
    cut employees and benefits but
    give a bigger-than-expected
    role to Song, its low-cost
    unit, in a widely anticipated
    but still unannounced
    overhaul, TheStreet.com has
    learned." + ], + [ + "By byron kho. A consortium of
    movie and record companies
    joined forces on Friday to
    request that the US Supreme
    Court take another look at
    peer-to-peer file-sharing
    programs." + ], + [ + "DUBLIN -- Prime Minister
    Bertie Ahern urged Irish
    Republican Army commanders
    yesterday to meet what he
    acknowledged was ''a heavy
    burden quot;: disarming and
    disbanding their organization
    in support of Northern
    Ireland's 1998 peace accord." + ], + [ + "Mayor Tom Menino must be
    proud. His Boston Red Sox just
    won their first World Series
    in 86 years and his Hyde Park
    Blue Stars yesterday clinched
    their first Super Bowl berth
    in 32 years, defeating
    O'Bryant, 14-0. Who would have
    thought?" + ], + [ + "While reproductive planning
    and women #39;s equality have
    improved substantially over
    the past decade, says a United
    Nations report, world
    population will increase from
    6.4 billion today to 8.9
    billion by 2050, with the 50
    poorest countries tripling in" + ], + [ + "Instead of the skinny black
    line, showing a hurricane
    #39;s forecast track,
    forecasters have drafted a
    couple of alternative graphics
    to depict where the storms
    might go -- and they want your
    opinion." + ], + [ + "South Korea have appealed to
    sport #39;s supreme legal body
    in an attempt to award Yang
    Tae-young the Olympic
    gymnastics all-round gold
    medal after a scoring error
    robbed him of the title in
    Athens." + ], + [ + "BERLIN - Volkswagen AG #39;s
    announcement this week that it
    has forged a new partnership
    deal with Malaysian carmaker
    Proton comes as a strong euro
    and Europe #39;s weak economic
    performance triggers a fresh
    wave of German investment in
    Asia." + ], + [ + "AP - Johan Santana had an
    early lead and was well on his
    way to his 10th straight win
    when the rain started to fall." + ], + [ + "ATHENS-In one of the biggest
    shocks in Olympic judo
    history, defending champion
    Kosei Inoue was defeated by
    Dutchman Elco van der Geest in
    the men #39;s 100-kilogram
    category Thursday." + ], + [ + "Consumers in Dublin pay more
    for basic goods and services
    that people elsewhere in the
    country, according to figures
    released today by the Central
    Statistics Office." + ], + [ + "LIBERTY Media #39;s move last
    week to grab up to 17.1 per
    cent of News Corporation
    voting stock has prompted the
    launch of a defensive
    shareholder rights plan." + ], + [ + "NBC is adding a 5-second delay
    to its Nascar telecasts after
    Dale Earnhardt Jr. used a
    vulgarity during a postrace
    interview last weekend." + ], + [ + "LONDON - Wild capuchin monkeys
    can understand cause and
    effect well enough to use
    rocks to dig for food,
    scientists have found.
    Capuchin monkeys often use
    tools and solve problems in
    captivity and sometimes" + ], + [ + "San Francisco Giants
    outfielder Barry Bonds, who
    became the third player in
    Major League Baseball history
    to hit 700 career home runs,
    won the National League Most
    Valuable Player Award" + ], + [ + "The blue-chip Hang Seng Index
    rose 171.88 points, or 1.22
    percent, to 14,066.91. On
    Friday, the index had slipped
    31.58 points, or 0.2 percent." + ], + [ + "BAR's Anthony Davidson and
    Jenson Button set the pace at
    the first Chinese Grand Prix." + ], + [ + "Shares plunge after company
    says its vein graft treatment
    failed to show benefit in
    late-stage test. CHICAGO
    (Reuters) - Biotechnology
    company Corgentech Inc." + ], + [ + "WASHINGTON - Contradicting the
    main argument for a war that
    has cost more than 1,000
    American lives, the top U.S.
    arms inspector reported
    Wednesday that he found no
    evidence that Iraq produced
    any weapons of mass
    destruction after 1991..." + ], + [ + "The key to hidden treasure
    lies in your handheld GPS
    unit. GPS-based \"geocaching\"
    is a high-tech sport being
    played by thousands of people
    across the globe." + ], + [ + "AFP - Style mavens will be
    scanning the catwalks in Paris
    this week for next spring's
    must-have handbag, as a
    sweeping exhibition at the
    French capital's fashion and
    textile museum reveals the bag
    in all its forms." + ], + [ + "Canadian Press - SAINT-
    QUENTIN, N.B. (CP) - A major
    highway in northern New
    Brunswick remained closed to
    almost all traffic Monday, as
    local residents protested
    planned health care cuts." + ], + [ + "The U.S. information tech
    sector lost 403,300 jobs
    between March 2001 and April
    2004, and the market for tech
    workers remains bleak,
    according to a new report." + ], + [ + " NAJAF, Iraq (Reuters) - A
    radical Iraqi cleric leading a
    Shi'ite uprising agreed on
    Wednesday to disarm his
    militia and leave one of the
    country's holiest Islamic
    shrines after warnings of an
    onslaught by government
    forces." + ], + [ + "Saudi security forces have
    killed a wanted militant near
    the scene of a deadly shootout
    Thursday. Officials say the
    militant was killed in a
    gunbattle Friday in the
    northern town of Buraida,
    hours after one" + ], + [ + "Portsmouth chairman Milan
    Mandaric said on Tuesday that
    Harry Redknapp, who resigned
    as manager last week, was
    innocent of any wrong-doing
    over agent or transfer fees." + ], + [ + "This record is for all the
    little guys, for all the
    players who have to leg out
    every hit instead of taking a
    relaxing trot around the
    bases, for all the batters
    whose muscles aren #39;t" + ], + [ + "Two South Africans acquitted
    by a Zimbabwean court of
    charges related to the alleged
    coup plot in Equatorial Guinea
    are to be questioned today by
    the South African authorities." + ], + [ + "Charlie Hodgson #39;s record-
    equalling performance against
    South Africa was praised by
    coach Andy Robinson after the
    Sale flyhalf scored 27 points
    in England #39;s 32-16 victory
    here at Twickenham on
    Saturday." + ], + [ + "com September 30, 2004, 11:11
    AM PT. SanDisk announced
    Thursday increased capacities
    for several different flash
    memory cards. The Sunnyvale,
    Calif." + ], + [ + "MOSCOW (CP) - Russia mourned
    89 victims of a double air
    disaster today as debate
    intensified over whether the
    two passenger liners could
    have plunged almost
    simultaneously from the sky by
    accident." + ], + [ + "US blue-chip stocks rose
    slightly on Friday as
    government data showed better-
    than-expected demand in August
    for durable goods other than
    transportation equipment, but
    climbing oil prices limited
    gains." + ], + [ + "BASEBALL Atlanta (NL):
    Optioned P Roman Colon to
    Greenville (Southern);
    recalled OF Dewayne Wise from
    Richmond (IL). Boston (AL):
    Purchased C Sandy Martinez
    from Cleveland (AL) and
    assigned him to Pawtucket
    (IL). Cleveland (AL): Recalled
    OF Ryan Ludwick from Buffalo
    (IL). Chicago (NL): Acquired
    OF Ben Grieve from Milwaukee
    (NL) for player to be named
    and cash; acquired C Mike ..." + ], + [ + "Australia #39;s prime minister
    says a body found in Fallujah
    is likely that of kidnapped
    aid worker Margaret Hassan.
    John Howard told Parliament a
    videotape of an Iraqi
    terrorist group executing a
    Western woman appears to have
    been genuine." + ], + [ + "roundup Plus: Tech firms rally
    against copyright bill...Apple
    .Mac customers suffer e-mail
    glitches...Alvarion expands
    wireless broadband in China." + ], + [ + "BRONX, New York (Ticker) --
    Kelvim Escobar was the latest
    Anaheim Angels #39; pitcher to
    subdue the New York Yankees.
    Escobar pitched seven strong
    innings and Bengie Molina tied
    a career-high with four hits,
    including" + ], + [ + "Business software maker
    PeopleSoft Inc. said Monday
    that it expects third-quarter
    revenue to range between \\$680
    million and \\$695 million,
    above average Wall Street
    estimates of \\$651." + ], + [ + "Sep 08 - Vijay Singh revelled
    in his status as the new world
    number one after winning the
    Deutsche Bank Championship by
    three shots in Boston on
    Monday." + ], + [ + "Reuters - Enron Corp. ,
    desperate to\\meet profit
    targets, \"parked\" unwanted
    power generating barges\\at
    Merrill Lynch in a sham sale
    designed to be reversed,
    a\\prosecutor said on Tuesday
    in the first criminal trial
    of\\former executives at the
    fallen energy company." + ], + [ + " NEW YORK (Reuters) - The
    dollar rebounded on Monday
    after a heavy selloff last
    week, but analysts were
    uncertain if the rally could
    hold as the drumbeat of
    expectation began for to the
    December U.S. jobs report due
    Friday." + ], + [ + "AP - Their first debate less
    than a week away, President
    Bush and Democrat John Kerry
    kept their public schedules
    clear on Saturday and began to
    focus on their prime-time
    showdown." + ], + [ + "Many people in golf are asking
    that today. He certainly wasn
    #39;t A-list and he wasn #39;t
    Larry Nelson either. But you
    couldn #39;t find a more solid
    guy to lead the United States
    into Ireland for the 2006
    Ryder Cup Matches." + ], + [ + "Coles Myer Ltd. Australia
    #39;s biggest retailer,
    increased second-half profit
    by 26 percent after opening
    fuel and convenience stores,
    selling more-profitable
    groceries and cutting costs." + ], + [ + "MOSCOW: Us oil major
    ConocoPhillips is seeking to
    buy up to 25 in Russian oil
    giant Lukoil to add billions
    of barrels of reserves to its
    books, an industry source
    familiar with the matter said
    on Friday." + ], + [ + "Australian Stuart Appleby, who
    was the joint second-round
    leader, returned a two-over 74
    to drop to third at three-
    under while American Chris
    DiMarco moved into fourth with
    a round of 69." + ], + [ + "PARIS Getting to the bottom of
    what killed Yassar Arafat
    could shape up to be an ugly
    family tug-of-war. Arafat
    #39;s half-brother and nephew
    want copies of Arafat #39;s
    medical records from the
    suburban Paris hospital" + ], + [ + "Red Hat is acquiring security
    and authentication tools from
    Netscape Security Solutions to
    bolster its software arsenal.
    Red Hat #39;s CEO and chairman
    Matthew Szulik spoke about the
    future strategy of the Linux
    supplier." + ], + [ + "With a doubleheader sweep of
    the Minnesota Twins, the New
    York Yankees moved to the
    verge of clinching their
    seventh straight AL East
    title." + ], + [ + "Global Web portal Yahoo! Inc.
    Wednesday night made available
    a beta version of a new search
    service for videos. Called
    Yahoo! Video Search, the
    search engine crawls the Web
    for different types of media
    files" + ], + [ + "Interactive posters at 25
    underground stations are
    helping Londoners travel
    safely over Christmas." + ], + [ + "Athens, Greece (Sports
    Network) - The first official
    track event took place this
    morning and Italy #39;s Ivano
    Brugnetti won the men #39;s
    20km walk at the Summer
    Olympics in Athens." + ], + [ + " THE HAGUE (Reuters) - Former
    Yugoslav President Slobodan
    Milosevic condemned his war
    crimes trial as a \"pure farce\"
    on Wednesday in a defiant
    finish to his opening defense
    statement against charges of
    ethnic cleansing in the
    Balkans." + ], + [ + "US Airways Group (otc: UAIRQ -
    news - people ) on Thursday
    said it #39;ll seek a court
    injunction to prohibit a
    strike by disaffected unions." + ], + [ + "Shares in Unilever fall after
    the Anglo-Dutch consumer goods
    giant issued a surprise
    profits warning." + ], + [ + "SAN FRANCISCO (CBS.MW) - The
    Canadian government will sell
    its 19 percent stake in Petro-
    Canada for \\$2.49 billion,
    according to the final
    prospectus filed with the US
    Securities and Exchange
    Commission Thursday." + ], + [ + "Champions Arsenal opened a
    five-point lead at the top of
    the Premier League after a 4-0
    thrashing of Charlton Athletic
    at Highbury Saturday." + ], + [ + "The Redskins and Browns have
    traded field goals and are
    tied, 3-3, in the first
    quarter in Cleveland." + ], + [ + " HYDERABAD, India (Reuters) -
    Microsoft Corp. <A HREF=\"ht
    tp://www.investor.reuters.com/
    FullQuote.aspx?ticker=MSFT.O t
    arget=/stocks/quickinfo/fullqu
    ote\">MSFT.O</A> will
    hire several hundred new staff
    at its new Indian campus in
    the next year, its chief
    executive said on Monday, in a
    move aimed at strengthening
    its presence in Asia's fourth-
    biggest economy." + ], + [ + "According to Swiss
    authorities, history was made
    Sunday when 2723 people in
    four communities in canton
    Geneva, Switzerland, voted
    online in a national federal
    referendum." + ], + [ + " GUWAHATI, India (Reuters) -
    People braved a steady drizzle
    to come out to vote in a
    remote northeast Indian state
    on Thursday, as troops
    guarded polling stations in an
    election being held under the
    shadow of violence." + ], + [ + "AFP - Three of the nine
    Canadian sailors injured when
    their newly-delivered,
    British-built submarine caught
    fire in the North Atlantic
    were airlifted Wednesday to
    hospital in northwest Ireland,
    officials said." + ], + [ + "Alitalia SpA, Italy #39;s
    largest airline, reached an
    agreement with its flight
    attendants #39; unions to cut
    900 jobs, qualifying the
    company for a government
    bailout that will keep it in
    business for another six
    months." + ], + [ + "BAGHDAD, Iraq - A series of
    strong explosions shook
    central Baghdad near dawn
    Sunday, and columns of thick
    black smoke rose from the
    Green Zone where U.S. and
    Iraqi government offices are
    located..." + ], + [ + "Forget September call-ups. The
    Red Sox may tap their minor
    league system for an extra
    player or two when the rules
    allow them to expand their
    25-man roster Wednesday, but
    any help from the farm is
    likely to pale against the
    abundance of talent they gain
    from the return of numerous
    players, including Trot Nixon
    , from the disabled list." + ], + [ + "P amp;Os cutbacks announced
    today are the result of the
    waves of troubles that have
    swamped the ferry industry of
    late. Some would say the
    company has done well to
    weather the storms for as long
    as it has." + ], + [ + "Sven-Goran Eriksson may gamble
    by playing goalkeeper Paul
    Robinson and striker Jermain
    Defoe in Poland." + ], + [ + "Foreign Secretary Jack Straw
    has flown to Khartoum on a
    mission to pile the pressure
    on the Sudanese government to
    tackle the humanitarian
    catastrophe in Darfur." + ], + [ + " Nextel was the big story in
    telecommunications yesterday,
    thanks to the Reston company's
    mega-merger with Sprint, but
    the future of wireless may be
    percolating in dozens of
    Washington area start-ups." + ], + [ + "Reuters - A senior U.S.
    official said on
    Wednesday\\deals should not be
    done with hostage-takers ahead
    of the\\latest deadline set by
    Afghan Islamic militants who
    have\\threatened to kill three
    kidnapped U.N. workers." + ], + [ + "I have been anticipating this
    day like a child waits for
    Christmas. Today, PalmOne
    introduces the Treo 650, the
    answer to my quot;what smart
    phone will I buy?" + ], + [ + "THOUSAND OAKS -- Anonymity is
    only a problem if you want it
    to be, and it is obvious Vijay
    Singh doesn #39;t want it to
    be. Let others chase fame." + ], + [ + "Wikipedia has surprised Web
    watchers by growing fast and
    maturing into one of the most
    popular reference sites." + ], + [ + "It only takes 20 minutes on
    the Internet for an
    unprotected computer running
    Microsoft Windows to be taken
    over by a hacker. Any personal
    or financial information
    stored" + ], + [ + "TORONTO (CP) - Russia #39;s
    Severstal has made an offer to
    buy Stelco Inc., in what #39;s
    believed to be one of several
    competing offers emerging for
    the restructuring but
    profitable Hamilton steel
    producer." + ], + [ + "Prices for flash memory cards
    -- the little modules used by
    digital cameras, handheld
    organizers, MP3 players and
    cell phones to store pictures,
    music and other data -- are
    headed down -- way down. Past
    trends suggest that prices
    will drop 35 percent a year,
    but industry analysts think
    that rate will be more like 40
    or 50 percent this year and
    next, due to more
    manufacturers entering the
    market." + ], + [ + "Walt Disney Co. #39;s
    directors nominated Michael
    Ovitz to serve on its board
    for another three years at a
    meeting just weeks before
    forcing him out of his job as" + ], + [ + "The European Union, Japan,
    Brazil and five other
    countries won World Trade
    Organization approval to
    impose tariffs worth more than
    \\$150 million a year on
    imports from the United" + ], + [ + "Industrial conglomerate
    Honeywell International on
    Wednesday said it has filed a
    lawsuit against 34 electronics
    companies including Apple
    Computer and Eastman Kodak,
    claiming patent infringement
    of its liquid crystal display
    technology." + ], + [ + "Sinn Fein leader Gerry Adams
    has put the pressure for the
    success or failure of the
    Northern Ireland assembly
    talks firmly on the shoulders
    of Ian Paisley." + ], + [ + "Australia #39;s Computershare
    has agreed to buy EquiServe of
    the United States for US\\$292
    million (\\$423 million),
    making it the largest US share
    registrar and driving its
    shares up by a third." + ], + [ + "David Coulthard #39;s season-
    long search for a Formula One
    drive next year is almost
    over. Negotiations between Red
    Bull Racing and Coulthard, who
    tested for the Austrian team
    for the first time" + ], + [ + "Interest rates on short-term
    Treasury securities were mixed
    in yesterday's auction. The
    Treasury Department sold \\$18
    billion in three-month bills
    at a discount rate of 1.640
    percent, up from 1.635 percent
    last week. An additional \\$16
    billion was sold in six-month
    bills at a rate of 1.840
    percent, down from 1.860
    percent." + ], + [ + "Two top executives of scandal-
    tarred insurance firm Marsh
    Inc. were ousted yesterday,
    the company said, the latest
    casualties of an industry
    probe by New York's attorney
    general." + ], + [ + "AP - Southern California
    tailback LenDale White
    remembers Justin Holland from
    high school. The Colorado
    State quarterback made quite
    an impression." + ], + [ + "TIM HENMAN last night admitted
    all of his energy has been
    drained away as he bowed out
    of the Madrid Masters. The top
    seed, who had a blood test on
    Wednesday to get to the bottom
    of his fatigue, went down" + ], + [ + "USDA #39;s Animal Plant Health
    Inspection Service (APHIS)
    this morning announced it has
    confirmed a detection of
    soybean rust from two test
    plots at Louisiana State
    University near Baton Rouge,
    Louisiana." + ], + [ + " JAKARTA (Reuters) - President
    Megawati Sukarnoputri urged
    Indonesians on Thursday to
    accept the results of the
    country's first direct
    election of a leader, but
    stopped short of conceding
    defeat." + ], + [ + "ISLAMABAD: Pakistan early
    Monday test-fired its
    indigenously developed short-
    range nuclear-capable Ghaznavi
    missile, the Inter Services
    Public Relations (ISPR) said
    in a statement." + ], + [ + "While the US software giant
    Microsoft has achieved almost
    sweeping victories in
    government procurement
    projects in several Chinese
    provinces and municipalities,
    the process" + ], + [ + "Mexican Cemex, being the third
    largest cement maker in the
    world, agreed to buy its
    British competitor - RMC Group
    - for \\$5.8 billion, as well
    as their debts in order to
    expand their activity on the
    building materials market of
    the USA and Europe." + ], + [ + "Microsoft Corp. has delayed
    automated distribution of a
    major security upgrade to its
    Windows XP Professional
    operating system, citing a
    desire to give companies more
    time to test it." + ], + [ + "The trial of a man accused of
    murdering York backpacker
    Caroline Stuttle begins in
    Australia." + ], + [ + "Gateway computers will be more
    widely available at Office
    Depot, in the PC maker #39;s
    latest move to broaden
    distribution at retail stores
    since acquiring rival
    eMachines this year." + ], + [ + "ATHENS -- US sailors needed a
    big day to bring home gold and
    bronze medals from the sailing
    finale here yesterday. But
    rolling the dice on windshifts
    and starting tactics backfired
    both in Star and Tornado
    classes, and the Americans had
    to settle for a single silver
    medal." + ], + [ + "Intel Corp. is refreshing its
    64-bit Itanium 2 processor
    line with six new chips based
    on the Madison core. The new
    processors represent the last
    single-core Itanium chips that
    the Santa Clara, Calif." + ], + [ + "The world's largest insurance
    group pays \\$126m in fines as
    part of a settlement with US
    regulators over its dealings
    with two firms." + ], + [ + "BRUSSELS: The EU sought
    Wednesday to keep pressure on
    Turkey over its bid to start
    talks on joining the bloc, as
    last-minute haggling seemed
    set to go down to the wire at
    a summit poised to give a
    green light to Ankara." + ], + [ + "AP - J. Cofer Black, the State
    Department official in charge
    of counterterrorism, is
    leaving government in the next
    few weeks." + ], + [ + "For the first time, broadband
    connections are reaching more
    than half (51 percent) of the
    American online population at
    home, according to measurement
    taken in July by
    Nielsen/NetRatings, a
    Milpitas-based Internet
    audience measurement and
    research ..." + ], + [ + "AP - Cavaliers forward Luke
    Jackson was activated
    Wednesday after missing five
    games because of tendinitis in
    his right knee. Cleveland also
    placed forward Sasha Pavlovic
    on the injured list." + ], + [ + "The tobacco firm John Player
    amp; Sons has announced plans
    to lay off 90 workers at its
    cigarette factory in Dublin.
    The company said it was
    planning a phased closure of
    the factory between now and
    February as part of a review
    of its global operations." + ], + [ + "AP - Consumers borrowed more
    freely in September,
    especially when it came to
    racking up charges on their
    credit cards, the Federal
    Reserve reported Friday." + ], + [ + "AFP - The United States
    presented a draft UN
    resolution that steps up the
    pressure on Sudan over the
    crisis in Darfur, including
    possible international
    sanctions against its oil
    sector." + ], + [ + "AFP - At least 33 people were
    killed and dozens others
    wounded when two bombs ripped
    through a congregation of
    Sunni Muslims in Pakistan's
    central city of Multan, police
    said." + ], + [ + "Bold, innovative solutions are
    key to addressing the rapidly
    rising costs of higher
    education and the steady
    reduction in government-
    subsidized help to finance
    such education." + ], + [ + "Just as the PhD crowd emerge
    with different interpretations
    of today's economy, everyday
    Americans battling to balance
    the checkbook hold diverse
    opinions about where things
    stand now and in the future." + ], + [ + "The Brisbane Lions #39;
    football manager stepped out
    of the changerooms just before
    six o #39;clock last night and
    handed one of the milling
    supporters a six-pack of beer." + ], + [ + "Authorities here are always
    eager to show off their
    accomplishments, so when
    Beijing hosted the World
    Toilet Organization conference
    last week, delegates were
    given a grand tour of the
    city's toilets." + ], + [ + "Cavaliers owner Gordon Gund is
    in quot;serious quot;
    negotiations to sell the NBA
    franchise, which has enjoyed a
    dramatic financial turnaround
    since the arrival of star
    LeBron James." + ], + [ + "WASHINGTON Trying to break a
    deadlock on energy policy, a
    diverse group of
    environmentalists, academics
    and former government
    officials were to publish a
    report on Wednesday that
    presents strategies for making
    the United States cleaner,
    more competitive" + ], + [ + "After two days of gloom, China
    was back on the winning rails
    on Thursday with Liu Chunhong
    winning a weightlifting title
    on her record-shattering binge
    and its shuttlers contributing
    two golds in the cliff-hanging
    finals." + ], + [ + "One question that arises
    whenever a player is linked to
    steroids is, \"What would he
    have done without them?\"
    Baseball history whispers an
    answer." + ], + [ + "AFP - A series of torchlight
    rallies and vigils were held
    after darkness fell on this
    central Indian city as victims
    and activists jointly
    commemorated a night of horror
    20 years ago when lethal gas
    leaked from a pesticide plant
    and killed thousands." + ], + [ + "Consider the New World of
    Information - stuff that,
    unlike the paper days of the
    past, doesn't always
    physically exist. You've got
    notes, scrawlings and
    snippets, Web graphics, photos
    and sounds. Stuff needs to be
    cut, pasted, highlighted,
    annotated, crossed out,
    dragged away. And, as Ross
    Perot used to say (or maybe it
    was Dana Carvey impersonating
    him), don't forget the graphs
    and charts." + ], + [ + "The second round of the
    Canadian Open golf tournament
    continues Saturday Glenn Abbey
    Golf Club in Oakville,
    Ontario, after play was
    suspended late Friday due to
    darkness." + ], + [ + "A consortium led by Royal
    Dutch/Shell Group that is
    developing gas reserves off
    Russia #39;s Sakhalin Island
    said Thursday it has struck a
    US\\$6 billion (euro4." + ], + [ + "Major Hollywood studios on
    Tuesday announced scores of
    lawsuits against computer
    server operators worldwide,
    including eDonkey, BitTorrent
    and DirectConnect networks,
    for allowing trading of
    pirated movies." + ], + [ + "A massive plan to attract the
    2012 Summer Olympics to New
    York, touting the city's
    diversity, financial and media
    power, was revealed Wednesday." + ], + [ + "A Zimbabwe court Friday
    convicted a British man
    accused of leading a coup plot
    against the government of oil-
    rich Equatorial Guinea on
    weapons charges, but acquitted
    most of the 69 other men held
    with him." + ], + [ + "But will Wi-Fi, high-
    definition broadcasts, mobile
    messaging and other
    enhancements improve the game,
    or wreck it?\\<br />
    Photos of tech-friendly parks\\" + ], + [ + "An audit by international
    observers supported official
    elections results that gave
    President Hugo Chavez a
    victory over a recall vote
    against him, the secretary-
    general of the Organisation of
    American States announced." + ], + [ + "Canadian Press - TORONTO (CP)
    - The fatal stabbing of a
    young man trying to eject
    unwanted party guests from his
    family home, the third such
    knifing in just weeks, has
    police worried about a
    potentially fatal holiday
    recipe: teens, alcohol and
    knives." + ], + [ + "NICK Heidfeld #39;s test with
    Williams has been brought
    forward after BAR blocked
    plans for Anthony Davidson to
    drive its Formula One rival
    #39;s car." + ], + [ + "MOSCOW - A female suicide
    bomber set off a shrapnel-
    filled explosive device
    outside a busy Moscow subway
    station on Tuesday night,
    officials said, killing 10
    people and injuring more than
    50." + ], + [ + "Grace Park closed with an
    eagle and two birdies for a
    7-under-par 65 and a two-
    stroke lead after three rounds
    of the Wachovia LPGA Classic
    on Saturday." + ], + [ + "ABIDJAN (AFP) - Two Ivory
    Coast military aircraft
    carried out a second raid on
    Bouake, the stronghold of the
    former rebel New Forces (FN)
    in the divided west African
    country, a French military
    source told AFP." + ], + [ + "Carlos Beltran drives in five
    runs to carry the Astros to a
    12-3 rout of the Braves in
    Game 5 of their first-round NL
    playoff series." + ], + [ + "On-demand viewing isn't just
    for TiVo owners anymore.
    Television over internet
    protocol, or TVIP, offers
    custom programming over
    standard copper wires." + ], + [ + "Apple is recalling 28,000
    faulty batteries for its
    15-inch Powerbook G4 laptops." + ], + [ + "Since Lennox Lewis #39;s
    retirement, the heavyweight
    division has been knocked for
    having more quantity than
    quality. Eight heavyweights on
    Saturday night #39;s card at
    Madison Square Garden hope to
    change that perception, at
    least for one night." + ], + [ + "PalmSource #39;s European
    developer conference is going
    on now in Germany, and this
    company is using this
    opportunity to show off Palm
    OS Cobalt 6.1, the latest
    version of its operating
    system." + ], + [ + "The former Chief Executive
    Officer of Computer Associates
    was indicted by a federal
    grand jury in New York
    Wednesday for allegedly
    participating in a massive
    fraud conspiracy and an
    elaborate cover up of a scheme
    that cost investors" + ], + [ + "Speaking to members of the
    Massachusetts Software
    Council, Microsoft CEO Steve
    Ballmer touted a bright future
    for technology but warned his
    listeners to think twice
    before adopting open-source
    products like Linux." + ], + [ + "MIAMI - The Trillian instant
    messaging (IM) application
    will feature several
    enhancements in its upcoming
    version 3.0, including new
    video and audio chat
    capabilities, enhanced IM
    session logs and integration
    with the Wikipedia online
    encyclopedia, according to
    information posted Friday on
    the product developer's Web
    site." + ], + [ + "Honeywell International Inc.,
    the world #39;s largest
    supplier of building controls,
    agreed to buy Novar Plc for
    798 million pounds (\\$1.53
    billion) to expand its
    security, fire and
    ventilation-systems business
    in Europe." + ], + [ + "San Francisco investment bank
    Thomas Weisel Partners on
    Thursday agreed to pay \\$12.5
    million to settle allegations
    that some of the stock
    research the bank published
    during the Internet boom was
    tainted by conflicts of
    interest." + ], + [ + "AFP - A state of civil
    emergency in the rebellion-hit
    Indonesian province of Aceh
    has been formally extended by
    six month, as the country's
    president pledged to end
    violence there without foreign
    help." + ], + [ + "Forbes.com - Peter Frankling
    tapped an unusual source to
    fund his new business, which
    makes hot-dog-shaped ice cream
    treats known as Cool Dogs: Two
    investors, one a friend and
    the other a professional
    venture capitalist, put in
    more than #36;100,000 each
    from their Individual
    Retirement Accounts. Later
    Franklin added #36;150,000
    from his own IRA." + ], + [ + "Reuters - Online DVD rental
    service Netflix Inc.\\and TiVo
    Inc., maker of a digital video
    recorder, on Thursday\\said
    they have agreed to develop a
    joint entertainment\\offering,
    driving shares of both
    companies higher." + ], + [ + "A San Diego insurance
    brokerage has been sued by New
    York Attorney General Elliot
    Spitzer for allegedly
    soliciting payoffs in exchange
    for steering business to
    preferred insurance companies." + ], + [ + "The European Union agreed
    Monday to lift penalties that
    have cost American exporters
    \\$300 million, following the
    repeal of a US corporate tax
    break deemed illegal under
    global trade rules." + ], + [ + "US Secretary of State Colin
    Powell on Monday said he had
    spoken to both Indian Foreign
    Minister K Natwar Singh and
    his Pakistani counterpart
    Khurshid Mahmud Kasuri late
    last week before the two met
    in New Delhi this week for
    talks." + ], + [ + "NEW YORK - Victor Diaz hit a
    tying, three-run homer with
    two outs in the ninth inning,
    and Craig Brazell's first
    major league home run in the
    11th gave the New York Mets a
    stunning 4-3 victory over the
    Chicago Cubs on Saturday.
    The Cubs had much on the
    line..." + ], + [ + "AFP - At least 54 people have
    died and more than a million
    have fled their homes as
    torrential rains lashed parts
    of India and Bangladesh,
    officials said." + ], + [ + "LOS ANGELES - California has
    adopted the world's first
    rules to reduce greenhouse
    emissions for autos, taking
    what supporters see as a
    dramatic step toward cleaning
    up the environment but also
    ensuring higher costs for
    drivers. The rules may lead
    to sweeping changes in
    vehicles nationwide,
    especially if other states opt
    to follow California's
    example..." + ], + [ + " LONDON (Reuters) - European
    stock markets scaled
    near-2-1/2 year highs on
    Friday as oil prices held
    below \\$48 a barrel, and the
    euro held off from mounting
    another assault on \\$1.30 but
    hovered near record highs
    against the dollar." + ], + [ + "Tim Duncan had 17 points and
    10 rebounds, helping the San
    Antonio Spurs to a 99-81
    victory over the New York
    Kicks. This was the Spurs
    fourth straight win this
    season." + ], + [ + "Nagpur: India suffered a
    double blow even before the
    first ball was bowled in the
    crucial third cricket Test
    against Australia on Tuesday
    when captain Sourav Ganguly
    and off spinner Harbhajan
    Singh were ruled out of the
    match." + ], + [ + "AFP - Republican and
    Democratic leaders each
    declared victory after the
    first head-to-head sparring
    match between President George
    W. Bush and Democratic
    presidential hopeful John
    Kerry." + ], + [ + "THIS YULE is all about console
    supply and there #39;s
    precious little units around,
    it has emerged. Nintendo has
    announced that it is going to
    ship another 400,000 units of
    its DS console to the United
    States to meet the shortfall
    there." + ], + [ + "Annual global semiconductor
    sales growth will probably
    fall by half in 2005 and
    memory chip sales could
    collapse as a supply glut saps
    prices, world-leading memory
    chip maker Samsung Electronics
    said on Monday." + ], + [ + "NEW YORK - Traditional phone
    systems may be going the way
    of the Pony Express. Voice-
    over-Internet Protocol,
    technology that allows users
    to make and receive phone
    calls using the Internet, is
    giving the old circuit-
    switched system a run for its
    money." + ], + [ + "AP - Former New York Yankees
    hitting coach Rick Down was
    hired for the same job by the
    Mets on Friday, reuniting him
    with new manager Willie
    Randolph." + ], + [ + "Last night in New York, the UN
    secretary-general was given a
    standing ovation - a robust
    response to a series of
    attacks in past weeks." + ], + [ + "FILDERSTADT (Germany) - Amelie
    Mauresmo and Lindsay Davenport
    took their battle for the No.
    1 ranking and Porsche Grand
    Prix title into the semi-
    finals with straight-sets
    victories on Friday." + ], + [ + "Reuters - The company behind
    the Atkins Diet\\on Friday
    shrugged off a recent decline
    in interest in low-carb\\diets
    as a seasonal blip, and its
    marketing chief said\\consumers
    would cut out starchy foods
    again after picking up\\pounds
    over the holidays." + ], + [ + "There #39;s something to be
    said for being the quot;first
    mover quot; in an industry
    trend. Those years of extra
    experience in tinkering with a
    new idea can be invaluable in
    helping the first" + ], + [ + "JOHANNESBURG -- Meeting in
    Nigeria four years ago,
    African leaders set a goal
    that 60 percent of children
    and pregnant women in malaria-
    affected areas around the
    continent would be sleeping
    under bed nets by the end of
    2005." + ], + [ + "AP - The first U.S. cases of
    the fungus soybean rust, which
    hinders plant growth and
    drastically cuts crop
    production, were found at two
    research sites in Louisiana,
    officials said Wednesday." + ], + [ + "Carter returned, but it was
    running back Curtis Martin and
    the offensive line that put
    the Jets ahead. Martin rushed
    for all but 10 yards of a
    45-yard drive that stalled at
    the Cardinals 10." + ], + [ + " quot;There #39;s no way
    anyone would hire them to
    fight viruses, quot; said
    Sophos security analyst Gregg
    Mastoras. quot;For one, no
    security firm could maintain
    its reputation by employing
    hackers." + ], + [ + "Symantec has revoked its
    decision to blacklist a
    program that allows Web
    surfers in China to browse
    government-blocked Web sites.
    The move follows reports that
    the firm labelled the Freegate
    program, which" + ], + [ + " NEW YORK (Reuters) - Shares
    of Chiron Corp. <A HREF=\"ht
    tp://www.investor.reuters.com/
    FullQuote.aspx?ticker=CHIR.O t
    arget=/stocks/quickinfo/fullqu
    ote\">CHIR.O</A> fell
    7 percent before the market
    open on Friday, a day after
    the biopharmaceutical company
    said it is delaying shipment
    of its flu vaccine, Fluvirin,
    because lots containing 4
    million vaccines do not meet
    product sterility standards." + ], + [ + "The nation's top
    telecommunications regulator
    said yesterday he will push --
    before the next president is
    inaugurated -- to protect
    fledgling Internet telephone
    services from getting taxed
    and heavily regulated by the
    50 state governments." + ], + [ + "Microsoft has signed a pact to
    work with the United Nations
    Educational, Scientific and
    Cultural Organization (UNESCO)
    to increase computer use,
    Internet access and teacher
    training in developing
    countries." + ], + [ + "DENVER (Ticker) -- Jake
    Plummer more than made up for
    a lack of a running game.
    Plummer passed for 294 yards
    and two touchdowns as the
    Denver Broncos posted a 23-13
    victory over the San Diego
    Chargers in a battle of AFC
    West Division rivals." + ], + [ + "DALLAS -- Belo Corp. said
    yesterday that it would cut
    250 jobs, more than half of
    them at its flagship
    newspaper, The Dallas Morning
    News, and that an internal
    investigation into circulation
    overstatements" + ], + [ + "AP - Duke Bainum outspent Mufi
    Hannemann in Honolulu's most
    expensive mayoral race, but
    apparently failed to garner
    enough votes in Saturday's
    primary to claim the office
    outright." + ], + [ + "roundup Plus: Microsoft tests
    Windows Marketplace...Nortel
    delays financials
    again...Microsoft updates
    SharePoint." + ], + [ + "The Federal Reserve still has
    some way to go to restore US
    interest rates to more normal
    levels, Philadelphia Federal
    Reserve President Anthony
    Santomero said on Monday." + ], + [ + "It took all of about five
    minutes of an introductory
    press conference Wednesday at
    Heritage Hall for USC
    basketball to gain something
    it never really had before." + ], + [ + "Delta Air Lines (DAL.N: Quote,
    Profile, Research) said on
    Wednesday its auditors have
    expressed doubt about the
    airline #39;s financial
    viability." + ], + [ + "POLITICIANS and aid agencies
    yesterday stressed the
    importance of the media in
    keeping the spotlight on the
    appalling human rights abuses
    taking place in the Darfur
    region of Sudan." + ], + [ + "AP - The Boston Red Sox looked
    at the out-of-town scoreboard
    and could hardly believe what
    they saw. The New York Yankees
    were trailing big at home
    against the Cleveland Indians
    in what would be the worst
    loss in the 101-year history
    of the storied franchise." + ], + [ + "The Red Sox will either
    complete an amazing comeback
    as the first team to rebound
    from a 3-0 deficit in
    postseason history, or the
    Yankees will stop them." + ], + [ + "\\Children who have a poor diet
    are more likely to become
    aggressive and anti-social, US
    researchers believe." + ], + [ + "OPEN SOURCE champion Microsoft
    is expanding its programme to
    give government organisations
    some of its source code. In a
    communique from the lair of
    the Vole, in Redmond,
    spinsters have said that
    Microsoft" + ], + [ + "The Red Sox have reached
    agreement with free agent
    pitcher Matt Clement yesterday
    on a three-year deal that will
    pay him around \\$25 million,
    his agent confirmed yesterday." + ], + [ + "Takeover target Ronin Property
    Group said it would respond to
    an offer by Multiplex Group
    for all the securities in the
    company in about three weeks." + ], + [ + "Canadian Press - OTTAWA (CP) -
    Contrary to Immigration
    Department claims, there is no
    shortage of native-borne
    exotic dancers in Canada, says
    a University of Toronto law
    professor who has studied the
    strip club business." + ], + [ + "HEN Manny Ramirez and David
    Ortiz hit consecutive home
    runs Sunday night in Chicago
    to put the Red Sox ahead,
    there was dancing in the
    streets in Boston." + ], + [ + "Google Inc. is trying to
    establish an online reading
    room for five major libraries
    by scanning stacks of hard-to-
    find books into its widely
    used Internet search engine." + ], + [ + "HOUSTON--(BUSINESS
    WIRE)--Sept. 1, 2004-- L
    #39;operazione crea una
    centrale globale per l
    #39;analisi strategica el
    #39;approfondimento del
    settore energetico IHS Energy,
    fonte globale leader di
    software, analisi e
    informazioni" + ], + [ + "The European Union presidency
    yesterday expressed optimism
    that a deal could be struck
    over Turkey #39;s refusal to
    recognize Cyprus in the lead-
    up to next weekend #39;s EU
    summit, which will decide
    whether to give Ankara a date
    for the start of accession
    talks." + ], + [ + " WASHINGTON (Reuters) -
    President Bush on Friday set a
    four-year goal of seeing a
    Palestinian state established
    and he and British Prime
    Minister Tony Blair vowed to
    mobilize international
    support to help make it happen
    now that Yasser Arafat is
    dead." + ], + [ + "WASHINGTON Can you always tell
    when somebody #39;s lying? If
    so, you might be a wizard of
    the fib. A California
    psychology professor says
    there #39;s a tiny subculture
    of people that can pick out a
    lie nearly every time they
    hear one." + ], + [ + " KHARTOUM (Reuters) - Sudan on
    Saturday questioned U.N.
    estimates that up to 70,000
    people have died from hunger
    and disease in its remote
    Darfur region since a
    rebellion began 20 months
    ago." + ], + [ + "Type design was once the
    province of skilled artisans.
    With the help of new computer
    programs, neophytes have
    flooded the Internet with
    their creations." + ], + [ + "RCN Inc., co-owner of
    Starpower Communications LLC,
    the Washington area
    television, telephone and
    Internet provider, filed a
    plan of reorganization
    yesterday that it said puts
    the company" + ], + [ + "MIAMI -- Bryan Randall grabbed
    a set of Mardi Gras beads and
    waved them aloft, while his
    teammates exalted in the
    prospect of a trip to New
    Orleans." + ], + [ + "<strong>Letters</stro
    ng> Reports of demise
    premature" + ], + [ + "TORONTO (CP) - With an injured
    Vince Carter on the bench, the
    Toronto Raptors dropped their
    sixth straight game Friday,
    101-87 to the Denver Nuggets." + ], + [ + "The US airline industry,
    riddled with excess supply,
    will see a significant drop in
    capacity, or far fewer seats,
    as a result of at least one
    airline liquidating in the
    next year, according to
    AirTran Airways Chief
    Executive Joe Leonard." + ], + [ + "Boeing (nyse: BA - news -
    people ) Chief Executive Harry
    Stonecipher is keeping the
    faith. On Monday, the head of
    the aerospace and military
    contractor insists he #39;s
    confident his firm will
    ultimately win out" + ], + [ + "While not quite a return to
    glory, Monday represents the
    Redskins' return to the
    national consciousness." + ], + [ + "Australia #39;s biggest
    supplier of fresh milk,
    National Foods, has posted a
    net profit of \\$68.7 million,
    an increase of 14 per cent on
    last financial year." + ], + [ + "Lawyers for customers suing
    Merck amp; Co. want to
    question CEO Raymond Gilmartin
    about what he knew about the
    dangers of Vioxx before the
    company withdrew the drug from
    the market because of health
    hazards." + ], + [ + "Vijay Singh has won the US PGA
    Tour player of the year award
    for the first time, ending
    Tiger Woods #39;s five-year
    hold on the honour." + ], + [ + "New York; September 23, 2004 -
    The Department of Justice
    (DoJ), FBI and US Attorney
    #39;s Office handed down a
    10-count indictment against
    former Computer Associates
    (CA) chairman and CEO Sanjay
    Kumar and Stephen Richards,
    former CA head of worldwide
    sales." + ], + [ + "AFP - At least four Georgian
    soldiers were killed and five
    wounded in overnight clashes
    in Georgia's separatist, pro-
    Russian region of South
    Ossetia, Georgian officers
    near the frontline with
    Ossetian forces said early
    Thursday." + ], + [ + "Intel, the world #39;s largest
    chip maker, scrapped a plan
    Thursday to enter the digital
    television chip business,
    marking a major retreat from
    its push into consumer
    electronics." + ], + [ + "PACIFIC Hydro shares yesterday
    caught an updraught that sent
    them more than 20 per cent
    higher after the wind farmer
    moved to flush out a bidder." + ], + [ + "The European Commission is
    expected later this week to
    recommend EU membership talks
    with Turkey. Meanwhile, German
    Chancellor Gerhard Schroeder
    and Turkish Prime Minister
    Tayyip Erdogan are
    anticipating a quot;positive
    report." + ], + [ + "The US is the originator of
    over 42 of the worlds
    unsolicited commercial e-mail,
    making it the worst offender
    in a league table of the top
    12 spam producing countries
    published yesterday by anti-
    virus firm Sophos." + ], + [ + " NEW YORK (Reuters) - U.S.
    consumer confidence retreated
    in August while Chicago-area
    business activity slowed,
    according to reports on
    Tuesday that added to worries
    the economy's patch of slow
    growth may last beyond the
    summer." + ], + [ + "Intel is drawing the curtain
    on some of its future research
    projects to continue making
    transistors smaller, faster,
    and less power-hungry out as
    far as 2020." + ], + [ + "England got strikes from
    sparkling debut starter
    Jermain Defoe and Michael Owen
    to defeat Poland in a Uefa
    World Cup qualifier in
    Chorzow." + ], + [ + "The Canadian government
    signalled its intention
    yesterday to reintroduce
    legislation to decriminalise
    the possession of small
    amounts of marijuana." + ], + [ + "A screensaver targeting spam-
    related websites appears to
    have been too successful." + ], + [ + "Titleholder Ernie Els moved
    within sight of a record sixth
    World Match Play title on
    Saturday by solving a putting
    problem to overcome injured
    Irishman Padraig Harrington 5
    and 4." + ], + [ + "If the Washington Nationals
    never win a pennant, they have
    no reason to ever doubt that
    DC loves them. Yesterday, the
    District City Council
    tentatively approved a tab for
    a publicly financed ballpark
    that could amount to as much
    as \\$630 million." + ], + [ + "Plus: Experts fear Check 21
    could lead to massive bank
    fraud." + ], + [ + "By SIOBHAN McDONOUGH
    WASHINGTON (AP) -- Fewer
    American youths are using
    marijuana, LSD and Ecstasy,
    but more are abusing
    prescription drugs, says a
    government report released
    Thursday. The 2003 National
    Survey on Drug Use and Health
    also found youths and young
    adults are more aware of the
    risks of using pot once a
    month or more frequently..." + ], + [ + " ABIDJAN (Reuters) - Ivory
    Coast warplanes killed nine
    French soldiers on Saturday in
    a bombing raid during the
    fiercest clashes with rebels
    for 18 months and France hit
    back by destroying most of
    the West African country's
    small airforce." + ], + [ + "Unilever has reported a three
    percent rise in third-quarter
    earnings but warned it is
    reviewing its targets up to
    2010, after issuing a shock
    profits warning last month." + ], + [ + "A TNO engineer prepares to
    start capturing images for the
    world's biggest digital photo." + ], + [ + "Defensive back Brandon
    Johnson, who had two
    interceptions for Tennessee at
    Mississippi, was suspended
    indefinitely Monday for
    violation of team rules." + ], + [ + "Points leader Kurt Busch spun
    out and ran out of fuel, and
    his misfortune was one of the
    reasons crew chief Jimmy
    Fennig elected not to pit with
    20 laps to go." + ], + [ + " LONDON (Reuters) - Oil prices
    extended recent heavy losses
    on Wednesday ahead of weekly
    U.S. data expected to show
    fuel stocks rising in time
    for peak winter demand." + ], + [ + "(CP) - The NHL all-star game
    hasn #39;t been cancelled
    after all. It #39;s just been
    moved to Russia. The agent for
    New York Rangers winger
    Jaromir Jagr confirmed Monday
    that the Czech star had joined
    Omsk Avangard" + ], + [ + "PalmOne is aiming to sharpen
    up its image with the launch
    of the Treo 650 on Monday. As
    previously reported, the smart
    phone update has a higher-
    resolution screen and a faster
    processor than the previous
    top-of-the-line model, the
    Treo 600." + ], + [ + "GAZA CITY, Gaza Strip --
    Islamic militant groups behind
    many suicide bombings
    dismissed yesterday a call
    from Mahmoud Abbas, the
    interim Palestinian leader, to
    halt attacks in the run-up to
    a Jan. 9 election to replace
    Yasser Arafat." + ], + [ + "Secretary of State Colin
    Powell is wrapping up an East
    Asia trip focused on prodding
    North Korea to resume talks
    aimed at ending its nuclear-
    weapons program." + ], + [ + "HERE in the land of myth, that
    familiar god of sports --
    karma -- threw a bolt of
    lightning into the Olympic
    stadium yesterday. Marion
    Jones lunged desperately with
    her baton in the 4 x 100m
    relay final, but couldn #39;t
    reach her target." + ], + [ + "kinrowan writes quot;MIT,
    inventor of Kerberos, has
    announced a pair of
    vulnerabities in the software
    that will allow an attacker to
    either execute a DOS attack or
    execute code on the machine." + ], + [ + "The risk of intestinal damage
    from common painkillers may be
    higher than thought, research
    suggests." + ], + [ + "AN earthquake measuring 7.3 on
    the Richter Scale hit western
    Japan this morning, just hours
    after another strong quake
    rocked the area." + ], + [ + "Vodafone has increased the
    competition ahead of Christmas
    with plans to launch 10
    handsets before the festive
    season. The Newbury-based
    group said it will begin
    selling the phones in
    November." + ], + [ + "Reuters - Former Pink Floyd
    mainman Roger\\Waters released
    two new songs, both inspired
    by the U.S.-led\\invasion of
    Iraq, via online download
    outlets Tuesday." + ], + [ + "A former assistant treasurer
    at Enron Corp. (ENRNQ.PK:
    Quote, Profile, Research)
    agreed to plead guilty to
    conspiracy to commit
    securities fraud on Tuesday
    and will cooperate with" + ], + [ + "Britain #39;s Prince Harry,
    struggling to shed a growing
    quot;wild child quot; image,
    won #39;t apologize to a
    photographer he scuffled with
    outside an exclusive London
    nightclub, a royal spokesman
    said on Saturday." + ], + [ + "UK house prices fell by 1.1 in
    October, confirming a
    softening of the housing
    market, Halifax has said. The
    UK #39;s biggest mortgage
    lender said prices rose 18." + ], + [ + "Pakistan #39;s interim Prime
    Minister Chaudhry Shaujaat
    Hussain has announced his
    resignation, paving the way
    for his successor Shauket
    Aziz." + ], + [ + "A previously unknown group
    calling itself Jamaat Ansar
    al-Jihad al-Islamiya says it
    set fire to a Jewish soup
    kitchen in Paris, according to
    an Internet statement." + ], + [ + "More than six newspaper
    companies have received
    letters from the Securities
    and Exchange Commission
    seeking information about
    their circulation practices." + ], + [ + "THE 64,000 dollar -
    correction, make that 500
    million dollar -uestion
    hanging over Shire
    Pharmaceuticals is whether the
    5 per cent jump in the
    companys shares yesterday
    reflects relief that US
    regulators have finally
    approved its drug for" + ], + [ + "The deadliest attack on
    Americans in Iraq since May
    came as Iraqi officials
    announced that Saddam
    Hussein's deputy had not been
    captured on Sunday." + ], + [ + "AP - Kenny Rogers lost at the
    Coliseum for the first time in
    more than 10 years, with Bobby
    Crosby's three-run double in
    the fifth inning leading the
    Athletics to a 5-4 win over
    the Texas Rangers on Thursday." + ], + [ + "A fundraising concert will be
    held in London in memory of
    the hundreds of victims of the
    Beslan school siege." + ], + [ + "Dambulla, Sri Lanka - Kumar
    Sangakkara and Avishka
    Gunawardene slammed impressive
    half-centuries to help an
    under-strength Sri Lanka crush
    South Africa by seven wickets
    in the fourth one-day
    international here on
    Saturday." + ], + [ + "Fresh off being the worst team
    in baseball, the Arizona
    Diamondbacks set a new record
    this week: fastest team to
    both hire and fire a manager." + ], + [ + "Nebraska head coach Bill
    Callahan runs off the field at
    halftime of the game against
    Baylor in Lincoln, Neb.,
    Saturday, Oct. 16, 2004." + ], + [ + "NASA has been working on a
    test flight project for the
    last few years involving
    hypersonic flight. Hypersonic
    flight is fight at speeds
    greater than Mach 5, or five
    times the speed of sound." + ], + [ + "AFP - The landmark trial of a
    Rwandan Roman Catholic priest
    accused of supervising the
    massacre of 2,000 of his Tutsi
    parishioners during the
    central African country's 1994
    genocide opens at a UN court
    in Tanzania." + ], + [ + "The Irish government has
    stepped up its efforts to free
    the British hostage in Iraq,
    Ken Bigley, whose mother is
    from Ireland, by talking to
    diplomats from Iran and
    Jordan." + ], + [ + "AP - Republican Sen. Lincoln
    Chafee, who flirted with
    changing political parties in
    the wake of President Bush's
    re-election victory, says he
    will stay in the GOP." + ], + [ + "AP - Microsoft Corp. announced
    Wednesday that it would offer
    a low-cost, localized version
    of its Windows XP operating
    system in India to tap the
    large market potential in this
    country of 1 billion people,
    most of whom do not speak
    English." + ], + [ + "Businesses saw inventories
    rise in July and sales picked
    up, the government reported
    Wednesday. The Commerce
    Department said that stocks of
    unsold goods increased 0.9 in
    July, down from a 1.1 rise in
    June." + ], + [ + " EAST RUTHERFORD, N.J. (Sports
    Network) - The Toronto
    Raptors have traded All-Star
    swingman Vince Carter to the
    New Jersey Nets in exchange
    for center Alonzo Mourning,
    forward Eric Williams,
    center/forward Aaron Williams
    and two first- round draft
    picks." + ], + [ + "AP - Utah State University has
    secured a #36;40 million
    contract with NASA to build an
    orbiting telescope that will
    examine galaxies and try to
    find new stars." + ], + [ + "South Korean President Roh
    Moo-hyun pays a surprise visit
    to troops in Iraq, after his
    government decided to extend
    their mandate." + ], + [ + "As the European Union
    approaches a contentious
    decision - whether to let
    Turkey join the club - the
    Continent #39;s rulers seem to
    have left their citizens
    behind." + ], + [ + "What riot? quot;An Argentine
    friend of mine was a little
    derisive of the Pacers-Pistons
    eruption, quot; says reader
    Mike Gaynes. quot;He snorted,
    #39;Is that what Americans
    call a riot?" + ], + [ + "All season, Chris Barnicle
    seemed prepared for just about
    everything, but the Newton
    North senior was not ready for
    the hot weather he encountered
    yesterday in San Diego at the
    Footlocker Cross-Country
    National Championships. Racing
    in humid conditions with
    temperatures in the 70s, the
    Massachusetts Division 1 state
    champion finished sixth in 15
    minutes 34 seconds in the
    5-kilometer race. ..." + ], + [ + "Shares of Genta Inc. (GNTA.O:
    Quote, Profile, Research)
    soared nearly 50 percent on
    Monday after the biotechnology
    company presented promising
    data on an experimental
    treatment for blood cancers." + ], + [ + "I spend anywhere from three to
    eight hours every week
    sweating along with a motley
    crew of local misfits,
    shelving, sorting, and hauling
    ton after ton of written
    matter in a rowhouse basement
    in Baltimore. We have no heat
    nor air conditioning, but
    still, every week, we come and
    work. Volunteer night is
    Wednesday, but many of us also
    work on the weekends, when
    we're open to the public.
    There are times when we're
    freezing and we have to wear
    coats and gloves inside,
    making handling books somewhat
    tricky; other times, we're all
    soaked with sweat, since it's
    90 degrees out and the
    basement is thick with bodies.
    One learns to forget about
    personal space when working at
    The Book Thing, since you can
    scarcely breathe without
    bumping into someone, and we
    are all so accustomed to
    having to scrape by each other
    that most of us no longer
    bother to say \"excuse me\"
    unless some particularly
    dramatic brushing occurs." + ], + [ + " BAGHDAD (Reuters) - A
    deadline set by militants who
    have threatened to kill two
    Americans and a Briton seized
    in Iraq was due to expire
    Monday, and more than two
    dozen other hostages were
    also facing death unless rebel
    demands were met." + ], + [ + "Alarmed by software glitches,
    security threats and computer
    crashes with ATM-like voting
    machines, officials from
    Washington, D.C., to
    California are considering an
    alternative from an unlikely
    place: Nevada." + ], + [ + "Some Venezuelan television
    channels began altering their
    programs Thursday, citing
    fears of penalties under a new
    law restricting violence and
    sexual content over the
    airwaves." + ], + [ + "SBC Communications Inc. plans
    to cut at least 10,000 jobs,
    or 6 percent of its work
    force, by the end of next year
    to compensate for a drop in
    the number of local-telephone
    customers." + ], + [ + "afrol News, 4 October -
    Hundred years of conservation
    efforts have lifted the
    southern black rhino
    population from about hundred
    to 11,000 animals." + ], + [ + "THE death of Philippine movie
    star and defeated presidential
    candidate Fernando Poe has
    drawn some political backlash,
    with some people seeking to
    use his sudden demise as a
    platform to attack President
    Gloria Arroyo." + ], + [ + " WASHINGTON (Reuters) - Major
    cigarette makers go on trial
    on Tuesday in the U.S.
    government's \\$280 billion
    racketeering case that
    charges the tobacco industry
    with deliberately deceiving
    the public about the risks of
    smoking since the 1950s." + ], + [ + "More Americans are quitting
    their jobs and taking the risk
    of starting a business despite
    a still-lackluster job market." + ], + [ + "AP - Coach Tyrone Willingham
    was fired by Notre Dame on
    Tuesday after three seasons in
    which he failed to return one
    of the nation's most storied
    football programs to
    prominence." + ], + [ + "COLLEGE PARK, Md. -- Joel
    Statham completed 18 of 25
    passes for 268 yards and two
    touchdowns in No. 23
    Maryland's 45-22 victory over
    Temple last night, the
    Terrapins' 12th straight win
    at Byrd Stadium." + ], + [ + "Manchester United boss Sir
    Alex Ferguson wants the FA to
    punish Arsenal good guy Dennis
    Bergkamp for taking a swing at
    Alan Smith last Sunday." + ], + [ + "VIENTIANE, Laos China moved
    yet another step closer in
    cementing its economic and
    diplomatic relationships with
    Southeast Asia today when
    Prime Minister Wen Jiabao
    signed a trade accord at a
    regional summit that calls for
    zero tariffs on a wide range
    of" + ], + [ + "SPACE.com - With nbsp;food
    stores nbsp;running low, the
    two \\astronauts living aboard
    the International Space
    Station (ISS) are cutting back
    \\their meal intake and
    awaiting a critical cargo
    nbsp;delivery expected to
    arrive \\on Dec. 25." + ], + [ + "British judges in London
    Tuesday ordered radical Muslim
    imam Abu Hamza to stand trial
    for soliciting murder and
    inciting racial hatred." + ], + [ + "Federal Reserve policy-makers
    raised the benchmark US
    interest rate a quarter point
    to 2.25 per cent and restated
    a plan to carry out" + ], + [ + " DETROIT (Reuters) - A
    Canadian law firm on Tuesday
    said it had filed a lawsuit
    against Ford Motor Co. <A H
    REF=\"http://www.investor.reute
    rs.com/FullQuote.aspx?ticker=F
    .N target=/stocks/quickinfo/fu
    llquote\">F.N</A> over
    what it claims are defective
    door latches on about 400,000
    of the automaker's popular
    pickup trucks and SUVs." + ], + [ + "Published reports say Barry
    Bonds has testified that he
    used a clear substance and a
    cream given to him by a
    trainer who was indicted in a
    steroid-distribution ring." + ], + [ + "SARASOTA, Fla. - The
    devastation brought on by
    Hurricane Charley has been
    especially painful for an
    elderly population that is
    among the largest in the
    nation..." + ], + [ + " ATHENS (Reuters) - Christos
    Angourakis added his name to
    Greece's list of Paralympic
    medal winners when he claimed
    a bronze in the T53 shot put
    competition Thursday." + ], + [ + " quot;He is charged for having
    a part in the Bali incident,
    quot; state prosecutor Andi
    Herman told Reuters on
    Saturday. bombing attack at
    the US-run JW." + ], + [ + "Jay Fiedler threw for one
    touchdown, Sage Rosenfels
    threw for another and the
    Miami Dolphins got a victory
    in a game they did not want to
    play, beating the New Orleans
    Saints 20-19 Friday night." + ], + [ + " NEW YORK (Reuters) - Terrell
    Owens scored three touchdowns
    and the Philadelphia Eagles
    amassed 35 first-half points
    on the way to a 49-21
    drubbing of the Dallas Cowboys
    in Irving, Texas, Monday." + ], + [ + "AstraZeneca Plc suffered its
    third setback in two months on
    Friday as lung cancer drug
    Iressa failed to help patients
    live longer" + ], + [ + "Virgin Electronics hopes its
    slim Virgin Player, which
    debuts today and is smaller
    than a deck of cards, will
    rise as a lead competitor to
    Apple's iPod." + ], + [ + "Researchers train a monkey to
    feed itself by guiding a
    mechanical arm with its mind.
    It could be a big step forward
    for prosthetics. By David
    Cohn." + ], + [ + "Bruce Wasserstein, the
    combative chief executive of
    investment bank Lazard, is
    expected to agree this week
    that he will quit the group
    unless he can pull off a
    successful" + ], + [ + "A late strike by Salomon Kalou
    sealed a 2-1 win for Feyenoord
    over NEC Nijmegen, while
    second placed AZ Alkmaar
    defeated ADO Den Haag 2-0 in
    the Dutch first division on
    Sunday." + ], + [ + "The United Nations called on
    Monday for an immediate
    ceasefire in eastern Congo as
    fighting between rival army
    factions flared for a third
    day." + ], + [ + "What a disgrace Ron Artest has
    become. And the worst part is,
    the Indiana Pacers guard just
    doesn #39;t get it. Four days
    after fueling one of the
    biggest brawls in the history
    of pro sports, Artest was on
    national" + ], + [ + "Allowing dozens of casinos to
    be built in the UK would bring
    investment and thousands of
    jobs, Tony Blair says." + ], + [ + "The shock here was not just
    from the awful fact itself,
    that two vibrant young Italian
    women were kidnapped in Iraq,
    dragged from their office by
    attackers who, it seems, knew
    their names." + ], + [ + "Tehran/Vianna, Sept. 19 (NNN):
    Iran on Sunday rejected the
    International Atomic Energy
    Agency (IAEA) call to suspend
    of all its nuclear activities,
    saying that it will not agree
    to halt uranium enrichment." + ], + [ + "A 15-year-old Argentine
    student opened fire at his
    classmates on Tuesday in a
    middle school in the south of
    the Buenos Aires province,
    leaving at least four dead and
    five others wounded, police
    said." + ], + [ + "Dr. David J. Graham, the FDA
    drug safety reviewer who
    sounded warnings over five
    drugs he felt could become the
    next Vioxx has turned to a
    Whistleblower protection group
    for legal help." + ], + [ + "AP - Scientists in 17
    countries will scout waterways
    to locate and study the
    world's largest freshwater
    fish species, many of which
    are declining in numbers,
    hoping to learn how to better
    protect them, researchers
    announced Thursday." + ], + [ + "AP - Google Inc.'s plans to
    move ahead with its initial
    public stock offering ran into
    a roadblock when the
    Securities and Exchange
    Commission didn't approve the
    Internet search giant's
    regulatory paperwork as
    requested." + ], + [ + "Jenson Button has revealed
    dissatisfaction with the way
    his management handled a
    fouled switch to Williams. Not
    only did the move not come
    off, his reputation may have
    been irreparably damaged amid
    news headline" + ], + [ + "The Kmart purchase of Sears,
    Roebuck may be the ultimate
    expression of that old saying
    in real estate: location,
    location, location." + ], + [ + "Citing security risks, a state
    university is urging students
    to drop Internet Explorer in
    favor of alternative Web
    browsers such as Firefox and
    Safari." + ], + [ + "Redknapp and his No2 Jim Smith
    resigned from Portsmouth
    yesterday, leaving
    controversial new director
    Velimir Zajec in temporary
    control." + ], + [ + "Despite being ranked eleventh
    in the world in broadband
    penetration, the United States
    is rolling out high-speed
    services on a quot;reasonable
    and timely basis to all
    Americans, quot; according to
    a new report narrowly approved
    today by the Federal
    Communications" + ], + [ + "Sprint Corp. (FON.N: Quote,
    Profile, Research) on Friday
    said it plans to cut up to 700
    jobs as it realigns its
    business to focus on wireless
    and Internet services and
    takes a non-cash network
    impairment charge." + ], + [ + "As the season winds down for
    the Frederick Keys, Manager
    Tom Lawless is starting to
    enjoy the progress his
    pitching staff has made this
    season." + ], + [ + "Britain #39;s Bradley Wiggins
    won the gold medal in men
    #39;s individual pursuit
    Saturday, finishing the
    4,000-meter final in 4:16." + ], + [ + "And when David Akers #39;
    50-yard field goal cleared the
    crossbar in overtime, they did
    just that. They escaped a
    raucous Cleveland Browns
    Stadium relieved but not
    broken, tested but not
    cracked." + ], + [ + "NEW YORK - A cable pay-per-
    view company has decided not
    to show a three-hour election
    eve special with filmmaker
    Michael Moore that included a
    showing of his documentary
    \"Fahrenheit 9/11,\" which is
    sharply critical of President
    Bush. The company, iN
    DEMAND, said Friday that its
    decision is due to \"legitimate
    business and legal concerns.\"
    A spokesman would not
    elaborate..." + ], + [ + "Democracy candidates picked up
    at least one more seat in
    parliament, according to exit
    polls." + ], + [ + "The IOC wants suspended
    Olympic member Ivan Slavkov to
    be thrown out of the
    organisation." + ], + [ + " BANGKOK (Reuters) - The
    ouster of Myanmar's prime
    minister, architect of a
    tentative \"roadmap to
    democracy,\" has dashed faint
    hopes for an end to military
    rule and leaves Southeast
    Asia's policy of constructive
    engagement in tatters." + ], + [ + "San Antonio, TX (Sports
    Network) - Dean Wilson shot a
    five-under 65 on Friday to
    move into the lead at the
    halfway point of the Texas
    Open." + ], + [ + "Now that Chelsea have added
    Newcastle United to the list
    of clubs that they have given
    what for lately, what price
    Jose Mourinho covering the
    Russian-funded aristocrats of
    west London in glittering
    glory to the tune of four
    trophies?" + ], + [ + "AP - Former Seattle Seahawks
    running back Chris Warren has
    been arrested in Virginia on a
    federal warrant, accused of
    failing to pay #36;137,147 in
    child support for his two
    daughters in Washington state." + ], + [ + "The anguish of hostage Kenneth
    Bigley in Iraq hangs over
    Prime Minister Tony Blair
    today as he faces the twin
    test of a local election and a
    debate by his Labour Party
    about the divisive war." + ], + [ + "Says that amount would have
    been earned for the first 9
    months of 2004, before AT
    amp;T purchase. LOS ANGELES,
    (Reuters) - Cingular Wireless
    would have posted a net profit
    of \\$650 million for the first
    nine months" + ], + [ + "NewsFactor - Taking an
    innovative approach to the
    marketing of high-performance
    \\computing, Sun Microsystems
    (Nasdaq: SUNW) is offering its
    N1 Grid program in a pay-for-
    use pricing model that mimics
    the way such commodities as
    electricity and wireless phone
    plans are sold." + ], + [ + " CHICAGO (Reuters) - Goodyear
    Tire Rubber Co. <A HREF=\"
    http://www.investor.reuters.co
    m/FullQuote.aspx?ticker=GT.N t
    arget=/stocks/quickinfo/fullqu
    ote\">GT.N</A> said
    on Friday it will cut 340 jobs
    in its engineered products and
    chemical units as part of its
    cost-reduction efforts,
    resulting in a third-quarter
    charge." + ], + [ + " quot;There were 16 people
    travelling aboard. ... It
    crashed into a mountain, quot;
    Col. Antonio Rivero, head of
    the Civil Protection service,
    told." + ], + [ + "Reuters - Shares of long-
    distance phone\\companies AT T
    Corp. and MCI Inc. have
    plunged\\about 20 percent this
    year, but potential buyers
    seem to be\\holding out for
    clearance sales." + ], + [ + "Reuters - Madonna and m-Qube
    have\\made it possible for the
    star's North American fans to
    download\\polyphonic ring tones
    and other licensed mobile
    content from\\her official Web
    site, across most major
    carriers and without\\the need
    for a credit card." + ], + [ + "President Bush is reveling in
    winning the popular vote and
    feels he can no longer be
    considered a one-term accident
    of history." + ], + [ + "Russian oil giant Yukos files
    for bankruptcy protection in
    the US in a last ditch effort
    to stop the Kremlin auctioning
    its main production unit." + ], + [ + "British Airways #39; (BA)
    chief executive Rod Eddington
    has admitted that the company
    quot;got it wrong quot; after
    staff shortages led to three
    days of travel chaos for
    passengers." + ], + [ + "It #39;s official: US Open had
    never gone into the third
    round with only two American
    men, including the defending
    champion, Andy Roddick." + ], + [ + "Canadian Press - TORONTO (CP)
    - Thousands of Royal Bank
    clerks are being asked to
    display rainbow stickers at
    their desks and cubicles to
    promote a safe work
    environment for gays,
    lesbians, and bisexuals." + ], + [ + "The chipmaker is back on a
    buying spree, having scooped
    up five other companies this
    year." + ], + [ + "The issue of drug advertising
    directly aimed at consumers is
    becoming political." + ], + [ + "WASHINGTON, Aug. 17
    (Xinhuanet) -- England coach
    Sven-Goran Eriksson has urged
    the international soccer
    authorities to preserve the
    health of the world superstar
    footballers for major
    tournaments, who expressed his
    will in Slaley of England on
    Tuesday ..." + ], + [ + " BAGHDAD (Reuters) - A car
    bomb killed two American
    soldiers and wounded eight
    when it exploded in Baghdad on
    Saturday, the U.S. military
    said in a statement." + ], + [ + "Juventus coach Fabio Capello
    has ordered his players not to
    kick the ball out of play when
    an opponent falls to the
    ground apparently hurt because
    he believes some players fake
    injury to stop the match." + ], + [ + "AP - China's economic boom is
    still roaring despite efforts
    to cool sizzling growth, with
    the gross domestic product
    climbing 9.5 percent in the
    first three quarters of this
    year, the government reported
    Friday." + ], + [ + "Soaring petroleum prices
    pushed the cost of goods
    imported into the U.S. much
    higher than expected in
    August, the government said
    today." + ], + [ + "Anheuser-Busch teams up with
    Vietnam's largest brewer,
    laying the groundwork for
    future growth in the region." + ], + [ + "You #39;re angry. You want to
    lash out. The Red Sox are
    doing it to you again. They
    #39;re blowing a playoff
    series, and to the Yankees no
    less." + ], + [ + "TORONTO -- There is no
    mystique to it anymore,
    because after all, the
    Russians have become commoners
    in today's National Hockey
    League, and Finns, Czechs,
    Slovaks, and Swedes also have
    been entrenched in the
    Original 30 long enough to
    turn the ongoing World Cup of
    Hockey into a protracted
    trailer for the NHL season." + ], + [ + "Sudanese authorities have
    moved hundreds of pro-
    government fighters from the
    crisis-torn Darfur region to
    other parts of the country to
    keep them out of sight of
    foreign military observers
    demanding the militia #39;s
    disarmament, a rebel leader
    charged" + ], + [ + "CHARLOTTE, NC - Shares of
    Krispy Kreme Doughnuts Inc.
    #39;s fell sharply Monday as a
    79 percent plunge in third-
    quarter earnings and an
    intensifying accounting
    investigation overshadowed the
    pastrymaker #39;s statement
    that the low-carb craze might
    be easing." + ], + [ + "The company hopes to lure
    software partners by promising
    to save them from
    infrastructure headaches." + ], + [ + "BAGHDAD, Iraq - A car bomb
    Tuesday ripped through a busy
    market near a Baghdad police
    headquarters where Iraqis were
    waiting to apply for jobs on
    the force, and gunmen opened
    fire on a van carrying police
    home from work in Baqouba,
    killing at least 59 people
    total and wounding at least
    114. The attacks were the
    latest attempts by militants
    to wreck the building of a
    strong Iraqi security force, a
    keystone of the U.S..." + ], + [ + "The Israeli prime minister
    said today that he wanted to
    begin withdrawing settlers
    from Gaza next May or June." + ], + [ + "Indianapolis, IN (Sports
    Network) - The Indiana Pacers
    try to win their second
    straight game tonight, as they
    host Kevin Garnett and the
    Minnesota Timberwolves in the
    third of a four-game homestand
    at Conseco Fieldhouse." + ], + [ + "OXNARD - A leak of explosive
    natural gas forced dozens of
    workers to evacuate an
    offshore oil platform for
    hours Thursday but no damage
    or injuries were reported." + ], + [ + "AP - Assets of the nation's
    retail money market mutual
    funds rose by #36;2.85
    billion in the latest week to
    #36;845.69 billion, the
    Investment Company Institute
    said Thursday." + ], + [ + "Peter Mandelson provoked fresh
    Labour in-fighting yesterday
    with an implied attack on
    Gordon Brown #39;s
    quot;exaggerated gloating
    quot; about the health of the
    British economy." + ], + [ + "Queen Elizabeth II stopped
    short of apologizing for the
    Allies #39; bombing of Dresden
    during her first state visit
    to Germany in 12 years and
    instead acknowledged quot;the
    appalling suffering of war on
    both sides." + ], + [ + "JC Penney said yesterday that
    Allen I. Questrom, the chief
    executive who has restyled the
    once-beleaguered chain into a
    sleeker and more profitable
    entity, would be succeeded by
    Myron E. Ullman III, another
    longtime retail executive." + ], + [ + " In the cosmetics department
    at Hecht's in downtown
    Washington, construction crews
    have ripped out the
    traditional glass display
    cases, replacing them with a
    system of open shelves stacked
    high with fragrances from
    Chanel, Burberry and Armani,
    now easily within arm's reach
    of the impulse buyer." + ], + [ + " LONDON (Reuters) - Oil prices
    slid from record highs above
    \\$50 a barrel Wednesday as the
    U.S. government reported a
    surprise increase in crude
    stocks and rebels in Nigeria's
    oil-rich delta region agreed
    to a preliminary cease-fire." + ], + [ + "Rocky Shoes and Boots makes an
    accretive acquisition -- and
    gets Dickies and John Deere as
    part of the deal." + ], + [ + "AP - Fugitive Taliban leader
    Mullah Mohammed Omar has
    fallen out with some of his
    lieutenants, who blame him for
    the rebels' failure to disrupt
    the landmark Afghan
    presidential election, the
    U.S. military said Wednesday." + ], + [ + "HAVANA -- Cuban President
    Fidel Castro's advancing age
    -- and ultimately his
    mortality -- were brought home
    yesterday, a day after he
    fractured a knee and arm when
    he tripped and fell at a
    public event." + ], + [ + " BRASILIA, Brazil (Reuters) -
    The United States and Brazil
    predicted on Tuesday Latin
    America's largest country
    would resolve a dispute with
    the U.N. nuclear watchdog over
    inspections of a uranium
    enrichment plant." + ], + [ + "Call it the Maximus factor.
    Archaeologists working at the
    site of an old Roman temple
    near Guy #39;s hospital in
    London have uncovered a pot of
    cosmetic cream dating back to
    AD2." + ], + [ + "It is a team game, this Ryder
    Cup stuff that will commence
    Friday at Oakland Hills
    Country Club. So what are the
    teams? For the Americans,
    captain Hal Sutton isn't
    saying." + ], + [ + "Two bombs exploded today near
    a tea shop and wounded 20
    people in southern Thailand,
    police said, as violence
    continued unabated in the
    Muslim-majority region where
    residents are seething over
    the deaths of 78 detainees
    while in military custody." + ], + [ + "BONN: Deutsche Telekom is
    bidding 2.9 bn for the 26 it
    does not own in T-Online
    International, pulling the
    internet service back into the
    fold four years after selling
    stock to the public." + ], + [ + "Motorola Inc. says it #39;s
    ready to make inroads in the
    cell-phone market after
    posting a third straight
    strong quarter and rolling out
    a series of new handsets in
    time for the holiday selling
    season." + ], + [ + "Prime Minister Junichiro
    Koizumi, back in Tokyo after
    an 11-day diplomatic mission
    to the Americas, hunkered down
    with senior ruling party
    officials on Friday to focus
    on a major reshuffle of
    cabinet and top party posts." + ], + [ + "Costs of employer-sponsored
    health plans are expected to
    climb an average of 8 percent
    in 2005, the first time in
    five years increases have been
    in single digits." + ], + [ + "NEW YORK - A sluggish gross
    domestic product reading was
    nonetheless better than
    expected, prompting investors
    to send stocks slightly higher
    Friday on hopes that the
    economic slowdown would not be
    as bad as first thought.
    The 2.8 percent GDP growth in
    the second quarter, a revision
    from the 3 percent preliminary
    figure reported in July, is a
    far cry from the 4.5 percent
    growth in the first quarter..." + ], + [ + "After again posting record
    earnings for the third
    quarter, Taiwan Semiconductor
    Manufacturing Company (TSMC)
    expects to see its first
    sequential drop in fourth-
    quarter revenues, coupled with
    a sharp drop in capacity
    utilization rates." + ], + [ + " SEOUL (Reuters) - A huge
    explosion in North Korea last
    week may have been due to a
    combination of demolition work
    for a power plant and
    atmospheric clouds, South
    Korea's spy agency said on
    Wednesday." + ], + [ + "The deal comes as Cisco pushes
    to develop a market for CRS-1,
    a new line of routers aimed at
    telephone, wireless and cable
    companies." + ], + [ + "Many people who have never
    bounced a check in their life
    could soon bounce their first
    check if they write checks to
    pay bills a couple of days
    before their paycheck is
    deposited into their checking
    account." + ], + [ + " LUXEMBOURG (Reuters) -
    Microsoft Corp told a judge on
    Thursday that the European
    Commission must be stopped
    from ordering it to give up
    secret technology to
    competitors." + ], + [ + "SEPTEMBER 14, 2004 (IDG NEWS
    SERVICE) - Sun Microsystems
    Inc. and Microsoft Corp. next
    month plan to provide more
    details on the work they are
    doing to make their products
    interoperable, a Sun executive
    said yesterday." + ], + [ + "MANCHESTER United today
    dramatically rejected the
    advances of Malcolm Glazer,
    the US sports boss who is
    mulling an 825m bid for the
    football club." + ], + [ + "Dow Jones Industrial Average
    futures declined amid concern
    an upcoming report on
    manufacturing may point to
    slowing economic growth." + ], + [ + "AP - Astronomy buffs and
    amateur stargazers turned out
    to watch a total lunar eclipse
    Wednesday night #151; the
    last one Earth will get for
    nearly two and a half years." + ], + [ + "Reuters - U.S. industrial
    output advanced in\\July, as
    American factories operated at
    their highest capacity\\in more
    than three years, a Federal
    Reserve report on
    Tuesday\\showed." + ], + [ + "As usual, the Big Ten coaches
    were out in full force at
    today #39;s Big Ten
    Teleconference. Both OSU head
    coach Jim Tressel and Iowa
    head coach Kirk Ferentz
    offered some thoughts on the
    upcoming game between OSU" + ], + [ + "Sir Martin Sorrell, chief
    executive of WPP, yesterday
    declared he was quot;very
    impressed quot; with Grey
    Global, stoking speculation
    WPP will bid for the US
    advertising company." + ], + [ + "The compact disc has at least
    another five years as the most
    popular music format before
    online downloads chip away at
    its dominance, a new study
    said on Tuesday." + ], + [ + "New York Knicks #39; Stephon
    Marbury (3) fires over New
    Orleans Hornets #39; Dan
    Dickau (2) during the second
    half in New Orleans Wednesday
    night, Dec. 8, 2004." + ], + [ + "AP - Sudan's U.N. ambassador
    challenged the United States
    to send troops to the Darfur
    region if it really believes a
    genocide is taking place as
    the U.S. Congress and
    President Bush's
    administration have
    determined." + ], + [ + "At the very moment when the
    Red Sox desperately need
    someone slightly larger than
    life to rally around, they
    suddenly have the man for the
    job: Thrilling Schilling." + ], + [ + "Does Your Site Need a Custom
    Search Engine
    Toolbar?\\\\Today's surfers
    aren't always too comfortable
    installing software on their
    computers. Especially free
    software that they don't
    necessarily understand. With
    all the horror stories of
    viruses, spyware, and adware
    that make the front page these
    days, it's no wonder. So is
    there ..." + ], + [ + "Dale Earnhardt Jr, right,
    talks with Matt Kenseth, left,
    during a break in practice at
    Lowe #39;s Motor Speedway in
    Concord, NC, Thursday Oct. 14,
    2004 before qualifying for
    Saturday #39;s UAW-GM Quality
    500 NASCAR Nextel Cup race." + ], + [ + "Several starting spots may
    have been usurped or at least
    threatened after relatively
    solid understudy showings
    Sunday, but few players
    welcome the kind of shot
    delivered to Oakland" + ], + [ + "TEMPE, Ariz. -- Neil Rackers
    kicked four field goals and
    the Arizona Cardinals stifled
    rookie Chris Simms and the
    rest of the Tampa Bay offense
    for a 12-7 victory yesterday
    in a matchup of two sputtering
    teams out of playoff
    contention. Coach Jon Gruden's
    team lost its fourth in a row
    to finish 5-11, Tampa Bay's
    worst record since going ..." + ], + [ + "AFP - A junior British
    minister said that people
    living in camps after fleeing
    their villages because of
    conflict in Sudan's Darfur
    region lived in fear of
    leaving their temporary homes
    despite a greater presence now
    of aid workers." + ], + [ + "US Deputy Secretary of State
    Richard Armitage (L) shakes
    hands with Pakistani Foreign
    Minister Khurshid Kasuri prior
    to their meeting in Islamabad,
    09 November 2004." + ], + [ + "Spammers aren't ducking
    antispam laws by operating
    offshore, they're just
    ignoring it." + ], + [ + "AP - Biologists plan to use
    large nets and traps this week
    in Chicago's Burnham Harbor to
    search for the northern
    snakehead #151; a type of
    fish known for its voracious
    appetite and ability to wreak
    havoc on freshwater
    ecosystems." + ], + [ + "BAE Systems PLC and Northrop
    Grumman Corp. won \\$45 million
    contracts yesterday to develop
    prototypes of anti-missile
    technology that could protect
    commercial aircraft from
    shoulder-fired missiles." + ], + [ + "AP - Democrat John Edwards
    kept up a long-distance debate
    over his \"two Americas\"
    campaign theme with Vice
    President Dick Cheney on
    Tuesday, saying it was no
    illusion to thousands of laid-
    off workers in Ohio." + ], + [ + "Like introductory credit card
    rates and superior customer
    service, some promises just
    aren #39;t built to last. And
    so it is that Bank of America
    - mere months after its pledge
    to preserve" + ], + [ + "A group of 76 Eritreans on a
    repatriation flight from Libya
    Friday forced their plane to
    change course and land in the
    Sudanese capital, Khartoum,
    where they" + ], + [ + "Reuters - Accounting firm KPMG
    will pay #36;10\\million to
    settle charges of improper
    professional conduct\\while
    acting as auditor for Gemstar-
    TV Guide International Inc.\\,
    the U.S. Securities and
    Exchange Commission said
    on\\Wednesday." + ], + [ + "Family matters made public: As
    eager cousins wait for a slice
    of the \\$15 billion cake that
    is the Pritzker Empire,
    Circuit Court Judge David
    Donnersberger has ruled that
    the case will be conducted in
    open court." + ], + [ + "Users of Microsoft #39;s
    Hotmail, most of whom are
    accustomed to getting regular
    sales pitches for premium
    e-mail accounts, got a
    pleasant surprise in their
    inboxes this week: extra
    storage for free." + ], + [ + "AP - They say that opposites
    attract, and in the case of
    Sen. John Kerry and Teresa
    Heinz Kerry, that may be true
    #151; at least in their public
    personas." + ], + [ + "The weekly survey from
    mortgage company Freddie Mac
    had rates on 30-year fixed-
    rate mortgages inching higher
    this week, up to an average
    5.82 percent from last week
    #39;s 5.81 percent." + ], + [ + "The classic power struggle
    between Walt Disney Co. CEO
    Michael Eisner and former
    feared talent agent Michael
    Ovitz makes for high drama in
    the courtroom - and apparently
    on cable." + ], + [ + "ATHENS France, Britain and the
    United States issued a joint
    challenge Thursday to Germany
    #39;s gold medal in equestrian
    team three-day eventing." + ], + [ + "LONDON (CBS.MW) -- Elan
    (UK:ELA) (ELN) and partner
    Biogen (BIIB) said the FDA has
    approved new drug Tysabri to
    treat relapsing forms of
    multiple sclerosis." + ], + [ + "IT services provider
    Electronic Data Systems
    yesterday reported a net loss
    of \\$153 million for the third
    quarter, with earnings hit in
    part by an asset impairment
    charge of \\$375 million
    connected with EDS's N/MCI
    project." + ], + [ + "ATLANTA - Who could have
    imagined Tommy Tuberville in
    this position? Yet there he
    was Friday, standing alongside
    the SEC championship trophy,
    posing for pictures and
    undoubtedly chuckling a bit on
    the inside." + ], + [ + "AP - Impoverished North Korea
    might resort to selling
    weapons-grade plutonium to
    terrorists for much-needed
    cash, and that would be
    \"disastrous for the world,\"
    the top U.S. military
    commander in South Korea said
    Friday." + ], + [ + "International Business
    Machines Corp. said Wednesday
    it had agreed to settle most
    of the issues in a suit over
    changes in its pension plans
    on terms that allow the
    company to continue to appeal
    a key question while capping
    its liability at \\$1.7
    billion. <BR><FONT
    face=\"verdana,MS Sans
    Serif,arial,helvetica\"
    size=\"-2\"\\ color=\"#666666\">
    <B>-The Washington
    Post</B></FONT>" + ], + [ + "Dubai: Osama bin Laden on
    Thursday called on his
    fighters to strike Gulf oil
    supplies and warned Saudi
    leaders they risked a popular
    uprising in an audio message
    said to be by the Western
    world #39;s most wanted terror
    mastermind." + ], + [ + "Bank of New Zealand has frozen
    all accounts held in the name
    of Access Brokerage, which was
    yesterday placed in
    liquidation after a client
    fund shortfall of around \\$5
    million was discovered." + ], + [ + "Edward Kozel, Cisco's former
    chief technology officer,
    joins the board of Linux
    seller Red Hat." + ], + [ + "Reuters - International
    Business Machines\\Corp. late
    on Wednesday rolled out a new
    version of its\\database
    software aimed at users of
    Linux and Unix
    operating\\systems that it
    hopes will help the company
    take away market\\share from
    market leader Oracle Corp. ." + ], + [ + "NASA #39;s Mars Odyssey
    mission, originally scheduled
    to end on Tuesday, has been
    granted a stay of execution
    until at least September 2006,
    reveal NASA scientists." + ], + [ + "ZDNet #39;s survey of IT
    professionals in August kept
    Wired amp; Wireless on top
    for the 18th month in a row.
    Telecommunications equipment
    maker Motorola said Tuesday
    that it would cut 1,000 jobs
    and take related" + ], + [ + "John Thomson threw shutout
    ball for seven innings
    Wednesday night in carrying
    Atlanta to a 2-0 blanking of
    the New York Mets. New York
    lost for the 21st time in 25
    games on the" + ], + [ + "BEIJING -- Chinese authorities
    have arrested or reprimanded
    more than 750 officials in
    recent months in connection
    with billions of dollars in
    financial irregularities
    ranging from unpaid taxes to
    embezzlement, according to a
    report made public yesterday." + ], + [ + "Canadian Press - GAUHATI,
    India (AP) - Residents of
    northeastern India were
    bracing for more violence
    Friday, a day after bombs
    ripped through two buses and a
    grenade was hurled into a
    crowded market in attacks that
    killed four people and wounded
    54." + ], + [ + "Vikram Solanki beat the rain
    clouds to register his second
    one-day international century
    as England won the third one-
    day international to wrap up a
    series victory." + ], + [ + "Some 28 people are charged
    with trying to overthrow
    Sudan's President Bashir,
    reports say." + ], + [ + "Hong Kong #39;s Beijing-backed
    chief executive yesterday
    ruled out any early moves to
    pass a controversial national
    security law which last year
    sparked a street protest by
    half a million people." + ], + [ + "BRITAIN #39;S largest
    financial institutions are
    being urged to take lead roles
    in lawsuits seeking hundreds
    of millions of dollars from
    the scandal-struck US
    insurance industry." + ], + [ + "Bankrupt UAL Corp. (UALAQ.OB:
    Quote, Profile, Research) on
    Thursday reported a narrower
    third-quarter net loss. The
    parent of United Airlines
    posted a loss of \\$274
    million, or" + ], + [ + "A three-touchdown point spread
    and a recent history of late-
    season collapses had many
    thinking the UCLA football
    team would provide little
    opposition to rival USC #39;s
    march to the BCS-championship
    game at the Orange Bowl." + ], + [ + " PHOENIX (Sports Network) -
    Free agent third baseman Troy
    Glaus is reportedly headed to
    the Arizona Diamondbacks." + ], + [ + "The European Union #39;s head
    office issued a bleak economic
    report Tuesday, warning that
    the sharp rise in oil prices
    will quot;take its toll quot;
    on economic growth next year
    while the euro #39;s renewed
    climb could threaten crucial
    exports." + ], + [ + "Third-string tailback Chris
    Markey ran for 131 yards to
    lead UCLA to a 34-26 victory
    over Oregon on Saturday.
    Markey, playing because of an
    injury to starter Maurice
    Drew, also caught five passes
    for 84 yards" + ], + [ + "Though Howard Stern's
    defection from broadcast to
    satellite radio is still 16
    months off, the industry is
    already trying to figure out
    what will fill the crater in
    ad revenue and listenership
    that he is expected to leave
    behind." + ], + [ + " WASHINGTON (Reuters) - The
    United States set final anti-
    dumping duties of up to 112.81
    percent on shrimp imported
    from China and up to 25.76
    percent on shrimp from Vietnam
    to offset unfair pricing, the
    Commerce Department said on
    Tuesday." + ], + [ + "Jay Payton #39;s three-run
    homer led the San Diego Padres
    to a 5-1 win over the San
    Francisco Giants in National
    League play Saturday, despite
    a 701st career blast from
    Barry Bonds." + ], + [ + "The optimists among Rutgers
    fans were delighted, but the
    Scarlet Knights still gave the
    pessimists something to worry
    about." + ], + [ + "Beer consumption has doubled
    over the past five years,
    prompting legislators to
    implement new rules." + ], + [ + "BusinessWeek Online - The
    jubilation that swept East
    Germany after the fall of the
    Berlin Wall in 1989 long ago
    gave way to the sober reality
    of globalization and market
    forces. Now a decade of
    resentment seems to be boiling
    over. In Eastern cities such
    as Leipzig or Chemnitz,
    thousands have taken to the
    streets since July to protest
    cuts in unemployment benefits,
    the main source of livelihood
    for 1.6 million East Germans.
    Discontent among
    reunification's losers fueled
    big gains by the far left and
    far right in Brandenburg and
    Saxony state elections Sept.
    19. ..." + ], + [ + "NEW YORK -- When Office Depot
    Inc. stores ran an electronics
    recycling drive last summer
    that accepted everything from
    cellphones to televisions,
    some stores were overwhelmed
    by the amount of e-trash they
    received." + ], + [ + "In a discovery sure to set off
    a firestorm of debate over
    human migration to the western
    hemisphere, archaeologists in
    South Carolina say they have
    uncovered evidence that people
    lived in eastern North America
    at least 50,000 years ago -
    far earlier than" + ], + [ + "AP - Former president Bill
    Clinton on Monday helped
    launch a new Internet search
    company backed by the Chinese
    government which says its
    technology uses artificial
    intelligence to produce better
    results than Google Inc." + ], + [ + "The International Monetary
    Fund expressed concern Tuesday
    about the impact of the
    troubles besetting oil major
    Yukos on Russia #39;s standing
    as a place to invest." + ], + [ + "Perhaps the sight of Maria
    Sharapova opposite her tonight
    will jog Serena Williams #39;
    memory. Wimbledon. The final.
    You and Maria." + ], + [ + "At least 79 people were killed
    and 74 were missing after some
    of the worst rainstorms in
    recent years triggered
    landslides and flash floods in
    southwest China, disaster
    relief officials said
    yesterday." + ], + [ + "LinuxWorld Conference amp;
    Expo will come to Boston for
    the first time in February,
    underscoring the area's
    standing as a hub for the
    open-source software being
    adopted by thousands of
    businesses." + ], + [ + "BANGKOK Thai military aircraft
    dropped 100 million paper
    birds over southern Thailand
    on Sunday in a gesture
    intended to promote peace in
    mainly Muslim provinces, where
    more than 500 people have died
    this year in attacks by
    separatist militants and" + ], + [ + "Insurgents threatened on
    Saturday to cut the throats of
    two Americans and a Briton
    seized in Baghdad, and
    launched a suicide car bomb
    attack on Iraqi security
    forces in Kirkuk that killed
    at least 23 people." + ], + [ + "Five days after making the
    putt that won the Ryder Cup,
    Colin Montgomerie looked set
    to miss the cut at a European
    PGA tour event." + ], + [ + "Sun Microsystems plans to
    release its Sun Studio 10
    development tool in the fourth
    quarter of this year,
    featuring support for 64-bit
    applications running on AMD
    Opteron and Nocona processors,
    Sun officials said on Tuesday." + ], + [ + "Karachi - Captain Inzamam ul-
    Haq and coach Bob Woolmer came
    under fire on Thursday for
    choosing to bat first on a
    tricky pitch after Pakistan
    #39;s humiliating defeat in
    the ICC Champions Trophy semi-
    final." + ], + [ + "Scottish champions Celtic saw
    their three-year unbeaten home
    record in Europe broken
    Tuesday as they lost 3-1 to
    Barcelona in the Champions
    League Group F opener." + ], + [ + "Interest rates on short-term
    Treasury securities rose in
    yesterday's auction. The
    Treasury Department sold \\$19
    billion in three-month bills
    at a discount rate of 1.710
    percent, up from 1.685 percent
    last week. An additional \\$17
    billion was sold in six-month
    bills at a rate of 1.950
    percent, up from 1.870
    percent." + ], + [ + "Most IT Managers won #39;t
    question the importance of
    security, but this priority
    has been sliding between the
    third and fourth most
    important focus for companies." + ], + [ + "AP - Andy Roddick searched out
    Carlos Moya in the throng of
    jumping, screaming Spanish
    tennis players, hoping to
    shake hands." + ], + [ + "AP - The Federal Trade
    Commission on Thursday filed
    the first case in the country
    against software companies
    accused of infecting computers
    with intrusive \"spyware\" and
    then trying to sell people the
    solution." + ], + [ + "While shares of Apple have
    climbed more than 10 percent
    this week, reaching 52-week
    highs, two research firms told
    investors Friday they continue
    to remain highly bullish about
    the stock." + ], + [ + "Moody #39;s Investors Service
    on Wednesday said it may cut
    its bond ratings on HCA Inc.
    (HCA.N: Quote, Profile,
    Research) deeper into junk,
    citing the hospital operator
    #39;s plan to buy back about
    \\$2." + ], + [ + "States are now barred from
    imposing telecommunications
    regulations on Net phone
    providers." + ], + [ + "Telekom Austria AG, the
    country #39;s biggest phone
    operator, won the right to buy
    Bulgaria #39;s largest mobile
    phone company, MobilTel EAD,
    for 1.6 billion euros (\\$2.1
    billion), an acquisition that
    would add 3 million
    subscribers." + ], + [ + "Shares of ID Biomedical jumped
    after the company reported
    Monday that it signed long-
    term agreements with the three
    largest flu vaccine
    wholesalers in the United
    States in light of the
    shortage of vaccine for the
    current flu season." + ], + [ + "ENGLAND captain and Real
    Madrid midfielder David
    Beckham has played down
    speculation that his club are
    moving for England manager
    Sven-Goran Eriksson." + ], + [ + "The federal government closed
    its window on the oil industry
    Thursday, saying that it is
    selling its last 19 per cent
    stake in Calgary-based Petro-
    Canada." + ], + [ + "Strong sales of new mobile
    phone models boosts profits at
    Carphone Warehouse but the
    retailer's shares fall on
    concerns at a decline in
    profits from pre-paid phones." + ], + [ + " NEW YORK (Reuters) - IBM and
    top scientific research
    organizations are joining
    forces in a humanitarian
    effort to tap the unused
    power of millions of computers
    and help solve complex social
    problems." + ], + [ + " NEW YORK, Nov. 11 -- The 40
    percent share price slide in
    Merck #38; Co. in the five
    weeks after it pulled the
    painkiller Vioxx off the
    market highlighted larger
    problems in the pharmaceutical
    industry that may depress
    performance for years,
    according to academics and
    stock analysts who follow the
    sector." + ], + [ + "Low-fare carrier Southwest
    Airlines Co. said Thursday
    that its third-quarter profit
    rose 12.3 percent to beat Wall
    Street expectations despite
    higher fuel costs." + ], + [ + "Another shock hit the drug
    sector Friday when
    pharmaceutical giant Pfizer
    Inc. announced that it found
    an increased heart risk to
    patients for its blockbuster
    arthritis drug Celebrex." + ], + [ + "AFP - National Basketball
    Association players trying to
    win a fourth consecutive
    Olympic gold medal for the
    United States have gotten the
    wake-up call that the \"Dream
    Team\" days are done even if
    supporters have not." + ], + [ + " BEIJING (Reuters) - Secretary
    of State Colin Powell will
    urge China Monday to exert its
    influence over North Korea to
    resume six-party negotiations
    on scrapping its suspected
    nuclear weapons program." + ], + [ + "Reuters - An Israeli missile
    strike killed at least\\two
    Hamas gunmen in Gaza City
    Monday, a day after a
    top\\commander of the Islamic
    militant group was killed in a
    similar\\attack, Palestinian
    witnesses and security sources
    said." + ], + [ + "England will be seeking their
    third clean sweep of the
    summer when the NatWest
    Challenge against India
    concludes at Lord #39;s. After
    beating New Zealand and West
    Indies 3-0 and 4-0 in Tests,
    they have come alive" + ], + [ + "AP - The Pentagon has restored
    access to a Web site that
    assists soldiers and other
    Americans living overseas in
    voting, after receiving
    complaints that its security
    measures were preventing
    legitimate voters from using
    it." + ], + [ + "The number of US information
    technology workers rose 2
    percent to 10.5 million in the
    first quarter of this year,
    but demand for them is
    dropping, according to a new
    report." + ], + [ + "Montreal - The British-built
    Canadian submarine HMCS
    Chicoutimi, crippled since a
    fire at sea that killed one of
    its crew, is under tow and is
    expected in Faslane, Scotland,
    early next week, Canadian
    naval commander Tyrone Pile
    said on Friday." + ], + [ + "AP - Grizzly and black bears
    killed a majority of elk
    calves in northern Yellowstone
    National Park for the second
    year in a row, preliminary
    study results show." + ], + [ + "Thirty-five Pakistanis freed
    from the US Guantanamo Bay
    prison camp arrived home on
    Saturday and were taken
    straight to prison for further
    interrogation, the interior
    minister said." + ], + [ + "AP - Mark Richt knows he'll
    have to get a little creative
    when he divvies up playing
    time for Georgia's running
    backs next season. Not so on
    Saturday. Thomas Brown is the
    undisputed starter for the
    biggest game of the season." + ], + [ + "Witnesses in the trial of a US
    soldier charged with abusing
    prisoners at Abu Ghraib have
    told the court that the CIA
    sometimes directed abuse and
    orders were received from
    military command to toughen
    interrogations." + ], + [ + "Insurance firm says its board
    now consists of its new CEO
    Michael Cherkasky and 10
    outside members. NEW YORK
    (Reuters) - Marsh amp;
    McLennan Cos." + ], + [ + "Michael Phelps, the six-time
    Olympic champion, issued an
    apology yesterday after being
    arrested and charged with
    drunken driving in the United
    States." + ], + [ + "AP - Italian and Lebanese
    authorities have arrested 10
    suspected terrorists who
    planned to blow up the Italian
    Embassy in Beirut, an Italian
    news agency and the Defense
    Ministry said Tuesday." + ], + [ + "CORAL GABLES, Fla. -- John F.
    Kerry and President Bush
    clashed sharply over the war
    in Iraq last night during the
    first debate of the
    presidential campaign season,
    with the senator from
    Massachusetts accusing" + ], + [ + "GONAIVES, Haiti -- While
    desperately hungry flood
    victims wander the streets of
    Gonaives searching for help,
    tons of food aid is stacking
    up in a warehouse guarded by
    United Nations peacekeepers." + ], + [ + "As part of its much-touted new
    MSN Music offering, Microsoft
    Corp. (MSFT) is testing a Web-
    based radio service that
    mimics nearly 1,000 local
    radio stations, allowing users
    to hear a version of their
    favorite radio station with
    far fewer interruptions." + ], + [ + "AFP - The resignation of
    disgraced Fiji Vice-President
    Ratu Jope Seniloli failed to
    quell anger among opposition
    leaders and the military over
    his surprise release from
    prison after serving just
    three months of a four-year
    jail term for his role in a
    failed 2000 coup." + ], + [ + "ATHENS Larry Brown, the US
    coach, leaned back against the
    scorer #39;s table, searching
    for support on a sinking ship.
    His best player, Tim Duncan,
    had just fouled out, and the
    options for an American team
    that" + ], + [ + "Sourav Ganguly files an appeal
    against a two-match ban
    imposed for time-wasting." + ], + [ + "Ziff Davis - A quick
    resolution to the Mambo open-
    source copyright dispute seems
    unlikely now that one of the
    parties has rejected an offer
    for mediation." + ], + [ + "Greek police surrounded a bus
    full of passengers seized by
    armed hijackers along a
    highway from an Athens suburb
    Wednesday, police said." + ], + [ + "Spain have named an unchanged
    team for the Davis Cup final
    against the United States in
    Seville on 3-5 December.
    Carlos Moya, Juan Carlos
    Ferrero, Rafael Nadal and
    Tommy Robredo will take on the
    US in front of 22,000 fans at
    the converted Olympic stadium." + ], + [ + "BAGHDAD: A suicide car bomber
    attacked a police convoy in
    Baghdad yesterday as guerillas
    kept pressure on Iraq #39;s
    security forces despite a
    bloody rout of insurgents in
    Fallujah." + ], + [ + "Slot machine maker
    International Game Technology
    (IGT.N: Quote, Profile,
    Research) on Tuesday posted
    better-than-expected quarterly
    earnings, as casinos bought" + ], + [ + "Fixtures and fittings from
    Damien Hirst's restaurant
    Pharmacy sell for 11.1, 8m
    more than expected." + ], + [ + "Last Tuesday night, Harvard
    knocked off rival Boston
    College, which was ranked No.
    1 in the country. Last night,
    the Crimson knocked off
    another local foe, Boston
    University, 2-1, at Walter
    Brown Arena, which marked the
    first time since 1999 that
    Harvard had beaten them both
    in the same season." + ], + [ + "Venezuelan election officials
    say they expect to announce
    Saturday, results of a partial
    audit of last Sunday #39;s
    presidential recall
    referendum." + ], + [ + "About 500 prospective jurors
    will be in an Eagle, Colorado,
    courtroom Friday, answering an
    82-item questionnaire in
    preparation for the Kobe
    Bryant sexual assault trial." + ], + [ + "Students take note - endless
    journeys to the library could
    become a thing of the past
    thanks to a new multimillion-
    pound scheme to make classic
    texts available at the click
    of a mouse." + ], + [ + "Moscow - The next crew of the
    International Space Station
    (ISS) is to contribute to the
    Russian search for a vaccine
    against Aids, Russian
    cosmonaut Salijan Sharipovthe
    said on Thursday." + ], + [ + "Locked away in fossils is
    evidence of a sudden solar
    cooling. Kate Ravilious meets
    the experts who say it could
    explain a 3,000-year-old mass
    migration - and today #39;s
    global warming." + ], + [ + "By ANICK JESDANUN NEW YORK
    (AP) -- Taran Rampersad didn't
    complain when he failed to
    find anything on his hometown
    in the online encyclopedia
    Wikipedia. Instead, he simply
    wrote his own entry for San
    Fernando, Trinidad and
    Tobago..." + ], + [ + "Five years after catastrophic
    floods and mudslides killed
    thousands along Venezuela's
    Caribbean coast, survivors in
    this town still see the signs
    of destruction - shattered
    concrete walls and tall weeds
    growing atop streets covered
    in dried mud." + ], + [ + "How do you top a battle
    between Marines and an alien
    religious cult fighting to the
    death on a giant corona in
    outer space? The next logical
    step is to take that battle to
    Earth, which is exactly what
    Microsoft" + ], + [ + "Sony Ericsson Mobile
    Communications Ltd., the
    mobile-phone venture owned by
    Sony Corp. and Ericsson AB,
    said third-quarter profit rose
    45 percent on camera phone
    demand and forecast this
    quarter will be its strongest." + ], + [ + "NEW YORK, September 17
    (newratings.com) - Alcatel
    (ALA.NYS) has expanded its
    operations and presence in the
    core North American
    telecommunication market with
    two separate acquisitions for
    about \\$277 million." + ], + [ + "Four U.S. agencies yesterday
    announced a coordinated attack
    to stem the global trade in
    counterfeit merchandise and
    pirated music and movies, an
    underground industry that law-
    enforcement officials estimate
    to be worth \\$500 billion each
    year." + ], + [ + "BEIJING The NBA has reached
    booming, basketball-crazy
    China _ but the league doesn
    #39;t expect to make any money
    soon. The NBA flew more than
    100 people halfway around the
    world for its first games in
    China, featuring" + ], + [ + "The UN nuclear watchdog
    confirmed Monday that nearly
    400 tons of powerful
    explosives that could be used
    in conventional or nuclear
    missiles disappeared from an
    unguarded military
    installation in Iraq." + ], + [ + " NEW YORK (Reuters) - Curt
    Schilling pitched 6 2/3
    innings and Manny Ramirez hit
    a three-run homer in a seven-
    run fourth frame to lead the
    Boston Red Sox to a 9-3 win
    over the host Anaheim Angels
    in their American League
    Divisional Series opener
    Tuesday." + ], + [ + "AP - The game between the
    Minnesota Twins and the New
    York Yankees on Tuesday night
    was postponed by rain." + ], + [ + "Oil giant Shell swept aside
    nearly 100 years of history
    today when it unveiled plans
    to merge its UK and Dutch
    parent companies. Shell said
    scrapping its twin-board
    structure" + ], + [ + "Caesars Entertainment Inc. on
    Thursday posted a rise in
    third-quarter profit as Las
    Vegas hotels filled up and
    Atlantic City properties
    squeaked out a profit that was
    unexpected by the company." + ], + [ + "Half of all U.S. Secret
    Service agents are dedicated
    to protecting President
    Washington #151;and all the
    other Presidents on U.S.
    currency #151;from
    counterfeiters." + ], + [ + "One of India #39;s leading
    telecommunications providers
    will use Cisco Systems #39;
    gear to build its new
    Ethernet-based broadband
    network." + ], + [ + "Militants in Iraq have killed
    the second of two US civilians
    they were holding hostage,
    according to a statement on an
    Islamist website." + ], + [ + "PARIS The verdict is in: The
    world #39;s greatest race car
    driver, the champion of
    champions - all disciplines
    combined - is Heikki
    Kovalainen." + ], + [ + "Pakistan won the toss and
    unsurprisingly chose to bowl
    first as they and West Indies
    did battle at the Rose Bowl
    today for a place in the ICC
    Champions Trophy final against
    hosts England." + ], + [ + "Shares of Beacon Roofing
    Suppler Inc. shot up as much
    as 26 percent in its trading
    debut Thursday, edging out
    bank holding company Valley
    Bancorp as the biggest gainer
    among a handful of new stocks
    that went public this week." + ], + [ + "More than 4,000 American and
    Iraqi soldiers mounted a
    military assault on the
    insurgent-held city of
    Samarra." + ], + [ + "Maxime Faget conceived and
    proposed the development of
    the one-man spacecraft used in
    Project Mercury, which put the
    first American astronauts into
    suborbital flight, then
    orbital flight" + ], + [ + "The first weekend of holiday
    shopping went from red-hot to
    dead white, as a storm that
    delivered freezing, snowy
    weather across Colorado kept
    consumers at home." + ], + [ + " MEXICO CITY (Reuters) -
    Mexican President Vicente Fox
    said on Monday he hoped
    President Bush's re-election
    meant a bilateral accord on
    migration would be reached
    before his own term runs out
    at the end of 2006." + ], + [ + " LONDON (Reuters) - The dollar
    fell within half a cent of
    last week's record low against
    the euro on Thursday after
    capital inflows data added to
    worries the United States may
    struggle to fund its current
    account deficit." + ], + [ + " BRUSSELS (Reuters) - French
    President Jacques Chirac said
    on Friday he had not refused
    to meet Iraqi interim Prime
    Minister Iyad Allawi after
    reports he was snubbing the
    Iraqi leader by leaving an EU
    meeting in Brussels early." + ], + [ + "China has pledged to invest
    \\$20 billion in Argentina in
    the next 10 years, La Nacion
    reported Wednesday. The
    announcement came during the
    first day of a two-day visit" + ], + [ + "NEW YORK - Former President
    Bill Clinton was in good
    spirits Saturday, walking
    around his hospital room in
    street clothes and buoyed by
    thousands of get-well messages
    as he awaited heart bypass
    surgery early this coming
    week, people close to the
    family said. Clinton was
    expected to undergo surgery as
    early as Monday but probably
    Tuesday, said Democratic Party
    Chairman Terry McAuliffe, who
    said the former president was
    \"upbeat\" when he spoke to him
    by phone Friday..." + ], + [ + "Toyota Motor Corp., the world
    #39;s biggest carmaker by
    value, will invest 3.8 billion
    yuan (\\$461 million) with its
    partner Guangzhou Automobile
    Group to boost manufacturing
    capacity in" + ], + [ + "Poland will cut its troops in
    Iraq early next year and won
    #39;t stay in the country
    quot;an hour longer quot; than
    needed, the country #39;s
    prime minister said Friday." + ], + [ + "AP - Boston Red Sox center
    fielder Johnny Damon is having
    a recurrence of migraine
    headaches that first bothered
    him after a collision in last
    year's playoffs." + ], + [ + "The patch fixes a flaw in the
    e-mail server software that
    could be used to get access to
    in-boxes and information." + ], + [ + "Felix Cardenas of Colombia won
    the 17th stage of the Spanish
    Vuelta cycling race Wednesday,
    while defending champion
    Roberto Heras held onto the
    overall leader #39;s jersey
    for the sixth day in a row." + ], + [ + "AP - Being the biggest dog may
    pay off at feeding time, but
    species that grow too large
    may be more vulnerable to
    extinction, new research
    suggests." + ], + [ + "A car bomb exploded at a US
    military convoy in the
    northern Iraqi city of Mosul,
    causing several casualties,
    the army and Iraqi officials
    said." + ], + [ + "Newest Efficeon processor also
    offers higher frequency using
    less power." + ], + [ + "BAE Systems has launched a
    search for a senior American
    businessman to become a non-
    executive director. The high-
    profile appointment is
    designed to strengthen the
    board at a time when the
    defence giant is" + ], + [ + "AP - In what it calls a first
    in international air travel,
    Finnair says it will let its
    frequent fliers check in using
    text messages on mobile
    phones." + ], + [ + "Computer Associates
    International is expected to
    announce that its new chief
    executive will be John
    Swainson, an I.B.M. executive
    with strong technical and
    sales credentials." + ], + [ + "Established star Landon
    Donovan and rising sensation
    Eddie Johnson carried the
    United States into the
    regional qualifying finals for
    the 2006 World Cup in emphatic
    fashion Wednesday night." + ], + [ + "STOCKHOLM (Dow
    Jones)--Expectations for
    Telefon AB LM Ericsson #39;s
    (ERICY) third-quarter
    performance imply that while
    sales of mobile telephony
    equipment are expected to have
    dipped, the company" + ], + [ + "ATHENS, Greece - Michael
    Phelps took care of qualifying
    for the Olympic 200-meter
    freestyle semifinals Sunday,
    and then found out he had been
    added to the American team for
    the evening's 400 freestyle
    relay final. Phelps' rivals
    Ian Thorpe and Pieter van den
    Hoogenband and teammate Klete
    Keller were faster than the
    teenager in the 200 free
    preliminaries..." + ], + [ + " JERUSALEM (Reuters) - Israeli
    Prime Minister Ariel Sharon
    intends to present a timetable
    for withdrawal from Gaza to
    lawmakers from his Likud Party
    Tuesday despite a mutiny in
    the right-wing bloc over the
    plan." + ], + [ + "Search Engine for Programming
    Code\\\\An article at Newsforge
    pointed me to Koders (
    http://www.koders.com ) a
    search engine for finding
    programming code. Nifty.\\\\The
    front page allows you to
    specify keywords, sixteen
    languages (from ASP to VB.NET)
    and sixteen licenses (from AFL
    to ZPL -- fortunately there's
    an information page to ..." + ], + [ + " #147;Apple once again was the
    star of the show at the annual
    MacUser awards, #148; reports
    MacUser, #147;taking away six
    Maxine statues including
    Product of the Year for the
    iTunes Music Store. #148;
    Other Apple award winners:
    Power Mac G5, AirPort Express,
    20-inch Apple Cinema Display,
    GarageBand and DVD Studio Pro
    3. Nov 22" + ], + [ + " KABUL (Reuters) - Three U.N.
    workers held by militants in
    Afghanistan were in their
    third week of captivity on
    Friday after calls from both
    sides for the crisis to be
    resolved ahead of this
    weekend's Muslim festival of
    Eid al-Fitr." + ], + [ + "AP - A sweeping wildlife
    preserve in southwestern
    Arizona is among the nation's
    10 most endangered refuges,
    due in large part to illegal
    drug and immigrant traffic and
    Border Patrol operations, a
    conservation group said
    Friday." + ], + [ + "ROME, Oct 29 (AFP) - French
    President Jacques Chirac urged
    the head of the incoming
    European Commission Friday to
    take quot;the appropriate
    decisions quot; to resolve a
    row over his EU executive team
    which has left the EU in
    limbo." + ], + [ + "The Anglo-Dutch oil giant
    Shell today sought to draw a
    line under its reserves
    scandal by announcing plans to
    spend \\$15bn (8.4bn) a year to
    replenish reserves and develop
    production in its oil and gas
    business." + ], + [ + "SEPTEMBER 20, 2004
    (COMPUTERWORLD) - At
    PeopleSoft Inc. #39;s Connect
    2004 conference in San
    Francisco this week, the
    software vendor is expected to
    face questions from users
    about its ability to fend off
    Oracle Corp." + ], + [ + "Russia's parliament will
    launch an inquiry into a
    school siege that killed over
    300 hostages, President
    Vladimir Putin said on Friday,
    but analysts doubt it will
    satisfy those who blame the
    carnage on security services." + ], + [ + "Tony Blair talks to business
    leaders about new proposals
    for a major shake-up of the
    English exam system." + ], + [ + "KUALA LUMPUR, Malaysia A
    Malaysian woman has claimed a
    new world record after living
    with over six-thousand
    scorpions for 36 days
    straight." + ], + [ + "PBS's Charlie Rose quizzes Sun
    co-founder Bill Joy,
    Monster.com chief Jeff Taylor,
    and venture capitalist John
    Doerr." + ], + [ + "Circulation declined at most
    major US newspapers in the
    last half year, the latest
    blow for an industry already
    rocked by a scandal involving
    circulation misstatements that
    has undermined the confidence
    of investors and advertisers." + ], + [ + "Skype Technologies is teaming
    with Siemens to offer cordless
    phone users the ability to
    make Internet telephony calls,
    in addition to traditional
    calls, from their handsets." + ], + [ + "AP - After years of
    resistance, the U.S. trucking
    industry says it will not try
    to impede or delay a new
    federal rule aimed at cutting
    diesel pollution." + ], + [ + "INDIANAPOLIS - ATA Airlines
    has accepted a \\$117 million
    offer from Southwest Airlines
    that would forge close ties
    between two of the largest US
    discount carriers." + ], + [ + "could take drastic steps if
    the talks did not proceed as
    Tehran wants. Mehr news agency
    quoted him as saying on
    Wednesday. that could be used" + ], + [ + "Wizards coach Eddie Jordan
    says the team is making a
    statement that immaturity will
    not be tolerated by suspending
    Kwame Brown one game for not
    taking part in a team huddle
    during a loss to Denver." + ], + [ + "EVERETT Fire investigators
    are still trying to determine
    what caused a two-alarm that
    destroyed a portion of a South
    Everett shopping center this
    morning." + ], + [ + "Umesh Patel, a 36-year old
    software engineer from
    California, debated until the
    last minute." + ], + [ + "Sure, the PeopleSoft board
    told shareholders to just say
    no. This battle will go down
    to the wire, and even
    afterward Ellison could
    prevail." + ], + [ + "Investors cheered by falling
    oil prices and an improving
    job picture sent stocks higher
    Tuesday, hoping that the news
    signals a renewal of economic
    strength and a fall rally in
    stocks." + ], + [ + "AP - Andy Roddick has yet to
    face a challenge in his U.S.
    Open title defense. He beat
    No. 18 Tommy Robredo of Spain
    6-3, 6-2, 6-4 Tuesday night to
    move into the quarterfinals
    without having lost a set." + ], + [ + "Now that hell froze over in
    Boston, New England braces for
    its rarest season -- a winter
    of content. Red Sox fans are
    adrift from the familiar
    torture of the past." + ], + [ + " CHICAGO (Reuters) - Wal-Mart
    Stores Inc. <A HREF=\"http:/
    /www.investor.reuters.com/Full
    Quote.aspx?ticker=WMT.N target
    =/stocks/quickinfo/fullquote\"&
    gt;WMT.N</A>, the
    world's largest retailer, said
    on Saturday it still
    anticipates September U.S.
    sales to be up 2 percent to 4
    percent at stores open at
    least a year." + ], + [ + " TOKYO (Reuters) - Tokyo's
    Nikkei stock average opened
    down 0.15 percent on
    Wednesday as investors took a
    breather from the market's
    recent rises and sold shares
    of gainers such as Sharp
    Corp." + ], + [ + "AP - From now until the start
    of winter, male tarantulas are
    roaming around, searching for
    female mates, an ideal time to
    find out where the spiders
    flourish in Arkansas." + ], + [ + "Israeli authorities have
    launched an investigation into
    death threats against Israeli
    Prime Minister Ariel Sharon
    and other officials supporting
    his disengagement plan from
    Gaza and parts of the West
    Bank, Jerusalem police said
    Tuesday." + ], + [ + "NewsFactor - While a U.S.
    District Court continues to
    weigh the legality of
    \\Oracle's (Nasdaq: ORCL)
    attempted takeover of
    \\PeopleSoft (Nasdaq: PSFT),
    Oracle has taken the necessary
    steps to ensure the offer does
    not die on the vine with
    PeopleSoft's shareholders." + ], + [ + "NEW YORK : World oil prices
    fell, capping a drop of more
    than 14 percent in a two-and-
    a-half-week slide triggered by
    a perception of growing US
    crude oil inventories." + ], + [ + "SUFFOLK -- Virginia Tech
    scientists are preparing to
    protect the state #39;s
    largest crop from a disease
    with strong potential to do
    damage." + ], + [ + "It was a carefully scripted
    moment when Russian President
    Vladimir Putin began quoting
    Taras Shevchenko, this country
    #39;s 19th-century bard,
    during a live television" + ], + [ + "China says it welcomes Russia
    #39;s ratification of the
    Kyoto Protocol that aims to
    stem global warming by
    reducing greenhouse-gas
    emissions." + ], + [ + "Analysts said the smartphone
    enhancements hold the
    potential to bring PalmSource
    into an expanding market that
    still has room despite early
    inroads by Symbian and
    Microsoft." + ], + [ + "The Metropolitan
    Transportation Authority is
    proposing a tax increase to
    raise \\$900 million a year to
    help pay for a five-year
    rebuilding program." + ], + [ + "AFP - The Canadian armed
    forces chief of staff on was
    elected to take over as head
    of NATO's Military Committee,
    the alliance's highest
    military authority, military
    and diplomatic sources said." + ], + [ + "AFP - Two proposed resolutions
    condemning widespread rights
    abuses in Sudan and Zimbabwe
    failed to pass a UN committee,
    mired in debate between
    African and western nations." + ], + [ + " NEW YORK (Reuters) -
    Citigroup Inc. <A HREF=\"htt
    p://www.investor.reuters.com/F
    ullQuote.aspx?ticker=C.N targe
    t=/stocks/quickinfo/fullquote\"
    >C.N</A> the world's
    largest financial services
    company, said on Tuesday it
    will acquire First American
    Bank in the quickly-growing
    Texas market." + ], + [ + " SINGAPORE (Reuters) - U.S.
    oil prices hovered just below
    \\$50 a barrel on Tuesday,
    holding recent gains on a rash
    of crude supply outages and
    fears over thin heating oil
    tanks." + ], + [ + "THE South Africans have called
    the Wallabies scrum cheats as
    a fresh round of verbal
    warfare opened in the Republic
    last night." + ], + [ + "The return of noted reformer
    Nabil Amr to Palestinian
    politics comes at a crucial
    juncture." + ], + [ + "An overwhelming majority of
    NHL players who expressed
    their opinion in a poll said
    they would not support a
    salary cap even if it meant
    saving a season that was
    supposed to have started Oct.
    13." + ], + [ + "Tony Eury Sr. oversees Dale
    Earnhardt Jr. on the
    racetrack, but Sunday he
    extended his domain to Victory
    Lane. Earnhardt was unbuckling
    to crawl out of the No." + ], + [ + "AP - After two debates, voters
    have seen President Bush look
    peevish and heard him pass the
    buck. They've watched Sen.
    John Kerry deny he's a flip-
    flopper and then argue that
    Saddam Hussein was a threat
    #151; and wasn't. It's no
    wonder so few minds have
    changed." + ], + [ + "(Sports Network) - The New
    York Yankees try to move one
    step closer to a division
    title when they conclude their
    critical series with the
    Boston Red Sox at Fenway Park." + ], + [ + "US retail sales fell 0.3 in
    August as rising energy costs
    and bad weather persuaded
    shoppers to reduce their
    spending." + ], + [ + "The United States says the
    Lebanese parliament #39;s
    decision Friday to extend the
    term of pro-Syrian President
    Emile Lahoud made a
    quot;crude mockery of
    democratic principles." + ], + [ + "Kurt Busch claimed a stake in
    the points lead in the NASCAR
    Chase for the Nextel Cup
    yesterday, winning the
    Sylvania 300 at New Hampshire
    International Speedway." + ], + [ + "TOKYO (AFP) - Japan #39;s
    Fujitsu and Cisco Systems of
    the US said they have agreed
    to form a strategic alliance
    focusing on routers and
    switches that will enable
    businesses to build advanced
    Internet Protocol (IP)
    networks." + ], + [ + "GEORGETOWN, Del., Oct. 28 --
    Plaintiffs in a shareholder
    lawsuit over former Walt
    Disney Co. president Michael
    Ovitz's \\$140 million
    severance package attempted
    Thursday to portray Ovitz as a
    dishonest bumbler who botched
    the hiring of a major
    television executive and
    pushed the release of a movie
    that angered the Chinese
    government, damaging Disney's
    business prospects in the
    country." + ], + [ + "US stocks gained ground in
    early trading Thursday after
    tame inflation reports and
    better than expected jobless
    news. Oil prices held steady
    as Hurricane Ivan battered the
    Gulf coast, where oil
    operations have halted." + ], + [ + "It #39;s a new internet
    browser. The first full
    version, Firefox 1.0, was
    launched earlier this month.
    In the sense it #39;s a
    browser, yes, but the
    differences are larger than
    the similarities." + ], + [ + "LOUDON, NH - As this
    newfangled stretch drive for
    the Nextel Cup championship
    ensues, Jeff Gordon has to be
    considered the favorite for a
    fifth title." + ], + [ + "PepsiCo. Inc., the world #39;s
    No. 2 soft- drink maker, plans
    to buy General Mills Inc.
    #39;s stake in their European
    joint venture for \\$750
    million in cash, giving it
    complete ownership of Europe
    #39;s largest snack-food
    company." + ], + [ + "Microsoft will accelerate SP2
    distribution to meet goal of
    100 million downloads in two
    months." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks opened little changed
    on Friday, after third-
    quarter gross domestic product
    data showed the U.S. economy
    grew at a slower-than-expected
    pace." + ], + [ + "International Business
    Machines Corp., taken to court
    by workers over changes it
    made to its traditional
    pension plan, has decided to
    stop offering any such plan to
    new employees." + ], + [ + "NEW YORK - With four of its
    executives pleading guilty to
    price-fixing charges today,
    Infineon will have a hard time
    arguing that it didn #39;t fix
    prices in its ongoing
    litigation with Rambus." + ], + [ + "By nick giongco. YOU KNOW you
    have reached the status of a
    boxing star when ring
    announcer extraordinaire
    Michael Buffer calls out your
    name in his trademark booming
    voice during a high-profile
    event like yesterday" + ], + [ + "Canadian Press - OTTAWA (CP) -
    The prime minister says he's
    been assured by President
    George W. Bush that the U.S.
    missile defence plan does not
    necessarily involve the
    weaponization of space." + ], + [ + "AP - Even with a big lead,
    Eric Gagne wanted to pitch in
    front of his hometown fans one
    last time." + ], + [ + " NEW YORK (Reuters) - Limited
    Brands Inc. <A HREF=\"http:/
    /www.investor.reuters.com/Full
    Quote.aspx?ticker=LTD.N target
    =/stocks/quickinfo/fullquote\"&
    gt;LTD.N</A> on
    Thursday reported higher
    quarterly operating profit as
    cost controls and strong
    lingerie sales offset poor
    results at the retailer's
    Express apparel stores." + ], + [ + "General Motors and
    DaimlerChrysler are
    collaborating on development
    of fuel- saving hybrid engines
    in hopes of cashing in on an
    expanding market dominated by
    hybrid leaders Toyota and
    Honda." + ], + [ + "ATHENS, Greece -- Larry Brown
    was despondent, the head of
    the US selection committee was
    defensive and an irritated
    Allen Iverson was hanging up
    on callers who asked what went
    wrong." + ], + [ + "Fifty-six miners are dead and
    another 92 are still stranded
    underground after a gas blast
    at the Daping coal mine in
    Central China #39;s Henan
    Province, safety officials
    said Thursday." + ], + [ + "BEIRUT, Lebanon - Three
    Lebanese men and their Iraqi
    driver have been kidnapped in
    Iraq, the Lebanese Foreign
    Ministry said Sunday, as
    Iraq's prime minister said his
    government was working for the
    release of two Americans and a
    Briton also being held
    hostage. Gunmen snatched
    the three Lebanese, who worked
    for a travel agency with a
    branch in Baghdad, as they
    drove on the highway between
    the capital and Fallujah on
    Friday night, a ministry
    official said..." + ], + [ + "PORTLAND, Ore. - It's been
    almost a year since singer-
    songwriter Elliott Smith
    committed suicide, and fans
    and friends will be looking
    for answers as the posthumous
    \"From a Basement on the Hill\"
    is released..." + ], + [ + "AP - Courtney Brown refuses to
    surrender to injuries. He's
    already planning another
    comeback." + ], + [ + " GAZA (Reuters) - Israel
    expanded its military
    offensive in northern Gaza,
    launching two air strikes
    early on Monday that killed
    at least three Palestinians
    and wounded two, including a
    senior Hamas leader, witnesses
    and medics said." + ], + [ + " TOKYO (Reuters) - Nintendo
    Co. Ltd. raised its 2004
    shipment target for its DS
    handheld video game device by
    40 percent to 2.8 million
    units on Thursday after many
    stores in Japan and the
    United States sold out in the
    first week of sales." + ], + [ + "It took an off-the-cuff
    reference to a serial
    murderer/cannibal to punctuate
    the Robby Gordon storyline.
    Gordon has been vilified by
    his peers and put on probation" + ], + [ + "By BOBBY ROSS JR. Associated
    Press Writer. play the Atlanta
    Hawks. They will be treated to
    free food and drink and have.
    their pictures taken with
    Mavericks players, dancers and
    team officials." + ], + [ + "ARSENAL #39;S Brazilian World
    Cup winning midfielder
    Gilberto Silva is set to be
    out for at least a month with
    a back injury, the Premiership
    leaders said." + ], + [ + "INDIANAPOLIS -- With a package
    of academic reforms in place,
    the NCAA #39;s next crusade
    will address what its
    president calls a dangerous
    drift toward professionalism
    and sports entertainment." + ], + [ + "Autodesk this week unwrapped
    an updated version of its
    hosted project collaboration
    service targeted at the
    construction and manufacturing
    industries. Autodesk Buzzsaw
    lets multiple, dispersed
    project participants --
    including building owners,
    developers, architects,
    construction teams, and
    facility managers -- share and
    manage data throughout the
    life of a project, according
    to Autodesk officials." + ], + [ + "AP - It was difficult for
    Southern California's Pete
    Carroll and Oklahoma's Bob
    Stoops to keep from repeating
    each other when the two
    coaches met Thursday." + ], + [ + "Reuters - Sudan's government
    resumed\\talks with rebels in
    the oil-producing south on
    Thursday while\\the United
    Nations set up a panel to
    investigate charges
    of\\genocide in the west of
    Africa's largest country." + ], + [ + "Andy Roddick, along with
    Olympic silver medalist Mardy
    Fish and the doubles pair of
    twins Bob and Mike Bryan will
    make up the US team to compete
    with Belarus in the Davis Cup,
    reported CRIENGLISH." + ], + [ + "NEW YORK - Optimism that the
    embattled technology sector
    was due for a recovery sent
    stocks modestly higher Monday
    despite a new revenue warning
    from semiconductor company
    Broadcom Inc. While
    Broadcom, which makes chips
    for television set-top boxes
    and other electronics, said
    high inventories resulted in
    delayed shipments, investors
    were encouraged as it said
    future quarters looked
    brighter..." + ], + [ + "A report indicates that many
    giants of the industry have
    been able to capture lasting
    feelings of customer loyalty." + ], + [ + "It is the news that Internet
    users do not want to hear: the
    worldwide web is in danger of
    collapsing around us. Patrick
    Gelsinger, the chief
    technology officer for
    computer chip maker Intel,
    told a conference" + ], + [ + " WASHINGTON (Reuters) - U.S.
    employers hired just 96,000
    workers in September, the
    government said on Friday in a
    weak jobs snapshot, the final
    one ahead of presidential
    elections that also fueled
    speculation about a pause in
    interest-rate rises." + ], + [ + "Cisco Systems CEO John
    Chambers said today that his
    company plans to offer twice
    as many new products this year
    as ever before." + ], + [ + "While the world #39;s best
    athletes fight the noble
    Olympic battle in stadiums and
    pools, their fans storm the
    streets of Athens, turning the
    Greek capital into a huge
    international party every
    night." + ], + [ + "Carlos Barrios Orta squeezed
    himself into his rubber diving
    suit, pulled on an 18-pound
    helmet that made him look like
    an astronaut, then lowered
    himself into the sewer. He
    disappeared into the filthy
    water, which looked like some
    cauldron of rancid beef stew,
    until the only sign of him was
    air bubbles breaking the
    surface." + ], + [ + "The mutilated body of a woman
    of about sixty years,
    amputated of arms and legs and
    with the sliced throat, has
    been discovered this morning
    in a street of the south of
    Faluya by the Marines,
    according to a statement by
    AFP photographer." + ], + [ + "Every day, somewhere in the
    universe, there #39;s an
    explosion that puts the power
    of the Sun in the shade. Steve
    Connor investigates the riddle
    of gamma-ray bursts." + ], + [ + "Ten months after NASA #39;s
    twin rovers landed on Mars,
    scientists reported this week
    that both robotic vehicles are
    still navigating their rock-
    studded landscapes with all
    instruments operating" + ], + [ + "EVERTON showed they would not
    be bullied into selling Wayne
    Rooney last night by rejecting
    a 23.5million bid from
    Newcastle - as Manchester
    United gamble on Goodison
    #39;s resolve to keep the
    striker." + ], + [ + "SAN FRANCISCO (CBS.MW) -- The
    magazine known for evaluating
    cars and electronics is
    setting its sights on finding
    the best value and quality of
    prescription drugs on the
    market." + ], + [ + "After months of legal
    wrangling, the case of
    <em>People v. Kobe Bean
    Bryant</em> commences on
    Friday in Eagle, Colo., with
    testimony set to begin next
    month." + ], + [ + "(SH) - In Afghanistan, Hamid
    Karzai defeated a raft of
    candidates to win his historic
    election. In Iraq, more than
    200 political parties have
    registered for next month
    #39;s elections." + ], + [ + "Lyon coach Paul le Guen has
    admitted his side would be
    happy with a draw at Old
    Trafford on Tuesday night. The
    three-times French champions
    have assured themselves of
    qualification for the
    Champions League" + ], + [ + "British scientists say they
    have found a new, greener way
    to power cars and homes using
    sunflower oil, a commodity
    more commonly used for cooking
    fries." + ], + [ + "A mouse, a house, and your
    tax-planning spouse all factor
    huge in the week of earnings
    that lies ahead." + ], + [ + "The United States #39; top
    computer-security official has
    resigned after a little more
    than a year on the job, the US
    Department of Homeland
    Security said on Friday." + ], + [ + "Reuters - Barry Bonds failed
    to collect a hit in\\his bid to
    join the 700-homer club, but
    he did score a run to\\help the
    San Francisco Giants edge the
    host Milwaukee Brewers\\3-2 in
    National League action on
    Tuesday." + ], + [ + " SAN FRANCISCO (Reuters) -
    Intel Corp. <A HREF=\"http:/
    /www.reuters.co.uk/financeQuot
    eLookup.jhtml?ticker=INTC.O
    qtype=sym infotype=info
    qcat=news\">INTC.O</A>
    on Thursday outlined its
    vision of the Internet of the
    future, one in which millions
    of computer servers would
    analyze and direct network
    traffic to make the Web safer
    and more efficient." + ], + [ + "Margaret Hassan, who works for
    charity Care International,
    was taken hostage while on her
    way to work in Baghdad. Here
    are the main events since her
    kidnapping." + ], + [ + "Check out new gadgets as
    holiday season approaches." + ], + [ + "DALLAS (CBS.MW) -- Royal
    Dutch/Shell Group will pay a
    \\$120 million penalty to
    settle a Securities and
    Exchange Commission
    investigation of its
    overstatement of nearly 4.5
    billion barrels of proven
    reserves, the federal agency
    said Tuesday." + ], + [ + "After waiting an entire summer
    for the snow to fall and
    Beaver Creek to finally open,
    skiers from around the planet
    are coming to check out the
    Birds of Prey World Cup action
    Dec. 1 - 5. Although everyones" + ], + [ + "TOKYO, Sep 08, 2004 (AFX-UK
    via COMTEX) -- Sony Corp will
    launch its popular projection
    televisions with large liquid
    crystal display (LCD) screens
    in China early next year, a
    company spokeswoman said." + ], + [ + "I confess that I am a complete
    ignoramus when it comes to
    women #39;s beach volleyball.
    In fact, I know only one thing
    about the sport - that it is
    the supreme aesthetic
    experience available on planet
    Earth." + ], + [ + "Manchester United Plc may
    offer US billionaire Malcolm
    Glazer a seat on its board if
    he agrees to drop a takeover
    bid for a year, the Observer
    said, citing an unidentified
    person in the soccer industry." + ], + [ + "PHOENIX America West Airlines
    has backed away from a
    potential bidding war for
    bankrupt ATA Airlines, paving
    the way for AirTran to take
    over ATA operations." + ], + [ + "US stock-index futures
    declined. Dow Jones Industrial
    Average shares including
    General Electric Co. slipped
    in Europe. Citigroup Inc." + ], + [ + "For a guy who spent most of
    his first four professional
    seasons on the disabled list,
    Houston Astros reliever Brad
    Lidge has developed into quite
    the ironman these past two
    days." + ], + [ + "AFP - Outgoing EU competition
    commissioner Mario Monti wants
    to reach a decision on
    Oracle's proposed takeover of
    business software rival
    PeopleSoft by the end of next
    month, an official said." + ], + [ + "Former Bengal Corey Dillon
    found his stride Sunday in his
    second game for the New
    England Patriots. Dillon
    gained 158 yards on 32 carries
    as the Patriots beat the
    Arizona Cardinals, 23-12, for
    their 17th victory in a row." + ], + [ + "If legend is to be believed,
    the end of a victorious war
    was behind Pheidippides #39;
    trek from Marathon to Athens
    2,500 years ago." + ], + [ + " WASHINGTON (Reuters) -
    Satellite companies would be
    able to retransmit
    broadcasters' television
    signals for another five
    years but would have to offer
    those signals on a single
    dish, under legislation
    approved by Congress on
    Saturday." + ], + [ + " BEIJING (Reuters) - Wimbledon
    champion Maria Sharapova
    demolished fellow Russian
    Tatiana Panova 6-1, 6-1 to
    advance to the quarter-finals
    of the China Open on
    Wednesday." + ], + [ + "Sven Jaschan, who may face
    five years in prison for
    spreading the Netsky and
    Sasser worms, is now working
    in IT security. Photo: AFP." + ], + [ + "Padres general manager Kevin
    Towers called Expos general
    manager Omar Minaya on
    Thursday afternoon and told
    him he needed a shortstop
    because Khalil Greene had
    broken his" + ], + [ + "Motorsport.com. Nine of the
    ten Formula One teams have
    united to propose cost-cutting
    measures for the future, with
    the notable exception of
    Ferrari." + ], + [ + "Reuters - The United States
    modified its\\call for U.N.
    sanctions against Sudan on
    Tuesday but still kept\\up the
    threat of punitive measures if
    Khartoum did not
    stop\\atrocities in its Darfur
    region." + ], + [ + "Human rights and environmental
    activists have hailed the
    award of the 2004 Nobel Peace
    Prize to Wangari Maathai of
    Kenya as fitting recognition
    of the growing role of civil
    society" + ], + [ + "Federal Reserve Chairman Alan
    Greenspan has done it again.
    For at least the fourth time
    this year, he has touched the
    electrified third rail of
    American politics - Social
    Security." + ], + [ + "An Al Qaeda-linked militant
    group beheaded an American
    hostage in Iraq and threatened
    last night to kill another two
    Westerners in 24 hours unless
    women prisoners were freed
    from Iraqi jails." + ], + [ + "TechWeb - An Indian Institute
    of Technology professor--and
    open-source evangelist--
    discusses the role of Linux
    and open source in India." + ], + [ + "Here are some of the latest
    health and medical news
    developments, compiled by
    editors of HealthDay: -----
    Contaminated Fish in Many U.S.
    Lakes and Rivers Fish
    that may be contaminated with
    dioxin, mercury, PCBs and
    pesticides are swimming in
    more than one-third of the
    United States' lakes and
    nearly one-quarter of its
    rivers, according to a list of
    advisories released by the
    Environmental Protection
    Agency..." + ], + [ + "Argentina, Denmark, Greece,
    Japan and Tanzania on Friday
    won coveted two-year terms on
    the UN Security Council at a
    time when pressure is mounting
    to expand the powerful
    15-nation body." + ], + [ + "Description: Scientists say
    the arthritis drug Bextra may
    pose increased risk of
    cardiovascular troubles.
    Bextra is related to Vioxx,
    which was pulled off the
    market in September for the
    same reason." + ], + [ + "Steve Francis and Shaquille O
    #39;Neal enjoyed big debuts
    with their new teams. Kobe
    Bryant found out he can #39;t
    carry the Lakers all by
    himself." + ], + [ + "The trial of a lawsuit by Walt
    Disney Co. shareholders who
    accuse the board of directors
    of rubberstamping a deal to
    hire Michael Ovitz" + ], + [ + "Australia, by winning the
    third Test at Nagpur on
    Friday, also won the four-
    match series 2-0 with one
    match to go. Australia had
    last won a Test series in
    India way back in December
    1969 when Bill Lawry #39;s
    team beat Nawab Pataudi #39;s
    Indian team 3-1." + ], + [ + "LOS ANGELES Grocery giant
    Albertsons says it has
    purchased Bristol Farms, which
    operates eleven upscale stores
    in Southern California." + ], + [ + "ISLAMABAD: Pakistan and India
    agreed on Friday to re-open
    the Khokhropar-Munabao railway
    link, which was severed nearly
    40 years ago." + ], + [ + "Final Score: Connecticut 61,
    New York 51 New York, NY
    (Sports Network) - Nykesha
    Sales scored 15 points to lead
    Connecticut to a 61-51 win
    over New York in Game 1 of
    their best-of-three Eastern
    Conference Finals series at
    Madison Square Garden." + ], + [ + "NEW YORK - Stocks moved higher
    Friday as a stronger than
    expected retail sales report
    showed that higher oil prices
    aren't scaring consumers away
    from spending. Federal Reserve
    Chairman Alan Greenspan's
    positive comments on oil
    prices also encouraged
    investors..." + ], + [ + "The figures from a survey
    released today are likely to
    throw more people into the
    ranks of the uninsured,
    analysts said." + ], + [ + "Bill Gates is giving his big
    speech right now at Microsofts
    big Digital Entertainment
    Anywhere event in Los Angeles.
    Well have a proper report for
    you soon, but in the meantime
    we figured wed" + ], + [ + "Spinal and non-spinal
    fractures are reduced by
    almost a third in women age 80
    or older who take a drug
    called strontium ranelate,
    European investigators
    announced" + ], + [ + "JB Oxford Holdings Inc., a
    Beverly Hills-based discount
    brokerage firm, was sued by
    the Securities and Exchange
    Commission for allegedly
    allowing thousands of improper
    trades in more than 600 mutual
    funds." + ], + [ + "BERLIN: Apple Computer Inc is
    planning the next wave of
    expansion for its popular
    iTunes online music store with
    a multi-country European
    launch in October, the
    services chief architect said
    on Wednesday." + ], + [ + "Charlotte, NC -- LeBron James
    poured in a game-high 19
    points and Jeff McInnis scored
    18 as the Cleveland Cavaliers
    routed the Charlotte Bobcats,
    106-89, at the Charlotte
    Coliseum." + ], + [ + "Goldman Sachs reported strong
    fourth quarter and full year
    earnings of \\$1.19bn, up 36
    per cent on the previous
    quarter. Full year profit rose
    52 per cent from the previous
    year to \\$4.55bn." + ], + [ + "Saddam Hussein lives in an
    air-conditioned 10-by-13 foot
    cell on the grounds of one of
    his former palaces, tending
    plants and proclaiming himself
    Iraq's lawful ruler." + ], + [ + "Lehmann, who was at fault in
    two matches in the tournament
    last season, was blundering
    again with the German set to
    take the rap for both Greek
    goals." + ], + [ + "Calls to 13 other countries
    will be blocked to thwart
    auto-dialer software." + ], + [ + "AP - Brazilian U.N.
    peacekeepers will remain in
    Haiti until presidential
    elections are held in that
    Caribbean nation sometime next
    year, President Luiz Inacio
    Lula da Silva said Monday." + ], + [ + "com September 27, 2004, 5:00
    AM PT. This fourth priority
    #39;s main focus has been
    improving or obtaining CRM and
    ERP software for the past year
    and a half." + ], + [ + "Jeju Island, South Korea
    (Sports Network) - Grace Park
    and Carin Koch posted matching
    rounds of six-under-par 66 on
    Friday to share the lead after
    the first round of the CJ Nine
    Bridges Classic." + ], + [ + "SPACE.com - The Zero Gravity
    Corporation \\ has been given
    the thumbs up by the Federal
    Aviation Administration (FAA)
    to \\ conduct quot;weightless
    flights quot; for the general
    public, providing the \\
    sensation of floating in
    space." + ], + [ + "A 32-year-old woman in Belgium
    has become the first woman
    ever to give birth after
    having ovarian tissue removed,
    frozen and then implanted back
    in her body." + ], + [ + "Argosy Gaming (AGY:NYSE - news
    - research) jumped in early
    trading Thursday, after the
    company agreed to be acquired
    by Penn National Gaming
    (PENN:Nasdaq - news -
    research) in a \\$1." + ], + [ + "No sweat, says the new CEO,
    who's imported from IBM. He
    also knows this will be the
    challenge of a lifetime." + ], + [ + "Iraq is quot;working 24 hours
    a day to ... stop the
    terrorists, quot; interim
    Iraqi Prime Minister Ayad
    Allawi said Monday. Iraqis are
    pushing ahead with reforms and
    improvements, Allawi told" + ], + [ + "AP - Their discoveries may be
    hard for many to comprehend,
    but this year's Nobel Prize
    winners still have to explain
    what they did and how they did
    it." + ], + [ + "The DuPont Co. has agreed to
    pay up to \\$340 million to
    settle a lawsuit that it
    contaminated water supplies in
    West Virginia and Ohio with a
    chemical used to make Teflon,
    one of its best-known brands." + ], + [ + "Benfica and Real Madrid set
    the standard for soccer
    success in Europe in the late
    1950s and early #39;60s. The
    clubs have evolved much
    differently, but both have
    struggled" + ], + [ + "AP - Musicians, composers and
    authors were among the more
    than two dozen people
    Wednesday honored with
    National Medal of Arts and
    National Humanities awards at
    the White House." + ], + [ + "More than 15 million homes in
    the UK will be able to get on-
    demand movies by 2008, say
    analysts." + ], + [ + "Wynton Marsalis, the trumpet-
    playing star and artistic
    director of Jazz at Lincoln
    Center, has become an entity
    above and around the daily
    jazz world." + ], + [ + "A well-researched study by the
    National Academy of Sciences
    makes a persuasive case for
    refurbishing the Hubble Space
    Telescope. NASA, which is
    reluctant to take this
    mission, should rethink its
    position." + ], + [ + "BUENOS AIRES: Pakistan has
    ratified the Kyoto Protocol on
    Climatic Change, Environment
    Minister Malik Khan said on
    Thursday. He said that
    Islamabad has notified UN
    authorities of ratification,
    which formally comes into
    effect in February 2005." + ], + [ + "Reuters - Jeffrey Greenberg,
    chairman and chief\\executive
    of embattled insurance broker
    Marsh McLennan Cos.\\, is
    expected to step down within
    hours, a newspaper\\reported on
    Friday, citing people close to
    the discussions." + ], + [ + "Staff Sgt. Johnny Horne, Jr.,
    and Staff Sgt. Cardenas Alban
    have been charged with murder
    in the death of an Iraqi, the
    1st Cavalry Division announced
    Monday." + ], + [ + "LONDON -- In a deal that
    appears to buck the growing
    trend among governments to
    adopt open-source
    alternatives, the U.K.'s
    Office of Government Commerce
    (OGC) is negotiating a renewal
    of a three-year agreement with
    Microsoft Corp." + ], + [ + "WASHINGTON - A Pennsylvania
    law requiring Internet service
    providers (ISPs) to block Web
    sites the state's prosecuting
    attorneys deem to be child
    pornography has been reversed
    by a U.S. federal court, with
    the judge ruling the law
    violated free speech rights." + ], + [ + "The Microsoft Corp. chairman
    receives four million e-mails
    a day, but practically an
    entire department at the
    company he founded is
    dedicated to ensuring that
    nothing unwanted gets into his
    inbox, the company #39;s chief
    executive said Thursday." + ], + [ + "The 7100t has a mobile phone,
    e-mail, instant messaging, Web
    browsing and functions as an
    organiser. The device looks
    like a mobile phone and has
    the features of the other
    BlackBerry models, with a
    large screen" + ], + [ + "President Vladimir Putin makes
    a speech as he hosts Russia
    #39;s Olympic athletes at a
    Kremlin banquet in Moscow,
    Thursday, Nov. 4, 2004." + ], + [ + "Apple Computer has unveiled
    two new versions of its hugely
    successful iPod: the iPod
    Photo and the U2 iPod. Apple
    also has expanded" + ], + [ + "Pakistan are closing in fast
    on Sri Lanka #39;s first
    innings total after impressing
    with both ball and bat on the
    second day of the opening Test
    in Faisalabad." + ], + [ + "A state regulatory board
    yesterday handed a five-year
    suspension to a Lawrence
    funeral director accused of
    unprofessional conduct and
    deceptive practices, including
    one case where he refused to
    complete funeral arrangements
    for a client because she had
    purchased a lower-priced
    casket elsewhere." + ], + [ + "AP - This year's hurricanes
    spread citrus canker to at
    least 11,000 trees in
    Charlotte County, one of the
    largest outbreaks of the
    fruit-damaging infection to
    ever affect Florida's citrus
    industry, state officials
    said." + ], + [ + "AFP - Hundreds of Buddhists in
    southern Russia marched
    through snow to see and hear
    the Dalai Lama as he continued
    a long-awaited visit to the
    country in spite of Chinese
    protests." + ], + [ + "British officials were on
    diplomatic tenterhooks as they
    awaited the arrival on
    Thursday of French President
    Jacques Chirac for a two-day
    state visit to Britain." + ], + [ + " SYDNEY (Reuters) - A group of
    women on Pitcairn Island in
    the South Pacific are standing
    by their men, who face
    underage sex charges, saying
    having sex at age 12 is a
    tradition dating back to 18th
    century mutineers who settled
    on the island." + ], + [ + "Two aircraft are flying out
    from the UK on Sunday to
    deliver vital aid and supplies
    to Haiti, which has been
    devastated by tropical storm
    Jeanne." + ], + [ + "A scare triggered by a
    vibrating sex toy shut down a
    major Australian regional
    airport for almost an hour
    Monday, police said. The
    vibrating object was
    discovered Monday morning" + ], + [ + "IBM moved back into the iSCSI
    (Internet SCSI) market Friday
    with a new array priced at
    US\\$3,000 and aimed at the
    small and midsize business
    market." + ], + [ + "Ron Artest has been hit with a
    season long suspension,
    unprecedented for the NBA
    outside doping cases; Stephen
    Jackson banned for 30 games;
    Jermaine O #39;Neal for 25
    games and Anthony Johnson for
    five." + ], + [ + "The California Public
    Utilities Commission on
    Thursday upheld a \\$12.1
    million fine against Cingular
    Wireless, related to a two-
    year investigation into the
    cellular telephone company
    #39;s business practices." + ], + [ + "Barclays, the British bank
    that left South Africa in 1986
    after apartheid protests, may
    soon resume retail operations
    in the nation." + ], + [ + "AP - An Indiana congressional
    candidate abruptly walked off
    the set of a debate because
    she got stage fright." + ], + [ + "Italian-based Parmalat is
    suing its former auditors --
    Grant Thornton International
    and Deloitte Touche Tohmatsu
    -- for billions of dollars in
    damages. Parmalat blames its
    demise on the two companies
    #39; mismanagement of its
    finances." + ], + [ + "Reuters - Having reached out
    to Kashmiris during a two-day
    visit to the region, the prime
    minister heads this weekend to
    the country's volatile
    northeast, where public anger
    is high over alleged abuses by
    Indian soldiers." + ], + [ + "SAN JUAN IXTAYOPAN, Mexico --
    A pair of wooden crosses
    outside the elementary school
    are all that mark the macabre
    site where, just weeks ago, an
    angry mob captured two federal
    police officers, beat them
    unconscious, and set them on
    fire." + ], + [ + "Three shots behind Grace Park
    with five holes to go, six-
    time LPGA player of the year
    Sorenstam rallied with an
    eagle, birdie and three pars
    to win her fourth Samsung
    World Championship by three
    shots over Park on Sunday." + ], + [ + "update An alliance of
    technology workers on Tuesday
    accused conglomerate Honeywell
    International of planning to
    move thousands of jobs to low-
    cost regions over the next
    five years--a charge that
    Honeywell denies." + ], + [ + "NSW Rugby CEO Fraser Neill
    believes Waratah star Mat
    Rogers has learned his lesson
    after he was fined and ordered
    to do community service
    following his controversial
    comments about the club rugby
    competition." + ], + [ + "American Technology Research
    analyst Shaw Wu has initiated
    coverage of Apple Computer
    (AAPL) with a #39;buy #39;
    recommendation, and a 12-month
    target of US\\$78 per share." + ], + [ + "washingtonpost.com - Cell
    phone shoppers looking for new
    deals and features didn't find
    them yesterday, a day after
    Sprint Corp. and Nextel
    Communications Inc. announced
    a merger that the phone
    companies promised would shake
    up the wireless business." + ], + [ + "ATHENS: China, the dominant
    force in world diving for the
    best part of 20 years, won six
    out of eight Olympic titles in
    Athens and prompted
    speculation about a clean
    sweep when they stage the
    Games in Beijing in 2008." + ], + [ + " NEW YORK (Reuters) - Northrop
    Grumman Corp. <A HREF=\"http
    ://www.investor.reuters.com/Fu
    llQuote.aspx?ticker=NOC.N targ
    et=/stocks/quickinfo/fullquote
    \">NOC.N</A> reported
    higher third-quarter earnings
    on Wednesday and an 11
    percent increase in sales on
    strength in its mission
    systems, integrated systems,
    ships and space technology
    businesses." + ], + [ + "JAPANESE GIANT Sharp has
    pulled the plug on its Linux-
    based PDAs in the United
    States as no-one seems to want
    them. The company said that it
    will continue to sell them in
    Japan where they sell like hot
    cakes" + ], + [ + "DUBLIN : An Olympic Airlines
    plane diverted to Ireland
    following a bomb alert, the
    second faced by the Greek
    carrier in four days, resumed
    its journey to New York after
    no device was found on board,
    airport officials said." + ], + [ + "New NavOne offers handy PDA
    and Pocket PC connectivity,
    but fails to impress on
    everything else." + ], + [ + "TORONTO (CP) - Shares of
    Iamgold fell more than 10 per
    cent after its proposed merger
    with Gold Fields Ltd. was
    thrown into doubt Monday as
    South Africa #39;s Harmony
    Gold Mining Company Ltd." + ], + [ + "The Marvel deal calls for
    Mforma to work with the comic
    book giant to develop a
    variety of mobile applications
    based on Marvel content." + ], + [ + " LONDON (Reuters) - Oil prices
    tumbled again on Monday to an
    8-week low under \\$46 a
    barrel, as growing fuel stocks
    in the United States eased
    fears of a winter supply
    crunch." + ], + [ + " WASHINGTON (Reuters) - The
    United States said on Friday
    it is preparing a new U.N.
    resolution on Darfur and that
    Secretary of State Colin
    Powell might address next week
    whether the violence in
    western Sudan constitutes
    genocide." + ], + [ + "The Houston Astros won their
    19th straight game at home and
    are one game from winning
    their first playoff series in
    42 years." + ], + [ + "washingtonpost.com - The
    Portable Media Center -- a
    new, Microsoft-conceived
    handheld device that presents
    video and photos as well as
    music -- would be a decent
    idea if there weren't such a
    thing as lampposts. Or street
    signs. Or trees. Or other
    cars." + ], + [ + "Alcoa Inc., one of the world
    #39;s top producers of
    aluminum, said Monday that it
    received an unsolicited
    quot;mini-tender quot; offer
    from Toronto-based TRC Capital
    Corp." + ], + [ + "The European Commission is to
    warn Greece about publishing
    false information about its
    public finances." + ], + [ + "The Air Force Reserve #39;s
    Hurricane Hunters, those
    fearless crews who dive into
    the eyewalls of hurricanes to
    relay critical data on
    tropical systems, were chased
    from their base on the
    Mississippi Gulf Coast by
    Hurricane Ivan." + ], + [ + " LONDON (Reuters) - U.S.
    shares were expected to open
    lower on Wednesday after
    crude oil pushed to a fresh
    high overnight, while Web
    search engine Google Inc.
    dented sentiment as it
    slashed the price range on its
    initial public offering." + ], + [ + "The eighth-seeded American
    fell to sixth-seeded Elena
    Dementieva of Russia, 0-6 6-2
    7-6 (7-5), on Friday - despite
    being up a break on four
    occasions in the third set." + ], + [ + "TUCSON, Arizona (Ticker) --
    No. 20 Arizona State tries to
    post its first three-game
    winning streak over Pac-10
    Conference rival Arizona in 26
    years when they meet Friday." + ], + [ + "NAJAF, Iraq : Iraq #39;s top
    Shiite Muslim clerics, back in
    control of Najaf #39;s Imam
    Ali shrine after a four-month
    militia occupation, met amid
    the ruins of the city as life
    spluttered back to normality." + ], + [ + "Embargo or not, Fidel Castro's
    socialist paradise has quietly
    become a pharmaceutical
    powerhouse. (They're still
    working on the capitalism
    thing.) By Douglas Starr from
    Wired magazine." + ], + [ + "AP - Phillip Fulmer kept his
    cool when starting center
    Jason Respert drove off in the
    coach's golf cart at practice." + ], + [ + "GOALS from Wayne Rooney and
    Ruud van Nistelrooy gave
    Manchester United the win at
    Newcastle. Alan Shearer
    briefly levelled matters for
    the Magpies but United managed
    to scrape through." + ], + [ + "The telemarketer at the other
    end of Orlando Castelblanco
    #39;s line promised to reduce
    the consumer #39;s credit card
    debt by at least \\$2,500 and
    get his 20 percent -- and
    growing -- interest rates down
    to single digits." + ], + [ + " NEW YORK (Reuters) - Blue-
    chip stocks fell slightly on
    Monday after No. 1 retailer
    Wal-Mart Stores Inc. <A HRE
    F=\"http://www.investor.reuters
    .com/FullQuote.aspx?ticker=WMT
    .N target=/stocks/quickinfo/fu
    llquote\">WMT.N</A>
    reported lower-than-expected
    Thanksgiving sales, while
    technology shares were lifted
    by a rally in Apple Computer
    Inc. <A HREF=\"http://www.i
    nvestor.reuters.com/FullQuote.
    aspx?ticker=AAPL.O target=/sto
    cks/quickinfo/fullquote\">AA
    PL.O</A>." + ], + [ + "Reuters - House of
    Representatives
    Majority\\Leader Tom DeLay,
    admonished twice in six days
    by his chamber's\\ethics
    committee, withstood calls on
    Thursday by rival\\Democrats
    and citizen groups that he
    step aside." + ], + [ + "WPP Group Inc., the world
    #39;s second- largest
    marketing and advertising
    company, said it won the
    bidding for Grey Global Group
    Inc." + ], + [ + "AFP - Australia has accounted
    for all its nationals known to
    be working in Iraq following a
    claim by a radical Islamic
    group to have kidnapped two
    Australians, Foreign Affairs
    Minister Alexander Downer
    said." + ], + [ + "A new worm can spy on users by
    hijacking their Web cameras, a
    security firm warned Monday.
    The Rbot.gr worm -- the latest
    in a long line of similar
    worms; one security firm
    estimates that more than 4,000
    variations" + ], + [ + " NEW YORK (Reuters) - Stocks
    were slightly lower on
    Tuesday, as concerns about
    higher oil prices cutting into
    corporate profits and
    consumer demand weighed on
    sentiment, while retail sales
    posted a larger-than-expected
    decline in August." + ], + [ + "The separation of PalmOne and
    PalmSource will be complete
    with Eric Benhamou's
    resignation as the latter's
    chairman." + ], + [ + "AP - Two-time U.S. Open
    doubles champion Max Mirnyi
    will lead the Belarus team
    that faces the United States
    in the Davis Cup semifinals in
    Charleston later this month." + ], + [ + "The hurricane appeared to be
    strengthening and forecasters
    warned of an increased threat
    of serious flooding and wind
    damage." + ], + [ + "AP - New York Jets safety Erik
    Coleman got his souvenir
    football from the equipment
    manager and held it tightly." + ], + [ + " SINGAPORE (Reuters) - Oil
    prices climbed above \\$42 a
    barrel on Wednesday, rising
    for the third day in a row as
    the heavy consuming U.S.
    Northeast feels the first
    chills of winter." + ], + [ + "\\\\\"(CNN) -- A longtime
    associate of al Qaeda leader
    Osama bin Laden surrendered
    to\\Saudi Arabian officials
    Tuesday, a Saudi Interior
    Ministry official said.\"\\\\\"But
    it is unclear what role, if
    any, Khaled al-Harbi may have
    had in any terror\\attacks
    because no public charges have
    been filed against him.\"\\\\\"The
    Saudi government -- in a
    statement released by its
    embassy in Washington
    --\\called al-Harbi's surrender
    \"the latest direct result\" of
    its limited, one-month\\offer
    of leniency to terror
    suspects.\"\\\\This is great! I
    hope this really starts to pay
    off. Creative solutions
    to\\terrorism that don't
    involve violence. \\\\How
    refreshing! \\\\Are you paying
    attention Bush
    administration?\\\\" + ], + [ + " NEW YORK (Reuters) - Merck
    Co Inc. <A HREF=\"http://www
    .investor.reuters.com/FullQuot
    e.aspx?ticker=MRK.N target=/st
    ocks/quickinfo/fullquote\">M
    RK.N</A> on Thursday
    pulled its arthritis drug
    Vioxx off the market after a
    study showed it doubled the
    risk of heart attack and
    stroke, a move that sent its
    shares plunging and erased
    \\$25 billion from its market
    value." + ], + [ + "AT T Corp. is cutting 7,400
    more jobs and slashing the
    book value of its assets by
    \\$11.4 billion, drastic moves
    prompted by the company's plan
    to retreat from the
    traditional consumer telephone
    business following a lost
    court battle." + ], + [ + "The government on Wednesday
    defended its decision to
    radically revise the country
    #39;s deficit figures, ahead
    of a European Commission
    meeting to consider possible
    disciplinary action against
    Greece for submitting faulty
    figures." + ], + [ + "Ivan Hlinka coached the Czech
    Republic to the hockey gold
    medal at the 1998 Nagano
    Olympics and became the coach
    of the Pittsburgh Penguins two
    years later." + ], + [ + "Four homeless men were
    bludgeoned to death and six
    were in critical condition on
    Friday following early morning
    attacks by unknown assailants
    in downtown streets of Sao
    Paulo, a police spokesman
    said." + ], + [ + "OPEC ministers yesterday
    agreed to increase their
    ceiling for oil production to
    help bring down stubbornly
    high prices in a decision that
    traders and analysts dismissed
    as symbolic because the cartel
    already is pumping more than
    its new target." + ], + [ + "Williams-Sonoma Inc. said
    second- quarter profit rose 55
    percent, boosted by the
    addition of Pottery Barn
    stores and sale of outdoor
    furniture." + ], + [ + "Celerons form the basis of
    Intel #39;s entry-level
    platform which includes
    integrated/value motherboards
    as well. The Celeron 335D is
    the fastest - until the
    Celeron 340D - 2.93GHz -
    becomes mainstream - of the" + ], + [ + "The entertainment industry
    asks the Supreme Court to
    reverse the Grokster decision,
    which held that peer-to-peer
    networks are not liable for
    copyright abuses of their
    users. By Michael Grebb." + ], + [ + " HONG KONG/SAN FRANCISCO
    (Reuters) - China's largest
    personal computer maker,
    Lenovo Group Ltd., said on
    Tuesday it was in acquisition
    talks with a major technology
    company, which a source
    familiar with the situation
    said was IBM." + ], + [ + "Phone companies are not doing
    enough to warn customers about
    internet \"rogue-dialling\"
    scams, watchdog Icstis warns." + ], + [ + "Reuters - Oil prices stayed
    close to #36;49 a\\barrel on
    Thursday, supported by a
    forecast for an early
    cold\\snap in the United States
    that could put a strain on a
    thin\\supply cushion of winter
    heating fuel." + ], + [ + "AUBURN - Ah, easy street. No
    game this week. Light
    practices. And now Auburn is
    being touted as the No. 3 team
    in the Bowl Championship
    Series standings." + ], + [ + "Portsmouth #39;s Harry
    Redknapp has been named as the
    Barclays manager of the month
    for October. Redknapp #39;s
    side were unbeaten during the
    month and maintained an
    impressive climb to ninth in
    the Premiership." + ], + [ + "India's main opposition party
    takes action against senior
    party member Uma Bharti after
    a public row." + ], + [ + "Hewlett-Packard will shell out
    \\$16.1 billion for chips in
    2005, but Dell's wallet is
    wide open, too." + ], + [ + "WASHINGTON (CBS.MW) --
    President Bush announced
    Monday that Kellogg chief
    executive Carlos Gutierrez
    would replace Don Evans as
    Commerce secretary, naming the
    first of many expected changes
    to his economic team." + ], + [ + "Iron Mountain moved further
    into the backup and recovery
    space Tuesday with the
    acquisition of Connected Corp.
    for \\$117 million. Connected
    backs up desktop data for more
    than 600 corporations, with
    more than" + ], + [ + "Three directors of Manchester
    United have been ousted from
    the board after US tycoon
    Malcolm Glazer, who is
    attempting to buy the club,
    voted against their re-
    election." + ], + [ + "The Russian military yesterday
    extended its offer of a \\$10
    million reward for information
    leading to the capture of two
    separatist leaders who, the
    Kremlin claims, were behind
    the Beslan massacre." + ], + [ + "AP - Manny Ramirez singled and
    scored before leaving with a
    bruised knee, and the
    streaking Boston Red Sox beat
    the Detroit Tigers 5-3 Friday
    night for their 10th victory
    in 11 games." + ], + [ + "A remote attacker could take
    complete control over
    computers running many
    versions of Microsoft software
    by inserting malicious code in
    a JPEG image that executes
    through an unchecked buffer" + ], + [ + "Montgomery County (website -
    news) is a big step closer to
    shopping for prescription
    drugs north of the border. On
    a 7-2 vote, the County Council
    is approving a plan that would
    give county" + ], + [ + "Israels Shin Bet security
    service has tightened
    protection of the prime
    minister, MPs and parliament
    ahead of next weeks crucial
    vote on a Gaza withdrawal." + ], + [ + "The news comes fast and
    furious. Pedro Martinez goes
    to Tampa to visit George
    Steinbrenner. Theo Epstein and
    John Henry go to Florida for
    their turn with Pedro. Carl
    Pavano comes to Boston to
    visit Curt Schilling. Jason
    Varitek says he's not a goner.
    Derek Lowe is a goner, but he
    says he wishes it could be
    different. Orlando Cabrera ..." + ], + [ + "The disclosure this week that
    a Singapore-listed company
    controlled by a Chinese state-
    owned enterprise lost \\$550
    million in derivatives
    transactions" + ], + [ + "Reuters - Iraq's interim
    defense minister
    accused\\neighbors Iran and
    Syria on Wednesday of aiding
    al Qaeda\\Islamist Abu Musab
    al-Zarqawi and former agents
    of Saddam\\Hussein to promote a
    \"terrorist\" insurgency in
    Iraq." + ], + [ + "AP - The Senate race in
    Kentucky stayed at fever pitch
    on Thursday as Democratic
    challenger Daniel Mongiardo
    stressed his opposition to gay
    marriage while accusing
    Republican incumbent Jim
    Bunning of fueling personal
    attacks that seemed to suggest
    Mongiardo is gay." + ], + [ + "The lawsuit claims the
    companies use a patented
    Honeywell technology for
    brightening images and
    reducing interference on
    displays." + ], + [ + "The co-president of Oracle
    testified that her company was
    serious about its takeover
    offer for PeopleSoft and was
    not trying to scare off its
    customers." + ], + [ + "VANCOUVER - A Vancouver-based
    firm won #39;t sell 1.2
    million doses of influenza
    vaccine to the United States
    after all, announcing Tuesday
    that it will sell the doses
    within Canada instead." + ], + [ + "An extremely rare Hawaiian
    bird dies in captivity,
    possibly marking the
    extinction of its entire
    species only 31 years after it
    was first discovered." + ], + [ + "Does Geico's trademark lawsuit
    against Google have merit? How
    will the case be argued?
    What's the likely outcome of
    the trial? A mock court of
    trademark experts weighs in
    with their verdict." + ], + [ + "Treo 650 boasts a high-res
    display, an improved keyboard
    and camera, a removable
    battery, and more. PalmOne
    this week is announcing the
    Treo 650, a hybrid PDA/cell-
    phone device that addresses
    many of the shortcomings" + ], + [ + "FRED Hale Sr, documented as
    the worlds oldest man, has
    died at the age of 113. Hale
    died in his sleep on Friday at
    a hospital in Syracuse, New
    York, while trying to recover
    from a bout of pneumonia, his
    grandson, Fred Hale III said." + ], + [ + "The Oakland Raiders have
    traded Jerry Rice to the
    Seattle Seahawks in a move
    expected to grant the most
    prolific receiver in National
    Football League history his
    wish to get more playing time." + ], + [ + "consortium led by the Sony
    Corporation of America reached
    a tentative agreement today to
    buy Metro-Goldwyn-Mayer, the
    Hollywood studio famous for
    James Bond and the Pink
    Panther, for" + ], + [ + "International Business
    Machines Corp.'s possible exit
    from the personal computer
    business would be the latest
    move in what amounts to a long
    goodbye from a field it
    pioneered and revolutionized." + ], + [ + "Leipzig Game Convention in
    Germany, the stage for price-
    slash revelations. Sony has
    announced that it #39;s
    slashing the cost of PS2 in
    the UK and Europe to 104.99
    GBP." + ], + [ + "AP - Florida coach Ron Zook
    was fired Monday but will be
    allowed to finish the season,
    athletic director Jeremy Foley
    told The Gainesville Sun." + ], + [ + "The country-cooking restaurant
    chain has agreed to pay \\$8.7
    million over allegations that
    it segregated black customers,
    subjected them to racial slurs
    and gave black workers
    inferior jobs." + ], + [ + "Troy Brown has had to make a
    lot of adjustments while
    playing both sides of the
    football. quot;You always
    want to score when you get the
    ball -- offense or defense" + ], + [ + " PORT LOUIS, Aug. 17
    (Xinhuanet) -- Southern
    African countries Tuesday
    pledged better trade and
    investment relations with
    China as well as India in the
    final communique released at
    the end of their two-day
    summit." + ], + [ + "In yet another devastating
    body blow to the company,
    Intel (Nasdaq: INTC) announced
    it would be canceling its
    4-GHz Pentium chip. The
    semiconductor bellwether said
    it was switching" + ], + [ + "Jenson Button will tomorrow
    discover whether he is allowed
    to quit BAR and move to
    Williams for 2005. The
    Englishman has signed
    contracts with both teams but
    prefers a switch to Williams,
    where he began his Formula One
    career in 2000." + ], + [ + "Seagate #39;s native SATA
    interface technology with
    Native Command Queuing (NCQ)
    allows the Barracuda 7200.8 to
    match the performance of
    10,000-rpm SATA drives without
    sacrificing capacity" + ], + [ + "ARSENAL boss Arsene Wenger
    last night suffered a
    Champions League setback as
    Brazilian midfielder Gilberto
    Silva (above) was left facing
    a long-term injury absence." + ], + [ + "BAGHDAD - A militant group has
    released a video saying it
    kidnapped a missing journalist
    in Iraq and would kill him
    unless US forces left Najaf
    within 48 hours." + ], + [ + "18 August 2004 -- There has
    been renewed fighting in the
    Iraqi city of Al-Najaf between
    US and Iraqi troops and Shi
    #39;a militiamen loyal to
    radical cleric Muqtada al-
    Sadr." + ], + [ + "You #39;re probably already
    familiar with one of the most
    common questions we hear at
    iPodlounge: quot;how can I
    load my iPod up with free
    music?" + ], + [ + " SINGAPORE (Reuters) -
    Investors bought shares in
    Asian exporters and
    electronics firms such as
    Fujitsu Ltd. on Tuesday,
    buoyed by a favorable outlook
    from U.S. technology
    bellwethers and a slide in oil
    prices." + ], + [ + "Ace Ltd. will stop paying
    brokers for steering business
    its way, becoming the third
    company to make concessions in
    the five days since New York
    Attorney General Eliot Spitzer
    unveiled a probe of the
    insurance industry." + ], + [ + "Vice chairman of the office of
    the chief executive officer in
    Novell Inc, Chris Stone is
    leaving the company to pursue
    other opportunities in life." + ], + [ + "Wm. Wrigley Jr. Co., the world
    #39;s largest maker of chewing
    gum, agreed to buy candy
    businesses including Altoids
    mints and Life Savers from
    Kraft Foods Inc." + ], + [ + "American improves to 3-1 on
    the season with a hard-fought
    overtime win, 74-63, against
    Loyala at Bender Arena on
    Friday night." + ], + [ + "Australia tighten their grip
    on the third Test and the
    series after dominating India
    on day two in Nagpur." + ], + [ + " #39;Reaching a preliminary
    pilot agreement is the single
    most important hurdle they
    have to clear, but certainly
    not the only one." + ], + [ + "Bee Staff Writers. SAN
    FRANCISCO - As Eric Johnson
    drove to the stadium Sunday
    morning, his bruised ribs were
    so sore, he wasn #39;t sure he
    #39;d be able to suit up for
    the game." + ], + [ + "The New England Patriots are
    so single-minded in pursuing
    their third Super Bowl triumph
    in four years that they almost
    have no room for any other
    history." + ], + [ + "TORONTO (CP) - Canada #39;s
    big banks are increasing
    mortgage rates following a
    decision by the Bank of Canada
    to raise its overnight rate by
    one-quarter of a percentage
    point to 2.25 per cent." + ], + [ + " SEOUL (Reuters) - North
    Korea's two-year-old nuclear
    crisis has taxed the world's
    patience, the chief United
    Nations nuclear regulator
    said on Wednesday, urging
    communist Pyongyang to return
    to its disarmament treaty
    obligations." + ], + [ + "washingtonpost.com - Microsoft
    is going to Tinseltown today
    to announce plans for its
    revamped Windows XP Media
    Center, part of an aggressive
    push to get ahead in the
    digital entertainment race." + ], + [ + "GROZNY, Russia - The Russian
    government's choice for
    president of war-ravaged
    Chechnya appeared to be the
    victor Sunday in an election
    tainted by charges of fraud
    and shadowed by last week's
    terrorist destruction of two
    airliners. Little more than
    two hours after polls closed,
    acting Chechen president
    Sergei Abramov said
    preliminary results showed
    Maj..." + ], + [ + "Because, while the Eagles are
    certain to stumble at some
    point during the regular
    season, it seems inconceivable
    that they will falter against
    a team with as many offensive
    problems as Baltimore has
    right now." + ], + [ + "AP - J.J. Arrington ran for 84
    of his 121 yards in the second
    half and Aaron Rodgers shook
    off a slow start to throw two
    touchdown passes to help No. 5
    California beat Washington
    42-12 on Saturday." + ], + [ + "BAGHDAD, Sept 5 (AFP) - Izzat
    Ibrahim al-Duri, Saddam
    Hussein #39;s deputy whose
    capture was announced Sunday,
    is 62 and riddled with cancer,
    but was public enemy number
    two in Iraq for the world
    #39;s most powerful military." + ], + [ + "AP - An explosion targeted the
    Baghdad governor's convoy as
    he was traveling through the
    capital Tuesday, killing two
    people but leaving him
    uninjured, the Interior
    Ministry said." + ], + [ + "GENEVA: Rescuers have found
    the bodies of five Swiss
    firemen who died after the
    ceiling of an underground car
    park collapsed during a fire,
    a police spokesman said last
    night." + ], + [ + "John Kerry has held 10 \"front
    porch visit\" events an actual
    front porch is optional where
    perhaps 100 people ask
    questions in a low-key
    campaigning style." + ], + [ + "AP - Most of the turkeys
    gracing the nation's dinner
    tables Thursday have been
    selectively bred for their
    white meat for so many
    generations that simply
    walking can be a problem for
    many of the big-breasted birds
    and sex is no longer possible." + ], + [ + "OTTAWA (CP) - The economy
    created another 43,000 jobs
    last month, pushing the
    unemployment rate down to 7.1
    per cent from 7.2 per cent in
    August, Statistics Canada said
    Friday." + ], + [ + "The right-win opposition
    Conservative Party and Liberal
    Center Union won 43 seats in
    the 141-member Lithuanian
    parliament, after more than 99
    percent of the votes were
    counted" + ], + [ + " TOKYO (Reuters) - Japan's
    Nikkei average rose 0.39
    percent by midsession on
    Friday, bolstered by solid
    gains in stocks dependent on
    domestic business such as Kao
    Corp. <A HREF=\"http://www.i
    nvestor.reuters.com/FullQuote.
    aspx?ticker=4452.T target=/sto
    cks/quickinfo/fullquote\">44
    52.T</A>." + ], + [ + " FRANKFURT (Reuters) -
    DaimlerChrysler and General
    Motors will jointly develop
    new hybrid motors to compete
    against Japanese rivals on
    the fuel-saving technology
    that reduces harmful
    emissions, the companies said
    on Monday." + ], + [ + " SEATTLE (Reuters) - Microsoft
    Corp. <A HREF=\"http://www.r
    euters.co.uk/financeQuoteLooku
    p.jhtml?ticker=MSFT.O
    qtype=sym infotype=info
    qcat=news\">MSFT.O</A>
    is making a renewed push this
    week to get its software into
    living rooms with the launch
    of a new version of its
    Windows XP Media Center, a
    personal computer designed for
    viewing movies, listening to
    music and scrolling through
    digital pictures." + ], + [ + "KHARTOUM, Aug 18 (Reuters) -
    The United Nations said on
    Wednesday it was very
    concerned by Sudan #39;s lack
    of practical progress in
    bringing security to Darfur,
    where more than a million
    people have fled their homes
    for fear of militia ..." + ], + [ + "SHANGHAI, China The Houston
    Rockets have arrived in
    Shanghai with hometown
    favorite Yao Ming declaring
    himself quot;here on
    business." + ], + [ + "Charleston, SC (Sports
    Network) - Andy Roddick and
    Mardy Fish will play singles
    for the United States in this
    weekend #39;s Davis Cup
    semifinal matchup against
    Belarus." + ], + [ + "AFP - With less than two
    months until the November 2
    election, President George W.
    Bush is working to shore up
    support among his staunchest
    supporters even as he courts
    undecided voters." + ], + [ + "The bass should be in your
    face. That's what Matt Kelly,
    of Boston's popular punk rock
    band Dropkick Murphys, thinks
    is the mark of a great stereo
    system. And he should know.
    Kelly, 29, is the drummer for
    the band that likes to think
    of itself as a bit of an Irish
    lucky charm for the Red Sox." + ], + [ + "Chile's government says it
    will build a prison for
    officers convicted of human
    rights abuses in the Pinochet
    era." + ], + [ + "Slumping corporate spending
    and exports caused the economy
    to slow to a crawl in the
    July-September period, with
    real gross domestic product
    expanding just 0.1 percent
    from the previous quarter,
    Cabinet Office data showed
    Friday." + ], + [ + "US President George W. Bush
    signed into law a bill
    replacing an export tax
    subsidy that violated
    international trade rules with
    a \\$145 billion package of new
    corporate tax cuts and a
    buyout for tobacco farmers." + ], + [ + "The Nikkei average was up 0.37
    percent in mid-morning trade
    on Thursday as a recovery in
    the dollar helped auto makers
    among other exporters, but
    trade was slow as investors
    waited for important Japanese
    economic data." + ], + [ + "Jennifer Canada knew she was
    entering a boy's club when she
    enrolled in Southern Methodist
    University's Guildhall school
    of video-game making." + ], + [ + "RICKY PONTING believes the
    game #39;s watchers have
    fallen for the quot;myth
    quot; that New Zealand know
    how to rattle Australia." + ], + [ + "MILWAUKEE (SportsTicker) -
    Barry Bonds tries to go where
    just two players have gone
    before when the San Francisco
    Giants visit the Milwaukee
    Brewers on Tuesday." + ], + [ + "Palestinian leader Mahmoud
    Abbas reiterated calls for his
    people to drop their weapons
    in the struggle for a state. a
    clear change of strategy for
    peace with Israel after Yasser
    Arafat #39;s death." + ], + [ + "The new software is designed
    to simplify the process of
    knitting together back-office
    business applications." + ], + [ + "SYDNEY (Dow Jones)--Colorado
    Group Ltd. (CDO.AU), an
    Australian footwear and
    clothing retailer, said Monday
    it expects net profit for the
    fiscal year ending Jan. 29 to
    be over 30 higher than that of
    a year earlier." + ], + [ + "NEW YORK - What are the odds
    that a tiny nation like
    Antigua and Barbuda could take
    on the United States in an
    international dispute and win?" + ], + [ + "AP - With Tom Brady as their
    quarterback and a stingy,
    opportunistic defense, it's
    difficult to imagine when the
    New England Patriots might
    lose again. Brady and
    defensive end Richard Seymour
    combined to secure the
    Patriots' record-tying 18th
    straight victory, 31-17 over
    the Buffalo Bills on Sunday." + ], + [ + "FRANKFURT, GERMANY -- The
    German subsidiaries of
    Hewlett-Packard Co. (HP) and
    Novell Inc. are teaming to
    offer Linux-based products to
    the country's huge public
    sector." + ], + [ + "SBC Communications expects to
    cut 10,000 or more jobs by the
    end of next year through
    layoffs and attrition. That
    #39;s about six percent of the
    San Antonio-based company
    #39;s work force." + ], + [ + " BAGHDAD (Reuters) - Iraq's
    U.S.-backed government said on
    Tuesday that \"major neglect\"
    by its American-led military
    allies led to a massacre of 49
    army recruits at the weekend." + ], + [ + "SiliconValley.com - \"I'm
    back,\" declared Apple
    Computer's Steve Jobs on
    Thursday morning in his first
    public appearance before
    reporters since cancer surgery
    in late July." + ], + [ + "BEIJING -- Police have
    detained a man accused of
    slashing as many as nine boys
    to death as they slept in
    their high school dormitory in
    central China, state media
    reported today." + ], + [ + "Health India: London, Nov 4 :
    Cosmetic face cream used by
    fashionable Roman women was
    discovered at an ongoing
    archaeological dig in London,
    in a metal container, complete
    with the lid and contents." + ], + [ + "Israeli Prime Minister Ariel
    Sharon #39;s Likud party
    agreed on Thursday to a
    possible alliance with
    opposition Labour in a vote
    that averted a snap election
    and strengthened his Gaza
    withdrawal plan." + ], + [ + "Another United Airlines union
    is seeking to oust senior
    management at the troubled
    airline, saying its strategies
    are reckless and incompetent." + ], + [ + "Approaching Hurricane Ivan has
    led to postponement of the
    game Thursday night between
    10th-ranked California and
    Southern Mississippi in
    Hattiesburg, Cal #39;s
    athletic director said Monday." + ], + [ + "Global oil prices boomed on
    Wednesday, spreading fear that
    energy prices will restrain
    economic activity, as traders
    worried about a heating oil
    supply crunch in the American
    winter." + ], + [ + "Custom-designed imported
    furniture was once an
    exclusive realm. Now, it's the
    economical alternative for
    commercial developers and
    designers needing everything
    from seats to beds to desks
    for their projects." + ], + [ + "SAN DIEGO (Ticker) - The San
    Diego Padres lacked speed and
    an experienced bench last
    season, things veteran
    infielder Eric Young is
    capable of providing." + ], + [ + "This is an eye chart,
    reprinted as a public service
    to the New York Mets so they
    may see from what they suffer:
    myopia. Has ever a baseball
    franchise been so shortsighted
    for so long?" + ], + [ + "Short-term interest rate
    futures struggled on Thursday
    after a government report
    showing US core inflation for
    August below market
    expectations failed to alter
    views on Federal Reserve rate
    policy." + ], + [ + "MacCentral - Microsoft's
    Macintosh Business Unit on
    Tuesday issued a patch for
    Virtual PC 7 that fixes a
    problem that occurred when
    running the software on Power
    Mac G5 computers with more
    than 2GB of RAM installed.
    Previously, Virtual PC 7 would
    not run on those computers,
    causing a fatal error that
    crashed the application.
    Microsoft also noted that
    Virtual PC 7.0.1 also offers
    stability improvements,
    although it wasn't more
    specific than that -- some
    users have reported problems
    with using USB devices and
    other issues." + ], + [ + "Samsung is now the world #39;s
    second-largest mobile phone
    maker, behind Nokia. According
    to market watcher Gartner, the
    South Korean company has
    finally knocked Motorola into
    third place." + ], + [ + "Don't bother buying Star Wars:
    Battlefront if you're looking
    for a first-class shooter --
    there are far better games out
    there. But if you're a Star
    Wars freak and need a fix,
    this title will suffice. Lore
    Sjberg reviews Battlefront." + ], + [ + "While the American forces are
    preparing to launch a large-
    scale attack against Falluja
    and al-Ramadi, one of the
    chieftains of Falluja said
    that contacts are still
    continuous yesterday between
    members representing the city
    #39;s delegation and the
    interim" + ], + [ + "UN Security Council
    ambassadors were still
    quibbling over how best to
    pressure Sudan and rebels to
    end two different wars in the
    country even as they left for
    Kenya on Tuesday for a meeting
    on the crisis." + ], + [ + "HENDALA, Sri Lanka -- Day
    after day, locked in a cement
    room somewhere in Iraq, the
    hooded men beat him. They told
    him he would be beheaded.
    ''Ameriqi! quot; they shouted,
    even though he comes from this
    poor Sri Lankan fishing
    village." + ], + [ + "THE kidnappers of British aid
    worker Margaret Hassan last
    night threatened to turn her
    over to the group which
    beheaded Ken Bigley if the
    British Government refuses to
    pull its troops out of Iraq." + ], + [ + " TOKYO (Reuters) - Tokyo
    stocks climbed to a two week
    high on Friday after Tokyo
    Electron Ltd. and other chip-
    related stocks were boosted
    by a bullish revenue outlook
    from industry leader Intel
    Corp." + ], + [ + "More than by brain size or
    tool-making ability, the human
    species was set apart from its
    ancestors by the ability to
    jog mile after lung-stabbing
    mile with greater endurance
    than any other primate,
    according to research
    published today in the journal" + ], + [ + " LONDON (Reuters) - A medical
    product used to treat both
    male hair loss and prostate
    problems has been added to the
    list of banned drugs for
    athletes." + ], + [ + "AP - Curtis Martin and Jerome
    Bettis have the opportunity to
    go over 13,000 career yards
    rushing in the same game when
    Bettis and the Pittsburgh
    Steelers play Martin and the
    New York Jets in a big AFC
    matchup Sunday." + ], + [ + "<p></p><p>
    BOGOTA, Colombia (Reuters) -
    Ten Colombian police
    officerswere killed on Tuesday
    in an ambush by the National
    LiberationArmy, or ELN, in the
    worst attack by the Marxist
    group inyears, authorities
    told Reuters.</p>" + ], + [ + "Woodland Hills-based Brilliant
    Digital Entertainment and its
    subsidiary Altnet announced
    yesterday that they have filed
    a patent infringement suit
    against the Recording Industry
    Association of America (RIAA)." + ], + [ + "AFP - US President George W.
    Bush called his Philippines
    counterpart Gloria Arroyo and
    said their countries should
    keep strong ties, a spokesman
    said after a spat over
    Arroyo's handling of an Iraq
    kidnapping." + ], + [ + "A scathing judgment from the
    UK #39;s highest court
    condemning the UK government
    #39;s indefinite detention of
    foreign terror suspects as a
    threat to the life of the
    nation, left anti-terrorist
    laws in the UK in tatters on
    Thursday." + ], + [ + "Sony Ericsson and Cingular
    provide Z500a phones and
    service for military families
    to stay connected on today
    #39;s quot;Dr. Phil Show
    quot;." + ], + [ + "Reuters - Australia's
    conservative Prime
    Minister\\John Howard, handed
    the most powerful mandate in a
    generation,\\got down to work
    on Monday with reform of
    telecommunications,\\labor and
    media laws high on his agenda." + ], + [ + " TOKYO (Reuters) - Typhoon
    Megi killed one person as it
    slammed ashore in northern
    Japan on Friday, bringing the
    death toll to at least 13 and
    cutting power to thousands of
    homes before heading out into
    the Pacific." + ], + [ + "Cairo: Egyptian President
    Hosni Mubarak held telephone
    talks with Palestinian leader
    Yasser Arafat about Israel
    #39;s plan to pull troops and
    the 8,000 Jewish settlers out
    of the Gaza Strip next year,
    the Egyptian news agency Mena
    said." + ], + [ + " NEW YORK (Reuters) - Adobe
    Systems Inc. <A HREF=\"http:
    //www.investor.reuters.com/Ful
    lQuote.aspx?ticker=ADBE.O targ
    et=/stocks/quickinfo/fullquote
    \">ADBE.O</A> on
    Monday reported a sharp rise
    in quarterly profit, driven by
    robust demand for its
    Photoshop and document-sharing
    software." + ], + [ + "Dow Jones amp; Co., publisher
    of The Wall Street Journal,
    has agreed to buy online
    financial news provider
    MarketWatch Inc. for about
    \\$463 million in a bid to
    boost its revenue from the
    fast-growing Internet
    advertising market." + ], + [ + "The first American military
    intelligence soldier to be
    court-martialed over the Abu
    Ghraib abuse scandal was
    sentenced Saturday to eight
    months in jail, a reduction in
    rank and a bad-conduct
    discharge." + ], + [ + " TOKYO (Reuters) - Japan will
    seek an explanation at weekend
    talks with North Korea on
    activity indicating Pyongyang
    may be preparing a missile
    test, although Tokyo does not
    think a launch is imminent,
    Japan's top government
    spokesman said." + ], + [ + "Michigan Stadium was mostly
    filled with empty seats. The
    only cheers were coming from
    near one end zone -he Iowa
    section. By Carlos Osorio, AP." + ], + [ + "The International Rugby Board
    today confirmed that three
    countries have expressed an
    interest in hosting the 2011
    World Cup. New Zealand, South
    Africa and Japan are leading
    the race to host rugby union
    #39;s global spectacular in
    seven years #39; time." + ], + [ + "Officials at EarthLink #39;s
    (Quote, Chart) R amp;D
    facility have quietly released
    a proof-of-concept file-
    sharing application based on
    the Session Initiated Protocol
    (define)." + ], + [ + "Low-fare airline ATA has
    announced plans to lay off
    hundreds of employees and to
    drop most of its flights out
    of Midway Airport in Chicago." + ], + [ + " BEIJING (Reuters) - Iran will
    never be prepared to
    dismantle its nuclear program
    entirely but remains committed
    to the non-proliferation
    treaty (NPT), its chief
    delegate to the International
    Atomic Energy Agency said on
    Wednesday." + ], + [ + " WASHINGTON (Reuters) -
    Germany's Bayer AG <A HREF=
    \"http://www.investor.reuters.c
    om/FullQuote.aspx?ticker=BAYG.
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    has agreed to plead guilty
    and pay a \\$4.7 million fine
    for taking part in a
    conspiracy to fix the prices
    of synthetic rubber, the U.S.
    Justice Department said on
    Wednesday." + ], + [ + "MIAMI (Ticker) -- In its first
    season in the Atlantic Coast
    Conference, No. 11 Virginia
    Tech is headed to the BCS.
    Bryan Randall threw two
    touchdown passes and the
    Virginia Tech defense came up
    big all day as the Hokies
    knocked off No." + ], + [ + "(CP) - Somehow, in the span of
    half an hour, the Detroit
    Tigers #39; pitching went from
    brutal to brilliant. Shortly
    after being on the wrong end
    of several records in a 26-5
    thrashing from to the Kansas
    City" + ], + [ + "<p></p><p>
    SANTIAGO, Chile (Reuters) -
    President Bush on
    Saturdayreached into a throng
    of squabbling bodyguards and
    pulled aSecret Service agent
    away from Chilean security
    officers afterthey stopped the
    U.S. agent from accompanying
    the president ata
    dinner.</p>" + ], + [ + " quot;It #39;s your mail,
    quot; the Google Web site
    said. quot;You should be able
    to choose how and where you
    read it. You can even switch
    to other e-mail services
    without having to worry about
    losing access to your
    messages." + ], + [ + "The US oil giant got a good
    price, Russia #39;s No. 1 oil
    company acquired a savvy
    partner, and Putin polished
    Russia #39;s image." + ], + [ + "With the Huskies reeling at
    0-4 - the only member of a
    Bowl Championship Series
    conference left without a win
    -an Jose State suddenly looms
    as the only team left on the
    schedule that UW will be
    favored to beat." + ], + [ + "Darryl Sutter, who coached the
    Calgary Flames to the Stanley
    Cup finals last season, had an
    emergency appendectomy and was
    recovering Friday." + ], + [ + "The maker of Hostess Twinkies,
    a cake bar and a piece of
    Americana children have
    snacked on for almost 75
    years, yesterday raised
    concerns about the company
    #39;s ability to stay in
    business." + ], + [ + "Iranian deputy foreign
    minister Gholamali Khoshrou
    denied Tuesday that his
    country #39;s top leaders were
    at odds over whether nuclear
    weapons were un-Islamic,
    insisting that it will
    quot;never quot; make the
    bomb." + ], + [ + " quot;Israel mercenaries
    assisting the Ivory Coast army
    operated unmanned aircraft
    that aided aerial bombings of
    a French base in the country,
    quot; claimed" + ], + [ + "Athens, Greece (Sports
    Network) - For the second
    straight day a Briton captured
    gold at the Olympic Velodrome.
    Bradley Wiggins won the men
    #39;s individual 4,000-meter
    pursuit Saturday, one day
    after teammate" + ], + [ + "AFP - SAP, the world's leading
    maker of business software,
    may need an extra year to
    achieve its medium-term profit
    target of an operating margin
    of 30 percent, its chief
    financial officer said." + ], + [ + "Tenet Healthcare Corp., the
    second- largest US hospital
    chain, said fourth-quarter
    charges may exceed \\$1 billion
    and its loss from continuing
    operations will widen from the
    third quarter #39;s because of
    increased bad debt." + ], + [ + "AFP - The airline Swiss said
    it had managed to cut its
    first-half net loss by about
    90 percent but warned that
    spiralling fuel costs were
    hampering a turnaround despite
    increasing passenger travel." + ], + [ + "Vulnerable youngsters expelled
    from schools in England are
    being let down by the system,
    say inspectors." + ], + [ + "Yasser Arafat was undergoing
    tests for possible leukaemia
    at a military hospital outside
    Paris last night after being
    airlifted from his Ramallah
    headquarters to an anxious
    farewell from Palestinian
    well-wishers." + ], + [ + "The world #39;s only captive
    great white shark made history
    this week when she ate several
    salmon fillets, marking the
    first time that a white shark
    in captivity" + ], + [ + "Nepal #39;s main opposition
    party urged the government on
    Monday to call a unilateral
    ceasefire with Maoist rebels
    and seek peace talks to end a
    road blockade that has cut the
    capital off from the rest of
    the country." + ], + [ + "Communications aggregator
    iPass said Monday that it is
    adding in-flight Internet
    access to its access
    portfolio. Specifically, iPass
    said it will add access
    offered by Connexion by Boeing
    to its list of access
    providers." + ], + [ + "The London-based brokerage
    Collins Stewart Tullett placed
    55m of new shares yesterday to
    help fund the 69.5m purchase
    of the money and futures
    broker Prebon." + ], + [ + "BOSTON - The New York Yankees
    and Boston were tied 4-4 after
    13 innings Monday night with
    the Red Sox trying to stay
    alive in the AL championship
    series. Boston tied the
    game with two runs in the
    eighth inning on David Ortiz's
    solo homer, a walk to Kevin
    Millar, a single by Trot Nixon
    and a sacrifice fly by Jason
    Varitek..." + ], + [ + "A Steffen Iversen penalty was
    sufficient to secure the
    points for Norway at Hampden
    on Saturday. James McFadden
    was ordered off after 53
    minutes for deliberate
    handball as he punched Claus
    Lundekvam #39;s header off the
    line." + ], + [ + "Reuters - The country may be
    more\\or less evenly divided
    along partisan lines when it
    comes to\\the presidential
    race, but the Republican Party
    prevailed in\\the Nielsen
    polling of this summer's
    nominating conventions." + ], + [ + " PARIS (Reuters) - European
    equities flirted with 5-month
    peaks as hopes that economic
    growth was sustainable and a
    small dip in oil prices
    helped lure investors back to
    recent underperformers such
    as technology and insurance
    stocks." + ], + [ + " NEW YORK (Reuters) - U.S.
    stocks looked to open higher
    on Friday, as the fourth
    quarter begins on Wall Street
    with oil prices holding below
    \\$50 a barrel." + ], + [ + "LONDON : World oil prices
    stormed above 54 US dollars
    for the first time Tuesday as
    strikes in Nigeria and Norway
    raised worries about possible
    supply shortages during the
    northern hemisphere winter." + ], + [ + "AP - Ailing St. Louis reliever
    Steve Kline was unavailable
    for Game 3 of the NL
    championship series on
    Saturday, but Cardinals
    manager Tony LaRussa hopes the
    left-hander will pitch later
    this postseason." + ], + [ + "Company launches free test
    version of service that
    fosters popular Internet
    activity." + ], + [ + "Outdated computer systems are
    hampering the work of
    inspectors, says the UN
    nuclear agency." + ], + [ + " In Vice President Cheney's
    final push before next
    Tuesday's election, talk of
    nuclear annihilation and
    escalating war rhetoric have
    blended with balloon drops,
    confetti cannons and the other
    trappings of modern
    campaigning with such ferocity
    that it is sometimes tough to
    tell just who the enemy is." + ], + [ + "MADRID: A stunning first-half
    free kick from David Beckham
    gave Real Madrid a 1-0 win
    over newly promoted Numancia
    at the Bernabeu last night." + ], + [ + "MacCentral - You Software Inc.
    announced on Tuesday the
    availability of You Control:
    iTunes, a free\\download that
    places iTunes controls in the
    Mac OS X menu bar.
    Without\\leaving the current
    application, you can pause,
    play, rewind or skip songs,\\as
    well as control iTunes' volume
    and even browse your entire
    music library\\by album, artist
    or genre. Each time a new song
    plays, You Control:
    iTunes\\also pops up a window
    that displays the artist and
    song name and the
    album\\artwork, if it's in the
    library. System requirements
    call for Mac OS X\\v10.2.6 and
    10MB free hard drive space.
    ..." + ], + [ + "Favourites Argentina beat
    Italy 3-0 this morning to
    claim their place in the final
    of the men #39;s Olympic
    football tournament. Goals by
    leading goalscorer Carlos
    Tevez, with a smart volley
    after 16 minutes, and" + ], + [ + "Shortly after Steve Spurrier
    arrived at Florida in 1990,
    the Gators were placed on NCAA
    probation for a year stemming
    from a child-support payment
    former coach Galen Hall made
    for a player." + ], + [ + "The US Secret Service Thursday
    announced arrests in eight
    states and six foreign
    countries of 28 suspected
    cybercrime gangsters on
    charges of identity theft,
    computer fraud, credit-card
    fraud, and conspiracy." + ], + [ + "US stocks were little changed
    on Thursday as an upbeat
    earnings report from chip
    maker National Semiconductor
    Corp. (NSM) sparked some
    buying, but higher oil prices
    limited gains." + ] + ], + "hovertemplate": "label=Other
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"white", + "zerolinewidth": 2 + } + } + }, + "title": { + "text": "Nearest neighbors of the chipset security article" + }, + "width": 600, + "xaxis": { + "anchor": "y", + "domain": [ + 0, + 1 + ], + "title": { + "text": "Component 0" + } + }, + "yaxis": { + "anchor": "x", + "domain": [ + 0, + 1 + ], + "title": { + "text": "Component 1" + } + } + } + } + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "# a 2D chart of nearest neighbors of the chipset security article\n", + "chart_from_components(\n", + " components=tsne_components,\n", + " labels=chipset_security_labels,\n", + " strings=article_descriptions,\n", + " width=600,\n", + " height=500,\n", + " title=\"Nearest neighbors of the chipset security article\",\n", + " category_orders={\"label\": [\"Other\", \"Nearest neighbor (top 5)\", \"Source\"]},\n", + ")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "For the chipset security example, the 4 closest nearest neighbors in the full embedding space remain nearest neighbors in this compressed 2D visualization. The fifth is displayed as more distant, despite being closer in the full embedding space." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Should you want to, you can also make an interactive 3D plot of the embeddings with the function `chart_from_components_3D`. (Doing so will require recomputing the t-SNE components with `n_components=3`.)" + ] + } + ], + "metadata": { + "interpreter": { + "hash": "365536dcbde60510dc9073d6b991cd35db2d9bac356a11f5b64279a5e6708b97" + }, + "kernelspec": { + "display_name": "Python 3.9.9 64-bit ('openai': virtualenv)", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.9.9" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/Regression_using_embeddings.ipynb b/examples/Regression_using_embeddings.ipynb new file mode 100644 index 0000000..cf99894 --- /dev/null +++ b/examples/Regression_using_embeddings.ipynb @@ -0,0 +1,109 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Regression using the embeddings\n", + "\n", + "Regression means predicting a number, rather than one of the categories. We will predict the score based on the embedding of the review's text. We split the dataset into a training and a testing set for all of the following tasks, so we can realistically evaluate performance on unseen data. The dataset is created in the [Obtain_dataset Notebook](Obtain_dataset.ipynb).\n", + "\n", + "We're predicting the score of the review, which is a number between 1 and 5 (1-star being negative and 5-star positive)." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Babbage similarity embedding performance on 1k Amazon reviews: mse=0.38, mae=0.39\n" + ] + } + ], + "source": [ + "import pandas as pd\n", + "import numpy as np\n", + "\n", + "from sklearn.ensemble import RandomForestRegressor\n", + "from sklearn.model_selection import train_test_split\n", + "from sklearn.metrics import mean_squared_error, mean_absolute_error\n", + "\n", + "df = pd.read_csv('output/embedded_1k_reviews.csv')\n", + "df['babbage_similarity'] = df.babbage_similarity.apply(eval).apply(np.array)\n", + "\n", + "X_train, X_test, y_train, y_test = train_test_split(list(df.babbage_similarity.values), df.Score, test_size = 0.2, random_state=42)\n", + "\n", + "rfr = RandomForestRegressor(n_estimators=100)\n", + "rfr.fit(X_train, y_train)\n", + "preds = rfr.predict(X_test)\n", + "\n", + "\n", + "mse = mean_squared_error(y_test, preds)\n", + "mae = mean_absolute_error(y_test, preds)\n", + "\n", + "print(f\"Babbage similarity embedding performance on 1k Amazon reviews: mse={mse:.2f}, mae={mae:.2f}\")" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Dummy mean prediction performance on Amazon reviews: mse=1.77, mae=1.04\n" + ] + } + ], + "source": [ + "bmse = mean_squared_error(y_test, np.repeat(y_test.mean(), len(y_test)))\n", + "bmae = mean_absolute_error(y_test, np.repeat(y_test.mean(), len(y_test)))\n", + "print(f\"Dummy mean prediction performance on Amazon reviews: mse={bmse:.2f}, mae={bmae:.2f}\")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can see that the embeddings are able to predict the scores with an average error of 0.39 per score prediction. This is roughly equivalent to predicting 2 out of 3 reviews perfectly, and 1 out of three reviews by a one star error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "You could also train a classifier to predict the label, or use the embeddings within an existing ML model to encode free text features." + ] + } + ], + "metadata": { + "interpreter": { + "hash": "be4b5d5b73a21c599de40d6deb1129796d12dc1cc33a738f7bac13269cfcafe8" + }, + "kernelspec": { + "display_name": "Python 3.7.3 64-bit ('base': conda)", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.7.3" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/Semantic_text_search_using_embeddings.ipynb b/examples/Semantic_text_search_using_embeddings.ipynb new file mode 100644 index 0000000..259355a --- /dev/null +++ b/examples/Semantic_text_search_using_embeddings.ipynb @@ -0,0 +1,185 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Semantic text search using embeddings\n", + "\n", + "We can search through all our reviews semantically in a very efficient manner and at very low cost, by simply embedding our search query, and then finding the most similar reviews. The dataset is created in the [Obtain_dataset Notebook](Obtain_dataset.ipynb)." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": {}, + "outputs": [], + "source": [ + "import pandas as pd\n", + "import numpy as np\n", + "\n", + "\n", + "df = pd.read_csv('output/embedded_1k_reviews.csv')\n", + "df['babbage_search'] = df.babbage_search.apply(eval).apply(np.array)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Remember to use the documents embedding engine for documents (in this case reviews), and query embedding engine for queries. Note that here we just compare the cosine similarity of the embeddings of the query and the documents, and show top_n best matches." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Jamaican Blue beans: Excellent coffee bean for roasting. Our family just purchased another 5 pounds for more roasting. Plenty of flavor and mild on acidity when roasted to a dark brown bean and befor\n", + "\n", + "Good Buy: I liked the beans. They were vacuum sealed, plump and moist. Would recommend them for any use. I personally split and stuck them in some vodka to make vanilla extract. Yum!\n", + "\n", + "Fantastic Instant Refried beans: Fantastic Instant Refried Beans have been a staple for my family now for nearly 20 years. All 7 of us love it and my grown kids are passing on the tradition.\n", + "\n" + ] + } + ], + "source": [ + "from openai.embeddings_utils import get_embedding, cosine_similarity\n", + "\n", + "# search through the reviews for a specific product\n", + "def search_reviews(df, product_description, n=3, pprint=True):\n", + " embedding = get_embedding(product_description, engine='text-search-babbage-query-001')\n", + " df['similarities'] = df.babbage_search.apply(lambda x: cosine_similarity(x, embedding))\n", + "\n", + " res = df.sort_values('similarities', ascending=False).head(n).combined.str.replace('Title: ','').str.replace('; Content:', ': ')\n", + " if pprint:\n", + " for r in res:\n", + " print(r[:200])\n", + " print()\n", + " return res\n", + "res = search_reviews(df, 'delicious beans', n=3)\n" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rustichella ROCKS!: Anything this company makes is worthwhile eating! My favorite is their Trenne.
    Their whole wheat pasta is the best I have ever had.\n", + "\n", + "sooo good: tastes so good. Worth the money. My boyfriend hates wheat pasta and LOVES this. cooks fast tastes great.I love this brand and started buying more of their pastas. Bulk is best.\n", + "\n", + "Wonderful: Came quickly. Was plentiful and delicious and cheaper than in the store. You will enjoy it if you like thick pasta.\n", + "\n" + ] + } + ], + "source": [ + "res = search_reviews(df, 'whole wheat pasta', n=3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can search through these reviews easily. To speed up computation, we can use a special algorithm, aimed at faster search through embeddings." + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "great product, poor delivery: The coffee is excellent and I am a repeat buyer. Problem this time was with the UPS delivery. They left the box in front of my garage door in the middle of the drivewa\n", + "\n" + ] + } + ], + "source": [ + "res = search_reviews(df, 'bad delivery', n=1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "As we can see, this can immediately deliver a lot of value. In this example we show being able to quickly find the examples of delivery failures." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Extremely dissapointed: Hi,
    I am very disappointed with the past shipment I received of the ONE coconut water. 3 of the boxes were leaking and the coconut water was spoiled.

    Thanks." + ] + }, + "metadata": { + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "import matplotlib.pyplot as plt\n", + "import statsmodels.api as sm\n", + "\n", + "\n", + "correlation = X_test[['percentile_cosine_similarity', 'Score']].corr().values[0,1]\n", + "print('Correlation between user & vector similarity percentile metric and review number of stars (score): %.2f%%' % (100*correlation))\n", + "\n", + "# boxplot of cosine similarity for each score\n", + "X_test.boxplot(column='percentile_cosine_similarity', by='Score')\n", + "plt.title('')\n", + "plt.show()\n", + "plt.close()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can observe a weak trend, showing that the higher the similarity score between the user and the product embedding, the higher the review score. Therefore, the user and product embeddings can weakly predict the review score - even before the user receives the product!\n", + "\n", + "Because this signal works in a different way than the more commonly used collaborative filtering, it can act as an additional feature to slightly improve the performance on existing problems." + ] + } + ], + "metadata": { + "interpreter": { + "hash": "be4b5d5b73a21c599de40d6deb1129796d12dc1cc33a738f7bac13269cfcafe8" + }, + "kernelspec": { + "display_name": "Python 3.7.3 64-bit ('base': conda)", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.7.3" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/Visualizing_embeddings_in_2D.ipynb b/examples/Visualizing_embeddings_in_2D.ipynb new file mode 100644 index 0000000..acb6b88 --- /dev/null +++ b/examples/Visualizing_embeddings_in_2D.ipynb @@ -0,0 +1,142 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Visualizing the embeddings in 2D\n", + "\n", + "We will use t-SNE to reduce the dimensionality of the embeddings from 2048 to 2. Once the embeddings are reduced to two dimensions, we can plot them in a 2D scatter plot. The dataset is created in the [Obtain_dataset Notebook](Obtain_dataset.ipynb)." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### 1. Reduce dimensionality\n", + "\n", + "We reduce the dimensionality to 2 dimensions using t-SNE decomposition." + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "(1000, 2)" + ] + }, + "execution_count": 19, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "import pandas as pd\n", + "from sklearn.manifold import TSNE\n", + "\n", + "# Load the embeddings\n", + "df = pd.read_csv('output/embedded_1k_reviews.csv')\n", + "\n", + "# Convert to a list of lists of floats\n", + "matrix = df.babbage_similarity.apply(eval).to_list()\n", + "\n", + "# Create a t-SNE model and transform the data\n", + "tsne = TSNE(n_components=2, perplexity=15, random_state=42, init='random', learning_rate=200)\n", + "vis_dims = tsne.fit_transform(matrix)\n", + "vis_dims.shape" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### 2. Plotting the embeddings\n", + "\n", + "We colour each review by its star rating, ranging from red to green." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can observe a decent data separation even in the reduced 2 dimensions. There seems to be a cluster of mostly negative reviews." + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Text(0.5, 1.0, 'Amazon ratings visualized in language using t-SNE')" + ] + }, + "execution_count": 20, + "metadata": {}, + "output_type": "execute_result" + }, + { + "data": { + "image/png": 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GdV0CngDT2Wl8po9dg7vonehlbcvaV7vLFc6RipCv8JojHopTNZmk6Qc/Irq/F0/BAk0jUqfTbGSgvgCR1WA0w/zJwFgM3vlOmJiA3l7IZkHToLsb3vte6LrQ8reXP+lCmvHUOIVSAdd1KZQKcy/bol2kUCqQzCdxXZfj08crQv51REXIV3htkcthHjvG7b/cR2rrLhAaOZ8XD3mWjRXwPfwcrK+DJb3gvQL8NywU9NdeC62tyoWyvx/q62HzZli9euF6FRZgaAYFq0DAG8B2bRBwotaEQOC6Lj7TR6FUQNf0V7m3Fc6HipCv8NphagoefBDSaWp39RBJO2SrA8hAkJAfjFwAcTQNPTOw/Goo7gVPNxh1J9sQQgn5D3/41TuO1yGxUIx4OM50dpqqYBVO2sExHKSU+Ewfhm4Q8oaoCdbQGet8tbtb4TyoCPkKrx1+9SvlCdPcDKaJ6fVTZQGuBKEpAS4dKDjqOzrYYwuFfIULwufxcdfau5jOTpOzcmSLWYLeILlSjrA3TFWwisZII90N3XTFK2av1xMVIV/htUE2qwKWmpuVd0xrK4yMKC+ZXB58LuRsqA/B0hOapARxDgFSFc6Jm5bfxHBymIf3P0w8GGcoNUSVrwpd1/F7/KxsXsnHb/w4HsPzane1wnlQEfIVXhvoZTuv66rJ0je/GQYHVWqCtAWGA1VeuOEqWL0U3AwIE8zWV7fflxEBb4CPXv9RVjWs4ukjTzOTnWEiPUFVoIrbVt7G5iWbqQnWvNrdrHCeVIR8hdcGPh8sWwZHj0JjIzQ1Kbv6Qw8pH/fGWljvh9s6wJwCghC6GzT/q93zywohBNcuvZarO68mVUjhNbwEvcFXu1sVLoKKkK/w2uH66yGXU2kJtLIN/k//VHnGSKleBO4sSBf0mrJdvsLLgaEbFa39MqEi5Cu8skipcsmMjKh0wI2Nyv7u86nP296mUgQXClBVBcFTtEi9IngqVDgfKkK+witHJgP/+q/w9NNqktXjUYJ93Tr4tV+DpUuV9l5b+2r3tEKFy4aKkK/wylAowFe/Co88AtPT6ns2qzJFTk7C6Cj80R9BQ8Or3dPzZrRUZNqxCWga9YaHYCVYqMJriIqQr/DK0NMDjz+uAp6GhiCRUKabZBLicZXz/cc/hk9+8tXu6TmTcmweSk+xr5gj5zr4hKDTE+COUDUdnsqEcIXXBhUhX+GVYds2NamayajkYbquJldP/DY7C0eOqO+BwMJtpQQcQD/31ASyBFYvWMdAC4B3JRiXbpRgSZdfZmZ4MZ8m6dhql8CsmyZhl/iN2mYieuXxqvDqU7kLK7wypFLKBl8squ96WWCfSAvsuup32164XfEoFF4ENwVaFfivAc9LhNXLEqR/CvYIaFFwJsHqgcCt4F1xSQ5nomRxpJgn4doENB1DCCSStOMw4VjsK2a5LhB96YYqVHiZuWQ+aEIIXQixUwjxUPl7pxDiRSFErxDi34QQlTC5NzLLlilhHg6rvO/ForLL6zr4/cqLprp6Yc735DYY+DJMD4NbpX7L/gysgbPvy+oHZwREHCaHYeQ4ZFzIP6NeAJeArOuScm2QYJRHFwKBKQRFJJO2dUn2U6HCxXIpNfnfAXqAckke/jfweSnlt4UQfwd8DPh/l3B/FV5PbNwIK1Yo4Z7PQzqthH51tZp8jUbh7ruVCQdgdBsc/RswNBApmOiDlisgVAWFbeBpO/O+7CHI5GDvV8HKghSAC3UtcPXd4Gm86MOJ6jqGJtAEOFKiC4GUEgeJLqFWNy96HxUqXAouiSYvhGgB7ga+Uv4ugFuB75VX+RrwjkuxrwqvU5qa4AMfgLY26OiAlhZV4KOxETo74YEH4MorVZGPpx6G//d5+MVR2FcAEQZ/FEb2gmuAO332fWkh6HlEmYAiTRBthHAjjB6G0d2X5HDqDA9LTT8RzSDj2mQdh7TrYCLo8PhZ5g28dCMVKrwCXCpN/gvAp4Fw+XstkJBSnjCwDgHNi20ohPgE8AmAtrazaGcVXv9s2qQE/JNPwoEDylzT2Qm3365+F0Llgd/1K6gKg1EPQxMwk4PbV6pslNlhqF5+9v1YIcjMQLSc10ZKIAv+BhjdDu13XvShaELw9nAMHUFPMUfKdfBpgnrN5O5wDfWVJF4VXiNctJAXQrwNmJBSbhdC3HK+20sp/wH4B4CNGzfKi+1PhQvAzYB1HErHQRbBaAJvN+hVl35fdXVw331K8J7qKTMzo3LXNDfD1Cx4mqEmAxNJGJmG6gKIIviuPvs+9Kg6Blxw0+o3rRp07yU9lGrD5D3ROKO2xYxdIqDpdHi8BLSKP0OF1w6X4m68HrhXCHEX4EPZ5L8IVAkhjLI23wIMX4J9VbhUyBK4eSgNQ+5RyO8AZ1YtM1rAbICqD4LZ8fLsfzFXyGxW/e6vLXve+MCzEjy9MDMJ9Z0Q/7DS8M9GqAmiKyA/BcEaQAc0yPRC6y2X9DD8mk6Xx09XxS++wmuUixbyUsr/AvwXgLIm//tSyvuFEN8F3gN8G3gA+PHF7qvCJUBKKO5Wk5duFrLPKE2+1A9Y4Lgg9oFvrZrAjP0pmJegKIe0wR4HHNDji2ePjERU/wwf1K2Hid1qu3wQYlfBqgfAe459WfebsOV/Q3II5cGOEvCN11z8sVSo8Dri5RxX/gHwbSHEnwM7ga++jPuqcK5YPZB7WhXBdlLK1dDqA4QS9gDSA6XDoOnKZTH6wMXVR3WmIfOzcvtCfQI3n+6zHo3CypWwf7+Kgm24Hkb7YdV6uPVD4F3kxeA4qpZrb6/yw1++XJl7ou3wps/BxE6wMhDtguolF34MFSq8TrmkQl5K+STwZPn/Y8CmS9l+hUtAYTvoDargRmkQpKXs8LLAnFmDAjh5KA2AM6PS+15o9kfpQuYXyo3RaCkHP+Uh+6gq23dquzfeqBKU7dmjAqOuuUV53Xj9Slgn+8EuQLgJ/PVqovbgQTUKcBw1qRsIqJdEVxesXw9NlaCkCm9cKjNEbySkVCYafZ7Qc3OACeRAeABNmUgogfArbf9iPG3tSbCT4GlVAnpsJ1hF0AU0x6H5voWjBF2HtWvVZz6pIej/WXnCVoOxF8Gug4Nj0Nau/O6PHFETt16vctE8elRp+e95z8Igq0wGdu5UyzRNjR7WrFEjgZc6HMdGSolpVPzgK7w+qAj5NxJCgNkC9rTSoLW4yusipgEPUAKpqf+FoTRtzatSA5wvrqu08R1PQeIFaNIh2gtVITANsLMw8BglzwrSkRoMbMJmHUJfpAqR68DgY+CJgln2P5cStj4HdlQJ7GPH1CcQUMtSKSXoR0fh0CHYsEFp+i++CN/5jsppHw6r9MbJpCo1eM89J4OxTqFQKrClbws9oz0AtNe2c92S64j4I4uuX6HCa4WKkH+j4dsM6R+CMw5GXGnvepMyySBB5pQGb8RAlEvsXYg9/oUXlPCtqwNPAIZfgBEDNjcpW79RYlpapI58jmTnKhA6fqHRGr2NgP+KhW0VZqGUB988044QEAzDzuehL6G08NlZME1l5jl2TAn5sAemt0JewqFJeHqXirhdskS9iI4cUXb8gweVaWeRWA0pJY8ceISRxAj1kXqEEIwkRvjJ7p/w3o3vrRS2rvCapiLk32gYMYjcB8X94ExB5CNQ2Ar2qErkJSLKXu9ZCcFrwbv6/PeRyyktvrlZmV9kMwRtVZC7/yg0hcgGuph2k0QcQVFvQuomWafIcPJhlhr1CHOem6Smw0wKdk9AMgONMVjWDh4Jx/oh2KTMMYmE0tDjcWWScWcg8CTE6qCgw/CvoCMCg6iXRDarNPiBPqgxIL8X3vVuWHIdBE/ufyozxeDMIK01J4uGx8NxhmeHGZwZZEndKzuh67oufVN95KwcHbUdhP3hl96owhuWipB/I6JHIXCd+l9KKG6C/FaVtdGZAqMLQreAV2nY580Jf3e9vK1WpcxD5ihkbMj5sDLjBH0pnPAaZDl4yK97yZQMssUjhOYL+UMD8JXHIZOGcBRiUTjSD9U5WLEORmfUi8XjUVq8rpcneF9Q/7dsAGnApA7hCQhpYFUpjxyfAeYEeEPg0+D5n0FpCJa+DaIq22XOyqEtYsbRNZ1UPrXoKci5DsdLBZKOQ5Wu02768F+CYiITqQk+/+jn6Z/qR0ND13Xev+n9vHXtWy+67QqXJxUh/0ZHCPBdqTR2WQThuzDBPp9wWbN0nLKg95SFbkAFJ3miOPkEgZkpprpWLTQHCQ1XWiqHzd69sGULfPvbEI1AQ0yZbYaHIJeATDssaYDMFARM6FwOKRv27IWGKERrofsWZXPfswdmBiAxBnoBDkyAa6qJZykhXAvNLZCYhaIBQ8+oCNnsGNGShVsq4EoXbV7x8JJTIhaOLTj0lGMzaBXYWcgQ0HQius6kXaLPKnBjsIrQRQr6v37irxmZHWFp3VJAzRX887P/TFtNG6ubL2DUVeGypyLkKyiEoT6XAp8PrrpKTXLG48pDp1AHXg/Ew+Ak8frDJPzLQD+ZycKWLqbME/QsgcceU3b16Wn1sijZMBuAzi6oLUEiD0UJqR6o8cJUSlWXmshArA2uvgbGn4WhR+H4ABhFiFqga1AVBHcIcq6aqO1eBquWg3GiKIkJk7sgNwneMFWuw5rSEHtHs9TUdqELnanMFM3VzTRVNc31/1Axy8Finv5SgaxjE9R1lgs/tYZJwrE5VMyx4SJMK0MzQxwZP8KS+EnzkM/0EfAG+NWhX73uhHzJLrF/ZD/5Up4l8SXURS5B0F2F06gI+QovDxs3Kt/1Xbsgk4em5bB8DXgtQCeoVZGa3kPezSBKQ7joCCyaglehJ31KwLe2Ku+YQEB9MlkouhCphskk1EjwN4GZBV8IJhLQUgV3vUVlnzxaAnM/mFXgLcKsA+1BkJ3QNQtmDpatBb8HdA84dtkDyYKZEWi6di7fzQ3eKHVT/ewTOrZ0ubbrWlY3r0Yva+azTomeYo5a3aSvpLJU2kh6SwXW6QYRTWf0InPMF+0iQgjEKRPhpmaSKWYuqu0TTKWneGj3Q+wc3EnEH+GutXdx7ZJrL0nb8zk+dZy/+OVfMJ1VGUUFgndd9S7es/E9l3xfb3QqQr7CJUFKuVD4CAHd3eqTn4FD/6YEsa48UbTCLI3RbjxNd5Ao9uHBIuZrI2g2wVTfSRNOTY3KOX+iJmyxqGz+hRysWAMzLhwbgcExCAbg+hXYAQd7YABvohMRyUJoCkIlCHnhkAndXdCVULb5lAVWCrQk6FlY2QKpHSpCdl5CM80TYkW4ihXL3wSB+GnHP2FbGEKgC4FHCGwkptDIuzZZ18EjNPxicffMc6W1ppWQN0SqkCLiO+m6mcwn2dix8aLaBkjkEvzxg39MMpckHo4zmZnkc498jg8kP8A7r3rnRbd/Atd1+dJjX6JoF+fMTpZt8Z1t32FF4wrWNK+5ZPuqUBHyFS4CKSUDpSKHrRx56VKve1jhDRA9tbapvwbaboXBXwGuEtbeKFrnncS91cS9p0S9BoPl9MAoN8i+PvX/6KjKVJnNqkyWh38JMghrlkGsGmfXQYZHkuxZ30ndyGH84RpaJ1ZRNTIGtTOQC8FEHlpyqr2VyyC6BiaPQ/ZH4PeCdwawlMCXrgq8mjtgF7JjkOwDM6gmZst++9qJdA1Ak+HlqJUnoqnvLpB0HTb45gVkXQAew8PHb/g4X3z8i0xnpvEYHjKFDGub13LDshsuqm2ARw88ykx2Zk7whggR9ob50a4fccfqOwhcohz5fVN9jCRPziuAOja/x88zR56pCPlLTEXIV7hgjloF9hSzVOs6tcJg1inxTC7JzYtNMNZ0Q6RdZYbUDAjULRSg86mrU6aaoSFoaIDNm2H7dqiKwLVrYdNmtX3PMxB2QEioiTLYUEeiIIknS3hrajEOHuRQPMaaQCfBp74DzhBYJuyyYfMVygTUdTU0DpFyP0jCNtENk5iVxHv0J1AaP1lFKjcJmWEYfAoML0gHRl+AJfdAoI46w8OBYg5bSuK6Scl06S8VsKXElpLV3gCthmfxFMvnwaauTXy26rM8feRpErkE69rWsbF94yXx1T80dojoKXVpfaYP27EZTY5eMldR27ERnH4OTM2k6BQvyT4qnKQi5CtcELaUHLJyxHRjrsZpRDeYdWz6rQJrfItErho+CLec0lABxg/D6BAE49DYAuTg2lVwuAYO9KigpbtWw8qssp+zDXolhNshXITMMCUpGF/WSe3BMZypWfJVUfyaht8fYHJ2hmDVCiiMwHoBjTbM7oZEM3J2Nwc1jUOiDiEk2KDrQa5vWE/19EHwuIAEuxyMFZ0XLGWlYOBx6H4fUd3gSl+QPcUsrpR4hMZKb5B1vhD1motZ3An5/aotzwrwbVTRxhdAa00rH7jmAxe07dloiDZwaPwQzBtw2K5K41AdqL5k+1kSX0LYH2Y2NzvXruM6JAtJNnduvmT7qaCoCPkKF0RRujhSzgn4E/iEIOGeY7Hs3CQ8+VV4bo+yaeTGVY60298EDbVQHYUPvge0FGQeVAnOXAF7++Cp52DnFHRfB8s3YocCULIR1OJcfx1ubS0zb34zYudOPI8+Dt0roGUz6HtBCyp//STkBn/FVFhSXxNB6kq6FaTgxVA3t8c2o4sloJswth2Ks3Ndl0DJE8FID6NZafBG6PD4qTc8JBwbTQhqdVM9YJmHwB7G1eJMZ2ZJjv8M3XyB2oaPEwlcYOK3l4HbV93Ok4eeZCozRSwUw7It+qb6uHn5zdSELl0/DcPgN27+DT73yOeYzkyjazq2Y3Pj0hvZ0Lbhku2ngqIi5CuoQKLdu1W4//Llyg5+NmQJX3EP7dln8clZLK2KrOdKXHMJeQI0nksRaymh56ew5TA0tIHMwuw0FG144Qjct0q5Xg7+Chq9SqoW98Khg3A4C50dMJ2Gvbth9248V19NyGeQXr0Ke7kqD+gCk1deQe1MQnnkZHdAwYLCMShOQDZESWugPZOm5B+gaHaS9KzHhyDjuiR9K6jxNiClZNbwkyuk8KNhITggvGSkwOuJ0+3YdJQnnv2avjDoyZ6A0iCO0cSugV2MpcbxGCZeeZznhv+BN6398AI3zFeTtto2/uCtf8DXnvsavRO9mLrJW9e+lfdf/f5Lvq91bev43H2fY8uxLaSKKdY0r2FF/QoM4+wiKZFL8NM9P2X78e1EfVHuXHsn13RVagScjYqQf6Nz5Aj8xV+oLI4neNvb4P77F1/fzcHs36MXttBmzTDgVBPUXYKFFxgyNqIH30x7cO3i287HSsFQv0qA5jEhMQ26DwLA7JQq/9dcB+khyDrg7FDBVEPT0GqAfVj1xRuHyWn0Xbuove5anl9/Jf5CEs/gQdKTE4SEl9apceXVU+iH7BaVtsEoQfU43rFZtOgySloDXruPsCyQMq9AelcijFpsKdleSDPqb0NkZ8jiZ1R4WEOReHEay1fLLtuFUoHOxapDuRkQGhOpCcaSY8QjyjPHi0NG13is5zHuv+b+RSNqXw3WNK/hL977FyRyCfymH695fiUTXddlODHM8OwwAU+ArngXoTNMOMfCMe668i4OjBzgi498kV1Du/B7/NxzxT381pt+C5/Ht2D9TCHDnzz4J4ynxolH4oymRvnLh/+S9218X8X18ixUhPwbGduGL31JZV5cuvTkbz/+sUr1e8UpicJkCVLfgtxz4MxQZR3DK1xyjpeCFqITSSx3CH/4z0F7icAWoSlV+9T5NylVxO0J7xokyKT6t+QDR4dSEIYPg0/C0uuhtQjRKA3xOm7evo1jDJMbTbLSLdE+M4Z3KgEjI9BwQOXPF35oleAJY7gF/Jlhkt73I2UHfneMCd8mAlqYqGZw1MozVkhTRwk8YWYKSSwzyLSTI+wJ44mvoUpo7CtkaTO86KcKay0CuIylxvHPewkIHEwzTjaTJZFPUBN87ZhtAKoCVee9jeM6PLL/EY5NHcNn+ig5JV7se5G3XfE2GqsaF91maHaI3/zGb2LZFs3VzZTsEt948RuMJkb5y/f95YJ1nzr0FGPJMZbWl+9VH0T8EX68+8fcsfqOSg6fM1AR8m9k+vpgauqkgAcwDOXC+Nxzpwt5axAKPWAPQmkIIfMEZI4ABrhBcEpgrFZFQqo+fPZ9e8LQ0Q07joJdrSY1C1NQkhBogliVyj4ZqFWTn3ajyjFTSMPeQZWvJhCBffuUH31HB9TUUPPcD6jRHWhtV9Gt8SYI6JAeVwFRRlBF9paSMDaLqRkEghHSuCD8+KQXR8LVvjBabpy+iUNEk8eUz7ymkdeCVHnDTHg7aQ3GmHBLDNs5Uq6DIQTrfCEa52u/ei2YSwlrR5iRBQQ+vCQpylrysgoYx7hMCn/3TfZxdPIobbUnJ6czxQyPH3yc9296/6Kjle9v+z7ZYpaueBcApm7SFeviqSNP0TfZR2e8c27dA2MHCPsWCnKv4cV1XYYSQ6z0r3yZjuz1zeVxd1W4MFz3zO58jrPI+pNKwLt5wAHK/uaUgHJ1KXtEpTF2s2qC03GUFj06qnLatLcr10WA1W+D4RF4fpv6nrXVJOcN7dDfAyUNVt8DIgu+K0DMQOm4KjoSrYW8HyxLtV9To37PzkKsEco+6kgLjFEQk2CGQctCYkIJ7VoJKR1/wWJ9oUQ2JNDMRtYEGzGGnoDpA8h8FiE0MPwkq7sZMcIMlxx03aJYSOICEc2gVjfwoPFsLskSj5+MdDDR6PT4qAu8iaqYTt/s18E1SYilpGUHE+lpWmpaLpuc9Ecnj56mTYe8IYZnh0kVUouODg6PHybgWehlpJcT2w3ODC4Q8g2RBnYP7l6wruM6F+f9Y9swPKyUnVJJ3a9+/5zScDlQEfJvZJYsgaoqFWB04oZ2HBVsdM0ik1muA2jKxCLd8o8aSuALJVAJlKtN6eqhefhhVYHJ61UPlNerinPEYkqbv/v34KpeGBkEbxWEfPDYLyGZg0A1jDwDnTZsSEImoOq0Nrgwsh8ekVBfr7JPJpOQkLC8HaYLJ/tcGoT8CGQt8HfD5HbwhiFtgjUNoSQk6/E++m28b7kTGt4PiV6YPQJGkA5nhoORToqOxVOuyQQ+MprEdGxmcynqTA8dpp81viAeTWOkaDFsW6zxBsjj8lwuyRpvkGWxN9NsNfJM7zNIV+LIMZqiTbyp+00v19V9xZBSknBtZhFMWgVCfhezHAMhpURKOZf+4VSW1y9na9/WBb/Zjg1Ae6x9we+3dN/CwwceZjIzSTwUx3Zs+qf72di+kYZow/l3vFiEb30LenpUTMbIiFJ8mppUrMb998Om138F04qQfyNjGPBbvwV/+ZdK0IOyhd92m0owdiqeFqWdC5+ya8sCcKI2rKc8oVkDRj1oPjjUo0xC7fMe1kQCfvUreNe71ChCaCqvTZPyiOHxx0GGYfmyk9v0D0J9HUQGwZeESCN4Pgw3ZlSxj8FBKBRU9afudTD8U0j4IeyF5AHI5iDcCFXAeAmsCSX0aYH0FaCHcY6XyD0XxXxXLb6ZZ5X5yM7RZU0z5Dbxy/ASxjUfRQQ6EgnkkEyXLJpLRQqFJDl/lKIUBOd52PiFRk8xT5vpY1XTKrriXcxmZ/EYHmqCNSdTQUgJ0qJouxRsi6A3iHFq5PBrEFdK9hQy9NtFClVNHBrYQcLwsjoQJajpjKfHqY/Wn6atn+A9G97Dj3f9mIHpARqiDVi2xWhilNtW3UZ77UIh31LTwh/c+Qf84zP/SO9EL4YwuGnZTTxw3QMX1vmf/AQeekh5l+3apZSQ6mpVY7hUgr/9W+jsVEn2Xse89u+iCqczugWOP6Zsyx23Q/26C29r9Wr44hdhxw51s3d3Kw1/MYwm8G8CZxrccgFwEQEsNcFo1IEWhuAdav3e3tOHvFVVaniczS6suwpzlZqcujq0+blw4nVwoADv/BQ8/jWY8sHMbDmHTUFNHBeLah7hSAya62HgEJi2Sl5W0wSrq8CbBbMesmlIOqCvgOJyEtkMhzpbGQtGEWMDtGh+1jpFTG81pqYRs1KYrgRdEMbBIx2k8DNtW0xNHWVHMUsiP8OMaxNpXs+m1pPeRboQSCHJSgcvGj7Td/okpDWEk32arf072T0+DkYrpreT65fdQHdD94Ve2VeECadEX6lInW7gSBeP6/DcnofY7QuzOlKPZVu0xdr42nNf46q2q7ii5YoFtvnmmmb+34f+H1967EvsGNhBwAzwwPUP8KmbPrXo/lY3r57z/jmRgfOCKBSUg8HMDBw/frIeweSkSnHd3q5Gh1u3wl13Xdg+XiNUhPzrjRc/C4e/q1IDSAmH/w1WPQAbfvvC24xE4JZbXno9YUD4HWBPqbKBZgPYY6A3lL93QvheMDvU+idK8c3HLZt5bFvVYQ0G54qLTJYsZqwCiVwKj2nSbHiJGybCdZUgD4YhEoMf/Ug9kH196uHUdfXC8PmUsF96Ayyvg7oE1N0M1kHQp0CvgiYB+8cg3QiBJLnCJH3eIFZnJ/GChetKBkNtMDHNVaZBpu5q9hQLTAsdCw1TumAGEJpOKTeL8IbRpaTG46VYzHGs/0VuruuCsieNlFIFuZ4phYM9CdkH2Tk8w/axFM2RBnQ5SVGP8tCehxiaHSLgCZAr5hhPjWNLm+X1y7my5coF3jqvFkOlIgFNMDzZxwsHHyEWreeW6rs5NHaE/WMHuGv1HcprxinxbO+zSClZ375+QRvdDd38zf1/c3qSuzOgadrFB2dNTak01smkMiGapvoLSnkYHVW2+RMj3NcxFSH/emJqHxz6LkTa5rI54hThwL9A113KXn0OOFIy69hY0iWiG+dXyMKog9rfh9xjUNgF0gY3AWYXBN8Kno6Tk7mrViltvkpAoA/MSRhMQL8L/3Grmohdtgze+lZm21t5tpimecUyansOkWtsoLeUBympm5iAm25SWtbMjLLpn7Chzs6qz5o16oUxPQjPfBfuugJSY1BbhHAdJIZBpMHnwpJu6NNBWoy2VzN15TUEvT5wHLRolFq9mqFigq7Z/bzgqcEORAjrBlkzQAaQwgArj+vaREwvtTkLqel0+kLYhSAj08epD8dwpWTGtWkwPQvO8WR6kj2De5jOTtPsHWVVzGDX6DiNkRo16ShryOcOcWDYoW+yj6AnyLHpYyyNL2V923p2DexiaGaIt697O6ZxDoFnLyMa4LqS/ce3Uh2K4z3xckNSE6hiPD1Oc3Uzpm7SWNXIjoEdrG1Zu6gp6lwE/CXDNJW5MpdT9+HkZNlkJtWyqSkVFNj92h5JnQsVIf96YuRFZcM+IeBBeYkICaMvvqSQz7sOB4o5nsklKLkSrxBIBJ1eHzf6o1Sdq8DQPBB6KwTfXA72CaqgplNpbYXNa2Doy5DWISmhrwfqXAivg3EXdu+CTIbj73sPvtoq7I0bKSVSBIeH8QqYcl1iq69EW7lSTZDpunoos1n1IJ4Y+g8OQkMNGBnIeKHgh+hyKE1BwYaGZeD6oZiF1VdAfQF27mBqyQo1iZxOwbp1YJpogKzppj/SjlVM0Wl4GXJ1LCvPrGORLeUxnCLeYopYYpprDQ81PmW3NbxBNOky5ZRACtpML6vm5fEZmhniwd0PEvAECHgC9Iz1sG/YJmtL6sNVAEh09owMEPQuBaGRKqboru9mKjPFTHaG5upmBmcGGZwdnHM9fNlxc+AmQQRU+cgyraaXI/kUx6ePk7dLCCBW3QKlHDWBapI5FeOwf2g/z/c9z0hihJ7RHj64+YOsa1t31l1atkXPSA/7R/YT8UdY17qOlprTo7Fd12UqM4XP8BEJnKOnUjyuPGiGhpSHVjSqNHtNUyNCx1HzUqtWneMJeu1SEfKvJ3SPEuin4roqM+JZKLouT2WT7MinyLoO006JnJTU6ybjjsWMbfOWUPVCH++XQnhU7dYz8fyfQOYAbL4J8gHY/iz4DQhVQfUEFBtAexpy+3D2deN7081Ir8HsXXdiTkyi5XJMBH10/9M34Wc/g09+UmlaiYTSvGIxmJhQLpmTk5Acg6X10BCBRArW3QBRE4afUYLesKBlPVS3wIYJ6LybmtRKjvs9BON1c66dlnTxIsiaQSy7RF4YXOMPoCMYzsySlII6TZDMjHJFaoBqmWc00k3vzBCjyVHeFY6zWTOoCkQWmGmklDx39DmqAlVz/t7+qi7GZvaRKZRIFXJEfAHyVppsyUVoUFVO5CWEIOANMJYco7m6Ga/hZSoztbiQz2Tg0CF1burqlDZ66vzHuSIlFLZBYTvKrcoFcykEbwFhEtNNjh96mt0TvXgML5pmcGx2kA6hMWt46Ih3sGdoDz/b+zNC3hCNkUbShTSf+fln+KO3/dEZ0wqX7BJfffqrPH/seUzdxHVdHut5jAeue2DBy6FntIevPPUVRpOjCAQbOjbw0Rs++tLBXJoGH/mIEvLDw8pkGYspgV9VpTxr3v1uNTp8nVMR8q8n2m6FnX8DVlq5HwIUk2D4ofmms256xMqxLZ/meKlA2nXIuy4CSLkOVZqBKyUBXeNdRhz9UgybpYRiAvZ+V+WgWbIJ8j0QyoBWAsMG/zB4dsHMGqrTGcZdF4+ugRCU6usouC6mdNESCRWZWxiFVQ2Q0ABbTRBPTCgTzswMxAIQckE4sLpbuVdqGtSvgq7bQE+AfQjkLDnPeh6Nt7M7UmTCsQjYOTZYOiFdx5KSjlKOx3qfY1T34Hcs/J4AV9Uv4crCCBO+GLfKHFSFeXimwJaJY/QOHCUQamJN4xqmMlP8bNePeO/G9+LxnhQSJac0p4nPYbZTHThO0Z4hm0/glkaR0iVpBajyClzH5cDwAQJepfmvaFgBgOVYVPmrTj/viQT88IfKOyQYVCOc3bvhne9UniOn4EqJjcRAoC123a2jkH9RJYc7EYlc6oV8EALXcWzyGDv2/IgN4Rh9M4P4TR9e1+XYdD8dsQ5CZohHDj5CwAwghKCtto3qYDVSSn6w4wdnFPI7ju/gycNP0lHbgcfw4EqXydQk39zyTVY2rsRreplITfDZn38Wv8fPkrolOK7D9uPbSRfT/PE9f/zS9+j69fDf/7tyozx2TJlpliyB97xHBQK+kuajl5GKkH89EW6Ga/4bbP2MyuAIKn3vDX+m0vSehS25NI50SboORdchV3YD9EhB1rUpSo3DxRwz/hJx8+JzkyME3PJ5KA3D3u+pHDSBdpgtgq2DNQCeKcheBVMbqF/WzQEg6dgENZ2idMm6Dpv8EcRf/g8o7oa//w588DqlrU8MQ/UKpbXOzMCqJlgVUTb3G1pg5VIQebC2gDcBjh9ELQTfCUaM7yfH6bcK1OkmdbrBsF1iWyHNOyO1dAidn+76EfX+CFagCgnIYoadQ7tYFvKxWRZox4FIjPtW3MBfJ4bYHO8k3rKRgC+ElDCVGufQ2CGuaj/pimpoBl7TS9Eu4j0x8tIC5PS1XNHYw1W1M4xlimTtMEXHwzf27CZVnnB1pIOpmRSsAn6Pn47ajtNcDAGVd19KaCx78EQiyqy1bRvcfvuCVQetAgesHEXp4hUaKz0B2k7JF0NxjxqtnSjuLoSaaLf2g/8aDo8fRghBY7AGv9AYTYySL+UJmAHevu7tNFc38/2d36e9tp36SP3cCCYaiDI4PXjG2+f5Y88T8oXm8uRrQiMeidM31cd4apy22jaeOvwUtmMTD6l7X9d0uuJd9Iz00D/ZT0e846Xv0XXrlIdZMqm+R6PKVn8ZcXkdzRuBZW+Dlhth9Fn14DVfD56zD8VzroMlJWnXxpWSEsrkowM2kpKUZByXqA4Jx744Ie9YkBqAQgL8tXDTF2BmOwzuh+g0eP1QnIZSCvLN0NMF164huHoNN3lMjlh5Jm2LsG6wzhcmrtkw9dfw35rAWQtfeQ4+ugk2L4EHdygPm2uXwpolEA7Bje3gmYDBJ6DeUuatxjeBWQ1OErI/ZTTwLo5ZBZp1E1G26bd7dEZLRaSEdGoU27WJevx47RzjmoekL0TKyrIsM0y32QKamr+Q0iXm8RFvvoIhzeRgPoMUyrBRNTu0QMhrmsaGtg08dfgpGqsa8RgeclYOWRrjio44ofAtLK01SOUT7Bn5MgFtjD3TMxSsAlJKAp4Ah8YPEfaH+Xc3/rvFk4cdOzbn1+1Kl77sOIdKg7BzB91XdtEZ60TTNEZKRbYW0tToBmHNxJIu2wtpdCFont+uLAKnztXoqmgKjvLwKVsQqwJVc2aS3oleltYtZXPXZlWaULIgJUEil6C1tvWMt5GUEiEXatICAZK5wKrJ9OSi50ATGsl88oxtn4ZpKlPNZUpFyL8e8UeVN815ENZ1ClLVHRVSInFxUWXrdCFwAL8Q6NpFDFGtNBz9CRRTysXTsVQO9uxy5dmSHDu5brYVikvhvffClQ3Q9wMinggb6tcvLCySfVaZB8wO+PMuFYj15S3AFrX8lnXwjtXQ1AXLgqCNQdEP2WlVVKT6GpXdEtSEoT1EwRpEE8E5AX8CQwimHJuoU5qrXOSTknanCE6RoXyCWHM3WqZXmcg0A39hBhFqpFeYJN0SEU1DCMGolWPI8JF3nQWph9e2rEUi2X58O5ZtEfKGeMvSBqqCYeWiChydOs7jfbME9SzSdQl5Q0ihhF7ADFAsFRmYGaAuskgSuGBQuZEGAjw9fYC9qX7Crs5MbpLvPPRntNW0cd/V9zEWqieq63NzBh6hEdF0DhdzC4W8ZzkUtoJ20sSUzg8wmMwxMPw4Vf4qTMNkJjszl2RtOjNNyBviypYrEULwnvXv4a+e/Csc6RD1R5nOTJPJp3h3y43KlBQIkIhFODTbz2xulqZoE1e2Xsm249vIWTkCngBSSsZSY7RUt9AYVaOUlY0reeLQEwtvQdtCIGitOfML5I1GRci/AfALjbhuEjNMCq5LXrg4UmlGvrIttko3qdVNas4lF/yZGN0CpZwyK4FKWLb3EdhjQeRKqHrs5Lq+9bDpZkjsgkMh8i3LmLDGyU59k1z77bTGVlFv6GAdAj2uvHjcGfivcfjyvH3+3Z9CfgyMYbD7gYjyd3ZGwZhd6IkEIAXVQvnuO667IGukJSUthkl9pB5HOjiuM6c1ngi1r2q7AewrVdoDx8LbfAPdVeN84/BTtEUbwPCQziXwahpN9csYtS265mefFIIrW69kTfMaLNvCa3jR0t8CTgrWwxPDWFLi1Sy8hoFpeLGljWVZFEoFHOkwPDu8ePHuq66CRx9lOh5hf/o4zWY1w727GG6OUheuYzgxzE/3/JRifCm3rnzzgk29QmPWPSVnkXc1lI6BPQTCz3R6hN3DB+nLryXrHsYqWaxrXUfPSA9HJ44CUB2s5vff8vtzgUo3rbgJ0zD5/s7vMzAzQGdNG7/tv4rVz/cgdZ2jyUGentzP7Ma11LR00T/Vj8/wsalzE3uH9pLIJSjaRWqDtXz8xo/PBVNt7trML/f9kiPjR1SRE8diNjvLuze8+5IWOXm9c9FCXgjRCvwLUI8auP2DlPKLQoga4N+ADqAfuE9KOXumdt5oSClxpXvGnB6XEiEE1/jD7C1maTLBKkl8UmnvJoJW3cSva6z1hYheaH+kq3K+BOflEJE2jFjg0yA6tHD9sWfAbIdAhtJkjtH+CWav3oDPH8c78jzPBprZ6PPTBqA3QvZBIAh/dmRhO1/4IXxsGTAGRnlewsqBtxPEBNjTJ3+XLgiXKm8TV7mCrfk01RgYCKZdm5hhsMoXxpdOc+Osh74XHqZUHyPRUkfB1Lhp+U0EvUHwBhcc5+pgM2udEuMj+0lmp6mramZ1x9UIX4TMqUKzjK7pJ4OZzCXK9q2V7ejCQ8SjM5r3YLlFisUMUkosxyLv5LFs68ypibu7IZdj5tmfoSVTFCgwGA/i71LumF7DS8Qf4ehMPyPpCVoi9XObZqVD7an+65pPBcBZx3FLozwzPMqh2S76po8hEEgkAsHHb/o4EX8EDY22mjaKThHLtuZs6tcs2cyqzqsBCPf2oj36GG5rC09M7uXnqS0IN4+290UGZZpNnZtIFVKsa11HU7SJJw8/Scgb4vql1xMLnzSr+Dw+/uvb/iuPHHiEF/teJOQN8dHrP8rmJZUSgvO5FJq8DfxHKeUOIUQY2C6EeAT4CPCYlPKzQoj/DPxn4A8uwf5e9xwZP8KWvi2kC2lioRibuzYv6v97UcgSFHvUB4h6V/GBcBeP5tP4hUZWOkhAl1BrerjKF+Zqf/giAlJE2fvCOVmgW/dCqgDeHhCjoK0D7TqYeBCah8B8BoJvJhHUEFNT1A4Ok1/Rja84S0xaHLB0mo0OdOsplZ/+T7bBP/bDrzfBn78H/nQ//N0/QeEO+FgUTF0Jct2E9k0g9yotVGiABm4KvGtBj3FHUBLXTLYXM1iuZLMvzHXBKL7paYrf+zeiqQliJYHsOU73ZIn6D32CSF0zrusqTVK6YI+CPUxEemhvWMKVzWsRnEzGNWGXqDmXNMK+tWVteRhEgFXxCC8c8zFt1SHlAEW7iECoIh66d26CcfHLIOCqq/A1RXC3/4iE6SEzeYiaEwnDkHgNL42uy2w+SThYi1/TyLsuJSSbTkkT4LgOfZPH6Z3spVgq8szADKlCilgoNqdRD80O8WjPo/zhXX/I7sHdfGPLN7AdG13TWd+6niUtV7CjmCUjHQQQGB9mY6yW6ew4e1N9BKRGtLoJM5nieMliz/Ae1reu57mjzxHxRbil+xY8hofZ7Cw/2qm8lk7Y9wPeIHeuu5e717/9zFHFb3AuWshLKUeB0fL/aSFED9AMvB24pbza14AnqQh5Do0d4pEDjxAPx2mubiaVT/Hg7gd594Z3Uz9Pq7oopITsY2AdU/nMAXJP0egZ4Z3hNzNQKjBcsrCkS61u0uH1U6+b5yXgXbtEIdmPR9oYgVrwxyG+FsZ3KJu668KBA+A9CKFRGGuGiQZYloOeTrgiBuyEtEsu8GYIhzEnJigs60JqHgzDj+VKrMA1+DOPKIH+j/3wsZXwP98OgU3wv1aAdxl86e9Avw7+6/UqXiAUA8MDVjsENipTj3TAezMYrSAEhhBsCkbZFIwuOK7ck4+xY3Qvab+Oz++jaEJ29Ch9P/4Hhla24zW9rG9ZxxXVk2jWNigN4nGTrJKt7PG9g5C3G8N1SUuHGt2g4VziDrQghN+l3BWdUZa1LuOK6UYOp35KLJvDdm1cXEKeEA2RBlY1vnSATmO8g+pYM8OzwziOQ8F1mMwnMXQPfn+Uol3kTdFGMoaHGcem3vCwxOMnOk+TL1pFvr3t2xwYOUB9uJ6gL8gLx16guaZ5wXxA2BdmJjNDz0gPT/c+TXOVinDNF/M8eeQptttFuhpXEi+bAnO6xvPVIZzBw1SZYZJCx3UdkAJD9zI8O0xtoJaRxAgrVqyYi46tDdUymhzl8NhhNnRsYKJksaeYIee6CATtHi+rvMHT6g6/0bmkNnkhRAewHngRqC+/AADGUOacxbb5BPAJgLa2tsVWuWyQUrK1byt14To1VM9PE5nci5MaZoc9xVtv/F3wXWRucWca8lsg+wh416jhNhpobWAdJeRbzypfHat8L9nSGekd3sNzL/w9ufwMHs1gQ7yVK5begmi5SfnGJ47C9Cwc/RqERiBzBYy2QakIh1Pgb4GWW4HvQfKXxFwY065Dmjpmbox0w2ZsoWMIB48WgT8fgK/2wCdvgc98WGW6lK6aqPz8FwCP8qM3vPC//73S3O1RMOvAt+Gk+99LUSoxvH8rOZ9JbVD5lGtCY3dxnNiBDM3XXYdlWzx7+CHs+Dgb6x0VAWrGWOKkCRe+zjHzU5SMNtaYQVo93nMXOJoPfKuB1fiA9169Fts1+d6O72E7Nj6Pj2V1y1jfup5cKUfJOXuxdEM3uPuKu3ms5zFeHN5Lz/gR6qtbaK5fwTPT/ayvbmVpVRM7Bnaw6+CTFOwC13Zdy43Lb1ReP8Uc//L8v/DEoSeIhWKkCikivgj10XoGpgdorW7FZ/qwbAvLtmiJt7B9YDvxUBxXuuwe2M1IcoS0dMkmJ/hUbQeUywAGmluY3LOLgteLVhTUReqZHOqlv8rD0fEDZIoZZrIzxEKxsqPvSYKeIBPpCVKOzQv5JMK2CGg6PtNHX6mAKyXrzrNClOM6TKenCXqCBP2v/+CnU7lkQl4IEQK+D/yulDI1XyuUUkohFgvVBCnlPwD/ALBx48ZF17lcsB2bTDFDc6AZUoMq0ZiUBDWD6d6fgT4N1/4x+C9w0qh4BHKPgDOjEl/JLSqIxbsWlQdeU6HpvERpvrMwNDPEL577MnWaTXVsKSXH5umZafSjv2JNuAU67oD8DDz+c/AtB+8K6HgA6iaUv3Ymo/LQ9PdD0ydBunhLabzZEbJr1pFvvJ5kTTczTonV3qCaGK1dAp98M3zmfjUycYuqMEngepVO4QtfABwI5UBOqZkh7wrwbVS/OxnQAioV8tnQdcatJOHgyaChycwkEc1LSncJuA4ew0NTSLJrqJcr48swTWXeEHqYejlOvfM0RH7jpMnqAgn5QrzjqndQcko0VjeiCQ1NaNiOTd7OEw8vjIvIFDLsGNjBkYkj+Awf69rWsbJhJbddeQ/J+hWMDe9mJjkGdp6O+m6iLev41tbv8OOdPyDij2BoBn//1N/zYt+LfPqOT7Pj+A6GE8PEw/G5ghwz2Rkao41MZ6YZS40R9oYxDZOl8aUsqVvCaHKUvJVn2/FtuNItBzJ5mchM8eKxF7mx+yY8ugnxOmRbO61bB9g6PkTWLrDLm+GoaxEpVhMNRLl95e08efhJHtz5IDWhGmKhGJ2xTjLFDCsbV7InMca2od3IQhoQVAWqWNu8lgFgpRvEe441c3+x9xf89RN/zXhqHFM3uWP1HXz6zk+/JpK/XSouiZAXQpgoAf8NKeUPyj+PCyEapZSjQohGYOJS7OsVw80qQSlMlR9dXPypMnSDqD9KtpglOPCosll7o6QKOdpqGyA3Acd+Bqs/eAH9LULuV6DXq1zvpWHQapVXhNFSNtu4SvO8CHYe30aVm8UfVpOEpm7QEKpmW2KGVVP70GKr1EvKUweh+yAShuwIiEGo0yDgh2s3q9DxfftAvgvD1AnfvJqjK7pJSIlXwpW+EB1mebjxp59RmS8LW1VlKC0IwVvBUy73JgR84a/U37liJkBxF+S3l4/bAN/V6oV3Ju1a08itXk7Nrh7cpibQNQq5NOGsxcBVnTSVBbehe3HcLAVHx5z33rCkh76SyUR2Go/mpdYt4Slm8Xv8xEKx857viIViXNF6BbsHd+P3+JFSUigVuGHZDWoCuEyhVOC7u3/CpOsSDDci7AJPHHySZC5JU9t6qsNxlq95K7ZjI4RA13SOzI7wo10/Ynmsc25ytDZYy67BXewa2sXBsYPUh+uZykzN7Sfqj5LIJVjesJzmaDPVwWo0oWHqJrrQ+fnenzOTnUEiqfJX0TPaQ311C3XVrRQdi8nUBI3RRo6O9rB9egf18Rzf69/PeG4aSzpISxKTJd636X0YwgAJB8cPsjm4meHZYfYN7WN9+3psx+b/PfdPFDWTqOGhubqZZG6W77/4dUzdw5A/zNUtV7Kudd1ZE7ht79/OHz/4x4S9YTpqOyg5JX6060cUSgU+8+7PnNe1ei1zKbxrBPBVoEdK+bl5ix4EHgA+W/7744vd1yuCMwOp70P+GXBtlZ/FqIPo/WcXEOeAEILNXZv52c7vYSdHCERbSBVzlFyXdTX1oDkwvu0ChfwM4Kj+atUqv7uTAAwojQMl0OvAWLyg8rmSyM0SOMUDw6ubTNkWjusypz91d6uCHoWj0PRTZNxl8MUVHJMJPD3fYuna+6i76iqVBTDgIzrzCW5O5nCCd6N5liOMDQvPtRFTSdGkXPB70VXJwGzpUqObhE/0rbBPFRw3mkEYWG4JN/cMGj4077IzmlE633QvO2fHWDY6jaZpxEo6j7f5aV61bk5IF2UtHsNHQC8CyjQwaLn8RF7LpN2MN5clPbqViZF9NJleGgwPjZFG3rL6LeeV/1wIwfVLr6cz1knfVB+a0FhSt+S0uZvtE8fYp3upCkWYyiWYyCfxlYqM7f85769bNne+5md+HJ8eQCDnBDyoYC2/6efA8AF0TacmVIOpm0op8QZxXId8Mc/VnVeztnktjnSoClaxb2gfjx18jHg4TiqfYjg5jKmbavSRHOGm5bcwYniZsguMDGxnf99WVta2saX3WWbtHHVVjTiugyY0xpJj/POz/0x1sJpYMEZHbQeWbXFs6hi2bbN7cDcvHHsBvaqFqtYr0QoZjkwcwc0lSOYSrGhdT8xfxdb+rUxmJnnrmree8eX6zRe/iaEb1IbVvJVX89IZ6+TxQ48zlZ5a4MnzeuZSaPLXAx8C9gohdpV/+0OUcP+OEOJjwHHgvkuwr5cXZxZm/w6yT6shvpsAdCUYS8NQ/e8heHHuWZ3xTt5x1bvYPvhTZoo5mkLVXFXTQK0voFIV+C+0Co3BXOih0JWpwjqkcqnLKjA3gf+aizYjtMW66J3YR10xpcr1AalilpghMOtOFsugtRVWtsFjzyAbioi2fcxkJ+jP3YVz/AV2ZXPcvPpuVjetguH3Qub7ELgLXauB0kFVKzbyLvXSms+8B3baLvF8PoWDnDv0FV4/3d4gFHeCXo+Fwb6Slz7Hz3RJxyzsJRiMsswTYLUveJpHxtLGbjL3fIidh59HFAtkjKWY+Wl8po+iXaRQKpDMFblt5cfQeQhKWaZFLU+5q8hqVTR74kwkxjhwfDudNW3Yuk7MF2IyPclzR5/jtlW3ndf5FkLQXN28MN/NPBwp2ZZLEhKCmen+chSon5w3zOGxQ2zteZSalbdhSXfuWAuuS9gXwlxE+FmOcs9sqm7ihWMvcHXH1ewb3sdkZpLRxCh+j5+MlWHP8B4M3aCztpN0MY3t2jRVNRHxR7COWWQLWVY2rMTv9bPU60cmxqiuauLFvq1srG2jwRfmy8O7qY/U40oXDY3p7DRBb5Dx1DixcIzR1ChtNW34TB8rG1fi0T0cGDlA0BdkfPQA0YZuHF8YJ59kaHaYulgnjZqO3/TSWtNK/1Q/U5mp00xbJxhODp9WscrUTaSrAq8qQr6MlPIZ4Ezq7ZvP8Ptrk9yzSvuzywWrpYVyvUuD2wCZH4NvFegXNznaHFtC84b3w9ATEG1XQtmxoDAFqz90YY3qMWWScabVX80Hnm7QqyHya2DUXlSfT7CudR1Hxw4wPrGLsDVCrmRRdErcuu7tUDMv97YQsH45eDYxlrqKieF/Yt3KQ0RSu9k/chUlf5hnjjxNN1/EyHwfArdD5EPldAGNysxkHVfeM4vgSMnWfIqA0PCV7a+OlPQUc9TpHqrdLFKLsq3kY9LVyJWy+HCR0mHMtjDQybou1wciStNzi1AaRLgJ1jfFWdX4ABmrQMAToOSU2Du0l/6pPqLCYXVjK1lPMzt8n6ZVO0SflccVLZh6I8KoYnxyBzWBKCnp4MMgL10sTecfdz/EM/k0y6INvLlxJQ3Bqou+HhnXwesLkZ9IMZmeKhcFF2gCRE0LM8lR1pcKDIoAjnRAgAfBezs3cnR3C4MzgzRXNaNpGtOZaQxfhMb2qwh6wzRnphmeOEJHrINUPkWukMNreumf6sdjeOiKd/GL/b9gaXwpXsNLySkR9oVZ0bCC3oleTNPEY3joHT/CVHoSWUwyPrIXKzFEob4b13URQmAIg0KpgKEbOK6D5VgUS0WCniDFUpFUIUVjVGWvlELSGGlkZHYEc/QQ3toW0nYBx8pSW0jRNs9bShMa6UL6jEJ+bfNafrzzxwtiDnKFHD7TR3vNInmBXqdUIl7nk3u2XARjCoQNIqT8zWUKqAWrF+zxixbyAKz6ENhZGNtW1k416L5P5aK5EIRQZfeyDysBiVBacOhtl0zAg0os9Z5N97NvaC3Do3vp8AdZ2341sfiK001ZZgAiAUa0PDvG30w4HKUrsgXsPPu1TVzr+zJG+nnQW0BvgvyvwGjC9axCw6teWCwu5JOujSUlEf2kJq6XXSPHHItqs42UNcOkG8Iri9gCwjJPyagrlx2XTDklEq5NNXnI/ET50WOCtPAaDXhDd4Hmw4+f65dex/VBjX1TRzgMBFLHkUmX49Uryfpj+NCQThEAx7XRNB0JuEgGU5Nsmeovh9zDnuQ4+9MT/Pbym2m8SEGvC0FNqIZtmVn6p/sIekOEvEE0X5SuUC1BTccopHhLTTNjJYuSdGk2ffg0jU/f8Wn++om/5sDwAWzXpqp+Bddvvp8R3Qu2Ba3rua5lLTW2xb7hfWwf2E5TVROmYWLZFvtH9uPRPYwkR+iKd7FveB+1wVqaqpso2CrfzvKG5QgEnfFO6sJ1DEwP4DN8HBg9QGesk6OTR4n41MRvbaiWydQkqxpWcf2y6xlNjJIpZnBdl3wpT9EusqxuGVJKqoPVTGfGaNI1jNQ4ZnaGFctvJOQ9mcfJle6CfDmn8qHNH+Kxnsfon+wnFo6Rs3Ik8gk+ddOnCJ+nh85rmYqQn4/MKZsvDsobBZAqjZfS6gW4+UuzL08ANv5/kBlT+V2CzRfvPqlHIPxucGfVy0qvuSQTxqcS8Ue4btkNsOyGs68YbAJvFd7ZKRzXZf/Mm8Eu0lWzmy4+CkDWqSGomVDYSYYljGWPMVHYgWnWUF/fTtsFzBNrAP5rsIoPIdwUNi4eNwOaScHsQJMCCxc/OkUpofCC0uSNeQFp9jAU94N/g/qenyI5sZfeyBLqkGofrk14Zj8TdddimD5MBHnXoT62hH29TxGM+gki2D87hOFY1ETiRH1hosBYIc3jYwe5/yKjM0OazsxUP5o/RNATxJUuM7kUVb4oqyP1ZNMT+LxB9hayDNkFhBQctPKs8QZpDtZwy/JbCJpBigKS8aWIYoZqXUcTAltKjruwvDpO3+6fEPFF5iYyPYaHsDfMTG6G2lAtIW+IJfElHJs6RiKXoCZQw4ev/TDdDd18a+u35vLNdNd3s2d4DwCrm1YzPDvMSGIETWjkS3nqInW8/5r3Ew/HOeQ5xPbj25nNzVIdrGZTxyZ0Xee5o88R9oZZ17aO8aQy7Wxo20DAG8B2bCSSseQYLdEWbEelg/B4Tk+611bbxlcf+Cr/8PQ/sPP4Thqjjfze7b/H265420Vdk9caFSE/H89yKOwGrUrZhCkobw3Np1zw9ACYFzdxeRqhBvUBleslOwJ2ESwDfvEUbNmiKtXcdhu85S0vnQZViLMX8ngl0U3oupsO8SjPjRwkl7bZr72Lrprdc6v4Q9cDaYrFUXL5FymxhpZAgiGriYcO7uVOo5OuutOjO6OagU/TyLkOgXKEqSMltoQGwwt6kFDk7cjUEUx3hoxei+VpxRV+XNcmKHRsICSkijbVT7muWq2azzgh5DMjpHQvAoF2YgJAM9CAWjePK3zUGR4mbAtZ1YQZbcKfmSJoBhhPTVAXiLC05cqT/Tc89GUvPstHKp/CGt7LslgnFlC0CgQ9PpyxIxx3S9TXLeWoJ0jaKtBgqIA3W0p2FNIcnjjKkbGDdDd0M1DKkRQah8d6iPpCtNW0zU1Oz9glpJRzfvEnJmtNXaU+fv+m93Nw9CD5Up6GSANhb5jGqkYOjh1UJqJ5tNW24TE87B3aixCCD1/7YfYM76HklMgUM3h1L08dfgoXl/7JfjUBG4oxnZ4mU8wQ9oVpr2kn6A1SG6xlfet6rmy9kpAvxI7jO+gZ7UETGhPJCb6z7Ttkf5YlYAZ436b38cmbPqnKK86jM97JZ951+XjSLEZFyM8n+CYo7ABLKK3eKaiiE3qjymDov0XZvl8OCrNw9CEopcEqwU8eVHHE0SuhZMM//iMMDMCnFq9i/5rFGyG04l3cXbuORw89zkr/Nxcs1koHwLOeWWsWr5YkIhLkaSCpX09NSGNL/5ZFhbwuBJv8EZ7PJZm0S3OzQmt9wbmoTb9ZxfLQKl5MjLN/+hhJax814To6Iw0UpMsyT4CQZpaDpVROzpO4LEixq3tUEe/Tpp8kpq5zVSBCwZVMOhaGP8K/2/QBnOwUw4lhDgloq2kjOG+SL+fYNPmjLIZlWxybPMbQ7BBV/iqWNywv29pPJ11I4xWCDseiMVjDodxRpsYPkpEuk7XrWdJxDc/lkmRnh9GzE4Q0k/baNqqjLTw1sJ2VwWqe6X2G4VKBQm0HRjFL3irwgWs+AMBsdpYn+rdwbOoYrusynZnGY3gQCFLFFNd0XUN3QzfdDd0cmzzGQ3seYkXjCgIelS1z2/FtFEoFMoUMoXIwVH2knmJDkRuX3chzR5/j3RveTckucWD0AC8cfYGdAzupDlazpnkNjVXKL/9EcrVbV9zKbStvo7m6+TSvmeuWXsd1S6/j+9u+z492/4jmaDP1kXqyxSxffvrLeHUvH7vpY4uex8uZipCfj2cFRO6H5D8rX2wnqTR4LQSBWyDyjrO6UBZcF0u6BDT9/EKrpYSen8BgL2SlqjU5loO2KNg2yChuKERix3YKw4MEGxoXhJ+/Iriu6ssiw95zoTnWyYecvWiJJyh43ozX04QobofSARAaWSeKrleTkW0k5ApswgS9Ki/KXL6YU6jSDW4P1TDtlHCkpFo3FqT1BXBnh9ly4GEy3iiWrtM7eQzLF+KtG97LEl9QXU/PapUgzCh7sEgXnEnli3+CcCu18hm8TpGsbhJEQilLRg/g91bRaKjo1mXMC6LxttBS08K07uXhyaMYbgmvZpIpFcm4Nrc0Lj/tmAqlAg/uepDpjPI06ZvqY8fADu5ddy8N0YbT1g/7wriuC1ISNUw2Na4gX7eUF+wSG5tWU+8N8OzIPgrpCUxvmIhmcmCkh+rMLPlSke3929k1uAvTE8AN1OBaOdKFNL/Y93NGUlOMF5Jc6Vgsr1/Os73P4kqXrngXtmOzxLeE+6+5f64v2/u3k8glsGyL1ppWAp4AjdFGBmcG6RntYSI9QcATIBaKsalzE9UB5WcvpWRr/1YyhQwIiEfi5C1lg9eFTnWgmnwpz9L4UkLeED1jPewY2EFXvIvl9csXuIECfP2FrxMLxgj51Usl6A3SGGnk21u/XRHyb3iEBqFbVXrVwhYVGi984Luy7CO/uPvhWKnIU7kkw3aRKs0gpntY6wvQca5RcyN98NwvQatRaXJ7emA2CQ3VYExTlNW82N7EbH0VYnYCN+yj1fSxzhe6NKX6zobrqpzfO3eq+pfxONxwgyqtd65ICRO/h5b4ElT/Dr7Ib8DMn4H/zaooRekIjWaMlF1FkRpmpQpyShfSxEPxRQX8CQwhqDcWf/E4rsO/9TxGxPTTUY5aktF6js4McnzqGMva16sV/RuVu2ypH6XNS3XNPStONuYJYXbeybUDj7ND+JkUBsL0EYldwYZA1Vlf6ne0rEUCz073Y8kcEcPP/W3rWVnVdNq6PSM9jKfGKTmluapLNcEaHj/4OO/f9P7TtNeIP8KKxhUcGDlAQ7QBXdMZKaTxB+M0R+uZTY2Ty0wRCcVI5FMcmB1ClPIcTo2zUggeO/YcDZEGPIYHOznMbLCO4fwsmeO7MIVATBzmAC6uY/GWNW9h9+BubMfm5u6bWduyds4zZc/gHv7qib866TGjG9y+8va5vq1qWkXYH2YqPUXvRC8SyfHp42zrV9GxxyaPoWs6Y6kxClYBXdMZTU9SE19C0BsilxplOKHs90vrl2LqJk8ffprD44e554p7FgQ9LeY2GfaHOTpx9Iz2+ZdipFTg+VyatFOi2xtgoy+MeYrp57VKRcgvhlkP5j3ntGpvMc/DmRmSrk1E05lxbGwpsQouQU0nfgYBtIAtW0A3IByFsRmYKsKxGWVGWB6mp6mGpMckPpyCYBipmwyUitToBp0vd/j11q2qdFxjo6qgk0zCj38M733vojVDT6Ms4Jn9IlT/DtR9HpAQeS8kvwue9SBtAhxnVNbzfGEztaafgpUkmUty77p7z9x2Nqv+nqHY8nQuwWQxR8e8os5CCGp9EXZM9HLbCSEvPBC6S3nzuDkVSKYvYkqJtBJd+QFuyU+RQSD8tQQ14yUjWXVd5+72ddzesoa8YxEyfAvy2M/n8MRhDo0domAXiIfiSCRDs0OMJce4c9WdjKXHVBbKWNecYLtx2Y1U+avYM7SHol2kvWEFTfElmLpJrpghkk+RNoOMF9LUml68vjBWcpTxmQEKpQKFUgHbtXHyadJD+8m7NnWxdjylImFvgFQhxfGp46xoWMGmjk1z+0zmk2SLWaSUfP7Rz1MbqMWWtjLV2EV+uf+X5EvKUWFZ/TJyVo7BmUF8po/BmUGGE8McGT+CEILx9Di60EGC3xvA9gQZLGapNf34AzUEfFF6+7dw56rb5l4sYV+YwZlBjk8fZ2n90rlzuLRuKccmj9FYdXKeZTI9SWt16wUJ+O25FP+cGKckXTQheC6f5klPkt+tacH/OhD0FSF/ERRdl72FDEXpEtNNNCHwoeqURqXBMavw0kLecWBsFiKNcLgXBhJQXQPBYRgcxbGbGGt0aNi5C62lFau6GoQgqukcLxVfXiFfLCotvrkZTtzM0SgUCmq0cd11L92GEKBFIfgxKL4TRp6F6BIIvx38m6B4iIGSjZP6BhlfhClvJ8dzCa7wV/HOFbfStIi2SyoFTz4JIyPqJdLUBLfcovo2D4/hRVDO2z8vSVnJLhJczLVOrz2ZtfNM6CYi1MiFONh5dAPPWcxsI4kRnjj4BDsHd1ITqCFTzNBR20FtqJbt/dv57W//NlKqSd9YOMZ/uPU/sLR+KYZusL59PevLL62C6/JIdpaSdPF7g5hOgeLIPvy+ELqVQxaS+DLTtMY66U204AvW4NN0wrpH+aInJqkRnRQ0gStdfKaPmdwMRbtYPg4PX3/h6+QtJcCLpSI5K0dTdRP7h/eTLWaJ+qMUS0V2DezitpUqAGxgZgDXdakN1XJ04ihew8vVnVezc2Anlm0R9UeRSMLhehLCIDPdR2L6OLrrYAZr8VY3URuqxXZsRmZHODJxhKyVxWt4Fwj533zTb/Lb3/xthmeHifqjpAtp8qU8/+3u/3be18ySLt9JT2IKiJXr8krpctTK82Q2wVsjl849+eXiDSXki1aRrzz9FX6y9ycUS0VuWHoDv/mm31zwxj8f0q6DWZimeWI38cIUeV81yaplZP1qYi8/P4/KmdA0CATA7oSBbRA1QDjQ3QoDaegdpi3/C8z2VkwnjfjXL5G481as+tW4p9XevMQUCkqInqqtBAJq3uBcKd0JU9vBPFbW7HdB47XQsIEZEWOHk6C2ZgO6pvEmVE1aF2gILjJSsG346U9V35qa1Etkehoeegjuu4/5yWSqfCHWNK3gwMhBWqqb0IRGoZgjZRe5tnn1hZyR8ybrOkyW8iAtYmaI0KmVqsqk8ike2v0Q8VAcj+4h4AmQyWfonegl7A0zMDtAXbiO9rgK0plMT/J/H/m/fO69n8PvXfii92ka631BdhQyyFAMzV/N6MgBfLkUuWKKvJXH7wlixLoIdlyDZmXIlPLkXZu8XcTn8dFa00qmkGFgdgCv4UVKSbaY5fD4YRzXYXXTapqrm5FS8tThp+gd7yWVT2HoBqlCitncLAFPgHVt6+by7MxmZ/F7/LjSJWflqA5U4zN91IXrkEg8ukflu69uxs3M4K1pZUl8KSvb19EQivOrnhwFp8TWYy+w4/gOADLFDP1T/QD8+vW/jqZpbOrcxN/e/7f80zP/RO9kLysaVvDA9Q9w/dKFMSiW69Jr5clJl7hu0mp4TjMNDltFko5Nw7yKaUJohDWdXcUMb6Ui5F9TfPr7n+apI0/REGkg5A3xi/2/YOfATr7x775BNLC4p8MZkRLv6PN07/86WqIPnAKalKSDzRxruYnprntpPJeACiFg/XolpDyd4PeDUwStAG9dh751K+GlHeSWS7zFAcwjk0T/9FGC16/Ge92vw+pr1Yvi5SAYVELTssDjwQFmDJ2SbRFpaebs5cPL5KdhYieEW0/OabgOjG2B6mWMo2MKscB8EdB0Ju0SKdehar7m6zjw/PPwzGMQD0OpFmrroLoKRqaVZt++MFLxAyvezFcdlyPjhxEIPKaXt629iyurL3GRlkUYKObZlT2MLPWDdJCYrAstoSN0ej74E6XzuuJdHBo/RM7K4UqXidQERpWBz/BRGzopUOLhOL0Tvewb2cfV5YpL82kxfdTqJtOOzZr172S89zmOpkaJ+qN0xbswo40cGN5HlS+E5guj2SXyQsMbjOE3vBweP0xbdRvNVc3sHdpLXaSOntEeTN0knU+z/fh2lsSX0N3QTUdtB+NplcXRNExqgjVoQmM2O8s7172Trce3kilmiPgjDM0OkSlmiIfjTKQmmM0rN9L2mnYcx2EgMYBpl4iFYmSyM/ikjd+V1ARrqK9tp2dkP3sH9+I3/Ri6gdfw0t3QzaMHHmVz12ZWl1/eGzo2sKFjwxmvzZRt8c3kBLOufWIWhm7TzzsjcTzz7kWPEKckO1Yo36vXR976N4yQPzB8gGd7n2VJbMmcr2xXvIveiV5+uuenfGDzB864rZTydLtrZpjQ0Z8Qn9qNnZ9l3FODjoORHaFm9EVC4VY6qhYGVcw4JY7lJ8i6LjWeGro8PoKaDmvWwOws7N+vCmAbHli7Rgk1TaO+SmOiOICzawaRLuE4PsKJNLFf/gvkDLjmmpc8fqeQp+/hHzD5wpPYjkX4qs0su/M+gtGzaCKGAZs3w+OPk6mt4cW6GjJWARpj0FrPsmKWlZ7A2W3SuQnm0hyfQNMBAflJXF8DZ8qKsSCXuOvCL38Gj34bcj2QycGBEjS2QeMKcGLKvHQKVR4/v7vuHgbzKdKlPI2BamoNz0VUwDo38q7D7uxRoqUDZKhlEB8ZV3IkeZz3YtIRWhjJmyvlMA2T6kA1XbEuEvkEftNPspDEdmzCvvCcRuy4DtPZaYZmhnim9xmaq5sXNWv5NZ0WTackNFY3LMevG/g8Pry6l0cHdjMxdZRZXwifGUDTDXTdg0RQ7Qngui47BncQD8W5b+N9+Ewf9dF6Do0eYtqYJuANcHTyKPXReo7PHidoBpnMTGLqJmPJMQzdYHPnZtpq26iP1vOrQ7/C1Exms7NU+atwpctkZpKiU6Q52ozf4ydn52iraePgRB/HE4P4DT+TiRFePPQ4S5rXc8+m97F/eA+jqVHioThhPczy+uUEvUH8Hj/b+rfNCfmX4ueZGfKuQ8sJE4zr0mPlaM+n2TwvNUKT6aXV8DLqWMR1E4Gg5Drkpcu1r5Oo2DeMkD82eUylWT3F9OAzffSM9Sy6zWipyMFijpR0qNYMVnoCxM3ycHtqH2L2MGErQ8kpgJMmq/sQxRlqciY1k1vwtF0PAeVXP14Yoy/5C8LuFGEhSBXqec57A9eFW5Wgv/lmJaT27lXJvbxe2LEDTBNPnUXjlEEpW8KOVWHOZjCjDWjhAmx/Ub0kzjD5CICUHPzaXzHRswNPUwumrpPe8iw7+g6z6Xf/J17fWcJKV62CUIhdxw5jlYrE6xuhowPX6+VQMUdMN6k727yD7oVFdSEJmodG08NhK4crJVpZ8OZdB7+mEZlfOm9kBHY+CfUmaCWVHM3UYHgcajvAPQKBxQW3LgQdgShwnqO1i2DWsXHt42So5qAbICBcqnXBpOPnicwgd3vbqJtXNaqluoVdA7uIhWJs7NzI0fGj9E33UXJK3L7ydh7c9SCO6yCE4MjEERLZBI50MDSDH+z4AW/qftOiAs51XY5NHePo5FHyVp5EPkEim2A6OUxjVSuZ/Awew4PlWOQKWVZ1XUONU6Qp0kTYH6bklHjH+nfwaM+j+EwfsXCM4cQwIV8IQzcYT41zdOIoTdVNrG5azUR6Akc6eDUvrbWtADRXN/P+a95PtphlKjPF5x/5PJqjsappFVJKSm6JTCHDDUtv4LEDj+HaebBtpvKjzGQmCPlryBcLzIz1sLqxG13oSCTVgeo533sp5VnTCs8n49r0lwo0avNMMJpGFQZ7rOwCIS+E4KNVjfzN7DCjJQuE8r+60R/hmvMd/b9KvGGEfHN1M650T/O5LpaKdNR0nLb+SKnIi/kUEU0nrptkXYdn8ylu0qLU6CYUpkG66E4BXfdgAtVOAc3OIEo+KCbBzgEgXYup5A8JA5rZggtUOdPohYc55vk11p4IdLn1VuWxsncvzMwoARsOgz6Lnsih+zxgu+AxIRwArZxqIZU6q5DPDBxjumcn/s4laGWN2tveSeHoEYb2b2XJhpvPeu6yLc1MVwWIz3uINMAvNAZLxbML+VCzqthkpcBTPs5iAswQhBqp0QxWeQMcLOaR6mgwhWBzILLQPXRkGKxJqAqB5YWZIpgeHFuQHd7HROsSStPbaa6tP2Pg0CuJQIIsMuTG8AsXT/lQvEIjSIZDVv40Id8R66B/qp+IP0JdtI6gL8itK25lRcMK8laexw8+TsktMTw7jMfwcNOym+iIqTzozx19jmX1y07zGX/6yNM8efBJ+qb6yBazaJrGRGqCoCdAojBDsZjB6wlQsLLk7CKz030UERRKBe5eezeJfIJMuZA4qECmmlANU+kpinaRyfQkUkoaog1EA1Gqy/MoU5kpSnZpzhNGCEHIF6JoF1nVuIqGaAOT6Ul6xnrIFrP4TT+Hxw9TcArkrTyWlSVgeLFdh3xumoRTwGMXaF51Kw3RBkpOieHkMEF/ENu28ZpeNrZvnDvuoZkh0sU07TXt55XeeTF1pMnj5U/iHfQUc6Rchw7TS7PnIkqrvcK8YYT8+vb1XNF8BXuH99Ja3ar8cJOjVAWqePu6t5+2fk8xNxc6D+BHIDWNw8UcmwNRCJdLFWomOEV0gco/L1HpD4QGXvWmL9lD2G4aj3myvKGt1xKyh5ksDoB/jfrRMODqq2HDhpMTnjt3wtM/gewsTOSgNgLrloKWUcWthaEmQs9CbmZCaSCn+Pkbpofk5NBLnrsTwvdUNCFw5VmKedk2JDMQuwVmn4f0sGrNVwPttyvTFNDtDdJs+ph1bAyg1jBPL8rs90FxTEX/BnPg0XEyMJ3MMx4JkIp4GB3Zz9OJFPdcec8FTabbUjJaKjLhlPAJjRbTe8FBZ7WGB12LkrJLVJcn7WwJQhaJmVGS7sJwf13TuWP1HfRP9XN08ig+00d3Q/dcANTHb/o4Gzs28vUXvk6Vv4oNHRtoq1H3k6mbuNIlkUssqL06lZ5iz+AepnJTrG5aTf90v8omaeUoZGcIh+Ogm2SKWZLZaaxiDlHVgqubJLIJnjv2HO017XMphFP5FBF/hI3tGxmcGeTw+GGu6biGmmANQgr6p/vnbpRiqchNV9xEoVTg4NhBBmYGiPqjtNe2I5Ekcgm2Hd9GxB8hHo4znZ3m0NghpjPTFEoFNE3DY5h4MMkUMlglC8d1KJQK3LvuXn6y+ydMpic5MHSAlY0redsVb2NJ3RISuQRffOyLHBw9qOZgDA/3X3M/t6++fe68hDSDDtPHSKlITFMvRem6JBybzWfIH2VoGmv95zQL9ZrjDSPkAb7wa1/g//zy/yiNyCmxrmUd/+mO/0QssjBVgSMlGdchbpj0Du/jkR3fZ2xmgEigho1r7mLzhndA7AqVWjc3CZlRlZZAGOAJg68amq6dE/K6W6AkYdIuoiMIaTpBTVNZFDXn9I7On0hdvx462uFgC/zyQYgGIVoENwSz1bBs6Wmug6fir6lDSnna3IJtl4jUvfQEZFBoRHSdjOsQKkeUSinJui5X+M6g0Rw9qvzpczllelq1EjbfBIGgOj+n2MRDmj7X9qI0W+ArQVqHSBCERk5Ok68WiK7VhHUf6fhKSq6YCxw6WxDVqdhS8kIuxZRTwq8JSlLSa+XZ5A/TeC7FuE/BIzQ2hVfSO72XKcfFKww0LJZpeWzPCmoWeXkYusHS+qUL3AFPoGs6Gzo2IIRg2/Ftcwm/gLlr6zMXXovp7DR5O4/jOFSFqljVuIpEPsFoahQhBI2BatANUrkUWdcGTcNneLBdm654FzPZGRrCDcRCMe5ccyc/2/szhmaHEAh8Hh+/detvkbfyPHfsOUYTo9RF6miuasZv+kGoBGRffvrLZItZ2mvbmc3N0jPaQ8AMsGtoFz7TN5ei2HVdltUvYzw5PjcXI6Vyf5VINKEhEIS8IZqqmvjYDR9TBcaj9dx75b3Ew3GEEPzt439L71gvXbEuNE0jZ+X4ytNfoaWmhZWNK+fOzVtDNXwjOc6wXZzTYro9Aa56ndjZz4c3lJCPBqL8z3f+T0qlEi4u3jM8vLoQRDSdg6M9fOvRzxP0hWmsaSdZSPHI819jncfL2ua1jIavIB4YpF7z4Reu0uRDzdB6Myx711x7x90QBddh2inh1zQSjk1Qg7h06fCdg8ZZXQPXfhSW3QLPPgEDE+CtgfVXwMaNL7l5uK2LmpXrmO7ZibepDU3TsEZHcBvqaVl9umfGqQghWO8Lz+WJEYAroMP0UqcvYgcdHITPflaNRkxT/R0bA8eFe88S3HQ29ANw37vgx7+AUQHeBJa3gNUWRgqTmapVlMwwIWB4dphUIUXVvCCol2K0VGTKKVE3zyRlSZcd+TR3Gp4Liiyu8zVxX63Oo+k+/DJNTI9SMteQw8PG8zAhzGdp3VK29W+bywXjSpex5Bidsc7TzFQ+04eGNic0NU0j7A3j1b1YusrZHjG8uI6FkILWmlayxSyxUIySU8Kje+id7OUrT3+F5fXLuffKe8laWRzXIewL82LfixwcPcimzk30TvTSO9HLRHqCm5fdzOqm1fzhD/+QQ+OH8Jt+qoPV3L32bmpDtWQKGQzNYDY3q9IvC6FMTbqHLce24DW8FKwC2WIWgSDsDRPyhqgOVs9VxdI0jWggyp2r75wbvYwlx9g3so/OWOfcCz7gCRDwBni85/EFQj5mePiN6uaXdKG8HHhDCfkTmOZLT9Cs9Ab46p6fohs+osFaSki8/ghNHj9//6u/56r2qxhPjlPKC1pLJm9pWcKmZW9CxNdA9TIo+0PnXYe9tp923xqWW/uZcJWfME6a6sj1VHvOo6h2rAve3gWlUjlx2jlG2wnBqgd+hyMPf5/JF5/EsUuENl1D9x3vw3eOwiaqG9waqmbKLlGULlW6QdUZoj2dH/6QXGqGTF01HlMj6g1jzMzC9u1qgvklRh6nIaVK8dzSDp/6KIwcByvN8el+jucT+JrejG0Ey6sqgWYu9vI5C8OlApOzg+yb7sd2bKqqWzGijcwicYH1/hAt5vnbYZt89dxt1HDIypFwbKp1g43egJrXuQAi/gj3XHkPTx56kuHZYUBFk96w9PS0z03RJuoj9RyfOU4qn8LUTXYPqQygYW+YiD+Cruu8fd3bGUmMYEubt655K4ZusH94P/3T/bRVt1EXqaN3opfR5CjvWP8OtvRt4ed7f85TR54i5A2xrnUdV3deTXdDN2OJMZbWLeVfX/xXxtJjtFa3YhomiVyCb235FnesuYOZzAwb2zcymhzFdu05E1WxVKS9ph2hCTRdo1TOfhn2hYlH4qxuXM1MZoYZMYOUkvVt6xdUzDoRnHWqoPYaXhK5xGnnx6NprPKdxWHhMuENKeQBXNdhxhpitjiGX/NTF+jEo58cqjWYXoz0BNFAFUUkPqHRZJi4SPqm+6gN1lIVrKImtIy01cbnRo/zO6tauS620MMh5ToqY5/3OvKiAb/dS1AzmdWXUPJ1n9qtc+McXlKnovv8rLj3g6y494OLu4SeAx6h0fQSpotSLsvxLY+RsZO4qQKu62IaHpZHW/GPjyuf+/NFCDDbVcEWoxbalPthY02I53sGaRSeubCw8dQ4HbGOBYWuFyOVT3Fo7BBTmSnCvjAPHtvC/tlBYr4wrub5/9n77yi57vO+H3/dMr3P7Mxs7wssdtELUUiQFChWkaKoZolm7Eh2ZMeyY+f4RIlPnPiX49hOvrblyFZiW3YiF0mWLFGUxN5JgATRO7Zge9+Z2Z1e79zy++MCQ6xQCDaLJPjm4SF35+6dO3dmns/zeZ73834zP30Cr9PPxt7bsNg9HCrlAd5SoA/JFnbJ176wGYZBVasii/JlM8sGfwOfu+Fz5Ct5LJIFu2y7rHCeRbZw74Z7EUSBvcN7ORk7iSAIrGteh6qpNAWaKCpFvE6vyWhVSqacsKYwl5nDYXHQXd+NJEpEfVHmUnP808F/4uFjDzOeGGe5sIxFsjC5NEksE+O+jfcRcAU4PnOcTDFDvbfeXFww/V5nkjMcGDuAz+FjKjnFXGqORC5Ry8R1TWdL+xaCriCJfMIcBLN76Ax3cu+6e2mva2csMcaJ6RPm47FRHBYH65rWYZEtNAWacNlc5Mq5FWYhqWKK+zdc2ne7XnBdBnlNr3IsvZe5SgZZkNAEDVtxjF2BHXitr5dP+sPdDMYGabpo8nIym0DXdZw2Z80f0ma14XP42Deyj+2d25Euqi1bBIG0pvKaUqSg+4CtSIZAl2Cj52e0NXw3OeJD84MULBrRko3qeZeeYqXEQmKaTn8L+P1v7cSO7ZB75LxonAuMPA2+ELf238j+idM15lRzsJlbVl2dLZTIJfjx8R+bOyoBXhh8gcnULHWtWzg7fpBMtYy3rotkLk45G2Pdjf8av7+J4UqRJtn2rt0/RVV4deRVXhh6gbySp9nfzEfXfJR1zesuCfaCIOCxKFDaB/lpED1g32yKqgkCqvZ6htzkb+K3bvstHjn5CE3+JrwOL+limoGFAapalZMzJ9nZtRO31U08GyddSqNrOlu7thJyrZyjePjYw8yl5rBZbPjsPgRRoKgUOTZzjPUt63Hb3aY7kwART4Sl/BJCVSCRMzn0Va3K2ua1rIquYmp5irA3jE224ba5aW9rx+fwMZ+e57fv+G0mEhM8duoxikqRxewi7aF2BuYHyFVytARb0A2dA+MHWMovcUf/HVhlK1+88Yv8xQt/QbKQxCbbyJQzrI6u5pbVV/9MfJBxXQb5ueIos0qaqPX17CqvlzmROcLNdffWsqJPbP4EJ35yglgmRsgTIlvKspRfojnQjOOnNGMkUUI3dCrVygrKlhuRM5UCZV0jdJ7eVtQ1zihFPna5i1sehJm9oJWgYTvUb6uxUN4PGE5PEl7fj3Z4AEsyg2634VE1tNgiyud+Getb2IUAphGK57OmObmWAHkNWFfT73fRXb+JVDGFTbbVKHxXw/7R/VhlKwFXgDNzZ3BYHNhEiULsHEWtim73kisuE/I3U8rFePLYD/nXt/07ls7LLbwbklTlapl/2P8P/ODoD0jkEjVzjkOTh/i1W3/t0iClZSD7iNnslxqpVHPMTP0TOWENwcAuTs+fZmppCq/Di6ZrnJk7g6ZpeB1eLJKFsCfMbvdunjnzTE0WAMBmsfGFnV/Aa/cS9a1UGo1n4+TLeZMa6QkiiiKZYgZJlMiVczw/9Dz9jf3c0HYDVbWK3WKnJ9zDcGyYglKgqlfpb+qnJ9JDVa2SzCcB6G/qpzXUisvmYj49z3hinK899zV+ePyHpkmLIPKjYz9iS8cW1jWuo72uvXZNLcEWRuOjbGnbQsgdYmf3Tup99bx07iVShRQbWzayq2sX9mugPMYyMV4dexURkZtX3UzQ/R4x33mbeP9Ej3cQs8oCHnEln9gt2kmoOUpqBofFD5i1zt+993f5/pHvMxobJeKL8JU7v8JTZ55iJjmD1+E1DRQEAVmUafA1XMJwWNCq2EUJhyhR1M8bKYsCEdHCWKVIx8UfvrHHYPBb5gCRYDHFvOq3w5bfet8EekGUON7fRiKdxT8fJ5gr4bBYmL15M+233/7GJ7gaJI8pC/xTsFlsl9VavxxUTWUhs1CbEF3KL+Fz+rDJNubTM1hFC3aLnVI5j6WSIeptZHZ5ioX8MnXu0FUbsLquo6jKNQWUn8bg/CDPnH2G2dQsdosdr8NLUSkyGh/lm69+k23t21byvSuDJl1RCjGfSfKXrz7BsdlzlCrfRba34bK6+fz2z9eSEa/Dy1x6jpnkDB11HQiCwOzyLK+Nv8aWti21+5cqpPjO4e/w6c2f5lz8HBFPBEmUWMovEXKHCHvCxLIxdF3HZXUhizLz6XmqWhXBEOit7yVVSuGyuzg7f5awJ0ydp47FzCKbWjbRHekmW85yYvoE2XIWq2QlkU8wsTRR2yFVlAo/OPoDvA6vKcEgWdA0jX3n9uGyuFYEeXjdsPuC7ENHuIOOcMebuv/fPfhdvvb816hqpgGN7Wkbv3vv73L3urvf9Hv5XsP7I3K8w5CwoLNSPMzQDTAMBGFlnramYQ3/9b7/Wvu5Uq0wHBvmldFXUHUVWZLx2D3s7tnNTd03XbKtVtGRBWiSbVQNHQOwILKkKZQvFjArp2Hon8DdAvL5IGFEYeGwKehV/8YsmvcEgu2cXRymePtO/MsZ1GKJJaHEzRt2Y7G/y7LI1wBRELHKVpM9IltrNVyf08f08jSCLGAr5yiIEharC6tsQbDYSVYr3HwFxU9N0/jJyZ/w+KnHKSgFmgPN/Ksd/4r1Leuv+brOzJ1hIbOAVbTW+gleh5dkPslyfpmJpYmVE61aHAQXhUqZP3r+n9k/fgbd0LCJOonsOSTRxnODz3Hv+nsRBAGLZKEl0ELUG2UuPYcoiByZOkLEG6kxVgACrgCjsVHq3HVEvVFOzZ5C0RT6G/tZ07CGRC7BSHyETCmDy+ZC1VQkQcLn8tFe187pudP4HX42NG9AEiQUTSFfzlNSSsiSzInpEyxmF1nKL6Fpmik7vDxDVauSLCZZ37Qer9eLJEpYJFMGIeKNIEkSDouDgYUB9qzZs+Le6Ya+wsD7zWIiPsFXn/sqEbc5gAZmz+b3H/t9trVvo87zLrnB/Qvhg8cXuga0OdrIayr6RQMpaT1PvdWHXT7fsNGyUD4BxdegOmO6BQGnZk9hYPDlj3yZj/Z9lPXN6+mq66K3vvey/OYm2Y4VgbKuYRFErIKIYeiUDZ1VF2dm6VGTRSJflAUKovlz4vS7cRveceQ0FcHfxJaWTSQLCSYcKtNBCbW1k/bWzT/rywNM5sWmlk0sZhbRdI3OcCf5ch6nxcnalrUk80kWlqdotNhodpkluuZgKze5fCsmVC/G949+n+8c/A4+h4/uSDe5So7/8dT/YCQ2cs3XpaNjkS2oqLXfGYbBhX90/acUTaUoGDmOzo1xdmEC0Am53Xidblw2D/lKnsGFwRqrRNVVErkERaWI2+ZmVXQV2zq2EXaHL+0xCOZ9Wt+ynod2PsQXb/oiN3bfSNBlmnNv79iO3WInW86i6Rp2q52IL4JFtuCwOEiX0pyLnUPRFH77jt+mt6GXu9bdRdQbxTAMBuYGODt3tibTMJueZWhxCF3XaQ+314KqTbaRq+RqlyVLMgFHwGTlaCpVrcpsapbOus4V4m1vFs8PP49u6LUAD+YCq6gKL597+S2f972C6zKTD1ubWO1YZrQyiaAbGAb4ZTfrveeFvpQZKDx5/mgZKsfB0g2ujzIwP0DEE8EqW9nRuQMwBaPiufhlWSt2UeQuV5DHCkkEXUPEoGIYrLO56b64tCM74Kd2FwAYVXPA6n2AkqEjiyLru3bQ1dRHrpjBZnVg2H1Ur1FX5N3Ehfdnfct6KlqFk9MmnbAj1MH48jiCItIUbDIDiFKglBhlQ0Mv66Nd9PouP89QVso8deYp2uvaa6W6sDtMpVrhidNP8JvR37yma9vQvAGP3XROKiklrLKVfMXkk3fWdV5SH8fWB8pZppbGqFbLWCUJiSoKXtx2B4n8MulCmlg2xuDCIEenjlLVqnjtXpqCTQwuDGLohmkYopk7UoBsOYvT6mR19PLMr6gvyoM7HiTijZAsJOmOdHN65jR2m73GaPFIHjRdYz49T7KQJFPK0BRoot5Tz9Gpo2iGRtQbxWVzISBQKBcoV8vohs5IbIT2YDtOq9Ms+1jNwFtSSuiGzr/76L9D13UGFweRRImtbVvZ2LLxbTXDVV298mPalR97v+C6CvKJbIKvPvtV9o3sA2BXzy5+cffnafA34rect5kzNCg+bxpdiE6qepXTM6eYX/oJE+VniJcdrGtct+LOGYaBcBXZ0Y1ODw1WG2fLBSqGRo/VSafFvrK0E1wNzgbIz5kDVWDqvSBC0zWYc7wH4BQlDMz74bJ7cZ0fEU+oVXzvYE9B1VTylTw22XZJA/xK+Lv0Inld48uBRiRRYkfnDja2bKSklChUCjxy/EeMNPbTqmt4xl5hNjlLQSmwvWM7e3r3XHFIJl1Ko2jKJb2YCzXwC1BUhefOPseL514ETGelO/ruqNXv1zev57be23i8+jiL6UUKlQIum4vWQCv3bbzvEjs7JA94HiDgniPokJjOVbEZITQccL58IUkSr429ht1qR5Zk1jSsYSg2hMPmoCXYwkxyhpu6b+LAxAEwTNVPWZT5xKZPcHDyIHXuOrrCXbXXli/neeT4IxSVIv1N/abGjKoQdAeZTc+SLWVxWB3UueqoVCsE/Wbj0jhv3DISH2Hvub0YGKSKKVO6QLLidXjJlXM1bfm59Byf2vwpvn3o2+RKOcbiY8iSzK/e/Kvs6ja/CzetunQu4K3ipu6b+Ju9f1NrdoO5qIiieIkG/fsR102QrygV/u23/y3TS9M0Bsym24sDLzKZmOQ7v/wdFE1B0iUkPcWCZqCKXqKGwk+OP2JaibltKJVh9o8XGU+M87kbPlc7dywbY33z+qtmEwFJZpvDg+NKJt+iDNt+G47+OaRNbXEsTtj0m+C+jDvSexBuUaLVYmOyWiEgSYgIZDQNryRdXcTsTWA0Nsq+0X0oqoJhGPQ19rGzc+dVFQiN8zIVD+eWAPhyoBFBELBb7NgtduZSc5yJdHPOGWRTJcvmti1saN7AXHqOvoa+qy4kQVcQl8VFsVJc0RhNFVLcvOpmwGzI/vlzf86hiUMmJ1yAbx/4NmfmzvCf7v5PiKKIKIp86eYvsSq6iv2j+0mWTEmB+zfez7rmdQiCQKqQYmp5Ct3QaQm2EPaE2bz6lwgNTDBdPE6iUMYq6pTUkqlv07qFda3rEBA4M38Gr8OL3WJnJDZCk78Ji2xhR+cOPr7x4wwuDKIbOrOpWZbyS2a5Z36Q41PH+fjGj+N1eDkzf4aSUqo1rd02N+lCmuMzx0lkE1TUCjpmQ/butXfT19BH0BXEa/dyLnaO0fgobrsbl9WFVbJSVErkKeN0BXF6IqxuXk+qmCJbzlLnruMvPvcXGBgs5ZZQDZWZ5Az/48n/wZbWLdzQccOb94C4AtY2reWh7Q/x7YPfXvEd/vJHvkxz8N33HXi3cd0E+ZfOvcREYmJF3bwz0snQwhD//Yn/TtQbpSxZSdT3YLM3IUoulHIWpWyw1t+E21KkLATpa2jnyOQRTkyfIOwJo6PT7G9mS9vlDQoypSyPTx7meHwch91DT/M6doU76bTYSSllEieOYTt1moCq4+ldg7jld0FfBk0BX4epLf8+wnq7G58kM6qU0AydDqudHqvjqkbX14rFzCLPDDxDxBPB5jY1T07PnkYSJG7suXLGJQgCXz6/sD+cW0I1DG6xOTkyd5rF+Cij7gjTwVY2ljPcWs4gYNZ/JVG6RNXxp2GVrXxyyyf55qvfpM5dh8vqIp6PI4kSd/XfBcBwbJjDU4fpifbUdgQ+u49Ts6cYWBhgbZMpUOe0Obln/T3ctuY2DFZq0ZyaOcW3DnyLWC4GmBz0T276JDu7d/JrH/k1/uL5v2BqeYq8kqfV28qe3j2E3CGCriClaskkFpy/3lw5h6qrqKqK2+ausVFeGnoJSZKo99aTL+eJ5+IMLgxSUSv8/I6fZ3p5+hKpiKnlKRYzi0S8EeyyuWPIFrOcnjvNr976q4iiyNrGtXztua8Ry8ZMqrGu0xBoxOcIMp+cwuEK4nL4CbVvI6yrJGIj7Fmzh1tX38piZpH//eL/Zu/wXjKVDKIg8tzAc9zRdwe/fPMv11QuL4czc2c4MnEEh8XBnjV7CHvDVzz239/x77ltzW28dO4lZEHmtr7bWF3/FocV32O4boL8TGoGQVwZaFRNJV1KM5eaY1PbFl6yeUmXSjQqaTqDGmOVDLORW+nKvoLLSHFwLslCPk5VrxLPxrmx+0b6m/qJeCKXzeLz5Tx/feh7TFZy1DsDVEtpjp5+gtyqW9gYXY364gsEh0coRMMsSjINg2donZ9H+PSnTVGv9yEkQaDT6qDzXfCeHZgfwGl1ouoqZ6bOEM/GEQSBmdQMG1o21LTFL4cLgV41DH6cX+bw4jB1s8eJN65n0RvFNnUEW2IU4Xw5IFlI4rV7r4maefe6u/HavTx2+jGWcktsbtnMA5sfqGWB8+l5REFcUfIRRdG89uRMLchfwE9rKuVLef7q5b9C1VUafeZilSqm+PvX/p6eaA9b27fy9Qe/zsDCAEXFNN5oDbay99xeplPThFwhQu4Qy4VlnFanaUZSzOC0OWkLve6kNRIfIewJkywkOThxsKbb/qMTPyJdTFPvq0fTtRU7m5OzJ3FYHBiGwWJmkapexWszy3Reu5fFzCJf+cFXiOViOGwOqtUqhWrBpBMbOqpaxiZK7F59C+VimucT4wSLaU7PnSZbyjKaGOXl4Zdr125gMJOa4Zmzz9Ad6eaTWz7J5fBHj/8RDx97GAOzlPrnL/w5v/+J37+EmXMx1resf1OMqPcLrpsg3+RvqumaXEC+bOpkt4XaSIkyRVEmYrWTKtmoaCI+uYxVrJJ0RDgztECiZNZa85U8baE2hmPDbGrdRMHQGSsXiWtV3KJEj9VJnWzhzNwZZss5WvwNSIKA3erAJsvMjj+BXxll1/AZyi19SKKM3TCYDwUJLaVwT07C6g9GFvFOIlvOIggCBycOoukaAWeAgmL6jr4w+AIf33R18TNBELjN6edcPsmgu475XpO331JYpkFTiGdjDC0O4ba6iXgj3LbmthXTy1fDjT03XnE34XP4LvnsgVlGuhYRtbMLZ0kVU3SFu2qKjGFPmLH4GGfnznJL7y14HB62d650CNvYupHRxCjJQpJ1Tes4NXuKscQYaxvXUuepY3fP7hULit1iR1EVzsydwWlxUlErjMRGUDWVI1NHCLlC+Bw+NrZuxG6xo2oqsWys1kiNeqPohk6ulKOiVlB1lX987R9ZLizTFe4iW87icDvwV/zkRZmNjb00rv4IAU+Ixcw8Y/ExRIudlkg3NtnGzPIMT55+kkwlg8fueX1XZUCqlOLVsVcvG+T3ndvH949+n/a69pqGUaqY4r89+t/Y2bXzmvs4HxRcN0H+ttW38X+D/5ex+JgpamSYW82oJ0pfYx9JBAQDs4EqWFClTrzeVjLJIxxZDpCIlzGMEgvpBRp8DXRHuknkE5yJj5GtMwcv3KJITtfYV0yzzebixMxJCrpKvpTFY/ciUaFOO8lMfg7/chGnNo6sFMlbN4FoQxIgJ0locxNozWECzsC7blP3fkJ7qJ1Hjj9CpVrBJts4M3+m5oX64xM/ZkPLBtrq2q56jiVdZbOSY9D6etbfn4+Tc3jZ2LaFm7tvpL3OHK+/2r1XVIWB+QGqapXV9atN/ZcrYH3zehp8DUwtT9EceD27D3vCbGrZ9IavO1vKksglyJayNUekpkATBgYFrUxB1zAMg2WtimJASJYJiDIhd4hPbv4kRyaPsJBZYFvHNn5p9y/R6G+8pFEMsLFlI8+efZZ8OY/f6WdgYQDd0OmKdOF3+jF0A6vFSiK3RMHQqCDQ0rSWc1NHMQwDRVPIlXOkCqmacuSRqSMEHAE8To/5WDGFLMpUKgXsjgB33fBZ3A4f+0f2spCNoSCSzMdJZeaQBVOpUtM0ROvruyBBEDAwKFVLl71fzw48i1W2rhCpCzgDjMXHeG3statm8x9EXDdB3mq18n/+1f/hfz37v9h7bi8IsKd3D211bVgkCz69ioCBYugIgM1iRxJE2us3cujEEwzMnkYSJZxWJ36nn1QxhSRKTOtV/Bg1YwiLAIau8q2RV5mcPcFYbolxq5OIO8TOejuaVqRqCWAEW9BYQtJz2NQpytZVFEpZBkcPsOCMkRUnCDqD3N5/+1XrjtcTeht6kU5IpEoplrPL5JQchXKhZkn3/aPf59/f/u+vmn27BZFj1pXiZQPuCG3ZBURBoMHf8IbZ9fDCMF997qtkS1kwzDr3F3Z9gVvX3HrZ462yld+553f45ivf5NTsKQDWNK7hizd98Q2nYw3DMM2vy3lcPhc22Ua2lCVTymKNdLEYaOPh9CKnczHslQJtziABdx3dNhfr7S7CnvAlU5u6rrOcXwbM4HehjNTf2E+ykOTI9BGy5SyFSoH2unbCnjDFShGf00fYU0+wcwd+uxeHxUYp1E5StjE+9hqZwpKpNX9+YveJ008QdZvCZtaqqVOv6zp5JU9V19nYtwfZ7kEQBBaTc5Q1jWZ/mLBWBSBdSGOVrOiaTrFSxGVzYWBQVsr4nX7WN12ltHIZLxsB4QMpJfxGeNeDvCAIdwFfw5T8+FvDMP7Hu/2cV0LEG+EPP/WHtcESTTcnFWdTswRdQVo1lbMWFw2BRjKGTrZaRkhNY80vs7ZpLT6nD6fVyXJ+mSdOP8GtvbficYdNj9aLsLg8zVhiFEUpU8osUBAlFpfGcRZ1gsEuNrWuoj4UJLuqAe/QLPa6KZJaE7GTe2nwB/D1rcdrMeVZHzv1GJ/f9vlr9q98T8M4r7kvWi6rmvhGsFvsfGrLp/j6C19nIjFhltrq2rDLdhL5BCdnTjK1PEVnuPPyT28YvFzKMIhAYzZGc2yIhcZ+pl116NUSm+1uIp6rSz8rqsJXn/sqkiDRHTGb+MVKkW+88g26o91XZGNEvBH+4z3/0XQ5UhV8Tt81lYKW88sUlSI7unZwevY0ZamMgEBCU9lcv4aQZOU7I3vJl/IIkpWzmVfoDbWj995Ko8VKWLaSKqRMsw9BwGFx8NLwS6SLaRxWB0FXkNv7bifkDiGKIresvgXd0HnqzFPYLXYa/Y1UtSqlaokNoQ0sAqIgsf68rssafyOl9u1YRAvtiTFcdhcemwe/y08sF+PmVTezb3QfqWIKn8OH3WInkU2wsWENcmwUJdxDQddJlzO4RLkW4AGsFitRb5SA87yyZSkDmBrxW9u3cs+6ey57z+5ceyePnXpsBSUyVTClFrZ3vLHp/QcN72qQF0yNgP8N3A7MAocFQfiJYRgD7+bzvhEurOaiKHLv+nsZXBhkND7KOsnCrdFOknYfc7lllOkjHDv8z6jVMggwl5xDEiUEBGLZGJ/d+lnCVhfDS1PIahm/M4BNsnJ48hgzc6dpEiVckpViIYmiVTldTvBvWtfxqY4+EEXGdvax5JVxnZ2inJzCaI1Q2HETxnkRL7/Tz2xyloXMAq2h1qu9pPc2DB2WzkDsGGhl0/CkcRd43zw9bVV0FU3+Jg5PHqbB14CISK6Soz3UTkWtMJGYuGyQNwyD/52a5/F8krtcAVa7/JxWCnhmTtBcv4ZZfzOTF4ubGcZlF6Kz82fJlrK1AA/UqJOHJg9dlXKXLqbZP7af6eVpJFGiv7Gfre1br8rgKSgFJEni5p6bafQ3Mrw4jGYY+LxRtkZ7eHZkLwUgaHejiRJioJnBhbMEA00sODzEYufYe24vkiBRrBR59PSj+Ow+GvymzlJ7uJ3HTz++IpHY3bMbQzf45mvfZCG9gNPmrA1rjegaay8S7moJtjCaGAd3iEZdRVPLSJJEX0Mfqq4S8oT4pRt/ia+/+HUSuQSCKLC+eT2f2/Y5lgvL9FaLhIMtJANNDJXSJHMJrLIVTdfQdI0bu2/k7rV388NjP2RgYQCbxcb6pvXcv/H+K34nbuy+kc9u+yw/OPKD2u9sFht/8MAfXHf1eHj3M/kbgFHDMMYBBEH4LnA/8DMN8hfDIllMc2JXkJA7hMvmolKt8I8nf0yvw8+A3Y2mVShVSxQU0zVHEARKaomyWub02aeYcvixYzAzsg9F10grJZKxcyzkErSHzDFtVVPRKxaE0jiirmGTLPQ73FQ2hFB3fIJy2s6pcy/hdqzcvguCYIomvZ+ROAVzr4CrHqQQKDkYfxRWfRqcV6a1XQ4W2UJ/Yz//dPifmE5OIwoiLcEWIp4Ii9nFS8xos6UsU8tTPKzrvGzApzx1fDnQiAHcFWxBNHQkBP4ys8jDuSVkFL5sO4egToPoAvsWsK6pBXxVvfwEpCiIlJVy7Wdd11nKL6EbOiFXCM3QTHljdBr9jWi6xsnZk+Qree7ov+OKr9fn8KHrem2YaU3DGqq6zjOLw3jsXmaT07gb+kCrIhgGxnk9nqH4CLujPRwb2UeDrwFJlHj67NNU1SpltYzH7kE3dIYXhumOdK9IJGRJZk/fHlrrWnni9BOI59VPlvPLbOzZvaKeb5Es7Oi4gYVyngYBPFYn9b567BY7i5lF3DY3H+3/KIZg1CSFPect9oSiOULoFiW2tW0mW0zRE+5hubiMTbahGzo7u3aytnkt/U39VNQKhmGe543KLv/p7v/E/Rvu5/DkYRxWB7euuvWqFMoPMt7tIN8EzFz08yywYr8kCMKXgC8BtLa+tWy1VCkxsTRBRa0Q8UbMDO8aam+ZYoYnzzxJqphCwGzm7OzcScgdQtVUNEOjs66TkdgIVc2UTnVYHIiiSIO3gZeGX2Jn1042W2ycKecpSzYEQaVZyxIvptEMjeXCMvW+eorVIhapgYlMhaPjP6Gzrps6dxC7Yz041hPV0zVPywuG29r5hlqd+30skKRrZgbvaiSjFJlaHiZfzhO1ytTPH8DVfd+1n0vTiI+cYnzgAB3uRqqyiEW2kCvnGFocYl3zOpr9r2fSI7ERnh98HoAlfws9osiNGAjBJgTALghckG/6cqARjApuZQhBToBk/kzhBfO/drNBurp+NbIkU1SKNT8BVVepqlU2tm4EzGD49NmnzZo9Zk2+PdhOqVqqORnJkkyTv4mxxBjpovneTy5NouoqLUFTSEwQBPxOP32NfZydO0udpw5JkIjn4mYgddch6iqSpqKJMrogYlfyaKKEjoBYStaeK1fOsVxYNoO7LJMUJbyOIFpmgUQ2cdlEojvSza/c/CssZBbQDZ16bz0pUeRQKYddEBHPL3xVi5ONwWYkrUK9rx5BMDXmNV2jO9KNKIjIkkydp67WDK2oFURRrImj9db3spRb4uz8WbwOL7qu01nXyebzmkcXhtfeDNY0rmFN45o3PvADjp9549UwjG8A3wDYunXrZdolV/1bXht7jf+77/8ym55F1cwvyJ7ePXz+hqvXsQ3D4LnB5yhXyzXGg6qpvDr6Kuub13NyxnTR0Q2dcrXMfGYezdBMmVpfhHv6buTQ1BkMDDyGhjh/msZqGdEwsElWwu46zsXOkS6mWcovmR90UcZh6+VEqpdzBYOoP8zta29CFiRC7hCbWjdxdOqo2WAyTPbADe3v3GTfzwRaBXSFpWKGg5OH0KxOFKuLmWKS0LnvsyUoEnD1grXN1Ea/EgoFePxxBif2E1CW2aU4GJBzaNEAAWeAolKkJ9pTk6EtKSVeGHyBsCeMVbbShE5VVXhtfD/tdW2X6M4LgsCX7TEEaR4kU0irpApYxAiW0lGwrQXBgtfp5Qu7vsDf7vtbBEEwd1pqldv7b6evsQ9N13ji9BMIglAL6CWlxFNnn7pEIlcQBAQEzs6d5dTcKSRBQhRFDk0equmgC4LA7p7dhD1hzsyeoaJV2N6xnUhkFUerZULBVlLpWcS6ThBEBMMgrWl8OthEQLbVqJuqpuK0OEk5RLRINzlvBEW2UpEdiKX0pbIJ52Gz2FZct1Ep0mjoLGhVRAR0oE6ycHP3Tg7oKhNLExgYuGwuPrb+Y7X7fHPPzTU5kQs00I+u+WitfCKJEresvoWNLRvJlrO4bK4PCQfvEN7tID8HtFz0c/P5370jGFoc4i9f+ksG5gcoqSUEBBaziyxmFnFYHHxm22eu+LfpYppELrHCI1KWZGyyjWcHnkXRFAKuwHn7OgsCAj67l3q3A1kocmz8CTRNRVYnwYggAKKhm6p8Di/3rL+H0edHkZGJeqK1YQ5VV4kEV2OTbYwvTzO1PEVXpAuAHZ07aA22MpYYQ0CgK9JFwxWEsd4vMCQbKcHK3oVh0p4GDNlCQM9RL8eYdNTz9PI0n2YKWWkB991XDvQHDkA2Sy7oxVaV6JNb8cydY0KTqXr9RDwR7lhzR433HcvGMDBW1LstkulJu5BZuKy5iKAnQHAxn1lm7/hZ0sUCkiiyPuJmiyuHbDWDzp41e+iOdHNo4hAVtcLGlo01GeBYNlbT1Tk4fpBStUTEE8EiWVjILNASfP3roBs6Va3KseljNPoba9eqGzonpk/QHekm6o3W6vf9jSutJb2aA2v3TfzzyUfJj+7DZmjklTLbI1383CqzgSpLMuVqGbfNjd3mwRbpIbY0iVZIAVBUy4TX3AaWq9eqy9Uyr4y8wkh8xGTQOPxs7rmJFl+05vV759o7KVQKZmPZ4Vuxm+5v6qcl2MJ8et5cAP1NlwyvlatlcuUcVtmK3+G/6vV8iGvHux3kDwM9giB0YAb3zwEPvhMnNgyDF4deZCQ+gqIpBJ1BBEEgXzHHsZ8ZeIa71919xSlI3biM4iNmoytTznDr6ls5MnWEkdgIiWzC1I4XNHx2EY81ylBiia3NXSwun2JsOclMpsh8eh4MWNWwimQhSX9DP4VKgYArgGaYDvdOi7MmZuayuZhNz9aC/IXs7+KF5/2OkWqFEd8aGN+PZGlAMUBRUyyKIYaDW6modrbgo0cdB2UcbKsuPYmqwsgINDTQni+xL5nAa3HSHu6g3YBizxYqamXFdKooiBiX49HBlVktUj3J3Kv85MwIPoeTJn8IVatwdG4WzX2GXd031w5tDbVetvGnaiqxTIyF7IKp0SJbmU3NkimZNnQzyRmskpVytUxFq9AdNuvhFy9GFzTvZ5OzK7TeS0qJ2dQsiqoQ9Uap89RxT2MvO3wRDk8eZjm/TFddFwjw3NBzNPubuW31bbx47kVUTcXtazCVKK0O7BYHmqHRXteGLEq8OneWu9s2XvF9fHHoRaaT0zT6GxEFkWwxy6Mnfkzvuo/hs3tZY3fSZLHjsrmu6K/rdXjxOi4/TzC0MMTL516uySsHnAHuWnvXVems8WycU7OnSBaSNPmbagy4D7ES72qQNwxDFQTh14GnMSmU/88wjLPvxLlVTSWeiVNRKzV3JqA2sl1WyyQLySsG+YAzgMfuIV/O144xDINMKUPUE8Xv9LO7ezdnZs9gtVhpcTaDnmI2m8dt1XDbHawONzOUmKSkDlHSwwzMD6Cjs1RYYim/RMAZ4Odu+Dlsko1jM8doDbSSq+RQNJPapWgKHtv7Q0b4rUAxdIYrJbyh1ZyO7sSnJLCqeSY8HZQ99aZSumRnr+Kg3ebFUp28fJAHs/FpGPS4mhjKzTJbWsKrS5QNFSWf5J5196zIHOt99dhkW03NEUzDF1EQafJfYREVmhmbGsWlJnFavFS0IoaawmJr5/FTTxP1NtFR13HVfk/QFWRiaYKwN4xVslKoFFA1lVw5R0+kh4mlCYZiQ1gkC43+RmwW22WnYS9k+YPzg6Z4niBxYPyA6cB0voy4qXUTOzp3EHQFubP/TtLFNH/81B8ztjSGRbKgqiqr61fz6x/5dUpqiWVd5zszJ8inZpFFGb/Tj9/pJ4vI0OIAd7asv+xryxQzTC5P0uRvqn3PhgU4XswyP3eaxqb1vFbKcrsrwA7X5YOsqqkIgnDZBXYpt8QLQy9Q76uv1eyThSTPnH2Gz2z9zGWH0maTszx68lEcVgdOq5PBxUGGY8N8avOnPgz0P4V3vSZvGMYTwBPv9HllScbr9OKyukgX07UmWFWrIogCQWfwqtQ0URS5bc1tPHbqMdKlNJIoUdWqtIfaOTlzkuHFYWRRxuMwx6ktFgvNrhCy6CRdKRL1+MlUCtzavRZVU/iHk2P0N/ejaRq5co6IN0Iim+DA2AF6oj20BlpJlVJYJSsOi4NCpYBhGHSFu97pW/OeQUHXMDCwiDKRxs28nIkj2J24xWUkTceppAg5XCBAXIOmi01ULoYsQ08PjI1hq6/nvvrtjBbmmJ4exLN5J71bP3aJe49VtnLPuntWNNZlSeaOvjsuXfgNAxYOQvwEtqUy7aUkGS3NsKORU/EiBXUem1LmR8d/RG9DL3f03XHFfo9u6ES8UU7OnmQ2NYOmazitThp9jfzwxA/Zs3oPn9xsjuJrulYzFkkX07WstVKtkCwkOTR5CKtkxcBg38g+Ao4Aq+tX0xRoQpZkjk0doy3UVlOF/N7h7zGxNEF3uNucCjUMhmJDPHHmCX5h1y8QNXTsc2cIhLtwnJd+1gBZEJHLeTRdQxRFDMOoTdf6HD7KqsnNvxBs46rCaKVEyGLHpVapl61UdZ1ni2n6bC688uthJVvKsn9sP5NLk0iiRF9jH9vat634bo4vjWORLCsmVIOuILPJWZKF5CWGIIZh8OrYq/id/tp76bA6iGVjnJw9WVP//BAmfuaN17cKQRC4bc1tvDzyMtlStuaAY+gGEV+EjS0b33Cwpd5Xz4M3PMh0cpql/BJ7h/eyb3gf2XKWY9PHsEm2mg2Zrhloho1UIUm6VKHFV4fDascpKyxpYUrVMzT4Gohn4yiqQne0m6pWJZ6LmyPoVgNZkKlz1zGbnGW5sEzEE+HY1DG2tG+54jb2/Qy7IIIAumGwqq6DSQyGSnkEXSRk5PB7G3HKNoJCiaKugvUqej07dkAqBbOz2ASBfl2if8snYM8ecxG4DOp99Ty0/aFafT7iiVwi/gVAehxih8HTiq1uHWdmDVp0ATVWQCGAZJVJK2UUVeGVkVeoc9fVdGIUVUFAwCJbKOkaR0tZTlcKZO1ebM4QPtmCgIFFtLCUW6KoFGtPK4kSHrsHn8NHqVpiLmW2qyRBQhZlot4oTquT41PHOTVzilK1xNHpo/gcPu7uvxuv08t4YpxGf6MZ+EZfpSXYUgvGF2rf+0b38Qu7fgGrILLd4WdfLo7m9GMAgmEQKCzR6Aljkc0BvOcHn6+JvwWcAXav2l1LgiyShVhVQQBUtUzgfM/IIoqgGcyoFfrPvx+VaoWfnPgJiqbQ4G9A13VOz54mU8qsGGSqatXLZviCIKBd5N528fGpQuqSsqbf6WcmOXPJ8dc73rdBHmBDywa+eOMX+dGxHzEaHwUB/A4/t66+lU9v/fQ10SidNidd4S72jexjMDZIR9jcji+kF4jlzCba57d9njNzZzg1dxxRL2GTBcIuC5NLY4i0kyeIqpoj2+lSGrfdjSRKhN1hc8Tbbpoyf+Wur6BoCn/xwl8Qy8YYS4yxb2Qfj556lN+67bdorXtjCmlZ1xlRiqS0KvWyjU6LHfk9OqrtECVaZBtT1TIhycKuQBsla4qyFiIqpQiQoU1cpGDY8Hl2gnwVHrPTCQ88AIuLUCyCzwd1dW84OWuRLTQHm1nMLHJg/ADlapmucBftde01NySWz4A9CIKI0xPBYnEQU/J4sxOIbg/T8TFkm5OJzCx6tcI3X/0mPoePyeVJppamQICOcDfV+l6mc3Ga/I3E5s/i9jdhlSTslTyJfILWQCsj8REk0WTR1LnqalIZ92+8n0Q+gaZr6LrOY6cew2l1kill2DuyF93QTe368+bZj51+jI+t+1gtOC5mFplPz5MupAm4A0Q9UWwWm5mBXzQ8cHPrehaPPkKmnMFlc0M5h6BV2bXpflRN5fFTj6Pqam2oK1PK8OzZZ9nWvo1XR18150hUhUxmkfZgC6HARcNfAiskpaeT0+QquRp7TZREmgJNvDryKjPLMzQFmrh11a20Bds4MX1ihbNaUSlis9guy7CRRRm7xU5FNTWMLqCoFAm53roN4AcV7+sgD/DxDR9nc+tmRmIjFKtFesI9rKpf9aaEveYz88yn5/HYPLUv/oUtca6co2pUiXqj3Oi4mZKSp9XnoT3g529fe5IXJw8TcI0xmzabaxbZpNlpukZZLXP3urvpCncxGh/lldFXeObsMwwsDNS23ZIgMZ+e5zuHv8NX7vzKVRemhKrw7XSMrK4hCqAZ0Gqx8nlfPfb3aKBfZ3djE0XGlTI6BhscbjDchOUGrJTIaAp1Fi8hxzXQ5UQRGt+8gcrZubO8NPwSTpsTWZQZjY/SEe7gzv47kUQJVSmSKqaoFvMkdQ2/y0+ikKRUyaPay/jdQbz+JtwWK4KsYOgGX3v+a2xp3UKDvwEDg1OxYZaSMzSFWmmtX83M/FlKSoFU1aDJYqO7rpuckuPM7JmaC5Nu6EQ9UXZ17UIURbx201e0Uq3UmsZTS1MomoLX4aWklBAEAbfdTS6d41zsHA/ueJCppSkeO/UYjf5GJhITaLpGspCkN9rLbGqWu9e+rl3jsXv43NZPMbQwxFx6jlD9Kvoa+gi4AswkZ8iWs7WgDOYw1kxyBp/DxwObH2BwfhBrOUsq1ElnpAvpgm2gpuIUJDou4rJnipkVJRhN0/j2oW9zcPwgfqcfq2Ql4AzwJ5/9E9Y2reXs/FksksUsGwkid6+7+/WFeMXHQGRr+1ZeGnqJBn8DVtlKUSmSK+W4qfsmUoVUbfG83N9fb3jf3wFBEGgJtqygpr1ZFCtFZFG+hI0hizKd4U5a/C2cWzhH0B1kVXQzrcFWvnv4u1gsXlpcDuyyHQzTHCLijTCbnMVr97K1dSs9kR7GlsZYLiwT9UZJFpLmUEkhha6bCn9BV5DJxCTpUvqq3OCn8knKhk7TRSWH6WqZQ6UsN7v8b/n1v5uQBYE+m4vVVic6BhICs9UKE9UymuFkrSNIm9WGdI2LcqVaQRCENzTzuIBytcyrY6/S4G+oBRy/08/k0iRzqTncdjevzI/hT55lWXQwHB9Ds3tY46sj62tnMjlDU6gdQQALAumyKTN9evY0TpsTQRD4p4P/RLyUpWfVzbQEmtE0jVVtW4klp7G4AoTVKk8f/wFlpUxvfS8VtYLdakdRFFKlFAFngBeHXmR4cRgwtVk0TSNVTJlNVySinigTyxO1ermiKXSGO4l4IvzTwX8i5A5x99q7efjYw8SzcapaleOV4+zs2skDmx9YcU9cNhdb2rewhZVGN4qqXPYeCoJARa3QEe6o1f83lHI8mk+RUysYgEeQ+Kw3jPWiZKPOU7finIenDnNk8ghBV5DOcCcWycJiZpHf+/Hv8fC/fZjV9auZS81hs9hoC7XVPGMvh/7GfjRd4+jUURRVwWl1EnKF+If9/8DU8hRVvUpnqJO71t3FtvZtHwztp7eI932QfycQcJlMG4tkoVQt1UwQkoUkjf5GHtj8AOVqmcaASR+LZWPEc3F8Th8em4eoN4qqq3SFuwi4AnjtXrx2L70NvYzGRxlYGKgZPkiShKqpuGxmw7iiVgDQ0ZGv4oNa1FUmlTL10soPa1CSOVUuvGeD/AVIgoB0vmzQarXT+gbqixdDMXRO55Z5fu4sy/kEAaXMpkADN3bsIJ433YtShRTNgWbWNq1d0YS9sJhafuq+2S12ppPTLGQWqDgaqQ+UsC+eotkiUSgvk3f5UDrvQkz/hIHZ02xoXkvW4iTqi1JUiqRLaZbzyyaDRinwwpkniJez7F79ETKVAl5vlGwxSSwbY9/px5lNTtPkb6LR30hBKRD1RdnatpWCUuCps0+RLWVp8DcgCiJFpUglV6klHwWlYFr1dezAZrGRLCTRdI01DWtMo+xyppZ9P7j9QSYSEyxmFwk4A/yHO//DNQe4kDuEYRhoulYrA+mGftmp636Hhy6bi9lqGVEQsBczPHnw25ycOYnP6eOedfewrX0b9b565lJzhNwhXht7DUmQCLqCtfcj6o0yvjTOSHyE3obeazJpAXPh2dCygbVNa83a/8mf8IOjP2Axs0i5WkYUREbjo4wvjfOZbZ/h/o33X9N5P4j4MMhjftB6G3opVApMJ6eZWpri1NwpU49ELfOnT/8p65rXsZhZpMHXgKpraIJEBWh3hxhfGscqW/E6vTT4Gnhg0wMMx4YRBZEfHPkByWKSilrh1ZFXqffV12qKAFX1vMtUz414HV6KlSKvjb3GmYUz1LnquGXVLTQHmxERL2sVrgPWD7DkvGYY7M+leG7iMGK1TJ3TT9ElcSAb57Vn/xSX1cnE0gRFpWjKBvibeGDTA9zYfSOiKJoUxcvw5ata1dRgzy/TFGgi5gpyeGGWqCtEStXJSA3E4xNE3HUUsnGESpGsqpIpZRhaHCJbzvLC0As0+hv5+PqPYxgGj516jH986f/wmw/8IZPL06h2NwPP/S9mk9Ps7NzJ+qb1hL1hqlqVTCmDy+ZiJjnDTHKGbe3banIWVsnKcmGZeDZOwBVgR+cOppPTFJQCk8uTpglI8zqmlqeYT81TrpZriosWycKq+lXUeepo8DVcU4DXdZOu6XP42Ny2mSOTR3Db3AiCQCwTo8HXQLqUxmFxkCqmSBVTeB1emvxNdNucJPNJ/svjv0+mmCHsCZPIJ/jqs1/loRse4p5193B27ixDi0OAmVBdvGAIgoBgXL7BenD8II+deoxMMcOGlg3ct+E+0yP3IkiiRCwb40fHf0SqmGI6OU2+nK89l12y8/3D32db27aat/P1husyyA/OD/LjEz8mkU+wq2sX96y7hz29e2gNtnJk8gh/s+9vWB1dTX9jP6IgMps22TB39d/FWDbOvM1LxRvB5Y6QsLooqAoBu5uKUqEn0oMgCPjsPv7ypb+kzl1HwB1gMbOITbYxl5pjTcMaYpkYFa2CO+Wmr6mPn9/+8+TLef7gsT9gfHkcj91DpVrhqTNP8R/u/A+sb1lPt9XBWLVMvWiWKgxdJ6Vp3On64DFzLiCpVRkrLCGUswTPBwevoROzWBlZHKbVG0WWZFqCLRiGwVJ+if2j+2n0N9IV6SLgDNDobySWiRHxmjaNF+irHruHfCV/Xi9IouKsZ8LQUEWVVGbRtP8LtRK0WMgWs4zGR7FIFlqDrciSzGJ6EY/dw4HxA9y66lYskpVHXvt7XBY7X37gjzh06FscPfcSe3r38KXdX2LvyF6qahWLbKFQKfDoyUcpV8s1rZdNLZuo89Rxeu4086l5WkItrGlcQ8gdqpUjT0yf4N7199IaajW15tOzzKXnUHWVzrpOrLLJzc+X86zvv7qVnWEYnJ0/y+HJw5SrZYLOIPXeegQETs+eNgenHF7cdjdPnn6SocUhmvxNhNwhNF0j6Apy34b7eGHoBZKFZE2Z040bj83DIyce4fb+283SUPsWrJKVP3ziD9ENHUkwdwrxbJyIN0Jvfe+Ka/vxsR/z7UPfJugO4pBNeeSjU0f575/47wTdK0uap+dOk8glWMotkS/naxr96WIai2TB7/JzYvbEh0H+esHTp5/m937yeyZ/W7bw4tCL/PDYD/nrh/6a3oZe4tk4Db4GeqI9tb9pDjQzEh8hGmii2rqZVr1KqzPAt1/8C6bySXL5JUJWBxua19fMf8cT4xSrRYLuILqu47A6KFVLyOdH6z/S+xG6w91s69zG6uhqRFHkxyd+zPjyOKuirw8EJQtJ/mbf3/C1z32Ne9whvpONMVctAwIG0Gdzss3xwR2oyuka6vk6/AUIQKmcQwXi+TgdIdOZ64KWjCzLDC8O0xXpQhAEbl9zOy8Nv8TUssmE0XUdzdB4bfw1hheHGUuMsb5pPUFXkOHFYXNIyzBw290ki0m2tm7l7PxZbBYby+e9BXRDx2P3kC6aui9N/ib+1Y6HyJdz/HDvX/PDvX8NwEPbH2JX1y6cNicbmzdybGGAarXI4MIQ3eEu9vTu4fj0cQwEDsycYFvLBhbSC9itdlNoTxCp99Wj6ioOi4M71t5B0BUkU8pwePIwiqaQzCcxdIOSUiLijeC1ebl3/b3YLXYG5geQROmyMgKnZ0+zd2Qv9b566tx1nJk7w8PHH+ajaz5KZ7iTHx77Ibqhs75lPU6rk7JSJpaN0dfYZ2bQmRgHJw4ytDh0yQCS3WJH0RQWMgu1ie77N97PvtF9vDryKpJkGnq77C5+7+O/hyS9TqEsVoo8cuIR2oJttYDttrsZj4/z/NDzfGbrSrmSC9pQFa2yYuDKMIyagF1JubyL1PWA6yrIK4rC//f0/4fP6VsxLj24MMj3j32fL9z4BdLFdG3bfDEEBOYqeaRgK0HBwdnMPNu6biRRWOLk1HEoZ1B1lYJSMD9waqVWkhFFkQZfA7lKjmK5iNfu5Vdv/dVaE+sCDk8cvqT2GXQFGY2PsphdpNHfyC/7G5hSK+Q0lbBsoelNKvO93+ASJRw2V83o5QIkyQK6hs1y6Qi9RbSskK1w2pzcs/4ecuUc+XKex089jsvmwm13Iwoi/7D/H3hx6EX8Dj8GBqujq6loFebSc9hlO7OpWbNcITtqi7QkSEQ8EcLuMF6Hl+HYMJ3hTn7lll/h2cFna8/99Qe/zncOfYecqpCv6yBa18F8ao6QI0iz1c7J2ZMsKkWm4mO4fVEysVGy+SXWhjtX1KetkpWKWqmxT45NHcMiWvDavWDAptZNpEtp7lt/H43+Rk7NnuKJ00+AQG2Q6fY1t9MZMbX2NV3jyNQRGnwmO0XVVObT80Q8EaaXpxmJj4AADtlBMp/kdPo0bcE2FE0hW8qaZRdPHSOxEaLeKOdi5+CiNUTVVQRMnn3tNVis/NnP/RkHxw5ybPoYQXeQO/ruuGSQLZaNmaqvP9W38Tl9DC4MXvJ+Nwea8dg9NXP0crWMYJguUFbRil22r2AMXW94b/Lu3iUMxYbIlrKX6GH4nD5eGnoJgM5IJ7qhrwgq2vnJzXpfEwICS5l5cqUUdf4GkCz0tG+lKdjGxNIEPzz2Q4YXh/nsts8SdAZJFU0hKEmUcFldOG1OvnTLly4J8AAem+cShoOmawiCUJvolUWRLquDjQ7PBz7Ag6lw2OwK4PY3EsvFKVZLLGlVrLpKf7Qbt91NvmLWYLOlLE6L0wzU9SsHq1KFVM0hqapVcdvdZqCbPIJVthJ0BemKdNHf2I+qq7QH2lF1FbfdjYFZuz86dZSSUqqVR7LlLGFPmMmlSdpCbfgcPv7+tb9f8by/95Pf47bVtzGo6ywU02j5JXyCgc/mYsriRPI24KrvZVXzepR8Eqco4Qq2EGlcS0zXOFcpMl+tkK+W2NiykbJiynWUqiUcVgeZUoagK4jP6cMm21jMLJIupnl11Oz/NAeaaQo0EXKFeH7oecpVU/NeUZUVzkkVtWJm1lYX08lpNF3DIppTqBfkNxYyC6YEw0U9oAtqkhf6GxfONZ4Y56bumy4prUiixK6eXfz6bb/Og9sfvCTAAzXxOFVfqd2fr+Qv+725oeMGswfhbaDeW28qvsqmMGCDr4GuSBdrGq5fyeHrKpN32VwYgoGu6YjS6+tbVavW6Fp9DX1sbdvKoclDtVJLuphmz5o9rA2383Q2wanpE5ydG0SQLTisTryuAE1N/cScPlqCLbSF2tjZtZP/et9/5Xd/9Lsk88nac31848e5dfWtl72+j/Z/lKNPHiXgCpimCbrO1PIUm1o3vaHv6PsRxYrJUrFb7FekjkqCwC6nH3/LRg67gsQyizTqVW7p2E7D+nt58syTPDfwHLFMDJ/TR2u0lbVNa+kMd6KoCmfmzvCTkz8hlo3R6GukolbIlrPU++rJlk1DEb/TT7laJuAyZYvHEmNMJCdo8DVQqpbIFDPky3ny1TxqVmVyeRK7bGd983p6o720BlsJuoL87St/y6MnH+W+DffxmS2f4a9e/iu+9vzXyCpF7r7v/4dVKSIKIqIo8Tev/D8kdx1jooRYyeF0BYmE2ljVtBbR4mDfzAlW+RuxizJjhSUCriCrWzbgsDh4/NTjxDJmthtyhdjQvMG8WYLJhplPz5Ov5Dk5c5JsOVujLKqaSjwbpzXUik224bF7arr4F4w4cuWceY+KWeYy5gRuW7CNxkAjZ+bOIAiCuXsAFrOLrGtcR0e4g/9493/k7/f/PWPxMSyyhbvX3s2D29+aFqHf6Wd3z25eHH6RtlAbNtkskxmGwe1rbr/s8b9y86/wrYPf4szcGRr9jZTVMlbRSke4gy/v+fIHcqL8WnFdBfmuSBdr6tcwGh+tKQhWFJOqdkFPRBRFfuujv8XL517mlZFXkCSJX9j5C+zq2kVRKbL30LcZyqdIFJYoGwYW2cZGXcNid+OwOOit7yVTyqDrOh/p/QiP/NojPDf4HEWlyLa2bWxo3XDF69vStoXP3/B5fnj8hzXDkP6Gfn5l96/8i9yffykYhsGJ6RMcmjxUM0ppC7Vx25rbLmsMYRdFNrt8bHKuhea1K+rzP7/j57ln7T0sZBcQEPDYPcSyMb514FucnDlJpmSW0UKuEDPpGZr9zcSzccYSY7Xg7nP4EBBwnTf4lkWZbCnLLatvYTGzyN6RvXRHus0mntNPwBlgKbdE1BPl1t5bOTVzir9++a954swT3Lv+Xvas3sNTZ55iY8tGREHkm/v+hsPxUb7y6T+lJdRCRati8dZTUBVUrYqslEgj4A82I+g6oZZNtMsOXKlpNLXMxpaN1EVXM6lWcVkdZkIiSixrBhabh0VdI6oqVKvVGof/+PRx6r2mQ1M8F2chs0BHXUft3omiyK6uXTxx+gk8Dg9Oq5OAK8Dg/CBFpYgiGPi8ETK5BOlSGjAI2iVuqk/hrTxCWg3R6t/B1vatAKxtWssff+aPyRaz2K32a55juBK+eNMXsVvsvDj8IlW1SkOggV/7yK/RVtd22eO3tJuDaftH9zOwOIBNsrG2aS3b2rddwsj5WUPVVIZjwxydPMrk8iQOi4P+pn42t26+Zgrpm4FwOQW8nxW2bt1qHDly5F19junlaX77+7/NRGKi1qR5aPtD/Pptv/6Gf/vIsUf4p8Pfpa1xLUPJaabjY6jlHA4MNrdupiXUwqroKlRN5ed3/PxbvsZsMctMagaP3fP+9na9AqaXp/nJyZ/QHGiuNcnm0/N0R7q5bc1tb/m8uq7z+OnHmUnOYGBwfOo440vjhFwhVtevxsAgVUwRcUcYiY+wu2c3f/fq3yEIAp3hTjrqzAbuWGIMq2zlgY0PML40zisjr5AsJBFFke5wN1FftFYmWdu4lkdPPco/vPYP3NJzC31NfZyaOYVhGPQ19mEYBoenjnBq9iQbe/fw2T2/RXfrZp4Y2QeyFS0Xo1QtE7A6Wcol2NO5k1y0m7Booc/+er9BMwxmMoukh1/A741yRq1ydvIQ5XIOl8NH1BPhs6tu5tbO7fz9/r9n37l9RLyR2sRnIp/AY/Xwh5/8wxW0yvn0PCdnTpIsJGkONDOUGOfHY/tZSM3jsNoJuupw61Xy6XP8hxs3s6ltHcWqilMq4nFEEbyfBvHdKxsqqkJRKX5gdrJlpcz/e+X/cWDiALFsDJfVRUkpIcsyDb4G/vXOf83O7p1vamIfQBCEo4ZhbL3cY9dVJg+mDvj3f/X7HJ08SqaUYW3jWiK+a1vpD04cJOIOY9cU1nnCuMt5BorL5JQyHXUddIY7SeQS3N536ZbyzcDr9NLv7H/jA9+nuGDxdrEoVb2vnpHYCDd23/imbd4uYDG7yExyhpZgC0MLQzhtTlw2F9lSlqJSNCWHDWgJtNSeoynQxEh8hKnlKZbyS0Q8EZr8TaxrXscLwy+QKqZYLiyzmFnEKllrVL9CpYCAwP6x/bisLj61+VM1RVSX1cWq6CoGFgYIuULs6NjOcilLtlrhZGKUkt1NpGENqZljBLUqi8BofARdsnK2kERKzrIqZGasVUMnrakUNZ3pxDhhi52szY3LYeHG3ttI5+Ik88vctO4ebIEmVE2lXC1zQ+cNnJo9VWtAS4JEa6j1Et58o7+xVuc2DIOjhRR9rZu4oWM7s4kxlnMJSrLE+gYvfc1b8DqD1Aof6hwoY2B/9z6rVtn6tncF7xVU1SrfPvhtnh9+nkwxw0xypmaS4rF7sEk2vrH3G7hsrqvu+N8srrsgfwFb2re88UE/BafVSbJg1tclUaI72o3f6efk7ElsVhtltcxta25bQYH8aaiaSjwXR9M1wp7wWw5o72eU1TIWcWWwudjX9q0iU8zUzuOyudB0jZArVPP/ddlcGBjklBztde0s5Zd4aMdDjCfGOTZzjFQhhcfu4dc/8uuU1TLHp46bjkpW0wwj4Awwl5pD13XG4mM0+htpDbbSE+khno0T9UWxSBaOTh3FYXVgk23MZ+ZJlpL0RbppivaglzKMjLzCzds+yy2rbmVk7hQTgy/gsbmoa91Mv81FITXN0XKGna1bmdAUqoZBQdcoynYkhx9NlLEbOpJsIxJoQbI4CTt8JNQqgl3CZXNht9j5SO9HamwxAeENyxZ5XcPmiUJiHGcgyKoWM9AUKnHc2Ry+n6bqCk7Q4sAHNyF5JzGdnObQ5CFimRgTSxNU1Sqlaun12Q0Mwu4wzw4+y5rGNe/Y4nbdBvm3gtv7budPn/lTAs5A7Q3IlDN8avOn+LVbfw2rZL2iwFhVrXJg4gDfP/x9MxiJoumDue5j3NZ32wcmW7kWrIqu4uXhl1fwtjOlDHXuuhqLCMzMMpaNmdOhVhcN/oYruzphim/pmJlrxBthODaMx+bBZrGRKWXIlrLmgJO3HofVgdfhRZZkVtWvoifaY7o65WJohsbjpx5HFEV6I71s79jOiekTHJ85zkJmgXQpTaO/kZA7RH9TP16Hl4GFASyyhb7GPqyylapWRcds2rc729F0jTqrk0oxiUWyYixNslTXxrmFIdMuUBQonHuJXKCJNY19nMonOFRYwmlxoJeyNIgSDleAY6U0IQREQMIwlRgtNqxWJ6qhIwoC29q38ezAs9T76ol6o+QreZL5ZM0U+0rQgfq6NlLxc8RT83hcflRVIVVIs7u1Dav0U/feKIL0oerjtWJoYYhENmH2PFSFglKoSVUbGGSKGepcdQwuDJIr5y7R0X+r+DDIvwls79jOpzZ/isdOPlZTEeyt7+WXdv/SVTPyC7Xibx34FtlilkQ+gaEbhDwhZtOzJHIJPrf9c9dNoF8VWcVofJTp5DQOiwNFU7CIFu7sv9N0PdJ1ppPTPHryUZZyS4Q8ISyShTp3Hfesu+eK9nIN/gYi7ggLmQUinghb27bW9N/r3HW0Bdu4c+2d9ER6eHbgWURBRDMMsrpKVtOwiQLT6Xl+dOxHzKRnWMgs1GQPbu29ld6GXp46/RSJfKJWrnny9JOkS2kq1QrDi8M0B5vZ3LaZo5NHMQyDvJI3dY3qzMGsqlZlbeNa/Eqe8eGXiKXnKJWyhESZZqePyaVJot4oXl1DLaRILu6nXMqQFgREQUZ0BcnoVSSrG0u1iKHrbG7bTErX6Lc5EQWhRh89NHmI5fwyfqefe9ffS4P/6n7BHlHCbbGzte8OYktjzC1N4Lb7aO+6kU3+JVAHQKoHZNCTINjB2v1Ofzw+sMhX8lhkC9lSFkmQamqbhilLSlU3dfXL1XKN6vpO4LoL8hf4vBW1gtfuxWVzXZPuPJiMhM9v/zx39t/JdHIan8NHR7jjDf9uNjnLY6ceYy49x3xqnmw5C0CikKAl0MKzg8+yrWMbaxqvDy6vRbbwsXUfYyZpBlKP3UNnuBOXzYWqqTwz8AyvjLzC5NIkPoePTDnDDR03kC6mOThxkD29ey57XkmUuGfdPRyZOsLQwhAlpUS9r55V9lVYZSuyJJMsJNENnVXRVTxx9inmBYm0rmERBFLZJJOVEmtkG6liioX0Ag6rwzR+8Zm1+zpvHQFXgIXsApPLkxQrRZqDzVS1Koqq8MTpJ7ip+6aaHEHEHcEiWzAwUDWVrW1bTRkFDAZG9+OTLKSKSdLVPIWshaZAE3OpOVyeOsYmD1FndVB/3hxDqZZZyC5ya8/NKEqRZW+EiDeKZHPSaXHQZTXNuAVBoLehl9X1q1E1tTbA9UYQBYGtdg+vGQb++l5C9b2ohkGbxU7AtgYUP5RPgFEGSyc4bgDxCm5eH+IS1HvrqWpVAq4A+Uoeq2ylrJQRRRGLZMFlcVFWy4TdYRzWqxurvxlcN0HeMAwmlyZ5+OjDLGQXKJQLZMtZuiPdfGz9x9jcuvma1fqC7uAlQx5Xw+DCIIlsgmwpS76Sx2k1zbyL1SKZUoZUMcWxmWPXTZAH076xI9xxySI5Gh9lYmkCRVNoCbZgla3ky3nOzJ3hho4bGImNcMuqW65YtnHanNy86mZu6r6JH5/4MT6nr6a1MhIf4f+89H9Y27iW3vpebK4wZ5KT+GUrimEgGwZ+XwPjmkJAtuNz+EiX0pSVMs8OPkvEG+GOvjvYN7KPUqJUG5gzDAOH1UHEE6En0kOmmOEjvR9h9ZbVxLIxnjrzFD6HD5/TV8viSkoJA4PGQCOKrpDIJSirZWaWTaPvVfWrmS7PIF0ktyvLNlSg0WLlxq4bUAydkq5jF0SsgmAO0kly7d4IgvCmJXZDsoXbXH4WVQXF0KmTrAQvLBL2TWDbCBhwmalwMLV1njz9JMlCkg0tG7hr7V2XHXi6HrG6YTXdkW6yxSzLVnNwzCbZEARzOjfgChB0BtnQsgG/w/+OPe91E+QPThzk7179O5ZyS8xn5qlUK4RcIeYz80wtT3FH/x1vi/Z4NeQqOQQEqmoVhPNNxvP/VVQFURDJFXPvynP/S0PVVMbiY5yLn8MiWVjTsIbWYOs1U8LOxc4RcAaYYqqmHum2u83dV7VyzddRUArEsrGaRdzgwiDnYuewS3ZylRx5Jc+CaOPG/jsRlSI2iwOX08/f7v0blrKL5HIJrKLJndd1HY/Nw8bmjTisDkKuEHXeOoYWhhAQyFfy9DX0IQhCjce9q3uXee1WN/etu49T86fIlrO017WzqWUT3zvyPZr8TeTKOVqDrXjtXlPCOhvnU5s/RVtdG4v5BFlEMppmDpkKUIfB8NQx0uk5mvxNrK5fzVIhySujJs3TJtvY3LqZ9c2XmnLrus5ofJTTc6dRVIVV0VX0N/VfUmp0iBIdV8okBYEVI68X4fmzz/ONfd/A6/DitDl54vQTvDb+Gr9//++/qaTog4qoN8rda+9mNjmL2+5mcnmSUqWERbbgtDnpCHXQ4G/gvg33XXN14VpwXQT5eDbOw0cfZmB+gGQxyVJuCYCZ1AwBZwCbZOPHx39MZ10nO7t3vuPP3xpsRZIkvA4vqVJqRb3N6/CCQE1T5P0MXdd5buC52qBRsVrk+cHn6Qn38InNn7imoRRZlNF0jZZgC2fnz1LnrqsF+6X8Ev2N/Vdtvl7AxfMfhUqB/WP7qapVqlqVWC6Gw+KAui7Khs66xn4MA7LlLMuZefLlPGGbAwumRk25WsYqm4bauXKOTClDsWJazRmGwdqmtbjsHlKlLCVVoS3QzPDiMH/y1J9wYvYEsiSzZ/UevnLXV2pCXi6ri9UNqzk7d5ZkIYmAgM/hY1PLJnav2k1FreCQLDTIFqqihIZBuZjixeljhNq3YbPYmEnN8NrYayiaQtgTpjlglo1eGX2FQqXAqvpVeO3emq/twYmDHJ08SsgdQpZkDk8eZnJ5ko9v+PjbNtVQVIVvH/42jf7GWkPda/cylhjj6bNP8/ntn39b5/8g4IIwod1i5ycnfoLb7qZcLaPrOvW+enoiPTy086F3fOdzXQT5A+MHGI+PU6gUWM4vU6gUqKiVmkmHqqv01veyb3QfW9u3vuMuMqvrV7O+eT2ZUoZytUy6lMYwDPxOPx67h76GvjdkPrwfsJBZYHxpnNZQK7OpWX507EcUlSIvDr/Ic4PP8eCOB/nEpk9c9RxrGtbw2GnTyu5CXfzCaH6Dv6FmoP1G8Ng9hD1hU2N8eZpMMUO915Qy6KjrYCG9QEi2kwy1UdV1BufP1HRtlHKOsewiUW8Ul81Fa6iVgCuA2+bmsZOPmV9Otcx0cpqSViVvcyM5vLSE2jleSBFt3sivfevXKFVLtAXbUA2VpweeZj49zze/+E3AFBR7afgldnbtJFfOkS1lyZQyfHLTJ9ENnUQuQcgV4vTsaSLeCFbJysHx1+iOdNMd6TblBRxeXhp6CYfVQWfYTBIEBJYLy/zly3/Jrq5dWCQLN3TcQGddJyemT9ASaqnRTJsCTaaefWqm9vdvFfFsnLJSpsm/0lw74AwwsDDwts79QcPOrp30N/Yzk5xhKbeEzWKj3ldPk7/pXXGwui6C/FhiDEEQWCosoekaiqpgYFDVq5SVMk6rk1QxRTKXJFfJEZTf2a2lz+njF3f9IqliCrfNTaqYQhTE2kTrL+z8hQ9E3TKejWORLFS1Ko+eeBRREmkMNJIr5Qi6gnz30Hfpa+hjVf2V5wja69rZ1r6NE9MnakboLpuLu/ruqpmsXwsuZE2PnnyUoYUhDMMgU84Q8UbwOX0mg2f+DA+tu5uR9ALnMjECniht/kbKSoVlZxEBgag3isPiYE39GmRJpq+xj3Oxc1RUBV+wBbGUYykzT50o4g62sa7nJp4depF4KUVfxJSrlpDoDHVyau4Ux6aOsbltM30NfZSUEocmD3Fq5hRjiTG8di+VaoWmYFNNJtdj95DIJdjUuonuaDf9Df0rSl+KpjA+P06ykDRpm4YOhrnIhVwhLLKFvSN7KSklRFG8RGHVbrETz8bfdpD32r0IglBr9F5AUSmypv766TVdK7wOL/1N/zLzBddFkLfLdlLFFD67j1zZrH1fcIaXRZmSUqLOXcdycdn0a30X0F7Xzh898EemhnlsGE3T6I50s6trF17nB0M8yWl1ouqmZG1eydPgMyl7BqZyYaqQ4m/2/g1fvOmLdEW6LksZFQSBHZ076G/sJ1VIYbfYCXvCb3rMG0yZ5s/f8HkWM4s0B5pZyCwQcodQNZVsKYtVtnJztIcnBp6j19DwqyX8oTZOFdM0B+qJZ+Osa1yHZmhsaNnA4anDrG1aSzwXZ33nDpJAplJCByLBFrobeqm6AmTSM6iCVPuMAQiigCiIzKfm2dy2GVEU2daxjbMLZ1nOL7Mqugq33c3QwhDPDD7Dr9z8K0S8EfzuMGfSixwopsnLDlyVAq02F9L5Rutsapal/BIIkC/nmUhMYLVYzcadYdaB6331TCxNXHbQTFGVd0QywOv0cnPPzTw/9DwddR1YZSupYopKtcKda+982+f/EG8d10WQX92w2pSkLeeRRRmbxUZFrWCVrEiyRJ27DotkoTnQjNP27lHCHDYHe9bsYc+ay1MA3+9oq2vDMe5gNjULmAtpUSkCMLk8Sa6SI1VM8dzAc7w88jK3rrqV5kDzJWYWYGaiVzNyvgBFNZkpgiAQ8URWZJFgjsXf1H0TZ+bP0BXpYmp5irJaxu/01+qjFgxcuorD0GkNtlBWSowvjSOKpiuYy+biyNQRDo0fwmM3JXedgWZk0WycuxxeHHY3saVxZE+YxrpuDl6kKQ+gaabg3AUDDTAndPeP7qc11Po69180k5Jj08e4rf8OBitFdLsbJRenp2ENZ2ZPUg20s8rhYi41h4iIVbaSKWRw2BxgwHJuGa/Ni9ViJVlIMhYfo6+xj9ZQKzPJGRr9plfxcmEZh9VBe137m3mbr4hf3PWLSJLEy+deRtM0Aq4Av/XR31phwPMh/uVxXQT5rW1bzXqXYKFYLZIupqlqVWRRxmqxmtQlV5BPbPzEz/pS33WUdZ3Bcp5FrUpYlOlzuHBexUD8zcBusXPfhvt4+szTqKpKIpcg6ouayoSynbyQpyPcwUhixFRGTC9Q76tnd89u1jatXXGurKaSPc9fr5MsSJfJ5KeXp3l24FkUTcEwzOnPm7pvYk3DmhULx6bWTcyl51jKL9Fe105VNbnKN3bfCMCa+jU8dfYp03hCEOlt6MVlczG5PEnEE8EqWzk0eYiF7AIz6RmUqoK1WsIZaEbUFZyeCJPTp5Bkyaylixb89gATSxM0+hpRNIX59HyNP1+sFCkoBXIl08Qk6o3WrlUUTF/aWDZGWlOpYOARRSqCQH/LJiyInJw/g1xysJhdoK+xj8GFQRayCywnl8mUM6ZssCCykF5AN8zBMkk05Q4uLLqiINIaamVX1653TFrDbrXzy7t/mQdveJB8JU+du+4dZYl8iLeG6yLIex1efmPPb/DNV79JUS3SHmwnWUqiaRrNgWbsVjsf3/jxt10jy5fyJPIJmnxNWK3vvenVrKryj9lFllXV5FUbBq+Usjzkj1L3Dk3bhtwhHtzxIE3+Jv56719T1sos5c3m0sbWjcSzcayylUafmU1GvVH2nttLg6+BkDuEbhicrRQYU0wGkgA4RZGdTh/ui1g1hUqBp84+hd/hRzM0jk4eJVlMcmz6GDe038Dunt1sbN0ImNz5BzY9wExqhmQ+ScAVoDVoinXNJmcZjg2TyCWYXp6mOdCMLMnmtXmidEY62Teyj5ArRL23nsH5QSpihXhmEUWQWNN9EyPxUVKFJbpX3UrE6cOjV7m3/3byxRSHJw5TUAr01vfSE+3hfz75P5FFGbfdTVWtki1nKVaKtR1k2BNmanmK7nA3JV1HBlLZOM3hbuxWB+u7dlDXtJaNspVKNs53Dn6HyeXJ2uSkgdl7mE3OslRYqslBNAYa6Qh3MJeeY13TOm5ov+FdafJduN/v5o74Q7w5XBdBHmBN4xp+86O/yYvDL/LKyCu0CC0EnAFC7hB39t9JX2PfW6r7gmkr+NXnv8qjJx5F0RRcNhdf3PVFfuHGX3iHX8Xbw75SmqSm0XSeUgcQVxWey6f4nD96lb9887il9xb6GvvYP7qfJ888yfqW9bhtbl4bfw2vw0u6mMZutWORTPehqeUpQu4Qca3KqFKmXijgUU5i02bIGU4G1M3c4H9dmW8uNYema9gtdvaP7kfVVZr8TeaCItvYP7afqDdaG+W3yBY6w50rGoxDC0M8P/g8HoeHvsY+ZpIzqLrK3evupqJW2Deyj6papVAp1GwZ6/319Nb3EsvGWMjF8egqopJna8/NhOxuWiwOolY/ScNgc/+d7F61m1g2ht9bz0AmwdlsnIAo8tG6dqyiRKO/kbPzZ+mOdmOTbZSrZaLeKGFPmOXULItalc5gC+s6bgBANwwsFiuNriCSw8tkcpJ0KY3P4aNYKaLpGk6LE1ESsct2nFanyes/39iPeCIMLQ7VdjEf4oOP6ybIg0kZe2jHQ/z89p8nXUxjYBBwBt5ycL+Ar7/4db536Hu0BluxW+3ky3n+7Pk/I+QJ8bH1H3uHrv7tY6BSpO6nOOZ1osxotYSq68hvZ2tdqUAqBTYbBEz7trA3zP2b78dldzGaGEUzNATMhqGqqTT7X/fdvMBrn1HKeClQV3oUDFBFPy7KVIpPULKJOBzrgNfVKguVAplyphaEBQQEUcAu2RlLjK3Qa1nMLHJw/CALmQX8Tj/Ty9N01HVQqBZYSC/gsrrIlXNMLU/R6G/EMAwkUUISJZNjr1dRqgo+pw+fw8fOzp3Ueeqo5hZRl0eJl0uMlDOE3CHqPfV4HV6ShSTBUBsjsoMxdRZvsIWMVuW4pnGDbGVX1y40TWMxs4ggCPREe9jesZ2iUiRbKVDnidDWvgNRtlLRddK6SodsZzk9z2h8lEwpw6roKqaWpygqRWyyjaJSfN3BSTAHwy7ca8MwEN/Drp+GYbCQXuDI1BHm0/M0eBvY3L6Z5kDz2/6eXq+4roL8BQiCUPORfLtQFIUfnfgRzcHmFc7ygWqAf3jtH95TQd6KgApcXJjRAPntfu3PnoX9+0HTzH+bm+GOO8BhTk3u7tmNIAgMLgzWJFW3tm/F6/Ci6RpVrbrCHMVVHUZAoyqZuwtdcFOSJMTSQbCvAUGm3lePYRhUtSrC+QlMVVMRBAG/w0+hUljhERrLxvjhsR/idXhp9DcSy8Y4OnW0Nv0sCAKzqVmWC8sMLw6zqW0TiVzC1BSxuTgwdgDDMLBb7AzMD2CX7SQLSfaP7Wf/6H68Ni+CKGCVrYwlxohlY6i6StgbYUa0YlQr6KU0JU0hnpgg4/RRDTZTTEyi6Ar3bbiPdCnN0PwQEU+EjnAHAgIL2TjTIy/iXvcxHJKV9VYX4xP7eXVhGE3XqFQrVPUq29q3cWjiEAAehwfRELFbTVbZTT034bA6yFfyDMwPsK5xHVW1+q6Va94qDMNg/+h+/u7Vv2MoZmoP2WQbG5o38ODOB7mp+6af9SW+L3FdBvl3EiW1REkpXaLw57F5WM4t/4yu6vLY5PDwQiGNAwFBFMEwiGtVtjrcb71BtrgIL70E4TBMT8PUFLz2GgwMwK//OnhMqd/dPbtpCbTQ19DH0OIQiqqY2uyGzo7OHYQ9YQCaLTYmc4uowutKk0Vdwy05sAkZ0IsgeQm4Auzs3Mn+sf1U1Apz6Tlsso11TeuwylZi2Rhd4deZLMemjuG2ufE5zInToCuIgcGp2VPs7NxJLGt6pjosDspqmeZAM5Vqxczsl6bwu/wYhkHEHWE+NY8oiHxsw8eYWp5iQ8sGDowdQBAEOuo6KBtl7BY7UV+UU/ODKJINylkmF4fIZBbxuvwEpTqOzp6mkhijt76XI1NH0DSN+fQ8UW8UMWy+H42+KHpylrXVEu2+KNPL05xbHKYl2EJVM+3+5tJzjMRHEAVT6ErURdY3r6c12Mr+0f0UK0VeHHqRY9PHMAyD4dgw+0b38Rt7fuOqMwv/0ojn4nzvyPc4NHkIVVexSBbKpTL7x/ejGRr99f0E3O9McnY94W0FeUEQ/hi4D1CAMeALhmGkzz/2O8AvYSaL/84wjKff3qW+N+Fz+qj31ZPMJQl6Xh+iiufjbGvb9jO8skuxw+ElrioMVIoImqkf3mW1s+cKJtrXhKEhcDrN/y4smKUar9fM7v/5n+Ghh5jOx3hm4BlUTTXLBYLIuuZ1ZjPTV7+Cpx2VreStjeTKp1EwdwI2QaDTIoEhgfi6psqmtk20hlpZ3bCaV0dexW13I4kSc6k5+hr62Deyj5fOvUS5UkYztBXqlVbZitvmZjo5XdORv9BwtckmxbYl2MLA/AA3dd9ErpKr1exD7hALmQVUTTUlBdxhot4oBaVgmsmEu3HZXLhsLhbTsziiPTgFEbWSp4pOVdOQZTulapl0IUm+kqfeV89SfomlwhID8wNsat1UW3hFUaRQKQAwtTxlCtwJ5q5hdcNqDENH18sIhpOypuGy+XFZXWRKmdp05ZOnn6Q91E5ToAlREEnkEvzps3/KVz/zVVz2ldLNC0qF/aUMKU1ljc3JNrvnUi35dwHDC8Oci52jqlUJuoKm7LRhavJPJ6c5u3CWm3o+zObfLN5uJv8s8DuGYaiCIPxP4HeA/ygIQh/wOUzLmEbgOUEQVhmG8dZtf97D+I3bfoP//Mh/pqSV8Ng8pItpJEniS7d86Wd9aStgFUU+7YsQUxWW1Sp+SaLx7dLnymVQFDPAh0LnBawwA30mQ3F0iKeXD+Fz+GryqUWlyODCIA/teOgS+p4oCPR4t1LURylSRpL8eAUNSVsE524QVpYYQu4QH13zUW7supGZ5AwVtULUG+U7B77D/vH9tARaCDlDHJo4xHcOfocv3fylGvMj4okQy8aYXJ5kMbNIo7+RzkgnimYaOVyQJh5NjKKoCvW+RiSbm+H4KOl8AtXQ8Dv8lFVzaloWZVqDrWiGRtQbZTwxTk+km7CvgYFsHKtkob9xLcVqGavdTRAB2V2HTbYhiRJeuxeHxcFyYdnUGhdfd8sKnl+I7RY7mvb612hVuAFb9QylYh63pLEq3Eg0dANWa50p01zKoBs6sizTEmyp/V3YE2YsPsbJ2ZM1MTWAk6Uc30gtoGJgAQ6WcrxsTfPvgy043uVAn6vk0DSzb1MzHD8/oavqKuli+l19/g8q3laQNwzjmYt+PAB8+vz/3w981zCMCjAhCMIocAPw2tt5vvcq7ui/A6/Ny98f+HvmUnPs7t7NF276wntWOjgqW4m+UwYlnZ1w5IgZ3C8E+FLJrMd7PCwsjKHK6gp9bKfVSapg6tJcVo9f8uP0fYoDyyd4riiSxUaTZTUPOLrovcJlOKyOWulhNjnLwcmDdEe6a2JmG1o2sPfcXg5PHmb3qt3kSjkSuYRZ19YqVNQK5xbPgQH9Tf04rA5OzZxiKb/EcGyYOk+EycIZvJ4oNleA5aUJThdS9Nav5tjkIewWO8uFZUpKCYfNQYOvgYmlCTY0b6DR4cFhdZFNTuBwhqBaZHuoleHMPIeT07XJX5tsw2F1oKoqS/klPA4PyUKSrrou6n315u0Od3Jk8ojpCCXJiOVjuK1WPr3xo7QHwrw2eRqXOIFsayJdTNMV7iJfztf6FhdDFEUKSqH2s6ppfCsTxy6KeGuzEwYTSoUXCkk+5g1f9aOgqArPDzzPa+OvYZWtfKT3I+zs3HnNpcC2YFttl6KoClbZakqQGAY+h4/2UPs1nedDrMQ7WZP/IvC98//fhBn0L2D2/O8ugSAIXwK+BNDa2nq5Q94X2NG9gx3dO37Wl/Evj85OWL0azpwBWTYbr0Bu01peSRznGb3EcCnGmsY1bGzeiN/lB6g5a10JjxXgR2orflnEL4gs6Bp/nprnt8Vmut6Agx3PxREEYYVapcvmoifaQ1EpEs/G0Q2d5kAzmqExnhgn7A6zkFlgYGGAxkAjE/EJppJTbGjeQLqYJqNVQZRYWp6iIdTKmqZ1zCzP4gnLtARaWM4vU+euI1PO0BxsRhAFPrb+Y8xn5pGAHkkk64kynpig0eak2epiyR2izl2HLMksF5YRELix60ZzItflRxZkPrL6I6yKrqpltiF3iNv7buelcy+hKgmM8gIRXzu3r9qIx+4k6gkytHCcqmywvesOOuo6mFia4PvHvo+qqzVRvqJSxCpa6Qm/Po06qylkdZVG2XbR3RRwixInykU+dhX1DVVV+ZOn/4QT0yeo89Sh6Rr/67n/xWD/IL+8+5ev+n5dwOr61dzUfROpUsqUaiibmXzYHebGrhtXTAt/iGvHGwZ5QRCeA+ov89B/Ngzjx+eP+c+ACnz7zV6AYRjfAL4BsHXrVuMNDv8Q7zXIMnzyk6DrcPAgNDaSivj5n+M/5EQ1TqxssFxMMZOcYWB+gBs6bqBcLZPIJVgVWUW9r/4SOz9F03i2mKZOknGcD9QhUSKuKTyVW+DLehKqUyB6wb4O5JVN74gngmEYaLq2UpbYgPvW38fd6+/mpeGXKFaKxPNxdnTtIFlI4rQ5WUwvMrk0yZbWLaxvXo/H7qE52AyailrOo1hstEa6cDt89HVsZ6mQRpk7wY6OHWZdvlogV8yxvX07raFWHjn2CHOpOXxOH03eMIViiqAryEJ2gd76XoLOIA6rA4fVgYhIqpji1vZb2dC8wZxcvUwW3B3tpjXUSjJ1EktFJuhdXVsEmvwhmtx94NgJdjOAd0e6uXvt3Txx+gkcVgeSIKHqKnf230lL6PUSjhURA9B1DfGi+6YaOrY3oC+emjvFyZmT9ER7atccdAd5buA57uq/y7yHbwCLbOELN30Bv8vP/tH9JHIJPA4Pe3r38MCmB95Rt6TrCW8Y5A3D+OjVHhcE4V8D9wK3Ga+LeM8BLRcd1nz+dx/igwhJgk9/GtauJX/8MN+ceppTUp5zapZqydRwn02bE5gTSxNsa9/GlvYtjCXGWMwu8snNn1zxBU5rGhVDJyitrL970VkqjYI0A6If1AXIjoDrTrC97jXaHGxmR8cOXh1/lRa/6S41l55bIWUgizJ5JY8oiKaoWmoei2Qh5AkRdAUZWBwg4AjQEmgh6o0yl5wlHGikUMxQrpToauino6GPytxpSlYnyWKyxnBp8Dewb2QfX4h+gQc2P8DgwiDTyWminij3b7i/5lRllU1tmYPjBxmODVOoFHBYHBwcO8iJ6RO4bW52dO6gO3qpj6pVtlIfWgOZI1wgwgJg6ObP8uva/YIg8Iu7fpFNrZs4NHEIwzC4oeOGS7T5G602WmUbC5pCBPP3VV2jbBjsPr8DuxJGE6NYZMuKRUkWTUepyeXJawryAHWeOn7ppl/iE5s+QUkp4XP48Dq8H3Lk3wbeLrvmLuArwC2GYRQveugnwHcEQfgqZuO1Bzj0dp7rQ7y3ka+WeEmbYDi8zKPzC0znFilVS/idfgS7QLZkju/Lbpl1zetq9Mb51DwjsRHWt6yvncsnSVgQKesa9ouCUFnL0CEUQG48/xsnCE4ovQrWDhBeP/bffuTfEvVFeX7oeSq5Clvat/Bz236upvjZHenmpaGXqGpVUsUUsiRjYOCyufDavfSEezg1e4rOcCebWjdRtTo5GTuHUS2xp/cW1rdvJ2UYlJcmODh+kEq1gsfuIegOMpucJZ6LU6gU8Dq8tAZbWcwuci52jsXMIlvbt9ZEu/wOP3XuOo5NHSNfyXNq7hQem4c1DWsIuoI8ffZpbBbbiqZpDaIbHLvM148NEMAomjZ98sr6uSAIbGjZwIaWDZee5yL8sr+Br6fmmFMriIZ5yj1OP9vfQCk16AheVuUSTBniNwNRFGuU2g/x9vF2a/Jfx/x0PXt+pT1gGMavGoZxVhCEfwYGMMs4X/6gMms+hOkI9eSZJ8mVctS563BYHVRUs5lpYCAgYJEsCKKA07qynu6yu1jILrCe14O8TZL4iNPPk4VlghjYEcgaOoZR4lbHT1X0RDuoy6AXQHo9mFhlK5+74XN87obPrThc0zVOz57m1NwpymqZhcwCi9lF/A4/TosTn8NHnaeOpkAT+Uq+xsEvZWO4BImu/rtoaNpARhBpFiVemD7O5PIkjb5GMuUMS4UlWgOtJLIJUsUUVa3KD4//EJtso9HfSKla4pmBZ0gX06i6yo9P/JhTM6cIe8LMpedo9DcS8UQYiY0Q8UTwO/0cmzp2+SAPYN8Acj0o44AGlo6LFsE3j3qrjf9W18ZQtURO1+iQHUSvQYdpW8c2vnv4u8QyMaK+KLquM3feorCvoe8tX8+HePt4u+yaS/eRrz/2B8AfvJ3zf4j3B+K5OEv5JZoDzSiqYhp9WF3kK3lKlRJW2WpOuhoGSlXBbXtdIbKklAg5Q5ec835PEKsALxbTLBoaDbKVj8sqEWWWrB7F4/CYjBFDQ9V1ppfnyFUmCLlCppTuFRgd+0b2cXb+LFFPlC1tWwg4A+wd2YtVshL2humo66Ar3EWpWqLJ38T2ju1878j3aPY3sdkbIVUpkBh+gU9v+RS5bJyqZr4eQRBwWpwoqsKx6WOEXCG+d+h7pEop7LKdTa2bzGOsTnx2H9/Y+w2skpV0KU2pWiJZTGJgkMgnCDgDyJLMUn6JtlAbqWLq6m+AHDX/fYcgSRL90qXyz1eDz+njd+75Hf7q5b9iND4KmC5fv3rLryLLH85c/izx4d3/gEMzDJa1KiVdwynKhCQZ8R2ub1bUSo2iZ5WtbGndwlxyjrySp1ApkC6nccgOXBYXhmAwEhsh6jGblALCZacuJUniXl+Yj3nr0HSd0fgIh0ZPk7MeoaJZcRdlNqkeRGuGAyU3Zx0GsmxB1VRagi3c2X+naVxRSBHPxZFFGY/dw+D8IC2BllqNd1X9KjRdI1/J1zxRC4ppE3nHmjs4MHGArkhXbQcSdAWZT88zs3iObDlLW7CNQqVAoVIgV85RUkpU9Sp1njo6I50cGD/AYmaRel89jX4zw55Nz1KoFBDsAhFvpOYUdsHjYCm/RMAZQBIl0sU0bcG2t/zejMfH+auX/4rDU4dx29x8ZvNneHD7g+9K4O2KdPHHn/ljYtkYkiB9INzOPgj4MMh/gFHWdV4rZcjoGqJhTriGJQs3OD1YhHdOpCroCq5gs6xrXke6lGbvyF4yxQxW2YrD6qA73M3qhtWMLI5wYuYEW9u3smvtLtPM/AoQBIFUMckLQy8Q9baiiF5CU49Rzs8xbPfiWKrHMpxg3aYW0htMFv1scpbBhUE0XePA+AFThtcwyCt5dF2nWVjZBKzz1NHX2Iema8yl5wi6gtyz9h6ivigvnHvhEt9Sv9PP1PIUbaE2vE5TCydbyuK0ORmYG8BpcdLX2IdNthFxR8gWs4zGR2tBfi41Z+rLCCIum2sFT76qVWvnArMUtqlt01t6X+ZT83zpH79U06wvqSX+7Lk/Yyo5xX+577+8pXNeCy7Wx/8QP3t8GOQ/wBiqFCjoOiFBIq/kkQSRBAbjSonVP0VbfDvw2D1sbd/KwYmDeOweJFGiNdTKlxq+xImZE3htXur99bQGW5FEia5wF06Lk/s2fpzZaoWXC2kqmoqSnCIdP4euVVkdNc3PHVYH52LnsMk2rLKV8pLI/GAUKbKa40sJvNkqHS1N+IbGyXe1orqdBN1BDo4fpKpViXgiFKvFmiPYS0MvsSq6aoWDVEWtsCq6ilX1q1ZY9lXVqkk3/Cnf0pJSIhQI0R3p5ujUUfoa+1jOLzOXmqOgFNjctplmfzOxbIw6Tx1TySniuTiarlGulikqRXrre0kWkpSUEl2RLs4tnMPAwGMzm7frmtaxqWUT65rXrZB9yJfz5Ct5PHbPJdTTn8b3Dn+PTClTk1e2W+24LC4eP/04/2b3v6Hefzlm9If4oOHDIP8BhW4YzKgKaiHJi3OnqapVwMDjDCA2r3tHgzzA1vat1PvqGV4cRtVVdnTsoMHfgKqpNPgbVhhIq5qK1+tloFLknFLEJ0oMTh5kYPY0UU+YfoeX49PHmUnOcP/G+1FU5XWqXzYDksxsMc+h9DjWYoUBdYm1qoNwJgtuJ7qh14ytz8yfQdPMHYbf5cdtdzO4OGgGelFmKbeE3+Gn0d94STC3yBbWN6/nyNQRmvxNSKJESSlRqBRY17wOn9PHvRvu5cWhF3FanfTU9+Cyu0hkExwYNwXLDMPAYXUQ9UZJ5BJ4HV4evOFBBhcHWeNbw/Hp45QqJVx2Fx3hDnZ27uT+TfdfEsALlQIPH3mYo9NH8dg9hL3h/397bx4lyVXf+X5uREbue1bWmrVXd/Wu3hep1VoRkhAIGbGYxcYGM28O8PxmjIexfWwfzpgZ894MYA/P42EMnLHB8GQhEEIgCyG0t5beN1V37fuSWVW5r5Fx3x9RKnWrV0ml7upWfM6pkxn7L29m/eLG7/7u98e21m3s7Nh53vGH4xPHzzqPXbNjVA0GZwctJ/8uwXLy1zCFcoEjIwfwO7yLaWxzhTTHxo7y/mDDkuYeCyFoDjeflQWyqn4VPVM9NAYbF6er50o5OhtWc6RSoFbVKJQyjE310B5pIV01GMnEyaanOTJ6hKArSHu0nRMTJ0zRKqeLuVKaF4sT+GwuukMxEtkEJyoz1OcmaJN1zGZniXgiPLD/AVRFRREKLs1FpVpBURS2t21nfH6cctUsYn18/DiPHn0Um2LjppU38bFtHzPrpWLevF5TqzQMA7fDzV3r7loMSTQGG/nt7b9NpphBUzUePfooj8QfIeAO4NJc5Mo5JuYn+MB1H2BHxw5cdhdSSiLeCPuG99FV28Vcbo7dK3Zza/etNIYaz/pe0vk03/z1NzkxcYK6QB3z+XnShTSlcomAO8DaxnNXNGuraePY+LEz1um6DgIa/A3nPMbi2sNy8tcoihCo2Thl1Y79tGnqNk8IZkfMafiXYWDs+q7r0Q19MeNCUzVuW30bXl8topBGEYJcMYMQCopQmI6fom9ulJg7QMWo8K8n/pXbVt1mhkZG9jOfitOTPoyC4O7m64lqPoz5OWadLl5OnkKZi9Bd382jhx9lKD5EbaCWsCdMVVYZmRsh4o2wuWUzt6+5nWQuyZcf+jJlvUxLqAXd0Hns2GPMZGb48l1fBkBVVHZ27GRzy2ZKegmP3XNGz1lKiaIoBNwBipUis9lZ3rvuvYzOj5LMJwm6g5QqJR7Y9wA9Uz101XWxu2s3m1o3saZxDblyDrfdfc46q/lSnh++/EMePPAgJydP0lbTRkOggbAnTLqYZj43z+GRw+d18r+9/bd57NhjTKWmqPXVUq6WGZ0b5abum2iLti39l22xLLGc/DVMTWqW4NgEVUNDlVAOBgiHvCh6gUq1cllssNvs3L7mdnZ27KRYKRJwBdBsGpmqTsUwSMkKus1F1TAolHKMzw7T5a/DY9MolAu0R9oZSAywrnEdhmHgcHrwtnahxecYmukn6mmjY+0OIq2NOLPT3LPhHl7qf4nDY4eJ+qNkChnmc/ME3UF0Q8dhc/DK0Ctsb9/O3v69pAtpumrNTGBVVemq7eLQ6CGGE8O01rSe8Tnsp4m6DSWGeHnwZWZzs9T6a9nRtoOIN2LG1Z0+1jauJV/O82zvszg0BwF3gIZgA/0z/ZT1MnevvxuH5sChOc5qs9f4+uNf59jEMYRhVnmKZ+I81/sct66+FZ/DRyKbOENg7I101nby9Y98nW888Q36pvuwa3bu23Qf//49/34JvlmLqwXLyV+r9PSw8oePYOx/EsUZoKQoeBUV6XPQc8cOIrvPzk1/J/E6vXidr+de54wqU3qZOaOCW1GhcR2n+veiSIlXVZjNzePUXNQH6plKT/Gbk79hXdM67DY7IU+IgdAAyUqZ6aZ1xGrbsJWydLo70as6w/PDNIXMeq91/jrypTxjyTEC7gC6oTM0O8TQ7BCpQgqXdqYeiqKYTxSJXOIMJ386Q4khfn7k50S8EWKhGOlCmocPPcx9m++jMdDIXG7O1JxPTi5mHcWCMVPyINDA8OwwqXyKgDtw3vbqne7l6MRRVtat5MTECbwOL6VqiXw5z/j8OM3hZoqVIivrLlz0Y3vHdn74uR+SK+XQhHbFCsxPpaZIZBN0RjstDZrLjOXkr0XSafjnf6Zu72FmcnnGCiM4DIWMy4HTH+LWV5qw75mG5vPMonyHyRtVXilm6Ha4iFc1ZvQykdoVGIYkP32KV1PTeLxRosFGjqenKSYnqRoVylVTfrYl3ML4/DhzpRyjmUkcLg8lvcT7N7yfgcQAYU+YgCuAXtVJFpLk9TyVagVpSK7vvJ7GYCPZYpbR2dGzesJVo0pJLzE5P8kDMw8szgHQVM2cKFXbyY/3/5i53Bx6VUcNqvhdfiSSfUP7uKn7Jh45/Ahjc2OMzo+SLqTpjHYupk8KIRAIinqRAOd38olsAnVBpqEuUEcqn6JsmLVxZzIzVGSF9Y3r2dK65ZLa/GKZOO8UqXyKP//pn/PigClK69Jc/Js9/4aP7/r4FbHn3Yjl5K9FXn0Vjh5FyedZP5ym0aYy5TCwZ/I0ylp8fdOmBvwVcvIzehkJuBWVVkWl2eZAApOtmyAzhcwnKebnGRo9xMuZGTKpKfw2O8fHj2NTbaxuWE13fTejc6P4XX7aI+247C72De+jd7qXdDFNd303x8aP4XV6OTlxEpfmYkPzBtY0mlPsvU4vtf5akoUkAzMDNAQbqFQrjMyN4Hf6mUhNkClmODR6CEVR2Nq6lZnMDP/r2f/F6NwoYa85Kao/3s+uzl14HV7i2ThBd5CPbvsoo3OjNE82c3jsMOsa1y3G8SvVipnp4wpesI3q/fUY0sAwDMLuMG3RNkbnRsmVcoQ8Id6/7v3ct+W+K+a8L5Wv/OwrPN//PG2RNmyqjWwxy3974r8RC8fY073nSpv3rsBy8tcio6Om9G8qhVI1iDo8RHUgn4d0HhokDA5eMfOqkjNKWCjCLFgxmpujuXEtE73P0jt+FKfDhVEp0Ny0jvGRg+TLeer99RwdP8pEcoKOaAef2/M5jo0f45WhV4h4I0R9UQ4u7LujYwfDs8OMz4/TWtPKe9a8ZzGuXjUKrPUe5FNtCfYOj/HC1BC6up7dXbtx291EfVGOjh8lFophSIPBxCBOzcmRsSMoQsGQBqqi4il5ODFxguZIM7XeWkqVEhPJCQxpsKtjF1VZNaWGXQEq1Qq5Uo4bV9xISS9hSOO8oYv2aDvb2rbx0uBLNAWbCLlCZNwZOmo6+Op9XyXsfRslG8/D0fGjHB49jKZoXNd8HZ3RThTFnEh2emrppRJPx3mu77lFBw/mzdVVdPHDV35oOfnLhOXkr0V8PrN4h8MBimK+VxSzcpOimNLA/jenDLiURGwa1ZLEkBJFCBLZBPtGDjNmVHDNjzE51cOG9p0UKjmcmptUKU+Nv45cIYVQBMl8krA7jJBmSubRsaO0RFpQhEK2mKW9pp29/XvJFDN013Xzmd2fYWx+DIGgUq1QqBRYKb/DquAoPlsrd3dFubs9AfYsD003UTEEhUqBqlFddE6pQoojY0cIeUI4VAelaglN1Yhn4ozOjtIWbWNl3Ur29u8lFoqh2TSQsLl1MytqV9Af78eluYh4IhwbP8YL/S+AgK5oF7tX7D5nds0Xb/sirYdaebLnSTJ6hjvW3MGHtnzojMlRS0GpUuJvn/hbnh94Hr/TjypUnut7jsZQI2FXGCkkXdEudnTswOf0XfJ5E9kEkrNvEC6bi0QmsaSfweL8WE7+WmT1amhpgdlZsNvNHjyYJfkCAfN1+/bLb1c1A6WDBMt9rKiG6dW7MAjx4thRnA4P6/QSs/46pqZ6GJ45RU2wAYRCQS8RcXhwKja6ol0cmziGRNIz3cO3n/k2iqIQC8dIFVK82G9OQuqs7STsMQty3Nx9M9Ppaf75pX9maHaIgDZPfewEs+71+F7rEashKL9Ks2Mdx1J1+Jy+xULSUkozx14oGIZBJBDB7/Iv1ob1Or3sbN9JX7yPYqWI1+llU8smqkaV/UP7uX/r/axpXEO6kOZHL/8In9NHU6gJQxr0x/spV81smzdit9m5f+v93L/1/rO2LRWpfIq/fuyv+eWxX1Lvq8cwDNpq2jg1dYq+mT7u23wf9f56hmeHiWfi3L/lfvMGdgl01nTic/rIFDL4XK/fHObz89yx9o536iNZvAHLyV+LxGJw552QSJi9+qkpswcfDpvLd99t3gguJ0YBsg+DLIISZg0lGvTnOZxvJFhIscJmQxWSglAJhlsZHD1AfUOMqqMOf00H2UIKe2qKV4ZeoVKtEIlFzFdPhGd6n6Ez2knvdC821RQiS2QTtNe047A5+OWxX4I0M2fuXHcnNeIgwepJjs3M4nE4UYTKRKaAYpTxuyeYSFZpDDTSEm4x8/sFtNe0k8wlmcnOEHQH8Tg8BF1BM0feGeDw+GEG4gO0R9qZmJ9gVf0qXHYXmqYxmBik1l+7eK7XsozeTLbNO8F0apqv/uKrPNv7LKVKifHUOAoKfTN9OO1OIp4IqUKKhkADtf7axcHk12QSLobdbufzN3+e//LL/0K6mMZpc5IsJIn6onxq16fe4U9n8RqWk78WURS45x6orYWf/hTm5szevNdrluq78UbQLq03tmSU+8DIUlJjJAwVAxchzUlz4QX2ztsYzsaxKTaavVGidVG6lCzhUD+by3kSuRIPTpaZzhcpJsfxqxqPHH6EHe07SBfTOGwOToyfWJTpTRfTuDQXYU+YXx3/FQdGDqAoCpqqMZAY4MNrOtFU8Nrt/OLUCDndhsSgVJhiuPQymnc3/fF+op4otb5ahBAE3UFaa1ppr22nUC5Q0ksMxAdAwvrYelRFZVqbZjo9TcAVoFwt48KMt782gzVTymBXz0xhvNRsm6WmWCnyo1d+RKqQIuKOMDA7QKFsykJrqkYyl0TXdRzqaRPpVBuZYuZNXedDWz9EU7CJH73yI6bSU9y1/i4+sfMTi4XJLd55LCd/raIosHMnbNlihm0UBWpqzNd3kHgmzoHhA+hVnXVN617PNa/GGSm7eDxbIlepoGIwNtlLduoQBwbGCftaaAw0YrfBvR15gnWdhGvWgl5Gi5+iw7YXwxUjGWxCy89T0kscnThKwB1gPj+P3+knX85T0ks0hcxCFcfGj3F47PCiLo3P5WN0bpRfDXj5dGcN1coMB6eq3NzaRK40TsXlYNaxnVSpyJ1r72RsfowPbvwgLRFzNmyxXOSx4+aM2FKlxNjcGDXeGvwuP4Y0sGt29Kpu1ou1u6kaVcqVMm2RNgBiwRjHxo4R8b4+R+FSs22WmtG5UfKlPG6HG7/LT7Vqjj/ky3kUoVCsFPHYPWeMFVSMCmHPmx/wfdcWuV8mWE7+WkfToP7y9Jpe6H2Bv3vq76ga1UVxrg9u+iAf3f5RpgsKfz84jazGsRkVDvcfQFNVvN5aVsbqGZ4eYlgfZnNDmNFEH7E1n6DOG8GG4MhwHzU1rdhDN3Co/wgimyBcLTOfnmIgPkC2ZNZqXdWwimq1yqr6VdhUG/uG9+G0OWkONzOdnqaoF6n11nJo7BhPBveQSzyI15bB0CfJVmzExR0oaoCqTJAtZanx1dA700tHbQc21YZTc/KRrR8x89SrFRw2B+PJcRKZBB6HhzpfHSenTuJz+phOTwNwfef11PrNequtkVZi4Rijc6NnZNvc3H3zBWe+vhFjoUbApF5GE4IGm4Pgm8x+KetlU+543pRk8Dl9zGZnyZaz2FU7PocPA4O+eB8hT4hENkFjoHEx39/i6sFy8hZLQqaQ4X8++z+J+qKLMeeKXuEnB3/C1ratPD82A6qdOq1CIlPCoUCxmmO+5KEt2MxmV5jh2WHqg7U0+8tE7S6KhoFXUcnafKRCGxCai0i4jen8PDl7kPRMP7O5WVY3rMZj91Dvr+eF/hcYSAxgV+2Mzo2ytmEtYU/YDOlMnmAiPcHw/DAPS4nPuZKgQyM3HyWr214PIUgzXq4KlXK1fMbnVBRlcb8NsQ1oNg1DGkylzBKCnbWdNIeb2da2jTp/3Rla+TbVxt3r7qZvpo/+eD9uu5vVDatpCF66WJiUkqPFHAOVAg6hYCA5WSqwyeml1X52hs75iPqiizfGl4dexmV34XV5cdgcdNd3s7F5I33xPiaSE+RLeXa072Bd07ozCn9bXB1YTt5iSeiZ6qGkl86QLtBsGjbFxt7+vST0Mu5wN5RHqFQmQBhoWpR4SdDmYjE33e9px2Gbpt6mcSg9zzMTQxybyzKTnsRffz0NniAi3Eq6lEXzhNjRvoOAO8BsdpaeqR7cmpuWSAtrG9fisrs4MXmCjtoOXHYXLs3FZHqSkDtk6tVIM9MjX7VT1ovkCxkcmTy+Soka3cZwKUl7pJ3n+56nrJfpiHbQHGpenNi0pW0LY/NjZEtZumq7KFaKuO1u7tlwz3kLoWg2jdWNq1nd+NYGvucNncFKkVpVW4z161JyuJij3mbHcYnhuKgvyvqm9RweO8yWli0MJ4bJFrNsbN9IZ20n6UKaVQ2riIVi3LX+rsWnEYurD8vJWywJQgiEPFu6+LWJNM5SjqLQqNi78ASCGHEdm82FqKaw6SXkQg+xaDjIK6vJZ0c4eaoXYbMTdgfIZr0M9zxLY91KikYFj9NHXds2Qp4Q2WKWUqVEIpugUCmQyCXwOrzctPImhmeHOTJ2BJfNxfDsMH6Xn9tX307AHVhMjcxX8tjyJeTxXmS5yo22GL7en1PX3ca+qo53oYrTiYkTrGpYxS3dt1CoFJhJz7CldQv5cp5MKUPUF6WjpuMd1WZJ6BVsgjPkiG1CAJK0oRNVLk2bRgjBDV030BpppT/ej91m58TECYQiyJVzdEQ7aI+0M5OdwWG79FCSxfLDcvIW58UwDHRDRzut13g+1jaYPedUIUXAZWaJFCtFDMyZn70zvRyfHyUXakV4o3hDzYxMneS6YAPjM72kCinao+0EXUG2r/oIjx39IfUBiUNzkqpGaA34GEsMoDm8tDasZqqQpaf3KQbSM0SdfnRDJ1PM4NScNAYaGZwdJFvKcs+Ge6jxmNWZov4oFb2y+LQhhCDkDdEcjLHx0BgrV25DcbioViu4NSfiwNO4O7ooe8IoQiHsCdMz2YOmapyYOIEhDQQCm2rjvWvfS0uk5dLbVkqqSCrlIpVqBZ/Td97iH6ejCYFxjvUSUN9kfQBFUWiJtNASaWFL65bFHH6v07sYgmqLtF321E6LpcVy8hZnIaXk+MRx9g3to1gpEvFG2NWxi1g4dt5jXA4Xf3jbH/LNX3+TeCYOmHHt39n1O7TWtFLrryV34nFGp09R0hys8YTZ030TJ0YPE/aEaQ23EvaECXlCOO0uUmWVWGgrAI2VEsOVIrWhJmRVpyu2gZn4IKtX7GF09BAzhk6ilEeXVQLuADXeGgqVAgdHDhL2hPn09Z9me+d2njjxBIVKgd7pXiLeCIpQqOgVtEyeXa5Wgk2rFj/PbHaWnGowse8ZDrR6sak2OqOdADx65FF2dOxAU8001GKlyOMnHudTOz910QFUKSUjlRJH80kODewlER+k0WanzuXj5pU3n1f58jXclSJ9o4c4WcjQEGwgFu2gKFS8ikpQeev/zn6Xn/df935+0/MbxufHkUhW1K5g94rdb/mcFssDIaW80jYssnXrVrlv374rbca7niOjR3im9xnqA/U4bA4yxQzJfJIPb/0wUV/0gsdmChmOThxF13VWN6wm6n99fykls9lZipUiQXeQp089zXR6evGcUkpG50a5pfsWXh56Ga/THAiUUjJTLXM4MQI2J2mjglYpErJplCpFUplZUtUyMYebUn4OwzAYnh0mXUizrX0baxvXEnKH0A2dfDlPspBkKGFKDdd4avg/N32Srqf2m5PIFphKTvHwk9+l1NLEzMZVVI0qc/k5qlVzotSK+hX0TfcxX5inzleH2+7mt7b8Fq2RCzvp4XKR/cUMwwN7mZjpx+eLUkTQIQSVQooPb/3weYu5zKRnePjQw6T1CpMCsqUcXleA2zfcw+5AHd4lGBQ1DINsKYumapYk8FWEEGK/lHLrubZZPXmLM6gaVfYN76Mh0LAo5uVz+ijrZQ6PHub2Nbdf8Hify8f1ndcvLpelQVVKnEJBCLHowIqVIiNzIzQFmxb3FUIQ8oQ4NXOKrW1beerkU9T767GpNgaHDzId78Pp9DOl2GgPNxP2N6MIQaGUw61DW/Mm3Pk5fnnsl/gcPmq8Nezq2EXAHWBsfowtrVtIF9K80P8C+VKesDvMjo4deOqbINiPkUwybmQYiA8wMjuMkc0yEXahSINCpQASJlIT+J1+fnzwx8xn51EVlYPVg9ht9sUY9/mQUtJTzuOulpmY6ScSqDNr3xoG84pC2Gbn5NTJs5x8tpDlW099i++/+H0qeoUNsQ2877p7cAQbmUpOEJ4fwxtamtRGRVHOO2hscXViOXmLMyjrZcp6+YwqSGBmv8zl5s55TDwT5+DIQWYyM9T6atnUsomAN8KRYo4JvYQEgoqN65zei+ZzSylRhMLaxrUIBPuG93Fq6hSTqUlu676VA2NHiAbqSBRSlNJT1HijhELNzE8cI56aZI03TL2/HlVRiYVii/Hk1yordUY78Tq8bGndghCCudwcDx38KR/ZdRvpB37AyPhJvE4fgWSO/q4mDuhTlPpGALMUoEtzLcoP1/nrEEJQ0StMp6d5+tTTbGnbct40wypQNAwcVR1gsbi5JgQFw8ChOUgX02cd9+8e+He8MvQKSAi6g2blqrlhvvSeL9HkrWFodojtHVdAi8jiqsBy8hZn4LA58Dq85Mt53Hb34vp0McXquk6QVRCvO7Hp9DQ/OfATnJoTn9PH/sQg3x86SGjVzbQGGtjs8GJXVXJGlefzKW71BHEpKk7NSXtNO+Pz44vpeVJK5vPzbG/fjhCCtU1rWd2wmn/c+49c13wdNtWGClQQ5FwhcoUsdleInKETqF/DnpYNTE72UJVV1jeupyX8+kBoSS/hdXh5+tTTDM8OU9bLKIopOOa0O3k+1ER8cz2ta2JkDMmEPsdkbpjK1CliwRj1wXo0VaN/up+J5AQOu2MxJq8IhRV1K0hkE8zl5s4b0lIBv6JSsbux2xyUK0XsmpOiNAgpGpncLJtbNp9xzP6h/RwYOUBnTSfD88PYVBtRX5Sp1BQHRg6wLrZucaDbwuJcvLNz3C2uOhRF4frO64mn4yTzScp6mZn5VxGFfazzHqKS/B653D50owrAy4Mv43a4iXgjHC8XecowGBVwaGg/+wpZHsnOoRsGHkVFRzKpvz65aHfXbnxOU2pgfH6csfkx1jWtWxzgfM2e19IwNVUjHKgjX0hjR2IgqUqDQm6e+sY17O7YwWf3fJaPb/84elXHkGYeSr6cp1Qp0VXbxfGJ40gpqfHVEPaEqfHVkClk2D+0H0OzUWpuJNfaRE3LSrNKVKWEgYEhDY6MHqFULVGoFChXymiqRlukjfWx9XgcHjMLifNnuAghWOtwU0CwomMHs9kEE+kZcsUMRnaGqC96xmcHGJodQhEKNpsNn8NHPBNnOjVNrpzj+b7nGZ0b5brYdUv5E7C4xrB68hZn0VHbwX2b7+PQ6CGS2WG6vBOsj20l7QhyqpzDXniC2fw8Lf4bmEhOUB+op2BUeamYwSdUcAeZTk0TVlQm9TK95QKrnR40BHnj9QRAr9PL/VvuZyI5YcoIeGsW49GGYfBM7zM8fuxxM/3RF+XWVbfS2rSeUyMHycQH8LmDqLlZtjetZWXDKpK6Ts/wAXqnexlPjnN49DDtNe101HZw9/q7iXqjZIvZswpu2FQbumHeFAxpoAgFh+ZgY/NGBuODKEJhLjeH3xOksaaL2VKOQiFFSS8tHlcoF9jSuuWi2i61moMbFYVe2wo8Th/JeD/BaoU1tZ10RbvOys5pDjdjLLRZVZo3VoTZPqqiUqgUCHlCb/crt7iGsZy8xTlpCjXRFGqC7K9BdzMpvYxWiuiFLEOpOdTqgzxpK+IsF/GX8kwLlaqU2BWFYimHw+0HIdCEYLBSZLXTQ1ka1JymRS6lpHeml1cGXyFbyhJ0B9nVsYvWmlb+8y/+Mw8deAin5sSQBi/0vcCpqVPcsf2jNDauw1XTyer6biL+KEFPhERVp3eqh5ND+2gKNRELx8gUM0wkJ7hpxU201rSiV3Xao+2MzI0gpcRhc5Ar56gaVVbWrWRl/UoOd0fAOwAAF0BJREFUjRyiLlCHpmqoikpzuJk71tzBc4MvYbhCTKkqoeaNaMlx4okhUoWjrKpfxabmTdy78d5LynUPqxo73Bo73H6ov0gh7vbtrGtcx+Hxw6aWvSdCPBOnPlDPZ3Z/hkKlQN90H5taN73t79zi2sRy8hYXZC7Tx0BigJM4UI0quVKGkCtEwFbCW60wBMh4H+5QK1JApVKkWEjRvmIPJSkpI1Ex67rW2uxE1ded/KnpUzx+4nHq/fUE3AGypSyPHn2UNfVr+Omhn9IaaV0cAG70N3Jq5hRR1U44toFaTwS/Zm7LG1U0YGDkIA3BhsWBT5/TR32gniPjR2itacWm2tjdtZsXB14kW8ySLWUJe8I0BhrZ2bGTlfUr8dg9HB49TFEv0hXtYmvrVvYN72e4XMShFrBRZWW4hdq2rYwkJ3Hn5vjiTX9AU7DpLZXIeyO5Uo5DI4fonenFqTnZENvANz/2Tb7y86/wyOFHKFaKrG5Yzb0b7yXoCWLkDJKF5Nu+rsW1y5I4eSHEHwH/FYhKKRPCnB75N8DdQB74tJTywFJcy+LyMRgfZF9vL02eOXC2MRofRAiIOF1UcKGpXupqO4nMj+ATApFNkNTcrF6xh3CklZShU6hWWeNws9HpJaY5GUkM8eC+BzkxeYKp9BRb27YuSvF6HV70qs7Dhx8GyRkZPpqm4bQ7KZRzfLiui1cKaRLVCkhwKgpbHW6GF0rynY7T5jxDA31nx05S+RST6UnAfJpYVb+KlfUrURWVjS0b2diyESnl4izfpmgXvaqdVGKIlmj7Yv64imRd980XzY2/VEqVEj879DMyxQw1vhoq1QpP9jzJ1ratfO1DX6O7rpsabw3O04TI8qU8jQFLGdLi/LxtJy+EaAbuAEZOW30XsGLhbwfwPxZeLa4SpJS80P8CTtc6XNohosyR0BScsoheGmfc/V6KwkYUHafNySd3fZLrC2m+k06QMHSmqmWys6M0xk8Rx+BgsInZcAv//cn/bubLe2sYnB3k16/+GsMw2Nlp6o17HB4q1cp5jDJTOQOqjVs9IdILg79+RUURgqg3SrqQPiPPez4/z9rGtYvLLruLezfdy3R6mkK5QMAdOGcc/XQZB7/bz841t9Pz6q9J5+YoFDX0agW/J8SK2IalaG4ABhODJPNJYuEYT5x4gmd7nyVdTBNwBfjab32N67uu59neZwm4Amiqxnx+nqgvSnu0fclssLj2WIqe/DeA/wA8fNq6e4F/lOZ02heFEEEhRIOUcnIJrmdxGSjpJdLFNP5QjEm5C4cxikPLMV/x0JNtIuJpwobEnkvQHGo266q6A/wnh5c+vchLg6/wxP4fMicEOZuDmfQM/7LvX6gYFVbVm/IBEU+EUqXES4Mvsbl1M3abnXQhza2rbuXI2BFmM7NEfGaBjbnsHE67k9tXmZOxFCHOyrnfvWI3Dx96mEKlgNvuJlPM4La72fAGRyyEeFOViTxCIeLwsW393eRS02SLKbzOAMJXS6cn+DZa+UxmMjO47C5+fuTnPH78cULuELXeWhLZBH/84z/mH37nH7h3470cGz9GvpxnV8cuVjWsOmtOg4XF6bwtJy+EuBcYl1IefoOAVRMwetry2MI6y8lfJdhVO07NSUkv4bC5yRvdqLKOg1NH8PtdrChlUfNzoJfZ3Pp6breqqjSUqzz94j8RcLgJuoIY0iCejdMf76fOX7e4b1OoiVPTp8iVcszl5rCpNip6hZu6b6IuUMdfPvyXDMQHkFIScAX4qw/8FbWB80ve1gfq+cjWj3Bi8gRzuTlW1a+iu74bj8PzttpCCMEWl4+9+RS2YANhGjEEtNoc1KpLV0Yx4o1waPQQz/U+Z4ZlNCdSSnwuH5qi8b3nvse3PvktmsPNS3ZNi2ufizp5IcQTwLm6PX8G/ClmqOYtI4T4HPA5gJaWS1fxs3hnURSFbW3b+E3Pb2gINODQHHgUlS3eKG3hGHo5S0Okleuarzsr3NEz2UOxUlysIqQIhYgngk2xMZOZWdwv4ArQUdPBqalTKEKhJdTCxpaNRLwRbui6gUe/+Cj7hvchkWxp2YLLcXEtlZAnxA1dN5x3e6lSYjY3i6Zq1HhrLqqu+RpB1cZt3hAzlTJlJCHVRlCxnff4dD6NgUHQHbyk8wN01HTwePVx8oZBqG4VhtNLOT+Hv5jBLQUDiYFLPpeFxWtc1MlLKc8pViKEWA+0A6/14mPAASHEdmAcOL27EVtYd67zfxv4NpgCZW/GeIt3ltdi2fuH9pPIJgh7wnx8x8cvqEYJkC1l8Tl9lCqlxbxvIQR1/jri2Th9031maqRhkCln+Oyez/Kx7R876zwOu4MbVpzfYb8ZJpITPLT/IfYO7DUHa+1OWsItfGzbx1hZv/KSnL1dKMQuUn1pJj3Dd577DkfHjiKRdNd189kbP3vRNgNzvOCD2z7OD2b6KCCwSwN/tAu33cN87zNsql1xyZ/XwuI13nK4Rkp5FFh8dhZCDAFbF7JrfgZ8QQjxI8wB15QVj7/6EEKwrmkdaxrWXLKuPEBTsInGQCPjyXG80ruoZOnUnNy57k6e632OucQcdpudLa1buG3VbehVnfmFAt16Vcdus1Pnr1uScnMjsyN841ff4MDIAapGlWKliCIUJpITDCYG+cItX2Br+zkF/N4Uuq7ztV9+jUQuQUu4hWwxy7HxY/zFw3/Btz7+LdwO90XPMed0c1v3rTx55BE87gAuwyCdncMINfPpGz79tm20ePfxTuXJ/wIzfbIPM4Xy996h61hcBhRFwX6JFYcA2qPtrGpYhaqozOXnGJ4dJpVP0RxuJp6J84kdn8Dr8KKqKvO5eX5y8CfYVBsD8QEG4gP4nD6667up89fxvg3ve1MhjzcipeS7z32XYxPHyBayjCXHKFQKixOhMsUMD+x7gJX1K9+2+uKR8SOMJceIhWK8OvUqZb2MQDA0O8T3X/w+f7DnDy56k5zWy9yz5jb8qsozvc8ym0tQH2hkz7aPsql18wWPtbA4F0vm5KWUbae9l8Dnl+rcFlcGXdeZyc7gsrkIeS996rzdZucDGz/A0bGjPHjgQaZSUyDg6VNPowiFQyOHuKn7Jja1bEKzaTx27DF2d+1mNjtLW00buVKOeDZOxBPh8eOP8+GtH77k2PkbmcnM8Orkq+SKOWayM2RLWWyqjUK5QDwTx+Pw0Dfdx2BikOua354GTLqYRiAYjA+CYPGmkS1l2T+8n/en3n/Rot0eRaUsDW5ZdQu3rLoFMOWay1KivMU2sHh3Y814tTgn+4f2893nv8t8fh6Aba3b+Oyez+Jz+i7peKfmRBEK+VKeqlFlOjNNPBNHr+qU9TK/OPoLNEXD7XCjCIVkIWmqTCoqfpefRDaBTbUxm51lPj9/UU2Y85HKpxb1ZUqVEkIIc8KUNLXz86U8Ekkyn3xL5z+d5lAzJb1EvpQnuJBaaRgGEkljsJH+eP9FnfwKu4sXC2lsQqAJBV1K5qpVtpxWIN3C4s1gOXmLsxhKDPH1X32dsDtMZ7QT3dB5ZfgVyk+W+fLdX77o8YPxQf5l37/w0IGHmExOoigK6VIao2pQrpaZSk+hKArP9z9Pd103DcEGFKGcofsiEFSNKkKYr28Vt92Nx+FBCIFds5Ov5KlUK1RlFafqxKW5kMiLVry6FDprO9nWto0f7/8xqqoiEKQKKTqiHcRCsUVVzAvRqDnYIn2cKOWoyCo2IbjO6aHlImUFlxIpJRPzE4zOj6IbOk3BJlojrZeky2Ox/LCcvMVZPNXzFIpQFnujNsVGR00Hh8YOMZmcvGBvdGxujK888hUUoaCqKlWqTKensat2/C4/mq6RLWXJFDPMpGfYs2IPLocLv8PP0OwQhsOgalRRFRVFKLgd7rfciwdoCDbQGe1kNjNLvpxHr+oU9SIeu4eQO4TX6WVN45oztOffDl96z5dIFVK8OvkqmqqxpW0L6xrXLRYsuRRa7U5imoOyNLAL5U0X6H47xNNxvvv8d9k3tI9UIYVNteG1e9navpVP7fzURZ9ELJYflpO3OIt4Nn5WfU8hBIpQSBfTNHD+f/RHjzyKIQ2aw81Mp6cZYQRN0SjpJZBmzrxbc2NX7ebEqWADbrubmfQMIXeIocSQqSfTsIpipcj7NrzvbWXYqIrK793we8zmZsmWs4uyCQJBwB2gPlDPJ3d8Eu8ShUPsdjtfeu+X+PmRn1MsF0FAIptgc+vmxXkDl2S3ELjE288sejOk8in+/um/59neZ0nlUxQrRcrVMn6nn+nMNJlChj++84+t8oBXGZaTtziLdU3r2D+0/4zZqYVKAZtqIxa8cL53f6KfoCuIlBK/y0+xUgSgWq2SLqaxq3azEIiQRDwRZtIz2Gw2trdtZ11sHYWyKUlQ56+jJdxy3rTDYqVIWS/jdXgvGkZoibTw5Tu/zC+O/oKXBl4iV8rhd/tZVbuKD239EC2RpZ2EF/VF+fj2jzORnKBcLRP1Rd/W08jl4qXBl9g7sJdkPsl4cpxUIQWYA8iNwUYOjRziiVef4Lc2/9YVttTizWA5eYuz2LNiD0/2PEnvdC8Rb4SSXiJTyPCpXZ/C47ywREB7pN2U8i1lmcvN0RJpYWR2BEMauO1unJoTVaj4nD40VWN4bpiKXiGVT/Hn7/tzbLYL/yQreoW9A3t5deJVEKZy5Z6Vey461b8p1MRnb/wsn9jxCfLlPKqiEnQH33LWzsVwaI6rTjjsxYEXQcJcbo50Mb2o6JktZknmkoRdYQ4MH+C+Tfe9Y+1msfRYTt7iLDxOD39xz1/wxKtPsG94H0FXkPeseQ8bWzZe9Ni719/Ns73P0h/vp95fj10xRcdiwRg2m41ipYhbcxP1R1ndsBq7zY4hDU5OneTloZe5vuv6C57/ub7n6JnqoSFg6sbnSjkePfIoH9n2kYv2loUQuB3uS5qU9G6kXClT0AsoQlksqA5QkRUzU6qSRwiBXtXRbEun2WPxzmI5eYtz4nP5uG/zfdy3+b43dVxrTStfuPULfPXRrxLPxvHYPXx020fpjHYyNj9GupAmXUzjc/peV0+UEHAHODFxgl2du87bS8yX8vRM9dAYbFx0QB6Hh2wpS89kz0VvEBYXZmX9Sp7ufRqPw4PdZqdYKZopp4qGqqq4NBdNgaUpjmJx+bC+LYslZ1PLJj5w3QeI+qJn9PjK1TK7V+zme899b1EZ0jAMZnOzdNR0LOazq+cZcCzqRQRi0cG/hlM7szDIUjBUKvBMPsVMtUytaudGT4B2+8UF0q5mblxxI/96/F+ZycxQqBTIFDIY0qDGW0PUFyXijXDb6tusUM1VhpX4arHkuOwuNrZsZCI1QbFSNKWGM3EcNgfb27dzz3X3kCwkmcvOkSwk6aztJOqLEgvFLphJ43f6sdvsZqbOaWQKmSWV3+0vFfin9DTT1TIeoTJTLfNPyRn6S4Ulu8ZypD3azmd2f4aoL0pDoIG2aBurG1cT9ASJhWP8/u7fZ0W9JZJ2tWH15C3eEba1bcPj8HBw5CDzuXk6o51sbduKx+HhznV3kivnSBfSBF1BytWyWR2qY+cFz2lTbdy44kZ+deJXuO1u7DY7qUKKqC96yTnol8JT+Xk8QiWg2ihUClSKGSoSHkPy+dq2JbvOcuTW1bfSUdPBU6ee4uT0SSp6hTWNa7i5+2baa66ugWQLE2HKzCwPtm7dKvft23elzbC4DORKOU5NnWIqPUWNt4bu+u5Lzr+eSk1xYuIE2VKWtkgb3fXdi5LGS8FX48PUKjamM9NMJMdRUKgCeZuDv6ztoKO2Y8muZWGxFAgh9kspzymlavXkLa4IHoeHTa2b3tKx9YH6N1W+780SUm0kSnkmkuN4HT4UoVBE4DB0ft3za5pCTUt6U7GweCexYvIWFm/gRqef6XIeXWiLDj6vqHRXy1SN6hnVrSwsljuWk7eweAPr3T72gKlOqaggYEM5Q8woI5FnZfdYWCxnrHCNhcU5uL22nanBFwjrtbhtdgTmzE+35j5D7sHCYrljdUksLM6B3+XnjtXvIZObY2J+nPH5cUp6ibvW32VNBrK4qrB+rRYW56GrrotYOMZMegZVUanz11kO3uKqw/rFWlhcAKfmXHKVSguLy4kVrrGwsLC4hrGcvIWFhcU1jOXkLSwsLK5hLCdvYWFhcQ1jOXkLCwuLa5hlJVAmhIgDw1fQhBogcQWv/1awbL58XI12WzZfPq6k3a1Syui5NiwrJ3+lEULsO5+S23LFsvnycTXabdl8+ViudlvhGgsLC4trGMvJW1hYWFzDWE7+TL59pQ14C1g2Xz6uRrstmy8fy9JuKyZvYWFhcQ1j9eQtLCwsrmEsJ29hYWFxDWM5eUAI8UUhRI8Q4rgQ4v8+bf2fCCH6hBAnhRDvvZI2ng8hxB8JIaQQomZhWQgh/nbB7iNCiM1X2sbXEEL8PwvtfEQI8RMhRPC0bcu2rYUQdy7Y1SeE+I9X2p5zIYRoFkL8RghxYuF3/IcL68NCiF8JIXoXXkNX2tZzIYRQhRAHhRA/X1huF0K8tNDm/58Qwn6lbTwdIURQCPHgwu/5VSHEruXa1u96Jy+EuAW4F7hOSrkW+K8L69cAHwPWAncCfyeEUK+YoedACNEM3AGMnLb6LmDFwt/ngP9xBUw7H78C1kkpNwCngD+B5d3WC3b8v5jtugb47QV7lxs68EdSyjXATuDzC3b+R+DXUsoVwK8Xlpcjfwi8etry14BvSCm7gHngM1fEqvPzN8BjUspVwHWYti/Ltn7XO3ng3wJ/LaUsAUgpX6vSfC/wIyllSUo5CPQB26+QjefjG8B/AE4fPb8X+Edp8iIQFEI0XBHr3oCU8nEppb6w+CIQW3i/nNt6O9AnpRyQUpaBH2Hau6yQUk5KKQ8svM9gOp0mTFv/98Ju/xv44BUx8AIIIWLA+4B/WFgWwK3Agwu7LCu7hRABYA/wHQApZVlKmWSZtrXl5GElcOPCo+HTQohtC+ubgNHT9htbWLcsEELcC4xLKQ+/YdOytvs0fh/45cL75WzzcrbtnAgh2oBNwEtAnZRycmHTFLAcC9R+E7OzYiwsR4DkaR2C5dbm7UAc+N5CiOkfhBAelmlbvysqQwkhngDqz7HpzzDbIIz5iLsNeEAI0XEZzTsvF7H7TzFDNcuKC9kspXx4YZ8/wwwv/OBy2vZuQAjhBX4M/F9SyrTZKTaRUkohxLLKmRZC3APMSCn3CyFuvsLmXCo2YDPwRSnlS0KIv+ENoZnl1NbvCicvpbz9fNuEEP8WeEiaEwZeFkIYmEJD40DzabvGFtZdNs5ntxBiPWZv4vDCP3EMOCCE2M4VtvtCbQ0ghPg0cA9wm3x9ksYVb+sLsJxtOwMhhIbp4H8gpXxoYfW0EKJBSjm5ELabOf8Zrgg3AB8QQtwNOAE/Zrw7KISwLfTml1ubjwFjUsqXFpYfxHTyy7K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    " + ] + }, + "metadata": { + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "import matplotlib.pyplot as plt\n", + "import matplotlib\n", + "import numpy as np\n", + "\n", + "colors = [\"red\", \"darkorange\", \"gold\", \"turquoise\", \"darkgreen\"]\n", + "x = [x for x,y in vis_dims]\n", + "y = [y for x,y in vis_dims]\n", + "color_indices = df.Score.values - 1\n", + "\n", + "colormap = matplotlib.colors.ListedColormap(colors)\n", + "plt.scatter(x, y, c=color_indices, cmap=colormap, alpha=0.3)\n", + "for score in [0,1,2,3,4]:\n", + " avg_x = np.array(x)[df.Score-1==score].mean()\n", + " avg_y = np.array(y)[df.Score-1==score].mean()\n", + " color = colors[score]\n", + " plt.scatter(avg_x, avg_y, marker='x', color=color, s=100)\n", + "plt.title(\"Amazon ratings visualized in language using t-SNE\")" + ] + } + ], + "metadata": { + "interpreter": { + "hash": "be4b5d5b73a21c599de40d6deb1129796d12dc1cc33a738f7bac13269cfcafe8" + }, + "kernelspec": { + "display_name": "Python 3.7.3 64-bit ('base': conda)", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.7.3" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/Visualizing_embeddings_in_3D.ipynb b/examples/Visualizing_embeddings_in_3D.ipynb new file mode 100644 index 0000000..ebbfdf6 --- /dev/null +++ b/examples/Visualizing_embeddings_in_3D.ipynb @@ -0,0 +1,268 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "id": "983ef639-fbf4-4912-b593-9cf08aeb11cd", + "metadata": {}, + "source": [ + "# Visualizing the embeddings in 3D" + ] + }, + { + "cell_type": "markdown", + "id": "9c9ea9a8-675d-4e3a-a8f7-6f4563df84ad", + "metadata": {}, + "source": [ + "The example uses [PCA](https://scikit-learn.org/stable/modules/generated/sklearn.decomposition.PCA.html) to reduce the dimensionality fo the embeddings from 2048 to 3. Then we can visualize the data points in a 3D plot. The small dataset `dbpedia_samples.jsonl` is curated by randomly sampling 200 samples from [DBpedia validation dataset](https://www.kaggle.com/danofer/dbpedia-classes?select=DBPEDIA_val.csv)." + ] + }, + { + "cell_type": "markdown", + "id": "8df5f2c3-ddbb-4cc4-9205-4c0af1670562", + "metadata": {}, + "source": [ + "### 1. Load the dataset and query embeddings" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "id": "133dfc2a-9dbd-4a5a-96fa-477272f7af5a", + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Categories of DBpedia samples: Artist 21\n", + "Film 19\n", + "Plant 19\n", + "OfficeHolder 18\n", + "Company 17\n", + "NaturalPlace 16\n", + "Athlete 16\n", + "Village 12\n", + "WrittenWork 11\n", + "Building 11\n", + "Album 11\n", + "Animal 11\n", + "EducationalInstitution 10\n", + "MeanOfTransportation 8\n", + "Name: category, dtype: int64\n" + ] + }, + { + "data": { + "text/html": [ + "
    \n", + "\n", + "\n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + " \n", + "
    textcategory
    0Morada Limited is a textile company based in ...Company
    1The Armenian Mirror-Spectator is a newspaper ...WrittenWork
    2Mt. Kinka (金華山 Kinka-zan) also known as Kinka...NaturalPlace
    3Planning the Play of a Bridge Hand is a book ...WrittenWork
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    " + ], + "text/plain": [ + " text category\n", + "0 Morada Limited is a textile company based in ... Company\n", + "1 The Armenian Mirror-Spectator is a newspaper ... WrittenWork\n", + "2 Mt. Kinka (金華山 Kinka-zan) also known as Kinka... NaturalPlace\n", + "3 Planning the Play of a Bridge Hand is a book ... WrittenWork\n", + "4 Wang Yuanping (born 8 December 1976) is a ret... Athlete" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "import pandas as pd\n", + "samples = pd.read_json(\"dbpedia_samples.jsonl\", lines=True)\n", + "categories = sorted(samples[\"category\"].unique())\n", + "print(\"Categories of DBpedia samples:\", samples[\"category\"].value_counts())\n", + "samples.head()" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "id": "19874e3e-a216-48cc-a27b-acb73854d832", + "metadata": {}, + "outputs": [], + "source": [ + "from openai.embeddings_utils import get_embeddings\n", + "# NOTE: The following code will send a query of batch size 200 to /embeddings, cost about $0.2\n", + "matrix = get_embeddings(samples[\"text\"].to_list(), engine=\"text-similarity-babbage-001\")" + ] + }, + { + "cell_type": "markdown", + "id": "d410c268-d8a7-4979-887c-45b1d382dda9", + "metadata": {}, + "source": [ + "### 2. Reduce the embedding dimensionality" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "id": "f5410068-f3da-490c-8576-48e84a8728de", + "metadata": {}, + "outputs": [], + "source": [ + "from sklearn.decomposition import PCA\n", + "pca = PCA(n_components=3)\n", + "vis_dims = pca.fit_transform(matrix)\n", + "samples[\"embed_vis\"] = vis_dims.tolist()" + ] + }, + { + "cell_type": "markdown", + "id": "b6565f57-59c6-4d36-a094-3cbbd9ddeb4c", + "metadata": {}, + "source": [ + "### 3. Plot the embeddings of lower dimensionality" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "id": "b17caad3-f0de-4115-83eb-55434a132acc", + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + }, + { + "data": { + "application/vnd.jupyter.widget-view+json": { + "model_id": "864488447fdd46b4ae1f338d3b0afded", + "version_major": 2, + "version_minor": 0 + }, + "image/png": 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    \n", + " Figure\n", + "
    \n", + " \n", + "
    \n", + " " + ], + "text/plain": [ + "Canvas(toolbar=Toolbar(toolitems=[('Home', 'Reset original view', 'home', 'home'), ('Back', 'Back to previous …" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib widget\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "\n", + "fig = plt.figure(figsize=(10, 5))\n", + "ax = fig.add_subplot(projection='3d')\n", + "cmap = plt.get_cmap(\"tab20\")\n", + "\n", + "# Plot each sample category individually such that we can set label name.\n", + "for i, cat in enumerate(categories):\n", + " sub_matrix = np.array(samples[samples[\"category\"] == cat][\"embed_vis\"].to_list())\n", + " x=sub_matrix[:, 0]\n", + " y=sub_matrix[:, 1]\n", + " z=sub_matrix[:, 2]\n", + " colors = [cmap(i/len(categories))] * len(sub_matrix)\n", + " ax.scatter(x, y, zs=z, zdir='z', c=colors, label=cat)\n", + "\n", + "ax.set_xlabel('x')\n", + "ax.set_ylabel('y')\n", + "ax.set_zlabel('z')\n", + "ax.legend(bbox_to_anchor=(1.1, 1))" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "id": "a8868043-9889-4a0b-b23d-79bb3823bdc7", + "metadata": {}, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3 (ipykernel)", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.9.9" + } + }, + "nbformat": 4, + "nbformat_minor": 5 +} diff --git a/examples/Zero-shot_classification_with_embeddings.ipynb b/examples/Zero-shot_classification_with_embeddings.ipynb new file mode 100644 index 0000000..dee08b4 --- /dev/null +++ b/examples/Zero-shot_classification_with_embeddings.ipynb @@ -0,0 +1,224 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Zero-shot classification with embeddings\n", + "\n", + "In this notebook we will classify the sentiment of reviews using embeddings and zero labeled data! The dataset is created in the [Obtain_dataset Notebook](Obtain_dataset.ipynb).\n", + "\n", + "We'll define positive sentiment to be 4 and 5-star reviews, and negative sentiment to be 1 and 2-star reviews. 3-star reviews are considered neutral and we won't use them for this example.\n", + "\n", + "We will perform zero-shot classification by embedding descriptions of each class and then comparing new samples to those class embeddings." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": {}, + "outputs": [], + "source": [ + "import pandas as pd\n", + "import numpy as np\n", + "\n", + "from sklearn.metrics import classification_report\n", + "\n", + "df = pd.read_csv('output/embedded_1k_reviews.csv')\n", + "df['babbage_similarity'] = df.babbage_similarity.apply(eval).apply(np.array)\n", + "df['babbage_search'] = df.babbage_search.apply(eval).apply(np.array)\n", + "\n", + "df= df[df.Score!=3]\n", + "df['sentiment'] = df.Score.replace({1:'negative', 2:'negative', 4:'positive', 5:'positive'})" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Zero-Shot Classification\n", + "To perform zero shot classification, we want to predict labels for our samples without any training. To do this, we can simply embed short descriptions of each label, such as positive and negative, and then compare the cosine distance between embeddings of samples and label descriptions. \n", + "\n", + "The highest similarity label to the sample input is the predicted label. We can also define a prediction score to be the difference between the cosine distance to the positive and to the negative label. This score can be used for plotting a precision-recall curve, which can be used to select a different tradeoff between precision and recall, by selecting a different threshold." + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " precision recall f1-score support\n", + "\n", + " negative 0.67 0.88 0.76 136\n", + " positive 0.98 0.93 0.95 789\n", + "\n", + " accuracy 0.92 925\n", + " macro avg 0.82 0.90 0.86 925\n", + "weighted avg 0.93 0.92 0.92 925\n", + "\n" + ] + }, + { + "data": { + "image/png": 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wfO4OktpIKo/hWuDB9PUnwIGSmklqTtJB7VnszMxqUZYEUd4M1Qd4OiIWZDlxRKwELgOGkHy4PxURkyQNkHRMultv4CNJU4CtgRvT9YNJBgl8n6SfYnxEvJDlumZmVjOyjOb6oqQPgW+BiyW1BZZWcwwAEfEy8HKFdf1zXg8mSQYVjysFLsxyDTMzK4wsM8pdA/QimQdiBfANye2qZmbWgFVbg0j7AM4AfpQ+zPwGcG+B4zIzsyLL0sR0D8mDaneny2em684rVFBmZlZ8WRLEXhHRPWf5dUn1c4JVMzPLLEuCKJXUJSKmA0jaASgtbFhmZo3PitIy5i5ezorSMlaWBRtv0JStWhVv+p0sCeJqYLikGSTjIm0P/KygUZmZNRLLV5YxYdZ83p42lz+8NmW1bU2biH//78G03mSDosRWbYKIiGGSdgK+l676KCKWFTYsM7OGZ8oXi5j25WKGTv6CIRM/B1hjhNddtmnFuft35r1P5/P4vz5h8bKVdTdBpNOLXgLsTzLK6khJ91achtTMzPIbNjkZjLrvn95abf1RP9iGLm03QRL7dWlNj46bs0GzpkBSe3j8X5/Ueqy5sjQxPQwsAv6ULp8GPAL8pFBBmZk1JP3268zYT97j6sO/R4ctNmSPjluw3ZZ1f2LOLAli14jInehnuKQPKt3bzMxWc0z3bTmm+7bFDmOtZRmLaaykH5YvSNoHGF24kMzMrC7IUoPYE/inpPLGsI4kA+y9D0Q6XaiZmTUwWRLEEQWPwszM6pwst7l+XBuBmJlZ3ZKlD8LMzBohJwgzM8vLCcLMzPLKlCAk3V/VspmZNTxZaxD3VbNsZmYNTKYEERFjqlo2M7OGp9LbXCW9QDI4X14RcUxBIjIzszqhqucgbqu1KMzMrM6pNEFExBvlryVtCHSMiI9qJSozMyu6avsgJB0NjANeTZd7SHq+wHGZmVmRZemk/jWwNzAfICLGAZ0LFpGZmdUJWRLEiohYUGFdpZ3XZmbWMGQZzXWSpNOApunc1JcD/yxsWGZmVmxZahA/B7oBy4BBwELg/ytgTGZmVgdkGe57CXCdpFuSxVhU+LDMzKzYstzFtFc6e9wE4H1J4yXtWfjQzMysmLL0QTwAXBIRIwEk7Q/8FfBUo2ZmDViWPojS8uQAEBFvASsLF5KZmdUFlSYISXtI2gN4Q9J9knpLOlDS3cCILCeXdISkjyRNk3RNnu3bSxomaYKkEZI65GzrKOkfkiZL+kBSp7UvnplZ/VTy9bcA3DNietFiqKqJ6fYKy7/KeV3tcxCSmgJ/Bg4FZgGjJD0fER/k7HYb8HBEDJR0EHATcGa67WHgxogYKmkToKy6a5qZNRSbbdQcgCdGfcrNJxanRb+qsZh+vJ7n3huYFhEzACQ9ARwL5CaIrsCV6evhwLPpvl2BZhExNI1l8XrGYmZWr5y1byc++nwRQyZ9XrQYsnRSI6kPybMQLcvXRcSAag5rD3yaszwL2KfCPuOBE4A/AscDrSS1BnYG5kt6hmRYj9eAayKitEJcFwAXAHTs2DFLUczMLKMst7neC5xM8sCcgJ8A29fQ9a8CDpT0HnAgUAKUkiSuA9LtewE7AP0qHhwR90dEz4jo2bZt2xoKyczMINtdTL0i4izg64i4AdiX5Bt+dUqA7XKWO6TrVomIzyLihIjYHbguXTefpLYxLiJmRMRKkqanPTJc08zMakiWBPFt+nuJpG2BFUC7DMeNAnaS1FlSC+AUYLVhwiW1kVQew7XAgznHbi6pvFpwEKv3XZiZWYFlSRAvStocuBUYC8wkGZOpSuk3/8uAIcBk4KmImCRpgKTy6Up7Ax9JmgJsDdyYHltK0rw0LH2KW8BfshfLzMzWV5axmH6TvvybpBeBlnmG/67s2JeBlyus65/zejAwuJJjh+Kntc3MiqbSBCHphCq2ERHPFCYkMzOrC6qqQRxdxbYAnCDMzBqwqh6U+1ltBmJmZnVLlk5qMzMrgulzFvPV4uUsWV6c8VGdIMzM6qh3Z8wD4F//mVeU6ztBmJnVUc9duh8A4z6ZX5TrZxlqYyNJ10v6S7q8k6S+hQ/NzKxxm7dkOQB/HDa1KM1MWWoQfwWWkQyxAclwGb8tWERmZgZA752/G2NuxcpqZ1mocVkSRJeI+D3JEBtExBKSJ5vNzKyAJNG/b9eiXT9LglguaUPSSYIkdSGpUZiZWQOWZT6IXwOvAttJegzYjzxDb5uZWcOSZSymf0gaA/yQpGnpioj4quCRmZlZUVWbICS9ADwOPB8R3xQ+JDMzqwuy9EHcRjK72weSBks6SVLL6g4yM7P6LUsT0xvAG5Kakkzccz7JxD6bFjg2MzNLjZjyJUuWl3Jsj23ZqEWW7uP1l+kq6V1MR5PMTb0HMLCQQZmZWWJiSTL9zhVPjAOgZfMmHL97h1q5dpYnqZ8imRHuIOAukucifl7owMzMDHp03Hy15fvemMGcRbXzpEGWPogHSJLCRRExPCLKCh2UmZklztq3EzNv7sPwq3oD8OHni7jvjem1cu2qZpQ7KCJeBzYGjpVWf3jaM8qZmdWezm025oCd2jBy6ld8U0vjMlVVgzgw/X10nh8P1mdmVsseOXcfAAb9+9NauV5VM8r9Kn05ICL+k7tNUueCRmVmZkWXpQ/ib3nWDa7pQMzMrHpn77s9m2/UvFauVVUfxC5AN2AzSSfkbNoU8INyZmYNXFXPQXyPpK9hc5J+h3KLSB6WMzOzWvbOjLnMX7KCFyd8Rt/dti3otarqg3gOeE7SvhHxTkGjMDOzTJatTJ40ePa9kuIlCEn/nU4UdJqkUytuj4jLCxqZmZmt4Y2rf0yna17itclfFvxaVTUxTU5/jy54FGZmttbmL1nO5hu1KNj5q2pieiH9vWrcJUlNgE0iYmHBIjIzsypdd9T3ufHlyZSWFXae6ixjMT0uaVNJGwMTSYb9vrqgUZmZWaU2aJ7lCYX1l+UqXdMaw3HAK0Bn4MxCBmVmZtUr77AulCwJormk5iQJ4vmIWAEUtl5jZmaVGjk1mfX5mLveKuh1siSI+4CZJIP2vSlpe8B9EGZmRXL+ATsAsHhZYQftqzZBRMSdEdE+Io6KxMfAj7OcXNIRkj6SNE3SNXm2by9pmKQJkkZI6lBh+6aSZkm6K3OJzMwauL07bwnA0hVllMz/tmDXydJJvZmkOySNTn9uJ6lNVHdcU+DPwJFAV+BUSV0r7HYb8HBE7AYMAG6qsP03wJsZymFm1qjs2j6Z9bnk6yImCJL5pxcBP01/FgJ/zXDc3sC0iJgREcuBJ4BjK+zTFXg9fT08d7ukPYGtgX9kuJaZWaNy7ZHfL/g1siSILhHxq/SDfkZE3ADskOG49kDuoOWz0nW5xgPlAwEeD7SS1Dp93uJ24KqqLiDpgvKazZw5czKEZGZmWWVJEN9K2r98QdJ+QE3Vaa4CDpT0HskERSVAKXAJ8HJEzKrq4Ii4PyJ6RkTPtm3b1lBIZmYGVQ+1Ue4i4GFJm6XLXwNnZziuBNguZ7lDum6ViPiMtAYhaRPgxIiYL2lf4ABJlwCbAC0kLY6INTq6zcysMKpMEJJ6ADsCp5B+uK/FMBujgJ3S2edK0nOcVuH8bYB5EVEGXEvS30FEnJ6zTz+gp5ODmVntqrSJSVJ/4CngROAl4OS1GYMpIlYClwFDSAb+eyoiJkkaIOmYdLfewEeSppB0SN+4TqUwM7MaV1UN4mSgR0QskdQaeBX4y9qcPCJeBl6usK5/zuvBVDN9aUQ8BDy0Ntc1M7P1V1Un9bKIWAIQEXOr2dfMzBqYqmoQO0h6Pn0toEvOMhFxTP7DzMysIagqQVR8qO22QgZiZmZ1S1UTBr1Rm4GYmVndUtVdTC9IOjod6rvith3Su5HOKWx4ZmZWLFU1MZ0PXAn8/5LmAXOAlkAnYDpwV0Q8V/AIzcysKKpqYvoc+G/gvyV1AtqRDLExpfzuJjMza7iyDLVBRMwkmTTIzMwaCT/bYGZmeTlBmJlZXk4QZmaWV5YpR/eTNFTSFEkzJP1H0ozaCM7MzPJbvrIMgGlfLi7YNbLUIB4A7gD2B/YCeqa/zcysSJYsLwVg4D9nFuwaWRLEgoh4JSK+jIi55T8Fi8jMzKrVZ7d2AOzafrNq9lx3WW5zHS7pVuAZYFn5yogYW7CozMysWu0337Cg58+SIPZJf/fMWRfAQTUfjpmZZbVk+UoWfLuiYOevNkFExI8LdnUzM1tnXy9ZwWuTv6CsLGjSRDV+/ix3MW0m6Q5Jo9Of2yUVrtHLzMzWShTovFk6qR8EFgE/TX8WAn8tUDxmZpbRlYfuXNDzZ+mD6BIRJ+Ys3yBpXIHiMTOzOiJLDeJbSfuXL0jaj2RUVzMza8Cy1CAuBgam/Q4C5gH9ChmUmZkVX5a7mMYB3SVtmi4vLHRQZmZWfJUmCElnRMSjkq6ssB6AiLijwLGZmVkRVVWD2Dj93ao2AjEzs7qlqilH70t/31B74ZiZWV2R5UG530vaVFJzScMkzZF0Rm0EZ2ZmxZPlNtfD0o7pviTzUu8IXF3IoMzMrPiyJIjyZqg+wNMRsaCA8ZiZWR2R5TmIFyV9SPJw3MWS2gJLCxuWmZkVW7U1iIi4BugF9IyIFcA3wLGFDszMzIqrqucgDoqI1yWdkLMud5dnChmYmZkVV1U1iAPT30fn+emb5eSSjpD0kaRpkq7Js3379M6oCZJGSOqQru8h6R1Jk9JtJ69VqczMbL1V9RzEr9LfP1uXE0tqCvwZOBSYBYyS9HxEfJCz223AwxExUNJBwE3AmcAS4KyImCppW2CMpCERMX9dYjEzs7WX5TmI30naPGd5C0m/zXDuvYFpETEjIpYDT7Bm30VX4PX09fDy7RExJSKmpq8/A74E2ma4ppmZ1ZAst7kemfvNPSK+Bo7KcFx74NOc5VnpulzjgfI+juOBVpJa5+4gaW+gBTC94gUkXVA+092cOXMyhGRmZlllSRBNJW1QviBpQ2CDKvZfG1cBB0p6j6TPowQozblWO+AR4GcRUVbx4Ii4PyJ6RkTPtm1dwTAzq0lZnoN4DBgmqXya0Z8BAzMcVwJsl7PcIV23Stp8dAKApE2AE8trK+nw4i8B10XEuxmuZ2ZmNSjLfBC3SBoPHJKu+k1EDMlw7lHATpI6kySGU4DTcneQ1AaYl9YOriWZ/xpJLYC/k3RgD85aGDMzqzlZahAAk4GVEfGapI0ktYqIRVUdEBErJV0GDAGaAg9GxCRJA4DREfE80Bu4SVIAbwKXpof/FPgR0FpSv3Rdv3TyIjMzqwXVJghJ5wMXAFsCXUg6mu8FDq7u2Ih4GXi5wrr+Oa8HA2vUECLiUeDR6s5vZmaFk6WT+lJgP2AhQHr76VaFDMrMzIovS4JYlj7HAICkZkAULiQzM6sLsiSINyT9L7ChpEOBp4EXChuWmZkVW5YE8T/AHOB94EKSPoVfFjIoMzMrvio7qdPxlCZFxC7AX2onJDMzqwuqrEFERCnwkaSOtRSPmZnVEVmeg9gCmCTp3ySTBQEQEccULCozMyu6LAni+oJHYWZmdU5VM8q1BC4CdiTpoH4gIlbWVmBmZlZcVfVBDAR6kiSHI4HbayUiMzOrE6pqYuoaET8AkPQA8O/aCcnMzOqCqmoQK8pfuGnJzKzxqaoG0V3SwvS1SJ6kXpi+jojYtODRmZlZ0VSaICKiaW0GYmZmdUuWoTbMzKwRyjphUL20YsUKZs2axdKlS4sditUjLVu2pEOHDjRv3rzYoZgVVYNOELNmzaJVq1Z06tQJScUOx+qBiGDu3LnMmjWLzp07FzscsyqVfP0tAHO/WcZWrVrW+PkbdBPT0qVLad26tZODZSaJ1q1bu9Zp9cI7M+YCMGzylwU5f4NOEICTg601/5ux+uLRc/cp6PkbfIIwM2uoWjQr7Ee4E0SBff7555xyyil06dKFPffck6OOOoopU6Ywc+ZMdt111xq7Tv/+/XnttdcAGDlyJN26daNHjx6UlJRw0kknrde5I4KDDjqIhQsXrlr37LPPIokPP/xw1bqZM2ey4YYb0qNHD7p27cpFF11EWVnZel37zTffZI899qBZs2YMHjy40v3GjBnDD37wA3bccUcuv/xyIpJZcefNm8ehhx7KTjvtxKGHHsrXX38NwIsvvkj//v3XKzazhs4JooAiguOPP57evXszffp0xowZw0033cQXX3xR49caMGAAhxxyCACPPfYY1157LePGjaN9+/ZVfrBWtHLlmg/Nv/zyy3Tv3p1NN/3u2chBgwax//77M2jQoNX27dKlC+PGjWPChAl88MEHPPvss+tWoFTHjh156KGHOO2006rc7+KLL+Yvf/kLU6dOZerUqbz66qsA3HzzzRx88MFMnTqVgw8+mJtvvhmAPn368MILL7BkyZL1is+sIWvQdzHluuGFSXzw2cLqd1wLXbfdlF8d3a3S7cOHD6d58+ZcdNFFq9Z1794dSL5tl5s5cyZnnnkm33yTTLdx11130atXL2bPns3JJ5/MwoULWblyJffccw+9evXi3HPPZfTo0UjinHPO4Re/+AX9+vWjb9++zJ8/n6eeeoohQ4bwyiuvcOONN9K3b18mTpxIaWkp11xzDSNGjGDZsmVceumlXHjhhYwYMYLrr7+eLbbYgg8//JApU6asVo7HHnuMCy64YNXy4sWLeeuttxg+fDhHH300N9xwwxplb9asGb169WLatGnr9N6W69SpEwBNmlT+XWb27NksXLiQH/7whwCcddZZPPvssxx55JE899xzjBgxAoCzzz6b3r17c8sttyCJ3r178+KLL/LTn/50vWI0a6gaTYIohokTJ7LnnntWu99WW23F0KFDadmyJVOnTuXUU09l9OjRPP744xx++OFcd911lJaWsmTJEsaNG0dJSQkTJ04EYP78+aud67zzzuOtt96ib9++nHTSSaslogceeIDNNtuMUaNGsWzZMvbbbz8OO+wwAMaOHcvEiRPz3tr59ttvc999961afu655zjiiCPYeeedad26NWPGjFmjnEuWLGHYsGEMGDBgjfMdcMABLFq0aI31t91226pa0NooKSmhQ4cOq5Y7dOhASUkJAF988QXt2rUDYJtttlmt9tazZ09GjhzpBGFWiUaTIKr6pl9sK1as4LLLLmPcuHE0bdp01Tf4vfbai3POOYcVK1Zw3HHH0aNHD3bYYQdmzJjBz3/+c/r06bPqAz6Lf/zjH0yYMGFVk9OCBQuYOnUqLVq0YO+99670vv958+bRqlWrVcuDBg3iiiuuAOCUU05h0KBBqxLE9OnT6dGjB5I49thjOfLII9c438iRIzPHXJMkrXaH0lZbbcVnn31WlFjM6oNGkyCKoVu3bpna///whz+w9dZbM378eMrKymjZMnng5Uc/+hFvvvkmL730Ev369ePKK6/krLPOYvz48QwZMoR7772Xp556igcffDBTPBHBn/70Jw4//PDV1o8YMYKNN9640uOaNWtGWVkZTZo0Yd68ebz++uu8//77SKK0tBRJ3HrrrcB3fRBVqekaRPv27Zk1a9aq5VmzZtG+fXsAtt56a2bPnk27du2YPXs2W2211ar9li5dyoYbbrjW1zNrLNxJXUAHHXQQy5Yt4/7771+1bsKECWt8g16wYAHt2rWjSZMmPPLII5SWlgLw8ccfs/XWW3P++edz3nnnMXbsWL766ivKyso48cQT+e1vf8vYsWMzx3P44Ydzzz33sGJFMpL7lClTVvV7VOV73/seM2bMAGDw4MGceeaZfPzxx8ycOZNPP/2Uzp07r1WtYOTIkYwbN26Nn3VJDgDt2rVj00035d133yUiePjhhzn22GMBOOaYYxg4cCAAAwcOXLUekvLX5J1kZsXyzbLCzMjgBFFAkvj73//Oa6+9RpcuXejWrRvXXnst22yzzWr7XXLJJQwcOJDu3bvz4Ycfrvo2P2LECLp3787uu+/Ok08+yRVXXEFJSQm9e/emR48enHHGGdx0002Z4znvvPPo2rUre+yxB7vuuisXXnhh3ruWKurTp8+qjt5BgwZx/PHHr7b9xBNPXONuppoyatQoOnTowNNPP82FF15It27fNRX26NFj1eu7776b8847jx133JEuXbqsatq65pprGDp0KDvttBOvvfYa11xzzapjhg8fTp8+fQoSt1ltWLI8+f/725cmF+T8Kr9fvL7r2bNnjB49erV1kydP5vvf/36RImo4Zs+ezVlnncXQoUOLHUqN+eKLLzjttNMYNmxY3u3+t2P1QVlZsMP/vsx5+3fml327rtM5JI2JiJ75trkPwqrVrl07zj//fBYuXLjasxD12SeffMLtt3uadavfmjQRM28uXC3YCcIyaWi3gu61117FDsGszmvwfRANpQnNao//zZglCpogJB0h6SNJ0yRdk2f79pKGSZogaYSkDjnbzpY0Nf05e12u37JlS+bOnev/8JZZ+XwQ5bcamzVmBWtiktQU+DNwKDALGCXp+Yj4IGe324CHI2KgpIOAm4AzJW0J/AroCQQwJj3267WJoUOHDsyaNYs5c+bURJGskSifUc6ssStkH8TewLSImAEg6QngWCA3QXQFrkxfDweeTV8fDgyNiHnpsUOBI4C1upeyefPmnhXMzGwdFbKJqT3wac7yrHRdrvHACenr44FWklpnPBZJF0gaLWm0awlmZjWr2J3UVwEHSnoPOBAoAUqzHhwR90dEz4jo2bZt20LFaGbWKBWyiakE2C5nuUO6bpWI+Iy0BiFpE+DEiJgvqQToXeHYEQWM1czMKijYk9SSmgFTgINJEsMo4LSImJSzTxtgXkSUSboRKI2I/mkn9Rhgj3TXscCe5X0SlVxvDvDxeoTcBvhqPY6vjxpbmRtbecFlbizWp8zbR0TeJpiC1SAiYqWky4AhQFPgwYiYJGkAMDoiniepJdwkKYA3gUvTY+dJ+g1JUgEYUFVySI9ZrzYmSaMre9y8oWpsZW5s5QWXubEoVJkL+iR1RLwMvFxhXf+c14OBvONhR8SDQLZxrM3MrMYVu5PazMzqKCeI79xf/S4NTmMrc2MrL7jMjUVBytxghvs2M7Oa5RqEmZnl5QRhZmZ5NaoEkWF02Q0kPZlu/5ekTkUIs0ZlKPOVkj5IR9QdJmn7YsRZk6orc85+J0oKSfX+lsgsZZb00/RvPUnS47UdY03L8G+7o6Thkt5L/30fVYw4a4qkByV9KWliJdsl6c70/ZggaY98+62ViGgUPyTPYkwHdgBakIwD1bXCPpcA96avTwGeLHbctVDmHwMbpa8vbgxlTvdrRfLszbtAz2LHXQt/552A94At0uWtih13LZT5fuDi9HVXYGax417PMv+I5OHhiZVsPwp4BRDwQ+Bf63vNxlSDWDW6bEQsB8pHl811LDAwfT0YOFiSajHGmlZtmSNieEQsSRffJRnWpD7L8ncG+A1wC7C0NoMrkCxlPh/4c6RD5kfEl7UcY03LUuYAyufI3Qz4rBbjq3ER8SZQ1QPDx5JMnxAR8S6wuaR263PNxpQgsowQu2qfiFgJLABa10p0hZFpVNwc55J8A6nPqi1zWvXeLiJeqs3ACijL33lnYGdJb0t6V9IRtRZdYWQp86+BMyTNInlg9+e1E1rRrO3/92p5TmoDQNIZJBM0HVjsWApJUhPgDqBfkUOpbc1Impl6k9QS35T0g4iYX8ygCuxU4KGIuF3SvsAjknaNiLJiB1ZfNKYaRLWjy+bukw42uBkwt1aiK4wsZUbSIcB1wDERsayWYiuU6srcCtgVGCFpJklb7fP1vKM6y995FvB8RKyIiP+QDKS5Uy3FVwhZynwu8BRARLwDtCQZ1K6hyvT/fW00pgQxCthJUmdJLUg6oZ+vsM/zQPn81ycBr0fa+1NPVVtmSbsD95Ekh/reLg3VlDkiFkREm4joFBGdSPpdjomI0cUJt0Zk+bf9LOkQ+ukoyjsDM2oxxpqWpcyfkIwmjaTvkySIhjyz2PPAWendTD8EFkTE7PU5YaNpYopso8s+QFINnUbSGXRK8SJefxnLfCuwCfB02h//SUQcU7Sg11PGMjcoGcs8BDhM0gckk3JdHRH1tnacscz/BfxF0i9IOqz71ecvfJIGkST5Nmm/yq+A5gARcS9JP8tRwDRgCfCz9b5mPX6/zMysgBpTE5OZma0FJwgzM8vLCcLMzPJygjAzs7ycIMzMLC8nCFtnkkoljZM0UdILkjav4fPPTO/ZR9LiSvbZUNIbkppK6iTp2zSmDyTdmz45vTbX7CnpzvR1b0m9crZdJOms9SlTep5fS7qqmn0eknTSWpyzU2WjfNY0SceUj54q6ThJXXO2DUgfvFyX8z4hqT4/vNfgOEHY+vg2InpExK4kz41cWoQYzgGeiYjSdHl6RPQAdiMZwfO4tTlZRIyOiMvTxd5Ar5xt90bEw+sbcH0XEc9HxM3p4nEk73P5tv4R8do6nvoe4L/XMzyrQU4QVlPeIR0YTFIXSa9KGiNppKRd0vVbS/q7pPHpT690/bPpvpMkXbCW1z0deK7iynSwxX8CO6bfrl/Xd3NedEyv+5O09jNe0pvput6SXlQyF8hFwC/SGskB5d/8Je0i6d/l10rP/376es+0RjNG0hBVM5qmpPMljUpj+JukjXI2HyJptKQpkvqm+zeVdGt6zARJF67NmyVpsaQ/pO/1MElt0/U9lAziNyH9G22Rrr9c380X8kS6rp+ku9K/3zHArel71KW85qNkroanc67bW9KL6evDJL0jaaykpyVtku42Mi1zo3mAt65zgrD1JqkpyZAG5U8p3w/8PCL2BK4C7k7X3wm8ERHdSca1n5SuPyfdtydwuaRMI+gqGWJhh4iYmWfbRmlM7wN/AgZGxG7AY2kcAP2Bw9N4Vnt6PD3nvcAf0lrSyJxtHwItJHVOV50MPCmpeXqtk9LyPAjcWE0xnomIvdIYJpOMH1SuE8mw1n2AeyW1TLcviIi9gL2A83PiKC/7tpJeruR6G5M8adwNeIPkaVyAh4H/Sd+j93PWXwPsnq6/qMJ79E+Sv/nV6Xs0PWfza8A+kjZOl08GnlDSZPhL4JCI2AMYDVyZnq+M5Cng7pW/XVabnCBsfWwoaRzwObA1MDT9NtiLZOiOcSTjPJV/iz6IpBmBiCiNiAXp+ssljScZF2k7sg8i1waYX2Fdl/S6bwMvRcQrwL5A+QxqjwD7p6/fBh6SdD7JcA1r4ymSDz3S308C3yMZCHBoGsMvqX5+jV3TWtb7JLWhbrnXiIiyiJhKMm7SLsBhJOPtjAP+RTIc/WrvV0R8FhGVzZ5WlsYK8Ciwv6TNgM0j4o10/UCSyWkAJgCPKRntd2U1ZcmNYSXwKnB0WiPoQ1LT+yFJk9TbaRnOBnJnMfwS2DbrdaywXJWz9fFtRPRIv60PIemDeAiYn/YDVEtSb+AQYN+IWCJpBMmgapmun2ff6VmvHREXSdqH5MNrjKQ9M14Xkg/ZpyU9k5wqpkr6ATApIvZdi/M8BBwXEeMl9SMdUK88xIohk8wW9vOIGJK7Qes+PW51Y+30IUkWRwPXpWXM6gngMpL+qdERsUiSgKERcWolx7Qk+btaHeAahK23dEa6y0kGR1sC/EfST2DVPLnlTQbDSKY1LW9L34xkSPWv0+SwC8k3zKzX/Rpomja9VOWffDfw4ukkbd1I6hIR/4qI/iSjfG5X4bhFJMOD57v2dJJB767nu2/kHwFtlcw9gKTmkrrlOz5HK2B22jx1eoVtP5HURFIXkqk1PyJJxBen+yNp55xmnCyakIxUDHAa8FZak/ta0gHp+jOBN5TcAbZdRAwH/ofkb7VJhfNV+h6RNGHtQTKb3RPpuneB/STtmMa/saSdc47ZGaiVu7Gsek4QViMi4j2S5ohTST7ozk2bjSbx3VSQVwA/TptTxpA0NbwKNJM0GbiZ5ANkbfyD75qMKvNz4GeSJpB8+F2Rrr9V0vtKbg/9J8m8xrleAI5PO2APYE1PAmfw3ZwDy0k+fG9Jyz6OnLugKnE9SVPR28CHFbZ9AvybZJa/iyJiKfB/wAfA2DTu+6jQElBNH8Q3wN7psQcBA9L1Z5O8HxOAHun6psCj6d/rPeDOPBMMPQFcLem9NJGtkt5Z9iJwZPqbiJhDMlnToPRa75A0nSFpa5Ja6eeVxG61zKO5Wr2mZPrQX0TEmcWOpT6QtDgiKtYC6gQlw3IvjIgHih2LJVyDsHotIsYCw9M7qax+m0/SQW51hGsQZmaWl2sQZmaWlxOEmZnl5QRhZmZ5OUGYmVleThBmZpbX/wPov24If76fLgAAAABJRU5ErkJggg==", + "text/plain": [ + "
    " + ] + }, + "metadata": { + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "from openai.embeddings_utils import cosine_similarity, get_embedding\n", + "from sklearn.metrics import PrecisionRecallDisplay\n", + "\n", + "def evaluate_emeddings_approach(\n", + " labels = ['negative', 'positive'], \n", + " engine = 'text-similarity-babbage-001',\n", + "):\n", + " label_embeddings = [get_embedding(label, engine=engine) for label in labels]\n", + "\n", + " def label_score(review_embedding, label_embeddings):\n", + " return cosine_similarity(review_embedding, label_embeddings[1]) - cosine_similarity(review_embedding, label_embeddings[0])\n", + "\n", + " engine_col_name = engine.replace('-','_').replace('_query','')\n", + " probas = df[engine_col_name].apply(lambda x: label_score(x, label_embeddings))\n", + " preds = probas.apply(lambda x: 'positive' if x>0 else 'negative')\n", + "\n", + " report = classification_report(df.sentiment, preds)\n", + " print(report)\n", + "\n", + " display = PrecisionRecallDisplay.from_predictions(df.sentiment, probas, pos_label='positive')\n", + " _ = display.ax_.set_title(\"2-class Precision-Recall curve\")\n", + "\n", + "evaluate_emeddings_approach(labels=['negative', 'positive'], engine='text-similarity-babbage-001')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can see that this classifier already performs extremely well. We used similarity embeddings, and the simplest possible label name. Let's try to improve on this by using more descriptive label names, and search embeddings." + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " precision recall f1-score support\n", + "\n", + " negative 0.65 0.93 0.76 136\n", + " positive 0.99 0.91 0.95 789\n", + "\n", + " accuracy 0.92 925\n", + " macro avg 0.82 0.92 0.86 925\n", + "weighted avg 0.94 0.92 0.92 925\n", + "\n" + ] + }, + { + "data": { + "image/png": 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dgZU1vAaSQf62k9STJDGcBJycu4OkzYGFEVEGXAbcmW6aAwyTdA1JH8R+wH9nOKeZWaMy+9rBHH/Ly8xcsIxT7/gXAKVlwZrS4Ih+W3HqXt2LFluNVxARcSkwAOgfEauBL0k6m2t63RrgfGA0yQRDD0TEdEnDJR2Z7jYIeFfSe0An4Ddp+SiSYcbfJOmnmBIRj61LxczMGosJHyxiwRerGD/jM17790JWl5bx9sdLeWzKR0WNK8uEQS2AU4Dvpo8iPA/cmuXgEfEk8GSlsitzlkeRJIPKrysFzslyDjOzxu4Xg3fkuXc+5dS9unNIny1p1kycdPsrlBV5ErosTUy3kNxm+ud0/dS07KxCBWVmtiE5a+A2DfJJ6iwJYo+I6Juz/pykKYUKyMzMGoYsdzGVSupVviJpG6C0cCGZmVlDkOUK4hJgrKT3Se4o6g6cUdCozMys6GpMEBHxrKTtgG+nRe9GxKrChmVmZsWW5S6mVsCPgX1JxkgaL+nWytOQmplZ05KlielukkmC/piunwzcA/ygUEGZmVnxZUkQO0VE75z1sZLeqnJvMzNrErLcxTRJ0l7lK5K+A0woXEhmZtYQZLmC2B14WdKcdL0byfAYbwIREbtU/VIzM2ussiSIQwsehZmZrWXq3CUs/6qUiKBYMy5nuc31g/oIxMzMvrb8q+R55HmLV9B1s42LEkOWPggzM6tnvz8hGeGotIgj9jlBmJk1QNM/WgrAE2/OL1oMThBmZg3Q8bt3BWDJ8tVFiyFTgpB0e3XrZmZWt3bs3JYWJaKkWXE6qCH7FcRtNaybmVkTkylBRMTE6tbNzKzpqfI2V0mPkQzOl1dEHFnVNjMza/yqew7ihnqLwszMGpwqE0REPF++LKk10C0i3q2XqMzMjNWlwVdryop2/hr7ICQdAUwGnk7X+0l6tMBxmZkZ8L8v/rto587SSX0VsCewGCAiJgM9CxaRmZmtJSKY+ekXPDrlI2Z+uqzezptlsL7VEbGk0mBRxXv228xsA9F1s9bMXbSCnpc9WVE2oFcH/jZsr2peVXeyXEFMl3QyUCJpO0l/BF4ucFxmZhu8FiXJR/TgnTtzwf7bArBydWm9nT/LFcQFwOXAKmAkMBr4VSGDMjMzGHvxoLXWJ81ZxKrV9ddpnWW47+XA5ZKuS1bji8KHZWZmxZblLqY90tnjpgJvSpoiaffCh2ZmZsWUpYnpDuDHETEeQNK+wP8BnmrUzKwJy9JJXVqeHAAi4kVgTeFCMjOzhqC6sZh2Sxefl3QbSQd1ACcC4wofmpmZ5Xpp5udAcidTqxYlBT9fdU1MN1Za/2XOsp+DMDMrknmLV9Cr4yYFP091YzF9r+BnNzOzzG78QV/+48EpvPbvhcVNELkkDQb6AK3KyyJieKGCMjOzb3rv0+Qpg8seepMhe3Yr+Pmy3OZ6K0m/wwWAgB8A3bMcXNKhkt6VNFPSpXm2d5f0rKSpksZJ6pqzrZukf0p6W9JbknpkrZSZWVP080N2oF3rFrRpWfj+B8h2F9OAiDgNWBQRVwN7A9vX9CJJJcCfgMOA3sAQSb0r7XYDcHdE7AIMB67J2XY3cH1E7EgyWOCnGWI1M2uySpqJo/ttRYvmWWeLrp0sZ1mR/l4uaStgNdA5w+v2BGZGxPsR8RVwH3BUpX16A8+ly2PLt6eJpHlEjAGIiGXpE91mZlZPsiSIxyW1B64HJgGzSW55rUkX4MOc9blpWa4pwLHp8jHAppI6kFyhLJb0kKQ3JF2fXpGsRdLZkiZImrBgwYIMIZmZWVY1JoiI+FVELI6Iv5P0PewQEVfU0fkvBvaT9AawHzAPKCXpPB+Ybt8D2AYYmie22yOif0T079ixYx2FZGZmUP2DcsdWs42IeKiGY88Dts5Z75qWVYiIj0ivICRtAhwXEYslzQUmR8T76baHgb1Ihv0wM7N6UN1trkdUsy2AmhLE68B2knqSJIaTgJNzd5C0ObAwIsqAy4A7c17bXlLHiFgA7A9MqOF8ZmZWh6p7UO6M2hw4ItZIOp9k/ogS4M6ImC5pODAhIh4FBgHXSArgBeAn6WtLJV0MPKtkKruJwF9qE4+Zma2bTA/Kra+IeBJ4slLZlTnLo4BRVbx2DB4x1sxsLZ8sXcXi5atZtaaUjZoX9nmI+rmZ1szM6sTT0z8G4OVZnxf8XE4QZmaNyH1n7wXA6jWFn3o0y1AbG0u6QtJf0vXtJB1e8MjMzOwbNtmooD0Da8lyBfF/wCqSITYguSPp1wWLyMzMGoQsCaJXRPyOZIgN0iEvVNCozMwsr7JIpuOZ9tHSgp8rS4L4SlJr0kmCJPUiuaIwM7N61kzJ9/P/Hf9+wc+VpTHrKuBpYGtJ9wL7kGfYCzMzK7ydurRj62+1pnO71gU/V40JIiL+KWkiyVAXAn4aEZ8VPDIzM8vrw4Ur+HDhClZ8VUrrAs4NkeUupseAg4FxEfG4k4OZWXF1+9bGACz/ak1Bz5OlD+IGkpFV35I0StLxklrV9CIzMyuMswb2rJfzZGlieh54Pp2PYX9gGMmgem0LHJuZmRVRpicu0ruYjiCZm3o3YEQhgzIzs+KrMUFIeoBk+tCngZuB59Phuc3MrAnLcgVxBzAkIkoLHYyZmTUc1c0ot39EPAe0AY6S1n54OsOMcmZmVgDT5yVPUU+as5iDencq2Hmqu4tpv/T3EXl+PFifmVmR7Ny1HQDTP1pS0PNUN6PcL9PF4RHx79xt6TSiZmZWBMfv3pVfPDyNls0LO2NDlqP/PU9Z3lngzMys6aiuD2IHoA/QTtKxOZvaAn5QzsysiavuLqZvk/Q1tCfpdyj3BcnDcmZm1oRV1wfxCPCIpL0j4pV6jMnMzBqA6pqYfp5OFHSypCGVt0fEhQWNzMzMiqq6Jqa3098T6iMQMzNrWKprYnos/V0x7pKkZsAmEVH4ue7MzKxaI1+bw/G7dWWLtoW5byjLfBB/k9RWUhtgGsmw35cUJBozM6tROi01Hy5cwVWPTS/YebI8B9E7vWI4GngK6AmcWrCIzMysWq1blrBHj80AaNWiiDPKAS0ktSBJEI9GxGogChaRmZnV6MFzB9ClfWuEat55PWVJELcBs0kG7XtBUnfAfRBmZk1clhnlbgJuyin6QNL3CheSmZk1BFk6qdtJ+r2kCenPjSRXE2Zm1oRlaWK6k2R4jRPSn6XA/xUyKDMzK74sM8r1iojjctavljS5QPGYmVkDkeUKYoWkfctXJO0DrChcSGZm1hBkuYI4F7hbUrt0fRFweuFCMjOzhqDaKwhJ/YDtgJOAXYBdImLXiJia5eCSDpX0rqSZki7Ns727pGclTZU0TlLXStvbSpor6ebMNTIzszpRZYKQdCXwAHAc8ARw4rqMwSSpBPgTcBjQGxgiqXel3W4A7o6IXYDhwDWVtv8KeCHrOc3MrO5UdwVxItAvIoYAewBnr+Ox9wRmRsT7EfEVcB9wVKV9egPPpctjc7dL2h3oBPxzHc9rZmZ1oLoEsSoilgNExOc17JtPF+DDnPW5aVmuKUD5dKbHAJtK6pCOGnsjcHF1J5B0dvnzGQsWLFjH8MzMrDrVdVJvI+nRdFlAr5x1IuLIOjj/xcDNkoaSNCXNA0qBHwNPRsRcqepxRiLiduB2gP79+3t8KDOzOlRdgqjcHHTDOh57HrB1znrXtKxCRHxEegUhaRPguIhYLGlvYKCkHwObAC0lLYuIb3R0m5lZYVQ3YdDztTz268B2knqSJIaTgJNzd5C0ObAwIsqAy0ie2iYifpizz1Cgv5ODmVn9qu4upsckHZEO9V152zaShks6s6rXR8Qa4HxgNMn0pQ9ExPT0deXNU4OAdyW9R9Ih/Zta1MXMzOpQdU1Mw4CLgP+WtBBYALQCegCzgJsj4pHqDh4RTwJPViq7Mmd5FDCqhmPcBdxV3T5mZlb3qmti+hj4OfBzST2AziRDbLxXfneTmZk1XVmG2iAiZpNMGmRmZhuIdX22wczMGoh5i1fw90lziSjMXf5OEGZmjVxZgZ4Cc4IwM2ukLjpo+4Iev8Y+iHT+h6uA7un+AiIitiloZGZmVlRZOqnvAH4GTCQZBsPMzDYAWRLEkoh4quCRmJlZg5IlQYyVdD3wELCqvDAiJhUsKjMzK7osCeI76e/+OWUB7F/34ZiZWUNRY4KIiO/VRyBmZtaw1Hibq6R2kn5fPjGPpBsltauP4MzMrHiyPAdxJ/AFcEL6sxT4v0IGZWZmxZelD6JXRByXs361pMkFisfMzBqILFcQKyTtW76SPji3onAhmZlZQ5DlCuI8YETa7yBgITC0kEGZmVnxZbmLaTLQV1LbdH1poYMyM7PiqzJBSDolIv4q6aJK5QBExO8LHJuZmRVRdVcQbdLfm9ZHIGZm1rBUN+Xobenvq+svHDMzayiyPCj3O0ltJbWQ9KykBZJOqY/gzMyseLLc5npw2jF9OMm81NsClxQyKDMzK74sCaK8GWow8GBELClgPGZm1kBkeQ7icUnvkDwcd56kjsDKwoZlZmbFVuMVRERcCgwA+kfEauBL4KhCB2ZmZsVV3XMQ+0fEc5KOzSnL3eWhQgZmZmbFVV0T037Ac8ARebYFThBmZk1adc9B/DL9fUb9hWNmZg1Flucgfiupfc76ZpJ+XdCozMys6LLc5npYRCwuX4mIRcD3CxaRmZk1CFkSRImkjcpXJLUGNqpmfzMzawKyPAdxL/CspPJpRs8ARhQuJDMzawiyzAdxnaQpwIFp0a8iYnRhwzIzs2LLcgUB8DawJiKekbSxpE0j4otCBmZmZsWV5S6mYcAo4La0qAvwcJaDSzpU0ruSZkq6NM/27ukIsVMljZPUNS3vJ+kVSdPTbSdmrpGZmdWJLJ3UPwH2AZYCRMQMYIuaXiSpBPgTcBjQGxgiqXel3W4A7o6IXYDhwDVp+XLgtIjoAxwK/HfurbZmZlZ4WRLEqoj4qnxFUnOSJ6lrsicwMyLeT19/H98cw6k3ydPaAGPLt0fEe2kiIiI+Aj4FOmY4p5mZ1ZEsCeJ5Sf8FtJZ0EPAg8FiG13UBPsxZn5uW5ZoClI/1dAywqaQOuTtI2hNoCcyqfAJJZ0uaIGnCggULMoRkZmZZZUkQ/wksAN4EzgGeBH5RR+e/GNhP0hskYz/NA0rLN0rqDNwDnBERZZVfHBG3R0T/iOjfsaMvMMzM6lK1dzGl/QjTI2IH4C/reOx5wNY5613Tsgpp89Gx6bk2AY4rf2pbUlvgCeDyiHh1Hc9tZma1VO0VRESUAu9K6rYex34d2E5ST0ktgZOAR3N3kLS5pPIYLgPuTMtbAv8g6cAetR7nNjOzWsryHMRmwHRJr5FMFgRARBxZ3YsiYo2k84HRQAlwZ0RMlzQcmBARjwKDgGskBfACyR1TACcA3wU6SBqalg2NiMlZK2ZmZrWTJUFcsb4Hj4gnSfoscsuuzFkeRfKMReXX/RX46/qe18zMaq+6GeVaAecC25J0UN8REWvqKzAzMyuu6vogRgD9SZLDYcCN9RKRmZk1CNU1MfWOiJ0BJN0BvFY/IZmZWUNQ3RXE6vIFNy2ZmW14qruC6Ctpaboskiepl6bLERFtCx6dmZkVTZUJIiJK6jMQMzNrWLIMtWFmZhugrBMGNUqrV69m7ty5rFy5stihWCPSqlUrunbtSosWLYodillRNekEMXfuXDbddFN69OiBpGKHY41ARPD5558zd+5cevbsWexwzIqqSTcxrVy5kg4dOjg5WGaS6NChg686zWjiCQJwcrB15n8zZokmnyDMzGz9OEEU2Mcff8xJJ51Er1692H333fn+97/Pe++9x+zZs9lpp53q7DxXXnklzzzzDADjx4+nT58+9OvXj3nz5nH88cfX6tgRwf7778/SpUsryh5++GEk8c4771SUzZ49m9atW9OvXz969+7NueeeS1nZN+Z5WicvvPACu+22G82bN2fUqKpHfp84cSI777wz2267LRdeeCERyay4Cxcu5KCDDmK77bbjoIMOYtGiRQA8/vjjXHnllVUez8ycIAoqIjjmmGMYNGgQs2bNYuLEiVxzzTV88skndX6u4cOHc+CBBwJw7733ctlllzF58mS6dOlS7QdrZWvWfPOh+SeffJK+ffvStu3Xz0aOHDmSfffdl5EjR661b69evZg8eTJTp07lrbfe4uGHH16/CqW6devGXXfdxcknn1ztfueddx5/+ctfmDFjBjNmzODpp58G4Nprr+WAAw5gxowZHHDAAVx77bUADB48mMcee4zly5fXKj6zpqxJ38WU6+rHpvPWR0tr3nEd9N6qLb88ok+V28eOHUuLFi0499xzK8r69u0LJN+2y82ePZtTTz2VL79Mptu4+eabGTBgAPPnz+fEE09k6dKlrFmzhltuuYUBAwbwox/9iAkTJiCJM888k5/97GcMHTqUww8/nMWLF/PAAw8wevRonnrqKX7zm99w+OGHM23aNEpLS7n00ksZN24cq1at4ic/+QnnnHMO48aN44orrmCzzTbjnXfe4b333lurHvfeey9nn312xfqyZct48cUXGTt2LEcccQRXX331N+revHlzBgwYwMyZM9frvS3Xo0cPAJo1q/q7zPz581m6dCl77bUXAKeddhoPP/wwhx12GI888gjjxo0D4PTTT2fQoEFcd911SGLQoEE8/vjjnHDCCbWK0ayp2mASRDFMmzaN3Xffvcb9tthiC8aMGUOrVq2YMWMGQ4YMYcKECfztb3/jkEMO4fLLL6e0tJTly5czefJk5s2bx7Rp0wBYvHjxWsc666yzePHFFzn88MM5/vjj10pEd9xxB+3ateP1119n1apV7LPPPhx88MEATJo0iWnTpuW9tfOll17itttuq1h/5JFHOPTQQ9l+++3p0KEDEydO/EY9ly9fzrPPPsvw4cO/cbyBAwfyxRdffKP8hhtuqLgKWhfz5s2ja9euFetdu3Zl3rxkdttPPvmEzp07A7DllluudfXWv39/xo8f7wRhVoUNJkFU902/2FavXs3555/P5MmTKSkpqfgGv8cee3DmmWeyevVqjj76aPr168c222zD+++/zwUXXMDgwYMrPuCz+Oc//8nUqVMrmpyWLFnCjBkzaNmyJXvuuWeV9/0vXLiQTTfdtGJ95MiR/PSnPwXgpJNOYuTIkRUJYtasWfTr1w9JHHXUURx22GHfON748eMzx1yXJK11h9IWW2zBRx99VJRYzBqDDSZBFEOfPn0ytf//4Q9/oFOnTkyZMoWysjJatWoFwHe/+11eeOEFnnjiCYYOHcpFF13EaaedxpQpUxg9ejS33norDzzwAHfeeWemeCKCP/7xjxxyyCFrlY8bN442bdpU+brmzZtTVlZGs2bNWLhwIc899xxvvvkmkigtLUUS119/PfB1H0R16voKokuXLsydO7dife7cuXTp0gWATp06MX/+fDp37sz8+fPZYostKvZbuXIlrVu3XufzmTUUf/vXHADGvvMpB/buVOfHdyd1Ae2///6sWrWK22+/vaJs6tSp3/gGvWTJEjp37kyzZs245557KC0tBeCDDz6gU6dODBs2jLPOOotJkybx2WefUVZWxnHHHcevf/1rJk2alDmeQw45hFtuuYXVq5OR3N97772Kfo/qfPvb3+b9998HYNSoUZx66ql88MEHzJ49mw8//JCePXuu01XB+PHjmTx58jd+1ic5AHTu3Jm2bdvy6quvEhHcfffdHHXUUQAceeSRjBgxAoARI0ZUlENS/7q8k8ysvp2//7YAzP685v/H68MJooAk8Y9//INnnnmGXr160adPHy677DK23HLLtfb78Y9/zIgRI+jbty/vvPNOxbf5cePG0bdvX3bddVfuv/9+fvrTnzJv3jwGDRpEv379OOWUU7jmmmsyx3PWWWfRu3dvdtttN3baaSfOOeecvHctVTZ48OCKjt6RI0dyzDHHrLX9uOOO+8bdTHXl9ddfp2vXrjz44IOcc8459OnzdVNhv379Kpb//Oc/c9ZZZ7HtttvSq1eviqatSy+9lDFjxrDddtvxzDPPcOmll1a8ZuzYsQwePLggcZvVhwG9OgDw6yfeLsjxVX6/eGPXv3//mDBhwlplb7/9NjvuuGORImo65s+fz2mnncaYMWOKHUqd+eSTTzj55JN59tln8273vx1rDMrKgm3+60mGDezJ5YN7r9cxJE2MiP75trkPwmrUuXNnhg0bxtKlS9d6FqIxmzNnDjfe6GnWrXFr1kzMvrZwV8FOEJZJU7sVdI899ih2CGYNXpPvg2gqTWhWf/xvxizRpBNEq1at+Pzzz/0f3jIrnw+i/FZjsw1Zk25i6tq1K3PnzmXBggXFDsUakfIZ5cw2dE06QbRo0cKzgpmZracm3cRkZmbrzwnCzMzycoIwM7O8msyT1JIWAB/U4hCbA5/VUTiNxYZW5w2tvuA6byhqU+fuEdEx34YmkyBqS9KEqh43b6o2tDpvaPUF13lDUag6u4nJzMzycoIwM7O8nCC+dnvNuzQ5G1qdN7T6guu8oShInd0HYWZmefkKwszM8nKCMDOzvDaoBCHpUEnvSpop6dI82zeSdH+6/V+SehQhzDqVoc4XSXpL0lRJz0rqXow461JNdc7Z7zhJIanR3xKZpc6STkj/1tMl/a2+Y6xrGf5td5M0VtIb6b/v7xcjzroi6U5Jn0qaVsV2SbopfT+mStqt1ieNiA3iBygBZgHbAC2BKUDvSvv8GLg1XT4JuL/YcddDnb8HbJwun7ch1Dndb1PgBeBVoH+x466Hv/N2wBvAZun6FsWOux7qfDtwXrrcG5hd7LhrWefvArsB06rY/n3gKUDAXsC/anvODekKYk9gZkS8HxFfAfcBR1Xa5yhgRLo8CjhAkuoxxrpWY50jYmxELE9XXwUa+zjXWf7OAL8CrgNW1mdwBZKlzsOAP0XEIoCI+LSeY6xrWeocQPkcue2Aj+oxvjoXES8AC6vZ5Sjg7ki8CrSX1Lk259yQEkQX4MOc9blpWd59ImINsAToUC/RFUaWOuf6Eck3kMasxjqnl95bR8QT9RlYAWX5O28PbC/pJUmvSjq03qIrjCx1vgo4RdJc4EnggvoJrWjW9f97jZr0fBCWnaRTgP7AfsWOpZAkNQN+Dwwtcij1rTlJM9MgkqvEFyTtHBGLixlUgQ0B7oqIGyXtDdwjaaeIKCt2YI3FhnQFMQ/YOme9a1qWdx9JzUkuSz+vl+gKI0udkXQgcDlwZESsqqfYCqWmOm8K7ASMkzSbpK320UbeUZ3l7zwXeDQiVkfEv4H3SBJGY5Wlzj8CHgCIiFeAViSD2jVVmf6/r4sNKUG8DmwnqaekliSd0I9W2udR4PR0+XjguUh7fxqpGussaVfgNpLk0NjbpaGGOkfEkojYPCJ6REQPkn6XIyNiQnHCrRNZ/m0/THL1gKTNSZqc3q/HGOtaljrPAQ4AkLQjSYJoyvMPPwqclt7NtBewJCLm1+aAG0wTU0SskXQ+MJrkDog7I2K6pOHAhIh4FLiD5DJ0Jkln0EnFi7j2Mtb5emAT4MG0P35ORBxZtKBrKWOdm5SMdR4NHCzpLaAUuCQiGu3VccY6/wfwF0k/I+mwHtqYv/BJGkmS5DdP+1V+CbQAiIhbSfpZvg/MBJYDZ9T6nI34/TIzswLakJqYzMxsHThBmJlZXk4QZmaWlxOEmZnl5QRhZmZ5OUHYepNUKmmypGmSHpPUvo6PPzu9Zx9Jy6rYp7Wk5yWVSOohaUUa01uSbk2fnF6Xc/aXdFO6PEjSgJxt50o6rTZ1So9zlaSLa9jnLknHr8Mxe1Q1ymddk3Rk+eipko6W1Dtn2/D0wcv1Oe59khrzw3tNjhOE1caKiOgXETuRPDfykyLEcCbwUESUpuuzIqIfsAvJCJ5Hr8vBImJCRFyYrg4CBuRsuzUi7q5twI1dRDwaEdemq0eTvM/l266MiGfW89C3AD+vZXhWh5wgrK68QjowmKRekp6WNFHSeEk7pOWdJP1D0pT0Z0Ba/nC673RJZ6/jeX8IPFK5MB1s8WVg2/Tb9XP6es6Lbul5f5Be/UyR9EJaNkjS40rmAjkX+Fl6RTKw/Ju/pB0kvVZ+rvT4b6bLu6dXNBMljVYNo2lKGibp9TSGv0vaOGfzgZImSHpP0uHp/iWSrk9fM1XSOevyZklaJukP6Xv9rKSOaXk/JYP4TU3/Rpul5Rfq6/lC7kvLhkq6Of37HQlcn75HvcqvfJTM1fBgznkHSXo8XT5Y0iuSJkl6UNIm6W7j0zpvMA/wNnROEFZrkkpIhjQof0r5duCCiNgduBj4c1p+E/B8RPQlGdd+elp+Zrpvf+BCSZlG0FUyxMI2ETE7z7aN05jeBP4IjIiIXYB70zgArgQOSeNZ6+nx9Ji3An9Ir5LG52x7B2gpqWdadCJwv6QW6bmOT+tzJ/CbGqrxUETskcbwNsn4QeV6kAxrPRi4VVKrdPuSiNgD2AMYlhNHed23kvRkFedrQ/KkcR/geZKncQHuBv4zfY/ezCm/FNg1LT+30nv0Msnf/JL0PZqVs/kZ4DuS2qTrJwL3KWky/AVwYETsBkwALkqPV0byFHDfqt8uq09OEFYbrSVNBj4GOgFj0m+DA0iG7phMMs5T+bfo/UmaEYiI0ohYkpZfKGkKybhIW5N9ELnNgcWVynql530JeCIingL2BspnULsH2Dddfgm4S9IwkuEa1sUDJB96pL/vB75NMhDgmDSGX1Dz/Bo7pVdZb5JcDfXJPUdElEXEDJJxk3YADiYZb2cy8C+S4ejXer8i4qOIqGr2tLI0VoC/AvtKage0j4jn0/IRJJPTAEwF7lUy2u+aGuqSG8Ma4GngiPSKYDDJld5eJE1SL6V1OB3IncXwU2CrrOexwvKlnNXGiojol35bH03SB3EXsDjtB6iRpEHAgcDeEbFc0jiSQdUynT/PvrOynjsizpX0HZIPr4mSds94Xkg+ZB+U9FByqJghaWdgekTsvQ7HuQs4OiKmSBpKOqBeeYiVQyaZLeyCiBidu0HrPz1uTWPtDCZJFkcAl6d1zOo+4HyS/qkJEfGFJAFjImJIFa9pRfJ3tQbAVxBWa+mMdBeSDI62HPi3pB9AxTy55U0Gz5JMa1relt6OZEj1RWly2IHkG2bW8y4CStKml+q8zNcDL/6QpK0bSb0i4l8RcSXJKJ9bV3rdFyTDg+c79yySQe+u4Otv5O8CHZXMPYCkFpL65Ht9jk2B+Wnz1A8rbfuBpGaSepFMrfkuSSI+L90fSdvnNONk0YxkpGKAk4EX0yu5RZIGpuWnAs8ruQNs64gYC/wnyd9qk0rHq/I9ImnC2o1kNrv70rJXgX0kbZvG30bS9jmv2R6ol7uxrGZOEFYnIuINkuaIISQfdD9Km42m8/VUkD8Fvpc2p0wkaWp4Gmgu6W3gWpIPkHXxT75uMqrKBcAZkqaSfPj9NC2/XtKbSm4PfZlkXuNcjwHHpB2wA/mm+4FT+HrOga9IPnyvS+s+mZy7oKpwBUlT0UvAO5W2zQFeI5nl79yIWAn8L/AWMCmN+zYqtQTU0AfxJbBn+tr9geFp+ekk78dUoF9aXgL8Nf17vQHclGeCofuASyS9kSayCumdZY8Dh6W/iYgFJJM1jUzP9QpJ0xmSOpFclX5cRexWzzyaqzVqSqYP/VlEnFrsWBoDScsiovJVQIOgZFjupRFxR7FjsYSvIKxRi4hJwNj0Tipr3BaTdJBbA+ErCDMzy8tXEGZmlpcThJmZ5eUEYWZmeTlBmJlZXk4QZmaW1/8HwFhzzjh8/JsAAAAASUVORK5CYII=", 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    " + ] + }, + "metadata": { + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "evaluate_emeddings_approach(labels=['An Amazon review with a negative sentiment.', 'An Amazon review with a positive sentiment.'], engine='text-similarity-babbage-001')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Using the search embeddings and descriptive names leads to an additional improvement in performance." + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " precision recall f1-score support\n", + "\n", + " negative 0.77 0.79 0.78 136\n", + " positive 0.96 0.96 0.96 789\n", + "\n", + " accuracy 0.94 925\n", + " macro avg 0.87 0.88 0.87 925\n", + "weighted avg 0.94 0.94 0.94 925\n", + "\n" + ] + }, + { + "data": { + "image/png": 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    " + ] + }, + "metadata": { + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "evaluate_emeddings_approach(labels=['An Amazon review with a negative sentiment.', 'An Amazon review with a positive sentiment.'], engine='text-similarity-babbage-001')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "As shown above, zero-shot classification with embeddings can lead to great results, especially when the labels are more descriptive than just simple words." + ] + } + ], + "metadata": { + "interpreter": { + "hash": "be4b5d5b73a21c599de40d6deb1129796d12dc1cc33a738f7bac13269cfcafe8" + }, + "kernelspec": { + "display_name": "Python 3.7.3 64-bit ('base': conda)", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.7.3" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/book_translation/data/geometry_English.tex b/examples/book_translation/data/geometry_English.tex new file mode 100644 index 0000000..d3f36c6 --- /dev/null +++ b/examples/book_translation/data/geometry_English.tex @@ -0,0 +1,30259 @@ +\documentclass[11pt]{book} + +%PAPIR - US TRADE +\paperwidth 15.24cm +\paperheight 22.86cm + +%TEKST +\textwidth 11.9cm \textheight 19.4cm +\oddsidemargin=-0.5cm +\evensidemargin=-1.2cm +\topmargin=-15mm + +\headheight=13.86pt + +%\usepackage[slovene]{babel} +\usepackage[english]{babel} + +%\usepackage[cp1250]{inputenc} +\usepackage[utf8]{inputenc} + + +\usepackage[T1]{fontenc} +\usepackage{amsmath} +\usepackage{color} +\usepackage{amsfonts} +\usepackage{makeidx} +\usepackage{calc} +\usepackage{gclc} +%\usepackage[dvips]{hyperref} +\usepackage{amssymb} +\usepackage[dvips]{graphicx} +\usepackage{fancyhdr} + +%za slike +\usepackage{caption} +\DeclareCaptionFormat{empty} + +\def\contentsname{Content} + +\makeindex + +\newcommand{\ch}{\mathop {\mathrm{ch}}} +\newcommand{\sh}{\mathop {\mathrm{sh}}} +\newcommand{\tgh}{\mathop {\mathrm{th}}} +\newcommand{\tg}{\mathop {\mathrm{tg}}} +\newcommand{\ctg}{\mathop {\mathrm{ctg}}} +\newcommand{\arctg}{\mathop {\mathrm{arctg}}} +\newcommand{\arctgh}{\mathop {\mathrm{arcth}}} + +\def\indexname{Index} + +\definecolor{green1}{rgb}{0,0.5,0} +\definecolor{viol}{rgb}{0.5,0,0.5} +\definecolor{viol1}{rgb}{0.2,0,0.9} +\definecolor{viol3}{rgb}{0.3,0,0.6} +\definecolor{viol4}{rgb}{0.6,0,0.6} +\definecolor{grey}{rgb}{0.5,0.5,0.5} + + \def\qed{$\hfill\Box$} +\newcommand{\kdokaz}{\color{red}\qed\vspace*{2mm}\normalcolor} + +\newcommand{\res}[1]{\color{green1}\textit{#1}\normalcolor} + +\newtheorem{izrek}{Theorem}[section] +\newtheorem{lema}{Lemma}[section] +\newtheorem{definicija}{Definition}[section] +\newtheorem{aksiom}{Axiom}[section] +\newtheorem{zgled}{Exercise}[section] +\newtheorem{naloga}{Problem} +\newtheorem{trditev}{Proposition}[section] +\newtheorem{postulat}{Postulate} +\newtheorem{ekv}{E} + + +%BARVA + +\newcommand{\pojem}[1]{\color{viol4}\textit{#1}\normalcolor} + +%\newcommand{\pojemFN}[1]{\textit{#1}} + +\newcommand{\blema}{\color{blue}\begin{lema}} +\newcommand{\elema}{\end{lema}\normalcolor} + +\newcommand{\bizrek}{\color{blue}\begin{izrek}} +\newcommand{\eizrek}{\end{izrek}\normalcolor} + +\newcommand{\bdefinicija}{\begin{definicija}} +\newcommand{\edefinicija}{\end{definicija}} + +\newcommand{\baksiom}{\color{viol3}\begin{aksiom}} +\newcommand{\eaksiom}{\end{aksiom}\normalcolor} + +\newcommand{\bzgled}{\color{green1}\begin{zgled}} +\newcommand{\ezgled}{\end{zgled}\normalcolor} + +\newcommand{\bnaloga}{\color{red}\begin{naloga}} +\newcommand{\enaloga}{\end{naloga}\normalcolor} + +\newcommand{\btrditev}{\color{blue}\begin{trditev}} +\newcommand{\etrditev}{\end{trditev}\normalcolor} + + +\newcommand{\del}[1]{\chapter{#1}} +\newcommand{\poglavje}[1]{\section{#1}} +%\newcommand{\naloge}[1]{\color{red}\section*{#1}\normalcolor} +\newcommand{\naloge}[1]{\section{#1}} +\newcommand{\ppoglavje}[1]{\subsection{#1}} + +\setlength\arraycolsep{2pt} + +\author{Milan Mitrovi\'c} +\title{\textsl{\Huge{\textbf{Euclidean Plane Geometry}}}} + +\date{} + +%_________________________________________________________________________________________ + +\begin{document} +\pagestyle{fancy} +\lhead[\thepage]{\textsl{\nouppercase{\rightmark}}} +\rhead[\textsl{\nouppercase{\leftmark}}]{\thepage} +\cfoot[]{} + + + \vspace*{-12mm} + +\hspace*{24mm} \textsl{\Huge{\textbf{Euclidean }}}\\ + +\hspace*{24mm} \textsl{\Huge{\textbf{Plane}}}\\ + +\hspace*{24mm} \textsl{\Huge{\textbf{Geometry}}} + + \vspace*{8mm}\hspace*{60mm}Milan Mitrovi\'c +% \normalcolor + + \vspace*{0mm} + +\hspace*{-21mm} +\input{sl.NASL2A4.pic} + +%\color{viol1} + \hspace*{48mm}Sevnica + + \hspace*{49mm} +2013 +% \normalcolor + %slikaNova0-1-1 +%\includegraphics[width=120mm]{slikaNaslov.pdf} + + \setcounter{section}{0} + \thispagestyle{empty} + \newpage + + +%\pagecolor{white} + +%\color{viol1} +\vspace*{17mm} + \hspace*{77mm} \textit{To Boris and Jasmina} +%\normalcolor + %slikaNova0-1-1 +%\includegraphics[width=120mm]{slikaNaslov.pdf} + + + \thispagestyle{empty} +\newpage + +%________________________________________________________________________ + +%PREDGOVOR +%{\hypertarget{Vsebina}\tableofcontents} + %\printindex +\thispagestyle{empty} + + +%_______________________________________________________________________ + + +\chapter*{Preface} + +\thispagestyle{empty} + + +\thispagestyle{empty} + +The present book is the result of the experience I gained as a professor in teaching geometry at the Mathematical Gymnasium in Belgrade for many years and preparing students in Slovenia for the International Mathematical Olympiad (IMO). + +Formally, the substance is presented in such a way that it does not rely on prior knowledge of geometry. In the book, we will deal only with planar Euclidean geometry - all definitions and statements refer to the plane. + +The first two chapters deal with the history and axiomatic design of geometry. +The consequences of the axioms of incidence, congruence and parallelism are discussed in detail, while in the other two groups (axioms of order and continuity) the consequences are mostly not proven. +Chapters three and four deal with the relation of the congruence of figures, the use of the triangle congruence theorems, and a circle. +In the fifth chapter, a vectors are defined. Thales's theorem of proportion is proven. +Chapter six deals with isometries and their use. Their classification has been performed. +Chapters 7 and 8 deal with similarity transformations, figure similarity relation, and area of figures. The ninth chapter presents the inversion. +At the end of each chapter (except the introductory one) are exercises. Solutions and instructions can be found in the last, tenth chapter. + + +The book contains 341 theorems, 247 examples and 418 solved problems (28 of them from the IMO). In this sense, the book in front of you is at the same time a preparing guide for the IMO. + +For some well-known theorems and problems, are given brief historical remarks. That can help high school and college students to better understand the development of geometry over the centuries. + +A lot of help in writing the book was selflessly offered to me by Prof. Roman Drstvenšek, who read the manuscript in its entirety. With his professional and linguistic comments, he made a great contribution to the final version of the book. +In that work, he was assisted by Prof. Ana Kretič Mamič. I kindly thank both of them for the effort and time they have generously devoted to this book. + +I would especially like to thank Prof. Kristjan Kocbek, who read the partial manuscript + and with his critical remarks contributed to + significant improvement of the book. + +%I sincerely thank prof. Gordana Kenda Kozinc, who read and proofread the introductory chapter, and thanks also to prof. Alenka Brilej, who helped Roman with the language test with quite a few useful tips. + +I also thank Prof. Dr Predrag Janiči\'c, who wrote the wonderful software package \textit {GCLC} for \ LaTeX {}. Almost all pictures in this book were made with this package. + +Last but not least, I would like to thank the students of the Bežigrad Grammar School, the 1st Grammar School in Celje and the Brežice Grammar School for their inspiration and support. Students from these schools attended the renewed course of geometry which I have already taken before +years in Belgrade. + +\thispagestyle{empty} + +\vspace*{12mm} Sevnica, December 2013 \hfill Milan +Mitrovi\'c +%\newpage + +%________________________________________________________________________ + + \tableofcontents + + %\thispagestyle{empty} + +%\newpage + + + +% DEL 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% O DEDUKTIVNI IN INDUKTIVNI METODI +%________________________________________________________________________________ + + \del{Introduction} \label{pogUVOD} + +%________________________________________________________________________________ +\poglavje{Deductive and Inductive Method} \label{odd1DEDUKT} + +We learn a lot of geometric concepts in elementary school, such as: +triangle, circle, right angle, etc. Later we also learn some +propositions: propositions about the congruence of triangles, +Pythagoras' and Tales' proposition. At the beginning we do not +prove the propositions, but we verify the facts based on several +individual examples. This way of reasoning is called inductive +method. Inductive method (lat. inductio -- introduction) is thus a +way of reasoning, in which we come to general conclusions from +individual examples. Later we start proving individual +propositions. Through these proofs we first encounter the so-called +deductive way of reasoning, or deduction. Deduction (lat. deductio +-- deduction) is a way of reasoning, in which we come from general +to individual conclusions. The idea of this method is thus to +deduce a general conclusion by proving it and then use it in +individual examples. Since we cannot verify all examples using +inductive method, because their number is usually infinite, we can +also come to wrong conclusions using this method. With deductive +method we always get correct conclusions, if the assumptions we +use in the proof are correct. Let us analyze both of these methods +using the following example. We try to come to the conclusion: + \btrditev \label{TalesUvod} + The diameter of a circle subtends a right angle to any point on the circle. + \etrditev + + +\begin{figure}[!htb] +\centering +\input{sl.1.2.1.6.pic} +\caption{} \label{sl.sl.1.2.1.6.pic} +\end{figure} + + %slikaNova0-1-1 +%\includegraphics[width=50mm]{slikaNova0-1-1.pdf} + +If we used inductive method, we would verify whether this proposition +is true in some individual examples; for example in the case when +the apex of the angle is in the center of the circle and similar +(Figure \ref{sl.sl.1.2.1.6.pic}). If we only deduced the general +conclusion from these individual examples, of course we could not +be sure that the proposition is not true in some example we did not +verify. + +We will now use the deductive method. Let $AB$ be the radius of the circle with center $O$ and $L$ any point on this circle, different from points $A$ and $B$ (Figure \ref{sl.sl.1.2.1.6.pic}). We will prove that the angle $ALB$ is a right angle. Because $OA\cong OB\cong OL$, it follows that the triangles $AOL$ and $BOL$ are isosceles, therefore $\angle ALO\cong\angle LAO=\alpha$ and $\angle BLO\cong\angle LBO=\beta$. Then $\angle ALB=\alpha+\beta$. The sum of the interior angles in triangle $ALB$ is equal to $180^0$, therefore $2\alpha+2\beta=180^0$. From this it follows: + $$\angle ALB=\alpha+\beta=90^0$$ + +We notice that in the case of using the deductive method or in the proof of the statement, we have not considered a certain point $L$ on the circle, but an arbitrary point (in general position). This means that the statement is valid for every point on the circle (except $A$ and $B$), if the proof is correct, of course. But is the proof correct? In this proof, we used the following two statements: + \btrditev + If two sides in a triangle are congruent, then the angles opposite the congruent sides are congruent angles. + \etrditev + \btrditev + The sum of the interior angles of a triangle is equal to $180^0$. + \etrditev +We also used concepts such as: isosceles triangle, angle congruence; in the statement itself, we also used the concepts: diameter, circle, angle over the diameter and right angle. In order to be sure that the statement we proved is true, we must be sure that the two statements we used in the proof are also true. In our case, we assume that we have already proved the aforementioned two statements and that we have introduced all the concepts mentioned. It is clear that this problem arises with every statement - even with the two on which we relied in the proof. This requires a certain systematization of the entire geometry. The question arises as to how to start if we again refer to previously proven statements in the proof of each statement. This process could then continue indefinitely. Thus we come to the need for initial statements - \index{aksiomi} \pojem{aksiomih}. The same applies to concepts - we need t. i. \pojem{začetni pojmi}.\index{začetni pojmi} In this way, each geometry (there can be more of them), which we consider, depends on the choice of initial concepts and axioms. We call this approach to building a geometry \pojem{sintetični postopek}, and we call the geometry itself \pojem{sintetična geometrija}\index{geometrija!sinteti\v{c}na}. + + +%________________________________________________________________________________ +\poglavje{Basic Terms and Basic Theorems} \label{odd1POJMI} + +In some theory (like geometry) we introduce every new +concept with a \index{definition} \pojem{definition}, which describes this concept +with the help + of some initial or already defined concepts. + The connections between concepts and their appropriate properties are given by +statements, + which we call +\pojem{theory statements}. As we have already mentioned, we call the initial statements +\index{aksiomi} \pojem{aksiomi}, the statements derived from them +are \pojem{izreki} of this theory. Formally, +\pojem{proof} \index{dokaz izreka} of some statement $\tau$ is a sequence of statements, which logically follow one + from the other, each of which +is either an axiom or a statement derived from the axioms (izrek), and the last one in this +sequence is precisely the statement $\tau$. + + Although the choice of axioms is not uniquely determined, it cannot be arbitrary. + When making this choice +we must be careful not to lead to contradictory statements +or to a contradiction. This means that, for a given choice of +axioms, there is no such statement that both the statement and its +negation are izreki in this theory. We also need enough +axioms so that we can determine, for every statement that we can formulate in this +theory, whether it is true or not. This means that +either the statement or its negation is an izrek in this theory. For a system of axioms that satisfies the first requirement, we say that it is +\index{sistem aksiomov!neprotisloven} \pojem{neprotisloven}, for one that satisfies the second requirement, we say that it is \index{sistem +aksiomov!popoln} \pojem{popoln}. When choosing axioms, there is also a third requirement -- that the system of axioms is \index{sistem +aksiomov!minimalen} \pojem{minimalen}, which means that none of the axioms can be derived from the others. We mention that the last + requirement is not as important as the first two. + +We must also add that we do not build Euclidean geometry +independently from algebra and logic. We will use concepts such as +set, function, relation with properties that apply to them. We will +also use so-called rules of inference, such as +the method of contradiction. For mathematical disciplines that we use in this way to build geometry, we say that they are +\pojem{predpostavljene teorije}. + +%________________________________________________________________________________ +\poglavje{A Brief Historical Overview of the Development of Geometry} +\label{odd1ZGOD} + +People started dealing with geometry in early history. +At first, it was only the observation of characteristic shapes, such as +a circle or a square. Based on the drawings found on the walls of old caves, we conclude +that people in prehistory were interested in the symmetry of shapes. + +In further development, man was discovering various properties +of geometric shapes. This was due to practical needs, e.g. +measuring the area of ​​land - which is also how the word +"geometry" came about. In this period, geometry developed as +an inductive science. This means that geometric propositions were coming from experiences - by means of measurements and checking on +individual examples. In this sense, geometry was developed by +all ancient civilizations: Chinese, Indian and especially +Egyptian. + +In Egypt, geometry developed mainly as a study of +measurements. Because the Nile river often flooded, +the land had to be measured very often. In addition, +the knowledge of geometry was used in construction. They knew +e.g. the formula for calculating the volume of a pyramid and a truncated +pyramid, although they came to it empirically. So geometry was for +the Egyptians primarily a pragmatic discipline. +The oldest records date back to approximately 1700 BC. + +Geometry of three-dimensional space was not +discussed as much as in Egypt. + +There is not as much data on Chinese geometry as on Egyptian, +although we know that it was also very developed. In +the oldest preserved records we find a description of the calculation of +the volume of a prism, pyramid, cylinder, cone, truncated pyramid and +truncated cone. + +Indian geometry is much younger than the previous three. It dates +back to approximately the 5th century BC. In it we already see the first +attempts at proof. Later it developed parallel to +Greek geometry. + +The turning point in the development of geometry took place in Ancient Greece. This is when the deductive method was first used in geometry. The first geometric proofs are associated with Tales\footnote{The Ancient Greek philosopher and mathematician \textit{Tales} \index{Tales} from Miletus (640--546 BC).}. We connect his name with the well-known statement about the proportionality of segments in parallel lines. He also proved the statement that the angles above the diameter of a circle are right, even though this claim was known without proof to the Babylonians 1000 years earlier. This way of developing geometry was continued by other Ancient Greek philosophers, of which Pythagoras\footnote{The Ancient Greek philosopher and mathematician \textit{Pythagoras} \index{Pythagoras} from the island of Samos (ca. 580--490 BC).} was one of the most important. Of course, his \index{statement!Pythagorean}\textit{Pythagorean statement} is famous. However, the Egyptians knew this statement as a fact 3000 years BC (maybe the statement was known even before that), but Pythagoras gave the first known proof. Archimedes\footnote{The Ancient Greek philosopher and mathematician \textit{Archimedes} \index{Archimedes} from Syracuse (287--212 BC).} was the first to present a theoretical calculation of the number $\pi$, by considering the inscribed and circumscribed polygons with $96$ sides. The statements about the congruence of triangles were also proven. With the rapid progress of geometry, reflected in the large number of proven statements, the need for systematization and, with that, the need for axioms, became apparent. The need for axioms was first described by Plato\footnote{The Ancient Greek philosopher and mathematician \textit{Plato} \index{Plato} (427--347 BC).} and Aristotle\footnote{The Ancient Greek philosopher and mathematician \textit{Aristotle} \index{Aristotle} from Athens (384--322 BC).}. Plato is also known in mathematics for his research of regular polygons: the tetrahedron, the cube, the octahedron, the dodecahedron and the icosahedron, which is why we also call them Platonic solids after him. + +One of the first attempts at an axiomatic approach to geometry - and the only one that has been preserved from that time - was made by the most famous mathematician of that time, Euclid, Plato's student, in his well-known work \textit{The Elements}, which consists of 13 books. In it, he systematized all the existing knowledge of geometry. He divided the initial statements into axioms and so-called postulates, of which the latter are purely geometric content (today we also call them axioms). \textit{The Elements} became one of the most important and influential books in the history of mathematics. The geometry, which he developed in this way, with minor unimportant changes, is the one that is taught in schools today. Proofs, such as the one about the central and peripheral angle, have been preserved in practically unchanged form. We give the postulates as Euclid gave them (Figure \ref{sl.sl.1.3.1.9.pic}): +\color{viol3} +\begin{postulat} + We can draw a straight line from any point to any point. + \end{postulat} + \begin{postulat} +We can produce a finite straight line continuously in a straight line. + \end{postulat} + \begin{postulat} +We can describe a circle with any center and distance. + \end{postulat} + \begin{postulat} +All right angles are equal to one another. + \end{postulat} + \begin{postulat} +If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the straight lines, if produced indefinitely, will meet on that side on which the angles are less that two right angles. + \end{postulat} +\normalcolor + +\begin{figure}[!htb] +\centering +\input{sl.1.3.1.9.pic} +\caption{} \label{sl.sl.1.3.1.9.pic} +\end{figure} + + %slikaNova1-3-3 +%\includegraphics[width=100mm]{slikaNova1-3-3.pdf} + + +However, the system of axioms that Euclid gave was not perfect. In some proofs, he took certain parts as obvious and did not prove them. Of course, we should not be too critical, because this was a revolutionary work for those times. \textit{The Elements} were an example and inspiration for mathematicians for centuries and laid the foundations for the further development of geometry to this day. The last axiom, i.e. the \index{aksiom!fifth Euclid's axiom} \pojem{fifth Euclid's axiom}, was particularly important for the further development of geometry. The problem of its independence from the other axioms was open for the next 2000 years! + +The last in a series of great ancient Greek mathematicians were +Apolonius\footnote{The Greek mathematician \textit{Apolonius} +\index{Apolonius} from Perga (262--190 BC).}, +Menelaus\footnote{The Greek mathematician \textit{Menelaus} +\index{Menelaus} from Alexandria + (ca. 70--130).} and Pappus\footnote{The Greek mathematician \textit{Pappus} + \index{Pappus} from + Alexandria (ca. 290--350).}. In his book \textit{On the Cutting of a Cone}, Apolonius defined the ellipse, parabola and hyperbola as the intersections of a plane and a circular (infinite) cone. This allowed him to consider their properties simultaneously, which was quite a modern approach for the time. Menelaus and Pappus proved certain theorems which didn't become relevant until the 19th century with the development of projective geometry. So the ideas of these three mathematicians were quite modern and in a way we can say that they were on the brink of discovering the first non-Euclidean geometry. + +After the final fall of the Old Greece under the Roman Empire, +the period of the glorious ancient Greek geometry came to an end. Although the Old Romans took over a large part of the ancient Greek culture and built roads, +aqueducts and so on, it is interesting that they never really showed much interest in +the ancient theoretical mathematics. So their contribution to the development +of geometry is very modest. + +An important role in the further development of geometry was taken over by the Arabs. +First of all, we should say that all the works of the Ancient Greeks +including Euclid's \textit{Elements} are known to us today because they +were translated and thus preserved by the Arabs. After the foundation of +Baghdad in 762, in the next 100 years they translated most of the works +of ancient Greek and Indian mathematics. They also made a synthesis +of ancient Greek mostly geometric and Indian mostly +algebraic approach. We mention that the word itself \pojem{algebra} +is of Arabic origin. In addition, the Arabs continued the development +of \pojem{trigonometry}, which was designed by the Ancient Greeks. A. R. al-Biruni\footnote{The Arab mathematician \textit{ A. R. al-Biruni} \index{al-Biruni, A. R.} (973--1048).} proved the now known \pojem{sine theorem}. + +The development of geometry in Europe began in the 12th century, when +Arabic and Jewish mathematicians brought their knowledge to +Spain and Sicily. (Euclid's \textit{Elements} were translated from +Arabic into Latin around 1200); but Europe did not experience its +true flowering until the 16th century. In the Middle Ages, +mathematics developed very slowly. In the Middle Ages, +Western European mathematicians only learned the Greek geometric +heritage from Arabic translations, but this process was not +quick. When this knowledge was accumulated and social and +political conditions changed, a new era began in the +development of geometry. The first new results were given by Italian +mathematicians of that time, who paid a lot of attention to +constructions using a ruler and a compass. + +As we mentioned earlier, Euclid's fifth axiom had a very big impact on +the further development of geometry. Because of its +formulation, which is not as simple as the previous axioms, and +also because of its importance, many mathematicians of that time +thought that it did not need to be considered as an axiom, but it +could be proven as a theorem with the other axioms. If we read +Euclid's other initial propositions, it is really true that the +fifth axiom is more complex. The problem of the independence of the +fifth axiom from the others occupied many +mathematicians in the following centuries. Until the second half of +the 19th century the problem was not solved. In many attempts to +prove the fifth axiom from the others, propositions were used, +whose proof was omitted. +It was later shown that these propositions could not even be +proven from the other axioms if the fifth axiom was omitted from the +list. +Just as they follow from the fifth axiom, these propositions +also follow from the fifth axiom (of course using the other +axioms). +We therefore call them \pojem{equivalents of the fifth Euclidean +axiom}. +We give some examples of these equivalents (Figure +\ref{sl.sl.1.3.1.9a.pic}): + +\color{blue} +\begin{ekv} +If $ ABCD $ is a quadrilateral with two right angles on the side $BC$ +and the sides $AB$ and $CD$ are congurent, then the two remain angles +of this quadrilateral are also right angles.\footnote{This equivalent was set by +Italian mathematician \index{Saccheri, G. G.} \textit{G. G. +Saccheri} (1667--1733).} +\end{ekv} +\begin{ekv} +A line perpendicular to one arm of an acute angle intersects +his other arm. +\end{ekv} +\begin{ekv} +Every triangle can be circumscribed. +\end{ekv} +\begin{ekv} +If three angles of a quadrilateral are right angles, then the fourth angle is also a right angle.\footnote{\index{Lambert, J. H.}\textit{J. +H. Lambert} (1728--1777), French mathematician.} +\end{ekv} +\begin{ekv} +The sum of the interior angles in every triangle is $180^0$.\footnote{\index{Legendre, A. M.} \textit{A. M. Legendre} +(1752--1833), French mathematician.} +\end{ekv} +\begin{ekv} +For any given line $p$ and point $A$ not on $p$, in the plane containing both line $p$ and point $A$ there is just one line + through point $A$ that do not intersect line $p$\footnote{\index{Playfair, J.} \textit{J. Playfair} +(1748--1819), Scottish mathematician.}. + \end{ekv} +\normalcolor + + +\begin{figure}[!htb] +\centering +\input{sl.1.3.1.9a.pic} +\caption{} \label{sl.sl.1.3.1.9a.pic} +\end{figure} + +So mathematicians were able to prove the fifth Euclidean axiom with +the help of each of these assertions, but eventually it turned out that +none of them could be proved without the fifth axiom. Therefore, +these assertions, as we have already mentioned, are equivalent to the fifth axiom. +Today, Playfair's equivalent is most commonly used, which was later +added to Euclid's axioms instead of the fifth axiom. + +But how did mathematicians find out that the fifth Euclidean axiom could not +be derived from the other axioms? Just the fact that they were not able +to prove it did not mean that it was not possible. The answer to this +question came at the end of the 19th century and, as we will see, +brought much more to the development of geometry than just the fact of +the unprovability of the fifth axiom. + +The next breakthrough in the development of geometry was the discovery +\index{geometrija!neevklidska}\pojem{non-Euclidean geometries} in the 19th century. For the beginning of this development +we count N.~I.~Lobačevskega\footnote{\index{Lobačevski, N. +I.}\textit{N. I. Lobačevski} (1792--1856), Russian mathematician.}. He also dealt with the problem of the independence of the fifth Euclidean axiom. +Based on its negation or the negation of Playfair's equivalent statement, Lobačevski postulated that through a point that does not lie on a straight line, there are at least two straight lines that do not intersect with that line and are coplanar. In an attempt to come to a contradiction (thus the fifth axiom would be proven), he built a whole +sequence of new statements. One of them, for example, is that the sum of the interior angles +of a triangle is always less than the extended angle. But none of these statements +were in contradiction with the other axioms, if of course we exclude +the fifth Euclidean axiom from the list. From this he got the idea that it is possible to build +a completely new geometry that is non-contradictory and based on all +Euclidean axioms except the fifth, which we replace with its +negation. Today we call this geometry +\index{geometrija!hiperbolična} \pojem{hyperbolic geometry} +or \pojem{Lobačevski geometry}. + +Independently of Lobačevski, J. +Bolyai\footnote{\index{Bolyai, J.} \textit{J. Bolyai} (1802--1860), +Hungarian mathematician.}. As it often happens, the ideas of Lobačevski during his lifetime +unfortunately were not accepted. The complete confirmation of these ideas +or the proof of the non-contradiction of this new geometry was at the end of the 19th century, that is, only after the death of Lobačevski, presented by A. +Poincar\'{e}\footnote{\index{Poincar\'{e}, J. H.} \textit{J. H. +Poincar\'{e}} (1854--1912), French mathematician.}. +Poincar\'{e} built a model on +the basis of which he showed that a possible contradiction of Lobačevski geometry +would at the same time be a contradiction of Euclidean geometry. +Later, the discovery of other non-Euclidean geometries followed. + +Although at the end of the 19th and the beginning of the 20th century the system of Euclid's geometry axioms was already almost completely built, the first correct and complete system was given by D. Hilbert\footnote{\index{Hilbert, D.} \textit{D. Hilbert} (1862--1943), German mathematician.} in his famous book \textit{The Foundations of Geometry}, published in 1899. We use a very similar system of axioms in an almost unchanged form even today. + +Parallel to the research of the fifth Euclid's axiom and the development of non-Euclidean geometries, other important methods have also emerged in the study of geometry. As early as around 1637, R. Descartes\footnote{\index{Descartes, R.} \textit{R. Descartes} (1596--1650), French mathematician.} in his book \textit{Geometry} showed that every point in a plane can be described by a suitable pair of two real numbers and similarly in space as a triple of three real numbers. He connected this with the concept of a coordinate representation of the dependence of one quantity (function) on another (variable), which was known earlier. Today, we call such a way of determining points in space after him \pojem{Cartesian coordinate system}. Lines and planes can then be described as sets of solutions of appropriate linear equations, where the unknowns are coordinates of points. + +Thus, under the influence of the ideas of F. Vi\'{e}te\footnote{\index{Vi\'{e}te, F.} \textit{F. Vi\'{e}te} (1540--1603), French mathematician.}, Descartes and P. Fermat\footnote{\index{Fermat, P.} \textit{P. Fermat} (1601--1665), French mathematician.}, the two very important mathematical disciplines began to develop - first \index{geometry!analytical} \pojem{analytical geometry}, then \index{linear algebra} \pojem{linear algebra}, which represent the connection between algebra and geometry. The further development of these two disciplines allowed the development of \index{geometry!multidimensional} \pojem{multidimensional geometry}, in which spaces of dimensions greater than three can be considered, because in algebra there are no such limitations as we have in the geometric perception of space. Thus, we can define the so-called \pojem{polytope} - objects of multidimensional space, which are the analogy of two-dimensional polygons and three-dimensional polyhedrons. + +Later, the discovery of other non-Euclidean geometries followed. +In the 19th century, the so-called \index{geometrija!projektivna}\pojem{projective geometry} developed, but +its development was not axiomatic like in hyperbolic +geometry; the appropriate system of axioms was set only +later. In this geometry, there are no lines in the plane that do not intersect. + +One of the first motives for the beginning of the + development + of projective geometry originates from painting or from the desire to transfer + the feeling of three-dimensional space into a plane. Already in very + early painting, we encounter a very important property--that + in the picture, parallel lines are represented as lines that intersect. + + In the 15th century, Italian artists were very interested in + the geometry of space. The theory of perspective was first considered by F. + Brunellechi\footnote{\index{Brunellechi, F.} \textit{F. + Brunellechi} (1377--1446), Italian architect.} in 1425. + His work was continued by L. B. Alberti\footnote{\index{Alberti, L. B.} + \textit{L. B. + Alberti} (1404--1472), Italian mathematician and painter.} and A. + D\"{u}rer\footnote{\index{D\"{u}rer, A.} \textit{A. + D\"{u}rer} (1471--1528), German painter.}. Alberti's book from + 1435 represents the first presentation of central projection. + + For the beginning of the development of projective geometry as a mathematical + discipline, we consider the period when + J. Kepler\footnote{\index{Kepler, J.} \textit{J. Kepler} (1571--1630), + German astronomer.} and G. Desargues\footnote{\index{Desargues, G.} + \textit{G. Desargues} + (1591--1661), French architect.} + independently introduced the concept of points at infinity. + Kepler showed that a parabola has two foci, one of which + is a point at infinity. Desargues + was writing in 1639: ‘‘Two parallel lines have a common endpoint at + an infinite distance.’’ In 1636, he wrote a book on perspective, + and in 1639, he wrote about cones. The famous \textit{Desargues' Theorem} + was published in 1648. + +With the further development of projective geometry we connect French mathematics. + The genius B. Pascal\footnote{\index{Pascal, B.} \textit{B. Pascal} (1623--1662), + French philosopher and mathematician.} was + already as a sixteen year old when he proved an important theorem about + cones, which we today call \textit{Pascal's theorem}. + This theorem, which is one of the basic + theorems of projective geometry, was published in 1640. + G. Monge\footnote{\index{Monge, G.} \textit{G. Monge} + (1746--1818), French mathematician.} + was among the first mathematicians who we can consider a specialist; he + is in fact the first true geometer. He developed \pojem{descriptive geometry} as + a special discipline. In his research in descriptive geometry + we find many ideas of projective geometry. + The most original Monge's student was + J. V. Poncelet\footnote{\index{Poncelet, J. V.} + \textit{J. V. Poncelet} (1788--1867), French mathematician.}. + Although Pappus\footnote{\index{Pappus} \textit{Pappus of Alexandria} (3rd century), Greek mathematician.} + discovered the first projective theorems, Poncelet + with a completely projective way of reasoning proved them only in the 19th century. + In 1822 + Poncelet published his famous "Treatise on the projective properties of figures", + in which all the important concepts characteristic for + projective geometry appear: harmonic quadruple, perspectivity, projectivity, + involution, etc. Poncelet introduced a line at infinity for all + planes that are parallel to a given plane. Poncelet and J. D. + Gergonne\footnote{\index{Gergonne, J. D.} \textit{J. D. Gergonne} (1771--1859), French mathematician.} independently + from each other studied duality in projective geometry, + C.~J.~Brianchon\footnote{\index{Brianchon, C. J.} \textit{C. J. Brianchon} + (1783--1864), French mathematician.} however + proved a theorem which is dual to Pascal's theorem. + M.~Chasles\footnote{\index{Chasles, M.} \textit{M. Chasles} (1793--1880), French mathematician.} was the last of + the great + school of + French projective geometers of that time. + +A typical representative of so-called pure geometry + (today we would say synthetic geometry) was + J. Steiner\footnote{\index{Steiner, J.} \textit{J. Steiner} + (1796--1863), + Swiss mathematician.}. + Steiner developed projective geometry very systematically, + from perspective to projectivity and then to conics. + + In the middle of the 19th century, German mathematicians took over the lead in the development of projective geometry. They advocated a synthetic approach + h geometry. All mathematicians until then had designed projective geometry + based on Euclidean metric geometry -- by adding + points at infinity. But C.~G.~C.~Staudt + \footnote{\index{Staudt, K. G. C.} \textit{C. G. C. Staudt} (1798--1867), + German mathematician.} + was the first to try to make it independent and + design it only on incidence axioms, without the help of metric. + This led to the abolition of the difference between points at infinity and ordinary + points + or the transition from extended Euclidean to projective space. + + F. Klein\footnote{\index{Klein, F. C.} \textit{F. C. Klein} (1849--1925), German mathematician.} set the projective geometry on algebraic foundations in 1871 + with the help of + so-called \pojem{homogeneous coordinates}, which were discovered independently of each other in 1827, + by K. W. Feuerbach\footnote{\index{Feuerbach, K. W.} \textit{K. W. Feuerbach} + (1800--1834), German mathematician.} + and A. F. M\"{o}bius\footnote{\index{M\"{o}bius, A. F.} \textit{A. F. M\"{o}bius} + (1790--1868), German mathematician.}. + A. Cayley\footnote{\index{Cayley, A.} \textit{A. Cayley} (1821--1895), English mathematician.} and + Klein + found the use of projective geometry in other non-Euclidean geometries. + They discovered the model of hyperbolic geometry and models of other geometries + in projective. + + The first to completely axiomatically design projective geometry, were + G. Fano\footnote{\index{Fano, G.} \textit{G. Fano} (1871--1952), Italian mathematician.} + in 1892 and M. Pieri\footnote{\index{Pieri, M.} \textit{M. Pieri} (1860--1913), Italian mathematician.} + in 1899. + +Because of its relative simplicity, the development of classical +(synthetic) projective geometry was almost completed by the end of +the 19th century. Its development today continues within the framework +of other theories -- especially in algebra and algebraic geometry as +$n$-dimensional projective geometry. + +G. F. B. Riemann\footnote{\index{Riemann, G. F. B.} \textit{G. F. +B. Riemann} (1828--1866), German mathematician.} defined the space of +arbitrary dimension in his book \textit{On the hypotheses which lie at +the foundations of geometry}, which is not always of constant +curvature. After him, we today call it the \pojem{Riemann metric +space}\index{Riemann spaces}. Euclidean geometry is then obtained as a +special case: if the curvature is constant and equal to 0; hyperbolic +geometry is obtained if we choose that the curvature is constant and +negative. If the curvature is constant and positive, we obtain the so- +called \index{geometry!elliptic} \pojem{elliptic geometry}. The latter +geometry is actually projective geometry, if we add a metric to it. +This research was also the beginning of the development of a new +discipline in mathematics, namely the \index{geometry!differential} +\pojem{differential geometry}. + +If we think about non-Euclidean geometries, it may seem strange to us +that in mathematics there can be considered more different theories, +such as Euclidean geometry and hyperbolic geometry, which are in +contradiction with each other. For modern mathematics, it is most +important that both geometries are determined by systems of axioms, +which are (each for itself) consistent and complete. To the question +of which of these two geometries is valid, it is pointless to seek an +answer within mathematics. This is because it depends on which +axioms we have chosen. Such a question would be the same as the +question of which axioms are valid. But the axioms are assumed by +definition without proof. Of course, we can ask the question of what +the geometry of space is in the physical sense and how we can describe +it with axioms." + +To answer this question, a physical interpretation of the basic geometric concepts is needed. For example, it is most natural to interpret a line as a ray of light. In this sense, physical space is not Euclidean. It is not even determined by hyperbolic geometry. With the advent of Einstein's\footnote{\index{Einstein, A.} \textit{A. Einstein} (1879--1955), famous German physicist.} theory of relativity at the beginning of the 20th century, it turned out that in space of cosmic dimensions it is more convenient to use non-Euclidean geometry with variable curvature (Riemann metric spaces!). We can say that the geometry of the universe is locally different in each point, depending on the proximity and size of some mass. Einstein's theory also tells us that space and time are interrelated and that time (which is of course surprising) does not flow evenly in each point of the universe. In connection with the mentioned connection of space and time, the so-called \pojem{Minkowsky space of four dimensions}\footnote{This space was discovered by \index{Minkowsky, H.} \textit{H. Minkowsky} (1864--1909), German mathematician.} is important. + +Since the 20s of the 20th century and the development of the theory of the primordial soup, we know that the universe is not static and that it is expanding. Its destiny depends on what geometry globally describes it best. But we still do not know for sure what the shape of the universe is or what its destiny is. We do not even know if the universe is finite or infinite. As S. Hawking writes in his famous popular book from 1988 A Brief History of Time (\cite{KratkaZgodCasa}), the universe may even be finite and unlimited. The latter seems paradoxical, although we can imagine it if, instead of three-dimensional, we imagine a "two-dimensional universe". So the beings of this two-dimensional universe could once find out that their universe is actually not a plane but a sphere that is finite but unlimited. The sphere is part of three-dimensional space. So we can theoretically imagine the universe as a three-dimensional sphere in four-dimensional space. The three-dimensional sphere is one of the 3-manifolds. In this case, the geometry of the universe would be globally elliptical. + +Although the question of the shape and destiny of the universe is a question of theoretical physics and cosmology, we see how modern geometry (non-Euclidean geometry, geometry of multi-dimensional spaces, etc.) is closely related to this problem (\cite{Oblika}). It is important to understand that geometry (and every other mathematical discipline) develops and is treated as an abstract discipline in which physical interpretations are only inspiration - in this sense, only axioms and initial concepts remain on which we then build mathematical theory. + +At the end, we mention another one of the most important geometers of the 20th century H. S. M. Coxeter\footnote{\index{Coxeter, H. S. M.} +\textit{H. S. M. Coxeter} (1907--2003), Canadian mathematician. One +of the greatest geometers of the 20th century.}. Coxeter further researched polytopes in arbitrary dimensions, primarily regular polytopes. In addition, he +was mostly occupied with groups of isometries in hyperbolic geometry and +with multidimensional hyperbolic geometry. + +It needs to be added that with everything +that we have +said, the development of geometry is far from over. Quite the opposite -- contrary to the usual +perception -- geometry and mathematics in general are now developing +even faster than ever before. Even today, there are in mathematics (in geometry in particular from the non-Euclidean +geometries) many +problems that are still unresolved. + + + + + + + +% DEL 2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% AKSIOMI RAVNINSKE EVKLIDSKE GEOMETRIJE +%________________________________________________________________________________ + + +\del{Axioms of Planar Euclidean Geometry} \label{pogAKS} + +In what follows, we will illustrate the axiomatic design of planar Euclidean +geometry. We will list the initial concepts and initial statements - +axioms, and then derive some new concepts and statements. +We mention that we have chosen the axioms of the plane, because in this book +we will only deal with the geometry of the Euclidean plane. + +Let $\mathcal{S}$ be a non-empty set. Its elements are called +\index{point} \pojem{points} and we denote them with $A, B, C, \ldots$ +Certain subsets of the set $\mathcal{S}$ are called \index{line} +\pojem{lines} and we denote them with $a, b, c, \ldots$ The set +$\mathcal{S}$ (the set of all points) is also called the +\index{plane} \pojem{plane}. In addition to these basic concepts, +there are also two relations on the set $\mathcal{S}$. The first is +the \index{relation!$\mathcal{B}$} \pojem{relation $\mathcal{B}$} +and it applies to three points. The fact that points $A$, $B$ and $C$ +are in this relation, we will denote with $\mathcal{B}(A,B,C)$ and +read: Point $B$ is between points $A$ and $C$. The second is the +\index{relation!compatibility of point pairs} \pojem{relation of +compatibility of point pairs}; the fact that pairs of points $A, B$ +and $C, D$ are in this relation, we will denote with $(A,B) \cong +(C,D)$ and read: The pair of points $(A,B)$ is compatible with the +pair of points $(C,D)$. + + With the help of the aforementioned basic concepts, we can +also define the following derived concepts: + +If point $A$ belongs to line $p$ ($A\in p$), or line $p$ contains +point $A$ ($p\ni A$), we will say that + point $A$\index{relation!lies on a line} \pojem{lies on} line $p$, or that line $p$ \index{relation!goes through a point}\pojem{goes + through} point $A$. + For three or +more points we say that they are \index{collinear points}\pojem{collinear}, +if they lie on the same line, otherwise they are +\index{non-collinear points}\pojem{non-collinear}. Two +different lines \pojem{intersect}, if their intersection (the +intersection of two subsets) is not an empty set. We call their +intersection the \index{intersection of two lines} \pojem{intersection} of two lines. Any non-empty subset $\Phi$ of the set $\mathcal{S}$ ($\Phi\subset\mathcal{S}$) is called a \index{figure} \pojem{figure}. We say that figures $\Phi_1$ and $\Phi_2$ \index{figures!coincide}\pojem{coincide} (or they are \index{figures!identical}\pojem{identical}), if $\Phi_1=\Phi_2$. + + Now we will +also list some basic theorems - axioms. By their nature, they are +divided into five groups: + +\begin{enumerate} + \item incidence axioms (three axioms), + \item ordering axioms (four axioms), + \item congruence axioms (four axioms), + \item continuity axiom (one axiom), + \item Playfair's axiom (one axiom). +\end{enumerate} + + + +%________________________________________________________________________________ + \poglavje{Incidence Axioms} + \label{odd2AKSINC} + + Because lines as basic notions represent certain sets of points, + we can +consider appropriate relations between elements +and sets +of points and lines: $\in$ and $\ni$ - relations we also call +\index{relacija!incidencije}\pojem{relations of incidence}. These axioms describe just the basic properties of these relations (Figure +\ref{sl.aks.2.1.1.pic}): + +\vspace*{3mm} + + \baksiom \label{AksI1} For every pair of distinct points $A$ and $B$ + there is exactly one line $p$ such that $A$ and $B$ lie on $p$. + \eaksiom + + \baksiom \label{AksI2} + For every line there exist at least two distinct points such that both lie on. + \eaksiom + + \baksiom \label{AksI3} There exist three points that do not all lie on any one line. + \eaksiom + +\vspace*{3mm} + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.1.1.pic} +\caption{} \label{sl.aks.2.1.1.pic} +\end{figure} + + + + From the first two axioms \ref{AksI1} and \ref{AksI2} it follows that each line is determined by + its two distinct points. Therefore +the line $p$, which is determined by the points $A$ and $B$, we also call the line $AB$. + + From the first axiom \ref{AksI1} it follows that the intersection of two + lines that intersect is one point. If, for example, two lines + had one more common point, according to this axiom they would coincide (they would be identical), + but in the definition of lines that intersect, we required + that +they are different. + The fact that the lines $p$ and $q$ intersect in the point $A$, we will + write $p\cap q=\{A\}$ or shorter $p\cap q=A$ (Figure \ref{sl.aks.2.1.2.pic}). + + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.1.2.pic} +\caption{} \label{sl.aks.2.1.2.pic} +\end{figure} + + + +The third axiom \ref{AksI3} can also be expressed as follows: There exist +at least three points that are not collinear. + +So we deduced the first consequences of the incidence axioms; + for simplicity, we did not express them in the form of propositions. These are almost all the consequences that arise from the first group of axioms. + Because of this, geometry, which is based solely on the axioms of incidence, + is too simple. In it, we could only prove the existence of three points and three + lines. So we need new axioms. + + + +%________________________________________________________________________________ + \poglavje{Ordering Axioms} + \label{odd2AKSURJ} + +The axioms in this group describe the basic characteristics of the relation +$\mathcal{B}$, which we listed as the basic concept. +\vspace*{3mm} + + + \baksiom \label{AksII1} If $\mathcal{B} (A, B, C)$, then $A$, $B$ + and $C$ are three distinct collinear points, and also + $\mathcal{B} (C, B, A)$ (Figure \ref{sl.aks.2.2.1.pic}). + \eaksiom + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.1.pic} +\caption{} \label{sl.aks.2.2.1.pic} +\end{figure} + + + \baksiom \label{AksII2} If $A$, $B$ and $C$ are three distinct + collinear points, + exactly one of the relations holds: $\mathcal{B}(A,B,C)$, + $\mathcal{B}(A,C,B)$, $\mathcal{B}(C,A,B)$ + (Figure \ref{sl.aks.2.2.2.pic}). + \eaksiom + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.2.pic} +\caption{} \label{sl.aks.2.2.2.pic} +\end{figure} + + + + \baksiom \label{AksII3} Given a pair of distinct points $A$ and $B$ there is a point $C$ on line $AB$, so that + is $\mathcal{B}(A,B,C)$ + (Figure \ref{sl.aks.2.2.3.pic}). + \eaksiom + +\vspace*{-1mm} + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.3.pic} +\caption{} \label{sl.aks.2.2.3.pic} +\end{figure} + +\baksiom \label{AksPascheva}\index{aksiom!Paschev} + (Pasch's\footnote{\index{Pasch, M.} + \textit{M. Pasch} + (1843--1930), German mathematician, who introduced the concept of ordering points + in his 'Lectures on Modern Geometry' from 1882. These + axioms were later supplemented by Italian mathematician \index{Peano, G.} \textit{G. Peano} + (1858--1932), in 'Principles of Geometry', then by German mathematician + \index{Hilbert, D.}\textit{D. Hilbert} (1862--1943) in his famous book + 'Foundations of Geometry' from + 1899.} axiom) + Let $A$, $B$ and $C$ be three noncollinear points and $l$ be a line that does not contain point $A$. + If there is a point $P$ on $l$ that is $\mathcal{B}(B,P,C)$ then either $l$ contains a point $Q$ that is $\mathcal{B}(A,Q,C)$ or $l$ contains a point $R$ that is $\mathcal{B}(A,R,B)$ (Figure \ref{sl.aks.2.2.4.pic}). + \eaksiom + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.4.pic} +\caption{} \label{sl.aks.2.2.4.pic} +\end{figure} + + +In the previous axiom, we did not particularly emphasize that the line $l$ lies in the plane $ABC$, because we are building a plane Euclidean geometry, where all points lie in the same plane. + + At this point we will not derive all the consequences of the ordering axioms. +The formal derivation of all the facts is not so simple and would take up a lot of space. Most of the proofs can be found in \cite{Lucic}. + +We prove the first consequence of the ordering axioms. + + + \bizrek \label{izrekAksUrACB} + Given a pair of distinct points $A$ and $B$ there is a point $C$, so that + is $\mathcal{B}(A,C,B)$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.5.pic} +\caption{} \label{sl.aks.2.2.5.pic} +\end{figure} + +\textbf{\textit{Proof.}} By Axiom \ref{AksI1} there exists exactly one line that goes through points $A$ and $B$ - we'll mark it with $AB$. +By Axiom \ref{AksI3} there are at least three non-linear points. +Therefore, there is at least one point outside of line $AB$ - we'll mark it with $D$ + (Figure \ref{sl.aks.2.2.5.pic}). Next, by +Axiom \ref{AksII3} there is such a point $E$, that $\mathcal{B}(B,D,E)$ is true, and then such a point $F$, that $\mathcal{B}(A,E,F)$ is true. $A$, $B$ and $E$ are non-linear points, +because otherwise point $D$ would lie on line $AB$ (Axiom \ref{AksI1}). +Line $FD$ does not go through point $A$, because by Axiom \ref{AksI1} points $F$, $D$, $A$ and $E$ +would be linear, and so would point $B$. But that is not possible, because it would imply that point $D$ lies on line $AB$. +We'll now use Pasch's Axiom \ref{AksPascheva} on +points $A$, $B$ and $E$ and line $FD$. Line $FD$ intersects line $EB$ in such a point $D$, +that $\mathcal{B}(B,D,E)$ is true, and therefore it intersects either line $AE$ in such a point $F$, that $\mathcal{B}(A,F,E)$ is true, or +line $AB$ in such a point $C$, that $\mathcal{B}(A,C,B)$ is true. But since $\mathcal{B}(A,E,F)$ is already true, by Axiom \ref{AksII2} $\mathcal{B}(A,F,E)$ cannot be true as well. Therefore line $FD$ intersects +line $AB$ in such a point $C$, for which $\mathcal{B}(A,C,B)$ is true. +\kdokaz + +Relation $\mathcal{B}$ and the order axioms related to it, allow us to define new concepts. + + + Let $A$ and $B$ be any two different points. + \index{distance!open}\pojem{Open distance} $AB$ with notation $(AB)$ is the set of all points + $X$, for + which $\mathcal{B}(A,X,C)$ is true (Figure \ref{sl.aks.2.2.6.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.6.pic} +\caption{} \label{sl.aks.2.2.6.pic} +\end{figure} + +If we add points $A$ and $B$ to an open line segment $AB$, we get a \pojem{line segment} (or a \pojem{closed line segment}) $AB$, which we also denote with $[AB]$. Points $A$ and $B$ are its \pojem{endpoints}, and all other points on it are \pojem{interior points} of the line segment (Figure \ref{sl.aks.2.2.6.pic}). More formally: a line segment (or a closed line segment) is the union of an open line segment and the set $\{A,B\}$ or $[AB]=(AB)\cup \{A,B\}$. + +Similarly, we define a \pojem{half-open line segment}: $(AB]=(AB)\cup \{B\}$, or $[AB)=(AB)\cup \{A\}$ (Figure \ref{sl.aks.2.2.6a.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.6a.pic} +\caption{} \label{sl.aks.2.2.6a.pic} +\end{figure} + + From Axiom \ref{AksII1} it follows immediately that line segments $AB$ and $BA$ are the same. From the same axiom it also follows that line segment $AB$ is a subset of the line $AB$. Therefore, we say that line segment $AB$ \pojem{lies on the line} $AB$, and we call the line $AB$ the \pojem{line supporting the line segment} $AB$ (Figure \ref{sl.aks.2.2.6b.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.6b.pic} +\caption{} \label{sl.aks.2.2.6b.pic} +\end{figure} + +By Theorem \ref{izrekAksUrACB}, line segment $AB$ has, besides its endpoints $A$ and $B$, at least one more point $C_1$. In this way, we can get an infinite sequence of points $C_1$, $C_2$, ..., for which $\mathcal{B}(A, C_n, C_{n-1})$ is true ($n\in \{2,3,\cdots\}$) (Figure \ref{sl.aks.2.2.6c.pic}). At this point, we will not formally prove the fact that all points in the sequence are different and that all of them lie on line segment $AB$. From this statement it follows that every line segment (and consequently every line) has infinitely many points. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.6c.pic} +\caption{} \label{sl.aks.2.2.6c.pic} +\end{figure} + +Let's define the relation $\mathcal{B}$, which relates to more than three collinear points. We say that $\mathcal{B}(A_1,A_2,\ldots,A_n)$ ($n\in\{4,5,\ldots\}$), if for every $k\in\{1,2,\ldots,n-2\}$ it holds that $\mathcal{B}(A_k,A_{k+1},A_{k+2})$ (Figure \ref{sl.aks.2.2.6d.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.6d.pic} +\caption{} \label{sl.aks.2.2.6d.pic} +\end{figure} + + Let $S$ be a point that lies on the line $p$. On the set $p\setminus \{S\}$ (all points of the line $p$ without the point $S$) we define two relations. +We say that the points $A$ and $B$ ($A,B\in p\setminus \{S\}$) \index{relacija!na različnih straneh točke} \pojem{on different sides of the point} $S$ (which we denote by $A,B\div S$), if $B(A,S,B)$, otherwise the points $A$ and $B$ ($A,B\in p\setminus \{S\}$) \index{relacija!na isti strani točke} \pojem{on the same side of the point} $S$ (which we denote by $A,B\ddot{-} S$). So for points $A,B\in p\setminus \{S\}$ it holds that $A,B\ddot{-} S$, if it is not $A,B\div S$ (Figure \ref{sl.aks.2.2.7.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.7.pic} +\caption{} \label{sl.aks.2.2.7.pic} +\end{figure} + + +Let $A$ and $B$ be two different points. The set of all such points $X$, for which $B,X\ddot{-} A$ including the point $A$, we call \index{poltrak}\pojem{poltrak} $AB$ with \pojem{starting point} or \pojem{origin} $A$. The line $AB$ is +\index{nosilka!poltraka} \pojem{the carrier of the poltrak} $AB$ (Figure \ref{sl.aks.2.2.8.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.8.pic} +\caption{} \label{sl.aks.2.2.8.pic} +\end{figure} + +From the definition itself it follows that the poltrak is a subset of its carrier or that it lies on its carrier. From the relation $B,X\ddot{-} A$ it follows that $B$, $X$ and $A$ are collinear points, so the point $X$ lies on the line $AB$. + +We will not prove other important properties of lines and line segments that we will use later. Let's mention some of these properties. + +If $C$ is the an interior point of the line segment $AB$, then that line segment can be expressed as the union +the line segments $AC$ and $CB$. + + \bizrek \label{izrekAksIIDaljica} + If $C$ is an interior point of the line segment $AB$, then that line segment can be expressed as the union + the line segments $AC$ and $CB$ (Figure \ref{sl.aks.2.2.9.pic}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.9.pic} +\caption{} \label{sl.aks.2.2.9.pic} +\end{figure} + + + \bizrek \label{izrekAksIIPoltrak} + Each point lying on the line determines exactly + two rays on it. The union of these rays is equal to that line (Figure \ref{sl.aks.2.2.9.pic}). + \eizrek + +The proof of the previous statement is based on the fact that the relation $\ddot{-} A$ is equivalent to the relation that has two classes. Each of the classes is a suitable open ray. + +The rays from the previous statement, which are determined by the same initial point on the line, are called \index{poltrak!komplementarni}\pojem{complementary rays}. + +The concepts of line and ray allow us to define new concepts. + +Let $A_1$, $A_2$, ... $A_n$ be such points in the plane that no three of them are collinear. The union of the lines $A_1A_2$, $A_2A_3$,... $A_{n-1}A_n$ is called \index{lomljenka} \pojem{broken line} $A_1A_2\cdots A_n$ or \index{poligonska +črta}\pojem{polygonal line} $A_1A_2\cdots A_n$ (Figure \ref{sl.aks.2.2.10.pic}). The points $A_1$, $A_2$, ... $A_n$ are \index{oglišče!lomljenke} \pojem{vertices of the broken line}, +the lines $A_1A_2$, $A_2A_3$,... $A_{n-1}A_n$ are \index{stranica!lomljenke} \pojem{sides of the broken line}. The sides of the broken line with a common vertex are \index{sosednji stranici!lomljenke} \pojem{adjacent sides of the broken line}. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.10.pic} +\caption{} \label{sl.aks.2.2.10.pic} +\end{figure} + +If the sides of a broken line do not have common points, except for adjacent sides that have a common vertex, such a broken line is called a \index{lomljenka!enostavna} \pojem{simple broken line} (Figure \ref{sl.aks.2.2.10a.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.10a.pic} +\caption{} \label{sl.aks.2.2.10a.pic} +\end{figure} + +A broken line $A_1A_2\cdots A_nA_{n+1}$, for which $A_{n+1}=A_1$ and $A_n$, $A_1$ and $A_2$ are non-linear points, is called a \index{lomljenka!sklenjena} \pojem{closed broken line} $A_1A_2\cdots A_n$ (Figure \ref{sl.aks.2.2.10b.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.10b.pic} +\caption{} \label{sl.aks.2.2.10b.pic} +\end{figure} + +We will be particularly interested in \pojem{simple closed broken lines} (Figure \ref{sl.aks.2.2.10b.pic}). + +Let $p$ and $q$ be two semi-lines with a common starting point $O$ (Figure \ref{sl.aks.2.2.10c.pic}). The union of these two semi-lines is called a \index{kotna lomljenka} \pojem{angular broken line} $pq$ (or $pOq$). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.10c.pic} +\caption{} \label{sl.aks.2.2.10c.pic} +\end{figure} + + +For a figure $\Phi$ we say that it is \index{lik!konveksen}\pojem{convex}, if for any two of its points $A,B\in \Phi$ the distance $AB$ is a subset of this figure or if the following is true (Figure \ref{sl.aks.2.2.10d.pic}): + $$(\forall A)(\forall B)\hspace*{1mm} (A,B\in \Phi \Rightarrow [AB]\subseteq \Phi).$$ +For a figure that is not convex, we say that it is \index{lik!nekonveksen}\pojem{non-convex} (Figure \ref{sl.aks.2.2.10d.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.10d.pic} +\caption{} \label{sl.aks.2.2.10d.pic} +\end{figure} + +It follows directly from the definition that a straight line is a convex figure. As a result of the axioms of this group, it can be proved that a distance and a semi-line are convex figures. + +We say that a figure $\Phi$ is +\index{figure!connected}\pojem{connected}, if for every two of its points $A,B\in \Phi$ there exists a broken line $AT_1T_2\cdots T_nB$, which is a subset of this figure, or if the following is true (Figure \ref{sl.aks.2.2.10e.pic}): + $$(\forall A\in \Phi)(\forall B\in \Phi)(\exists T_1,T_2,\cdots , T_n)\hspace*{1mm} AT_1T_2\cdots T_nB\subseteq \Phi.$$ + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.10e.pic} +\caption{} \label{sl.aks.2.2.10e.pic} +\end{figure} + + A figure that is not connected, we call \index{figure!not connected}\pojem{not connected}. + + It is clear that every convex figure is also connected. For the broken line it is enough to take the distance $AB$. The converse is of course not true. There are figures that are connected, but not convex, which we will discover later. + + + +Now we will define two relations that are analogous to the relations $\ddot{-} S$ and $\div S$. + Let $p$ be a line that lies in the plane $\alpha$ (because we axiomatically build only the Euclidean geometry of the plane, in fact all the points that exist to us are in this plane). On the set $\alpha\setminus p$ (all points except the points of the line $p$) we define two relations. +We say that the points $A$ and $B$ ($A,B\in \alpha\setminus p$) are \index{relation!on different sides of the line} \pojem{on different sides of the line} $p$ (which we denote by $A,B\div p$), if the line $AB$ has a common point with the line $p$, otherwise the points $A$ and $B$ ($A,B\in \alpha\setminus p$) are \index{relation!on the same side of the line} \pojem{on the same side of the line} $p$ (which we denote by $A,B\ddot{-} p$). So for points $A,B\in \alpha\setminus p$ it is $A,B\ddot{-} p$, if it is not $A,B\div p$ (Figure \ref{sl.aks.2.2.11.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.11.pic} +\caption{} \label{sl.aks.2.2.11.pic} +\end{figure} + +Let $A$ be a point that does not lie on the line $p$. The set of all such points $X$, for which $A,X\ddot{-} p$, is called +\index{polravnina!odprta}\pojem{open half-line} $pA$. The union of the open half-lines $pA$ and the line $p$ is the \index{polravnina!zaprta}\pojem{closed half-line} or just +\index{polravnina}\pojem{half-line} $pA$. The line $p$ is the \index{rob!polravnine} \pojem{edge} of this half-line (Figure \ref{sl.aks.2.2.11a.pic}). If the points $B$ and $C$ lie on the edge $p$ of the half-line $pA$, we will call this half-line the half-line $BCA$. In addition, we will denote half-lines by Greek letters $\alpha$, $\beta$, $\gamma$,... + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.11a.pic} +\caption{} \label{sl.aks.2.2.11a.pic} +\end{figure} + +Similarly to the case of a segment, it can be shown (as a consequence of the axioms of this group), that each line $p$ in the plane determines two half-lines $\alpha$ and $\alpha'$, which have the line $p$ as an edge (Figure \ref{sl.aks.2.2.11a.pic}). We say that in this case $\alpha$ and $\alpha'$ are \index{polravnina!komplementarna}\pojem{complementary half-lines}. +It turns out that the union of two complementary half-lines is the whole plane. Similarly to the case of a segment, the proof of these statements is based on the fact that the relation $\ddot{-} p$ is equivalent to the relation with two classes. Each of the classes is the appropriate open half-line. + +Let $pq$ or $pOq$ be an angle. Define a new relation on the set of all points of the plane except the points that lie on the angle. We say that the points $A$ and $B$ are on the same side of the angle $pq$ (which we denote by $A,B\ddot{-} pq$), if there exists an angle $AT_1T_2\cdots T_nB$, which does not intersect the angle $pq$ or does not have any common points with it (Figure \ref{sl.aks.2.2.12.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.12.pic} +\caption{} \label{sl.aks.2.2.12.pic} +\end{figure} + +The relation $\ddot{-} pq$ is also an equivalence relation that has two classes. The union of each of these two classes with the angular bracket $pq$ is called + \index{kot}\pojem{the angle} $pq$, which is denoted by $\angle pq$, or $\angle pOq$. + The angular bracket therefore determines two angles. We will soon resolve the dilemma of which angle is meant by the notation $\angle pOq$. + The segments $p$ and $q$ are +\index{krak!kota}\pojem{the sides of the angle} and the point $O$ \index{vrh kota}\pojem{the vertex of the angle}. +If $P\in p$ and $Q\in q$ are points that lie on the sides of the angle $pOq$ and differ from its vertex $O$, we will also call the angle $POQ$ and denote it by $\angle POQ$ (Figure \ref{sl.aks.2.2.12a.pic}). If we know which angle it is, we will denote it by its vertex: $\angle O$. We will also denote angles by Greek letters $\alpha$, $\beta$, $\gamma$,... + +All points of the angle $pOq$, which do not lie on either of the sides $p$ and $q$, are called \index{notranje točke!kota} \pojem{internal points of the angle}, the set of all these points is called \index{notranjost!kota}\pojem{the interior of the angle}. It is clear that these are points of the appropriate class determined by the relation $\ddot{-} pq$. The points of the other class are \index{zunanje!točke kota}\pojem{external points of the angle}, the whole class is \index{zunanjost!kota}\pojem{the exterior of the angle}. The points that lie on the sides, or on the angular bracket $pOq$, which determines the angle $pOq$, are \index{robne točke!kota}\pojem{the boundary points of the angle}, the whole bracket is \index{rob!kota}\pojem{the boundary of the angle}. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.12a.pic} +\caption{} \label{sl.aks.2.2.12a.pic} +\end{figure} + +If the sides of the angle are complementary segments, such an angle is called +\index{kot!iztegnjeni}\pojem{an extended angle} (Figure \ref{sl.aks.2.2.12b.pic}). As a set of points, this angle is essentially the same as the half-line with the boundary that is the carrier of both sides. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.12b.pic} +\caption{} \label{sl.aks.2.2.12b.pic} +\end{figure} + +If the angular bisector $pOq$ does not determine the extended angle or is not equal to the line, it turns out that $pOq$ determines two angles that represent a convex and a concave shape - we call them the \index{kot!konveksen}\pojem{convex (protruding) angle} and the \index{kot!nekonveksen}\pojem{concave (indented) angle}. We will omit the formal proof of this fact here. Unless otherwise stated, we will always mean the convex angle under the label $\angle pOq$ (or $\angle pq$ or $\angle POQ$). In this sense, it is clear from the definition of the angle that (convex) angle $pOq$ and $qOp$ represent the same angle. + +The angle $pOq$ and $qOr$, which have a common leg $q$, which is also their intersection (as a set of points), are \index{kot!sosednji}\pojem{adjacent angles} (Figure \ref{sl.aks.2.2.12c.pic}). If in addition the segments $p$ and $r$ are complementary (or determine the extended angle), we say that $pOq$ and $qOr$ are \index{kota!sokota}\pojem{sokota} (Figure \ref{sl.aks.2.2.12c.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.12c.pic} +\caption{} \label{sl.aks.2.2.12c.pic} +\end{figure} + +The angle $pOq$ and $rOs$ are \index{kota!sovršna}\pojem{sovršna kota}, if $p$ and $r$ or $q$ and $s$ are a pair of complementary (supplementary) segments (Figure \ref{sl.aks.2.2.12d.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.12d.pic} +\caption{} \label{sl.aks.2.2.12d.pic} +\end{figure} + + Let $A_1A_2\cdots A_n$ ($n\in \{3,4,5,\cdots\}$) be a simple closed curve. + Similarly to the angular bisector, on the set of all points in the plane except for the points that lie on the curve $A_1A_2\cdots A_n$, we can define the following relation: we say that the points $B$ and $C$ are on the same side of the simple closed curve $A_1A_2\cdots A_n$ (which we denote by $B,C\ddot{-} A_1A_2\cdots A_n$), if there exists a curve $BT_1T_2\cdots T_nC$, which does not intersect the simple closed curve $A_1A_2\cdots A_n$ or does not have any common points with it (Figure \ref{sl.aks.2.2.13.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.13.pic} +\caption{} \label{sl.aks.2.2.13.pic} +\end{figure} + + It can also be proven in this case that it is an equivalence relation with two classes - for one class there is a line that lies entirely within it, but for the other class there is no such line. The union of the class that does not contain any line (intuitively - the one that is limited) and the simple closed broken line, $A_1A_2\cdots A_n$ is called +\index{polygon}\pojem{polygon} $A_1A_2\cdots A_n$ or +\index{$n$-gon}\pojem{$n$-gon} $A_1A_2\cdots A_n$ (Figure \ref{sl.aks.2.2.13a.pic}). All points of the aforementioned class that do not contain any line are called \index{internal points!of a polygon} \pojem{internal points of a polygon}, the entire class is \index{interior!of a polygon}\pojem{interior of a polygon}. The points of the other class are \index{external!points of a polygon}\pojem{external points of a polygon}, the entire class is \index{exterior!of a polygon}\pojem{exterior of a polygon}. The points that lie on the broken line $A_1A_2\cdots A_n$, which determines the polygon, are \index{vertices!of a polygon}\pojem{vertices of a polygon}, the entire broken line is \index{border!of a polygon}\pojem{border of a polygon}. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.13a.pic} +\caption{} \label{sl.aks.2.2.13a.pic} +\end{figure} + + +An equivalent statement holds true for the internal points of a polygon. We will state the statement without proof. + + \bizrek + A point $N$ is an interior point of a polygon if and only + if any ray from the point $N$, that does not contain the vertices of the + polygon, intersects an odd number of sides of the polygon (Figure \ref{sl.aks.2.2.13b.pic}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.13b.pic} +\caption{} \label{sl.aks.2.2.13b.pic} +\end{figure} + +The points $A_1$, $A_2$,..., $A_n$ (the vertices of the broken line) are the \index{oglišče!večkotnika}\pojem{vertices of the polygon}, the segments $A_1A_2$, $A_2A_3$, ... $A_{n-1}A_n$, $A_nA_1$ are the \index{stranica!večkotnika}\pojem{sides of the polygon}. The lines $A_1A_2$, $A_2A_3$, ... $A_{n-1}A_n$, $A_nA_1$ are the \index{nosilka!stranice}\pojem{supports of the sides} $A_1A_2$, $A_2A_3$, ... $A_{n-1}A_n$, $A_nA_1$. The sides that contain a common vertex are the \index{stranica!sosednja}\pojem{adjacent sides}, otherwise the sides are \index{stranica!nesosednja}\pojem{non-adjacent}. If the vertices are at the same time the endpoints of the same side, we say that the vertices are \index{oglišče!sosednje}\pojem{adjacent}, otherwise the vertices are \index{oglišče!nesosednje}\pojem{non-adjacent}. +It is clear from the definition that each vertex has exactly two adjacent vertices. Similarly, each side has exactly two adjacent sides. + The segment determined by two non-adjacent vertices is called the +\index{diagonala!večkotnika}\pojem{diagonal of the polygon} (Figure \ref{sl.aks.2.2.13c.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.13c.pic} +\caption{} \label{sl.aks.2.2.13c.pic} +\end{figure} + +The following statement is true for the diagonals of the polygon. + + \bizrek + The number of diagonals of an $n$-gon is $\frac{n(n-3)}{2}$. + \eizrek + +We will give the proof of this statement in section \ref{odd3Helly}, where we will separately consider the combinatorial properties of sets of points in the plane. + +Let's define the angles of a polygon. Let $O$ be any vertex of a polygon, and $P$ and $Q$ be its two adjacent vertices. The line segments $OP$ and $OQ$ are denoted by $p$ and $q$. In this case, the angular bisector $pOq$ determines two angles. The angle for which it is true that every line segment with initial point $O$, which belongs to this angle and does not contain any other vertices of the polygon, intersects the edge of the polygon except at point $O$ in an odd number of points, is called the \index{angle!inner polygon}\pojem{inner angle of the polygon} or, more briefly, the \index{angle!inner}\pojem{angle of the polygon} at vertex $O$ (Figure \ref{sl.aks.2.2.13d.pic}). If the inner angle of the polygon is convex, its supplement is called the \index{angle!outer polygon}\pojem{outer angle of the polygon} (Figure \ref{sl.aks.2.2.13d.pic}). + The angles of the polygon are + \index{angle!adjacent}\pojem{adjacent angles}, if their vertices are adjacent vertices of the polygon, otherwise the angles are \index{angle!non-adjacent}\pojem{non-adjacent}. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.13d.pic} +\caption{} \label{sl.aks.2.2.13d.pic} +\end{figure} + +The most simple $n$-gon and at the same time one of the most used shapes in the geometry of a plane is the case $n=3$ - \index{trikotnik}\pojem{trikotnik}. +In the case of the triangle $ABC$ (we will mark it with $\triangle ABC$), the points $A$, $B$ and $C$ are its \pojem{oglišča}, and the lines $AB$, $BC$ and $CA$ are its +\index{stranica!trikotnika}\pojem{stranice} (Figure \ref{sl.aks.2.2.14.pic}). Obviously, every two sides of the triangle are adjacent. The same goes for every two vertices. The triangle therefore has no diagonal. We say that the vertex $A$ ($B$ and $C$) or the angle $BAC$ ($ABC$ and $ACB$) is the \index{oglišče!nasprotno trikotnika}\pojem{nasprotno oglišče} or the \index{kot!nasprotni trikotnika}\pojem{nasprotni kot} of the side $BC$ ($AC$ and $AB$) of the triangle $ABC$. And also the side $BC$ ($AC$ and $AB$) is the \index{stranica!nasprotna trikotnika}\pojem{nasprotna stranica} of the vertex $A$ ($B$ and $C$) or the angle $BAC$ ($ABC$ and $ACB$) of the triangle $ABC$. The angles (internal) of the triangle $ABC$ at the vertices $A$, $B$ and $C$ are often marked with $\alpha$, $\beta$ and $\gamma$, the corresponding external ones with $\alpha_1$, $\beta_1$ and $\gamma_1$. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.14.pic} +\caption{} \label{sl.aks.2.2.14.pic} +\end{figure} + +The triangle $ABC$ can also be defined as the intersection of the half-planes $ABC$, $ACB$ and $BCA$. We will not prove the equivalence of these two definitions here. + + Pasch's axiom in terms of triangles can now be expressed in a shorter form: + +If a line in the plane of a triangle intersects one of its sides +and does not pass through any of its vertices, then intersects exactly one more side of this triangle + + + + \bizrek \label{PaschIzrek} + If a line, not passing through any vertex of a triangle, intersects one side of the triangle + then the line intersects exactly one more side of the triangle (Figure \ref{sl.aks.2.2.14a.pic}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.14a.pic} +\caption{} \label{sl.aks.2.2.14a.pic} +\end{figure} + +In the case of $n=4$ for the $n$-gon we get +\index{štirikotnik}\pojem{štirikotnik}. Because each vertex of the quadrilateral has exactly one non-adjacent vertex, we will also call this vertex the \index{oglišče!nasprotno štirikotnika}\pojem{opposite vertex} of the quadrilateral. Similarly, the non-adjacent sides will be the \index{stranica!nasprotna štirikotnika}\pojem{opposite sides} of the quadrilateral. The diagonal of the quadrilateral is therefore determined by the opposite vertices. + From the definition itself, it is clear that the quadrilateral has two diagonals\index{diagonala!štirikotnika} (Figure \ref{sl.aks.2.2.14b.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.14b.pic} +\caption{} \label{sl.aks.2.2.14b.pic} +\end{figure} + +For the non-adjacent angle of the quadrilateral, we also say that they are the + \index{kot!nasprotni štirikotnika}\pojem{opposite angles} of the quadrilateral. The angles (internal) of the quadrilateral $ABCD$ at the vertices $A$, $B$, $C$ and $D$ are usually denoted by $\alpha$, $\beta$, $\gamma$ and $\delta$, and the corresponding external (those that exist) by $\alpha_1$, $\beta_1$, $\gamma_1$ and $\delta_1$ (Figure \ref{sl.aks.2.2.14c.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.14c.pic} +\caption{} \label{sl.aks.2.2.14c.pic} +\end{figure} + + + +As a result of the axioms of order, we will also introduce the concepts of orientation of a triangle and orientation of an angle in this section. + +The triangle $ABC$, for which the vertices are a triple $(A,B,C)$, is called the + \index{orientacija!trikotnika} \pojem{oriented triangle}. +We say that the oriented triangles $ABC$ and $BCA'$ are of the \pojem{same orientation}, if $A,A'\ddot{-} BC$, and of the opposite orientation, if $A,A'\div BC$ + (Figure \ref{sl.aks.2.2.15.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.15.pic} +\caption{} \label{sl.aks.2.2.15.pic} +\end{figure} + +When we talk about the orientation of two triangles, from now on we will always mean an oriented triangle (we will often omit the word oriented). Triangles $ABC$ and $A'B'C'$ are of the \pojem{same orientation} or are \pojem{equally oriented}, if there exists such a sequence of triangles: $\triangle ABC=\triangle P_1P_2P_3$, $\triangle P_2P_3P_4$, $\triangle P_3P_4P_5$, ..., $\triangle P_{n-2}P_{n-1}P_n=\triangle A'B'C'$, that in this sequence the number of changes in orientation of two adjacent triangles is even + (Figure \ref{sl.aks.2.2.15a.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.15a.pic} +\caption{} \label{sl.aks.2.2.15a.pic} +\end{figure} + +It can be proven that the relation of the same orientation of two triangles is equivalent to a relation that has two classes. For two triangles that are not in the same class, we say that they are of \pojem{different orientation} or are \pojem{differently oriented}. Each of the two classes determines the \index{orientation!plane}\pojem{orientation of the plane}. We call them the \pojem{positive orientation} and the \pojem{negative orientation}. For the sake of easier understanding, let us agree that the orientation that corresponds to the direction of the rotation of the hour hand is negative, and the opposite is the positive orientation + (Figure \ref{sl.aks.2.2.15b.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.15b.pic} +\caption{} \label{sl.aks.2.2.15b.pic} +\end{figure} + + +Let us also define the orientation of angles. Angles $ASB$ and $A'S'B'$, neither of which is a right angle, are of the \pojem{same orientation}, if: +\begin{itemize} + \item both are convex or both are concave, and triangles $ASB$ and $A'S'B'$ are of the same orientation (Figure \ref{sl.aks.2.2.16.pic}), + \item one angle is convex and the other is concave, and triangles $ASB$ and $A'S'B'$ are of opposite orientation (Figure \ref{sl.aks.2.2.16c.pic}). +\end{itemize} + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.16.pic} +\caption{} \label{sl.aks.2.2.16.pic} +\end{figure} + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.16c.pic} +\caption{} \label{sl.aks.2.2.16c.pic} +\end{figure} + + If $\angle ASB$ is an obtuse angle, $\angle A'S'B'$ is a convex angle, then the angles $ASB$ and $A'S'B'$ have the same orientation, if there is a point $C$ inside the angle $ASB$, such that the angles $ASC$ and $A'S'B'$ have the same orientation (Figure \ref{sl.aks.2.2.16d.pic}). We do the same if the angle $A'S'B'$ is obtuse or if both angles $ASB$ and $A'S'B'$ are obtuse. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.16d.pic} +\caption{} \label{sl.aks.2.2.16d.pic} +\end{figure} + + +It turns out that the relation of the same orientation of angles is an equivalent relation that has two classes. In this case, the positive orientation of the angle represents the class in which the triangle $ASB$ has a negative orientation for the convex angle $ASB$ from this class (Figure \ref{sl.aks.2.2.16a.pic}). In this sense, the angles $ASB$ and $BSA$ are oriented in the opposite direction. + \index{orientation!angle} \pojem{Oriented angle} $ASB$ we will denote with $\measuredangle ASB$. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.16a.pic} +\caption{} \label{sl.aks.2.2.16a.pic} +\end{figure} + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.16b.pic} +\caption{} \label{sl.aks.2.2.16b.pic} +\end{figure} + +If $C$ is an arbitrary point that does not lie on the edge of the angle $ASB$, we will define the sum of the oriented angles $\measuredangle ASC$ and $\measuredangle CSB$ + (Figure \ref{sl.aks.2.2.16b.pic}): + \begin{eqnarray} + \measuredangle ASC+\measuredangle CSB = \measuredangle ASB. + \label{orientKotVsota} + \end{eqnarray} + + +%________________________________________________________________________________ + \poglavje{Congruence Axioms} + \label{odd2AKSSKL} + + +The following axioms are needed to introduce the concept and properties of +the congruence of figures. With the previous axioms, we could introduce and +consider the concepts: distance, segment, angle, polygon, ... but not +the concepts related to congruence: circle, right angle, +congruence of triangles,~... + +The intuitive idea of the compatibility of shapes that we used in +elementary school is associated with the movement that the first +shape transforms into the other. We will now use this idea to +formally define the concept of compatibility and its properties. +We will first start with the basic, already mentioned, concept of +compatibility of pairs of points $(A,B)\cong (C,D)$ (Figure +\ref{sl.aks.2.3.1.pic}) and formally define the concept of +‘‘movement’’. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.1.pic} +\caption{} \label{sl.aks.2.3.1.pic} +\end{figure} + + Using the compatibility of pairs of points, we first define the compatibility of an $n$-tuple of points. + We say that +two $n$-tuples of points are compatible (Figure \ref{sl.aks.2.3.2.pic}) or +$$(A_1 , A_2,\ldots ,A_n ) \cong ( A'_1 , A'_2 ,\ldots , A'_n ),$$ +if: $(A_i,A_j)\cong (A'_i,A'_j)$ for every $i,j\in \{1,2,\ldots, +n\}$. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.2.pic} +\caption{} \label{sl.aks.2.3.2.pic} +\end{figure} + + +A bijective mapping of a plane into a plane +$\mathcal{I}:\mathcal{S}\rightarrow \mathcal{S}$ is +\index{izometrija}\pojem{izometrija} or \pojem{izometrijska +transformacija}, if it preserves the relation of compatibility of +pairs of points (Figure \ref{sl.aks.2.3.3.pic}) or if for every two +points $A$ and $B$ it holds: + $$(\mathcal{I}(A),\mathcal{I}(B))\cong (A,B).$$ + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.3.pic} +\caption{} \label{sl.aks.2.3.3.pic} +\end{figure} + + +With the following axioms, we will introduce the properties of the +newly defined mapping. + + +\vspace*{3mm} + + + \baksiom \label{aksIII1} Isometries preserve the relation + $\mathcal{B}$ (Figure \ref{sl.aks.2.3.4.pic}), which means that for every + isometry $\mathcal{I} $ holds: + $$\mathcal{I}: A, B,C\mapsto A',B',C'\hspace*{2mm}\wedge \hspace*{2mm} + \mathcal{B}(A,B,C) + \hspace*{1mm}\Rightarrow\hspace*{1mm} \mathcal{B}(A',B',C').$$ + \eaksiom + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.4.pic} +\caption{} \label{sl.aks.2.3.4.pic} +\end{figure} + + + \baksiom \label{aksIII2} If $ABC$ and $A'B'C'$ are two half-planes + (Figure \ref {asl.aks.2.3.5.pic}), then there is a single isometry + $\mathcal{I}$, which maps: + + \begin{itemize} + \item point $A$ to point $A'$, + \item half-line $AB$ in half-line $A'B'$, + \item half-plane $ABC$ to half-plane $A'B'C'$. + \end{itemize} + If $(A,B)\cong (A',B')$ holds, + then it is $\mathcal{I}(B)=B'$. + \\ If in addition + $(A,B,C)\cong (A',B',C')$ also holds, + then it is $\mathcal{I}(B)=B'$ and $\mathcal{I}(C)=C'$. + \eaksiom + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.5.pic} +\caption{} \label{asl.aks.2.3.5.pic} +\end{figure} + + \baksiom \label{aksIII3} For every two points $A$ and $B$ there exists + isometry such that holds $$\mathcal{I}: A, B\mapsto B,A.$$ + If $(S,A)\cong (S,B)$ and $S\in AB$, then for each isometry $\mathcal{I}$ with this property holds $\mathcal{I}(S)=S$ (Figure \ref{sl.aks.2.3.6.pic}). + \eaksiom + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.6.pic} +\caption{} \label{sl.aks.2.3.6.pic} +\end{figure} + + \baksiom \label{aksIII4} The set of all isometries with respect to the composition of mappings form a group, which means that: + \begin{itemize} + \item composition of two isometries $\mathcal{I}_2\circ \mathcal{I}_1$ is isometry, + \item identity map $\mathcal{E}$ is isometry, + \item if $\mathcal{I}$ is isometry, then its inverse transformation + $\mathcal{I}^{-1}$ is also isometry. + \end{itemize} + \eaksiom + +\vspace*{3mm} + +We mention that in the structure of a group the property of associativity is also required, i.e. $\mathcal{I}_1\circ (\mathcal{I}_2\circ \mathcal{I}_3)= + (\mathcal{I}_1\circ \mathcal{I}_2)\circ \mathcal{I}_3$ (for any isometries + $\mathcal{I}_1$, $\mathcal{I}_2$ and $\mathcal{I}_3$), which is automatically fulfilled + in the operation of the composition of functions. We also mention the \pojem{identity} \index{identity} + $\mathcal{E}$ from the previous axiom of the mapping, for which + $\mathcal{E}(A)=A$ for every point on the plane. The mapping + $\mathcal{I}^{-1}$ is the \pojem{inverse mapping} for the isometry + $\mathcal{I}$, if $\mathcal{I}^{-1}\circ \mathcal{I} + =\mathcal{I}\circ\mathcal{I}^{-1}=\mathcal{E}$. According to the previous + axiom, the identity and the inverse mapping are therefore also isometries of every isometry. + + + +We prove the first consequences of the compatibility axioms. First, we will +consider the following properties of isometries. + + + + \bizrek \label{izrekIzoB} Isometry maps a line to a line, a line segment to a line segment, a ray to a ray, + a half-plane to a half-plane, an angle to an angle and an $n$-gon to an $n$-gon. + \eizrek + +\textbf{\textit{Proof.}} + According to axiom \ref{aksIII1}, isometries preserve the relation + $\mathcal{B}$. Therefore, all points of the line $AB$ are mapped by isometry $I$ + into points that lie on the line $A'B'$, where $A'=\mathcal{I}(A)$ and + $B'=\mathcal{I}(B)$. Since the inverse mapping $\mathcal{I}^{-1}$ is also an isometry (axiom \ref{aksIII4}), each point of the line + $A'B'$ is the image of some point that lies on the line $AB$. So with + isometry $\mathcal{I}$ + the line $AB$ is mapped into the line $A'B'$. + + We defined the other shapes from the statement using the relation + $\mathcal{B}$, so the proof is similar to the one for the line. + \kdokaz + +From the proof of the previous statement it follows that the endpoints of the line $AB$ +are mapped by isometry into the endpoints of the image $A'B'$. In a similar +way, the starting point of the ray is mapped into the starting point of the ray, the edge of the half-plane into the edge of the half-plane, the vertex of the angle into the vertex of the angle and +the vertices of the polygon into the vertices of the polygon. + +Isometries are defined as bijective maps, which preserve the +congruence of pairs of points. But does it also hold true, that for every congruent pair of points +there is an isometry, which maps the first pair into the second one? Let us answer this question with the following theorem. + + + \bizrek \label{izrekAB} If $(A,B)\cong (A',B')$, then there is an isometry + $\mathcal{I}$, which maps the points $A$ and $B$ to the points + $A'$ and $B'$, + i.e.: + $$\mathcal{I}: A, B\mapsto A',B'.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.7.pic} +\caption{} \label{sl.aks.2.3.7.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Let $C$ be a point, which does not lie on the line $AB$, and $C'$ be a point, + which does not lie on the line $A'B'$ (Figure \ref{sl.aks.2.3.7.pic}). + By Axiom \ref{aksIII2} there is a single isometry $\mathcal{I}$, which + maps the point $A$ into the point $A'$, the segment $AB$ into the segment $A'B'$ + and the plane $ABC$ into the plane $A'B'C'$. Because of $(A,B)\cong (A',B')$ from the same Axiom \ref{aksIII2}, it follows that $\mathcal{I}(B)=B'$. + \kdokaz + +The proof of the following theorem, which will later be stated in a different form as the first theorem about the congruence of triangles, is similar. + + + + \bizrek \label{IizrekABC} Let $(A,B,C)$ and $(A',B',C')$ be + triplets of non-collinear points such that $$(A,B,C)\cong (A',B',C'),$$ + then there is a single isometry $\mathcal{I}$, that maps the points + $A$, $B$ and $C$ into the points $A'$, $B'$ and $C'$, i.e.: + $$\mathcal{I}: A, B,C\mapsto A',B',C'.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.5.pic} +\caption{} \label{sl.aks.2.3.5.pic} +\end{figure} + +\textbf{\textit{Proof.}} +By Axiom \ref{aksIII2} there exists a single isometry $\mathcal{I}$, +that maps point $A$ to point $A'$, line segment $AB$ to line segment +$A'B'$ and plane $ABC$ to plane $A'B'C'$ (Figure \ref{sl.aks.2.3.5.pic}). +Because, by the assumption $(A,B,C)\cong (A',B',C')$ from the same +Axiom \ref{aksIII2}, it follows that $\mathcal{I}(B)=B'$ and +$\mathcal{I}(C)=C'$. + + It is necessary to prove that $\mathcal{I}$ is the only such isometry. + Assume that there exists such an isometry $\mathcal{\widehat{I}}$, that + $\mathcal{\widehat{I}}: A, B,C\mapsto A',B',C'$. By + Theorem \ref{izrekIzoB} isometry $\mathcal{\widehat{I}}$ + also maps line segment $AB$ to line segment $A'B'$ and plane $ABC$ + to plane $A'B'C'$. From Axiom \ref{aksIII2} it follows that + $\mathcal{\widehat{I}}=\mathcal{I}$. + \kdokaz + +A direct consequence is the following theorem. + + + \bizrek \label{IizrekABCident} Let $A$, $B$ and $C$ be three non-collinear points, then the identity map + $\mathcal{E}$ is the only isometry that maps points $A$, $B$, and $C$ to the same points + $A$, $B$ and $C$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.5a.pic} +\caption{} \label{sl.aks.2.3.5a.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.aks.2.3.5a.pic}) + +First, the identical mapping $\mathcal{E}$, that maps points $A$, $B$ +and $C$ to the same points $A$, $B$ and $C$, is an isometry by Axiom +\ref{aksIII4}. From the previous theorem \ref{IizrekABC} it follows +that such an isometry is unique. + \kdokaz + +For point $A$ we say that it is \index{point!fixed} \pojem{fixed point} +(or \index{point!invariant} \pojem{invariant point}) of isometry +$\mathcal{I}$, if $\mathcal{I}(A)=A$. The previous theorem tells us +that the only isometries that have three fixed non-collinear points +are identities. + +We will discuss isometries in more detail in chapter +\ref{pogIZO}, but here we will use them primarily to help us +introduce the concept of congruence of figures. Two figures $\Phi$ +and $\Phi'$ are \index{figures!congruent}\poem{congruent} (we will +write $\Phi\cong \Phi'$), if there exists an isometry $I$, that +transforms figure $\Phi$ into figure $\Phi'$. + +A direct consequence of axiom \ref{aksIII4} is the following +proposition. + +\bizrek + Congruence of figures is an equivalence relation. \label{sklRelEkv} + \eizrek + +\textbf{\textit{Proof.}} + +\textit{Reflexivity.} For every figure $\Phi$ it holds that $\Phi \cong +\Phi$, because the identity transformation $\mathcal{E}$ is an isometry +(axiom \ref{aksIII4}) and $\mathcal{E}:\Phi\rightarrow\Phi$. + +\textit{Symmetry.} From $\Phi \cong \Phi_1$ it follows that there exists an isometry $\mathcal{I}$, that transforms figure $\Phi$ into figure $\Phi_1$. +The inverse transformation $\mathcal{I}^{-1}$, which is an isometry according to axiom \ref{aksIII4}, transforms figure $\Phi_1$ into figure $\Phi$, +so $\Phi_1 \cong \Phi$ holds. + +\textit{Transitivity.} From $\Phi \cong \Phi_1$ and $\Phi_1 \cong +\Phi_2$ it follows that there exist such isometries $\mathcal{I}$ and +$\mathcal{I}'$, that $\mathcal{I}:\Phi\rightarrow\Phi_1$ and +$\mathcal{I}':\Phi_1\rightarrow\Phi_2$ hold. +Then the composition $\mathcal{I}'\circ\mathcal{I}$, + which is an isometry according to axiom \ref{aksIII4}, transforms figure $\Phi$ +into figure $\Phi_2$, so $\Phi \cong \Phi_2$ holds. +\kdokaz + + + The concept of congruence of figures also applies to lines. We have intuitively + associated the congruence of lines with the congruence of pairs of points. + Now we will prove the equivalence of both relations. + + \bizrek \label{izrek(A,B)} $AB \cong A'B' \Leftrightarrow + (A,B)\cong (A',B')$ + \eizrek + +\textbf{\textit{Proof.}} + + ($\Rightarrow$) If $(A,B)\cong +(A',B')$, according to proposition \ref{izrekAB} there exists an isometry +$\mathcal{I}$, that transforms points $A$ and $B$ into points $A'$ and +$B'$. From proposition \ref{izrekIzoB} it follows that isometry $\mathcal{I}$ +transforms line $AB$ into line $A'B'$ or $AB \cong A'B'$ holds. + +($\Leftarrow$) If $AB \cong A'B'$, there exists an isometry +$\mathcal{I}$, which maps the line segment $AB$ to the line segment $A'B'$. By +the consequence of Theorem \ref{izrekIzoB}, the endpoint of the line segment is mapped to the endpoint of the line segment. This means that either +$\mathcal{I}:A,B\mapsto A',B'$ or $\mathcal{I}:A,B\mapsto +B',A'$ holds. From the first relation it follows that $(A,B)\cong (A',B')$ and from the second that $(A,B)\cong (B',A')$. But from the second example we also get +$(A,B)\cong (A',B')$, which is a consequence of Axioms \ref{aksIII3} and +\ref{aksIII4}. +\kdokaz + + Because of the previous theorem, in the following we will always write + the relation $(A,B)\cong (A',B')$ instead of $AB\cong + A'B'$. + + + \bizrek \label{ABnaPoltrakCX} + For each line segment $AB$ and each ray $CX$, there is exactly + one point $D$ on the ray + $CX$ that $AB\cong CD$ holds. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.5b.pic} +\caption{} \label{sl.aks.2.3.5b.pic} +\end{figure} + + + + \textbf{\textit{Proof.}} Let $P$ be a point that does not lie on the line $AB$ and $Q$ be a point that does not lie on the line $CX$ (Figure \ref{sl.aks.2.3.5b.pic}). + By Axiom \ref{aksIII2}, there is only one + isometry $\mathcal{I}$, which + maps the point $A$ to the point $C$, the line segment $AB$ to the line segment $CX$ + and the half-plane $ABP$ to the half-plane $CXQ$. + Let $D=\mathcal{I}(C)$, then $AB \cong CD$ holds. + + We assume that + on the line segment $CX$ there is another point $\widehat{D}$, for + which + $AB \cong C\widehat{D}$ holds. Because the line segments + $CX$ and $CD$ are congruent, and the isometry $\mathcal{I}$ maps the point + $A$ to the point $C$, the line segment $AB$ to the line segment $CD$ + and the half-plane $ABP$ to the half-plane $CDQ$, + from Axiom \ref{aksIII2} it follows that $\mathcal{I}(C)=\widehat{D}$ or + $\widehat{D}=D$. + \kdokaz + +\bizrek \label{izomEnaC'} Let $A$, $B$, $C$ be three non-collinear points + and $A'$, $B'$ points of the edge of a half-plane $\pi$ such that $AB \cong A'B'$. + Then there is exactly one point $C'$ in the half-plane $\pi$ such that $AC \cong A'C'$ and $BC \cong B'C'$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.11a.pic} +\caption{} \label{sl.aks.2.3.11a.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.aks.2.3.11a.pic}) + + By Axiom \ref{aksIII2} there is only one + isometry $\mathcal{I}$, which + maps the point $A$ into the point $A'$, the segment $AB$ into the segment $A'B'$ + and the plane $ABC$ into the plane $\pi$ and holds $\mathcal{I}(B)=B'$. + Let $C'=\mathcal{I}(C)$, then it holds $AC \cong A'C'$ and + $BC \cong B'C'$. We assume that there is such a point + $\widehat{C}'$, which lies in the plane $\pi$ and holds $AC \cong A'\widehat{C}'$ and + $BC \cong B'\widehat{C}'$. Because $AB \cong +A'B'$, by Theorem \ref{IizrekABC} there is only one isometry +$\mathcal{\widehat{I}}$, which maps the points $A$, $B$ and $C$ into +the points $A'$, $B'$ and $\widehat{C}'$. But this also maps +the segment $AB$ into the segment $A'B'$ and the plane $ABC$ into the plane +$A'B'\widehat{C}'=\pi$. By Axiom \ref{aksIII2} is +$\mathcal{\widehat{I}}=\mathcal{I}$ and therefore also +$\widehat{C}'=\mathcal{\widehat{I}}(C)=\mathcal{I}(C)=C'$. + \kdokaz + + + \bizrek \label{izoABAB} If $\mathcal{I}$ is an isometry that maps a points $A$ and $B$ into the same points + $A$ and $B$ (i.e. $\mathcal{I}(A)=A$ and $\mathcal{I}(B)=B$), then it also holds for each point $X$ on the line + $AB$ (i.e. $\mathcal{I}(X)=X$). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.8.pic} +\caption{} \label{sl.aks.2.3.8.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $X$ be an arbitrary point on the line +$AB$. Without loss of generality, we assume that the point $X$ lies +on the segment $AB$ (Figure \ref{sl.aks.2.3.8.pic}). We prove that +$\mathcal{I}(X)=X$. + +Let $P$ be a point that does not lie on the line $AB$ and +$P'=\mathcal{I}(P)$. The isometry $\mathcal{I}$ +maps the point $A$ to the point $A$, the segment $AB$ to the segment $AB$ + (or the segment $AX$ to the segment $AX$) + and the half-plane $ABP$ to the half-plane $ABP'$ + (or the half-plane $AXP$ to the half-plane $AXP'$). + By Axiom \ref{aksIII2} + from $AX\cong AX$ it follows that $\mathcal{I}(X)=X$. + \kdokaz + + We introduce new concepts related to distances. + +We say that the line $EF$ \index{vsota!daljic}\pojem{vsota daljic} +$AB$ and $CD$, which we denote $EF=AB+CD$, if there exists such a point $P$ +on the line $EF$, that $AB \cong EP$ and $CD \cong PF$ (Figure +\ref{sl.aks.2.3.9.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.9.pic} +\caption{} \label{sl.aks.2.3.9.pic} +\end{figure} + + +The line $EF$ is \index{razlika!daljic}\pojem{razlika daljic} $AB$ +and $CD$, which we denote $EF=AB-CD$, if $AB=EF+CD$ (Figure +\ref{sl.aks.2.3.9.pic}). + + In a similar way, we can also define + multiplication of a line by a natural and a positive rational + number. For the lines $AB$ and $CD$ it is $AB=n\cdot CD$ + ($n\in \mathbb{N}$), if there exist such points + $X_1$, $X_2$,..., $X_{n-1}$, that + $\mathcal{B}(X_1,X_2,\ldots,X_{n-1})$ and + $AX_1 \cong X_1X_2 \cong X_{n-1}B \cong CD$ (Figure +\ref{sl.aks.2.3.10.pic}). + In this case, it is also $CD=\frac{1}{n}\cdot AB$. + + At this point, we will not formally prove the fact that for every line $PQ$ and every natural number $n$ there exists a line $AB$, for which $AB=n\cdot PQ$, and a line $CD$, for which $CD=\frac{1}{n}\cdot PQ$. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.10.pic} +\caption{} \label{sl.aks.2.3.10.pic} +\end{figure} + +We introduce multiplication of a line segment with a positive rational number in the following way. For $q=\frac{n}{m} \in \mathbb{Q^+}$ we have: +$$q\cdot AB=\frac{n}{m}\cdot AB = n\cdot\left(\frac{1}{m}\cdot AB\right)$$ + +If for a point $P$ on the line segment $AB$ it holds that $AP=\frac{n}{m}\cdot PB$, we say that the point $P$ \pojem{divides} the line segment $AB$ in the \index{razmerje} \pojem{ratio} $n:m$, which we write as $AP:PB=n:m$. + +The line segment $AB$ is \index{relacija!urejenosti daljic}\pojem{longer} than the line segment $CD$, which we denote $AB>CD$, if there exists such a point $P\neq B$ on the line segment $AB$, that it holds $CD \cong AP$ (Figure \ref{sl.aks.2.3.11.pic}). In this case we also say that the line segment $CD$ is \pojem{shorter} than the line segment $AB$ (notation $CDCD$, $AB \angle cd$), if there exists a segment $s=SX$ in the angle $ab$, such that $\angle as \cong \angle cd$ (Figure \ref{sl.aks.2.3.17.pic}). In this case, the angle $cd$ is also \pojem{smaller} than the angle $ab$ ($\angle cd< \angle ab$). It is not difficult to prove that for two angles $ab$ and $cd$ one of the relations holds: $\angle ab > \angle cd$, $\angle ab < \angle cd$ or $\angle ab \cong \angle cd$. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.17.pic} +\caption{} \label{sl.aks.2.3.17.pic} +\end{figure} + + +The angles are \index{angles!supplementary}\pojem{supplementary}, if their sum is equal to the straight angle (Figure +\ref{sl.aks.2.3.18.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.18.pic} +\caption{} \label{sl.aks.2.3.18.pic} +\end{figure} + + +The segment $s=SX$ is the \index{bisector of an angle}\pojem{bisector of the angle} $\angle pSq=\alpha$ (Figure \ref{sl.aks.2.3.19.pic}), if it lies in this angle and +it holds $\angle ps \cong \angle sq$. The carrier of this bisector is the \index{simetrala!kota}\pojem{simetrala kota} $pSq$ (the line $s_{\alpha}$). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.19.pic} +\caption{} \label{sl.aks.2.3.19.pic} +\end{figure} + + + +Similarly to the center of a line, the following statement holds for the bisector of an angle. + + \bizrek \label{izrekSimetralaKota} + An angle has exactly one bisector. + %(oz. eno samo simetralo). + \eizrek + +\textbf{\textit{Proof.}} + Let $\alpha=pSq$ be an arbitrary angle, $P$ an arbitrary point, which lies on the + segment $Sp$ ($P\neq S$) and $Q$ a point, which lies on the segment $Sq$ and it holds + $SP\cong SQ$. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.20.pic} +\caption{} \label{sl.aks.2.3.20.pic} +\end{figure} + +Let the angle $\alpha$ be the extended angle + (Figure \ref{sl.aks.2.3.20.pic}), which determines + the line $\pi$. Let $A$ be any point on it. By the statement + \ref{izomEnaC'} in the line $\pi$ there is only one point $B$, + so that $(P,Q,A)\cong (Q,P,B)$. From the statement \ref{IizrekABC} it follows that there is only one izometry $\mathcal{I}$, which maps points $P$, $Q$ and + $A$ into points $Q$, $P$ and $B$. Let + $\mathcal{I}(B)=\widehat{A}$. Because + $(Q,P,B)\cong(P,Q,\widehat{A})$, by the statement \ref{izomEnaC'} + $\widehat{A}=A$. Therefore: + $$\mathcal{I}:P,Q,A,B\mapsto Q,P,B,A.$$ +Therefore, the centers $S$ and $L$ of the lines $PQ$ and $AB$ map into each other +(axiom \ref{aksIII4}), which then also holds for the line segment $s=SL$ and +every point on it (statement \ref{izoABAB}). Therefore, the izometry +$\mathcal{I}$ maps the line segment $pSs$ into the line segment $sSq$, so + the line segment $pSs\cong sSq$ or the line segment $s$ is the bisector of the angle $pSq$. + + We will prove that $s$ is the only bisector of the angle $\alpha$. Let + $\widehat{s}=S\widehat{L}$ + be a line segment that lies in the angle $\alpha$ and $pS\widehat{s}\cong + \widehat{s}Sq$. Then there is an izometry $\mathcal{\widehat{I}}$, which + maps the angle $pS\widehat{s}$ into the angle $\widehat{s}Sq$. This izometry + maps the point $S$ into the point $S$, the line segment $p$ into the line segment $q$ and + the line $\pi$ into the line $\pi$, so by the axiom + \ref{aksIII2} $\mathcal{\widehat{I}}=\mathcal{I}$. Therefore + $\mathcal{I}(\widehat{s})= + \mathcal{\widehat{I}}(\widehat{s})=\widehat{s}$. If $\widehat{L} \notin + s$, the izometry $\mathcal{I}$ maps three non-collinear points + $S$, $L$ and $\widehat{L}$ into itself and is the identical mapping + (statement \ref{IizrekABCident}), which is not possible. Therefore $\widehat{L} \in + s$ or $\widehat{s}=s$. + +If $\alpha$ is an unextended convex angle (Figure \ref{sl.aks.2.3.20.pic}), + then points $S$, $P$ and $Q$ + are nonlinear, so according to Theorem \ref{IizrekABC} there is only one + isometry $\mathcal{I}$, which maps points $P$, $S$ and $Q$ to + points $Q$, $S$ and $P$. With $L$ we denote the center of the line $PQ$. + By the Axiom \ref{aksIII3} we have $\mathcal{I}(L)=L$. Then also all + points of the ray $s=SL$ are mapped to themselves (Theorem \ref{izoABAB}). As +$\alpha$ is a convex angle, this means that the point $L$ and then also +the ray $s$ lie within this angle. + Therefore the isometry $\mathcal{I}$ maps angle $pSs$ to angle $sSq$, so + $pSs\cong sSq$ or the ray $s$ is the bisector of angle $pSq$. + +Similarly as in the previous example we prove that angle $\alpha$ has no other bisectors. + +If $\alpha$ is a non-convex angle, the bisector is obtained as the +complementary (supplementary) ray of the ray $s$. + \kdokaz + +We prove two theorems, which relate to right angles and perfect angles.\index{angle!right} \index{angle!perfect} + + + + \bizrek + The adjacent supplementary angles of two congruent angles are also congruent. \label{sokota} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.20a.pic} +\caption{} \label{sl.aks.2.3.20a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $\alpha'=\angle P'OQ$ and $\alpha_1'=\angle P_1'O_1Q_1$ be the supplement angles of two adjacent angles $\alpha=\angle POQ$ and $\alpha_1=\angle P_1O_1Q_1$ (Figure \ref{sl.aks.2.3.20a.pic}). By Axiom \ref{aksIII2}, there exists a single isometry $\mathcal{I}$, which maps point $O$ to point $O_1$, line segment $OP$ to line segment $O_1P_1$, and line $POQ$ to line $P_1O_1Q_1$. Let $Q_2=\mathcal{I}(Q)$. Then $\angle P_1O_1Q_2\cong \angle POQ$. Isometry $\mathcal{I}$ maps line $POQ$ to line $P_1O_1Q_1$, so point $Q_2$ (and also line segment $O_1Q_2$) lies on line $P_1O_1Q_1$. Since, by assumption, $\angle POQ\cong\angle P_1O_1Q_1$, by \ref{KotNaPoltrak} Theorem, $OQ_1$ and $OQ_2$ represent the same line segment. Therefore, point $Q_2$ lies on line segment $O_1Q_1$. Let $P_2'=\mathcal{I}(P')$. Since isometries map line segments to line segments (Theorem \ref{izrekIzoB}), point $P_2'$ lies on line segment $O_1P_1'$. From $\mathcal{I}:P',O,Q\mapsto P_2',O_1,Q_2$ it follows that isometry $\mathcal{I}$ maps angle $P'OQ$ to angle $P_2'O_1Q_2$ (Theorem \ref{izrekIzoB}), so + $\angle P'OQ\cong \angle P_2'O_1Q_2=\angle P_1'O_1Q_1$. +\kdokaz + + + \bizrek \label{sovrsnaSkladna} + Vertical angles are congruent. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.20b.pic} +\caption{} \label{sl.aks.2.3.20b.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Let $\alpha=\angle POQ$ and $\alpha'=\angle P'OQ'$ be two adjacent angles, where points $P$, $O$, $P'$ (or $Q$, $O$, $Q'$) are collinear (Figure \ref{sl.aks.2.3.20b.pic}). Angle $\beta=\angle QOP'$ is the supplement of both angles $\alpha$ and $\alpha'$. Since $\beta\cong\beta$, by the previous Theorem \ref{sokota}, $\alpha\cong\alpha'$. +\kdokaz + +\bizrek \label{sredZrcObstoj} + For each point $S$ there exists an isometry $\mathcal{I}$ such that $\mathcal{I}(S)=S$. + In addition, for each point $X\neq S$ the following holds:\\ if $\mathcal{I}(X)=X'$, then $S$ is the midpoint of the line segment $XX'$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.20c.pic} +\caption{} \label{sl.aks.2.3.20c.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Let $P$ be an arbitrary point different from $S$ (Figure \ref{sl.aks.2.3.20c.pic}). By Axiom \ref{AksII3}, there exists a point $Q$ on the line $SP$ such that $\mathcal{B}(P,S,Q)$. We mark the half-planes determined by the edge $SP$ with $\alpha$ and $\alpha'$. By Axiom \ref{aksIII2}, there exists (one and only one) isometry $\mathcal{I}$, which maps the point $S$ to the point $S$, the line segment $SP$ to the line segment $SQ$, and the half-plane $\alpha$ to the half-plane $\alpha'$. + +We mark the line $SP$ with $p$. +The point $P'=\mathcal{I}(P)$ lies on the line segment $SQ$ or on the line $p$. Therefore, since $\mathcal{I}:S,P \mapsto S,P'$, the line $SP$ is mapped to the line $SP'$ by Axiom \ref{aksIII1}, i.e. $\mathcal{I}:p\rightarrow p$. +The image of the half-plane $\alpha'$ with the edge $p$ is therefore a half-plane with the same edge (Proposition \ref{izrekIzoB}). This half-plane cannot be $\alpha'$, since the isometry $\mathcal{I}$ is a bijective mapping and it maps the half-plane $\alpha$ to the half-plane $\alpha'$. Therefore, $\mathcal{I}:\alpha'\rightarrow \alpha$. + +Now it is clear that, without loss of generality, it is enough to carry out the proof only for points that lie in the half-plane $\alpha$ (without the edge or only the line segment $SP$). + +Let $X\in \alpha\setminus p$ and $X'=\mathcal{I}(X)$. We immediately see that $X'\in \alpha'\setminus p$. By Axiom \ref{AksII3} there exists on the line $SX$ such a point $X_1$, that $\mathcal{B}(X,S,X_1)$ is true. Because $\angle PSX$ and $\angle P'SX_1$ are perfect angles, by Theorem \ref{sovrsnaSkladna} they are also compatible. But from $\mathcal{I}:S,P,X \mapsto S,P',X'$ it follows that $\angle PSX \cong \angle P'SX'$. Therefore $\angle P'SX_1\cong \angle P'SX'$ is true (Theorem \ref{sklRelEkv})), so by Theorem \ref{KotNaPoltrak} the line segments $SX_1$ and $SX'$ are identical. This means that the point $X'$ lies on the line segment $SX_1$ or $\mathcal{B}(X,S,X')$ is true. Because of $\mathcal{I}:S,X \mapsto S,X'$ it is also true that $SX\cong SX'$, so by definition the point $S$ is the center of the line $XX'$. + + Let in the end $Y$ be an arbitrary point on the line segment $SP$, which is different from the point $S$, and $Y'=\mathcal{I}(Y)$. The point $Y'$ lies on the line segment $SQ$, so $\mathcal{B}(Y,S,Y')$ is true. Because of $\mathcal{I}:S,Y \mapsto S,Y'$ it is also true that $SY\cong SY'$, so by definition the point $S$ is the center of the line $YY'$. +\kdokaz + +In section \ref{odd6SredZrc} we will discuss the isometry, which is mentioned in the previous Theorem \ref{sredZrcObstoj}, in more detail. + + + Let us define new types of angles. + A \index{kot!ostri}\pojem{acute angle} is a \index{kot!pravi} + \pojem{right angle} or a \index{kot!topi}\pojem{obtuse angle}, if it is + smaller, equal to or greater than its supplement (Figure \ref{sl.aks.2.3.21.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.21.pic} +\caption{} \label{sl.aks.2.3.21.pic} +\end{figure} + + + +From the definition it follows that acute (or obtuse) angles are those convex angles, which are smaller (or greater) than a right angle. + +From Theorem \ref{izrekSimetralaKota} it follows that a right angle +exists, since the bisector of an extended angle divides it into two compatible supplements. + +It is not difficult to prove that every two right angles are compatible and that an angle, which is compatible with a right angle, is also a right angle. + +If the sum of two angles is a right angle, we say that the angles are +\pojem{complementary} (Figure +\ref{sl.aks.2.3.22.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.22.pic} +\caption{} \label{sl.aks.2.3.22.pic} +\end{figure} + + + +We will now introduce an extremely important relation between lines. If +the lines $p$ and $q$ contain the segments of a right angle, we say that +$p$ and $q$ are \pojem{perpendicular}, which we denote $p \perp q$ +(Figure \ref{sl.aks.2.3.23.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.23.pic} +\caption{} \label{sl.aks.2.3.23.pic} +\end{figure} + +From the definition itself it is clear that the perpendicularity is a +symmetric relation, i.e. from $p \perp q$ it follows that $q \perp p$. If +$p \perp q$ and $p \cap q=S$, we say that the line $p$ is +\pojem{perpendicular} to the line $q$ at the point $S$ or that $p$ is a +\pojem{perpendicular} of the line $q$ at this point. + + + +The following theorem is the most important theorem that characterizes +the relation of perpendicularity. + + + + \bizrek \label{enaSamaPravokotnica} + For each point $A$ and each line $p$, there is a unique line $n$ + going through the point $A$, which is perpendicular on the line $p$. + \eizrek + +\textbf{\textit{Proof.}} +Assume that point $A$ does not lie on line $p$. Let $B$ and $C$ be any points that lie on line $p$ (Figure \ref{sl.aks.2.3.24.pic}). We denote the plane $BCA$ with $\pi$, and the complementary plane with $\pi_1$. By izreku \ref{izomEnaC'}, there exists only one point $A_1\in \pi_1$, for which $(A,B,C) \cong (A_1,B,C)$. From izreku \ref{IizrekABC} it follows that there exists only one izometrija $\mathcal{I}$, which maps points $A$, $B$ and $C$ into points $A_1$, $B$ and $C$. We denote the line $AA_1$ with $n$. Because $A$ and $A_1$ are on different sides of line $p$, line $n$ intersects line $p$ in some point $S$. From $\mathcal{I}:B,C \mapsto B,C$ it follows that $\mathcal{I}(S)=S$ (izrek \ref{izoABAB}). Therefore, izometrija $\mathcal{I}$ maps angle $ASB$ into angle $A_1SB$. It follows that $\angle ASB$ and $\angle A_1SB$ are complementary angles, therefore they are right angles. Therefore, $n \perp p$. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.24.pic} +\caption{} \label{sl.aks.2.3.24.pic} +\end{figure} + +We will prove that $n$ is the only perpendicular to line $p$ through point $A$. Let $\widehat{n}$ be a line, for which $A\in \widehat{n}$ and $\widehat{n} \perp p$. Let point $\widehat{S}$ be the intersection of lines $\widehat{n}$ and $p$. By the assumption, $\angle A\widehat{S}B$ is a right angle and is compatible with its complementary angle $\angle B\widehat{S}A_2$ ($A_2$ is such a point that $\mathcal{B}(A,\widehat{S},A_2)$), which is also a right angle. + +From $\mathcal{I}:B,C \mapsto B,C$ it follows +$\mathcal{I}(\widehat{S})=\widehat{S}$ (statement \ref{izoABAB}). +Therefore, the isometry $\mathcal{I}$ maps the angle $A\widehat{S}B$ into the angle +$A_1\widehat{S}B$. It follows that $\angle A\widehat{S}B$ and $\angle +A_1\widehat{S}B$ are congruent, so $\angle A_1\widehat{S}B$ +is a right angle. Therefore, the angle $A_1\widehat{S}B$ and $A_2\widehat{S}B$ +are right angles and are therefore congruent. It follows that the line segments +$\widehat{S}A_1$ and $\widehat{S}A_2$ are the same, so $A_1 \in +\widehat{S}A_2=\widehat{n}$ or $\widehat{n}=AA_1=n$. + + In the case when the point $A$ lies on the line $p$, the rectangle $n$ + is the symmetry of the corresponding extended angle (statement \ref{izrekSimetralaKota}). +\kdokaz + + +The previous statement has the following important consequence - the existence of pairs of disjoint lines in the plane - or those that do not have common points. This is the content of the following two statements. + + + \bizrek \label{absolGeom1} + Let $p$ and $q$ be a lines perpendicular on a line $PQ$ in the points $P$ and $Q$. + Then the lines $p$ and $q$ do not have a common points i.e. $p\cap q=\emptyset$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.25b.pic} +\caption{} \label{sl.aks.2.3.25b.pic} +\end{figure} + +\textbf{\textit{Proof.}} The statement is a direct consequence of the previous statement \ref{enaSamaPravokotnica}. Namely, if the lines $p$ and $q$ intersected at some point $S$, we would have two rectangles on the line $PQ$ from the point $S$ (Figure \ref{sl.aks.2.3.25b.pic}), which is in contradiction with the aforementioned statement. + \kdokaz + + + + \bizrek \label{absolGeom} + If $A$ is a point that does not lie on a line $p$, then there exists at least + one line (in the same plane) passing through the point $A$ and not intersecting the line + $p$ (Figure \ref{sl.aks.2.3.25a.pic}). + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.25a.pic} +\caption{} \label{sl.aks.2.3.25a.pic} +\end{figure} + +\textbf{\textit{Proof.}} By the statement \ref{enaSamaPravokotnica} there exists (exactly one) rectangle $n$ of the line $p$, which goes through the point $A$. We mark with $A'$ the intersection of the lines $p$ and $n$. From the same statement it follows that there exists another rectangle $q$ of the line $n$ in the point $A$. By the previous statement \ref{absolGeom1} the line $q$ goes through the point $A$ and doesn't have any common points with the line $p$. + \kdokaz + +The point $A'$ is the \index{node}\pojem{node} or the +\index{orthogonal projection}\pojem{orthogonal projection} of the +point $A$ on the line $p$, if the rectangle of the line $p$ through +the point $A$ intersects the line in the point $A'$. We will mark +it with $A'=pr_{\perp p}(A)$ (Figure \ref{sl.aks.2.3.25.pic}). From +the previous statement it follows that for every point and line +there exists only one orthogonal projection. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.25.pic} +\caption{} \label{sl.aks.2.3.25.pic} +\end{figure} + +The line, which goes through the center $S$ of the distance $AB$ and is perpendicular to the line $AB$, is called the \index{symmetry!of a line}\pojem{symmetry of the line} $AB$ and we mark it with $s_{AB}$ +(Figure \ref{sl.aks.2.3.26.pic}). The properties of the symmetry of a line we will discuss in the next chapter. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.26.pic} +\caption{} \label{sl.aks.2.3.26.pic} +\end{figure} + + +We say that the point $A$ is \index{symmetry!with respect to a line}\pojem{symmetric} to the point $B$ with respect to the line $s$, if the line $s$ is the symmetry of the line $AB$. The symmetry with respect to a line (as a mapping - i.e. the basic mirroring) we will discuss in more detail in the section \ref{odd6OsnZrc}. + +Let $S$ be a point and $AB$ a line. The set of all points $X$, for +which it holds that $SX \cong AB$, is called the +\index{circle}\pojem{circle} with +\index{center!of a circle}\pojem{center} $S$ and \index{radius +of a circle}\pojem{radius} $AB$; we mark it with $k(S,AB)$ (Figure +\ref{sl.aks.2.3.27.pic}) i.e.: + $$k(S,AB)=\{X;\hspace*{1mm}SX \cong AB\}.$$ + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.27.pic} +\caption{} \label{sl.aks.2.3.27.pic} +\end{figure} + + Of course, a circle is a set + of points in a plane, because in this book we only consider + planar geometry (all points and all figures belong to the same plane). + + From the definition it is clear that for the radius we can choose any distance + that is consistent with + the distance $AB$, that is, any distance $SP$, where $P$ is any + point on the circle. Since the radius is not tied to + a specific distance, we usually denote it with a small letter $r$. So we can also write the circle like this: + $$k(S,r)=\{X;\hspace*{1mm}SX \cong r\}.$$ + The set + +$$\{X;\hspace*{1mm}SX \leq r\}$$ +we call the \index{krog}\pojem{circle} with center $S$ and radius $r$ (Figure \ref{sl.aks.2.3.28.pic}) and denote it with $\mathcal{K}(S,r)$. +The set + $$\{X;\hspace*{1mm}SX < r\}$$ + is the \index{notranjost!kroga} + \pojem{interior of the circle} $\mathcal{K}(S, r)$, and its points are + \pojem{interior points of the circle}. + This means that the circle is actually the union of its interior and the corresponding circle. + +The set + $$\{X;\hspace*{1mm}SX > r\}$$ + we call the \index{zunanjost!kroga}\pojem{exterior of the circle} $\mathcal{K}(S, r)$ + and its points \pojem{exterior points of the circle}. + + For practical reasons, we will call the interior of the circle $\mathcal{K}(S, r)$ the \index{notranjost!krožnice}\pojem{interior} of the corresponding circle $k(S,r)$, the exterior of the circle $\mathcal{K}(S, r)$ the \index{zunanjost!krožnice}\pojem{exterior} of the corresponding circle $k(S,r)$. We define the \pojem{interior} and \pojem{exterior} points of the circle in the same way. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.28.pic} +\caption{} \label{sl.aks.2.3.28.pic} +\end{figure} + +If $P$ and $Q$ are two points on the circle $k(S, r)$, the distance $PQ$ + is called the \index{tetiva krožnice} \pojem{tetiva} of + the circle. If +the tether contains the center of the circle, it is called +\index{premer krožnice}\pojem{premer} or \index{diameter +krožnice}\pojem{diameter} of the circle (Figure +\ref{sl.aks.2.3.29.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.29.pic} +\caption{} \label{sl.aks.2.3.29.pic} +\end{figure} + +We prove the following statement. + + + \bizrek \label{premerInS} + The centre $S$ of the circle $k(S, r)$ is at the same time the midpoint of each + diameter of that circle. + \eizrek + +\textbf{\textit{Proof.}} + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.30.pic} +\caption{} \label{sl.aks.2.3.30.pic} +\end{figure} + + If $PQ$ is the diameter of the circle $k(S, r)$, the points $P$ and $Q$ lie +on the circle, which means: $SP \cong SQ \cong r$ (Figure +\ref{sl.aks.2.3.30.pic}). Because the point $S$ lies on the line $PQ$, +it follows that the point $S$ is the center of this line. + \kdokaz + + From the previous statement it follows that the diameter is equal to two radii, because: +$PQ = PS + SQ = 2\cdot PS = 2\cdot r$. This means +that all diameters of a circle are consistent with each other. + + +Let $P$ and $Q$ be any two points on the circle $k(S, r)$. The intersection +of the circle $k$ with one of the planes (in the plane of this circle) with the edge +$s=PQ$ is called the \index{krožni!lok} \pojem{krožni lok} $PQ$ (or +shorter \pojem{lok}) with the endpoints $P$ and $Q$. + +Thus, each tether $PQ$ on a circle $k$ determines two arcs. Assume that the center $S$ does not lie on the edge of the plane that generates the circular arc. +If this plane +contains the center $S$ of the circle $k$, +it is a \pojem{veliki lok} $PQ$, otherwise it is a \pojem{mali +lok} $PQ$. + But if + the center $S$ is on the edge $PQ$ of the plane, then each of the two arcs $PQ$ + \index{polkrožnica}\pojem{polkrožnica} $PQ$ (Figure +\ref{sl.skk.4.2.1.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.1.pic} +\caption{} \label{sl.skk.4.2.1.pic} +\end{figure} + +Since the locus is not uniquely determined by its endpoints, we must also know +at least one point on the circle that belongs or does not belong to this locus. + +In a similar way, we define certain parts of the circle. + +Let $P$ and $Q$ be arbitrary points on the circle $k(S, r)$. The intersection +of the circle $\mathcal{K}(S, r)$ with one of the planes (in the plane of this circle) with the edge +$s=PQ$ is called +\index{krožni!odsek} \pojem{circular segment}. + So each chord $PQ$ on some circle $k(S, r)$ determines on the circle $\mathcal{K}(S, r)$ two circular segments. Assume that the center $S$ does not lie on the edge of the plane that generates the circular segment. +If this plane +contains the center $S$ of the circle $k$, +it is a \pojem{major circular segment} $PQ$, otherwise it is a \pojem{minor +circular segment} $PQ$. + If the + center $S$ is on the edge $PQ$ of the plane, then each of the two circular segments +\index{polkrog}\pojem{arc} + is an \pojem{arc}. + From the definition it is clear that the edge of the circular segment is the union of the chord $PQ$ and the corresponding arc (Figure +\ref{sl.skk.4.2.1b.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.1b.pic} +\caption{} \label{sl.skk.4.2.1b.pic} +\end{figure} + + +We will also define one more term related to the circle. +Let $P$ and $Q$ be arbitrary points on the circle $k(S, r)$. The intersection +of the circle $\mathcal{K}(S, r)$ with one of the angles $PSQ$ is called +\index{krožni!izsek} \pojem{circular sector}. Again we have two circular sectors. If the angle $PSQ$ is an obtuse angle, we get two arcs, otherwise a convex and a concave circular sector, depending on whether the angle $PSQ$ is convex or concave (Figure +\ref{sl.skk.4.2.1c.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.1c.pic} +\caption{} \label{sl.skk.4.2.1c.pic} +\end{figure} + + + + + +%________________________________________________________________________________ + \poglavje{Continuity Axiom} \label{odd2AKSZVE} + +We have already learned in elementary school, when introducing the numerical line and the coordinate system, that it is possible to establish a connection in which each point of a line corresponds to a certain real number, and vice versa, each real number can be assigned a point that lies on that line. This is related to the following axiom. + +\baksiom \label{aksDed}\index{aksiom!Dedekind's} +(Dedekind's\footnote{\index{Dedekind, R.} +\textit{R. Dedekind} (1831--1916), +German mathematician.} +axiom) +Suppose that all points on open line segment $AB$ are divided into the union of two nonempty disjoint sets $\mathcal{U}$ and +$\mathcal{V}$ such that no point of $\mathcal{U}$ is +between two points of $\mathcal{V}$ and vice versa: no point of $\mathcal{V}$ is +between two points of $\mathcal{U}$. Then there is a unique point $C$ on open line segment $AB$ such that +$B(A',C,B')$ for any two points $A'\in +\mathcal{U}\setminus{C}$ and $B'\in \mathcal{V} \setminus {C}$ +(Figure \ref{sl.aks.2.4.1.pic}). +\eaksiom + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.4.1.pic} +\caption{} \label{sl.aks.2.4.1.pic} +\end{figure} + +Let us state without proof two important consequences of the axiom +of continuity\footnote{Until the 19th century, mathematicians did not +feel the need to prove these two statements, or the need to +introduce the axiom of continuity. Even \index{Evklid} Euclid from +Alexandria (3rd century BC) in his famous work +'Elements', does not prove that a certain circle intersects.}. + + +\bizrek \label{DedPoslKrozPrem} Let $k$ be a circle and $P$ a point inside that circle. +Then any line $p$ passing through the point $P$ and the circle $k$ has exactly two common points (Figure +\ref{sl.aks.2.4.2.pic}). +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.4.2.pic} +\caption{} \label{sl.aks.2.4.2.pic} +\end{figure} + + + +\bizrek \label{DedPoslKrozKroz} If $k$ and $l$ are circles such that $l$ contains at least one point inside and one point outside $k$, +then the circles has exactly two points (Figure \ref{sl.aks.2.4.3.pic}). +\eizrek + +\poglavje{The basics of Geometry} \label{osn9Geom} +We have already learned in elementary school, when introducing the numerical line and the coordinate system, that it is possible to establish a connection in which each point of a line corresponds to a certain real number, and vice versa, each real number can be assigned a point that lies on that line. This is related to the following axiom. + +\baksiom \label{aksDed}\index{aksiom!Dedekind's} +(Dedekind's\footnote{\index{Dedekind, R.} +\textit{R. Dedekind} (1831--1916), +German mathematician.} +axiom) +Suppose that all points on open line segment $AB$ are divided into the union of two nonempty disjoint sets $\mathcal{U}$ and +$\mathcal{V}$ such that no point of $\mathcal{U}$ is +between two points of $\mathcal{V}$ and vice versa: no point of $\mathcal{V}$ is +between two points of $\mathcal{U}$. Then there is a unique point $C$ on open line segment $AB$ such that +$B(A',C,B')$ for any two points $A'\in +\mathcal{U}\setminus{C}$ and $B'\in \mathcal{V} \setminus {C}$ +(Figure \ref{sl.aks.2.4.1.pic}). +\eaksiom + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.4.1.pic} +\caption{} \label{sl.aks.2.4.1.pic} +\end{figure} + +Let us state without proof two important consequences of the axiom +of continuity\footnote{Until the 19th century, mathematicians did not +feel the need to prove these two statements, or the need to +introduce the axiom of continuity. Even \index{Evklid} Euclid from +Alexandria (3rd century BC) in his famous work +'Elements', does not prove that a certain circle intersects.}. + + +\bizrek \label{DedPoslKrozPrem} Let $k$ be a circle and $P$ a point inside that circle. +Then any line $p$ passing through the point $P$ and the circle $k$ has exactly two common points (Figure +\ref{sl.aks.2.4.2.pic}). +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.4.2.pic} +\caption{} \label{sl.aks.2.4.2.pic} +\end{figure} + + + +\bizrek \label{DedPoslKrozKroz} If $k$ and $l$ are circles such that $l$ contains at least one point inside and one point outside $k$, +then the circles has exactly two points (Figure \ref{sl.aks.2.4.3.pic}). +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.4.3.pic} +\caption{} \label{sl.aks.2.4.3.pic} +\end{figure} + +Dedekind's axiom is used in a few different ways to establish the set of real numbers. This is reminiscent of the already mentioned connection between the set of points on a line and the set of real numbers. + +We have already defined the multiplication operation of line segment $AB$ by any positive rational number $q$. Now we can extend the concept of multiplication to any positive real number $\lambda$. The definition of line segment $\lambda\cdot AB$, ($\lambda \in \mathbb{R}^+$), which we will not formally derive here, is associated with two sets of points on line segment $CD$: + \begin{eqnarray*} +&& \{X\in CD;\hspace*{1mm}CX=q\cdot +AB,\hspace*{1mm}q\leq\lambda,\hspace*{1mm}q\in +\mathbb{Q}^+ \} \hspace*{1mm}\textrm{ in}\\ +&& \{X\in CD;\hspace*{1mm}CX=q\cdot +AB,\hspace*{1mm}q>\lambda,\hspace*{1mm}q\in \mathbb{Q}^+ \} + \end{eqnarray*} + and +Dedekind's axiom \ref{aksDed}. + + With the help of the axiom of continuity \ref{aksDed} we can also introduce the concepts of measuring line segments and angles. + + When measuring line segments we will assign each line segment $AB$ a positive real number $\textsl{m}(AB)$ in the following way. + Let $\mathcal{D}$ be the set of all line segments and $\mathbb{R}^+$ be the set of all positive real numbers. The mapping $\textsl{m}:\mathcal{D}\rightarrow\mathbb{R}^+$, which satisfies the following properties: + \begin{itemize} + \item $(\exists A_0B_0\in\mathcal{D})\hspace*{1mm}\textsl{m}(A_0B_0)=1$, + \item $(\forall AB, CD\in\mathcal{D})\hspace*{1mm}(AB\cong CD \Rightarrow\textsl{m}(AB)=\textsl{m}(CD))$, + \item $(\forall AB, CD, EF\in\mathcal{D})\hspace*{1mm}(AB+CD=EF\Rightarrow \textsl{m}(AB)+\textsl{m}(CD)=\textsl{m}(EF))$, + \end{itemize} + is called the \index{dolžina!daljice}\pojem{length of a line segment} or the \index{mera!daljice}\pojem{measure of a line segment}, and the triple $\textsl{M}=(\mathcal{D},\mathbb{R}^+,\textsl{m})$ is called the \index{sistem merjenja!daljic}\pojem{system of measuring line segments}. + +The length of the line $AB$ (or $\textsl{m}(AB)$) will usually be denoted by $|AB|$. + + It is intuitively clear that there are an infinite number of measuring systems, which depend on the choice of the unit length line $A_0B_0$ - the one that has a length of 1, or $\textsl{m}(A_0B_0)=1$. In one measuring system, the length of any line $AB$ is represented by a positive real number $x$, for which $AB=x\cdot A_0B_0$. So $\textsl{m}(AB)=x$ exactly when $AB=x\cdot A_0B_0$ (Figure \ref{sl.aks.2.4.4.pic}). Now it is clear why we need the axiom of connectivity - without it, we would have problems with the definition of the length of the diagonal of a square with a unit length side (of length 1)\footnote{The ancient Greeks always represented the length as a rational number, so they called the side +and the diagonal of the square \pojem{incomparable lines}.}. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.4.4.pic} +\caption{} \label{sl.aks.2.4.4.pic} +\end{figure} + + + \bizrek \label{meraDalj1} + For any measuring system of lengths it holds: + + (\textit{i}) $AB|AB|$. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.4.6.pic} +\caption{} \label{sl.aks.2.4.6.pic} +\end{figure} + +(\textit{ii}) Suppose that $AB\not\cong CD$. Without loss of generality, let $AB\alpha$ and $\gamma'>\beta$. The relations are a direct consequence of the proven inequality $\alpha+\beta=\gamma'$ from the previous \ref{zunanjiNotrNotr}. + \kdokaz + +Based on the internal angles, we can consider three types of triangles. + We have already proven + that in any triangle the sum of the internal angles is equal to $180^0$ + (the statement \ref{VsotKotTrik}). +This means that at most one of these angles is a right angle or a reflex angle, +or at least two are acute. So we have three possibilities (Figure +\ref{sl.aks.2.6.4a.pic}): +\begin{itemize} + \item The triangle is \index{trikotnik!ostrokotni} + \pojem{acute-angled}, if all of its internal angles are acute. + \item The triangle is \index{trikotnik!pravokotni}\pojem{right-angled}, + if it has one internal angle that is a right angle. The side opposite +the right angle of a right-angled triangle is called +\index{hipotenuza}\pojem{the hypotenuse}, the other two sides are +\index{kateta}\pojem{the catheti}. Because the sum of the internal angles in any + triangle is equal to $180^0$, the +angles at the hypotenuse of a right-angled triangle are complementary. + \item The triangle is \index{trikotnik!topokotni}\pojem{obtuse-angled}, + if it has one internal angle that is obtuse. + The other two internal angles are then acute. +\end{itemize} + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.6.4a.pic} +\caption{} \label{sl.aks.2.6.4a.pic} +\end{figure} + +%slika 33.1 + + We will now prove a statement that applies to any $n$-gon. + + + + \bizrek \label{VsotKotVeck} + The sum of the interior angles of any $n$-gon is equal to + $(n - 2) \cdot 180^0$. + \eizrek + + + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.5.5c.pic} +\caption{} \label{sl.aks.2.5.5c.pic} +\end{figure} + +\textbf{\textit{Proof.}} First, let's assume that $A_1A_2\ldots A_n$ is a convex $n $-gon (Figure \ref{sl.aks.2.5.5c.pic}). Its $n-3$ diagonals $A_1A_3$, $A_1A_4$, ... $A_1A_{n-1}$ divide this polygon into $n-2$ triangles $\triangle_1$, $\triangle_2$, ..., $\triangle_{n-2}$. Because each internal angle of the $n $-gon $A_1A_2\ldots A_n$ is divided into the appropriate internal angles of the aforementioned triangles, the sum of all internal angles + $n $-gon $A_1A_2\ldots A_n$ is equal to the sum of all angles of triangles $\triangle_1$, $\triangle_2$, ..., $\triangle_{n-2}$. By Theorem \ref{VsotKotTrik} in the end this sum is equal to exactly $(n - 2) \cdot 180^0$. + + At this point we will not prove the fact that even in the case of a non-convex $n $-gon $A_1A_2\ldots A_n$ it can be divided into $n-2$ triangles. + \kdokaz + + + \bizrek \label{VsotKotVeckZuna} + The sum of the exterior angles of any $n$-gon is equal to + $360^0$. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.5.5b.pic} +\caption{} \label{sl.aks.2.5.5b.pic} +\end{figure} + + \textbf{\textit{Proof.}} We mark with $\alpha_1$, $\alpha_2$,..., $\alpha_n$ the internal and $\alpha'_1$, $\alpha'_2$, ..., $\alpha'_n$ the corresponding external angles at the vertices $A_1$, $A_2$,...,$A_n$ of a convex $n $-gon $A_1A_2\ldots A_n$. (Figure \ref{sl.aks.2.5.5b.pic}). For the appropriate internal and external angle we have: + + \begin{eqnarray*} + & & \alpha_1+\alpha'_1=180^0\\ + & & \alpha_2+\alpha'_2=180^0\\ + & & \vdots\\ + & & \alpha_n+\alpha'_n=180^0 + \end{eqnarray*} + + If we add all the equalities and take into account the proven equality from the previous theorem \ref{VsotKotVeck} $$\alpha_1+\alpha_2+\cdots + \alpha_n=(n - 2) \cdot 180^0,$$ + we get $(n - 2) \cdot 180^0+\alpha'_1+\alpha'_2+\cdots + \alpha'_n=n\cdot 180^0$ or + $$\alpha'_1+\alpha'_2+\cdots + \alpha'_n=2 \cdot 180^0=360^0,$$ which was to be proven. \kdokaz + +From the statement \ref{VsotKotVeck} it follows that the sum of all the internal angles of an arbitrary quadrilateral is equal to $360^0$, and from the statement \ref{VsotKotVeckZuna} it also follows that the sum of all the external angles of a convex quadrilateral is equal to $360^0$ (Figure \ref{sl.aks.2.5.5a.pic}). + + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.5.5a.pic} +\caption{} \label{sl.aks.2.5.5a.pic} +\end{figure} + + + + Because a triangle (as the intersection of three planes) is a convex figure, the sum of all the external angles of an arbitrary triangle is equal to $360^0$ (Figure \ref{sl.aks.2.5.5d.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.5.5d.pic} +\caption{} \label{sl.aks.2.5.5d.pic} +\end{figure} + + +A similar claim, based on the statement \ref{KotaVzporKraki}, which refers to an angle with parallel sides, also applies to an angle with perpendicular sides. + + + + \bizrek \label{KotaPravokKraki} + Angles with perpendicular sides are either congruent or supplementary. + \index{kota!s pravokotnimi kraki} + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.5.5.pic} +\caption{} \label{sl.aks.2.5.5.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Let $\angle aSb$ and $\angle a'S'b'$ be such angles that $a\perp a'$ and $b\perp b'$ (Figure \ref{sl.aks.2.5.5.pic}). Let $A$ be the intersection of the sides of $a$ and $a'$, and $B$ be the intersection of the sides of $b$ and $b'$. In the quadrilateral $SAS'B$, the internal angles at the vertices $A$ and $B$ each measure $90^0$. According to the statement \ref{VsotKotVeck}, the sum of all the internal angles of this quadrilateral is equal to $360^0$. Therefore, the internal angle $BSA$ and $AS'B$ of this quadrilateral measure together $180^0$, which means that two angles, determined by the sides of $a$ and $b$ or $a'$ and $b'$, are supplementary. However, if we replace one of these angles with its supplement, the corresponding angles are congruent. + \kdokaz + + +%________________________________________________________________________________ +\naloge{Exercises} +\begin{enumerate} + +\item Let $P$, $Q$ and $R$ be the internal points of the sides of a triangle +$ABC$. Prove that $P$, $Q$ and $R$ are non-collinear. + +\item Let $P$ and $Q$ be points on sides $BC$ and $AC$ of triangle $ABC$ and at the same time different from its vertices. Prove that the line segments $AP$ and $BQ$ intersect in one point. + +\item Points $P$, $Q$ and $R$ are located in order on sides $BC$, $AC$ and $AB$ of triangle $ABC$ and are different from its vertices. Prove that the line segments $AP$ and $QR$ intersect in one point. + +\item A line $p$, which lies in the plane of the quadrilateral, intersects its diagonal $AC$ and does not pass through any vertex of this quadrilateral. Prove that the line $p$ intersects exactly two sides of this quadrilateral. + +\item Prove that the half-line is a convex figure. + +\item Prove that the intersection of two convex figures is a convex figure. + +\item Prove that any triangle is a convex figure. + +\item If $\mathcal{B}(A,B,C)$ and $\mathcal{B}(D,A,C)$, then $\mathcal{B}(B,A,D)$ is also true. Prove it. + +\item Let $A$, $B$, $C$ and $D$ be such collinear points that $\neg\mathcal{B}(B,A,C)$ and $\neg\mathcal{B}(B,A,D)$ are true. Prove that $\neg\mathcal{B}(C,A,D)$ is also true. + +\item Let $A_1A_2\ldots,A_{2k+1}$ be an arbitrary polygon with an even number of vertices. Prove that there is no line that intersects all of its sides. + +\item If izometry $\mathcal{I}$ maps figure $\Phi_1$ and $\Phi_2$ into figure $\Phi'_1$ and $\Phi'_2$, then the intersection $\Phi_1\cap\Phi_2$ is mapped by this izometry into the intersection $\Phi'_1\cap\Phi'_2$. Prove it. + +\item Prove that any two strips of a plane are compatible with each other. + +\item Prove that any two lines of a plane are compatible with each other. + + +\item Let $k$ and $k'$ be two circles of a plane with centers $O$ and $O'$ and radii $AB$ and $A'B'$. Prove the equivalence: $k\cong k' \Leftrightarrow AB\cong A'B'$. + +\item Let $\mathcal{I}$ be a non-identical izometry of a plane with two fixed points $A$ and $B$. If $p$ is a line of this plane, which is parallel to the line $AB$ and $A\notin p$. Prove that there are no fixed points of izometry $\mathcal{I}$ on the line $p$. + +\item Let $S$ be the only fixed point +of the isometry $\mathcal{I}$ in some plane. Prove that if this isometry +maps the line $p$ to itself, then $S\in p$. + +\item Prove that any two lines +of a plane either intersect or are parallel. + +\item If a +line in a plane intersects one of the two parallel lines of the same plane, + then it also intersects the other parallel line. Prove. + +\item Prove that every isometry maps a parallel line to a parallel line. + +\item Let $p$, $q$ and $r$ be such lines of a plane that +$p\parallel q$ and $r\perp p$. Prove that $r\perp q$. + +\item Prove that a convex $n$-gon cannot have more than three +acute angles. + +\end{enumerate} + + + +% DEL 3 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% SKLADNOST TRIKOTNIKOV. VEČKOTNIKI +%________________________________________________________________________________ + + \del{Congruence. Triangles and Polygons} \label{pogSKL} + +%________________________________________________________________________________ + \poglavje{Triangle Congruence Theorems} \label{odd3IzrSkl} + +From the general definition of the congruence of figures it follows that two +triangles are congruent if there exists an isometry that maps the first triangle to +the second. It is clear that from the congruence of two triangles follows the +congruence of the corresponding sides and interior angles. We are interested in the inverse problem: +When does the congruence of some of +the corresponding sides and angles imply the congruence of two triangles? This is +expressed by the following \index{statement!about the congruence +of triangles}\emph{triangle congruence theorems}\footnote{The first, third and fourth triangle congruence theorems are attributed to \index{Pythagoras} \textit{Pythagoras of Samos} (6th century BC), while the second theorem is assumed to have been known to \index{Thales} \textit{Thales of Miletus} (7th--6th century BC). All four +are mentioned by \index{Euclid} \textit{Euclid of Alexandria} +(3rd century BC) in the first book of his \textit{Elements}.}: + +\bizrek \label{SSS} (\textit{SSS}) + Triangles are congruent if three sides of one triangle are congruent + to the corresponding sides of the other triangle, + i.e. (Figure + \ref{sl.skl.3.1.1.pic}): + \begin{eqnarray*} + \left. + \begin{array}{l} + AB \cong A'B'\\ + BC \cong B'C'\\ + AC \cong A'C' + \end{array} + \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC \cong \triangle A'B'C' + \end{eqnarray*} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.1.pic} +\caption{} \label{sl.skl.3.1.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} + From the congruence of sides by \ref{izrek(A,B)} there is also $(A,B) + \cong (A',B')$, $(B,C) \cong (B',C')$ + and $(A,C) + \cong (A',C')$ or $(A,B,C) \cong (A',B',C')$. From + \ref{IizrekABC} it follows that there is an isometry $\mathcal{I}$, which + maps points $A$, $B$ and $C$ to points $A'$, $B'$ and $C'$. + It maps triangle $ABC$ to triangle $A'B'C'$, which is + a consequence of \ref{izrekIzoB}. So triangles $ABC$ and + $A'B'C'$ are congruent. + \kdokaz + +\bizrek \label{SKS} (\textit{SAS}) + Triangles are congruent if two pairs of sides and the included angle of one triangle + are congruent to the corresponding sides and angle of the other triangle, + i.e. (Figure + \ref{sl.skl.3.1.2.pic}): + \begin{eqnarray*} + \left. + \begin{array}{l} + AB \cong A'B'\\ + AC \cong A'C'\\ + \angle BAC \cong \angle B'A'C' + \end{array} + \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC \cong \triangle A'B'C' + \end{eqnarray*} + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.2.pic} +\caption{} \label{sl.skl.3.1.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Because angles $BAC$ and $B'A'C'$ are congruent, + there exists an isometry $\mathcal{I}$, which maps angle $BAC$ to angle $B'A'C'$. + This isometry maps point $A$ to point $A'$, and segments $AB$ and $AC$ to $A'B'$ and $A'C'$. + Let $\mathcal{I}(B)=\widehat{B}'$ and $\mathcal{I}(C)=\widehat{C}'$. + From this + it follows that $AB \cong A'\widehat{B}'$ and $AC \cong A'\widehat{C}'$, + but according to \ref{ABnaPoltrakCX} we have that $\widehat{B}'=B'$ and + $\widehat{C}'=C'$. + So the isometry $\mathcal{I}$ + maps points $A$, $B$ and $C$ to points $A'$, $B'$ and $C'$, + or triangle $ABC$ to triangle $A'B'C'$, which means that triangles $ABC$ and + $A'B'C'$ are congruent. +\kdokaz + +\bizrek \label{KSK} (\textit{ASA}) + Triangles are congruent if two pairs of angles and the included side of one triangle + are congruent to the corresponding angles and side of the other triangle + i.e. (Figure + \ref{sl.skl.3.1.3.pic}): + \begin{eqnarray*} + \left. + \begin{array}{l} + AB \cong A'B'\\ + \angle BAC \cong \angle B'A'C'\\ + \angle ABC \cong \angle A'B'C' + \end{array} + \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC \cong \triangle A'B'C' + \end{eqnarray*} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.3.pic} +\caption{} \label{sl.skl.3.1.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} + From axiom \ref{aksIII2} it follows that there exists an isometry $\mathcal{I}$, + that maps + point $A$ to point $A'$, line segment $AB$ to line segment $A'B'$ and + plane $ABC$ to plane $A'B'C'$. + Because $AB \cong A'B'$, from the same axiom it follows + $\mathcal{I}(B)=B'$. Let $\mathcal{I}(C)=\widehat{C}'$. + Then $\angle BAC \cong \angle B'A'\widehat{C}'$ and + $\angle ABC \cong \angle A'B'\widehat{C}'$. + Because according to the assumption also $\angle BAC \cong \angle B'A'C'$ and + $\angle ABC \cong \angle A'B'C'$, from + \ref{KotNaPoltrak} it follows that line segments $A'\widehat{C}'$ and + $A'\widehat{C}'$ (or $B'C'$ and + $B'\widehat{C}'$) are equal. Therefore $\widehat{C}'=A'\widehat{C}'\cap + A'\widehat{C}'=A'C'\cap + A'C'=C'$. So $\mathcal{I}:A,B,C \mapsto A',B',C'$, therefore triangles $ABC$ and $A'B'C'$ are congruent. + \kdokaz + +We will omit the proof of the fourth statement about congruent triangles. + +\bizrek \label{SSK} (\textit{SSA}) + Triangles are congruent if two pairs of sides and the angle opposite to the longer side of one triangle + are congruent to the corresponding sides and angle of the other triangle. + (Figure \ref{sl.skl.3.1.4.pic}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.4.pic} +\caption{} \label{sl.skl.3.1.4.pic} +\end{figure} + + +One of the most important consequences of theorems about the congruence of triangles is the following statement. + + + + \bizrek \label{enakokraki} + If two sides of a triangle are congruent, then angles opposite those sides are congruent.\\ + And vice versa:\\ + If angles opposite those sides are congruent, then two sides of a triangle are congruent. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.5.pic} +\caption{} \label{sl.skl.3.1.5.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + Let $ABC$ be such a triangle that $AB \cong AC$ + (Figure \ref{sl.skl.3.1.5.pic}). Because $AC \cong AB$ and $BC \cong CB$ still hold, from the \textit{SSS} theorem it follows that +the triangles $ABC$ and $ACB$ are congruent (these two triangles have +different orientations). Therefore, $\angle ABC \cong \angle CBA$. In the same way, we could have proven the converse statement. In this case, we would have used the \textit{ASA} theorem. + \kdokaz + + A triangle (as is the triangle $ABC$ from the previous theorem), + which has at least two +sides congruent, is called +\index{trikotnik!enakokraki}\pojem{an isosceles triangle}. Each of the +two congruent sides is a \index{krak!enakokrakega +trikotnika}\pojem{arm}, and the third side is +\index{osnovnica!enakokrakega trikotnika}\pojem{the base} of this +triangle. So, according to the previous theorem, the angles at the +base of an isosceles triangle are congruent. And vice versa - if two +angles of a triangle are congruent, then this triangle is isosceles. + +\bzgled + Let the $E$ and $F$ be a points lies on the line containing the hypotenuse $AB$ of a perpendicular + triangle $ABC$ and let $B(E,A,B,F)$, $EA\cong AC$ and + $FB\cong BC$. What is the measure of the angle $ACB$? + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.6.pic} +\caption{} \label{sl.skl.3.1.6.pic} +\end{figure} + + + \textbf{\textit{Solution.}} (Figure \ref{sl.skl.3.1.6.pic}) + + The internal angle of the triangle $ABC$ at the + vertices + $A$ and $B$ we denote by $\alpha$ and + $\beta$. + The triangle $EAC$ and $CBF$ are + isosceles triangles with the bases $CE$ and $CF$, so $\angle CEA \cong \angle ACE$ and + $\angle CFB \cong \angle BFC$ (statement \ref{enakokraki}). The + $\alpha$ is the external angle of the triangle $EAC$, so by the statement \ref{zunanjiNotrNotr}: + $\alpha = 2\angle ECA$ or $\angle ECA = \frac{1}{2} \alpha$. + Similarly, $\angle FCB = \frac{1}{2} \beta$. Therefore: +\begin{eqnarray*} + \angle ECF&=&\angle ECA+\angle ACB+\angle BCF=\\ + &=&\frac{1}{2} + \cdot\alpha+90^0+ + \frac{1}{2} \cdot\beta=90^0+ + \frac{1}{2} \cdot\left(\alpha+\beta\right)=\\ + &=&90^0+ + \frac{1}{2}\cdot 90^0=135^0, + \end{eqnarray*} +or $\angle ECF=135^0$. \kdokaz + +A triangle in which all sides are equal is called +\index{trikotnik!enakostranični}\pojem{an equilateral triangle} +(Figure \ref{sl.skl.3.1.7.pic}), which is a special case +of an isosceles triangle. Therefore, from the above statement it follows that +all the internal angles of an equilateral triangle are equal. Since their sum is equal to $180^0$, each of these angles measures $60^0$. The converse is also true: if at least two angles of a triangle are equal to $60^0$ (and therefore also the third one), then this triangle is equilateral. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.7.pic} +\caption{} \label{sl.skl.3.1.7.pic} +\end{figure} + +\bzgled +Let $ABC$ be an equilateral triangle and $P$, $Q$ and $R$ points such that $\mathcal{B}(A,B,R)$, $\mathcal{B}(B,C,Q)$, $\mathcal{B}(C,A,P)$ and +$BR\cong CQ\cong AP$. Prove that $PQR$ is also an equilateral triangle. +\ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.8.pic} +\caption{} \label{sl.skl.3.1.8.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.1.8.pic}) + + From the given conditions, it follows first that + $AR\cong BQ\cong CP$. The triangle $ABC$ is an equilateral triangle, + so all three internal angles measure $60^0$. From this it follows + that $\angle PAR\cong \angle RBQ\cong QCP$. By the \textit{SAS} theorem, + the triangles $PAR$, $RBQ$ and $QCP$ are similar and $PR\cong RQ\cong QP$. + This means that $PQR$ is also an equilateral triangle. + \kdokaz + + +In the previous chapter we defined the perpendicular bisector of a line segment as the line that is perpendicular to the line segment and goes through its center. Now we prove an equivalent definition of the perpendicular bisector, which will be very important in the sequel. \index{perpendicular bisector} + + + \bizrek \label{simetrala} + The perpendicular bisector of a line segment $AB$ is the set of all points $X$ + that are equidistant from its endpoints, i.e. $AX \cong BX$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.9.pic} +\caption{} \label{sl.skl.3.1.9.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Let $s$ be a line that is the perpendicular bisector of the line +segment $AB$ in some plane. By definition, $s$ is perpendicular to the line +$AB$ through its center - the point $S$. We denote by $\mathcal{M}$ +the set of all points $X$ in this plane, for which $AX \cong BX$. +It is necessary to prove that $s =\mathcal{M}$. We will prove this by two inclusions (Figure \ref{sl.skl.3.1.9.pic}). + +($s\subseteq \mathcal{M}$). Let $X \in s$. We will prove that +then $X \in \mathcal{M}$ is true. From the relations $AS \cong BS$, $XS \cong XS$ +and $\angle ASX \cong \angle BSX = 90^0$ it follows that the triangles +$ASX$ and $BSX$ are congruent (theorem \ref{SAS}). Therefore $AX \cong BX$ +or $X \in \mathcal{M}$. + +($\mathcal{M}\subseteq s$). Now let $X \in \mathcal{M}$. +We will prove that $X \in s$ is true. From $X \in \mathcal{M}$ it follows that $AX \cong BX$. +Now from $AS \cong BS$, $XS \cong XS$ and $AX \cong BX$ it follows that +the triangles $ASX$ and $BSX$ are congruent (theorem \ref{SSS}). Therefore the angles +$ASX$ and $BSX$ are congruent and the angle between the lines is both a right angle. This means that +the line $XS$ is perpendicular to the line segment $AB$ in its center. +Therefore, the line $XS$ is the perpendicular bisector of $s$ or $X \in s$. + \kdokaz + +The next problem is an example of multiple use of the theorem of an isosceles triangle (\ref{enakokraki}). + + \bnaloga\footnote{42. IMO USA - 2001, Problem 5.} + In a triangle $ABC$, let $AP$ bisect $\angle BAC$, with $P$ on $BC$, and let $BQ$ bisect $\angle ABC$, with $Q$ on $CA$. +It is known that $\angle BAC=60^0$ and that $|AB|+|BP|=|AQ|+|QB|$. +What are the possible measures of interior angles of triangle $ABC$? + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.9.IMO1.pic} +\caption{} \label{sl.skk.4.9.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let's mark the interior angles of triangle $ABC$ with +$\alpha=60^0$, $\beta$ and $\gamma$. Let $D$ and $E$ be such +points that $BD\cong BP$, $\mathcal{B}(A,B,D)$, $QE \cong QB$ +and $\mathcal{B}(A,Q,E)$ (Figure \ref{sl.skk.4.9.IMO1.pic}). From these +conditions it follows that $DBP$ and $BQE$ are isosceles triangles with +bases $DP$ and $BE$. From the given condition $|AB|+|BP|=|AQ|+|QB|$ +it also follows that $AD\cong AE$, which means that $ADE$ is also an isosceles triangle with base $DE$. + +Since $DBP$ is an isosceles triangle, from the statements \ref{enakokraki} and \ref{zunanjiNotrNotr} it follows: $\angle BDP\cong \angle BPD =\frac{1}{2}\angle ABC=\frac{1}{2}\beta$. +Since $BQE$ is also an isosceles triangle, $\angle QBE\cong\angle QEB$ follows. +From the congruence of triangles $ADP$ and $AEP$ (the statement \textit{SAS} \ref{SKS}) it follows that $\angle ADP\cong \angle AEP$ and $PD\cong PE$. + +If we connect the proven relations, it holds: + \begin{eqnarray*} +&& \angle AEP\cong \angle BDP +=\frac{1}{2}\beta\cong \angle QBP\\ +&&\textrm{ and } \angle AEB\cong\angle QBE + \end{eqnarray*} + + Let's first assume that $QB>QC$ or $\mathcal{B}(Q,C,E)$ holds. In this case: + \begin{eqnarray*} + \angle PEB &=&\angle AEB-\angle AEP=\\ + &=&\angle QBE-\angle QBP=\\ + &=&\angle PBE. + \end{eqnarray*} +This means that $PBE$ is an isosceles triangle with the base $BE$ or $PE\cong PB$. But from the already proven $PE\cong PD$ and the assumption $PB\cong BD$ it follows that $PD\cong PB\cong BD$, therefore $BDP$ is an equilateral triangle. From this it follows that $\beta=2\angle BDP=2\cdot 60^0=120^0$, or $\alpha+\beta=60^0+120^0=180^0$, which is not possible (the statement \ref{VsotKotTrik}). Therefore the relation $QB>QC$ is not possible. + +In a similar way, the relation $QB>QC$ leads to a contradiction. This means that only $QB\cong QC$ is possible. In this case $C=E$ and $\gamma=\angle ACB=\angle AEB\cong AEP=\frac{1}{2}\beta$ holds. From $\alpha+\beta+\gamma=180^0$, it follows that $60^0+\beta+\frac{1}{2}\beta=180^0$ or $\beta=80^0$. + +We have shown that from the conditions of the task it follows that $\beta=80^0$. So the only possible solution is $\beta=80^0$. It is still necessary to show that $\beta=80^0$ is a solution, or that from $\alpha=60^0$, $\beta=80^0$ it follows that $|AB|+|BP|=|AQ|+|QB|$. +First, from $\angle QCB=\gamma=\frac{1}{2}\beta=40^0=\angle QBC$ it follows + (from the statement \ref{enakokraki}) $QC\cong QB$ or + $|AQ|+|QB|=|AQ|+|QC|=|AC|$. +If we define the point $D$ in the same way as in the first part, we again +get $\angle ADP\cong \angle BPD=\frac{1}{2}\angle ABC=40^0=\angle ACB=\angle ACP$. +This means that the triangles +$ADP$ and $ACP$ are congruent (from the statement \textit{ASA} \ref{KSK}) or +$AD\cong AC$. Therefore, in the end: + $$|AB|+|BP|=|AB|+|BD|=|AD|=|AC|=|AQ|+|QB|,$$ which had to be proven. \kdokaz + + + +%________________________________________________________________________________ + \poglavje{Constructions in Geometry} \label{odd3NacrtNaloge} + +At the next example we will describe the so-called design tasks. + + + \bzgled \label{načrt1odd3} + Two congruent line segments $AB$ and $A'B'$ in a plane are given. + Construct a point $C$ such that $\triangle ABC \cong \triangle A'B'C$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.10.pic} +\caption{} \label{sl.skl.3.1.10.pic} +\end{figure} + + \textbf{\textit{Solution.}} We assume that $C$ is a point in the plane of the segments, for + which $\triangle ABC \cong \triangle A'B'C$. Then $AC\cong A'C$ + and $BC\cong B'C$ or the point $C$ lies on the perpendiculars of the segments $AA'$ + and $BB'$ (from the statement \ref{simetrala}). This fact allows us to construct + (Figure \ref{sl.skl.3.1.10.pic}). + + We draw the perpendiculars of the segments $AA'$ + and $BB'$. We get the point $C$ as their intersection. + + We prove that $C$ is the desired point or that it satisfies the conditions of + the task. By assumption, $AB\cong A'B'$ already. Because we got the point $C$ as + the intersection of the perpendiculars of the segments $AA'$ + and $BB'$, it is $AC\cong A'C$ and $BC\cong B'C$. From the statement \ref{SSS} (SSS) it follows that the triangles $ABC$ and $A'B'C$ + are congruent. + +The task has a solution (one) exactly when the lines of symmetry + $AA'$ + and $BB'$ intersect, or when the lines $AA'$ + and $BB'$ are not parallel. + \kdokaz + + The previous example is therefore called the \index{task!design} + \pojem{design task}, in which + for the given data it is necessary to plan or construct a new element or figure, + which in relation to the given data + satisfies certain conditions. The \pojem{planning} or \index{construction} + \pojem{construction} + means + the use of a ruler and a compass or the use of + \index{construction!elementary} + \pojem{elementary construction}:\label{elementarneKonstrukcije} +\begin{itemize} + \item for the given points $A$ and $B$ we draw: + \begin{itemize} + \item the line $AB$, + \item the line segment $AB$, + \item the midpoint $AB$; +\end{itemize} + \item we draw the circle $k$: +\begin{itemize} + \item with the center $S$, which goes through the given point $A$, + \item with the center $S$ and the radius, which is consistent with the given + line segment; +\end{itemize} +\item we draw the circular arc with the given center and radius, +\item we draw the intersection (or intersections): +\begin{itemize} + \item of two lines, + \item of a line and a circle, + \item of two circles. +\end{itemize} +\end{itemize} + + The solution to the design task (draw the figure $\Phi$, which satisfies the conditions + $\mathcal{A}$) is formally composed of four steps: +\begin{itemize} + \item \textit{analysis} - in which we assume that + the figure $\Phi$ is already designed and satisfies the conditions $\mathcal{A}$, then + we look for new conditions $\mathcal{B}$, which the figure satisfies. + These follow from the conditions $\mathcal{A}$ and are more favorable for + the construction of the figure $\Phi$. We prove + $\mathcal{A}\Rightarrow \mathcal{B}$. + \item \textit{construction} - we plan the figure $\Phi'$, which satisfies + the conditions + $\mathcal{B}$. We exactly describe the course of the design. + \item \textit{proof} - we prove that $\Phi' = \Phi$ or + $\mathcal{B}\Rightarrow \mathcal{A}$. + \item \textit{discussion} - we investigate + the number of task solutions, depending on the conditions $\mathcal{A}$. +\end{itemize} + +\bzgled \label{konstrTrik1} +Line segments $a$, $l$ and an angle $\alpha$ are given. Construct a triangle + $ABC$, such that the side $BC$ is congruent to the line segment $a$, the sum of the sides $AB + AC$ + equal to the line segment $l$ and the interior angle $BAC$ congruent to the angle $\alpha$ ($a$, $b+c$, $\alpha$). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.10a.pic} +\caption{} \label{sl.skl.3.1.10a.pic} +\end{figure} + + + \textbf{\textit{Analysis.}} Let $ABC$ be a triangle, where $BC \cong a$, the sum of $AB+AC$ is equal to the given line segment $l$ and +$\angle BAC\cong \alpha$ (Figure \ref{sl.skl.3.1.10a.pic}). Let $D$ be a point on the line segment $BA$, such that $AD\cong AC$ and points $B$ and $D$ are on different sides +of point $A$. Therefore $BD=BA+AD=AB+AC=l$ or $BD\cong l$. Triangle $ACD$ is isosceles, so the angles $ADC$ and $ACD$ +are congruent by Theorem \ref{enakokraki}. Because they are also the interior +angles of triangle $CAD$, both are equal to half of the exterior angle $BAC$ of this triangle (Theorem \ref{zunanjiNotrNotr}), or + $\angle BDC=\angle ADC\cong \angle ACD=\frac{1}{2}\angle BAC=\frac{1}{2}\alpha$. + This allows us +to construct triangle $BCD$. + +\textbf{\textit{Construction.}} First, let's construct triangle $BCD$, where +$\angle BDC=\frac{1}{2}\alpha$, + $BC\cong a$ and $BD\cong l$, +then point $A$ as the intersection of the line segment $CD$'s perpendicular with line segment $BD$. We will prove that $ABC$ +is the desired triangle. + +\textbf{\textit{Proof.}} First, $BC\cong a$ by construction. By construction, point $A$ lies on the perpendicular of line +segment $CD$, so $AD\cong AC$ (Theorem \ref{simetrala}). Therefore triangle $CAD$ is isosceles with the base $CD$, so it is (Theorem \ref{enakokraki}) also $\angle ADC\cong \angle ACD$. Because of this, (Theorem \ref{zunanjiNotrNotr}) $\angle BAC = 2 \cdot \angle BDC= 2\cdot\frac{1}{2}\alpha=\alpha$. From $AD\cong AC$ it follows + $AB + AC = AB + AD = BD \cong l$ +. + +\textbf{\textit{Discussion.}} The task has a solution (namely one or two) exactly when +the line segment $DC$ intersects the circle $k(B,a)$ +and the perpendicular of line segment $CD$ intersects line segment $BD$. + \kdokaz + +In the future, we will not carry out all the steps in every design task. +In most cases, we will only do the first step and thus indicate +the course of the solution. + + + + +%________________________________________________________________________________ + \poglavje{Triangle inequality} \label{odd3NeenTrik} + +First, we will prove two important expressions that are a consequence +of theorems about the congruence of triangles. + + \bizrek \label{vecstrveckot} + One side of a triangle is longer than another side of a triangle if and only if + the measure of the angle opposite the longer side is greater than the angle opposite the shorter side. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.1.pic} +\caption{} \label{sl.skl.3.2.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} Let $ABC$ be a triangle in which $AC > AB$ (Figure \ref{sl.skl.3.2.1.pic}). + We prove that then $\angle ABC > \angle ACB$. Because $AC > AB$, there is +such a point $B'$ between points $A$ and $C$, for which $AB \cong AB'$. +Then the triangle $BAB’$ is isosceles and $\angle ABB' \cong \angle AB'B$ (theorem \ref{enakokraki}). +The segment $BB'$ is inside the angle $ABC$, so $\angle ABC > \angle ABB '$. +Then $\angle AB'B$ is the external angle of the triangle $BCB'$. By + theorem \ref{zunanjiNotrNotrVecji} this angle is greater than its adjacent internal +angle $B'CB$. If we use what has been proven so far, we get: + $$\angle ABC > +\angle ABB ' \cong \angle AB'B > \angle B'CB \cong \angle ACB.$$ + Therefore $\angle ABC +> \angle ACB$. In a similar way, we prove that the converse is also true. +\kdokaz + + If we denote the lengths of the sides $BC$, $AC$ and + $AB$ of the triangle $ABC$ with $a$, $b$ and $c$, and the measures of the opposite angles at the vertices $A$, $B$ and $C$ with $\alpha$, $\beta$ and $\gamma$, we can write the previous theorem + in the form: + $$a > b \Leftrightarrow \alpha > \beta,$$ + the theorem about the isosceles triangle \ref{enakokraki} in the form: + $$a = b \Leftrightarrow \alpha = \beta.$$ + + + This means that both expressions $a-b$ and $\alpha-\beta$ are both positive, + both negative or both equal to zero. Thus we have proven the following property: + +\bzgled \label{vecstrveckotAlgeb} + For each triangle $ABC$ is: + $$(a-b)(\alpha-\beta)\geq 0, \hspace*{4mm} + (b-c)(\beta-\gamma)\geq 0, \hspace*{4mm} + (c-a)(\gamma-\alpha)\geq 0$$ + \ezgled + +The following proposition follows directly from +\ref{vecstrveckot}: + + + \bizrek \label{vecstrveckotHipot} + The hypotenuse of a right-angled triangle is longer than its + two legs. + The longest side of an obtuse triangle is the one opposite to the obtuse angle. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.2.pic} +\caption{} \label{sl.skl.3.2.2.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.2.2.pic}) + + The sum of the inner angles of a triangle is equal to $180^0$. Therefore, in +a right-angled triangle, the largest angle is a right angle. The hypotenuse of a right-angled triangle is the longest side of this triangle according to the previous proposition. We can similarly prove this for an obtuse triangle. + \kdokaz + +A line $AA'$ is the \index{height!of a triangle}\pojem{height} of a triangle $ABC$, if $AA'\perp BC$ and $A'\in BC$. The latter of the two relations means that the point $A'$ lies on the line $BC$, but not necessarily on the line segment $BC$. The relation $\mathcal{B}(B,A',C)$ is valid exactly when both of the internal angles at the vertices $B$ and $C$ are acute (Figure \ref{sl.skl.3.2.3.pic}). This is a consequence of the theorem about the sum of the internal angles of any triangle (theorem \ref{VsotKotTrik}). In the case that $\angle ABC\geq 90^0$ and $\mathcal{B}(B,A',C)$, the sum of the internal angles in the triangle $ABA'$ would be greater than $180^0$. So the height of a triangle is not always inside the triangle. In a right triangle, the heights from the two vertices with acute angles are equal to the corresponding catheti. The heights from the vertices $A$, $B$ and $C$ are usually denoted by $v_a$, $v_b$ and $v_c$. From the previous theorem \ref{vecstrveckotHipot} it follows that the length of the height of any triangle is less than or equal to the length of the opposite side of that triangle, e.g.: $v_a\leq b$, $v_a\leq c$, ... + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.3.pic} +\caption{} \label{sl.skl.3.2.3.pic} +\end{figure} + +Now we will solve a design problem in which the height of a triangle is given as data. + +\bzgled + Construct a triangle $ABC$ such that the sides $AB$, + $AC$ and the altitude from the vertex $B$ are congruent to the three given line segments $c$, $b$ and $v_b$. + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.4a.pic} +\caption{} \label{sl.skl.3.2.4a.pic} +\end{figure} + +\textbf{\textit{Analysis.}} Let $ABC$ be a triangle for which $AB\cong c$, + $AC\cong b$ and $AD\cong v_b$ (where $BD$ is the height of this triangle from the vertex $B$). In the right triangle $ABD$ +therefore the known hypotenuse $AB\cong c$ and the cathetus +$AD\cong v_b$ are known, which means that we can design it. The third vertex $C$ of the triangle $ABC$ lies on the line $AD$ (Figure \ref{sl.skl.3.2.4a.pic}). + +\textbf{\textit{Construction.}} Let's first draw a rectangular triangle +$ABD$ (with conditions: $AB \cong c$, $\angle ADB = 90^0$ and $BD \cong v_b$). On the line $AD$ then determine such a point $C$, +so that $AC \cong b$. Prove that $ABC$ is the desired triangle. + + +\textbf{\textit{Proof.}} First, $AB \cong c$ and + $AC \cong b$ already by construction. Since $\angle ADB = 90^0$, $BD$ is the height of the triangle $ABC$ from the vertex $B$ and is consistent with the distance $v_b$ by construction. + + +\textbf{\textit{Discussion.}} The task has a solution exactly when it is possible +construction of the triangle $ABD$ or $hb \leq c$. In the construction of point $C$ +there are two possibilities - on different sides of point $A$, which means that we have two solutions for triangle $ABC$. In the case $hb \cong c$, the solutions are a right triangle and a congruent triangle. + \kdokaz + + + + \bzgled + If $v_a$, $v_b$ and $v_c$ are altitudes corresponding + to the sides $a$, $b$ and $c$ of a triangle, then: + $$\frac{v_a}{b+c}+\frac{v_b}{a+c}+\frac{v_c}{a+b}<\frac{3}{2}.$$ + \ezgled + +\textbf{\textit{Proof.}} +By adding the inequalities $v_a \leq b$, +$v_a \leq c$ +we get $2v_a \leq b + c$ or $\frac{v_a}{b + c} \leq \frac{1}{2}$. +Similarly, we get $\frac{v_b}{a + c} \leq \frac{1}{2}$ and +$\frac{v_c}{a + b} \leq \frac{1}{2}$. Since all inequalities cannot be true at the same time, +by adding we get the desired inequality. + \kdokaz + +The next property of the triangle will be called \index{triangle +inequality} \pojem{triangle inequality}. + + + \bizrek \label{neenaktrik} + The sum of any two sides of a triangle is greater than the third side. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.4.pic} +\caption{} \label{sl.skl.3.2.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $ABC$ be an arbitrary triangle. We will prove that $AB + AC > BC$. We mark a point $D$ so that $\mathcal{B}(B,A,D)$ and $AD \cong AC$ (Figure \ref{sl.skl.3.2.4.pic}). By the \ref{enakokraki} theorem ($\triangle CAD$ is an isosceles triangle with the base $CD$), we also have that $\angle BDC=\angle ADC \cong \angle ACD$. The line segment $CA$ is inside the angle $DCB$, so $\angle ACD < \angle DCB$. Then $\angle BDC < \angle DCB$ as well. From \ref{vecstrveckot} (referring to the triangle $BCD$) it follows that: $$BC < BD = AB + AD = AB + AC,$$ which was to be proven. \kdokaz + +From the triangle inequality we obtain a criterion for the existence of such a triangle, that its sides are consistent with three given line segments. + + + \bzgled + Let $a$, $b$ and $c$ be three line segments. A triangle with sides $a$, + $b$ and $c$ exist if and only if: + $$b + c > a,\hspace*{2mm} + a + c > b \hspace*{1mm}\textrm{ in }\hspace*{1mm} + a + b > c.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.5.pic} +\caption{} \label{sl.skl.3.2.5.pic} +\end{figure} + +\textbf{\textit{Proof.}} If such a triangle exists, then the three relations are direct consequences of the triangle inequality \ref{neenaktrik}. So we assume that all three relations are true. Without loss of generality, let $a$ be the longest side of this triangle (it is enough that it is not shorter than any other side) and let $B$ and $C$ be any points for which $BC \cong a$ (Figure \ref{sl.skl.3.2.5.pic}). Because $b + c > a$, this means that the circles $k(B,c)$ and $k(C,b)$ intersect in some point $A$ (a consequence of Dedekind's axiom - \ref{DedPoslKrozKroz} theorem, because each of them contains the inner points of the other), which is not on the line segment $BC$. The triangle $ABC$ is then the desired triangle. \kdokaz + +If we know which of the three sides is the longest, + it is enough to check only one inequality, as + the other two are automatically fulfilled. The proof + of the previous statement can also be used for the following, equivalent + criterion. + + \bzgled \label{neenaktrik1} + Let $a$, $b$ and $c$ be three line segments, such that $a \geq b,c$. A triangle with sides $a$, + $b$ and $c$ exist if and only if $b + c > a$. + \ezgled + +For example, we can determine that a triangle exists with sides that have lengths 7, 5 and 3 (because $5+3>7$), but a triangle with sides that have lengths 9, 6 and 2 +does not exist (because $6+2$ is not greater than 9). + + Let's look at some consequences of the previous statements. + + + \bzgled + If $X$ is an arbitrary point of the side $BC$ of a triangle $ABC$, + then: + $$AX < AB + AC.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.6.pic} +\caption{} \label{sl.skl.3.2.6.pic} +\end{figure} + +\textbf{\textit{Proof.}} If we use the triangle inequality for +the triangles $ABX$ and $AXC$ (Figure \ref{sl.skl.3.2.6.pic}), we get: + $$AX < AB + BX \hspace*{1mm} \textrm{ and }\hspace*{1mm} AX < AC + CX.$$ +By adding these two inequalities and using the triangle inequality for the triangle $ABC$, we get: + $$2AX < AB + AC + +BC < 2(AB + AC),$$ which was to be proven. \kdokaz + +%% !!! Dosegel magično stran - 100!!! Wow Bravo!!! + +The next inequality is a generalization of the previous one. In this sense, the previous statement is its consequence and it was not necessary to prove it separately. + + + \bzgled + Let $X$ be an arbitrary point of the side $BC$, different from the vertices $B$ and $C$ of a triangle $ABC$. + Then the line segment $AX$ is shorter than at least one of the two line segments $AB$ and $AC$ i.e.: + $$AX < \max\{AB, AC\}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.7.pic} +\caption{} \label{sl.skl.3.2.7.pic} +\end{figure} + +\textbf{\textit{Proof.}} Because for the point $X$ it holds that $\mathcal{B}(B, X, C)$, +one of the sides $AXB$ and $AXC$ is not acute. Without loss of +generality, let it be $AXC$ (Figure \ref{sl.skl.3.2.7.pic}). Then +it is the largest angle in the triangle $AXB$, which means that \\ +$AX < AC$ (statement \ref{vecstrveckot}). Similarly, if the angle $AXB$ is not acute, +it holds that $AX < AB$. + \kdokaz + +We will especially consider the case of the distance $AX$, if the point $X$ is from +the previous two statements the center of the side $BC$ (Figure +\ref{sl.skl.3.2.8.pic}). Such a distance, which is determined by the vertex +and the center of the opposite side of the triangle, is called +\index{težiščnica trikotnika} \pojem{težiščnica} trikotnika. +Težiščnice, ki ustrezajo ogliščem $A$, $B$ in $C$ trikotnika $ABC$, +običajno označujemo s $t_a$, $t_b$ in $t_c$. Zadnji +dve trditvi lahko uporabimo tudi za težiščnice. Toda za težiščnice +velja še dodatna lastnost, ki jo bomo sedaj dokazali. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.8.pic} +\caption{} \label{sl.skl.3.2.8.pic} +\end{figure} + + + + \bzgled \label{neenTezisZgl} If $a$, $b$, $c$ are the sides and $t_a$ + the corresponding median of a triangle $ABC$, then: + $$\frac{b+c-a}{2} AB$ and +$AA_1 + A_1C > AC$ or: +$$t_a+\frac{a}{2}>c \hspace*{2mm} \textrm{ in } \hspace*{2mm} + t_a+\frac{a}{2}>b.$$ +If we add these two inequalities, we get $\frac{b+c-a}{2} AD = 2AA_1 = 2t_a,$$ which had to be proven. \kdokaz + + We will show some more examples of the use of the triangle inequality. + + + \bzgled + Let $M$ be an arbitrary point of the bisector of the exterior angle at + the vertex $C$ of a triangle $ABC$. Then + $$MA + MB \geq CA + CB.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.10.pic} +\caption{} \label{sl.skl.3.2.10.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Let $D$ be such a point on the line segment $BC$, that $AC \cong CD$ + and $\mathcal{B}(A,C,D)$ (Figure +\ref{sl.skl.3.2.10.pic}). The triangles $ACM$ and $DCM$ are congruent, +by statement \textit{SAS} \ref{SKS} ($AC \cong DC$, $CM \cong CM$, $\angle ACM \cong +\angle DCM$). Therefore $MA \cong MD$. If we use the triangle inequality now, we get: $$MA + MB = MD + MB \geq BD = DC + +CB = CA + CB.$$ + Of course, the equality holds when the points $B$, $M$ and $D$ are collinear or $M = C$. + \kdokaz + + + + \bzgled + If the bisector of the interior angle at + the vertex $A$ of a triangle $ABC$ intersects the side $BC$ in the point $E$, then + $$AB > BE \hspace*{2mm} \textrm{ in }\hspace*{2mm} AC > CE .$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.11.pic} +\caption{} \label{sl.skl.3.2.11.pic} +\end{figure} + + \textbf{\textit{Proof.}} Because $\mathcal{B}(B,E,C)$, the angle + $AEB$ is the external angle of triangle $AEC$ (Figure +\ref{sl.skl.3.2.11.pic}). Therefore $\angle BEA > \angle EAC \cong +\angle BAE$ (statement \ref{zunanjiNotrNotrVecji}). Opposite the larger +angle in triangle $BAE$ is the larger side, or $AB > BE$ +(statement \ref{vecstrveckot}). Similarly, we prove that the other of the two relations is also true. + \kdokaz + + + \bzgled \label{zgled3.2.9} + If $M$ is the interior point of a triangle $ABC$, then \\ + $BA + AC > BM + MC.$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.12.pic} +\caption{} \label{sl.skl.3.2.12.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $N$ be the intersection of lines $BM$ and +$CA$ (Figure \ref{sl.skl.3.2.12.pic}). Because $M$ is an interior point of +triangle $ABC$, $\mathcal{B}(B,M,N)$ and +$\mathcal{B}(A,N,C)$ are true. If we now use the triangle inequality (statement \ref{neenaktrik}) twice, we get: + \begin{eqnarray*} +\hspace*{-4mm}BM + MC &<& BM + (MN + NC) = (BM + MN) + NC = BN + NC\\ + \hspace*{-4mm}&<& (BA + +AN) + NC = BA + (AN + NC) = BA + AC. + \end{eqnarray*} + +Let's define two new concepts. The sum of all sides of a polygon is called its \index{obseg!večkotnika} \pojem{obseg}. Half of this sum is the \pojem{polobseg} of this polygon. + + + + \bzgled + If $M$ is the interior point and $s$ the semiperimeter of a triangle $ABC$, then + $$s < AM + BM + CM < 2s.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.13.pic} +\caption{} \label{sl.skl.3.2.13.pic} +\end{figure} + +\textbf{\textit{Proof.}} + We first get the inequality, if +we use the triangle inequality for the triangles $MAB$, +$MBC$ and $MCA$ and then add them up. The second inequality is obtained, +if we use the previous statement (Example \ref{zgled3.2.9}) +and add the corresponding inequalities (Figure \ref{sl.skl.3.2.13.pic}). + \kdokaz + + + \bzgled + In each convex pentagon there exist three diagonals, + which are congruent to the sides of a triangle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.14.pic} +\caption{} \label{sl.skl.3.2.14.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Let $AD$ be the longest diagonal +of the pentagon $ABCDE$ (not shorter than any other +diagonal). We prove that $AD$, $AC$ and $BD$ are the desired diagonals, +i.e. those for which there exists a triangle, whose sides + are congruent to these diagonals (Figure \ref{sl.skl.3.2.14.pic}). Because + $AD\geq AC$ and $AD\geq BD$, it is enough to prove (Example \ref{neenaktrik1}), +that $AC + BD > AD$. The pentagon $ABCDE$ is convex, +so its diagonals $AC$ and +$BD$ intersect in some point $S$. Then: $$AC + BD > AS + SD > AD,$$ which was to be proven. \kdokaz + +The following consequence of the theorems about the congruence of +triangles is very important. We will also need the triangle inequality in this proof. + + + \bizrek \label{SkladTrikLema} + Let $ABC$ and $A'B'C'$ triangles such that $AB \cong A'B'$ + and $AC \cong A'C'$. Then $BC > B'C'$ if and only if + $\angle BAC > \angle B' A'C'$ i.e. + $$BC > B'C' \Leftrightarrow \angle BAC > \angle B' + A'C'.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.15.pic} +\caption{} \label{sl.skl.3.2.15.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.2.15.pic}) + +($\Leftarrow$) Let $\angle BAC > \angle B' A'C'$. Then there exists +within the angle $BAC$ such a line segment $l$, that $\angle BA,l \cong \angle B'A'C'$. +With $C''$ we mark the point of the line segment $l$, for which +$AC'' \cong A'C'$. Then both triangle $ABC''$ and +$A'B'C'$ are congruent and $BC'' \cong B'C'$. It is enough to prove that +$BC > BC''$. If $C''$ lies on the side $BC$, it is trivially +fulfilled. We assume that the point $C''$ does not lie on the side $BC$. +Let the point $E$ be the intersection of the perpendicular bisector of the angle $CAC''$ and the side $BC$. +By the \textit{SAS} theorem, the triangles $ACE$ and $AC''E$ are congruent, therefore +$CE \cong C''E$. Now it is: +$$BC = BE + EC = BE + EC'' \hspace{0.1mm} > BC'' = B'C'.$$ + ($\Rightarrow$) Let $BC > B'C'$. The relation $\angle BAC \cong + \angle B' A'C'$ is not true, + because then (by the \textit{SAS} theorem) the triangle +$ABC$ and $A'B'C'$ would be congruent and then also $BC \cong B'C'$. If +it would be true that $\angle BAC < \angle B' A'C'$, then from what has already been proven it would follow that $BC < B'C'$. Therefore $\angle BAC > \angle B' A'C'$. + \kdokaz + + \bizrek \label{neenakIzlLin} + If $A_1A_2\ldots A_n$ ($n\in \mathbb{N}$, $n\geq 3$) is polygonal chain, then + $$|A_1A_2|+|A_2A_2|+\cdots +|A_{n-1}A_n|\geq |A_1A_n|.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.16.pic} +\caption{} \label{sl.skl.3.2.16.pic} +\end{figure} + +\textbf{\textit{Proof.}} We will carry out the proof by induction over $n$ +(Figure \ref{sl.skl.3.2.16.pic}). + + In the case $n=3$ we get the triangle inequality - izrek + \ref{neenaktrik}. + +Let's assume that the inequality is true for $n=k$ ($k\in \mathbb{N}$, $k> 3$) or + $|A_1A_2|+|A_2A_2|+\cdots +|A_{k-1}A_k|\geq |A_1A_k|.$ We will prove that + the inequality is also true for $n=k+1$ or + $|A_1A_2|+|A_2A_2|+\cdots +|A_kA_{k+1}|\geq |A_1A_{k+1}|.$ If + we first use the induction + assumption, and then the triangle inequality, we get: + \begin{eqnarray*} + && |A_1A_2|+|A_2A_2|+\cdots +|A_{k-1}A_k|+|A_kA_{k+1}|\geq\\ + && \geq|A_1A_k|+|A_kA_{k+1}|\geq |A_1A_{k+1}|, + \end{eqnarray*} + which is what needed to be proven. \kdokaz + +We will now prove another inequality that is true in any triangle. + + + \bzgled + If $a$, $b$, $c$ are the sides and $\alpha$, $\beta$, $\gamma$ + the opposite interior angles of a triangle, then + $$60^0\leq \frac{a\alpha+b\beta +c\gamma}{a+b+c} < 90^0.$$ + \ezgled + +\textbf{\textit{Proof.}} We will prove each of the inequalities separately. In doing so, we will use the statement about the sum of the interior angles of a triangle (statement \ref{VsotKotTrik}). First, we will prove the second inequality: + \begin{eqnarray*} + \frac{a\alpha+b\beta +c\gamma}{a+b+c} < 90^0 + &\Leftrightarrow& a\alpha+b\beta +c\gamma - 90^0(a+b+c)<0\\ + &\Leftrightarrow& a(\alpha-90^0)+b(\beta-90^0) +c(\gamma-90^0)<0\\ + &\Leftrightarrow& a(180^0-2\alpha)+b(180^0-2\beta) +c(180^0-2\gamma)>0\\ + &\Leftrightarrow& a(\beta+\gamma-\alpha)+b(\alpha+\gamma-\beta) + + c(\alpha+\beta-\gamma)>0\\ + &\Leftrightarrow& \alpha(b+c-a)+\beta(a+c-b) + + \gamma(a+b-c)>0 + \end{eqnarray*} + The last inequality is fulfilled because, according to the triangle inequality (statement \ref{neenaktrik}), $b+c-a>0$, $a+c-b>0$ and $a+b-c>0$ hold. + We will now prove the first inequality: + \begin{eqnarray*} + && \frac{a\alpha+b\beta +c\gamma}{a+b+c} \geq 60^0\Leftrightarrow\\ + &\Leftrightarrow& a\alpha+b\beta +c\gamma - 60^0(a+b+c)\geq 0\\ + &\Leftrightarrow& a(\alpha-60^0)+b(\beta-60^0) +c(\gamma-60^0)\geq 0\\ + &\Leftrightarrow& a(3\alpha-180^0)+b(3\beta-180^0) +c(3\gamma-180^0)\geq 0\\ + &\Leftrightarrow& a(2\alpha-\beta-\gamma)+b(2\beta-\alpha-\gamma) + + c(2\gamma-\alpha-\beta)\geq 0\\ + &\Leftrightarrow& a(\alpha-\beta+\alpha-\gamma)+b(\beta-\alpha+\beta-\gamma) + + c(\gamma-\alpha+\gamma-\beta)\geq 0\\ + &\Leftrightarrow& a(\alpha-\beta)+a(\alpha-\gamma)+ + b(\beta-\alpha)+b(\beta-\gamma) + + c(\gamma-\alpha)+c(\gamma-\beta)\geq 0\\ + &\Leftrightarrow& (a-b)(\alpha-\beta)+(a-c)(\alpha-\gamma)+ + (b-c)(\beta-\gamma) \geq 0 + \end{eqnarray*} + The last inequality is a consequence of statement \ref{vecstrveckotAlgeb}. + \kdokaz + + The next statement will be the motivation for defining the distance of a point from a line. + + \bizrek Let $A'=pr_{\perp p}(A)$ be the foot of the perpendicular from a point $A$ on a line $p$. + If $X\in p$ and $X\neq A'$, then $AX>AA'$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.17.pic} +\caption{} \label{sl.skl.3.2.17.pic} +\end{figure} + + \textbf{\textit{Proof.}} + By definition, $AA'\perp p$ +(Figure \ref{sl.skl.3.2.17.pic}), which means that $AA'X$ + is a right angled triangle with hypotenuse $AX$. From + \ref{vecstrveckotHipot} it follows that $AX>AA'$. + \kdokaz + + If $A'=pr_{\perp +p}(A)$, we say that the length of the line $AA'$ \index{distance!point + from a line} \pojem{distance of point $A$ from line $p$}. +We denote it by $d(A,p)$. So $d(A,p)=|AA'|$. + + + + +%________________________________________________________________________________ + \poglavje{Circle and Line} \label{odd3KrozPrem} + +In the following we will deal with a circle and the mutual +position of a circle and a line. We prove first a property of the +diameter of a circle, which is a simple consequence of the +triangle inequality. + + + \bizrek \label{premerNajdTetiva} + The longest chord of a circle is its diameter. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.1.pic} +\caption{} \label{sl.skl.3.3.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} Let +$AB$ be any chord of a circle that is not a diameter, and $C$ a +point on the line $AS$, for which $CS\cong SA$ and $\mathcal{B}(A,S,C)$ +(Figure \ref{sl.skl.3.3.1.pic}). Then point $C$ lies on +the circle $k$ and $AC$ is its diameter. We have already shown (a consequence +of \ref{premerInS}) that all diameters of a circle +are mutually congruent. So it is enough to show that $AC>AB$. This follows +from the triangle inequality (triangle $ASB$). It holds: + $$AC=AS+SC=AS+SB>AB,$$ which was to be proved. \kdokaz + +It follows another property of a chord of a circle, as a consequence of +\ref{vecstrveckotHipot}. + + + \bzgled \label{tetivaNotrTocke} + Every point that lies on a chord of a circle, except its endpoints, + is an interior point of the circle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.2.pic} +\caption{} \label{sl.skl.3.3.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $X$ be an inner point of the segment $AB$ +with shorter sides on the circle $k(S,r)$ (Figure \ref{sl.skl.3.3.2.pic}). The angle $AXS$ and +$BXS$ are adjacent angles, which means that they are not both acute angles. Without +loss of generality, assume that the angle $BXS$ is not an acute angle. Then in +the triangle $SXB$ the side $SB$ is the longest (by +\ref{vecstrveckotHipot}), which means that: + $$SX ST = r.$$ + Therefore, none of the points $T_1$ ($T_1 \neq T$), lying on +the line $PT$, lies on the circle $k$. This means that the line +$PT$ is a tangent of this circle. + \kdokaz + +From the proof of the previous statement ($\Leftarrow$) it follows that all +points lying on the tangent of the circle (except for its +point of contact), are external points of this circle. With this +property we will prove the following statement. + + +\bzgled \label{tangKrozEnaStr} +All points of a circle are on the one side of its tangent. +\ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.7.pic} +\caption{} \label{sl.skl.3.3.7.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $T$ be the point of tangency of the circle $k(S, r)$ and its tangent $t$ (Figure \ref{sl.skl.3.3.7.pic}). The tangent $t$ divides the plane in which $k$ and $t$ lie into two half-planes. The half-plane in which the point $S$ lies, we denote by $\alpha_1$, the other half-plane by $\alpha_2$. We prove that all points of the circle $k$ lie in the half-plane $\alpha_1$. Let $X$ be an arbitrary point of the half-plane $\alpha_2$. Since the points $S$ and $X$ are on different sides of the line $t$, it follows that the open line segment $SX$ intersects at some point $Y$. Then we have: + $$SX = SY + YX > SY \geq ST = r,$$ + which means that the point $X$ does not lie on the circle $k$ and is its external point. Therefore, none of the points of the half-plane $\alpha_2$ lies on the circle $k$, that is, all of them are in the half-plane $\alpha_1$ with the edge $t$. \kdokaz + +A direct consequence of the statement \ref{TangPogoj} is also that in each point of the circle we can draw only one tangent. If $X$ is an internal point of the circle $k(S, r)$, then no tangent passes through this point, since all lines through $X$ are intersecting, which is a consequence of Dedekind's axiom (statement \ref{DedPoslKrozPrem}). Later (statement \ref{tangentiKroznice}) we will find that through each external point of the circle we can draw exactly two tangents. For the time being, we prove the following statement (the reader will remember that this is a statement that we have already considered at the beginning of the introductory chapter - statement \ref{TalesUvod}). + +\bizrek \label{TalesovIzrKroz} \index{izrek!Talesov za krožnico} + Thales' theorem for a circle\footnote{Starogrški + filozof in matematik \textit{Tales} + \index{Tales} iz Mileta (640--546 pr. n. š.) + ni prvi, ki je odkril to trditev. Kot empirično dejstvo so jo poznali + že stari Egipčani in Babilonci. Izrek imenujemo po Talesu, ki ga je prvi dokazal. + V dokazu je uporabljal lastnosti enakokrakih trikotnikov in dejstvo, da je + vsota notranjih kotov trikotnika enaka vsoti dveh pravih kotov. Torej je + dokaz enak temu, ki ga bomo izpeljali tukaj.}:\\ + Let $AB$ be a diameter of a circle $k$. Then for any point $X$ of this circle different from $A$ and $B$ + ($X\in k$ in $X\neq A$ in $X\neq B$) is $\angle AXB=90^0$ + \index{izrek!Talesov za krožnico} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.8.pic} +\caption{} \label{sl.skl.3.3.8.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $O$ središče krožnice $k$ (Figure +\ref{sl.skl.3.3.8.pic}). Ker $A,B,X\in k$, je + $OA\cong OB\cong OX$. Torej sta + trikotnika $AOX$ in $BOX$ enakokraka, zato je (izrek +\ref{enakokraki}): + $\angle AXO\cong\angle XAO=\alpha$ in $\angle BXO\cong\angle XBO=\beta$. +Tedaj je $\angle AXB=\alpha+\beta$. + Vsota notranjih kotov v +trikotniku $AXB$ je enaka $180^0$ (izrek \ref{VsotKotTrik}), torej +je $2\alpha+2\beta=180^0$. Iz tega sledi + $$\angle AXB=\alpha+\beta=90^0,$$ kar je bilo treba dokazati. \kdokaz + + Dokažimo tudi obratno trditev. + + + \bizrek \label{TalesovIzrKrozObrat} + If $A$, $B$ and $X$ are non-collinear points such that + $\angle AXB=90^0$, then the point $X$ lies on a circle with the diameter $AB$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.9.pic} +\caption{} \label{sl.skl.3.3.9.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Let $O$ be the center of the line $AB$ and $k$ +the circle with center $O$ and radius $OA$ or diameter $AB$ +(Figure \ref{sl.skl.3.3.9.pic}). From $\angle AXB=90^0$ it follows from +the theorem \ref{VsotKotTrik}: + \begin{eqnarray} + \angle XAB+ \angle XBA = 90^0 \label{relacija336} + \end{eqnarray} + +We prove $X\in k$. We assume the contrary, i.e. that the point $X$ does not +lie on the circle $k$. In this case $OX\neq OA$. Let $X_1$ +be the point on the segment $OX$, for which $OX_1\cong OA$ (theorem +\ref{ABnaPoltrakCX}). This means that the point $X_1$ lies on the circle +$k$ and by the theorem \ref{TalesovIzrKroz} we have $\angle +AX_1B=90^0$. + +By our assumption $OX\neq OA$ it is clear that $X\neq X_1$. +We will consider two possibilities: + +\textit{1)} Let $OX_1OX$ or $\mathcal{B}(O,X,X_1)$. Similarly +as in the first case we get: + $$\angle X_1AB+ \angle X_1BA>\angle XAB+ \angle XBA = 90^0.$$ + In this case the sum of the angles in the triangle $AX_1B$ is greater than +$180^0$, which by the theorem \ref{VsotKotTrik} is not possible. + +It follows that $OX=OX_1$ or $X\in k$. + \kdokaz + + + Use the previous theorem for the design of tangents. + + + \bzgled \label{tangKrozKonstr} + Let $A$ be an exterior point of a circle $k(S,r)$. + Construct all tangents of this circle passing through the point $A$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.10.pic} +\caption{} \label{sl.skl.3.3.10.pic} +\end{figure} + + + \textbf{\textit{Solution.}} Let $l$ be a circle with diameter $SA$ +(Figure \ref{sl.skl.3.3.10.pic}). Because $S$ is an interior, $A$ is an exterior point of the given circle $k$, the circle $k$ and $l$ have exactly two common points $T_1$ and $T_2$ (statement \ref{DedPoslKrozKroz}). +By Tales' statement \ref{TalesovIzrKroz}, +$\angle ST_1A\cong \angle ST_2A=90^0$. Because $ST_1$ and $ST_2$ +are radii of the circle $k$, $AT_1$ and $AT_2$ are tangents of the circle $k$ +through the point $A$ (statement \ref{TangPogoj}). + + We prove that $AT_1$ and $AT_2$ are the only tangents of the circle $k$ + from the point $A$. If $AT$ is a tangent from the point $A$, which the circle $k$ + touches in the point $T$, by statement \ref{TangPogoj} $\angle ATS=90^0$. + This means that the point $T$ lies on the circle $l$ + (statement \ref{TalesovIzrKrozObrat}) or $T\in k\cap l$. Therefore $T$ is one + of the points $T_1$ and $T_2$, so $AT_1$ and $AT_2$ are the only tangents of the circle + $k$ from the point $A$. + \kdokaz + + The following statement follows from the previous construction. + + + + \bizrek \label{tangentiKroznice} + If $V$ is an exterior point of a circle $k(S,r)$, then there are exactly + two tangents of the circle $k$ through the point $V$. + \eizrek + + + We prove some more properties of a tangent of a circle. + + + \bzgled \label{TangOdsek} + If $VA$ and $VB$ are tangents of a circle $k(S,r)$ in a points + $A$ and $B$ of the circle, then the centre $S$ lies on the bisector + of the angle $AVB$ and $VA \cong VB$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.11.pic} +\caption{} \label{sl.skl.3.3.11.pic} +\end{figure} + +\textbf{\textit{Proof.}} From izrek \ref{TangPogoj} it follows: $VA +\perp AS$ and $VB \perp BS$ +(Figure \ref{sl.skl.3.3.11.pic}). Therefore $ASV$ and $BSV$ are right-angled +triangles with a shared hypotenuse $SV$. Because $SA \cong SB = r$, +these two triangles are congruent (izrek \textit{SSA} \ref{SSK}). Therefore the angles +$AVS$ and $BVS$ are also congruent, which means that the line $VS$ +is the bisector of angle $AVB$. From the congruence of these two triangles it also follows +that $VA \cong VB$. + \kdokaz + +The converse is also true. + + + \bzgled \label{SimKotaKraka} + If a point $S$ lies on the bisector of a convex angle, + then it is the centre of a circle touching both sides of this angle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.12.pic} +\caption{} \label{sl.skl.3.3.12.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $A$ and $B$ be the right-angled projections +of point $S$ on the sides of given angle with the vertex $V$ (Figure +\ref{sl.skl.3.3.12.pic}). Triangles $ASV$ and $BSV$ are congruent +(izrek \textit{ASA} \ref{KSK}), because they have a shared side $VS$ and +two pairs of congruent angles - from $\angle AVS\cong \angle BVS$ and $\angle +SAV\cong \angle SBV=90^0$ it follows that $\angle ASV\cong \angle BSV$. Therefore $SA\cong SB$ and $k(S,SA)$ is the desired circle. The sides +of given angle are tangents to the circle by izrek \ref{TangPogoj}. + \kdokaz + +Now we will prove another criterion for the mutual position of a line and +a circle in a plane. + + + + \bizrek \label{TangSekMimobKrit} + Let $P$ be the foot of the perpendicular from the centre of a circle $k(S,r)$ + on a line $p$ (lying in the plane of the circle). Then the line $p$ is: + + (i) secant, if and only if $SP < r$; + + (ii) tangent, if and only if $SP \cong r$; + + (iii) non-intersecting line, if and only if $SP> r$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.13.pic} +\caption{} \label{sl.skl.3.3.13.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.3.13.pic}) + +The statement (ii) follows directly from the criterion for tangency (the statement \ref{TangPogoj}). + +(i) In the proof of the direct direction of equivalence we use the fact that the hypotenuse of a right angled triangle is longer than either of the two shorter sides (the statement \ref{vecstrveckotHipot}). If $A$ and $B$ are the intersections of the secant $p$ and the circle $k$, then $SA$ is the hypotenuse of the right angled triangle $ASP$ and it holds: + $r \cong SA > SP$. + + In the proof of the inverse direction of equivalence we use the consequence of Dedekind's axiom (the statement \ref{DedPoslKrozPrem}). Because in this case $P$ is an inner point of this circle, each straight line of this plane that goes through +the point $P$ is a secant of the circle $k$. + +(iii) It follows from the proven (i) and (ii). Because if $SP > r$, +then neither $SP < r$ nor $SP \cong r$. From the equivalences (i) and (ii) it follows that the straight line $p$ is neither a secant nor a tangent. Therefore $p$ +is a parallel of the circle $k$. In the same way we prove the inverse direction of equivalence. + \kdokaz + +%________________________________________________________________________________ + \poglavje{Quadrilaterals} \label{odd3Stirik} + + +In section \ref{odd2AKSURJ} we introduced the concept of a quadrilateral as +a polygon with four sides and four vertices. +We defined the concepts of adjacent and opposite sides, adjacent and +opposite angles and diagonal. To a quadrilateral $ABCD$ with the lengths of its sides $AB$, $BC$, $CD$ and $DA$ we usually assign the letters $a$, +$b$, $c$ and $d$, and to the lengths of its diagonals $AC$ and $BD$ the letters $e$ and +$f$ (Figure \ref{sl.skl.3.4.1.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.1.pic} +\caption{} \label{sl.skl.3.4.1.pic} +\end{figure} + +In the same section, we introduced the concepts of internal and external angles of a quadrilateral. We also mentioned that the internal angles at vertices $A$, $B$, $C$ and $D$ of a quadrilateral $ABCD$ are usually denoted by $\alpha$, $\beta$, $\gamma$ and $\delta$, and its external angles by $\alpha'$, $\beta'$, $\gamma'$ and $\delta'$. We proved (as a consequence of the general statement \ref{VsotKotVeck}) that the sum of all four internal angles of an arbitrary quadrilateral is equal to $360^0$ (Figure \ref{sl.skl.3.4.1.pic}). The sum of external angles is also equal to $360^0$ (in a convex quadrilateral). So: + \begin{eqnarray*} + \alpha+\beta+\gamma+\delta=360^0,\\ + \alpha'+\beta'+\gamma'+\delta'=360^0 + \end{eqnarray*} + +Let us also add that we call the internal angle \pojem{adjacent} or \pojem{opposite}, if the corresponding vertices are adjacent or opposite. + + + +Now we will consider some types of quadrilaterals in more detail. + + A quadrilateral +$ABCD$ is a \pojem{trapezoid}, if $AB\parallel CD$ (Figure \ref{sl.skl.3.4.2.pic}). +Sides $AB$ and $CD$ are the \pojem{bases}, $BC$ and $AD$ are the +\pojem{legs} of this trapezoid. +The line $PQ$ ($P\in AB$, $Q\in CD$ and $PQ\perp AB$) is the \pojem{height} of the trapezoid. We often denote it by $v$. + + +A trapezoid is + \pojem{isosceles}, + if $BC \cong AD$ and $BC \not\parallel AD$, + or \pojem{right}, if at least +one of the internal angles is a right angle. + + + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.2.pic} +\caption{} \label{sl.skl.3.4.2.pic} +\end{figure} + + +Two internal angles at the same leg of a trapezoid are supplementary, because +they are angles with parallel legs (statement \ref{KotiTransverzala}). The supplementarity of these angles is also +a sufficient condition for a quadrilateral to be a trapezoid. It follows from this that a right trapezoid has at least two right internal angles. + +We get another group of quadrilaterals as a special type of trapezoids. These are \pojem{parallelograms}. They can be defined in different ways. We will choose one, and prove that the others are equivalent. + +Quadrilateral $ABCD$ is a \index{paralelogram} \pojem{parallelogram}, if $AB \parallel CD$ and $AD \parallel BC$ (Figure \ref{sl.skl.3.4.3.pic}). Line $PQ$ ($P\in AB$, $Q\in CD$ and $PQ\perp AB$) and $MN$ ($M\in BC$, $N\in AD$ and $MN\perp BC$) are the \index{višina!paralelograma}\pojem{heights of the parallelogram}. We often denote them with $v_a$ and $v_b$. + + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.3.pic} +\caption{} \label{sl.skl.3.4.3.pic} +\end{figure} + + A parallelogram is therefore a quadrilateral that has two pairs of parallel sides. The term congruence is not used in the definition of a parallelogram. We can also consider parallelograms (and trapezoids) in so-called \index{geometrija!afina} \pojem{affine geometry}. This is a geometry that is based on all the axioms of Euclidean geometry, with the exception of the third group of axioms - the axioms of congruence. + +We will now prove the aforementioned equivalents for the definition of a parallelogram. + +\bizrek \label{paralelogram} +Let $ABCD$ be a convex quadrilateral. +Then the following statements are equivalent: +\begin{enumerate} + \item The quadrilateral $ABCD$ is a parallelogram. + \item Any two adjacent interior angles of the quadrilateral $ABCD$ are supplementary. + \item Any two opposite interior angles of the quadrilateral $ABCD$ are congruent. + \item $AB \parallel CD$ and $AB \cong CD$\footnote{This equivalent + in a slightly different form is given by \index{Euclid} + \textit{Euclid of Alexandria} (3rd century BC) in + the first book of his 'Elements'.}. + \item $AB \cong CD$ and $AD \cong BC$. + \item The diagonals of the quadrilateral $ABCD$ bisect each other, i.e. + line segments $AC$ and $BD$ have a common midpoint. +\end{enumerate} + \eizrek + + \textbf{\textit{Proof.}} +It is enough to prove the equivalence of all the statements $(1)-(6)$. To avoid proving all equivalences (two implications - for example, with the statement (1), which would give 10 implications in total), we will simplify the proof a little, so that we prove implications according to the following scheme. + +\vspace*{5mm} +\hspace*{25mm} +$\begin{array}{ccccccc} + \textit{(1)} & \Leftarrow & \textit{(2)} & \Leftarrow & \textit{(3)} & & \\ + \Downarrow & & & & \Uparrow & & \\ + \textit{(4)} & & \Rightarrow & & \textit{(5)} & \Leftrightarrow & \textit{(6)} +\end{array}$ + +\vspace*{5mm} + + As we can see, this is enough because the implication $\textit{(1)} +\Rightarrow \textit{(2)}$ follows directly from: $\textit{(1)}\Rightarrow \textit{(4)}\Rightarrow +\textit{(5)}\Rightarrow \textit{(3)} \Rightarrow \textit{(2)}$. + +Let's mark with $\alpha$, $\beta$, $\gamma$ and $\delta$ the internal angles at +the vertices $A$, $B$, $C$ and $D$ of the quadrilateral $ABCD$. The quadrilateral +$ABCD$ is convex, which means that its diagonals intersect in +some point $S$. + +$\textit{(2)}\Rightarrow \textit{(1)}$. Let the angles $\alpha$ and +$\beta$ be complementary (Figure \ref{sl.skl.3.4.4.pic}). Then the angles at the transversal $AB$ are congruent to the angles $AD$ and $BC$, so $AD\parallel +BC$ (by Theorem \ref{KotiTransverzala}). Similarly, from the complementarity of the angles $\beta$ and $\gamma$ it follows that $AB\parallel CD$. Therefore, the quadrilateral $ABCD$ is a parallelogram. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.4.pic} +\caption{} \label{sl.skl.3.4.4.pic} +\end{figure} + +$\textit{(3)}\Rightarrow\textit{(2)}$. Let $\alpha =\gamma$ and $\beta =\delta$ (Figure \ref{sl.skl.3.4.4.pic}). +Since $\alpha + \beta +\gamma +\delta = 360°$ (the sum of all +internal angles in a quadrilateral is $360°$ - by Theorem \ref{VsotKotVeck}), it follows that +$\alpha + \beta =180°$ and $\beta +\gamma = 180°$. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.4a.pic} +\caption{} \label{sl.skl.3.4.4a.pic} +\end{figure} + +$\textit{(1)}\Rightarrow\textit{(4)}$. Let the quadrilateral $ABCD$ +be a parallelogram, i.e. let $AB \parallel CD$ and $AD \parallel BC$ +(Figure \ref{sl.skl.3.4.4a.pic}). We will prove that then also $AB \cong CD$. +The line $AC$ is a transversal of the parallels $AB$ and $CD$, which +means that the angles $CAB$ and $ACD$ are alternate angles at this +transversal and are therefore congruent. Similarly, from the parallelism of the lines +$AD$ and $BC$ it follows that the angles $ACB$ and $CAD$ are congruent. Since $AC \cong AC$, the triangles $ACB$ and $CAD$ are congruent (by Theorem \ref{KSK} - \textit{ASA}). Therefore, $AB \cong CD$. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.4b.pic} +\caption{} \label{sl.skl.3.4.4b.pic} +\end{figure} + + $\textit{(4)}\Rightarrow \textit{(5)}$. Let $ABCD$ be such a quadrilateral that + $AB \parallel CD$ and $AB \cong CD$ (Figure \ref{sl.skl.3.4.4b.pic}). + We will prove that then also $AD \cong BC$. + The line $AC$ is a transversal of the parallels $AB$ and +$CD$, which means that the angles $CAB$ and $ACD$ are alternate angles at this +transversal and are therefore congruent. Since $AC \cong AC$, the triangles $ACB$ and $CAD$ are congruent (by Theorem \ref{SKS} - \textit{SAS}). Therefore, $BC \cong AD$. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.4c.pic} +\caption{} \label{sl.skl.3.4.4c.pic} +\end{figure} + + $\textit{(5)}\Rightarrow \textit{(3)}$. Let $ABCD$ be a quadrilateral such that $AB \cong CD$ and + $AD \cong BC$ (Figure \ref{sl.skl.3.4.4c.pic}). We prove that +then $\beta =\delta$ and $\alpha =\gamma$. Because $AC \cong +AC$, the triangles $ACB$ and $CAD$ are congruent (\ref{SSS} - \textit{SSS}). +It follows that $\angle ABC \cong \angle CDA$ or $\beta =\delta$. In a +similar way we prove that $\alpha =\gamma$. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.4d.pic} +\caption{} \label{sl.skl.3.4.4d.pic} +\end{figure} + +$\textit{(5)}\Leftrightarrow \textit{(6)}$. Let $ABCD$ be a quadrilateral such that +$AB \cong CD$ and $AD \cong BC$ (Figure \ref{sl.skl.3.4.4d.pic}). We prove that the point $S$ +is the common center of its diagonals $AC$ and $BD$. Because $AC \cong +AC$, the triangle $ACB$ is congruent to the triangle $CAD$ (\ref{SSS} - \textit{SSS}). Therefore +$\angle ACB \cong \angle CAD$ or $\angle SCB \cong \angle +SAD$. From the congruence of the right angles $CSB$ and $ASD$, it follows that the +angles $SBC$ and $SDA$ are also congruent. Because $AD \cong BC$, it follows that +$\triangle CSB \cong \triangle ASD$ (\ref{KSK} - \textit{ASA}). Therefore $SB +\cong SD$ and $SC \cong SA$ or the point $S$ is the common center +of the diagonals $AC$ and $BD$. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.4e.pic} +\caption{} \label{sl.skl.3.4.4e.pic} +\end{figure} + + +Conversely, let $S$ be the common center of the diagonals $AC$ and $BD$ (Figure \ref{sl.skl.3.4.4e.pic}). Then +$SB \cong SD$ and $SC \cong SA$. The right angles $CSB$ and $ASD$ are also congruent, so +$\triangle CSB \cong \triangle ASD$ (\ref{SKS} - \textit{SAS}). It follows that $AD \cong BC$. In a similar way +we prove that $AB \cong CD$. + \kdokaz + +We recommend to the reader to prove the previous statement by using a +similar scheme. This will be a good exercise in using theorems about the congruence of triangles. + +Let us now define a type of quadrilateral for which it will +be shown that they are special cases of parallelograms (Figure +\ref{sl.skl.3.4.5.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.5.pic} +\caption{} \label{sl.skl.3.4.5.pic} +\end{figure} + +A quadrilateral with all sides congruent is called a \index{rhombus} +\pojem{rhombus}. + +A quadrilateral with all interior angles congruent (and therefore +equal to $90^0$, because their sum is $360^0$) is a \index{rectangle} +\pojem{rectangle}. + +A quadrilateral with all sides congruent and all interior angles +congruent (and equal to $90^0$) is called a \index{square} +\pojem{square}. + + +It is not difficult to prove that each of these quadrilaterals is also a parallelogram. This is a direct consequence of the previous statement. The rhombus is a parallelogram due to $\textit{(5)}\Rightarrow\textit{(1)}$; the rectangle is a parallelogram due to +$\textit{(2)}\Rightarrow\textit{(1)}$ (or $\textit{(3)}\Rightarrow\textit{(1)}$). For the square it is +clear that it is also a rectangle and a rhombus, so it is also +a parallelogram. + +From the previous statement \ref{paralelogram} - equivalent (\textit{5}) it follows that a parallelogram with two adjacent sides congruent is a rhombus. Similarly, + according to the same statement from the equivalents \textit{(2)} and \textit{(3)} it follows that a parallelogram with at least one right angle is a rectangle. + +The next statement provides additional criteria when a parallelogram is also a rhombus, +a rectangle or a square. This statement refers to diagonals. In +a parallelogram, the diagonals always intersect, but in a rhombus, +a rectangle or a square we will have additional properties. + + + + \bizrek \label{RombPravKvadr} $ $ (Figure \ref{sl.skl.3.4.5a.pic}) + + a) A parallelogram is a rhombus if and only if their diagonals are perpendicular. + + b) A parallelogram is a rectangle if and only if their diagonals are congruent. + + c) A parallelogram is a square if and only if their diagonals are perpendicular and congruent. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.5a.pic} +\caption{} \label{sl.skl.3.4.5a.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Let $ABCD$ be a parallelogram and $S$ the intersection of its diagonals $AC$ and $BD$. According to +the previous statement \ref{paralelogram}, $S$ is their common +center. From the same statement it also follows that $AB \cong CD$ and $AD \cong +BC$. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.6.pic} +\caption{} \label{sl.skl.3.4.6.pic} +\end{figure} + +\textit{a)} (Figure \ref{sl.skl.3.4.6.pic}) + + If $ABCD$ is a rhombus, all sides are congruent. Therefore the +triangles $ABS$ and + $ADS$ +are congruent (statement \ref{SSS} - \textit{SSS}). Then the angles $ASB$ +and $ASD$ are also congruent and are (as the supplement of the angle) both right angles. This means that the diagonals are perpendicular. + +If the diagonals of the parallelogram $ABCD$ are perpendicular, the triangles +$ABS$ and $ADS$ are congruent (statement \ref{SKS} - \textit{SAS}). Therefore the sides +$AB$ and $AD$ are congruent. In a similar way, we prove that all sides of this parallelogram are congruent, which means that the parallelogram is a rhombus. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.6a.pic} +\caption{} \label{sl.skl.3.4.6a.pic} +\end{figure} + + +\textit{b)} (Figure \ref{sl.skl.3.4.6a.pic}) + + If $ABCD$ is a rectangle, all interior angles are congruent and are right angles. Then +the triangles $ABC$ and $DCB$ are congruent (statement \ref{SKS} - \textit{SAS}). +Therefore $AC \cong DB$. + +If in the parallelogram $ABCD$ it holds that $AC \cong DB$, the triangles +$ABC$ and $DCB$ are congruent (statement \ref{SSS} - \textit{SSS}). From this it follows that +the interior angles at the vertices $B$ and $D$ of the parallelogram +$ABCD$ are congruent. According to the previous statement \ref{paralelogram} they are supplementary, which means that they are both right angles. Similarly, all angles of this parallelogram are right angles, which means that the parallelogram is a rectangle. + + \textit{c)} A parallelogram is a square if and only if it is a rhombus and a rectangle at the same time. + The latter is equivalent to +the fact that the diagonals are perpendicular and +congruent, which follows from the proven (\textit{a.} and \textit{b.}). + \kdokaz + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.7.pic} +\caption{} \label{sl.skl.3.4.7.pic} +\end{figure} + +Since the diagonals of a rectangle are perpendicular and intersect, there is a circle that contains all the vertices of this rectangle (Figure \ref{sl.skl.3.4.7.pic}). This is called the \index{circumscribed circle!rectangle} \pojem{circumscribed circle of the rectangle}. Its center is the intersection of its diagonals. If the point $S$ is the intersection of the diagonals of the rectangle $ABCD$, then from the previous equation \ref{RombPravKvadr} it follows: + $$SA \cong SC \cong SB \cong SD.$$ +The radius of this circle is equal to half the diagonal of the rectangle. Since the square is a special type of rectangle, it also has a circumscribed circle. + +We will now prove an important property of isosceles trapezoids. + + + \bizrek \label{trapezEnakokraki} + Interior base angles of an isosceles trapezium are congruent. + The diagonals of an isosceles trapezium are congruent line segments. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.8.pic} +\caption{} \label{sl.skl.3.4.8.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Let $ABCD$ be an isosceles trapezoid with the base $AB$ (Figure \ref{sl.skl.3.4.8.pic}). Without loss of generality, assume that $AB>CD$. Let $C'$ and $D'$ be the orthogonal projections of the vertices $C$ and $D$ onto the line $AB$. The quadrilateral $D'C'CD$ is a parallelogram with a right angle, so it is a rectangle. From the fact that $D'C'CD$ is a parallelogram, it follows that $CC'\cong DD'$ (equation \ref{paralelogram}). Since $\angle CC'B\cong \angle DD'A=90^0$, the triangles $CC'B$ and $DD'A$ are congruent (the \textit{SSA} equation \ref{SSK}), so $\beta=\angle CBC'\cong \angle DAD'=\alpha$. + + We will now prove that the diagonals $AC$ and $BD$ are congruent. This follows from the congruence of the triangles $ABC$ and $BAD$ (the \textit{SAS} equation \ref{SKS}). + \kdokaz + + + \bzgled + Let $ABCD$, $AEBK$ and $CEFG$ be equally oriented squares in a plane. + Then $B$, $D$ and $F$ are collinear points and the point $B$ is a midpoint of the line segment $DF$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.9.pic} +\caption{} \label{sl.skl.3.4.9.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.4.9.pic}) + + The triangles $CAE$ and $FBE$ are congruent, because $CE\cong FE$, $AE\cong BE$ and + $\angle AEC=90^0-\angle CEB=\angle BEF$ (statement \ref{SKS} - \textit{SAS}). Therefore $\angle EBF$ is a right angle and the points $D$, $B$ and $F$ are collinear. From + the congruence of these two triangles it also follows that $BF\cong AC\cong BD$. + \kdokaz + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.10.pic} +\caption{} \label{sl.skl.3.4.10.pic} +\end{figure} + +Except for trapezoids and parallelograms, we will define one more group +of quadrilaterals. A quadrilateral $ABCD$ is +\index{deltoid}\pojem{deltoid}, if its diagonals are perpendicular and +one of the diagonals bisects the other (Figure \ref{sl.skl.3.4.10.pic}). +The following statement is related to the deltoid and is equivalent to its +definition. + + + \bzgled + A quadrilateral is a deltoid if and only if it has two pairs of congruent adjacent sides. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.11.pic} +\caption{} \label{sl.skl.3.4.11.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.4.11.pic}) + + Let the quadrilateral $ABCD$ be a deltoid. Then the diagonals +$AC$ and $BD$ are perpendicular and one of the diagonals bisects the other. Without +loss of generality, let the diagonal $BD$ bisect the diagonal $AC$. +It follows that the right triangles $ABS$ and $CBS$ are congruent (statement +\ref{SKS} - \textit{SAS}). Then $AB \cong CB$. From the congruence of the triangles $ADS$ and $CDS$ it follows that $AD \cong CD$. + + Let $ABCD$ be a quadrilateral, in which $AB \cong CB$ and $AD \cong CD$ hold. + The triangles $ABD$ and $CBD$ are +congruent (statement \ref{SSS} - \textit{SSS}), so the angles $ADS$ +and $CDS$ are also congruent. From this it follows that the triangles $ADS$ and +$CDS$ are also congruent (statement \ref{SKS} - \textit{SAS}). Therefore $S$ is the center of the diagonal $AC$, and the angles $DSA$ +and $DSC$ are right angles, because they are congruent with the angles. + \kdokaz + +\bzgled + Let $k_1(S_1,r)$, $k_2(S_2,r)$, $k_3(S_3,r)$ be congruent circles and + $k_1\cap k_2=\{B,A_3\}$, $k_2\cap k_3=\{B,A_1\}$, $k_3\cap k_1=\{B,A_2\}$. + Prove that the lines $S_1A_1$, + $S_2A_2$ and $S_3A_3$ intersect at a single point . + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.12.pic} +\caption{} \label{sl.skk.4.2.12.pic} +\end{figure} + +\textbf{\textit{Proof.}} + We will prove that the lines $O_1A_1$, $O_2A_2$ and $O_3A_3$ + have the same center, or that the corresponding quadrilaterals +are parallelograms (Figure \ref{sl.skk.4.2.12.pic}). Because $k_1$, $k_2$ +and $k_3$ are congruent circles, the quadrilaterals $O_1A_2O_3B$ and $O_2A_1O_3B$ + are rhombuses. Because of this, the lines $O_1A_2$ and $O_2A_1$ are congruent and +parallel, which means that the quadrilateral $O_1A_2A_1O_2$ is a parallelogram +(statement \ref{paralelogram}). From the same statement it follows that its +diagonals $O_1A_1$ and $O_2A_2$ have a common center. In a similar way we prove that the lines $O_2A_2$ and $O_3A_3$ have a common center, which means that this is also true for all three lines $O_1A_1$, +$O_2A_2$ and $O_3A_3$ at the same time. \kdokaz + + + \bzgled + Construct a rectangle $ABCD$ if its diagonals and the difference of its sides + are congruent with the two given line segments $d$ and $l$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.10a.pic} +\caption{} \label{sl.skl.3.4.10a.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $ABCD$ be a rectangle, where $AC\cong d$ and $AB-BC=l$ (Figure \ref{sl.skl.3.4.10a.pic}). Let $E$ be a point on side $AB$ such that $EB\cong BC$. In this case, $AE=AC-EB=AC-BC=l$. Because $EBC$ is an isosceles right triangle, $\angle CEB\cong\angle ECB=45^0$ (\ref{enakokraki} and \ref{VsotKotTrik}) or $\angle AEC=135^0$. +This allows us to first construct the triangle $AEC$ ($AC\cong d$, $\angle AEC=135^0$ and $AE\cong l$), and then the rectangle $ABCD$. + \kdokaz + + + \bzgled + Construct a triangle with given $b$, $c$ and $t_a$ (sides $AC$, $AB$ and triangle median $AA_1$). + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.10b.pic} +\caption{} \label{sl.skl.3.4.10b.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $ABC$ be a triangle, where $AC\cong b$, $AB\cong c$ and $AA_1\cong t_a$, where $A_1$ is the center of line segment $BC$ (Figure \ref{sl.skl.3.4.10b.pic}). Let $D$ be a point on line segment $AA_1$ such that $DA_1\cong AA_1$ and $\mathcal{B}(A, A_1,D)$. This means that $A_1$ is the common center of line segments $BC$ and $AD$, so by \ref{paralelogram} quadrilateral $ABDC$ is a parallelogram. By the same \ref{paralelogram}, $CD\cong AB\cong c$. This allows us to first construct the triangle $ADC$ ($AC\cong b$, $CD\cong c$ and $AD=2t_a$), and then point $A$. + \kdokaz + +We will now introduce a shorter form of data entry for designing triangles. Similarly, as we had in the previous task with the notation: $b$, $c$, $t_a$, for the elements of the triangle $ABC$ we will usually use the following labels: + \begin{itemize} + \item $a$, $b$, $c$ - sides, + \item $\alpha$, $\beta$, $\gamma$ - internal angles, + \item $v_a$, $v_b$, $v_c$ - altitudes, + \item $t_a$, $t_b$, $t_c$ - centroids, + \item $l_a$, $l_b$, $l_c$ - distances that are determined by the vertex and the intersection of the internal angle's symmetry line at that vertex with the opposite side; + \item $s$ - semi-perimeter ($s=\frac{a+b+c}{2}$), + \item $R$ - radius of the circumscribed circle (see section \ref{odd3ZnamTock}), + \item $r$ - radius of the inscribed circle (see section \ref{odd3ZnamTock}), + \item $r_a$, $r_b$, $r_c$ - radii of the excircles (see section \ref{odd4Pricrt}). + \end{itemize} + + + \bzgled + Construct a trapezium if its sides are congruent with the four given line segments $a$, $b$, $c$ and $d$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.10c.pic} +\caption{} \label{sl.skl.3.4.10c.pic} +\end{figure} + +\textbf{\textit{Solution.}} Without loss of generality, we will first assume that $a\geq c$. Let $ABCD$ be a trapezium, in which sides $AB\cong a$, $BC\cong b$, $CD\cong c$ and $DA\cong d$ (Figure \ref{sl.skl.3.4.10c.pic}). In this case, $AB\geq CD$, so on the side $AB$ there exists a point $E$, such that $AE\cong CD$. Because $AB\parallel CD$, by \ref{paralelogram} the quadrilateral $AECD$ is a parallelogram, so by the same theorem $CE\cong DA\cong d$. It also holds that $EB=AB-AE=AB-CD=a-c$. This allows us to construct the triangle $EBC$ ($EB=a-c$, $BC\cong c$ and $CE\cong d$), and then the vertices $A$ and $D$ (from the condition $AE\cong CD\cong c$). + \kdokaz + + + +%________________________________________________________________________________ + \poglavje{Regular Polygons}\label{odd3PravilniVeck} + +The concept of a square fits into the general definition of a new type of polygons. + A polygon is +\index{pravilni!večkotniki} \pojem{regular}, if all +of its sides are congruent and all of its interior angles are congruent (Figure +\ref{sl.skl.3.5.1.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.1.pic} +\caption{} \label{sl.skl.3.5.1.pic} +\end{figure} + + A square is therefore a regular quadrilateral. An equilateral triangle + \index{trikotnik!pravilni}\pojem{regular triangle} +is also a regular polygon. +This is due to the fact that, in an equilateral triangle, +all angles are also equal. + +We have already established that the sum of the interior angles of any +$n$-gon is equal to $(n - 2) \cdot 180^0$ (Theorem \ref{VsotKotVeck}). Because +in a regular $n$-gon, all interior angles are congruent, we can calculate the interior angle +by dividing the sum of all angles by the number $n$. Thus, we have +proved the following theorem (Figure \ref{sl.skl.3.5.2.pic}). + + + \bizrek \label{pravVeckNotrKot} + The measure of each interior angle of a regular $n$-gon is: + $$\frac{(n - 2)\cdot 180^0}{n}.$$ + \eizrek + + Thus, the interior angle of a regular triangle measures $60^0$, the interior angle of a regular quadrilateral measures $90^0$, the interior angle of a regular pentagon measures $108^0$, the interior angle of a regular hexagon measures $120^0$, ... + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.2.pic} +\caption{} \label{sl.skl.3.5.2.pic} +\end{figure} + +We shall now prove two important properties of regular polygons. + + + \bizrek \label{sredOcrtaneKrozVeck} + For each regular polygon, there exists a circle passing through each of its vertices. + \eizrek + +\textbf{\textit{Proof.}} Let $A_1A_2\ldots A_n$ be a regular +$n$-gon (Figure \ref{sl.skl.3.5.2.pic}). Then all sides are +congruent and all internal angles are congruent and equal to $\frac{(n +- 2)\cdot 180^0}{n}$. Let $s_1$ and $s_2$ be the lines of symmetry of sides +$A_1A_2$ and $A_2A_3$ of this polygon and let $S$ be their +intersection point. From \ref{simetrala} it follows that $SA_1 \cong SA_2$ +and $SA_2 \cong SA_3$ or: + $$SA_1 \cong SA_2 \cong SA_2 \cong SA_3.$$ + It follows that the triangle $A_1SA_2$ and $A_2SA_3$ +are congruent (from \ref{SSS} - \textit{SSS}). Then the angles +$SA_1A_2$, $SA_2A_1$, $SA_2A_3$ and $SA_3A_2$ are congruent. From $\angle SA_2A_1 \cong +\angle SA_2A_3$ it follows that both angles are equal to half of the internal +angle of the polygon or $\frac{\alpha}{2}=\frac{(n-2)\cdot +180^0}{2n}$. Therefore: + $$\angle SA_3A_4 =\alpha - \frac{\alpha}{2}=\frac{\alpha}{2} = \angle + SA_3A_2.$$ +Thus the triangle $A_2SA_3$ and $A_3SA_4$ are congruent (from \ref{SKS} - \textit{SAS}). Because of this, $SA_3 \cong SA_4$ or: + $$SA_1 \cong SA_2 \cong SA_2 \cong SA_3\cong SA_4.$$ + If we continue this process, we get: + $$SA_1 \cong SA_2 \cong SA_2 \cdots \cong SA_n,$$ +which means that the point $S$ is the center of the circle $k(S, SA_1)$, which +contains all its vertices. + \kdokaz + +The circle from the previous theorem is called the \index{circumscribed circle!regular polygon} \pojem{circumscribed circle +of a regular polygon}. From the proof of the previous theorem it is clear that +its center lies at the intersection of the lines of symmetry of all its sides. + +We prove the following theorem in an analogous way. + + + \bizrek \label{sredVcrtaneKrozVeck} + For each regular polygon, there exists a circle touching each of its sides. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.3.pic} +\caption{} \label{sl.skl.3.5.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $A_1A_2\ldots A_n$ be a regular $n$-gon + (Figure \ref{sl.skl.3.5.3.pic}). +Define the point $S$ as in the proof of the previous statement. +We have shown that: $SA_1 \cong SA_2 \cong SA_2 \cdots \cong +SA_n$. From this, by statement \ref{SSS} - \textit{SSS}, it follows that the congruence of the isosceles triangles: + $$\triangle A_1SA_2 \cong \triangle A_2 SA_3 \cong \cdots \cong + \triangle A_{n-1}SA_n \cong \triangle A_nSA_1.$$ +Because of this, all angles at the bases of these triangles are also congruent. +Therefore, the lines $SA_1$, $SA_2$,..., $SA_n$ are the altitudes of the polygon $A_1A_2\ldots A_n$. Let $P_1$, $P_2$,..., +$P_n$ be the points of intersection of these altitudes with the polygon $A_1A_2\ldots A_n$. +From the congruence of the triangles $\triangle A_1SP_1$, $\triangle A_2SP_1$, +$\triangle A_2SP_2$, ..., $\triangle A_1SP_n$ (statements \ref{SSK} and +\ref{KSK}), it follows that the segments $SP_1$, $SP_2$,..., $SP_n$ are congruent. By statement \ref{TangPogoj}, the circle $k(S, SP_1)$ touches all sides +of the polygon $A_1A_2\ldots A_n$. + \kdokaz + +The circle from the previous statement is called the \index{circumscribed circle!of a regular polygon} \pojem{circumscribed circle of a regular +polygon}. From the proof of this statement, it is clear that the center of the circumscribed circle of a regular polygon lies at the intersection of the altitudes of all its internal angles. From the proof it is also clear that the points at which this circle touches the sides of the regular polygon are also the centers of these sides. The center of the circumscribed and inscribed circle is the same point and is therefore also called the \index{center!of a regular polygon}\pojem{center of a regular polygon}. + +We have also seen that all triangles, +determined by the center +of the regular $n$-gon and by its sides, are isosceles and all +congruent. The radius of the circumscribed and inscribed circle of the $n$-gon is equal +to the leg or. the height of each of these triangles. The angles at the top of these +triangles are also congruent and because there are a total of $n$ (the same as +the sides of the $n$-gon), each of them measures (Figure \ref{sl.skl.3.5.4.pic}): + $$\varphi = \frac{360^0}{n}.$$ + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.4.pic} +\caption{} \label{sl.skl.3.5.4.pic} +\end{figure} + + For a regular hexagon, or for $n = 6$, it holds: + $$\varphi = \frac{360^0}{6}=60^0.$$ + + This means that the aforementioned triangles are regular. +Therefore, a regular hexagon is composed of six +regular triangles (Figure \ref{sl.skl.3.5.4.pic}). + +In the following we will consider the properties of regular $n$-gons. + + Let $n$ be an even number and $k = \frac{n}{2}+1$. +We say that $A_k$ is the \pojem{opposite vertex} of vertex $A_1$ +of the regular $n$-gon $A_1A_2\ldots A_n$ (Figure +\ref{sl.skl.3.5.5.pic}). Analogously, $A_2$ and $A_{k+1}$, $A_3$ and +$A_{k+2}$, ... , $A_{k-1}$ and $A_n$ are opposite vertices of this +$n$-gon. Similarly, the sides $A_1A_2$ and $A_kA_{k+1}$, ... , +$A_{k-1} A_k$ and $A_nA_1$ of the polygon +$A_1A_2\ldots A_n$ are \pojem{opposite sides}. We notice that it holds: + $$\angle A_1SA_k=\frac{n}{2}\varphi = \frac{n}{2}\cdot + \frac{360^0}{n}=180^0,$$ +which means that the diagonal $A_1A_k$ of this $n$-gon contains its +center. Therefore, this diagonal represents the diameter of the circumscribed circle. +Analogously, this holds for all diagonals determined by opposite +vertices. Because of this, we call such diagonals \index{velika diagonala +pravilnega $n$-kotnika} \pojem{major diagonals} of a regular +$n$-gon. The radius of the circumscribed circle is + equal to half the major diagonal. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.5.pic} +\caption{} \label{sl.skl.3.5.5.pic} +\end{figure} + +It can be proven in a similar way that for an even number $n$ of centers of the opposite sides of a regular $n$-gon +$A_1A_2\ldots A_n$ determine the diameters of the inscribed circle of this +$n$-gon. If we consider the previous labels, we get: + \begin{eqnarray*} + \angle P_1SP_k&=&\angle P_1SA_2 + \angle A_2SA_{k-1} + \angle + A_{k-1}SP_k\\ + &=& \frac{\varphi}{2}+\frac{n-2}{2}\cdot \varphi+\frac{\varphi}{2} + =\frac{n}{2}\cdot\varphi + =180^0. + \end{eqnarray*} + + + The distances that are determined by a pair of centers of opposite sides of an $n$-gon + $A_1A_2\ldots A_n$ or the distances $P_1P_k$, +$P_2P_{k+1}$, ... , $P_{k-1}P_n$, are called the \pojem{heights} +\index{height!of a regular $n$-gon} of this $n$-gon. The radius of the inscribed circle is equal to half the height. + + So every regular $n$-gon, where $n$ is an even number, contains +$\frac{n}{2}$ big diagonals (equal to the diameter of the circumscribed circle) and +$\frac{n}{2}$ heights (equal to the diameter of the inscribed circle). Each of them +goes through the center of this $n$-gon. + + + Let $n$ be an odd number now (Figure \ref{sl.skl.3.5.6.pic}) and + $k=\frac{n+1}{2}+1$. + Then: +\begin{eqnarray*} + \angle P_1SA_k=\angle P_1SA_2 + \angle A_2SA_k= + \frac{\varphi}{2}+\frac{n-1}{2}\cdot \varphi + =\frac{n}{2}\cdot\varphi + =180^0. + \end{eqnarray*} + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.6.pic} +\caption{} \label{sl.skl.3.5.6.pic} +\end{figure} + +This means that the distance $P_1A_k$ contains the center $S$ of a regular +$n$-gon $A_1A_2\ldots A_n$. This distance is called the \index{height!of a regular $n$-gon} +\pojem{height} of this $n$-gon, the side +$A_1A_2$ and the point $A_k$ are \pojem{opposite}. We define the remaining $n$ heights and $n$ pairs of opposite sides and points in a similar way. In a similar way we can prove that the other heights of this $n$-gon also contain its center. + +In a right (isosceles) triangle, we therefore have three +heights, which intersect in its center (Figure +\ref{sl.skl.3.5.7.pic}). If this is true for +any triangle, we will find out later + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.7.pic} +\caption{} \label{sl.skl.3.5.7.pic} +\end{figure} + + In a square, its diagonals are also the main + diagonals and intersect in +its center (Figure \ref{sl.skl.3.5.7.pic}). The heights +of the square are consistent with its side, which is not difficult to prove. + +We have already mentioned that a regular hexagon is composed of six +triangles, which intersect in its center. We will consider a regular +hexagon in more detail later. We will also prove +some properties of a regular pentagon, heptagon, nonagon and dodecagon. The problem +of designing regular $n$-gons for any number $n$ will be particularly interesting. + +We will now prove an interesting property of a regular nonagon. + + \bzgled + If $a$ is a side and $d$ and $e$ are the shortest and longest + diagonal of a regular nonagon ($9$-gon), then $e - d = a$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.8.pic} +\caption{} \label{sl.skl.3.5.8.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $d = CE$ and $e = BF$ be the shortest and longest +diagonals of a regular nonagon $ABCDEFGHI$ with side +$a$ and $P$ the intersection of lines $BC$ and $FE$ (Figure +\ref{sl.skl.3.5.8.pic}). The internal angle of this nonagon measures + $\angle CDE=\frac{9-2}{9}\cdot 180^0=140^0$, +so $\angle ECD = \angle CED = 20^0$. From this it follows + $\angle BCE = \angle FEC = 120^0$, i.e.: + $$\angle ECP = \angle CEP = 60^0.$$ + Therefore, triangle $CPE$ is regular. Since $CB = EF=a$ +and $\angle BPF \cong \angle CPE = 60^0$, triangle $BPF$ is also regular. Therefore: + $$e = BF = BP = BC + CP = BC + CE = a + d,$$ which was to be proved. \kdokaz + + +%%________________________________________________________________________________ + \poglavje{Midsegment of Triangle} \label{odd3SrednTrik} + +We will now look at a very important property of a triangle, which we will use often. Let $P$ and $Q$ be the centers of the sides $AB$ and $AC$ of the triangle $ABC$. The distance $PQ$ is called the \index{midsegment!of a triangle} \pojem{midsegment of the triangle} $ABC$, which corresponds to the side $BC$ (Figure \ref{sl.skl.3.6.1.pic}). We also say that $PQ$ is the midsegment of the triangle $ABC$ with the base $BC$. We prove the basic property that relates to the midsegment. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.1.pic} +\caption{} \label{sl.skl.3.6.1.pic} +\end{figure} + + + + \bizrek \label{srednjicaTrik} + Let $PQ$ be the midsegment of a triangle $ABC$ corresponding to the side $BC$. Then: + $$ PQ = \frac{1}{2} BC\hspace*{2mm} + \textrm{ in } \hspace*{2mm} PQ \parallel BC.$$ + \eizrek + + + + \textbf{\textit{Proof.}} Let $R$ be a point such that $PQ \cong QR$ and $\mathcal{B}(P,Q,R)$ (Figure \ref{sl.skl.3.6.1.pic}). The line segments $AC$ and $PR$ have a common center, so the quadrilateral $APCR$ is a parallelogram (\ref{paralelogram}). Therefore, the line segments $AP$ and $RC$ are congruent and parallel. The point $P$ is the center of the line segment $AB$, so the line segments $PB$ and $RC$ are congruent and parallel. This means that the quadrilateral $PBCR$ is a parallelogram. It follows that the line segments $BC$ and $PR$ are congruent and parallel. The final conclusion follows from the fact that the point $Q$ is the center of the line segment $PR$. + \kdokaz + + + \bzgled + Let $AB$ and $A'B'$ be congruent line segments, $C$ and $D$ the midpoints of the line segments + $AA'$ and $BB'$. Suppose that $CD =\frac{1}{2} AB$. + What is a measure of the angle between the lines $AB$ and $A'B'$? + \ezgled + +\textbf{\textit{Solution.}} Let point $S$ be the center of line +$A'B$ (Figure \ref{sl.skl.3.6.2.pic}). Lines $CS$ and $DS$ are the +medians of triangles $A'AB$ and $BA'B'$, so: $$CS = \frac{1}{2}AB = +CD = \frac{1}{2}A'B'= DS,$$ or $SCD$ is an equilateral triangle. +Angles $\angle AB,A'B'$ and $\angle CSD$ have corresponding sides. +Therefore: $\angle AB, A'B' \cong \angle CSD = 60^0$. + \kdokaz + +The next consequence of \ref{srednjicaTrik} applies to a trapezium. + + + \bizrek \label{srednjTrapez} + Let $P$ and $Q$ be the midpoints of legs $BC$ and $DA$ of a trapezium $ABCD$. + Suppose that $M$ and $N$ are the midpoints of the diagonals $AC$ and $BD$ of that trapezium. + Then the points $M$ and $N$ lie on the line $PQ$, which is parallel to the bases $AB$ in $CD$ of the trapezium, + and also: + $$PQ = \frac{1}{2}( AB + CD) \hspace*{2mm} + \textrm{ in } \hspace*{2mm} MN=\frac{1}{2}( AB - CD).$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.3.pic} +\caption{} \label{sl.skl.3.6.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} The lines $PN$, $NQ$ and $PM$ are in turn medians of +the triangles +$DAC$, $ACB$ and $ADB$ for +the corresponding bases $DC$, $AB$ and $AB$ (Figure +\ref{sl.skl.3.6.3.pic}). Because of this, all three lines $PN$, $NQ$ +and $PM$ are parallel to the bases $CD$ and $AB$. Since through each point (first $N$, then $P$) there is only one parallel to the line +$AB$ (Playfair's\footnote{\index{Playfair, J.}\textit{J. +Playfair} (1748--1819), Scottish mathematician.} axiom +\ref{Playfair}), the points $P$, $N$, $M$ and $Q$ are collinear. It also holds (from the statement \ref{srednjicaTrik}): + $PN = \frac{1}{2} CD$ and + $NQ = PM = \frac{1}{2} AB$. From this it follows: + \begin{eqnarray*} + PQ&=& PN+NQ=\frac{1}{2}CD+ + \frac{1}{2}AB=\frac{1}{2}\left(AB+CD\right)\\ + NM&=& PM-PN=\frac{1}{2}AB- + \frac{1}{2}CD=\frac{1}{2}\left(AB-CD\right), + \end{eqnarray*} + which was to be proven. \kdokaz + +The line $PQ$ from the previous statement is called +\index{srednjica!trapeza} \pojem{median of the trapezoid}. + + The following statements apply to an arbitrary quadrilateral. + + + \bizrek \label{Varignon} + Let $ABCD$ be an arbitrary quadrilateral and $P$, $Q$, $K$ and $L$ + the midpoints of the sides $AB$, $CD$, $BC$ and $AD$, respectively. Then the quadrilateral $PKQL$ is + a parallelogram (so-called \index{paralelogram!Varignonov} + Varignon\footnote{\index{Varignon, P.} + \textit{P. Varignon} (1654--1722), + French mathematician, + who was the first to prove this property. However, the statement was not published until + after his death in 1731. Given the simplicity, it is + quite surprising that this statement "waited" so long to be + discovered.} parallelogram). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.4.pic} +\caption{} \label{sl.skl.3.6.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.6.4.pic}) +The lines $PK$ and $LQ$ are the medians of the triangles $ABC$ and $ADC$ for the same base $AC$, so they are congruent and +parallel. Therefore, the quadrilateral $PKQL$ is a parallelogram. + \kdokaz + +In a special case, Varignon's parallelogram can even be a rectangle, +rhombus or square. When is this possible? This question gives us an idea for +the next statement. We know that a parallelogram is a rectangle if it has +at least one internal angle right. The other (equivalent) condition is that it has congruent diagonals. A similar treatment can be used for rhombus and square. + + + \bzgled \label{VarignonPoslPravRomb} + Let $ABCD$ be an arbitrary quadrilateral and $P$, $K$, $Q$ and $L$ + the midpoints of the sides $AB$, $BC$, $CD$ and + $DA$, respectively. Then: + + a) $AC \perp BD \Leftrightarrow PQ \cong KL$; + + b) $AC \cong BD \Leftrightarrow PQ \perp KL$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.5.pic} +\caption{} \label{sl.skl.3.6.5.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.6.5.pic}) + From the previous statement \ref{Varignon} it follows that + the quadrilateral $PKQL$ is always a parallelogram – Varignon's +parallelogram. The lines $PL$ and $PK$ are the medians of the triangles $ABD$ +and $ABC$ for the bases $AD$ and $AC$. Therefore, we have: + $PL= \frac{1}{2}BD$ and $PL \parallel BD$ and $PK= \frac{1}{2}AC$ and $PK \parallel AC$. + Therefore, we have: + + a) $AC \perp BD \Leftrightarrow PL \perp PK +\Leftrightarrow PKQL \textrm{ rectangle} \Leftrightarrow PQ +\cong KL$; + + b) $AC \cong BD \Leftrightarrow PL \cong PK \Leftrightarrow + PKQL \textrm{ rhombus } \Leftrightarrow PQ \perp KL$. + \kdokaz + +If Varignon's parallelogram is a square, all four +conditions from the previous equivalences are fulfilled, or in this case: $AB \perp CD$, $AB \cong CD$, $PQ \perp KL$ and $PQ \cong KL$. + +Let's look at one more use of the property of Varignon's parallelogram. + +\bzgled \label{VagnanPosl} +Let $ABCD$ be an arbitrary quadrilateral. If $P$, $Q$, $K$, $L$, $M$ and +$N$ are the midpoints of the line segments $AB$, $CD$, $BC$, $AD$, $AC$ and $BD$, respectively, +then the line segments $PQ$, $KL$ and +$MN$ have a common midpoint. +\ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.6.pic} +\caption{} \label{sl.skl.3.6.6.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + (Figure \ref{sl.skl.3.6.6.pic}) Štirikotnik $PKQL$ je Varignonov +paralelogram (izrek \ref{Varignon}). Na podoben način dokazujemo, da +je tudi štirikotnik $LNKM$ paralelogram (srednjice trikotnikov $ADB$ +in $ACB$). Paralelograma $PKQL$ in $LPKQ$ imata skupno diagonalo +$LK$. Ker se diagonali poljubnega paralelograma razpolavljata (izrek +\ref{paralelogram}), imajo daljice $LK$, $PQ$ in $MN$ skupno +središče. + \kdokaz + +Točko iz prejšnjega primera, v kateri se daljice sekajo, imenujemo +\index{težišče!štirikotnika} \pojem{težišče štirikotnika}. Več +o tem bomo povedali v razdelku \ref{odd5TezVeck}. + +Naslednja trditev je lep primer kombiniranja neenakosti trikotnika +in srednjice trikotnika. Trditev je pravzaprav posplošitev izreka +o srednjici trapeza \ref{srednjTrapez}. + + + \bizrek + If $P$ and $Q$ are the midpoints of the sides $AB$ and $CD$ of + an arbitrary quadrilateral + $ABCD$, then: $$PQ \leq \frac{1}{2}\left( BC + AD\right).$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.7.pic} +\caption{} \label{sl.skl.3.6.7.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $S$ be the center of the diagonal $AC$ of the quadrilateral $ABCD$ (Figure \ref{sl.skl.3.6.7.pic}). If we use the theorem about the median of a triangle (\ref{srednjicaTrik}) and the triangle inequality (\ref{neenaktrik}), we get: $$BC + AD = 2PS + 2SQ = 2(PS + SQ) \geq 2PQ.$$ The equality holds when the points $P$, $S$ and $Q$ are collinear, i.e. when the quadrilateral $ABCD$ is a trapezoid with the base $BC$. \kdokaz + + We mention that the inequality from the previous example also holds when the points $A$, $B$, $C$ and $D$ are not in the same plane, i.e. when the $ABCD$ is a \pojem{tetrahedron}. + + + \bzgled \label{TezisceSredisceZgled} + Let $P$ be the midpoint of the median $AA_1$ of a triangle $ABC$ and + $Q$ + the intersection of the side $AC$ and the line $BP$. Determine the ratios + $AQ :QC$ and $BP : PQ$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.8.pic} +\caption{} \label{sl.skl.3.6.8.pic} +\end{figure} + + +\textbf{\textit{Solution.}} Let $R$ be the center of the line segment $QC$ (Figure \ref{sl.skl.3.6.8.pic}). The line segment $A_1R$ is the median of the triangle $BCQ$ for the base $BQ$, so (by the theorem \ref{srednjicaTrik}) $BQ = 2A_1R$ and $BQ\parallel A_1R$. From this parallelism and the definition of the point $P$ it follows that $PQ$ is the median of the triangle $AA_1R$ for the base $A_1R$, so (by the theorem \ref{srednjicaTrik} and Playfair's axiom \ref{Playfair}) the point $Q$ is the center of the line segment $AR$ and it holds $A_1R = 2PQ$. Therefore: $AQ \cong QR \cong RC$ i.e. $AQ:QC=1:2$. In the end it is also $BQ=2A_1R=4PQ$ i.e. $BP:PQ=3:1$. + \kdokaz + +\bnaloga\footnote{19. IMO Yugoslavia - 1977, Problem 1.} + Equilateral triangles $ABP$, $BCL$, $CDM$, $DAN$ are constructed inside the + square $ABCD$. Prove that the midpoints of the segments $LM$, $MN$, $NP$, $PL$, $AN$, $LB$, + $BP$, $CM$, $CL$, $DN$, $DM$ in $AP$ + are the twelve vertices of a regular dodecagon ($12$-gon). + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.IMO1.pic} +\caption{} \label{sl.skl.3.6.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} We mark with $a$ the length of the side and +with $O$ the center of the square $ABCD$ (Figure \ref{sl.skl.3.6.IMO1.pic}). + +We first prove that the quadrilateral $MNPL$ is also a square with the same +center $O$. Because $ABP$, $BCL$, $CDM$ and $DAN$ are all right +triangles, the diagonals $MP$ and $LN$ of the quadrilateral $MNPL$ are on +the similitudes of the sides $AB$ and $BC$ of the square $ABCD$. From this it follows that $MP \perp LN$. +Since $d(M,AB)=d(P,CD)=a-v$ (where $v$ is the length +of the height of the aforementioned right triangles), it also holds that $OM\cong OP$. +Similarly, $OL\cong ON$, which means that the quadrilateral $MNPL$ +is really a square with the same center~$O$. + +We now prove that $LAM$ is a right triangle. Because $AB\cong +AD\cong BL\cong DM=a$ and $\angle LBA\cong\angle MDA +=90^0-60^0=30^0$, the triangles $LBA$ and $MDA$ are similar (by the \textit{SAS} \ref{SKS}). This means that $LA\cong MA$ and $\angle +DAL= 90^0-\angle LAB=15^0$. Similarly, $\angle BAM=15^0$ or +$\angle LAM = 90^0-2\cdot 15^0=60^0$. Therefore, $LAM$ is a right triangle, so the side of the square $MNPL$ has length +$b=|LM|=|LA|$. + +Let's mark the center of the line $LM$ with $S$. The centers of the sides of the square $MNPL$ lie on the circle $k(O,\frac{b}{2})$, which is the inscribed circle of this square. We will prove that the point $T$ - the center of the line $AN$ - also lies on this circle. The line $OT$ is the median of the triangle $LAN$ for the base $LA$, so $OT\parallel LA$ and $|OT|=\frac{1}{2}|LA|=\frac{b}{2}$. Therefore, the point $T$ and, analogously, all the points defined in the problem of the $12$-gon lie on the circle $k(O,\frac{b}{2})$. + +We will also prove that the aforementioned $12$-gon is regular. Without loss of generality, it is enough to prove that $\angle SOT=\frac{360^0}{12}=30^0$. But from the already proven fact $OT\parallel LA$ it follows that $\angle SOT\cong \angle LAS=\frac{1}{2}\angle LAM=30^0$. + \kdokaz + +%________________________________________________________________________________ + \poglavje{Triangle Centers} \label{odd3ZnamTock} + + We will continue our research with a triangle - the most simple + polygon, + which is at the same time +a figure that has unexpectedly many interesting properties. We will +discuss some of them in this section, some of them later, when we will +deal with other concepts, such as isometries and similarity. + + Now we will consider four \index{značilne točke + trikotnika}\pojem{značilne točke + trikotnika}\footnote{These four points are mentioned by the Ancient Greeks, although they + (especially the centroid) were probably known long before that.} and + their use in quadrilaterals and polygons. + +We will start with the first among the characteristic points, which is related to + the medians of the triangle. We have already defined + the median in chapter \ref{odd3NeenTrik}. + + + + \bizrek \label{tezisce} + The medians of a triangle intersect at one point. + That point divides the medians in the ratio $2:1$ + (from the vertex to the midpoint of the opposite side). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.1.pic} +\caption{} \label{sl.skl.3.7.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Let $AA_1$, $BB_1$ and $CC_1$ be the altitudes of the triangle $ABC$ + (Figure \ref{sl.skl.3.7.1.pic}). +Because of Pasch's axiom, \ref{AksPascheva} with respect to the triangle +$BCB_1$ and the line $AA_1$, the line $AA_1$ intersects the line segment $BB_1$. +Similarly, the line $BB_1$ intersects the line $AA_1$, which means that the +altitudes $AA_1$ and $BB_1$ intersect at some point $T$. Let $A_2$ +and $B_2$ be the midpoints of the line segments $AT$ and $BT$. The line segments $A_1B_1$ and $A_2B_2$ +are the medians of the triangles $ABC$ and $ABT$ with the same base $AB$. This +means that the line segments $A_1B_1$ and $A_2B_2$ are parallel and equal +to half of the side $AB$. Therefore, the quadrilateral $B_2A_1B_1A_2$ +is a parallelogram (statement \ref{paralelogram}), which means that its +diagonals $A_1A_2$ and $B_1B_2$ intersect at their common center $T$. So it holds: + $A_1T\cong TA_2 \cong A_2A$ and $B_1T\cong TB_2 \cong B_2B$ + or $AT:TA_1=2:1$ and $BT:TB_1=2:1$ and + $$A_1T=\frac{1}{3}A_1A \textrm{ and } B_1T = \frac{1}{3}B_1B.$$ + In the same way + we prove that the altitudes $AA_1$ and $CC_1$ intersect at some +point $T'$, for which it holds $AT':T'A_1=2:1$ and $CT':T'C_1=2:1$ or: + $$A_1T'=\frac{1}{3}A_1A \textrm{ and } C_1T' = \frac{1}{3}C_1C.$$ +This means that $T$ and $T'$ are points on the line segment $A_1A$, for which it holds +$A_1T\cong A_1T'=\frac{1}{3}A_1A$, so according to \ref{ABnaPoltrakCX} $T = T'$, which means that the altitudes $AA_1$, +$BB_1$ and $CC_1$ intersect at the point $T$ and it holds +$AT:TA_1=BT:TB_1=CT:TC_1=2:1$. + \kdokaz + + +The point from the previous statement, in which all altitudes of the triangle intersect, is called the +\index{težišče!trikotnika}\pojem{center of the triangle}. + +The center of a triangle in the physical sense represents the point that is the center of mass of that triangle. This will be even more clear when we in section \ref{odd8PloTrik} prove the fact that the center divides the triangle into triangles with the same area. In the next chapter \ref{pogVEKT} (section \ref{odd5TezVeck}) we will consider the center of any polygon. + +In section \ref{odd3PravilniVeck} we found that for any regular polygon there exist circumscribed (which contains all its vertices) and inscribed (which touches all its sides) circle with the same center. This property is also transferred to regular or equilateral triangles. But how is it with any triangle? We will prove that there exist the mentioned circles for any triangle, just that in the general case they have different centers. + +\bizrek \label{SredOcrtaneKrozn} + The perpendicular bisectors of the sides of any triangle intersect at a single point, + which is the centre of a circle containing all its vertices. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.2.pic} +\caption{} \label{sl.skl.3.7.2.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Let $p$, $q$ and $r$ be the perpendicular bisectors of sides $BC$, $AC$ and $AB$ of triangle $ABC$ (Figure \ref{sl.skl.3.7.2.pic}). The perpendicular bisectors $p$ and $q$ are not parallel (because in that case by Playfair's axiom \ref{Playfair} the lines $BC$ and $AC$ would be parallel too) and they intersect in some point $O$. Because this point lies on the perpendicular bisectors $p$ and $q$ of sides $BC$ and $AC$, we have $OB \cong OC$ and $OC \cong OA$. From this follows $OA \cong OB$, which means that the point $O$ also lies on the perpendicular bisector $r$ of the line $AB$. So the perpendicular bisectors $p$, $q$ and $r$ intersect in one point. + +Because $OA \cong OB \cong OC$, the point $O$ is the center of the circle $k(O,OA)$, which contains all vertices of the triangle $ABC$. + \kdokaz + +The circle from the previous theorem, which contains all the vertices of the triangle, is called the \index{circumscribed circle!triangle} \pojem{circumscribed circle of the triangle}, and its center is the \index{center!circumscribed circle!triangle} \pojem{center of the circumscribed circle of the triangle}. + +\bizrek \label{SredVcrtaneKrozn} +The bisectors of the interior angles of any triangle intersect at a single point, which is the centre of a circle touching all its sides. +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.3.pic} +\caption{} \label{sl.skl.3.7.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Let $p$, $q$, and $r$ be the angle bisectors of the angles at the vertices $A$, $B$, and $C$ of the triangle $ABC$ (Figure \ref{sl.skl.3.7.3.pic}). We shall prove that the bisectors $p$ and $q$ are not parallel. Otherwise, by Theorem \ref{KotiTransverzala}, the sum of the halves of the angles at the vertices $A$ and $B$ would be equal to $180°$, which, by Theorem \ref{VsotKotTrik}, is not possible. Therefore, $p$ and $q$ intersect at some point $S$. Since the point $S$ lies on the angle bisectors of the angles at the vertices $A$ and $B$, it is equidistant from the sides $AC$ and $AB$ (Theorem \ref{SimKotaKraka}). From this it follows that the point $S$ is also equidistant from the sides $BA$ and $BC$, which means that it lies on the bisector $r$ of the angle at the vertex $C$. Therefore, the angle bisectors $p$, $q$, and $r$ intersect at the point $S$. + +With $P$, $Q$ in $R$ we denote the orthogonal projections of point $S$ onto +the sides $BC$, $CA$ and $AB$. Because $\frac{1}{2}\angle CBA<90^0$ +and $\frac{1}{2}\angle BCA<90^0$, it follows that $\mathcal{B}(B,P,C)$. Similarly, +$\mathcal{B}(C,Q,A)$ and $\mathcal{B}(A,R,B)$ also hold. Because of the already proven properties of point $S$, it follows that $SP \cong SQ \cong SR$. Therefore, the point $S$ is the center of the circle $l$, which passes through points $P$, $Q$ and +$R$. Because of the perpendicularity of the radii $SP$, $SQ$ and $SR$ to the respective sides, they are tangent to the circle $l$. Because $\mathcal{B}(B,P,C)$, $\mathcal{B}(C,Q,A)$ and $\mathcal{B}(A,R,B)$ hold, the circle $l$ is tangent to all + sides of the triangle $ABC$. + \kdokaz + + + The circle from the previous statement, which is tangent to all + sides of the triangle, is called \index{včrtana krožnica!trikotnika} \pojem{the inscribed circle of the triangle}, and its center + \index{središče!včcrtane krožnice!trikotnika} + \pojem{the center of the inscribed circle of the triangle}. + +There is one more of the four aforementioned characteristic points of the triangle. +It is related to the altitudes of the triangle. + + + \bizrek \label{VisinskaTocka} + The lines containing the altitudes of a triangle intersect at a single point. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.4.pic} +\caption{} \label{sl.skl.3.7.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Let $p$, $q$ and $r$ be the altitude lines + $AA'$, $BB'$ and $CC'$ + of the triangle $ABC$ (Figure \ref{sl.skl.3.7.4.pic}). +We denote by $a$, $b$ and $c$ the lines that are perpendicular to the respective heights in points $A$, $B$ and $C$. Because the lines $a$, $b$ and $c$ are parallel to the sides of the triangle $ABC$, each two of them intersect. We denote by $P$, $Q$ and $R$ the intersections of the lines $b$ and $c$, $a$ and $c$, and $a$ and $b$, in order. The quadrilateral $ABCQ$ and $RBCA$ are parallelograms, which means that $RA \cong BC \cong AQ$, or the point $A$ is the center of the line $RQ$. The line $AA'$ is therefore the perpendicular bisector of the side $RQ$ of the triangle $PQR$. Similarly, $BB'$ and $CC'$ are the perpendicular bisectors of the sides $PR$ and $PQ$ of the same triangle. By the theorem \ref{SredOcrtaneKrozn}, the perpendicular bisectors $AA'$, $BB'$ and $CC'$ of the triangle $PQR$ intersect in some point $V$. The point $V$ is therefore the intersection of the altitude lines + $AA'$, $BB'$ and $CC'$ + of the triangle $ABC$. + \kdokaz + +The point from the previous theorem, in which the altitude lines intersect, is called +\index{višinska točka trikotnika} + \pojem{the altitude point of the triangle}. The triangle $A'B'C'$, which is determined by the altitudes of the triangle $ABC$, is called +\index{trikotnik!pedalni} \pojem{the pedal triangle} of the triangle +$ABC$. + +We have found that every triangle has four characteristic points, +namely: the centroid, the center of the circumscribed circle, the center of the inscribed circle and the altitude point. But these are not the only characteristic points +of the triangle. We will mention some of them later. For +a point in the plane of the triangle in general, we say that it is its +\pojem{characteristic point}, if its definition is symmetrical with respect to +the vertices of this triangle. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.5.pic} +\caption{} \label{sl.skl.3.7.5.pic} +\end{figure} + + + It is clear that in any triangle the four characteristic points differ + (Figure \ref{sl.skl.3.7.5.pic}). + +In an isosceles triangle, the centroid, +the altitude, the line of symmetry of the base, and the line of symmetry of the internal angle opposite the base +have the same carrier. If $A_1$ is the center of the base +$BC$ of the isosceles triangle $ABC$, the triangles $ABA_1$ and +$ACA_1$ are congruent, which means that the angle at the vertex $A_1$ is a right angle +and the angles $BAA_1$ and $CAA_1$ are congruent. Therefore, the distance +$AA_1$ is both the centroid and the altitude, and the line $AA_1$ is both +the line of symmetry of the side $BC$ and the line of symmetry of the internal angle at the vertex $A$ +of the triangle $ABC$. It follows that all four characteristic points of this +triangle lie on one line $AA_1$ (Figure +\ref{sl.skl.3.7.6.pic}). + +If we use the already proven property of an isosceles triangle for +an equilateral triangle, we find that all the appropriate +centroids, altitudes, side lines of symmetry, and internal angle lines of symmetry +have the same carrier. This means that in an equilateral triangle +all four characteristic points coincide (Figure \ref{sl.skl.3.7.6.pic}). +This is actually already defined (section \ref{odd3PravilniVeck}) +as the center of this equilateral (or regular) triangle. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.6.pic} +\caption{} \label{sl.skl.3.7.6.pic} +\end{figure} + +We will also show the position of the characteristic points with respect to the type of +triangle. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.7.pic} +\caption{} \label{sl.skl.3.7.7.pic} +\end{figure} + +The centroids are always inside the triangle. Therefore, the center +is an internal point of every triangle (Figure \ref{sl.skl.3.7.7.pic}). +The same conclusion applies to the center of the inscribed circle (Figure +\ref{sl.skl.3.7.7s.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.7s.pic} +\caption{} \label{sl.skl.3.7.7s.pic} +\end{figure} + +In an acute triangle, the vertices of the perpendiculars from its vertices +lie on the sides of this triangle, which means (Pasch's axiom +\ref{AksPascheva}), that its altitudes intersect in the interior. So in +an acute triangle, the altitude point lies in its interior +(Figure \ref{sl.skl.3.7.7v.pic}). In a right-angled triangle, the altitude point is the vertex at the right angle. This is because its +catheti are at the same time altitudes of the triangle. The altitude point of an obtuse triangle lies in its exterior, because + not all of the altitudes are in its interior. The corresponding vertices +belong to the sides' supports, not to the sides themselves. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.7v.pic} +\caption{} \label{sl.skl.3.7.7v.pic} +\end{figure} + +The centre of the circumscribed circle is an interior or exterior point of the triangle, depending on whether the triangle is acute or obtuse (Figure +\ref{sl.skl.3.7.7o.pic}). We will omit the formal proof of this fact. We will only prove the following statement, which refers to right-angled triangles. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.7o.pic} +\caption{} \label{sl.skl.3.7.7o.pic} +\end{figure} + + + \bizrek The circumcentre of a right-angled triangle is at the same time the midpoint of + its hypotenuse. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.8.pic} +\caption{} \label{sl.skl.3.7.8.pic} +\end{figure} + + +\textbf{\textit{Proof.}} We mark with $O$ the centre of the hypotenuse $AB$ and +with $P$ the centre of the cathetus $AC$ of the right-angled triangle $ABC$ (Figure +\ref{sl.skl.3.7.8.pic}). The distance $OP$ is the median of this triangle, +which corresponds to the cathetus $BC$, so $OP\parallel BC$. From this it follows that $OP +\perp AC$. Therefore, the triangles $OPC$ and $OPA$ are congruent (the statement +\textit{SAS} \ref{SKS}) and then $OC \cong OA$. +Since $OB \cong OA$ as well, + the point $O$ is the centre of the circumscribed circle of this triangle. + \kdokaz + +The line $OC$ from the previous theorem is the median of the triangle. This means that the median of a right angled triangle is equal to the radius of the inscribed circle of that triangle, and also to half of its hypotenuse. + +The previous theorem is also connected to Thales’ theorem for a circle \ref{TalesovIzrKroz} and its converse \ref{TalesovIzrKrozObrat}. We will now state all of these theorems in one theorem. + + + \bizrek Thales’ theorem for a circle (several forms - Figure + \ref{sl.skl.3.7.9.pic}): + \index{theorem!Thales’ for a circle} + \label{TalesovIzrKroz2} + \begin{enumerate} + \item The circumcentre of a right-angled triangle is at the same time the midpoint of + its hypotenuse. + \item If $t_c$ is the median of a right-angled triangle for its hypotenuse $c$ and $R$ + the circumradius of that triangle, then + $R=t_c=\frac{c}{2}$. + \item If $AB$ is a diameter of a circle $k$, then for any point $X\in k$ ($X\neq A$ and $X\neq B$) is + $\angle AXB=90^0$. + \item If $A$, $B$ in $X$ are three non-collinear points, such that $\angle AXB=90^0$, then the point $X$ lies on a circle with the diameter $AB$. + \end{enumerate} + \index{theorem!Thales’ for a circle} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.9.pic} +\caption{} \label{sl.skl.3.7.9.pic} +\end{figure} + +Knowing the characteristic points of a triangle and the properties of a median of a triangle allows us to prove various other properties of both triangles and quadrilaterals and $n$-gons. + + + \bzgled + Let $CD$ be the altitude at the hypotenuse $AB$ of a right-angled triangle $ABC$. + If $M$ and $N$ are the midpoints of the line segments $CD$ and $BD$, then $AM \perp CN$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.10.pic} +\caption{} \label{sl.skl.3.7.10.pic} +\end{figure} + +\textbf{\textit{Proof.}} The line $NM$ is the median of the triangle +$BCD$, so by the \ref{srednjicaTrik} $NM \parallel BC$ +(Figure \ref{sl.skl.3.7.10.pic}). Because the angle at the vertex $C$ is a right +angle, $NM \perp AC$ as well. Because of this, the line $NM$ is the altitude of the triangle $ANC$. Since $CD$ is also the altitude of this triangle, $M$ is its height point. Therefore, the line $AM$ is the altitude of the third triangle of this triangle and $AM \perp CN$ applies. + \kdokaz + + + \bzgled + If $P$ and $Q$ are the midpoints of the sides $BC$ and $CD$ of a parallelogram $ABCD$, + then the lines $AP$ and $AQ$ + divide the diagonal $BD$ of this parallelogram into three congruent line segments. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.11.pic} +\caption{} \label{sl.skl.3.7.11.pic} +\end{figure} + +\textbf{\textit{Proof.}} We mark with $E$ and $F$ the intersections of the lines +$AP$ and $AQ$ with the diagonal $BD$ of the parallelogram $ABCD$ and with $S$ +the intersection of its diagonals $AC$ and $BD$ (Figure +\ref{sl.skl.3.7.11.pic}). The diagonals of the parallelogram are divided +(from \ref{paralelogram}), so the point $S$ is the common center of the lines +$AC$ and $BD$. This means that the points $E$ and $F$ are the centers of the triangles +$ACB$ and $ACD$, so they divide the altitudes $SB$ and $SD$ of these triangles +in the ratio $2:1$ (from \ref{tezisce}). Therefore: + \begin{eqnarray*} + BE &=& \frac{2}{3}BS = \frac{2}{3}DS = FD,\\ + EF&=& ES+ SF= \frac{1}{3}SB+ \frac{1}{3}SD= + \frac{1}{3}(SB +SD)=\frac{1}{3} BD, + \end{eqnarray*} + which is what we wanted to prove. \kdokaz + + + \bzgled + Let $BAKL$ and $ACPQ$ be positively oriented squares + in the same plane. Prove that the lines $BP$ and $CL$ intersect at a point lying + on the line containing the altitude $AA'$ of the triangle $ABC$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.12.pic} +\caption{} \label{sl.skl.3.7.12.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Let $AA'$ + be the altitude of the triangle $ABC$ (Figure \ref{sl.skl.3.7.12.pic}). We mark with $X$ + and $Y$ the intersections of the line $AA'$ with +the rectangles on the line $CL$ through the point $B$ and on the line $BP$ through the point $C$. +We prove that $X = Y$. The triangle $BLC$ and $ABX$ are similar according to the \textit{ASA} theorem \ref{KSK} because: $BL \cong AB$, +$\angle BLC\cong\angle ABX$ and $\angle BCL\cong\angle AXB$ (the angle with perpendicular sides - +theorem \ref{KotaPravokKraki}). Therefore $AX \cong BC$. Similarly, +the triangle $CPB$ and $ACY$ are similar, so $AY \cong BC$. +Therefore $AX \cong AY$ or $X = Y$. This means that the lines +$AA'$, $BP$ and $CL$ are the altitudes of the triangle $XBC$, so they intersect at +one point. + \kdokaz + + + \bzgled \label{zgledPravokotnik} + Let $K$ be the midpoint of the side $CD$ of a rectangle $ABCD$. + A point $L$ is the foot of the perpendicular from the vertex $B$ on the diagonal + $AC$ and $S$ is the midpoint of + the line segment $AL$. Prove that $\angle KSB$ is a right angle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.13.pic} +\caption{} \label{sl.skl.3.7.13.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $V$ be the center of the line $BL$ (Figure \ref{sl.skl.3.7.13.pic}). The line $SV$ is the median of the triangle $ABL$ for the base $AB$, so $SV\parallel AB$ and $SV =\frac{1}{2} AB$ (statement \ref{srednjicaTrik}). From the first relation and $BC\perp AB$ it follows that $SV\perp BC$ (statement \ref{KotiTransverzala}). This means that $BL$ and $SV$ are the altitude of the triangle $CSB$. Therefore, $V$ is the altitude point of this triangle, so $CV$ is the altitude of its third altitude (statement \ref{VisinskaTocka}) or $CV\perp SB$ is true. From $SV\parallel AB$ and $SV =\frac{1}{2} AB=KC$ it follows that the quadrilateral $SVCK$ is a parallelogram or $CV\parallel SK$ is true. From this and $CV\perp SB$ it finally follows (statement \ref{KotiTransverzala}) $SK\perp SB$, so $\angle KSB$ is a right angle. + \kdokaz + + + + \bzgled + Let $AP$, $BQ$ and $CR$ be the altitude, the median + and the bisector of the angle $ACB$ ($R\in AB$) of a triangle $ABC$. Prove that if + the triangle $PQR$ is regular, then the triangle $ABC$ is also regular. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.14.pic} +\caption{} \label{sl.skl.3.7.14.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $PQR$ be a right triangle or $PQ\cong QR\cong RP$ (Figure \ref{sl.skl.3.7.14.pic}). The point $Q$ is the center of the hypotenuse $AC$ of the right triangle $APC$, so $QA \cong QC \cong QP$ (statement \ref{TalesovIzrKroz2}. Because in this case $QR\cong QC\cong QP$, from the same statement it follows that $\angle ARC$ is a right angle. From the similarity of the triangles $ACR$ and $BCR$ (statement \ref{KSK}) we get that the point $R$ is the center of the side $AB$ and that $AC\cong BC$ is true. Because the point $R$ is the center of the hypotenuse $AB$ of the right triangle $APB$, $AB = 2RP = 2PQ = 2AQ = AC$. Therefore, $AB\cong AC\cong BC$ is true, which means that $ABC$ is a right triangle. + \kdokaz + +It is not difficult to prove that a triangle is equilateral if and only if the corresponding medians are concurrent. The same is true for altitudes. But is something similar true for so-called angle bisectors? + The line segments $BB'$ and $CC'$, where $BB'$ and $CC'$ are angle bisectors of the triangle $ABC$ and $B'\in AC$ and $C'\in AB$, are called + \index{angle bisector} \pojem{angle bisectors}. Angle bisectors are denoted by $l_a$, $l_b$ and $l_c$. + The aforementioned + statement is also true in this case, but +the proof is not so simple. This is the subject of the following well-known theorem. + + + + \bizrek \index{theorem!Steiner-Lemus} + (Steiner-Lehmus\footnote{\textit{D. C. L. Lehmus} (1780--1863),\index{Lehmus, D. C. L.} French + mathematician, who in 1840 sent this, at first glance simple + statement, to the famous Swiss geometer \index{Steiner, J.} \textit{J. Steiner} (1796--1863), + who derived a very extensive proof of this theorem. Then followed + several different solutions to this problem and one of them was published in 1908 + by + the French mathematician \index{Poincar\'{e}, J. H.} \textit{J. H. Poincar\'{e}} (1854--1912).}) + Let $BB'$ ($B' \in AC$) and $CC'$ ($C' \in AB$) be the bisectors + of the interior angles of a triangle $ABC$. Then: + $$AB \cong AC \Leftrightarrow BB'\cong CC'.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.15.pic} +\caption{} \label{sl.skl.3.7.15.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.7.15.pic}) + +($\Rightarrow$) From $AB \cong AC$ it follows that $\angle ABC \cong \angle +ACB$ (theorem \ref{enakokraki}) or $\angle B'BC \cong \angle C'CB$. +By the \textit{ASA} theorem \ref{KSK}, the triangles $B'BC$ and $C'CB$ +are congruent, so $BB'\cong CC'$. + +($\Leftarrow$) Let $BB'\cong CC'$. + We assume that $AB\not\cong AC$. Without loss of generality + let $AB < AC$. In this case $\angle ACB < \angle ABC$ + (by \ref{vecstrveckot}) or +$\angle ACC'< \angle ABB'$. This means that inside the angle $ABB'$ +there is a segment $p$ with endpoint $B$, which intersects the side +$AC$ in such a point $D$, that both $\mathcal{B}(A,D,B')$ and $\angle +DBB'\cong \angle ACC'$ are true. In the triangle $BCD$ is $\angle ACB < +\angle DBC$ and because of that also $BD < CD$ (by \ref{vecstrveckot}). +Therefore there is such a point $E$, which is between the points $C$ +and $D$, so that $BD \cong CE$. By the \textit{SAS} \ref{SKS} the +triangles $BDB'$ and $CEC'$ are similar, therefore the angles $BDB'$ +and $CEC'$ are similar. We prove that this is not possible. Because +of Pasch's axiom \ref{AksPascheva} (used for the triangle $AC'E$ and +the line $BD$) the line $BD$ intersects the line $C'E$ in some point +$S$. In the triangle $SDE$ is the angle $SEC$ (or the angle $CEC'$) +external and by \ref{zunanjiNotrNotrVecji} it can not be similar to +the adjacent internal angle $SDE$ (or the angle $BDB'$). This means +that the assumption $AB < AC$ (analogously $AB > AC$) is not possible. +Therefore $AB\cong AC$. + \kdokaz + + + \bzgled + Let $A_1$ be the midpoint of the side $BC$ of a triangle $ABC$. + Calculate the measure of the angle $AA_1C$, if + $\angle BAC=45^0$ and $\angle ABC=30^0$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.1a.pic} +\caption{} \label{sl.skl.3.7.1a.pic} +\end{figure} + +From $\angle CC'B = 90^0$ it follows first +$\angle C'CB=60^0$, then that the point $C'$ lies on the circle above +the diameter $CB$ and with the center $A_1$ (by \ref{TalesovIzrKroz}), so +$A_1C'\cong A_1C\cong A_1B$. Therefore, the triangle $CC'A_1$ +is isosceles, or by \ref{enakokraki} it holds $\angle CC'A_1 +\cong\angle C'CB=60^0$. This means that the triangle $CC'A_1$ +is equilateral and $C'C\cong C'A_1$. From the fact that $AC'C$ is an isosceles triangle ($\angle CAC'=\angle ACC'=45^0$), it follows that $AC'\cong C'C$. If we connect this with the previous relation, we get $AC'\cong C'A_1$, which +means that the triangle $AC'A_1$ is also isosceles. Therefore, it is (by \ref{enakokraki} and \ref{zunanjiNotrNotr}): + $$\angle C'A_1A\cong\angle C'AA_1=\frac{1}{2}\angle A_1C'B= + \frac{1}{2}\angle C'BA_1=\frac{1}{2}\cdot 30^0=15^0.$$ + In the end there is also: + $$\angle AA_1C=\angle C'A_1C-\angle C'A_1A =60^0-15^0=45^0,$$ which needed to be calculated. \kdokaz + + + \bzgled + Let $P$ be the midpoint of the side $BC$ of an isosceles triangle $ABC$ + and $Q$ the foot of the perpendicular from the point $P$ on the leg + $AC$ of that triangle. Let $S$ be the midpoint of the line segment $PQ$. Prove that + $AS \perp BQ$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.16.pic} +\caption{} \label{sl.skl.3.7.16.pic} +\end{figure} + +\textbf{\textit{Proof.}} From the congruence of the triangles $ABP$ and $ACP$ +(from the statement \textit{SSS} \ref{SSS}) it follows that the congruence of the sides $APB$ and +$APC$ or $AP\perp BC$ (Figure \ref{sl.skl.3.7.16.pic}). We mark with +$R$ the center of the line $QC$. The line $SR$ is the median of the triangle +$QPC$ for the base $PC$, so according to the statement \ref{srednjicaTrik} +$SR\parallel CP$. From this and $AP\perp BC$ it follows that $SR\perp AP$. So +$S$ is the altitude point of the triangle $APR$, so according to the statement +\ref{VisinskaTocka} also $AS\perp PR$. But the line $PR$ is +the median of the triangle $BQC$ for the base $BQ$, so $PR\parallel BQ$ +(from the statement \ref{srednjicaTrik}). From $AS\perp PR$ and $PR\parallel BQ$ +we get $AS\perp BQ$. + \kdokaz + + + \bzgled \label{kotBSC} + If $S$ is the incentre and $\alpha$, $\beta$, + $\gamma$ the interior angles at the vertices + $A$, $B$, $C$ of a triangle + $ABC$, then + $$\angle BSC=90^0+\frac{1}{2}\cdot\alpha.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.1c.pic} +\caption{} \label{sl.skl.3.7.1c.pic} +\end{figure} + +\textbf{\textit{Proof.}} According to the statement \ref{SredVcrtaneKrozn} the simetrals of the internal angles of the triangle $ABC$ intersect in the center of the inscribed circle - in the point $S$ (Figure \ref{sl.skl.3.7.1c.pic}). So +$\angle SBC =\frac{1}{2}\cdot \beta$ and $\angle SCB +=\frac{1}{2}\cdot \gamma$. Because according to the statement \ref{VsotKotTrik} in every +triangle the sum of the internal angles is equal to $180^0$, it follows: + $$\angle BSC = 180^0-\frac{1}{2}\cdot\left( \beta+ + \gamma\right)=180^0-\frac{1}{2}\cdot\left( 180^0- + \alpha\right)=90^0+\frac{1}{2}\cdot\alpha,$$ which had to be proven. \kdokaz + + + \bzgled + Construct a triangle with given $a$, $t_a$, $R$ (see the labels in section \ref{odd3Stirik}). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.16a.pic} +\caption{} \label{sl.skl.3.7.16a.pic} +\end{figure} + +\textbf{\textit{Proof.}} The construction can be carried out by first drawing the circumscribed circle $k(O,R)$, choosing an arbitrary point $B\in k$, planning the cord $BC\cong a$ of the circle, the center $A_1$ of the cord $BC$ and finally the point $A$ as the intersection of the circles $k(O,R)$ and $k_1(A_1,t_a)$ (Figure \ref{sl.skl.3.7.16a.pic}). It is clear that the task of the solution is exactly when $a\leq 2R$ and the intersection of the circles $k(O,R)$ and $k_1(A_1,t_a)$ is not an empty set. The number of solutions in this case depends on the number of intersections of the circles $k(O,R)$ and $k_1(A_1,t_a)$. + \kdokaz + + \bzgled + Construct a right-angled triangle if the hypotenuse and the altitude to that hypotenuse + are congruent to the given line segments $c$ and $v_c$. + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.16b.pic} +\caption{} \label{sl.skl.3.7.16b.pic} +\end{figure} + +\textbf{\textit{Analysis.}} +Let $ABC$ be a right-angled triangle with a right angle at the vertex $C$, +in which the hypotenuse $AB$ and the altitude $CC'$ are congruent to the line segments $c$ and $v_c$ (Figure \ref{sl.skl.3.7.16b.pic}). +By the \ref{TalesovIzrKroz2} theorem, the center $O$ of the hypotenuse $AB$ is also the center of the circumscribed circle of the triangle $ABC$. +Therefore, the vertex $C$ lies on the circle $k$ with diameter $AB$. Since $CC'\cong v_c$, the vertex $C$ also lies on the parallel $p$ to the line $AB$ at a distance $v_c$. The point $C$ is then the intersection of this parallel and the circle $k$. + + +\textbf{\textit{Construction.}} +First, we plan the line $AB$, which is congruent to the given line segment $c$, then the center $O$ of the line $AB$ and the circle $k(O,OA)$. Then we plan the parallel $p$ to the line $AB$ at a distance $v_c$. One of the intersections of the line $p$ and the circle $k(O,OA)$ is denoted by $C$. We prove that $ABC$ is the desired triangle. + +\textbf{\textit{Proof.}} + By construction, point $C$ lies on the circle with radius $AB$, so by \ref{TalesovIzrKroz2} $\angle ACB=90^0$, which means that $ABC$ is a right triangle with hypotenuse $AB$. By construction, it is consistent with the distance $c$. Let $CC'$ be the altitude of the triangle $ABC$. By construction, point $C$ lies on the line $p$, which is from the line $AB$ at a distance $v_c$, so also $|CC'|=d(C,AB)=d(p,AB)$ or $CC'\cong v_c$. + + +\textbf{\textit{Discussion.}} + The number of solutions is dependent on the number of intersections of the line $p$ and the circle $k(O,OA)$. + \kdokaz + + + + \bnaloga\footnote{2. IMO Romania - 1960, Problem 4.} + Construct triangle ABC, given $v_a$, $v_b$ (the altitudes from $A$ and $B$) and $t_a$, + the median from vertex $A$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.IMO1.pic} +\caption{} \label{sl.skl.3.7.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $ABC$ be a triangle, such that +$AA' \cong v_a$ and $BB' \cong v_b$ are its altitudes and $AA_1\cong +t_a$ is its median (Figure \ref{sl.skl.3.7.IMO1.pic}). We mark with +$A'_1$ the orthogonal projection of point $A_1$ on the line $AC$. +The distance $A_1A'_1$ is the median of the triangle $BB'C$ for the +base $BB'$, so by \ref{srednjicaTrik}: + $$|A_1A'_1|=\frac{1}{2}\cdot|BB'|=\frac{1}{2}\cdot v_b + \hspace*{1mm} \textrm{ in } \hspace*{1mm} A_1A'_1 +\parallel BB'.$$ From this it follows that the lines +$A_1A'_1$ and $AC$ are perpendicular in point $A'_1$, thus the line $AC$ +is tangent to the circle $k(A_1,\frac{1}{2} v_b)$ \ref{TangPogoj}. +The proven properties allow us to construct it. + +First, we can plan the rectangular triangle $AA'A_1$ ($AA'\cong v_a$, + $AA_1\cong t_a$ and $\angle AA'A_1 = 90^0$), then the circle + $k(A_1,\frac{1}{2} v_b)$. From the point $A$ we plan the tangents to + the circle $k(A_1,\frac{1}{2} v_b)$. The intersection of one of the tangents + with the line $A'A_1$ is denoted by $C$. In the end, we plan such a point + $B$, that + $BA_1 \cong CA_1$ and $\mathcal{B}(C,A_1,B)$. + + We prove that the triangle $ABC$ satisfies the given conditions. From + the construction it is $AA'\cong v_a$ the height and $AA_1 \cong t_a$ + the centroid (because $A_1$ is the center of the line $BC$) of the triangle + $ABC$. Let $BB'$ be the height of this triangle. We prove that $BB' \cong + v_b$. + The line $AC$ is by construction the tangent of the circle $k(A_1,\frac{1}{2} + v_b)$. Their point of contact is denoted by $A'_1$. By the theorem + \ref{TangPogoj} the lines +$A_1A'_1$ and $AC$ are perpendicular in the point $A'_1$, therefore the line +$A_1A'_1$ is the median of the triangle $BB'C$ for the base $BB'$ and it is valid +$|BB'|= 2\cdot |A_1A'_1|=2\cdot\frac{1}{2}\cdot v_b=v_b$. + + The task has no solution when $v_a>t_a$. If $v_a\leq + t_a$, the number of solutions depends on the number of tangents, which we + can plan from the point $A$ on the circle $k(A_1,\frac{1}{2} + v_b)$. In this case, the tangent must intersect the line $A'A_1$. + If $\frac{1}{2} v_b4. \label{teselRelHyp} +\end{eqnarray} + This inequality has infinitely many solutions in the set $\mathbb{N}^2$, which means +that we have in hyperbolic geometry infinitely many regular tilings. +Two of them are for example $(3,7)$ and $(4,5)$ (Figure +\ref{sl.skl.3.9.2H.pic}\footnote{http://math.slu.edu/escher/index.php/Category:Hyperbolic-Tessellations}). +In the latter, five squares touch around one vertex. This is +possible because in hyperbolic geometry the interior angle of a square is always +sharp and is not constant. It turns out that the square with the longer side +has a smaller interior angle. It is possible to choose such a side of the square +that the interior angle is equal to $\frac{360^0}{5} =72^0$, which just +corresponds to the tiling $(4,5)$. + +\begin{figure}[!htb] +\centering +\includegraphics[width=0.413\textwidth]{whyptess1.eps}\hspace*{4mm} + \includegraphics[width=0.387\textwidth]{whyptess.eps} +\caption{} \label{sl.skl.3.9.2H.pic} +\end{figure} + +In elliptic geometry, where the sum of the angles in a triangle is always +greater than $180^0$, the aforementioned relation for triangle $OSB$ becomes: +$\frac{360^0}{2n}+\frac{360^0}{2m}+90^0>180^0$, or: + \begin{eqnarray} +(n-2)(m-2)<4. \label{teselRelElipt} +\end{eqnarray} +This equation has solutions $(3,3)$, $(4,3)$, +$(3,4$), $(5,3)$ and $(3,5)$ in the set $\mathbb{N}^2$. Because elliptic geometry is realized +as a model on a sphere, these solutions represent tessellations of the sphere with spherical +polygons. The sides of these polygons are arcs of great circles of the sphere. +If in Euclidean space with distances we connect the appropriate vertices of these +tessellations, we get the so-called \index{pravilni!poliedri} \pojem{regular +polyhedra} (Figure +\ref{sl.skl.3.9.2E.pic}\footnote{http://www.upc.edu/ea-smi/personal/claudi/web3d/}): +\pojem{regular tetrahedron}, \pojem{cube} (or \pojem{regular +hexahedron}), \pojem{regular octahedron}, \pojem{regular dodecahedron} +and \pojem{regular icohedron}. For example, $(4,3)$ would represent a cube, +in which three squares (regular 4-gon) meet at one point. + + +\begin{figure}[!htb] +\centering + \includegraphics[bb=0 0 11cm 6cm]{wpoliedri.eps} +\caption{} \label{sl.skl.3.9.2E.pic} +\end{figure} + + +Let's go back to the Euclidean plane. If we allow the possibility that in +covering the plane we use more (finitely many) types of regular +polygons or more types of cells that are arranged the same way at each vertex, then in addition to the three regular +tessellations from the statement \ref{pravilnaTlakovanja} there are eight so-called +\index{tlakovanja!Arhimedova} \pojem{Archimedean tessellations} +\footnote{The proof of this statement was carried out by the German astronomer, mathematician and +physicist \index{Kepler, J.} \textit{J. Kepler} (1571--1630).} (Figure +\ref{sl.skl.3.9.2A.pic}\footnote{http://commons.wikimedia.org/wiki/File\%3AArchimedean-Lattice.png}). + + + +\begin{figure}[!htb] +\centering + \includegraphics[bb=0 0 10cm 7.7cm]{Archimedean.eps} +\caption{} \label{sl.skl.3.9.2A.pic} +\end{figure} + +In addition to regular and Archimedean tilings, there are also other +tilings with polygons that are not regular. The simplest example +is the tiling with congruent parallelograms (Figure +\ref{sl.skl.3.9.3a.pic}), which we get if we deform the regular +tiling $(4,4)$ so that instead of squares in one vertex, four +parallelograms meet. This tiling is determined by the grid +of two sets of parallel lines. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.9.3aa.pic} +\input{sl.skl.3.9.3aaa.pic} +\caption{} \label{sl.skl.3.9.3a.pic} +\end{figure} + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.9.3bb.pic} +\input{sl.skl.3.9.3cc.pic} +\caption{} \label{sl.skl.3.9.3bc.pic} +\end{figure} + +If we divide all parallelograms into +two triangles with diagonals that have the same direction, we get a tiling of the plane with congruent triangles +(Figure \ref{sl.skl.3.9.3a.pic}). The triangle (basic cell) can be +arbitrary, because two such (congruent) triangles can always be connected by a common +side into a parallelogram and thus get a tiling with +parallelograms. Tiling with arbitrary congruent triangles is +a generalization of the regular tiling $(3,6)$ with regular triangles. + + +%\vspace*{-1mm} + +As special cases of tiling with parallelograms, we get tiling with +rectangles and tiling with rhombuses (Figure \ref{sl.skl.3.9.3bc.pic}). +We will prove an even more general +statement, which may not be so obvious. It is possible that +there is also a tiling with arbitrary congruent quadrilaterals. + +%\vspace*{-1mm} + + + + \bzgled + Let $ABCD$ be an arbitrary quadrilateral. Prove that it is + possible to tessellate the plane with the cell $ABCD$ so that each vertex + is surrounded by four such quadrilaterals. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.9.4.pic} +\caption{} \label{sl.skl.3.9.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure +\ref{sl.skl.3.9.4.pic}) + Let $ABCD$ be an arbitrary quadrilateral in +the plane with internal angles $\alpha$, $\beta$, $\gamma$ and $\delta$. +Points $O$ and $S$ are the centers of its sides $AB$ and $BC$. With central +reflections + $\mathcal{S}_O$ and $\mathcal{S}_S$ (for the definition of central reflection + see section \ref{odd6SredZrc}) + the quadrilateral $ABCD$ is mapped into quadrilateral +$BAC_1D_1$ and $A_2CBD_2$. Here $\angle ABD_1\cong\alpha$ and +$CBD_2\cong\gamma$, therefore $\angle D_1BD_2\cong\delta$. Because +$BD_1\cong AD$ and $BD_2\cong CD$, there exists a point $E$, such +that quadrilaterals $D_1ED_2B$ and $ABCD$ are congruent. So around point +$B$ four quadrilaterals intersect, all of which are congruent to quadrilateral +$ABCD$. We can continue the process of tiling the plane, if +we use central symmetry with respect to the centers of sides +of newly formed quadrilaterals. + \kdokaz + + Therefore there exist tilings of the plane with any triangle and any + quadrilateral. It is clear that for any pentagon, hexagon,... + this property does not hold. +For a regular hexagon there exists a regular tiling $(6,3)$. If in +the previous statement we put together two appropriate adjacent quadrilaterals, +we get a tiling of the plane with congruent hexagons, which are not necessarily +regular, but are always centrally symmetric. + + + %_______________________________________________________________________________ + \poglavje{Sets of Points in a Plane. Sylvester's Problem} + \label{odd3Silvester} + +In this section we will investigate problems related to sets of points in the plane and the lines determined by these points. At the beginning we will consider some consequences of the first two groups of axioms (incidence and order). First we will define new concepts. Let $\mathfrak{T}$ be a set of $n$ ($n>2$) points in the plane. With $\mathcal{P}(\mathfrak{T})$ we denote the set of all lines, each of which goes through at least two points from the set $\mathfrak{T}$ (Figure \ref{sl.skl.3.10.1.pic}). Because the set $\mathfrak{T}$ contains at least two points, from the axioms of incidence it follows that the set $\mathcal{P}(\mathfrak{T})$ is not empty. The question arises, how many lines are in the set $\mathcal{P}(\mathfrak{T})$. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.1.pic} +\caption{} \label{sl.skl.3.10.1.pic} +\end{figure} + + + + \bizrek \label{stevPremic} + Let $\mathfrak{T}$ be a set of $n$ ($n>2$) points in the plane such that + no three of them are collinear. Then the number of lines of the set $\mathcal{P}(\mathfrak{T})$ is equal to + $$\frac{n(n-1)}{2}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.2.pic} +\caption{} \label{sl.skl.3.10.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} Through each of the $n$ points from the set $\mathfrak{T}$ there are exactly $n -1$ lines from the set $\mathcal{P}(\mathfrak{T})$. Because we count each line in this way twice (Figure \ref{sl.skl.3.10.2.pic}), we have to divide by 2. So there are exactly $\frac{n(n-1)}{2}$ lines in the set $\mathcal{P}(\mathfrak{T})$. + \kdokaz + +The previous statement can also be solved in the following way: through the first point there are $n -1$ lines, through the second point $n - 2$ lines (one less, because the line determined by these two points is not counted twice), $n - 3$ lines through the third point and so on until one line through the penultimate point. This is a total of $(n -1) + (n - 2) +\cdots+1$ lines. Of course, this is again equal to $\frac{n(n-1)}{2}$. If we take $n-1= k$, we get the known formula for the sum of the first $k$ natural numbers: + $$1+ 2+ \cdots + k=\frac{k(k+1)}{2}.$$ + + From the previous statement we can derive the formula for the number of diagonals of an arbitrary $n$-gon. + + + + \bizrek + If $D_n$ is the number of diagonals of an arbitrary $n$-gon, + then + $$D_n=\frac{n( n-3)}{2}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.3.pic} +\caption{} \label{sl.skl.3.10.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} We mark with $\mathfrak{O}$ the set of all vertices of an arbitrary $n$-gon. + The number of diagonals is equal to the number of all lines +from the set $\mathcal{P}(\mathfrak{O})$ (statement \ref{stevPremic}) +decreased by the number of its sides (Figure +\ref{sl.skl.3.10.3.pic}). Therefore: + $$D_n=\frac{n(n- 1)}{2}-n=\frac{n^2-3n}{2}=\frac{n(n-3)}{2},$$ which was to be proven. \kdokaz + +We mention that the formula for the number of diagonals of an $n$-gon could also be derived directly - with a similar treatment as in the proof of statement \ref{stevPremic}. From each of the $n$ vertices of an $n$-gon we can draw $n - 3$ diagonals. In this way we count each diagonal twice, so we have to divide by 2 and we get the previous formula. + +Statement \ref{stevPremic} referred to the number of lines determined by a set of points in a plane that are in such a position that no three of them are collinear. In the next example we will check what happens if we add new conditions. + +\bzgled + Let $\mathfrak{T}$ be a set of $n$ ($n > 2$) points in the plane which are + in such a position that $m$ ($m < n$) of them lies on the same line, but otherwise + no other three points are collinear. What is the number of lines + in the set $\mathcal{P}(\mathfrak{T})$? + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.4.pic} +\caption{} \label{sl.skl.3.10.4.pic} +\end{figure} + +\textbf{\textit{Solution.}} + (Figure \ref{sl.skl.3.10.4.pic}) + + Without the additional condition, there would be $\frac{n(n -1)}{2}$ lines in the set $\mathcal{P}(\mathfrak{T})$ + (statement \ref{stevPremic}). + The additional condition in the problem reduces this +number by one less than the number of lines determined +by the set of $m$ points in general position. This is because otherwise +these lines would be counted multiple times. So the number of lines in the set +$\mathcal{P}(\mathfrak{T})$ is equal to: +$$\frac{(n-1)}{2}-\frac{(m -1)}{2}+1.$$ + \kdokaz + +The next simple example will be an introduction to the very interesting problem of the relationship +between the set of points $\mathfrak{T}$ and the set of all lines +$\mathcal{P}(\mathfrak{T})$, which this set of points determines. + + + \bzgled + Construct nine points lying on ten lines + in such a way, that each of those ten lines contains exactly three of these nine points\footnote{\index{Newton, I.} + \textit{I. Newton} (1642--1727), znani angleški + matematik + in fizik, ki je zastavil ta problem v obliki: + ‘‘How can you plant 9 trees in a garden with 10 rows and each row containing exactly 3 trees?’’}. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.5.pic} +\caption{} \label{sl.skl.3.10.5.pic} +\end{figure} + +\textbf{\textit{Proof.}} One of the possibilities is the following. Let $ABCD$ +be an arbitrary rectangle, $P$ and $Q$ be the centers of sides $AB$ and $CD$, $S$ +be the intersection of the diagonals, $K$ be the intersection of lines $AP$ and $DQ$, and $L$ +be the intersection of lines $BP$ and $CQ$ (Figure \ref{sl.skl.3.10.5.pic}). Because +the rectangle $ABCD$ is a rectangle, the points $P$, $Q$ and $S$ are collinear. So are +the points $K$, $L$ and $S$ (points $K$ and $L$ are the centers of rectangles +$AQPD$ and $QBCP$). So we have nine points $A$, $B$, $C$, $D$, $P$, +$Q$, $S$, $K$ and $L$, of which three lie on each of the ten lines $AB$, +$CD$, $PQ$, $KL$, $AC$, $BD$, +$PA$, $PB$, $QC$ and $QD$. + \kdokaz + + If in the previous example we denote the set of nine points with $\mathfrak{T}$, we see that the set of ten lines is not the set $\mathcal{P}(\mathfrak{T})$. + In the set $\mathcal{P}(\mathfrak{T})$ we would have sixteen + lines - our ten and also the additional lines $AD$, $BC$ $DL$, $AL$, $CK$ and $BK$. + But each of these six lines contains +only two points. So the condition that they contain +exactly three points of the initial set is not fulfilled. It is now logical to ask +the following question: Is it possible in a plane to set a finite +set of non-collinear points $\mathfrak{T}$ so that each line from the set +$\mathcal{P}(\mathfrak{T})$ contains exactly three points from the set +$\mathfrak{T}$? It is clear that it is possible if we require +that each line from $\mathcal{P}(\mathfrak{T})$ contains exactly two +points from $\mathfrak{T}$. The most simple example for this are the vertices +of a triangle and its altitudes, or any set of points from +the statement \ref{stevPremic}. The aforementioned problem for three points is not so +simple, and we will find the answer in the continuation. We mention that +the answer is negative. Even more - the answer is negative even if we require that each line from $\mathcal{P}(\mathfrak{T})$ contains +at least three points from $\mathfrak{T}$. First, we prove one lemma +(an auxiliary statement). + +\bizrek \label{SylvesterLema} + Let $\mathfrak{T}$ be a finite set of points in the plane which do not all lie on the same line + and $\mathcal{P}=\mathcal{P}(\mathfrak{T})$. If $T_0\in \mathfrak{T}$ and $p_0\in \mathcal{P}$ ($T_0\notin p_0$) + are such that they determine the minimum distance, i.e. + $$(\forall T\in \mathfrak{T})(\forall p\in \mathcal{P}) + ( T\notin p\Rightarrow d(T,p)\geq d(T_0,p_0)),$$ + then the line $p_0$ contains exactly two points from the set $\mathfrak{T}$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.6.pic} +\caption{} \label{sl.skl.3.10.6.pic} +\end{figure} + +\textbf{\textit{Proof.}} We assume the opposite. Let the line $p_0$ +contain at least three different points $A$, $B$ and $C$ from the set +$\mathfrak{T}$ (Figure \ref{sl.skl.3.10.6.pic}). We mark with $T'_0$ the orthogonal projection of the point $T_0$ +on the line $p_0$. If the point $T'_0$ differs from the points $A$, $B$ and +$C$, then at least two of these three points (let it be $B$ and $C$) are on the line $p_0$ on the same +side of the point $T'_0$. Without loss of generality, let +$\mathcal{B}(T'_0,B,C)$. Because $T_0,C \in \mathfrak{T}$, then the line +$q= CT_0$ ($q \neq p_0$, because $T_0\notin p_0$) belongs to the set +$\mathcal{P}$. Let $B'$ be the orthogonal projection of the point $B$ on +the line $q$. It is not difficult to prove that in this case it holds: + $$d(B,q)=|BB'| BP > PA$, so $\angle APB$ is the largest angle in the triangle $APB$ and is therefore larger than $60^0$. Similarly, $CB > BP,BC$, so $\angle BPC > 60^0$. The same would hold for the angles $CPD$, $DPE$, $EPF$, and $EPA$, but that is not possible, because their sum is always equal to or even greater than $360^0$, regardless of whether $P$ is an inner or outer point of the hexagon $ABCDEF$. Therefore, the point $P$ is connected to no more than five points. + \kdokaz + + + \bzgled + What is the maximum number of regions in the plane that can be + divided by $n$ lines\footnote{Ta problem je rešil švicarski geometer + \index{Steiner, J.} \textit{J. Steiner} (1796--1863).}? + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.9.pic} +\caption{} \label{sl.skl.3.10.9.pic} +\end{figure} + +\textbf{\textit{Solution.}} We will describe the process of designing such +$n$ lines (Figure \ref{sl.skl.3.10.9.pic}). If $n = 1$, or +if there is only one line, the plane is divided into two areas. Two +lines, if they are not parallel, divide the plane into four areas. If +we add a third line $p_3$, which is not parallel to them and does not go through +their common point, it intersects the initial two lines in two points. These +two points divide the line $p_3$ into three parts, each of which is in +one of the three previous four areas of the plane. So with line +$p_3$ we get three more parts of the plane, a total of seven. We continue the process. If $n -1$ lines divide the plane into $k$ parts, by +adding the $n$-th line $p_n$ (which is not parallel to any of the previous $n -1$ lines and does not contain any of their intersections), +we first get $n -1$ intersections on that line, then $n$ of its +parts or $n$ new areas of the given plane. So the maximum possible +number of areas for $n$ lines is equal to: +\begin{eqnarray*} +2+2+3+\cdots+n&=&1+1+2+3+\cdots+n=\\ +&=&1+\frac{n(n+1)}{2}=\\&=&\frac{n^2+n+2}{2}. +\end{eqnarray*} +A formal proof of this fact could be derived by mathematical induction. + \kdokaz + + + %_______________________________________________________________________________ + \poglavje{Helly's Theorem} +\label{odd3Helly} + +The next important statement is a consequence of only the first two groups of axioms +or the axioms of incidence and the axioms of order. We will also use it +in tasks related to consistency. + +\bizrek \label{Helly} + Let $\Phi_1$, $\Phi_2$, ... , $\Phi_n$ ($n \geq 4$) be convex sets in the plane. + If every three of these sets have a common point, then all $n$ sets have a common point + \index{izrek!Hellyjev}(Helly's theorem\footnote{Austrian + mathematician \index{Helly, E.} \textit{E. Helly} (1884--1943) + discovered this statement in the general case of $n$-dimensional space $\mathbb{E}^n$ + in 1913, but published it only in 1923. Alternative proofs were + meanwhile given by Austrian mathematician \index{Radon, J. K. A.} + \textit{J. K. A. Radon} (1887-–1956) in 1921 and + Hungarian mathematician \index{Kőnig, D.} + \textit{D. Kőnig} (1884–-1944) in 1922.}). + \eizrek + +\begin{figure}[!htb] +\centering +\hspace*{10mm} +\input{sl.skl.3.11.1.pic} +\caption{} \label{sl.skl.3.11.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} We will prove this by induction on $n$. + +\textit{(A)} Let $n = 4$ and (Figure +\ref{sl.skl.3.11.1.pic}): +\begin{itemize} + \item $P_4\in \Phi_1 \cap \Phi_2 \cap \Phi_3$, + \item $P_3\in \Phi_1 \cap \Phi_2 \cap \Phi_4$, + \item $P_2\in \Phi_1 \cap \Phi_3 \cap \Phi_4$, + \item $P_1\in \Phi_2 \cap \Phi_3 \cap \Phi_4$. +\end{itemize} +We will prove that there exists a point that lies in each of the figures $\Phi_1$, +$\Phi_2$, $\Phi_3$ and $\Phi_4$. Based on the mutual position +of points $P_1$, $P_2$, $P_3$ and $P_4$ we will consider only two most +general cases (the proof in the other cases +is similar). + +\textit{1)} A quadrilateral, determined by points $P_1$, $P_2$, $P_3$ and +$P_4$, is non-convex. In this case, one of the points $P_1$, $P_2$, +$P_3$ and $P_4$ is an inner point of the triangle determined by the remaining +three points. Without loss of generality, let $P_4$ be the inner point +of the triangle $P_1P_2P_3$. The vertices of this triangle lie in the shape +$\Phi_4$. Because $\Phi_4$ is a convex shape, all sides and inner points of the triangle $P_1P_2P_3$, as well as the point $P_4$, lie in it. In this case, the point $P_4$ is the common point of shapes +$\Phi_1$, $\Phi_2$, $\Phi_3$ and $\Phi_4$. + +\textit{2)} The quadrilateral determined by points $P_1$, $P_2$, $P_3$ and $P_4$ is convex. Without loss of generality, let its diagonals be $P_1P_2$ and $P_3P_4$. Because the quadrilateral is convex, its diagonals intersect in a point $S$. Given shapes are convex, so from $P_1, P_2\in \Phi_3,\Phi_4$ it follows that the diagonal $P_1P_2$ lies entirely in shapes $\Phi_3$ and $\Phi_4$. Analogously, from $P_3, P_4\in \Phi_1,\Phi_2$ it follows that the diagonal $P_3P_4$ lies entirely in shapes $\Phi_1$ and $\Phi_2$. The point $S$, which lies on both diagonals +$P_1P_2$ and $P_3P_4$, lies in all four shapes $\Phi_1$, $\Phi_2$, +$\Phi_3$ and $\Phi_4$. + +With this we have proven that the statement is true for $n=4$. + +\textit{(B)} Let's now assume that the statement is true for $n = k$ ($k\in +\mathbb{N}$ and $k>4$). + We shall prove that the statement is also true for +$n = k +1$. Let $\Phi_1$, $\Phi_2$, $\ldots$ , $\Phi_{k-1}$, +$\Phi_k$ and $\Phi_{k+1}$ be such figures that every triplet of these figures +has at least one common point. Let $\Phi'=\Phi_k\cap\Phi_{k+1}$. We shall first prove that every triplet of figures $\Phi_1$, $\Phi_2$ ,$\ldots$, +$\Phi_{k-1}$, $\Phi'$ has a common point. For triplets of figures from +$\Phi_1$, $\Phi_2$, $\ldots$, $\Phi_{k-1}$ this is already fulfilled according to the assumption. Without loss of generality, it is enough to prove that +figures $\Phi_1$, $\Phi_2$ and $\Phi'=\Phi_k\cap\Phi_{k+1}$ +have a common point. This is true (based on the proven example for $n = 4$), +because every triplet of figures from $\Phi_1$, $\Phi_2$, $\Phi_k$ and +$\Phi_{k+1}$ has a common point. From the induction assumption +(for $n=k$) it follows that figures $\Phi_1$, $\Phi_2$, $\ldots$, +$\Phi_{k-1}$, $\Phi'$ have a common point, which at the same time lies in +each of figures $\Phi_1$, $\Phi_2$, $\ldots$, $\Phi_{k-1}$, +$\Phi_k$, $\Phi_{k+1}$. + \kdokaz + +In the continuation we shall consider some consequences of Helly's theorem. + + + + \bzgled + Let $\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$ + ($n > 3$) be half-planes covering a plane $\alpha$. + Prove that there are three of these half-planes that also cover the plane $\alpha$. + \ezgled + +\textbf{\textit{Proof.}} Let $\beta_1$, $\beta_2$, $\cdots$, +$\beta_n$ be open half-planes, which are determined by the half-planes +$\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$ as complementary +half-planes with respect to the plane $\alpha$ or +$\beta_k=\alpha\setminus\alpha_k$, $k\in\{1,2,\ldots,n\}$. +For every point $X$ of the plane $\alpha$ and every +$k\in\{1,2,\ldots,n\}$ the equivalence holds: + $$X\in \alpha_k \Leftrightarrow X\notin \beta_k.$$ +Assume the contrary, that none of the three half-planes $\alpha_1$, +$\alpha_2$, $\cdots$, $\alpha_n$ covers the plane $\alpha$. This +means that for every triple of them there is a point in the plane $\alpha$, +which does not lie on any of them, or for every triple of the half-planes +$\beta_1$, $\beta_2$, $\cdots$, $\beta_n$ there is a point of the plane +$\alpha$, which lies on each of them. Because the half-planes are convex +shapes, from Helly's theorem \ref{Helly} it follows that there is a point +$X$, which lies on each of the half-planes $\beta_1$, $\beta_2$, $\cdots$, +$\beta_n$. This point therefore lies in the plane $\alpha$, but does not lie in +any of the half-planes $\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$, +which is in contradiction with the basic assumption that the half-planes $\alpha_1$, +$\alpha_2$, $\cdots$, $\alpha_n$ cover the plane $\alpha$. + Therefore, there is at least one triple of the +half-planes $\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$, which +covers the plane $\alpha$. + \kdokaz + + + + \bzgled \label{lemaJung} + If for every three of $n$ ($n > 3$) points of a plane + there is such a circle with radius $r$ containing these three points, then it exists + a circle of equal radius containing all these $n$ points. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.11.2.pic} +\caption{} \label{sl.skl.3.11.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.11.2.pic}) + +Let $A_1$, $A_2$,$\ldots$, $A_n$ be points with given properties. With +$\mathcal{K}_i$ ($i\in\{1,2,\ldots,n\}$) we denote circles with centers $A_i$ and with +radius $r$. Let $A_p$, $A_q$ and $A_l$ be any points from +the set $\{A_1, A_2,\ldots, A_n\}$. By assumption, there exists a circle with +radius $r$, which contains these three points. We denote the center of this circle +with $O$. From this it follows that $|OA_p|, |OA_q|, |OA_l|\leq r$, which means that +point $O$ lies in each of the circles $\mathcal{K}_p$, $\mathcal{K}_q$ and $\mathcal{K}_l$. Therefore, +each three of the circles $\mathcal{K}_1$, $\mathcal{K}_2$,$\ldots$, $\mathcal{K}_n$ have at least one +common point. Because the circles are convex figures (statement \ref{KrogKonv}), by +Helly's theorem there exists a point $S$, which lies in each of the circles +$\mathcal{K}_1$, $\mathcal{K}_2$,$\ldots$, $\mathcal{K}_n$. From this it follows that $\mathcal{K}(S, r)$ is the desired +circle, since $|SA_1|, |SA_2|,\ldots |SA_n|\leq r$. + \kdokaz + + An interesting consequence of the last statement \ref{lemaJung} will be given in section + \ref{odd7Pitagora} (statement \ref{Jung}). + +%________________________________________________________________________________ +\naloge{Exercises} + +\begin{enumerate} + + \item Let $S$ be a point, which lies in the angle $pOq$, and to point $A$ and $B$ the orthogonal projection of + point $S$ on the sides $p$ and $q$ of this +angle. Prove that $SA\cong SB$ if and only when the line +$OS$ is the angle bisector of the angle $pOq$. + +\item Prove that the sum of the diagonals of a convex quadrilateral is greater than +the sum of its opposite sides. + + \item Prove that in every triangle + there is at most one side shorter than the corresponding altitude. + + \item Let $AA_1$ be the altitude of triangle $ABC$. Prove + that of the two angles, which altitude $AA_1$ determines with +sides $AB$ and $AC$, the larger one is the one, which altitude $AA_1$ determines with +the shorter side. + + \item Let $BB_1$ and $CC_1$ be the altitudes of triangle $ABC$ and + $AB \frac{3}{4}(a + b + c)$. + \end{enumerate} + +\item Let $p$ be a line that is parallel to a circle $k$. Prove that +all points of this circle are on the same side of the line $p$. + +\item If the circle $k$ lies in a convex figure $\Phi$, then the +circle determined by this circle also lies in this figure. Prove it. + +\item Let $p$ and $q$ be two different tangents to the circle $k$ that +touch it in points $P$ and $Q$. Prove the equivalence: $p \parallel q$ +exactly when $AB$ is the diameter of the circle $k$. + +\item If $AB$ is the chord of the circle $k$, then the intersection of +the line $AB$ and the circle determined by the circle $k$ is equal to +this chord. Prove it. + +\item Let $S'$ be the orthogonal projection of the center $S$ of the +circle $k$ onto the line $p$. Prove that $S'$ is an external point of +this circle exactly when the line $p$ does not intersect the circle. + +\item Let $V$ be the altitude of the triangle $ABC$, for which $CV \cong +AB$. Determine the size of the angle $ACB$. + +\item Let $CC'$ be the altitude of the right triangle $ABC$ ($\angle +ACB = 90^0$). If $O$ and $S$ are the centers of the inscribed circles +of the triangles $ACC'$ and $BCC'$, then the altitude of the internal +angle $ACB$ is perpendicular to the line $OS$. Prove it. + +\item Let $ABC$ be a triangle in which $\angle ABC = 15^0$ and $\angle +ACB = 30^0$. Let $D$ be such a point of the side $BC$ that $\angle BAD += 90^0$. Prove that $BD = 2AC$. + +\item Prove that there exists a pentagon that can be covered with +such pentagons that are congruent to it. + +\item Prove that there exists a decagon that can be covered with +such decagons that are congruent to it. + +\item In a plane, each point is painted red or black. Prove that +there exists a right triangle that has all its vertices the same +color. + +\item Let $l_1,l_2,\ldots, l_n$ ($n > 3$) be arcs, which all lie on the same +circle. The central angle of each arc is at most $180^0$. + Prove that there exists a point, which lies on each arc, + if every three arcs have at least one common point. + +%drugi del + +\item +Let $p$ and $q$ be rectangles, which intersect in the point $A$. If +$B, B'\in p$, $C, C'\in q$, $AB\cong AC'$, $AB'\cong AC$, +$\mathcal{B}(B,A,B')$ and $\mathcal{B}(C,A,C')$, then +the rectangle on the line $BC$ through the point $A$ goes through the center +of the line $B'C'$. Prove. + +\item +Prove that the altitudes of an inner angle of a rectangle, which is not +a square, intersect in points, which are the vertices of a square. + +\item + Prove that the altitudes of an inner angle of a parallelogram, which is not +a rhombus, intersect in points, which are the vertices of a rectangle. Prove also that the diagonals +of this rectangle are parallel to the sides of the parallelogram and are +equal to the difference of the adjacent sides of this parallelogram. + +\item +Prove that the altitudes of two sides of a triangle +are perpendicular to each other. + +\item Let $B'$ and $C'$ be the points of intersection of the altitudes from the vertices $B$ and $C$ of the triangle +$ABC$. Prove the equivalence $AB\cong AC \Leftrightarrow BB'\cong +CC'$. + +\item Prove that a triangle is right, +if the center of the circle drawn through the triangle and its altitude point coincide. +Is a similar statement true for any two characteristic +points of this triangle? + +\item Prove that a right triangle $ABC$ and a right triangle $A'B'C'$ are congruent +exactly when they have congruent altitudes $CD$ and $C'D'$, sides $AB$ and +$A'B'$ and angle $ACD$ and $A'C'D'$. + +\item If $ABCD$ is a rectangle and $AQB$ and $APD$ are right triangles with the same orientation, then the line $PQ$ +is congruent with the diagonal of this rectangle. Prove. + +\item Let $BB'$ and $CC'$ be the altitudes of the triangle $ABC$ ($AC>AB$) and + $D$ is such a point on the line segment $AB$, that $AD\cong AC$. The point +$E$ is the intersection of the line $BB'$ with the line, which goes through the point $D$ and is +parallel to the line $AC$. Prove that $BE=CC'-BB'$. + +\item Let $ABCD$ be a convex quadrilateral, for which it holds that + $AB\cong BC\cong CD$ and $AC\perp BD$. Prove that $ABCD$ + is a rhombus. + +\item Let $BC$ be the base of an isosceles triangle $ABC$. If $K$ and +$L$ are such points, that $\mathcal{B}(A,K,B)$, $\mathcal{B}(A,C,L)$ and $KB\cong LC$, then +the center of the line $KL$ lies on the base $BC$. Prove. + +\item Let $S$ be the center of the triangle $ABC$ of an inscribed circle. +A line, which goes through the point $S$ and is parallel to the side $BC$ +of this triangle, intersects the sides $AB$ and $AC$ in succession in the points +$M$ and $N$. Prove that $BM+NC=NM$. + +\item Let $ABCDEFG$ be a convex heptagon. Calculate the sum +of the convex angles, which are determined by the broken line $ACEGBDFA$. + +\item Prove that the centers of the sides and the vertex of an altitude of an arbitrary triangle, +in which no two sides are congruent, are the vertices +of an isosceles trapezoid. + + \item Let $ABC$ be a right triangle with a right angle at the vertex $C$. +The points $E$ and $F$ shall be the intersections of the internal angle bisectors at +the vertices $A$ and $B$ with the opposite sides, $K$ and $L$ shall +be the orthogonal projections of the points $E$ and $F$ on the hypotenuse of this +triangle. Prove that $\angle LCK=45^0$. + + +\item Let $M$ be the center of the side $CD$ of a square $ABCD$ and $P$ such a point + on the diagonal $AC$, for which it holds that $3AP=PC$. Prove that $\angle BPM$ +is a right angle. + + \item Let $P$, $Q$ and $R$ be the centers of the sides $AB$, $BC$ and $CD$ + of a parallelogram $ABCD$. The lines $DP$ and $BR$ shall intersect the line +$AQ$ in the points $K$ and $L$. Prove that $KL= \frac{2}{5} AQ$. + + \item Let $D$ be the center of the hypotenuse $AB$ of a right triangle $ABC$ ($AC>BC$). +The points $E$ and $F$ shall be the intersections of the angle trisectors of +the sides $CA$ and $CB$ with a line, which goes through $D$ and is orthogonal +to the line $CD$. The point $M$ shall be the center of the line $EF$. Prove that +$CM\perp AB$. + +\item Let $A_1$ and $C_1$ be the centers of sides $BC$ and $AB$ of triangle $ABC$. + The internal angle bisector at point $A$ intersects the line +$A_1C_1$ at point $P$. Prove that $\angle APB$ is a right angle. + + \item Let $P$ and $Q$ be points on sides $BC$ and $CD$ of square $ABCD$, + such that the line $PA$ is the angle bisector of angle $BPQ$. Determine the size of the angle + $PAQ$. + +\item Prove that the center of the circumscribed circle lies closest to the longest side +of the triangle. + + \item Prove that the center of the inscribed circle is closest to the vertex of +the largest internal angle of the triangle. + +\item Let $ABCD$ be a convex quadrilateral. Find a point $P$, such that +the sum $AP+BP+CP+DP$ is minimal. + + \item The diagonals $AC$ and $BD$ of the trapezoid $ABCD$ with the base $AB$ +intersect at point $O$ and $\angle AOB=60^0$. Points $P$, $Q$ +and $R$ are in order the centers of the lines $OA$, $OD$ and $BC$. Prove that $PQR$ is a right triangle. + +\item Let $P$ be an arbitrary internal point of triangle $ABC$, for which + $\angle PBA\cong \angle PCA$. Points $M$ and $L$ are the orthogonal +projections of point $P$ on sides $AB$ and $AC$, point $N$ is +the center of side $BC$. Prove that $NM\cong +NL$\footnote{Predlog za MMO 1982 (SL 9.)).}. + +\item Let $AD$ be the internal angle bisector at point $A$ ($D\in BC$) of triangle +$ABC$ and $E$ a point on side $AB$, such that $\angle +BDE\cong\angle BAC$. Prove that $DE\cong DC$. + +\item Let $O$ be the center of square $ABCD$ and $P$, $Q$ and $R$ points, + which divide its perimeter into three equal parts. Prove that the minimum of the sum $|OP|+|OQ|+|OR|$ is achieved, when one of these points +is the center of a side of the square. + +\item There is a given finite number of lines that divide the plane into areas. +Prove that the plane can be colored with two colors, so that each area is colored with one color, and adjacent areas are always colored with different colors. + + +\item Draw a triangle $ABC$, if the following data are given (see labels in section \ref{odd3Stirik}): + + (\textit{a}) $\alpha$, $\beta$, $s$; \hspace*{2mm} + (\textit{b}) $a-b$, $c$, $\gamma$; \hspace*{2mm} + (\textit{c}) $a$, $\beta-\gamma$, $b-c$; \hspace*{2mm} + + (\textit{d}) $a$, $\beta-\gamma$, $b+c$; \hspace*{2mm} + (\textit{e}) $b$, $c$, $v_a$; \hspace*{2mm} + (\textit{f}) $b$, $v_a$, $v_b$; \hspace*{2mm} + +(\textit{g}) $\alpha$, $v_a$, $v_b$; \hspace*{2mm} + (\textit{h}) $c$, $a+b$, $\gamma$; + (\textit{i}) $v_a$, $\alpha$, $\beta$; \hspace*{2mm} + + (\textit{j}) $b$, $a+c$, $v_c$; \hspace*{2mm} + (\textit{k}) $b-c$, $v_b$, $\alpha$; \hspace*{2mm} + (\textit{l}) $a$, $t_b$, $t_c$;\hspace*{2mm} + (\textit{m}) $b$, $c$, $t_a$; \hspace*{2mm} + + (\textit{n}) $t_a$, $t_b$, $t_c$; \hspace*{2mm} +(\textit{o}) $c$, $v_a$, $l_a$; \hspace*{2mm} + (\textit{p}) $c$, $v_a$, $t_b$; + (\textit{r}) $b$, $l_a$, $\alpha$; \hspace*{2mm} + + (\textit{s}) $v_a$, $v_b$, $t_a$; \hspace*{2mm} +(\textit{t}) $t_a$, $v_b$, $b+c$; \hspace*{2mm} + (\textit{u}) $a$, $b$, $\alpha-\beta$; + + \item Draw an isosceles triangle $ABC$, if the following are given: + \begin{enumerate} + \item the base and the sum of the leg and the height on the base, + \item the circumference and the height on the base, + \item both heights, + \item the angle at the base and the segment of its altitude, + \item the leg and the point of intersection of the corresponding altitude on it, + \item the leg and the corresponding altitude. + \end{enumerate} + + \item Draw a right triangle $ABC$ with a right angle at point $C$, if the following data are given: + +(\textit{a}) $\alpha$, $a+b$, \hspace*{2mm} + (\textit{b}) $\alpha$, $a-b$, \hspace*{2mm} + (\textit{c}) $a$, $b+c$, + + (\textit{d}) $c$, $a+b$,\hspace*{2mm} + (\textit{e}) $t_a$, $t_c$, \hspace*{2mm} + (\textit{f}) $a$, $c-b$, + +(\textit{g}) $a+v_c$, $\alpha$, \hspace*{2mm} + (\textit{h}) $t_c$, $v_c$,\hspace*{2mm} + (\textit{i}) $a$, $t_a$,\hspace*{2mm} + (\textit{j}) $v_c$, $l_c$. + + \item Draw a rectangle $ABCD$, if given: + \begin{enumerate} + \item the diagonal and one side, + \item the diagonal and the perimeter, + \item one side and the angle between the diagonals, + \item the perimeter and the angle between the diagonals. + \end{enumerate} + + \item Draw a rhombus $ABCD$, if given: + \begin{enumerate} + \item a side and the sum of the diagonals, + \item a side and the difference of the diagonals, + \item one angle and the sum of the diagonals, + \item one angle and the difference of the diagonals. + \end{enumerate} + + \item Draw a parallelogram $ABCD$, if given: + \begin{enumerate} + \item one side and the diagonals, + \item one side and the altitude, + \item one diagonal and the altitude, + \item side $AB$, the angle at point $A$ and the sum $BC+AC$. + \end{enumerate} + + \item Draw a trapezoid $ABCD$, if given: + \begin{enumerate} + \item the bases, the leg and the smaller angle that is not adjacent to this leg, + \item the bases and the diagonals, + \item the bases and the angle at the longer base, + \item the sum of the bases, the altitude and the angle at the longer base. + \end{enumerate} + + \item Draw a deltoid $ABCD$, if given: the diagonal $AC$, which lies on the slanted side of the deltoid, $\angle CAD$ and the sum $AD+DC$." + +\item Draw a quadrilateral $ABCD$, if it is given: + \begin{enumerate} + \item four sides and one angle, + \item four sides and an angle between the opposite sides, + \item three sides and an angle at the fourth side, + \item the centers of three sides and a distance that is consistent and parallel to the fourth side. + \end{enumerate} + + +\end{enumerate} + + + +%%% Do tu pregledala tudi Ana. + +% DEL 4 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% SKLADNOST TRIKOTNIKOV IN KROŽNICA +%________________________________________________________________________________ + + \del{Congruence and Circle} \label{pogSKK} + + +We have already looked at some of the properties of circles in the previous two chapters - certain properties of the radius, diameter, cord, the relationship between the circle and the line, and the properties of the tangent to the circle. We saw that for every triangle there is an inscribed and circumscribed circle. We proved that for regular polygons and some quadrilaterals (rectangle, square) there is a circumscribed circle, and for some there is also an inscribed circle. In this chapter we will look further into the properties of the circle that are a result of the congruence of triangles. + +%________________________________________________________________________________ + \poglavje{Two Circles} \label{odd4DveKroz} + + We will carry out a similar analysis of the position of the circles and lines that we did in section \ref{odd3KrozPrem}, but this time with two circles in the same plane. We will assume that the two circles we are dealing with are in the same plane from now on. + + First, let's define some terms that relate to two circles. The line that goes through the centers of two circles is the + \index{centrala dveh krožnic}\pojem{centrala} of those two circles. + The distance between the centers of two circles is called the + \index{središčna razdalja dveh krožnic} + \pojem{središčna razdalja dveh krožnic} + (Figure \ref{sl.skk.4.1.1.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.1.1.pic} +\caption{} \label{sl.skk.4.1.1.pic} +\end{figure} + +Circles in the same plane with the same +center (their central distance is equal to $0$) +we call +\index{circles!concentric} + \pojem{concentric circles} + (Figure \ref{sl.skk.4.1.1.pic}). If concentric circles have at least + one common point, the circles are identical (coincide). If $X$ is a common point + of concentric circles $k_1(S,r_1)$ and $k_2(S,r_2)$, it holds + $|SX|=r_1=r_2$ or $r_1=r_2$, which means that the circles + are identical. Concentric circles are either identical or have + no common points. Similarly to the circle and the line, here we raise the question of how many common points different circles can have and what is their mutual position. We will begin + with the following statement. + + + \bizrek + Two different circles lying in the same plane + have at most two common points. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.1.2.pic} +\caption{} \label{sl.skk.4.1.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} We assume the opposite. Let $A$, $B$ +and $C$ be three different common points of two circles $k(O,r_1)$ and +$l(S,r_2)$ (Figure \ref{sl.skk.4.1.2.pic}). These three points are not +collinear, which would mean that the line $AB$ intersects the circle (e.g. +$k$) in three different points, which according to izrek \ref{KroznPremPresek} is not +possible. But if $A$, $B$ and $C$ are non-collinear points, they determine +the triangle $ABC$, which according to izrek \ref{SredOcrtaneKrozn} means that $O=S$, or both points are +located in the intersection of the perpendiculars of this +triangle. The radii are also equal, because $r_1=|OA|=|SA|=r_2$, so +the circles $k(O,r_1)$ and $l(S,r_2)$ are identical - both represent +the circle circumscribed around the triangle $ABC$. + \kdokaz + +So, two different circles in the same plane can have two common points, one common point, or no common points. In the first case, we say that the \index{circles!intersect} \pojem{circles intersect}, in the second case the \index{circles!touch} \pojem{circles touch} in their \index{touching point!of two circles} \pojem{touching point}, and in the third case they are \index{circles!are non-intersecting}\pojem{non-intersecting circles} (Figure \ref{sl.skk.4.1.3.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.1.3.pic} +\caption{} \label{sl.skk.4.1.3.pic} +\end{figure} + + +If the circles do not intersect, then the interior of at least one of these two circles is either in the interior or in the exterior of the other circle. This is a consequence of Theorem \ref{DedPoslKrozKroz}. + +The line that is determined by the intersection + of two circles that +intersect is called the \pojem{secant, which is the tangent of their common chord}. In connection with this, we prove the following theorem. + + + + \bizrek \label{KroznPresABpravokOS} + If two circles intersect at two points $A$ and $B$, + then the line containing the centres of the two circles is perpendicular to the line $AB$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.1.4.pic} +\caption{} \label{sl.skk.4.1.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $A$ and $B$ be the intersection points of the circles $k_1(S_1,r_1)$ and $k_2(S_2,r_2)$ (Figure \ref{sl.skk.4.1.4.pic}). + Because $S_1A\cong S_1B\cong r_1$ and $S_2A\cong S_2B\cong r_2$, + the line $S_1S_2$ is the perpendicular bisector of the line segment $AB$ (Theorem \ref{simetrala}), + so $S_1S_2\perp + AB$. + \kdokaz + + + We will prove that the circles that touch have a common tangent at +their touching point. + + + + \bizrek \label{tangSkupnaDotikKrozn} + Let $k_1$ and $k_2$ be different circles with centres $S_1$ and + $S_2$ touching at a point $T$. Then: + + (i) $S_1$, $S_2$ and $T$ are collinear points; + +(ii) the tangent of the circle $k_1$ at the point $T$ is at the same time the tangent of the circle $k_2$ at + the same point. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.1.5.pic} +\caption{} \label{sl.skk.4.1.5.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.1.5.pic}). + +(\textit{i}) We assume that the points $S_1$, $S_2$ and $T$ are not +collinear. The line $S_1S_2$ divides the plane in which the circles +lie, into two half-planes. The half-plane that contains the point $T$, +we denote by $\pi_1$, the other by $\pi_2$. From the statement \ref{izomEnaC'} +it follows that in the half-plane $\pi_2$ there exists (one and only one) point $T'$, for +which $S_1T'\cong S_1T$ and $S_2T'\cong ST_2$. This would mean that even the point $T'$, which is different from the point $T$, lies on the circles +$k_1$ and $k_2$, which is not possible. Therefore, the points $S_1$, $S_2$ and $T$ +are collinear. + + (\textit{ii}) By the statement \ref{TangPogoj} the tangent +of the circle $k_1$ at the point $T$ is perpendicular to the radius $S_1T$. Similarly +the tangent of the circle $k_2$ at the same point $T$ is perpendicular to the radius $S_2T$. Because of (\textit{i}) the lines $S_1T$ and $S_2T$ coincide, +so do both perpendiculars or tangents. + \kdokaz + +By the statement \ref{tangKrozEnaStr} all points of the circle are on the same side +of each of its tangents - on the side where its center is. This +means that the circles $k_1$ and $k_2$ from the previous statement are either + on the same side or on different sides of their common tangent. When $B(S_1,T,S_2)$, the circles are on different sides +of their common tangent and we say that the circles $k_1$ +and $k_2$ \pojem{touch each other from the outside}. Otherwise the circles are on the same side +of this tangent and we say that they \pojem{touch each other from the inside}. In +the first case, the interior of one of these two circles is in the exterior of the other, in +the second case, however, it is in the interior of the other circle (Figure +\ref{sl.skk.4.1.6.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.1.6.pic} +\caption{} \label{sl.skk.4.1.6.pic} +\end{figure} + +When the circles $k_1(S_1,r_1)$ and $k_2(S_2,r_2)$ touch each other from the outside, +it follows from the previous statement that $|S_1S_2| = r_1 + r_2$. +If the circles touch each other from the inside, then $|S_1S_2| = |r_1 - r_2|$. +It is clear that the converse is also true. The condition $|S_1S_2| = r_1 + +r_2$ or $|S_1S_2| = |r_1 - r_2|$ is sufficient for the circles +to touch each other from the outside or from the inside. In a similar way, we obtain the other +criteria for the mutual position of two circles. + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.1.7.pic} +\caption{} \label{sl.skk.4.1.7.pic} +\end{figure} + + \bizrek + Let $k_1(S_1,r_1)$ + and $k_2(S_2,r_2)$ ($r_1\geq r_2$) be two circles. Then + (Figure \ref{sl.skk.4.1.7.pic}): + + (i) the circles $k_1(S_1,r_1)$ + and $k_2(S_2,r_2)$ are lying outside each other + if and only if $|S_1S_2|>r_1+r_2$; + + (ii) the circles $k_1(S_1,r_1)$ + and $k_2(S_2,r_2)$ are touching each other externally + if and only if $|S_1S_2|=r_1+r_2$; + + (iii) the circles $k_1(S_1,r_1)$ + and $k_2(S_2,r_2)$ are intersecting each other at two points + if and only if $r_1-r_2<|S_1S_2|\omega$. If $\mathcal{B}(A,N,M)$ holds, we can, by using a similar reasoning, conclude that in this case $\angle +AMB<\omega$, which means that for no point $M\notin l \cup +l'\setminus\{A,B\}$ it holds that $\angle AMB \cong \omega$. + \kdokaz + +From the proof of the previous theorem we also get the following conclusion. + + + \bizrek \label{obodKotGMTZunNotr} + Let $AB$ be an arc of a circle $k$, $\omega$ the corresponding circumferential angle + of this arc and $M$ a point of the half-plane with the edge $AB$ not containing this arc (Figure + \ref{sl.skk.4.2.8.pic}). Then: + +(i) $\angle AMB >\omega$ if and only if the point $M$ is an interior point of the circle $k$; + + (ii) $\angle AMB \cong\omega$ if and only if the point $M$ lies on the circle $k$; + + (iii) $\angle AMB <\omega$ if and only if the point $M$ is an exterior point of the circle $k$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.8.pic} +\caption{} \label{sl.skk.4.2.8.pic} +\end{figure} + +In the following examples we will see the use of the theorem about the peripheral and central angle and its consequences. + + + + \bzgled + Construct a triangle with given $a$, $\alpha$, $v_a$. \label{konstr_aalphava} + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.1d.pic} +\caption{} \label{sl.skk.4.3.1d.pic} +\end{figure} + + \textbf{\textit{Analysis.}} The point $A$ lies at the same time at the geometric location of the points from which the line $BC$ is seen at an angle $\alpha$ (the union of two circular arcs - Theorem \ref{ObodKotGMT}) and at the parallel of the line $BC$, which is $v_a$ away from it (Figure \ref{sl.skk.4.3.1d.pic}). So the vertex $A$ is the intersection of this parallel and the aforementioned geometric location of points. + +\textbf{\textit{Construction.}} First, let's draw the line $BC\cong a$, then the geometric location of points $\mathcal{L}$, from which this line is seen at an angle $\alpha$ (Theorem \ref{ObodKotGMT}). Then let's draw the parallel $p$ of the line $BC$ at a distance $v_a$. With $A$ we mark the intersection of the line $p$ and the aforementioned geometric location of points $\mathcal{L}$. We will prove that the triangle $ABC$ is the desired triangle. + +\textbf{\textit{Proof.}} By construction, it is clear that $BC\cong a$. By construction, the point $A$ lies on the geometric location of points from which the line $BC$ is seen at an angle $\alpha$, so $BAC\cong\alpha$. The altitude of the triangle $ABC$ from the vertex $A$ is consistent with the line $v_a$, because the point $A$ by construction lies on the line $p$, which is $v_a$ away from the line $BC$. + +\textbf{\textit{Discussion.}} The necessary condition is of course $\alpha<180^0$. The number of solutions to the task is equal to the number of intersections +of the line $p$ and the set $\mathcal{L}$. + \kdokaz + + + + \bzgled + Let $p$, $q$ and $r$ be lines in the plane intersecting + at one point and divide this plane into six congruent angles. Suppose that $P$, $Q$ and $R$ + are the foots of the perpendiculars from an arbitrary point $X$ of this plane on the lines $p$, $q$ and + $r$, respectively. Prove that $PQR$ is a regular triangle. + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.9.pic} +\caption{} \label{sl.skk.4.2.9.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.2.9.pic}). + Let $S$ +be the intersection of the lines $p$, $q$ and $r$. It is clear that the lines determine +the angles $60^0$. Because $\angle XPS \cong \angle XQS \cong \angle XRS = +90^0$, by \ref{TalesovIzrKroz2} the points $S$, $X$, $P$, $Q$ and +$R$ lie on the circle $k$ with diameter $SX$. + If we use \ref{ObodObodKot} for the appropriate arcs $PQ$ and $QR$, + we have $\angle PRQ \cong \angle PSQ = +60^0$ and $\angle QPR \cong \angle QSR = 60^0$. Because all angles +are equal to $60^0$, $PQR$ is a regular triangle. + \kdokaz + +Let $k(O,R)$ and $l(S,r)$ ($R = 2r$) be circles touching each other internally +and $P$ an arbitrary point on the circle $l$. Which curve +is described by the point $P$ if the circle $l$ rolls without slipping around the circle $k$\footnote{This problem was solved by the Polish astronomer + \index{Copernicus, N.} \textit{N. Copernicus} (1473--1543). + In the general case, when it is not necessarily $R = 2r$, the curve + which is described by the point $P$, is called \index{hipocikloida} + \pojem{hipocikloida}. In the case of the outer rolling of the circle + around the other circle, the curve is called \index{epicikloida} + \pojem{epicikloida}, in the case + of the rolling of the circle around the straight line, it is called \index{cikloida} + \pojem{cikloida}. The cikloida was first investigated by the German mathematician and + philosopher \index{Kuzanski, N.} \textit{N. Kuzanski} (1401--1464) + and later by the French mathematician and philosopher \index{Mersenne, M.} + \textit{M. Mersenne} (1588--1648). It was named by the Italian physicist, + mathematician, astronomer and philosopher \index{Galilei, G.} + \textit{G. Galilei} (1564--1642) in 1599.}? + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.10.pic} +\caption{} \label{sl.skk.4.2.10.pic} +\end{figure} + +\textbf{\textit{Solution.}} + Let +$P_0$ be the position of the point $P$ at the moment when it lies on the circle $k$ (Figure +\ref{sl.skk.4.2.10.pic}). At the moment when the point $P$ is in position $P_i$, +the circle $l$, which is in position $l_i$, touches the circle $k$ at some +point $T_i$. Since this is a "movement without slipping", the lengths +of the corresponding arcs $P_0T_i$ and $P_iT_i$ of the circles $k$ and $l$ are equal to each other. The radius of the circle $k$ is twice as large as the radius of the circle $l$, +so for the corresponding central angles of the aforementioned arcs we have $\angle +T_iS_iP_i= 2\angle T_iOP_0$. But $\angle T_iOP_i$ is the corresponding arc angle of the circle $l$ for the same arc $P_iT_i$, so $2\angle +T_iOP_i = \angle T_iS_iP_i$ (statement \ref{ObodObodKot}). From the two previous +relations we obtain $\angle T_iOP_i= \angle T_iOP_0$, which means that the point $P_i$ is collinear with the points $O$ and $P_0$, so the desired path +of the point $P$ represents the diameter $P_0P'_0$ of the circle $k$. + \kdokaz + + + + \bzgled + Let $C$ be the midpoint of an arc $AB$ and $D$ an arbitrary point of this + arc other than $C$. Prove that + $$AC + BC > AD + BD.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.11.pic} +\caption{} \label{sl.skk.4.2.11.pic} +\end{figure} + +\textbf{\textit{Proof.}} + (Figure +\ref{sl.skk.4.2.11.pic}). + Let $C'$ and $D'$ be such points on the line segments $AC$ and $AD$, + that: $CC'\cong CB \cong CA$, $DD'\cong DB$, +$\mathcal{B}(A,C,C')$ and $\mathcal{B}(A,D,D')$. The triangles $C'CB$ and +$D'DB$ are isosceles, therefore by \ref{enakokraki} and \ref{zunanjiNotrNotr}, +it follows: + $$\angle CC'B \cong \angle CBC' = \frac{1}{2}\angle ACB +\hspace*{2mm}\textit{ and }\hspace*{2mm} + \angle DD'B \cong \angle DBD' += \frac{1}{2}\angle ADB.$$ + Because the angles $ACB$ and $ADB$ are supplementary + (by \ref{ObodObodKot}). +The angles $AC'B$ and $AD'B$ are supplementary as well, and the points $C'$ and $D'$ lie +on the corresponding arc $l'$ over the segment $AB$ (by \ref{ObodKotGMT}). Because $CC'\cong CB \cong CA$, the distance $AC'$ +is the diameter of the circle that contains the arc $l'$, therefore the $\angle AD'C'$ is a right angle (by \ref{TalesovIzrKroz2}). The distance $AC'$ is therefore +the hypotenuse of the right triangle $AD'C'$ and by \ref{vecstrveckot} it follows: +$$AC + CB = AC'> AD'= AD + DB,$$ which was to be proven. \kdokaz + + + + \bnaloga\footnote{35. IMO Hong Kong - 1994, Problem 2.} + $ABC$ is an isosceles triangle with $BC \cong AC$. Suppose that: + + (i) $D$ is the midpoint of $AB$ and $E$ is the point on the line $CD$ such that + $EA\perp AC$; + + (ii) $F$ is an arbitrary point on the segment $AB$ different from $A$ and $B$; + + (iii) $G$ lies on the line $CA$ and $H$ lies on the line $CB$ such that $G$, $F$, $H$ are + distinct and collinear. + + Prove that $EF$ is perpendicular to $HG$ if and only if $GF\cong FH$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.IMO1.pic} +\caption{} \label{sl.skk.4.2.IMO1.pic} +\end{figure} + + +\textbf{\textit{Proof.}} First, from the similarity of the triangles $CAE$ and +$CBE$ (by \textit{SAS} \ref{SKS}) it follows that $\angle EBC\cong\angle +EAC=90^0$ and $EA\cong EB$ (Figure \ref{sl.skk.4.2.IMO1.pic}). +We prove the equivalence $EF\perp HG \Leftrightarrow GF\cong FH$ in both +directions. + +($\Rightarrow$) Let's assume that the lines $EF$ and $HG$ are +perpendicular, i.e. $\angle EFG\cong\angle EFH=90^0$. Let $k$ +and $l$ be circles with diameters $EG$ and $EH$. Because +$\angle EAG\cong\angle EFG=90^0$ and $\angle EBH\cong\angle +EFH=90^0$, by \ref{TalesovIzrKroz} we have $A,F\in k$ and $B,F\in +l$. First, from $EA\cong EB$ it follows that $\angle EAB\cong +\angle EBA$. From this and \ref{ObodObodKot} it follows: + $$\angle EGF\cong\angle EAF\cong\angle EBF\cong\angle EHF.$$ + Therefore, the triangle $EGH$ is equilateral, so its height $EF$ + is also the altitude (congruence of triangles $EFG$ and $EFH$, + \textit{ASA} \ref{KSK}) or $GF\cong FH$. + +($\Leftarrow$) Now let $GF\cong FH$, i.e. the point $F$ is the +center of the line $GH$. Let $k$ be a circle with diameter $EG$. +In addition to the point $A$ we mark with $\widehat{F}$ the +other intersection of this circle with the line $AB$. If the +circle $k$ touches the line $AC$, it follows that $G=A$ or $F=D$ +and $H=B$, so in this case $GF\cong FH$ is already fulfilled. + +Assume that $\widehat{F}\neq F$. Let $\widehat{H}$ be the +intersection of the lines $G\widehat{F}$ and $CB$. Because the +point $\widehat{F}$ lies on the circle $k$ with diameter $EG$, +$\angle G\widehat{F}E=90^0$, i.e. $E\widehat{F}\perp +G\widehat{H}$. Therefore, for the points $G$, $\widehat{F}$ and +$\widehat{H}$ the assumptions of the left side of the +equivalence are fulfilled, so from the already proven first part +of the statement ($\Rightarrow$) it follows that +$G\widehat{F}\cong \widehat{F}\widehat{H}$, i.e. the point +$\widehat{F}$ is the center of the line $G\widehat{H}$. In the +triangle $GH\widehat{H}$ the $\widehat{F}F$ is the median, so +$\widehat{F}F\parallel \widehat{H}H$ and $AB\parallel BC$, which +is not possible. Thus, the assumption $\widehat{F}\neq F$ +disappears, so $\widehat{F}= F$, so $\widehat{H}=H$. In the end, +from $E\widehat{F}\perp G\widehat{H}$ it follows that $EF\perp +GH$. + \kdokaz + + + + + +%________________________________________________________________________________ + \poglavje{More About Circumcircle and Incircle of a Triangle} + \label{odd4OcrtVcrt} + + +First, we will consider some important points that lie on the +circumcircle of a triangle. + +\bizrek \label{TockaN} + The bisector of the side $BC$ and the bisector of the interior angle $BAC$ of a + triangle $ABC$ ($AB\neq AC$) intersect at the Circumcircle $l(O,R)$ + of that triangle. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.1.pic} +\caption{} \label{sl.skk.4.3.1.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.3.1.pic}). + + Let the point $N$ be one of the intersections of the Circumcircle $l(O,R)$ of the triangle $ABC$ and + the bisector of the side $BC$ +(such that $A,N\perp BC$). Because the point $N$ lies on the bisector of the side +$BC$, it follows that $NB \cong NC$. Therefore, the triangle $BNC$ is an isosceles triangle, +so the angles $\angle NBC$ and $\angle NCB$ are congruent angles (by Theorem \ref{enakokraki}). Because the point $N$ also lies on the Circumcircle $l(O,R)$ +of the triangle $ABC$, by Theorem \ref{ObodObodKot} it follows that: + \begin{eqnarray*} + \angle BAN +\cong \angle BCN \textrm{ (obodna kota za krajši lok }BN \textrm{) }\\ + \angle NAC \cong \angle NBC \textrm{ (obodna kota za krajši lok } + CN \textrm{).} + \end{eqnarray*} + Therefore, $\angle BAN \cong \angle NAC$ or the line $AN$ is the bisector of the angle +$BAC$, thus the theorem is proven. + \kdokaz + + + The point $N$ from the previous theorem is the center of that arc $BC$ of the Circumcircle $l(O,R)$ of the triangle $ABC$ which does not contain the vertex $A$. + We shall now prove another important property of the point $N$. + + + + \bizrek \label{TockaN.NBNC} + For the point $N$ from the previous theorem is $NB\cong NS\cong NC$, + where $S$ is the incentre of the triangle $ABC$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.2.pic} +\caption{} \label{sl.skk.4.3.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let's mark with $\alpha$ and $\beta$ the internal +angles of the triangle $ABC$ at the vertices $A$ and $B$ (Figure +\ref{sl.skk.4.3.2.pic}). $BNS$ is an isosceles triangle (statement +\ref{enakokraki}), because the angles at the vertices $B$ and $S$ are equal. +If we use the statement \ref{zunanjiNotrNotr} and \ref{ObodObodKot}, +we get: + \begin{eqnarray*} + \angle BSN &=& \angle ABS + \angle BAS =\frac{1}{2}\alpha+\frac{1}{2}\beta,\\ +\angle SBN &=& \angle SBC +\angle CBN = \angle SBC + \angle CAN = +\frac{1}{2}\beta+\frac{1}{2}\alpha. + \end{eqnarray*} +Therefore $NB \cong NS$ and similarly $NC \cong NS$. +\kdokaz + + + + \bizrek \label{TockaNbetagama} + Let $AA'$ be the altitude from the vertices $A$ and $AE$ bisector of the interior + angle $BAC$ of a triangle $ABC$ ($A',E\in BC$). Suppose that $l(O,R)$ is the + circumcircle of that triangle. If $\angle CBA=\beta\geq\angle ACB=\gamma$, + then + $$\angle A'AE\cong \angle EAO=\frac{1}{2}\left( \beta-\gamma\right).$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.1b.pic} +\caption{} \label{sl.skk.4.3.1b.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Let $N$ be the point defined as in the previous statements + (Figure \ref{sl.skk.4.3.1b.pic}). The lines + $AA'$ and $ON$ are parallel, because they are both perpendicular to the line $BC$. + Because $OA\cong ON=R$, $AON$ is an isosceles triangle. Therefore, first of all: + $$\angle A' AE \cong \angle ANO \cong \angle NAO = +\angle EAO,$$ + and then: + $$\angle A'AE=\frac{1}{2}\alpha-\left(90^0-\beta \right)= + \frac{1}{2}\alpha-\left(\frac{\alpha+\beta+\gamma}{2}-\beta \right)= + \frac{1}{2}\left( \beta-\gamma\right),$$ which was to be proven. \kdokaz + +\bzgled \label{tockaNtockePQR} + Let $P$, $Q$ and $R$ be the midpoints of those arcs $BC$, $AC$ and $AB$ of + the circumcircle of a triangle $ABC$ not containing the vertices $A$, $B$ and $C$ + of that triangle. + If $E$ and $F$ are intersections of the line $QR$ with sides $AB$ and $AC$, respectively + and $S$ the incentre of this triangle, then: + + (i) $AP \perp QR$, + + (ii) the quadrilateral $AESF$ is a rhombus. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.2a.pic} +\caption{} \label{sl.skk.4.3.2a.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.3.2a.pic}) + +(\textit{i}) Let $L$ be the intersection of the lines $AP$ and $QR$. By +\ref{TockaN} points $P$, $Q$ and $R$ lie on the simetrals $AS$, $BS$ +and $CS$ of the internal angles of the triangle $ABC$ ($S$ is the center of the triangle +$ABC$ of the inscribed circle). If we denote with $\alpha$, $\beta$ and $\gamma$ the internal angles of the triangle $ABC$, then due to the similarity of the corresponding external angles (\ref{ObodObodKot}) we get: + \begin{eqnarray*} +\angle RPL &=& \angle RPA = \angle RCA =\frac{1}{2}\gamma,\\ + \angle PRL &=& \angle PRQ = \angle PRC + \angle CRQ = +\angle PAC + \angle CBQ = \frac{1}{2}\alpha+ \frac{1}{2}\beta. + \end{eqnarray*} +Therefore, the sum of the angles in the triangle $PRL$ (\ref{VsotKotTrik}) +$180^0 =\frac{1}{2}\alpha+ \frac{1}{2}\beta +\frac{1}{2}\gamma ++\angle RLP = 90° + \angle RLP$. So $\angle RLP = 90°$ or $AP +\perp QR$. + +(\textit{ii}) By +the statement \ref{TockaN.NBNC} it is $RA \cong RS$, therefore from the similarity of +right-angled triangles $ALR$ and $SLR$ (statement \textit{SSA} +\ref{SSK}) it follows that the point $L$ is the center of the diagonal $AS$ +of the quadrilateral $AESF$. From the similarity of triangles $AEL$ and $AFL$ (statement +\textit{ASA} \ref{KSK}) it follows that the point $L$ is also the center +of the diagonal $EF$, therefore the $AESF$ is a parallelogram (statement +\ref{paralelogram}). Because the diagonals $AS$ and $EF$ are also perpendicular, the +quadrilateral $AESF$ is a rhombus (statement \ref{RombPravKvadr}). + \kdokaz + +From the previous statement we directly get a consequence. + + + \bzgled \label{PedalniLemasPQR} + Let $P$, $Q$ and $R$ be the midpoints of those arcs $BC$, $AC$ and $AB$ of + the circumcircle of a triangle $ABC$ not containing the vertices $A$, $B$ and $C$ + of that triangle. + Prove that the incentre of the triangle $ABC$ is at the same time + orthocentre of the triangle $PQR$ (Figure \ref{sl.skk.4.3.2b.pic}). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.2b.pic} +\caption{} \label{sl.skk.4.3.2b.pic} +\end{figure} + + + +We prove some consequences of statement \ref{ObodObodKot}, which are related to +the altitude point. + + + + \bizrek \label{TockaV'} + Points that are symmetric to the orthocentre of an acute triangle + with respect to its sides lie on the circumcircle of this triangle. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.3.pic} +\caption{} \label{sl.skk.4.3.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $V$ be the orthocentre of a triangle +$ABC$, $l$ the circumcircle of the triangle $ABC$ and $V_a$ the other +intersection point of the circle $l$ with the altitude $AA'$ (Figure +\ref{sl.skk.4.3.3.pic}). We prove that the point $V_a$ is symmetric to +the point $V$ with respect to the line $BC$. It is enough to prove +that $VA'\cong V_aA'$. The angles $V_aBC$ and $V_aAC$ are complementary +(circumscribed angle for the chord $V_aC$ - izrek \ref{ObodObodKot}), +the angles $V_aAC$ and $CBV$ are complementary angles with +perpendicular legs (izrek \ref{KotaPravokKraki}). Therefore, the +angles $V_aBC$ and $CBV$ are also complementary, and so are the +triangles $V_aBA'$ and $VBA'$ or $VA'\cong V_aA'$. A similar statement +holds for the other two altitudes. + \kdokaz + + \bizrek \label{TockaV'a} + Let $V$ be the orthocentre of a triangle + $ABC$. The circumcircles of the triangles $VBC$, $AVB$ and $ACV$ are congruent to + the circumcircle of the triangle $ABC$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.3a.pic} +\caption{} \label{sl.skk.4.3.3a.pic} +\end{figure} + + +\textbf{\textit{Proof.}} A direct consequence of the previous izrek +\ref{TockaV'}, because the three aforementioned circumcircles are +symmetric to the circumcircle of the triangle $ABC$ with respect to +the altitude of its sides (Figure \ref{sl.skk.4.3.3a.pic}). + + + + \bizrek \label{TockaV1} + Points that are symmetric to the orthocentre of an acute triangle + with respect to the midpoints of its sides lie on the circumcircle of this triangle. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.3b.pic} +\caption{} \label{sl.skk.4.3.3b.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $V$ be the altitude point of the triangle $ABC$, +$l$ the circumscribed circle (with the center $O$) of the triangle $ABC$ and $V_{A_1}$ the point that is symmetrical to the point $A$ with respect to the point $O$ (Figure \ref{sl.skk.4.3.3.pic}). From the very definition of the point $V_{A_1}$ it is clear that it lies on the circle $l$. We shall prove that $V_{A_1}$ is symmetrical to the point $V$ with respect to the point $A_1$, which is the center of the line $BC$. Because $AV_{A_1}$ is the diameter of the circle $l$, according to the Theorem \ref{TalesovIzrKroz2} $\angle ACV_{A_1}=90^0$ or $V_{A_1}C\perp AC$. The line $BV$ is the altitude of the triangle $ABC$, therefore $BV\perp AC$. From the last two relations it follows that $V_{A_1}C\parallel BV$. Similarly +$V_{A_1}B\parallel CV$. Therefore the quadrilateral $V_{A_1}CVB$ is a parallelogram, thus its diagonals $VV_{A_1}$ and $BC$ have a common center. The center of the line $BC$ is the point $A_1$, which means that $V_{A_1}$ is symmetrical to the point $V$ with respect to the point $A_1$. The point $V_{A_1}$ but, according to the construction, lies on the circle $l$. +\kdokaz + + +We shall also prove some consequences of the Theorem \ref{ObodObodKot}, which are connected with the circumscribed +circle of a triangle. + + + \bzgled \label{zgledTrikABCocrkrozP} + Let $k$ be the circumcircle of a regular triangle $ABC$. + If $P$ is an arbitrary point lying + on the shorter arc $BC$ of the circle $k$, then + $$|PA|=|PB|+|PC|.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.4.pic} +\caption{} \label{sl.skk.4.3.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} Because $\angle ACP>60^0>\angle PAC$, by +the statement \ref{vecstrveckot} $AP>PC$ (Figure +\ref{sl.skk.4.3.4.pic}). Therefore, on the line $AP$ there exists such +a point $Q$, that $PQ\cong PC$. By the statement \ref{ObodObodKot} +$\angle CPQ=\angle CPA\cong\angle CBA=60^0$, which means that +$PCQ$ is an isosceles triangle, therefore also $CQ\cong CP$ and +$\angle PCQ=60^0$. From this it follows that $\angle ACQ=\angle +ACP-60^0=\angle BCP$. By the statement \textit{SAS} (statement +\ref{SKS}) the triangles $ACQ$ and $BCP$ are similar, therefore +also $AQ\cong BP$. + +In the end: $|PB|+|PC|=|AQ|+|PQ|=|AP|$. +\kdokaz + + + + \bzgled + Three circles of equal radii $r$ intersect at point $O$. + Furthermore each two of them intersect at one more point: $A$, $B$, and $C$. + Prove that the radius of the circumcircle of the triangle $ABC$ is also $r$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.5.pic} +\caption{} \label{sl.skk.4.3.5.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.3.5.pic}) + +We mark with $k$, $l$, $j$ and $o$ the circumcircles of the triangles +$OBC$, $OAC$, $OAB$ and $ABC$ and with $P$ an arbitrary point of the +circle $k$ so that the points $O$ and $P$ are on different sides of +the line $BC$. By the assumption the circles $k$, $l$ and $j$ are +similar. The angles $BAO$ and $BCO$ are also similar, because they +are the angles between the similar circumcircles $k$ and $j$ over +the chord $BO$ (statement \ref{SklTetSklObKot2}). Analogously the +angles $CAO$ and $CBO$ are similar. Because of this: + $$\angle BAC = \angle BAO + \angle CAO = \angle BCO ++ \angle CBO = 180° - \angle BOC = \angle BPC.$$ + Therefore the circles $k$ and $o$ have a similar +circumferential angle over the common chord $BC$, therefore they are +similar. + \kdokaz + + + + \bzgled \label{KvadratKonstr4tocke} + Construct a square $ABCD$ such that given points $P$, $Q$, $R$ and $S$ + lyes on the sides $AB$, $BC$, $CD$ and $DA$ of this square, respectively. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.1c.pic} +\caption{} \label{sl.skk.4.3.1c.pic} +\end{figure} + + +\textbf{\textit{Solution.}} +Since $\angle PAS$ and $\angle QCR$ are right angles, points $A$ and $B$ lie on the circles $k$ and $l$ with diameters +$PS$ and $QR$ (Figure \ref{sl.skk.4.3.1c.pic}). The carrier of the diagonal $AC$ of the square $ABCD$ is also the symmetry of the internal angles $BAD$ and $BCD$, so it goes through +the centers $N$ and $M$ of the corresponding arcs, which are determined by $k$ and $l$ (statement \ref{TockaN}). +The construction can therefore be carried out by first planning the circles $k$ and $l$, then the line $NM$, the points $A$ and $C$ and finally +the points $B$ and $D$. + \kdokaz + + + + + \bzgled + Construct a triangle $v_a$, $t_a$, $l_a$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.1e.pic} +\caption{} \label{sl.skk.4.3.1e.pic} +\end{figure} + + \textbf{\textit{Solution.}} + Let $ABC$ be a triangle in which the altitude $AA'$, +the median $AA_1$ and the section of the symmetry $AE$ of the internal angle $BAC$ are consistent +with the distances $v_a$, $t_a$ and $l_a$. With $O$ we mark the center of the triangle $ABC$ of the drawn +circle $k$. By statement \ref{TockaN} +the lines $AE$ and $OA_1$ intersect in the point $N$, which lies on +the circle $k$ (Figure \ref{sl.skk.4.3.1e.pic}). + +So we can first plan +a right triangle $AA'E$ with the leg $v_a$ and the hypotenuse $l_a$ and the point $A_1$ from the condition $AA_1\cong t_a$. Then +plan the point $N$ as the intersection of the line $AE$ and the perpendicular of the line $A'E$ through +the point $A_1$. The center $O$ is the intersection of the line $A_1N$ and +the symmetry of the distance $AN$ (because $AN$ is the chord of the circle $k$). The points $B$ and $C$ are the intersections +of the circle $k(O,OA)$ with the line $A'E$. + \kdokaz + + + + + \bzgled + Construct a triangle $R$, $r$, $a$. \label{konstr_Rra} + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.1a.pic} +\caption{} \label{sl.skk.4.3.1a.pic} +\end{figure} + + +\textbf{\textit{Solution.}} Let $ABC$ be a triangle such that $BC\cong a$ and $l(O,R)$ and $k(S,r)$ are its circumscribed and inscribed circle + (Figure \ref{sl.skk.4.3.1a.pic}). We denote with $\alpha$, $\beta$ and $\gamma$ its internal angles at vertices $A$, $B$ and $C$. By +\ref{SredObodKot} we have $\alpha = \angle BAC = \frac{1}{2}\cdot\angle BOC$. +From \ref{kotBSC} it follows that $\angle BSC=90^0+\frac{1}{2}\cdot\alpha$. +From two relations we obtain the equality +$\angle BSC=90^0+\frac{1}{4}\cdot\angle BOC$, which allows the construction. + +First we draw an isosceles triangle $BOC$ ($BC\cong a$ and $OB\cong +OC\cong R$). We obtain point $S$ as one of the intersections of the arc with the chord +$BC$ and the external angle $90^0+\frac{1}{4}\cdot\angle BOC$ and the line, +which is at a distance $r$ parallel to the line $BC$. Then we draw +the inscribed circle $k(S,r)$ and point $A$ as the intersection of the other two tangents +of this circle from points $B$ and $C$. + \kdokaz + + + + \bnaloga\footnote{47. IMO Slovenia - 2006, Problem 1.} + Let $ABC$ be a triangle with incentre $I$. A point $P$ in the interior of the + triangle satisfies + $$\angle PBA + \angle PCA = \angle PBC + \angle PCB.$$ + Show that $|AP| \geq |AI|$, and that equality holds if and only if $P = I$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skl.4.3.IMO1.pic} +\caption{} \label{sl.skl.4.3.IMO1.pic} +\end{figure} + +\textbf{\textit{Proof.}} We mark with $\alpha$, $\beta$ and $\gamma$ +the internal angles of the triangle $ABC$ at the vertices $A$, $B$ and $C$ +(Figure \ref{sl.skl.4.3.IMO1.pic}). The condition $\angle PBA + \angle PCA += \angle PBC + \angle PCB$ can be rewritten in the form +$\beta-\angle PBC + \gamma-\angle PCB = \angle PBC + \angle PCB$ +or: + $$\angle PBC + \angle PCB=\frac{1}{2}\left( \beta+\gamma\right).$$ +From this and the fact that the sum of the internal angles of each of the +triangles $BPC$ and $ABC$ is equal to $180^0$ (\izrekref{VsotKotTrik}), +it follows: + $$\angle BPC =180^0-\frac{1}{2}\left( \beta+\gamma\right)=90^0+ + \frac{1}{2} \alpha.$$ +But from \kotBSC it follows $\angle BIC =90^0+ + \frac{1}{2}\cdot \alpha$, so $\angle BPC\cong \angle BIC$. + Therefore, the points $P$ and $I$ lie on the same curve $\mathcal{L}$ with + the chord $BC$ and the central angle $90^0+ + \frac{1}{2} \alpha$. Let the point $N$ be the intersection + of the perpendicular bisector of the side $BC$ and the perpendicular + bisector of the internal angle $BAC$ of the triangle $ABC$. By + \izrekref{TockaN}, the point $N$ lies on the circumscribed circle + of the triangle $ABC$ and $NB\cong NI\cong NC$ (\izrekref{TockaN.NBNC}). This means that $N$ is the center of the curve + $\mathcal{L}$, so $NP\cong NI$ or $\angle NIP\cong\angle + NPI<90^0$. Because the points $A$, $I$ and $N$ are collinear (they lie on + the perpendicular bisector of the internal angle $BAC$), $\angle AIP =180^0-\angle + NIP>90^0$. From \izrekref{vecstrveckot} (for the triangle $API$) + it now follows that $|AP| \geq |AI|$ and the equality holds exactly when + the triangle $API$ is not, i.e. when $P=I$. + \kdokaz + + + \bnaloga\footnote{43. IMO United Kingdom - 2002, Problem 2.} + $BC$ is a diameter of a circle center $O$. $A$ is any point on + the circle with $\angle AOC>60^0$. $EF$ is the chord which is the perpendicular + bisector of $AO$. $D$ is the midpoint of the minor arc $AB$. The line through + $O$ parallel to $AD$ meets $AC$ at $J$. Show that $J$ is the incenter of triangle + $CEF$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.IMO4.pic} +\caption{} \label{sl.skk.4.3.IMO4.pic} +\end{figure} + +\textbf{\textit{Proof.}} Because +the points $E$ and $F$ lie on the line of symmetry $EF$, and also on +the circle with center $O$, we have $$AF\cong FO\cong AO\cong EO \cong EA.$$ This +means that the quadrilateral $EOFA$ is a rhombus, which is made up of two congruent triangles $AOF$ and $AEO$. + +Without loss of generality, we +assume that +$\angle COE>\angle COF$ (Figure \ref{sl.skk.4.3.IMO4.pic}). +First, from the condition $\angle + AOC>60^0$ it follows that $\angle COF=60^0- \angle + AOC>0^0$, so the points $A$ and $F$ are on the same side of the line + $BC$. + + Because $AOC$ is an isosceles triangle with the base $AC$, by the + \ref{enakokraki} and \ref{zunanjiNotrNotr} we have + $\angle ACO =\frac{1}{2}\angle AOB$. The point $D$ is the center + of the arc $BD$, so $\angle AOD\cong\angle DOB$ + or $\angle DOB=\frac{1}{2}\angle AOB$. This means that + $\angle ACO\cong\angle DOB$ and the lines $AC$ and $DO$ are parallel by + \ref{KotiTransverzala}. Because $AD\parallel JO$ by assumption, the + quadrilateral $ADOJ$ is a parallelogram, so $AJ\cong OD$. Therefore + $$AJ\cong OD\cong OE\cong AF\cong AE.$$ + From $AF\cong AE$ it follows that + $AJ$ is the line of symmetry of the internal angle at the vertex $C$ of the triangle $CEF$ + (\ref{SklTetSklObKot}). Because $AJ\cong AF\cong AE$, + by \ref{TockaN.NBNC} the point $J$ is the center + of the inscribed circle of this triangle. +\kdokaz + + +%________________________________________________________________________________ + \poglavje{Cyclic Quadrilateral} \label{odd4Tetivni} + +We say that a +\index{štirikotnik!tetiven}\index{večkotnik!tetiven}\pojem{tetiven}, +if there exists a circumscribed circle, or if there is a circle that +contains all of its vertices (Figure \ref{sl.skk.4.5.10.pic}). For +vertices in this case we say that they are \index{konciklične +točke}\pojem{konciklične točke}. We have already seen that every +triangle is tetiven (izrek \ref{SredOcrtaneKrozn}) and also that every +regular polygon is tetiven (izrek \ref{sredOcrtaneKrozVeck}). On the +other hand, it is clear that not all polygons are cyclic. For example, +a diamond is a quadrilateral that does not have a circumscribed +circle. In this section we will therefore deal with cyclic +quadrilaterals. + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.10.pic} +\caption{} \label{sl.skk.4.5.10.pic} +\end{figure} + +Since a square is a regular polygon, it is also a cyclic +quadrilateral. It is not difficult to prove that a rectangle is also a +type of cyclic quadrilateral - the center of the circumscribed circle +is the intersection of its diagonals, which are consistent and +bisect each other. But how would we generally determine if a +quadrilateral is cyclic? It is clear that in a cyclic quadrilateral +(generally also in a polygon) the altitudes of all its sides intersect +in one point (Figure \ref{sl.skk.4.5.10.pic}). This condition is +sufficient for the quadrilateral to be cyclic, but it is not +sufficiently operative in specific cases. For the cyclicity of +quadrilaterals there is a necessary and sufficient condition that is +more useful. + + + + \bizrek \label{TetivniPogoj} + A convex quadrilateral is cyclic if and only if + its opposite interior angles are supplementary. + Thus, if $\alpha$, $\beta$, $\gamma$ and $\delta$ are + the interior angles of a convex quadrilateral $ABCD$, + it is cyclic if and only if + $$\alpha+\gamma=180^0.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.11.pic} +\caption{} \label{sl.skk.4.5.11.pic} +\end{figure} + +\textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.5.11.pic}) + +($\Rightarrow$) First, let's assume that the quadrilateral $ABCD$ +is cyclic. Because it is convex, the vertices $A$ and $C$ are on +different sides of the line $BD$. By \ref{ObodObodKotNaspr} +$\alpha+\gamma=180^0$. + +($\Leftarrow)$ Now let's assume that the opposite angles of the +quadrilateral $ABCD$ are supplementary, i.e. $\alpha+\gamma=180^0$. +Let $k$ be the circle drawn through the triangle $ABD$. In this +case, the fourth vertex $C$ of the line $BD$ is seen under the +angle complementary to the angle at the vertex $A$, which means +that the point $C$ also lies on the circle $k$ (\ref{ObodKotGMT}). +\kdokaz + + A direct consequence is the following theorem. + + + + \bizrek \label{TetivniPogojZunanji} + A convex quadrilateral is cyclic if and only if + one of its interior angles is congruent to the opposite exterior angle. + Thus, if $\alpha$, $\beta$, $\gamma$ and $\delta$ are + the interior angles and + $\alpha'$, $\beta'$, $\gamma'$ in $\delta'$ the exterior angles + of a convex quadrilateral $ABCD$, + it is cyclic if and only if + $$\alpha\cong\gamma'.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.12.pic} +\caption{} \label{sl.skk.4.5.12.pic} +\end{figure} + +We use the criterion from \ref{TetivniPogoj} for a parallelogram +and a trapezoid. + + \bizrek \label{paralelogramTetivEnakokr} + A parallelogram is cyclic if and only if it is a rectangle. + \eizrek + + +\textbf{\textit{Proof.}} Let $\alpha$, $\beta$, $\gamma$ and $\delta$ +be the interior angles of the parallelogram $ABCD$ + (Figure \ref{sl.skk.4.5.13.pic}). + +($\Leftarrow$) If the parallelogram is a rectangle, +$\alpha+\gamma=90^0+90^0=180^0$, which means that $ABCD$ is a cyclic +quadrilateral (\ref{TetivniPogoj}). + +($\Rightarrow$) Let's assume that $ABCD$ is a trapezoidal parallelogram. + Because $ABCD$ is a parallelogram, according to Theorem \ref{paralelogram} + $\alpha\cong\gamma$. Because it is also a trapezoid, according to Theorem \ref{TetivniPogoj} +$\alpha+\gamma=180^0$. So $\alpha\cong\gamma=90^0$, therefore + $ABCD$ is a rectangle. + \kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.13.pic} +\caption{} \label{sl.skk.4.5.13.pic} +\end{figure} + + + + \bizrek \label{trapezTetivEnakokr} + A trapezium is cyclic if and only if it is isosceles. + \eizrek + +\textbf{\textit{Proof.}} Let $ABCD$ be a trapezoid with a base $AB$ and with +internal angles $\alpha$, $\beta$, $\gamma$ and $\delta$ + (Figure \ref{sl.skk.4.5.13.pic}). In any trapezoid + it holds that $\alpha+\delta=180^0$ and $\beta+\gamma=180^0$. + + +($\Leftarrow$) Let's assume that trapezoid $ABCD$ is isosceles, i.e. $AD +\cong BC$. According to Theorem \ref{trapezEnakokraki} in this case +$\alpha\cong\beta$. So $\alpha+\gamma=\beta+\gamma=180^0$, therefore +according to Theorem \ref{TetivniPogoj} $ABCD$ is a cyclic quadrilateral. + +($\Rightarrow$) Let trapezoid $ABCD$ be a cyclic quadrilateral and $k$ +its circumscribed circle. The bases $AB$ and $CD$ are parallel +chords of this circle, so they have a common perpendicular, which goes through +the center $S$ of the circle $k$ and is perpendicular to the chords $AB$ and +$CD$. This means that the legs $AD$ and $BC$ are symmetrical with respect to this perpendicular, so they are congruent and trapezoid $ABCD$ is isosceles. + \kdokaz + + Particularly interesting are cyclic quadrilaterals with perpendicular + diagonals\footnote{\index{Brahmagupta}\textit{Brahmagupta} (598--660), Indian mathematician, who + studied such quadrilaterals.}. + The following example applies to +such quadrilaterals. + + + \bzgled \label{TetivniLemaBrahm} + Suppose that the diagonals of a cyclic quadrilateral $ABCD$ are perpendicular and intersect + at a point $S$. Prove that the foot of the perpendicular from the point $S$ on the line $AB$ + contains the midpoint of the line $CD$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.14.pic} +\caption{} \label{sl.skk.4.5.14.pic} +\end{figure} + +\textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.5.14.pic}) + +Let $N$ and $M$ be the intersections of the rectangle with the line $AB$ through the point $S$ with the sides $AB$ and $CD$ of the quadrilateral $ABCD$. Then it holds: + \begin{eqnarray*} + \angle CDB &\cong& \angle CAB \hspace*{3mm} + \textrm{(external angle for the appropriate locus } CB + \textrm{ - izrek \ref{ObodObodKot}})\\ + &\cong& \angle NSB \hspace*{3mm} + \textrm{ (angle with +perpendicular arms - izrek \ref{KotaPravokKraki})}\\ + &\cong& \angle MSD \hspace*{3mm} + \textrm{(perfect angle)} + \end{eqnarray*} + Because $\angle CDB\cong \angle MSD$, $MD \cong MS$ (izrek \ref{enakokraki}). + Similarly, $MC \cong MS$. Therefore, + $MD \cong MC$, which means that $M$ is the center of the side $CD$. + \kdokaz + +We will consider one property of cyclic quadrilaterals with perpendicular diagonals +in the example \ref{HamiltonPoslTetiv}. + + + + \bzgled \label{TetŠtirZgl0} + Let $k$ be the circumcircle of a cyclic quadrilateral $ABCD$ + and $N$, $M$, $L$ and $P$ the midpoints of those arcs $AB$, $B$C, $CD$ and $AD$ + of the circle $k$, not containing the third vertices of this quadrilateral. + Prove that $NL\perp PM$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.0.pic} +\caption{} \label{sl.skk.4.5.0.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $S$ be the intersection of the lines $NL$ and $PM$ +(Figure \ref{sl.skk.4.5.0.pic}). + If we use the izrek \ref{ObodObodKot} and \ref{TockaN} twice, we get: + + \begin{eqnarray*} + \angle PNS &=& \angle PND +\angle DNL = +\angle PBD +\angle DBL =\\ + &=& \frac{1}{2} \angle ABD +\frac{1}{2}\angle CBD = \frac{1}{2}\angle + ABC. + \end{eqnarray*} + +We similarly prove that $\angle NPS = \frac{1}{2}\angle + ADC$. Therefore, according to the statement \ref{TetivniPogoj}: + $$\angle PNS +\angle NPS = \frac{1}{2} \left(\angle ABC+\angle + ADC\right)=90^0.$$ + If we use the statement \ref{VsotKotTrik} for the triangle $PSN$, we get + $\angle PSN = 90^0$. + \kdokaz + + + \bzgled \label{TetivniVcrtana} + Let $ABCD$ be a cyclic quadrilateral. + Prove that incentres of the triangles $BCD$, $ACD$, $ABD$ and $ABC$ + are the vertices of a rectangle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.1.pic} +\caption{} \label{sl.skk.4.5.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} We mark with $A_1$, $B_1$, $C_1$ and $D_1$ +the incentres of the triangles $BCD$, $ACD$, $ABD$ and $ABC$ +and with $N$, $M$, $L$ and $P$ the incentres of those arcs $AB$, $BC$, $CD$ and +$AD$ of the cyclic quadrilateral $ABCD$, which do not contain the other +vertices of this quadrilateral (Figure \ref{sl.skk.4.5.1.pic}). From the statement +\ref{TockaN} it follows that $BL$ and $DM$ are the angle bisectors of the angles $CBD$ and +$BDC$, therefore the point $A_1$ is the intersection of the lines $BL$ and $DM$. Similarly, +the point $B_1$ is the intersection of the lines $CP$ and $AL$. According to the statement +\ref{TockaN.NBNC}, + $LC\cong LA_1\cong LB_1\cong LD$, therefore $A_1LB_1$ is an isosceles triangle + with the base $A_1B_1$. From the statement \ref{TockaN} it also follows that + $LN$ is the angle bisector of the angle $ALB$ or $B_1LA_1$. In an isosceles + triangle $A_1LB_1$ the angle bisector of the angle $B_1LA_1$ contains the altitude + of this triangle from the point $L$. This means that $LN\perp + A_1B_1$ holds. Similarly, + $LN\perp C_1D_1$, $PM\perp A_1D_1$ + and + $PM\perp C_1B_1$ hold. From the previous statement \ref{TetŠtirZgl0} we know that $LN\perp PM$, + therefore the quadrilateral $A_1B_1C_1D_1$ is a rectangle. + \kdokaz + +We have already mentioned that a rectangle is a right-angled quadrilateral. Now we will +prove an interesting property of rectangles that relates to points +that lie on its circumscribed circle. + + + + \bzgled + Let $P$ be an arbitrary point on the shorter arc $AB$ of the circumcircle of a rectangle $ABCD$. + Suppose that $L$ and $M$ are the foots of the perpendiculars from the point $P$ on the + diagonals $AC$ and $BD$, respectively. Prove that the length of the line segment $LM$ does not depend + on the position of the point $P$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.15.pic} +\caption{} \label{sl.skk.4.5.15.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Let $O$ be the center of the circle $k$ (Figure \ref{sl.skk.4.5.15.pic}). +The quadrilateral $PMOL$ is a right-angled one, because $\angle OLP + \angle OMP +=90^0+90^0= 180^0$ (statement \ref{TetivniPogoj}). We mark with $l$ +the circumscribed circle of this quadrilateral. Because the angles $OLP$ and $OMP$ are both +right, the distance $OP$ (or the radius of the circle $k$) is the radius of the +circle $l$. Then $LM$ is the chord of the circle $l$, which belongs to the peripheral angle +$\angle LOM =\angle AOB$, which is constant. Regardless of the choice +of the point $P$, the distance $LM$ is the chord of the circle with a constant radius +$OA$, which belongs to a constant peripheral angle $AOB$ (or the corresponding +constant central angle of this circle). The chords, which belong +to the corresponding central angles of the corresponding circles, are proportional to each other, +so the length of the distance $LM$ does not depend on the position of the point $P$. + \kdokaz + + The properties of the right-angled quadrilateral are often used to prove + various +properties of triangles. + + + \bzgled \label{PedalniVS} + The orthocentre of an acute triangle is the incentre of its \index{trikotnik!pedalni} pedal triangle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.16.pic} +\caption{} \label{sl.skk.4.5.16.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Let $AA'$, $BB'$ and $CC'$ be the altitudes of the triangle $ABC$, which intersect + in the point $V$ of +the altitude of this +triangle (Figure \ref{sl.skk.4.5.16.pic}). If $A_1$ is the midpoint +of the side $BC$, the points $B'$ and $C'$ lie on the circle $k(A_1,A_1B)$ +(statement \ref{TalesovIzrKroz2}). Therefore, the quadrilateral $BC'B'C$ +is cyclic, so according to \ref{TetivniPogojZunanji} $\angle +AC'B'\cong \angle ACB = \gamma$. Similarly, we prove that the quadrilateral $AC'A'C$ is cyclic, so $\angle BC'A'\cong +\angle ACB = \gamma$. Therefore, the angles $AC'B'$ and $BC'A'$ are congruent. Because +$CC'\perp AB$, the angles $CC'B'$ and $CC'A'$ are also congruent. This +means that the line $C'C$ is perpendicular to the angle $A'C'B'$. Similarly, the lines $A'A$ and $B'B$ are perpendicular to the corresponding internal angles +of the triangle $A'B'C'$, so the point $V$ is the centre of the triangle $A'B'C'$ +of the inscribed circle. + \kdokaz + + From the proof of the previous statement (\ref{PedalniVS}) we can conclude that the angles, which are determined by the sides +of the pedal triangle $A'B'C'$ with the sides of the triangle +$ABC$, are equal to the corresponding angles of the triangle $ABC$. We will use this fact in the following example. + + + + \bzgled \label{PedalniLemaOcrtana} + Let $O$ be the circumcentre of a triangle $ABC$. + Prove that the lines $OA$, $OB$ and $OC$ are perpendicular to the corresponding sides of the + pedal triangle $A'B'C'$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.17.pic} +\caption{} \label{sl.skk.4.5.17.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.5.17.pic}). + +Let $L$ denote the intersection of the lines $OA$ and $B'C'$. It is enough to prove that the internal angle at the vertex $L$ of the triangle $C'LA$ is a right angle. Let us calculate the other two angles of this triangle. From the previous example \ref{PedalniVS} the angle at the vertex $C'$ is equal to $\gamma$. The triangle $AOB$ is isosceles and $\angle AOB=2\gamma$ (statement \ref{SredObodKot}). Therefore (statements \ref{enakokraki} and \ref{VsotKotTrik}) $\angle C' AL=\angle BAO =90^0-\gamma$, which implies that $\angle ALC'=90^0$. + \kdokaz + +A direct consequence of the statement \ref{PedalniVS} and \ref{PedalniLemasPQR} is the following statement. + + + + \bzgled \label{PedalniLemasLMN} + Let $P$, $Q$ and $R$ be the midpoints of those arcs $BC$, $AC$ and $AB$ + of the circumcircle of a triangle $ABC$ not containing the vertices $A$, $B$ and $C$. + Suppose that the point $S$ is the incentre of the triangle $ABC$ + and $L=SA\cap QR$, $M=SB\cap PR$ and $N=SC\cap PQ$. Then the triangles $LMN$ and $ABC$ have + the common incentre. + (Figure \ref{sl.skk.4.5.18.pic}). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.18.pic} +\caption{} \label{sl.skk.4.5.18.pic} +\end{figure} + +\bzgled \label{Miquelova točka} + Let $P$, $Q$ and $R$ be an arbitrary points on the sides $BC$, $AC$ and $AB$ + of the triangle $ABC$, respectively. Prove that the circumcircles of triangles + $AQR$, $PBR$ and $PQC$ intersect at in one point (so-called \index{točka!Miquelova} + \pojem{Miquel point}\color{green1}\footnote{The point is named after + the French mathematician \index{Miquel, A.} \textit{A. Miquel} (1816–-1851), who published + this statement in 1838 as an article in Liouville's + (\index{Liouville, J.}\textit{J. Liouville} (1809–-1882), French + mathematician) journal. But, as is often the case in mathematics, Miquel + was not the first to prove this statement. Ten years before him, this fact was + discovered and published by the famous Swiss mathematician + \index{Steiner, J.} \textit{J. Steiner} (1769--1863).}). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.2.pic} +\caption{} \label{sl.skk.4.5.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.5.2.pic}) + +We denote with $k_A$, $k_B$ and $k_C$ the circumcircles of triangles +$AQR$, $PBR$ and $PQC$ and the internal angles of triangle $ABC$ in +order with $\alpha$, $\beta$ and $\gamma$. Let $S$ be the other +intersection of the circles $k_B$ and $k_C$ (the proof is similar in +the case when $S = P$). Quadrilateral $BPSR$ and $PCQS$ are +tangential, so $\angle RSP = 180^0 - \beta$ and $\angle QSP = 180^0 +-\gamma$ (statement \ref{TetivniPogoj}). From this it follows that +$\angle RSQ = \beta +\gamma$ and then also $\angle RAQ + \angle RSQ +=\alpha + \beta +\gamma = 180^0$. Quadrilateral $ARSQ$ is also +tangential (statement \ref{TetivniPogoj}) or it has its own +circumcircle, which is actually the circle $k_A$, circumscribed to +triangle $AQR$. This means that the circles $k_A$, $k_B$ and $k_C$ +intersect in point $S$. + \kdokaz + +In this chapter \ref{pogINV} we will prove one generalization of the previous +statement (see example \ref{MiquelKroznice}). + + + + \bnaloga\footnote{45. IMO Greece - 2004, Problem 1.} + Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The +circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$, +respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of +the angles $\angle BAC$ and $\angle MON$ intersect at $L$. Prove that the circumcircles +of the triangles $BML$ and $CNL$ have a common point lying on the side +$BC$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.4.IMO1.pic} +\caption{} \label{sl.skk.4.4.IMO1.pic} +\end{figure} + + +\textbf{\textit{Proof.}} We mark with $E$ the intersection of the angle bisector +of $BAC$ with the side $BC$ of the triangle $ABC$ (Figure +\ref{sl.skk.4.4.IMO1.pic}). We prove that $E$ is the desired point i.e. +that it lies on the circumcircles +of both triangles $BML$ + and $CNL$. + +Because from the construction of the points $M$ and $N$ it follows that $OM\cong ON$, the triangle $OMN$ +is isosceles. This means that the bisector of $OL$ is also the bisector of the side $MN$ (follows from the similarity of the triangles $MSO$ and $NSO$, where $S$ is the center +of the segment $MN$). Therefore, the point $L$ lies on the bisector of the segment $MN$ + of the triangle, so by \ref{TockaN} it lies on the circumcircle + $k$ + of the triangle $AMN$. The condition $AB\neq AC$ tells us that the angle bisectors of $BAC$ and + the side $MN$ (or the angle bisector of $MON$) are different, so their intersection is a point. + + If we use \ref{TetivniPogojZunanji} and \ref{ObodObodKot}, + we get: + \begin{eqnarray*} + \angle BCA &\cong& AMN \cong\angle ALN,\\ + \angle ABC &\cong& ANM \cong\angle ALM. + \end{eqnarray*} +From these relations and \ref{TetivniPogojZunanji} it follows that $NLEC$ and $LMBE$ are tangential +quadrilateral. Therefore, the point $E$ lies on the circumcircles +of both triangles $BML$ + and $CNL$. + \kdokaz + +We say that a +\index{tetragon!tangential}\index{polygon!tangential}\pojem{tangential} +\index{tetragon!tangential}\index{polygon!tangential}\pojem{tangential}, +if there exists an inscribed circle, or if there is such a +circle that the normals of all the sides of the polygon are its tangents +(Figure \ref{sl.skk.4.6.1.pic}). We have already seen that every +triangle is tangential (\ref{SredVcrtaneKrozn}) and also +a regular polygon is tangential (\ref{sredVcrtaneKrozVeck}). On the other hand, it is clear that not all +polygons are tangential. For example, a rectangle is a tetragon which does not have +an inscribed circle. In this section we will focus on +tangential tetragons. + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.1.pic} +\caption{} \label{sl.skk.4.6.1.pic} +\end{figure} + +Since a square is a regular polygon, it is also a tangential tetragon. But how would we in general determine whether a tetragon is tangential? It is clear that in a tangential tetragon (in general also in a polygon) the simetrals of all of its internal angles intersect in one point (Figure +\ref{sl.skk.4.6.1.pic}). This condition is sufficient for the tangentiality of a polygon. +Unfortunately, this condition is not very useful in +specific cases. There is a more useful condition that is necessary and +sufficient for the tangentiality of tetragons. + + + + \bizrek \label{TangentniPogoj} + A quadrilateral $ABCD$ is tangential if and only if + $$|AB| + |CD| = |BC| + |AD|.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.2.pic} +\caption{} \label{sl.skk.4.6.2.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.6.2.pic}) + +($\Rightarrow$) First, let's assume that the quadrilateral $ABCD$ is tangent and + $k$ is its inscribed circle. + Let $P$, $Q$, $R$ and $S$ +be the points of tangency of sides $AB$, $BC$, $CD$ and $DA$ with the circle $k$. Because +the appropriate tangent lines are concurrent (by Theorem \ref{TangOdsek}), it holds: +$AP \cong AS$, $BP \cong BQ$, $CQ \cong CR$ and $DR \cong DS$. Therefore + \begin{eqnarray*} +|AB| + |CD|&=&|AP| + |PB| + |CR| + |RD| \\&=& |AS| + |SD| + |BQ| + +|QC|\\&=&|AD| + |BC|. + \end{eqnarray*} + + ($\Leftarrow$) We prove the converse statement. Let's assume that in the quadrilateral + $ABCD$ the sums of the pairs of opposite sides are equal, i.e. + $|AB| + |CD| = |BC| + |AD|$. There exists a circle $k$, which touches +sides $AB$, $BC$ and $DA$ of this quadrilateral (its center is +the intersection of the internal angle bisectors at vertices $A$ and $B$ of this +quadrilateral). We prove that this circle also touches side +$CD$ of the quadrilateral $ABCD$. Let $D'$ be the intersection of the other tangent from +point $C$ of the circle $k$ and the line $AD$. Let's assume that $D'\neq +D$. Without loss of generality, let $\mathcal{B}(A,D',D)$. Because the quadrilateral $ABCD'$ is tangent to the circle $k$, by the already proven part of the theorem it holds $|AB| + |CD'| = |AD'|+|BC|$. But since by the assumption also $|AB| + |CD| = |AD| + |BC|$, it also holds $|CD|-|CD'| = |DA| - |D'A|=|DD'|$ i.e. $|CD|= |CD'| + |DD'|$. But this is not possible due to the triangle inequality \ref{neenaktrik} (points $C$, +$D$ and $D'$ cannot be collinear, because otherwise it would hold $C\in +AD$). In a similar way we arrive at a contradiction also in the case when $\mathcal{B}(A,D,D')$. Therefore it holds $D'= D$, thus $ABCD$ is a tangential quadrilateral. + \kdokaz + + The following theorems are a direct consequence of the previous criterion. + + \bizrek \label{TangDeltoidRomb} + A rhombus, a deltoid, and a square are + tangential quadrilaterals (Figure \ref{sl.skk.4.6.3.pic}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.3.pic} +\caption{} \label{sl.skk.4.6.3.pic} +\end{figure} + + \bizrek \label{TangParalelogram} + A parallelogram is a tangential quadrilateral + if and only if it is a rhombus (Figure \ref{sl.skk.4.6.3.pic}). + \eizrek + +In the next two examples we will consider +tension and tangent quadrilaterals at the same time. + + + + \bzgled Let $k_A$, $k_B$, $k_C$ and $k_D$ circles with centres $A$, + $B$, $C$ and $D$, such that two in a row (also $k_A$ and $k_D$) + are touching each other externally. Prove that the quadrilateral defined by + the touching points of circles is cyclic, and the quadrilateral $ABCD$ is tangential. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.4.pic} +\caption{} \label{sl.skk.4.6.4.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Let $P$, $Q$, $R$ and $S$ be the touching points of +the given circles in order, and $p$ and $r$ the common tangents of +the corresponding circles at points $P$ and $R$ (Figure \ref{sl.skk.4.6.4.pic}). + +First, we have: + \begin{eqnarray*} + |AD| + |BC| &=& |AP| + |PD| + |BR| + |RC| =\\ + &=& |AQ| + |SD| + |QB| + |SC| =\\ + &=& |AB| + +|CD|. + \end{eqnarray*} + Therefore, $ABCD$ is a tangent quadrilateral (statement \ref{TangentniPogoj}). + +Tangents $p$ and $r$ divide the internal angle at vertices $P$ and $R$ +of the quadrilateral $PQRS$ into angles, each of which is equal to half +of the corresponding central angle (statement \ref{ObodKotTang}). The aforementioned +central angles are the internal angles of the quadrilateral $ABCD$. We denote +them with $\alpha$, $\beta$, $\gamma$ and $\delta$. Therefore, (statement +\ref{VsotKotVeck}): + \begin{eqnarray*} + \angle QPS+ \angle SRQ&=& \angle QP,p+\angle p,PS+ \angle SR,r+\angle r,RQ=\\ + &=& \frac{1}{2}\alpha+\frac{1}{2}\delta+\frac{1}{2}\gamma+\frac{1}{2}\beta=\\ + &=& \frac{1}{2}\left(\alpha+\delta+\gamma+\beta\right)=\\ + &=& \frac{1}{2}\cdot360^0=180^0, + \end{eqnarray*} +which means that $PQRS$ is a tension quadrilateral. + \kdokaz + +It is clear that the inscribed + circle of the quadrilateral $ABCD$ is also the circumscribed circle of the quadrilateral $PQRS$. Because according to the assumption + $PAQ$, $QBR$, $RCS$ and $SDP$ are equilateral triangles with bases + $PQ$, $QR$, $RS$ and $SP$, the simetrals of the internal angles of the quadrilateral $ABCD$ + are also the simetrals of the sides of the quadrilateral $PQRS$ (Figure + \ref{sl.skk.4.6.4a.pic}). This is also the second (simpler) + way to prove the second part of the previous example - the assertion that $PQRS$ + is a tangential quadrilateral. + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.4a.pic} +\caption{} \label{sl.skk.4.6.4a.pic} +\end{figure} + + + + \bzgled \label{tetivTangLema} + Let $L$ be the intersection of the diagonals of a cyclic quadrilateral $ABCD$. + Prove that the foots of the perpendiculars from the point $L$ on the sides of + this quadrilateral are the vertices of a tangential quadrilateral. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.5.pic} +\caption{} \label{sl.skk.4.6.5.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + Let $P$, $Q$, $R$ and $S$ be the perpendicular projections from the point $L$ on the sides + $AB$, $BC$, $CD$ and $DA$ +of the quadrilateral $ABCD$ (Figure \ref{sl.skk.4.6.5.pic}). +Because of the appropriate right angles, $PBQL$ and $APLS$ +are tangential quadrilaterals (statement \ref{TetivniPogoj}). According to the assumption, $ABCD$ is also tangential. If we +use this, we get the equality of the appropriate external angles (statement \ref{ObodObodKot}). Therefore: +$$\angle SPL \cong \angle SAL = \angle DAC \cong \angle DBC += \angle LBQ \cong \angle LPQ.$$ +From this it follows that the line $PL$ is the simetral of the internal angle at +the vertex $P$ of the quadrilateral $PQRS$. Similarly, the lines $QL$, $RL$ and $SL$ +are the simetrals of the other three internal angles of this quadrilateral. Therefore, $L$ is the center of the inscribed circle of the quadrilateral $PQRS$, +so this is tangential. + \kdokaz + +We will now prove another interesting property of tangential quadrilaterals. + +\bzgled + Prove that the incircles of triangles $ABC$ and $ACD$ touch each + other if and only if $ABCD$ is a tangential quadrilateral. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.6.pic} +\caption{} \label{sl.skk.4.6.6.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.6.6.pic}). + + First, we prove some relations that hold for any convex quadrilateral $ABCD$. +Let $P$, $Q$ and $X$ be the points in which the inscribed circle $k$ of triangle $ABC$ touches its sides +$AB$, $BC$ and $CA$, and $R$, $S$ and $Y$ be the points in which the inscribed circle $l$ of triangle $ACD$ touches its sides $CD$, $DA$ and $AC$. First, it holds (from \ref{TangOdsek}): + \begin{eqnarray*} + |AX| &=& |AP|=\frac{1}{2}\left(|AX|+|AP|\right)=\frac{1}{2}\left(|AC|-|CX|+|AB|-|BP|\right)\\ + &=& \frac{1}{2}\left(|AC|-|CQ|+|AB|-|BQ|\right)= + \frac{1}{2}\left(|AC|+|AB|-|BC|\right), + \end{eqnarray*} + therefore it holds: + $$|AX|=\frac{1}{2}\left(|AC|+|AB|-|BC|\right).$$ + In the same way, we prove that it also holds: + $$|AY|=\frac{1}{2}\left(|AC|+|AD|-|DC|\right).$$ + Now we can start with proving the equivalence. + +The circles $k$ and $l$ touch each other exactly when $X = Y$ or $|AX| = |AY|$. The last equality holds exactly when: $$\frac{1}{2}\left(|AC|+|AB|-|BC|\right)=\frac{1}{2}\left(|AC|+|AD|-|DC|\right)$$ +or $|AB| + |DC| = |AD| + |BC|$, which is fulfilled exactly when $ABCD$ is a tangential quadrilateral (from \ref{TangentniPogoj}). + \kdokaz + +The consequence of this is the following claim. + +\bzgled + Let $ABCD$ be a tangential quadrilateral. + Then the incircles of the triangles $ABC$ and $ACD$ touch each other + if and only if + the incircles of the triangles $ABD$ and $CBD$ touch each + other (Figure \ref{sl.skk.4.6.6a.pic}). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.6a.pic} +\caption{} \label{sl.skk.4.6.6a.pic} +\end{figure} + + + \textbf{\textit{Proof.}} The statements that the incircles of the triangles $ABC$ and $ACD$ or the triangles $ABD$ and $CBD$ touch, are + equivalent to the statement that the quadrilateral $ABCD$ is tangent. This means that the initial statements are equivalent. + \kdokaz + +%_______________________________________________________________________________ + \poglavje{Bicentric Quadrilateral} \label{odd4TetivniTangentni} + + Some quadrilaterals are both tangential and chordal. We call them \index{štirikotnik!tetivnotangentni}\pojem{tetivnotangentni} or \index{štirikotnik!bicentrični}\pojem{bicentrični} quadrilaterals. Which + quadrilaterals are these? The square is certainly one of them. Is it the only one? The answer is negative. The quadrilateral we get from the example \ref{tetivTangLema} is always tangent. In a certain case it will be chordal as well. + Namely, the following statement is true. + + + + \bizrek \label{tetivTangIzrek} + If $L$ is the intersection of the perpendicular diagonals of a cyclic quadrilateral + $ABCD$, then the foots of the perpendiculars from the point $L$ on the sides of + this quadrilateral are the vertices of a bicentric quadrilateral. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.7.2.pic} +\caption{} \label{sl.skk.4.7.2.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.7.2.pic}) + +Let's use the same notation as in the example \ref{tetivTangLema}. We have already proven that $PQRS$ is a tangent quadrilateral. We will now prove that it is also a bicentric quadrilateral. From the proof of the statement in the aforementioned example \ref{tetivTangLema} it follows: +\begin{eqnarray*} + \angle SPQ &=& \angle SPL+\angle LPQ = \angle SAL+\angle LBQ =\\ + &=& \angle DAC + \angle DBC + = 2\cdot\angle DBC +\end{eqnarray*} +or $\angle SPQ= 2\cdot\angle DBC$. Similarly, $\angle SRQ= 2\cdot\angle ACB$. + Because, by assumption, $AC\perp BD$, $CLB$ is a right angled triangle, therefore: + $$\angle SRQ+\angle SRQ=2\cdot(\angle DBC+\angle ACB)=2\cdot 90^0=180^0.$$ + By \ref{TetivniPogoj} $PQRS$ is a bicentric quadrilateral. + \kdokaz + + It is not difficult to convince oneself that the converse statement is also true. + + + + \bizrek + Let $PQRS$ be a bicentric quadrilateral. Suppose that point $L$ is + the incentre of this quadrilateral and at the same time the intersection of the diagonals + of a cyclic quadrilateral $ABCD$. If $P$, $Q$, $R$ and $S$ the foots of the perpendiculars + from the point $L$ on the sides of quadrilateral $ABCD$, then $AC\perp BD$. + \eizrek + +In the next exercise we will see that for three non-collinear points $A$, $B$ and $C$ there is only one point $D$, such that $ABCD$ is a bicentric quadrilateral. + + + \bnaloga\footnote{4. + IMO Czechoslovakia - 1962, Problem 5.} + On the circle $k$ there are given three distinct points $A$, $B$, $C$. Construct (using + only straightedge and compasses) a fourth point $D$ on $k$ such that a circle + can be inscribed in the quadrilateral thus obtained. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.IMO1.pic} +\caption{} \label{sl.skk.4.5.IMO1.pic} +\end{figure} + + +\textbf{\textit{Solution.}} Without loss of generality we can assume +that $AB\geq BC$. + +Let $D$ be a point that satisfies the conditions of the task, or such that $ABCD$ is a tangent-chord quadrilateral (Figure \ref {sl.skk.4.5.IMO1.pic}). We denote by $a$, $b$, $c$ and $d$ the sides of $AB$, $BC$, $CD$ and $DA$ of this quadrilateral, and by $\alpha$, $\beta$, $\gamma$ and $\delta$ its internal angles at vertices $A$, $B$, $C$ and $D$. From the condition of the tangency of the quadrilateral $ABCD$ (formula \ref {TetivniPogoj}) it follows that $\delta = 180^0 - \beta$, and from its tangency (formula \ref {TangentniPogoj}) that $a + c = b + d$ or $d-c = a-b$. In this way, the task is reduced to the design of the third vertex $D$ of the triangle $ACD$, where the sides $AC$, the angle $\angle ADC = 180^0 - \beta$ and the difference of sides $AD-CD = AB-BC = a-b$ are given. Let $E$ be a point on the line $AD$, for which $DE \cong DC$. Then $AE = AD-DE = AD-CD = a-b$. The triangle $ECD$ is isosceles, so by formula \ref {enakokraki} it follows that $\angle CED \cong \angle DCE$. Therefore, $\angle AEC = 180^0 - \angle DEC = 180^0 - \frac {1}{2} \beta$. This allows us to construct the triangle $ACE$ or the point $E$. + +First, we plan the point $E$ as the intersection of the arc $l$ (see the construction described in formula \ref {ObodKotGMT}) $180^0 - \frac {1}{2} \angle ABC$) and the circle $j (A, AB-BC)$ (if $AB \cong AC$, we assume $E = A$). The point $D$ can then be designed as the intersection of the strip $AE$ and the similitude $s_{EC}$ of the line $EC$ (in the case $AB \cong AC$, or $E = A$, $D$ is the second intersection of the similitude $s_{EC} = s_{AC}$ with the circle $k$). + +We prove that the point $D$ satisfies the conditions of the task, or that $ABCD$ is a tangent-chord quadrilateral. First, we will consider the case when $AB>BC$. + +By construction, the point $D$ lies on the line of symmetry $EC$, so $DE\cong DC$ and also (by statement \ref{enakokraki}) $\angle DEC\cong\angle DCE$. We have drawn the point $E$ so that it lies on the arc $l$ with the string $AC$ and the angular measure $180^0-\frac{1}{2}\angle ABC$, so $\angle AEC=180^0-\frac{1}{2}\angle ABC$. Because, by construction, $\mathcal{B}(A,E,D)$, we have $\angle DCE\cong\angle DEC=\frac{1}{2}\angle ABC$. From the isosceles triangle $EDC$ by statement \ref{VsotKotTrik} it follows that $\angle ADC=\angle EDC=180^0-\angle ABC$. Therefore, $\angle EDC+\angle ABC=180^0$, so by statement \ref{TetivniPogoj} the quadrilateral $ABCD$ is a string quadrilateral, or $D\in k$. + + We will now prove that the quadrilateral $ABCD$ is a tangent quadrilateral. In the first part of the proof (the string property) we have already seen that $DE\cong DC$. The point $E$ by construction lies on the circle $j(A,|AB-BC|)$, so $|AE|=|AB|-|BC|$. Because $\mathcal{B}(A,E,D)$ also holds, we have $|AD|-|CD|=|AD|-|DE|=|AE|=|AB|-|BC|$. From this it follows that $|AD|+|BC|=|AB|+|CD|$ and by statement \ref{TangPogoj} the quadrilateral $ABCD$ is tangent. + + If $AB\cong BC$, the point $D$ by construction already lies on the circle $k$. Because both points $B$ and $D$ lie on the line of symmetry $AC$, the quadrilateral $ABCD$ is a deltoid, so it is also tangent (by statement \ref{TangDeltoidRomb}). + + We will now investigate the number of solutions to the problem. The circle $k(A,AB-AC)$ and the arc $l$ always intersect in one point $E$. Because $ABC<180^0$, we have $\angle AEC=180^0-\frac{1}{2}\beta>90^0$. This means that the line of symmetry $s_{EC}$ always intersects the half-line $AE$ in one point $D$ and $\mathcal{B}(A,E,D)$ holds. This means that the problem always has one and only one solution. + \kdokaz + + + + + %______________________________________________________________________________ + \poglavje{Simson Line} \label{odd4Simson} + +We will first prove the basic statement. + +\bizrek \label{SimpsPrem} +The foots of the perpendiculars from an arbitrary point lying on the circumcircle of a triangle to the lines containing the sides of this triangle are three collinear points. The line containing these points is the so-called \pojem{Simson\footnote{Premico imenujemo po škotskem matematiku \index{Simson, R.} \textit{R. Simsonu} (1687--1768), čeprav je to lastnost prvi objavil škotski matematik \index{Wallace, W.} \textit{W. Wallace} (1768--1843) šele leta 1799.} line}\color{blue}. +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.7.1a.pic} +\caption{} \label{sl.skk.4.7.1a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $S$ be an arbitrary point of the circumcircle $k$ of the triangle $ABC$ and $P$, $Q$ and $R$ the orthogonal projections of the point $S$ on the lines containing the sides $BC$, $AC$ and $AB$ (Figure \ref{sl.skk.4.7.1a.pic}). Without loss of generality, we assume that $\mathcal{B}(B,P,C)$, $\mathcal{B}(A,Q,C)$ and $\mathcal{B}(A,B,R)$ hold. In this case, the points $Q$ and $R$ are on different sides of the line $BC$, so it is enough to prove $\angle BPR \cong \angle CPQ$. Because of the appropriate right angles and the position of the point $S$, the quadrilaterals $BRSP$, $ABSC$, $ARSQ$ and $SPQC$ are cyclic, so (from izrek \ref{TetivniPogoj}: +\begin{eqnarray*} + \angle BPR &=& \angle BSR = \angle RSC - \angle BSC=\\ +&=& \angle RSC - (180° - \angle BAC) =\\ +&=& \angle RSC - \angle RSQ = \angle CSQ = \angle CPQ, + \end{eqnarray*} +which means that the points $P$, $Q$ and $R$ are collinear. \kdokaz + +In the following we will consider further interesting properties of Simson's line. Because each point $X$, which lies on the circumcircle of some triangle, determines the Simson line, we will denote this line by $x$. In this way, each triangle determines one mapping $X\mapsto x$. + +\bzgled \label{SimsZgled1} + Let $P$ be an arbitrary point of the circumcircle $k$ of a triangle + $ABC$. Suppose that $P_A$ is the intersection of the perpendicular line of + the line $BC$ through the point $P$ with the circle $k$. + Prove that the line $AP_A$ is parallel to the Simson line $p$ + of the triangle at the point $P$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.7.1b.pic} +\caption{} \label{sl.skk.4.7.1b.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Let $X$, $Y$ and $Z$ be the orthogonal projections of the point $P$ on the lines + $BC$, $AC$ and $AB$ (Figure \ref{sl.skk.4.7.1b.pic}). By izreku \ref{SimpsPrem} the Simson line + $p$ is determined by the points $X$, $Y$ and $Z$. Similarly to izreku + \ref{SimpsPrem}, +the quadrilateral $PYXC$ is a trapezoid, therefore $\angle YXP \cong \angle ACP$. +By izreku \ref{ObodObodKot} the angles $ACP$ and $AP_AP$ above +the trapezoid $AP$ are supplementary, therefore $\angle YXP \cong \angle AP_AP$ +or $XY\parallel AP$ (izrek \ref{KotiTransverzala}). + \kdokaz + + + + \bzgled \label{SimsZgled2} + Let $P$ and $Q$ be arbitrary points lying on the circumcircle + $k(O,r)$ of a triangle $ABC$ and $p$ and $q$ their Simson lines. Prove that + $$\angle pq = \frac{1}{2}\angle POQ.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.7.1c.pic} +\caption{} \label{sl.skk.4.7.1c.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Let $X_P$ and $X_Q$ be the feet of the perpendiculars from points $P$ and $Q$ + on the line $BC$ and $P_A$ and $Q_A$ +the intersections of these perpendiculars with the circle $k$ (Figure +\ref{sl.skk.4.7.1c.pic}). The Simson lines $p$ and $q$ are parallel to the lines $AP_A$ and $AQ_A$ (example \ref{SimsZgled1}). +Therefore, the angle determined by the lines $p$ and $q$ is equal to the inscribed angle $Q_AAP_A$, which is equal to half of the central angle $Q_AOP_A$ +(by statement \ref{SredObodKot}) or half of the angle $QOP$ (because the trapezoid $PP_AQ_AQ$ is isosceles and by statement \ref{trapezTetivEnakokr} it is also equilateral, i.e. $PQ \cong P_AQ_A$). + \kdokaz + + + \bzgled \label{SimsZgled3} + Let $P$ be an arbitrary point of the circumcircle $k$ of a triangle + $ABC$, $p$ its Simson line and $V$ the orthocentre of this triangle. + Prove that the line $p$ bisects the line segment $PV$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.7.1d.pic} +\caption{} \label{sl.skk.4.7.1d.pic} +\end{figure} + + \textbf{\textit{Proof.}} +Let $P_A$ and $X$ be the points defined as in example +\ref{SimsZgled1} (Figure \ref{sl.skk.4.7.1d.pic}). Let $V'$ +and $P'$ be the points that are symmetric to the points $V$ and $P$ with respect to the line $BC$. The point $V'$ lies on the circumscribed circle $k$ of the triangle $ABC$ +(by statement \ref{TockaV'}). Because of the properties of symmetry +or the axis of reflection (see subsection \ref{odd6OsnZrc}), statement +\ref{KotiTransverzala}, statement \ref{ObodObodKot} and example +\ref{SimsZgled1} it holds: + $$\angle VP'P\cong\angle V'PP'\cong\angle AV'P + \cong\angle AP_AP\cong\angle p,PP'.$$ + Therefore, $\angle VP'P \cong \angle p,PP'$, so by statement + \ref{KotiTransverzala} the lines $VP'$ and $p$ are parallel. Because +the point $X$ is the midpoint of the segment $PP'$, the line $p$ contains the midpoint +of the triangle $PVP'$ (by statement \ref{srednjicaTrik}) or the midpoint of its +side $PV$. + \kdokaz + +In sections \ref{odd5Hamilton} and \ref{odd7SredRazteg} we will prove two more properties +of Simson lines (see \ref{HamiltonSimson} and \ref{SimsEuler}), which are related to Hamilton's theorem or +Euler's circle of a triangle. + + + + +%________________________________________________________________________________ + \poglavje{Torricelli Point} \label{odd4Torricelli} + +In this section we will give another famous point of a triangle. + + + \bizrek \label{izrekTorichelijev} + On each side of a triangle $ABC$ the equilateral triangles $BEC$, $CFA$ and $AGB$ + are externally erected. Prove: + \begin{enumerate} + \item $AE$, $BF$ and $CG$ are congruent line segments; + \item the lines $AE$, $BF$ and $CG$ intersect at one point + (so-called \pojem{Torricelli\footnote{Problem was first posed by French mathematician \index{Fermat, P.} \textit{P. + Fermat} (1601--1665) as a challenge to Italian mathematician and physicist \index{Torricelli, E.} \textit{E. + Torricelli} (1608--1647). Torricelli's solution was published by his student - Italian mathematician and physicist \textit{V. Viviani} (1622–-1703) - in 1659. We also call this point \index{point!Fermat's}\pojem{Fermat + point}.} + point} \color{blue}of this triangle) and every two of them determine an angle with measure $60^0$. + \end{enumerate} + \index{point!Torricelli's} + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.7.1.pic} +\caption{} \label{sl.skk.4.7.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.7.1.pic}) + + (\textit{i}) Triangles $AEC$ and $FBC$ are congruent by \textit{SAS} \ref{SKS} theorem ($AC \cong FC$ , $CE \cong CB$ and +$\angle ACE \cong \angle FCB = \angle ACB + 60°$), so $AE\cong BF$. Analogously is $AE\cong CG$. + +(\textit{ii}) Let $k$, $l$ and $j$ be the circumscribed circles of the triangles $BEC$, $CFA$ and $AGB$. We shall first prove that these circles intersect in one point. With $T$ we denote the second intersection point of the circles $k$ and $l$ ($T\neq C$). The quadrilaterals $BECT$ and $CFAT$ are cyclic, therefore (by the statement \ref{TetivniPogoj}) both the angles $BTC$ and $ATC$ measure $120^0$. Thus, also the angle $ATB$ measures $120^0$, which means that the quadrilateral $AGBT$ is cyclic (by the statement \ref{TetivniPogoj}) or that the point $T$ also lies on the circle $j$. + +We shall prove that each of the lines $AE$, $BF$, $CG$ goes through the point $T$. From the equality of the corresponding +circumscribed angles (by the statement \ref{ObodObodKot}) we obtain: + +\begin{eqnarray*} +\angle ATE&=&\angle ATF+\angle FTC+\angle CTE=\\ +&=&\angle ACF+\angle FAC+\angle CBE +=3\cdot 60^0=180^0. +\end{eqnarray*} + +Thus, $A$, $T$ and $E$ are collinear points, or the point $T$ lies on the line $AE$. Analogously, the point $T$ also +lies on the lines $BF$ and $CG$. It is also clear that: +$\angle AE,BF\cong\angle ATF\cong\angle ACF=60^0$. + \kdokaz + + In the section \ref{odd9MetrInv} (by the statement \ref{izrekToricheliFerma}) we shall prove another interesting property of the Torricelli's point. + + +%________________________________________________________________________________ + \poglavje{Excircles of a Triangle} \label{odd4Pricrt} + + We have already proved that for any triangle there exist + circumscribed and inscribed circle. The first one contains all the vertices +of the triangle, the second one touches all its sides. Now we shall +show that there also exist circles which touch one side +and two lines containing the sides of the triangle. + + + \bizrek + The bisector of the interior angle at vertex $A$ and the bisectors of + the exterior angles at vertices $B$ and $C$ of a triangle $ABC$ intersect at one point, + which is the centre of the circle touching the side $BC$ and the lines containing the sides $AB$ and + $AC$. It is so-called \index{pričrtane krožnice trikotnika} \pojem{excircle of the triangle}\color{blue}. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.27.1.94_veliki_zadatak_lema.pic} +\caption{} \label{sl.27.1.94_veliki_zadatak_lema.pic} +\end{figure} + +\textbf{\textit{Proof.}} We prove the statement similarly to the inscribed circle of a triangle. The simetrali +of the external angles at the vertices $B$ and $C$ are not parallel and they +intersect at some point - we mark it with $S_a$ (Figure +\ref{sl.27.1.94_veliki_zadatak_lema.pic}). Because the point $S_a$ lies on +these two simetrali, it holds $A,S_a\div BC$ and $S_a$ is equally distant from the lines $AB$, $BC$ and +$AC$. Therefore, $S_a$ also belongs to the simetral of the internal angle at +the vertex $A$ and is the center of the circle that touches the side $BC$ and +the lines $AB$ and $AC$. + \kdokaz + + Now we are ready to prove the so-called \index{velika naloga} + \pojem{‘‘velika naloga’’}, which is very useful in + designing triangles. + + + + + \bizrek \label{velikaNaloga} + Let $P$, $Q$, $R$ be the touching points of the incircle $k(S,r)$ + of a triangle $ABC$ with the sides $BC=a$, $AC=b$, $AB=c$ ($b>c$) and $P_i$, $Q_i$, $R_i$ + ($i\in \{a,b,c\}$) the touching points of the excircles + $k_i(S_i,r_i)$ with lines $BC$, $AC$ in $AB$. Let $l(O,R)$ + be the circumcircle of this triangle with the semiperimeter + $s=\frac{a+b+c}{2}$, $A_1$ the midpoint of the line segments $BC$, $M$ and $N$, + intersections of the line $OA_1$ with the circle $l$ ($N,A\div BC$) and + $M’$, $N’$ the foots of the perpendiculars from these points on the line $AB$ (Figure + \ref{sl.27.1.94_veliki_zadatak.pic}). Then: + \vspace*{2mm} + + (\textit{i}) $AQ_a\cong AR_a=s$, \hspace*{0.4mm} (\textit{ii}) + $AQ\cong AR=s-a$, \hspace*{0.4mm} (\textit{iii}) $QQa\cong RRa\cong a$, + % \vspace*{1mm} + +(\textit{iv}) $PPa=b-c$,\hspace*{1mm} + (\textit{v}) $P_bP_c=b+c$, + % \vspace*{1mm} + + (\textit{vi}) + $A_1$ is the midpoint of the line segment $PP_a$ in $P_bP_c$, + %\vspace*{1mm} + + (\textit{vii}) $A_1N= \frac{r_a- r}{2}$,\hspace*{2mm} + (\textit{viii}) $A_1M= \frac{r_b + r_c }{2} $,\hspace*{1mm} + + + (\textit{ix}) $r_a +r_b +r_c =4R+r$,\footnote{To lastnost + trikotnika je leta 1790 odkril francoski matematik \index{L'Huilier, S. A. J.} + \textit{S. A. J. L'Huilier} + (1750--1840).} + %\vspace*{1mm} + + (\textit{x}) $NN’= \frac{r_a +r}{2}$, \hspace*{1mm} (\textit{xi}) + $MM’= \frac{r_b -r_c }{2 }$,\hspace*{1mm} (\textit{xii}) + $N’B\cong AM’=\frac{ b - c}{2}$, + % \vspace*{1mm} + + (\textit{xiii}) $AN’\cong BM’= \frac{b + c}{2} $, \hspace*{1mm} (\textit{xiv}) $M’N’\cong b$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.27.1.94_veliki_zadatak.pic} +\caption{} \label{sl.27.1.94_veliki_zadatak.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Preden začnemo z dokazovanjem, omenimo, da po izreku \ref{TockaN} + točka $N$ leži na simetrali notranjega kota $BAC$ trikotnika + $ABC$. + +(\textit{i}) If we use the equality of tangent lines + (statement \ref{TangOdsek}), we get: + \begin{eqnarray*} + AQ_a&\cong &AR_a= \frac{1}{2}\left(AQ_a+AR_a\right)= + \frac{1}{2}\left(AB+BR_a+AC+CQ_a\right)\\ + &=& \frac{1}{2}\left(AB+BP_a+AC+CP_a\right)= +\frac{1}{2}\left(AB+BC+AC\right)=s. + \end{eqnarray*} + + (\textit{ii}) In a similar way, we get: + \begin{eqnarray*} + AQ&\cong &AR= \frac{1}{2}\left(AQ+AR\right)= + \frac{1}{2}\left(AB-BR+AC-CQ\right)\\ + &=& \frac{1}{2}\left(AB-BP+AC-CP\right)= +\frac{1}{2}\left(AB+AC-BC\right)=s-a. + \end{eqnarray*} + + (\textit{iii}) $QQ_a\cong AQ_a-AQ=a$. + + (\textit{iv}) We prove the equality by first calculating:\\ $BP$ and $CP_a\cong CQ_a$. + + (\textit{v}) $P_bP_c\cong CP_c+BP_b-a=2s-a=b+c$. + + (\textit{vi}) It follows from $BP\cong CP_a=s-b$. + + (\textit{vii}) Points $A_1$ and $N$ are the centers of the diagonal of trapezoid + $SPS_aP_a$ with + bases $SP\cong r$ and +$S_aP_a\cong r_a$. The equality follows +from statement \ref{srednjTrapez}. + + (\textit{viii}) Lines $NA$ and $S_cS_b$ + are perpendicular (the internal + and external angles at point +$A$ are symmetrical), so point $M$ lies on line $S_cS_b$. The desired equality follows +from the fact that line $A_1M$ is the median of trapezoid +$S_cP_cP_bS_b$ (statement \ref{srednjTrapez}). + + (\textit{ix}) It follows directly from $2\cdot R=NM=NA_1+A_1M$ and + (\textit{vii}) and (\textit{viii}). + +(\textit{x}) It follows from statement \ref{srednjTrapez} and the fact that +$NN'$ is the median of trapezoid $SRR_aS_a$. + +(\textit{xi}) Points $M$ and $M'$ are the centers of the diagonal of trapezoid +$R_cS_cR_bS_b$ with bases $R_cS_c\cong r_c$ and $R_bR_b\cong r_b$. The equality +follows from statement \ref{srednjTrapez}. + + (\textit{xii}) Point $N$ is the center of line $RR_a$, so: + $$N'B=AN'-AB= \frac{1}{2}\left(AR_a+AR\right)-c= + \frac{1}{2}\left(s+(s-a)\right)-c= \frac{1}{2}\left(b-c\right).$$ + +(\textit{xiii}) and (\textit{xiv}) follow + directly from the proven equality (\textit{xii}). + \kdokaz + +%KONSTRUKCIJA (VN) + + \bzgled + Construct a triangle $ABC$, with given: + + (\textit{a}) $a$, $b-c$, $r$, \hspace*{3mm} (\textit{b}) + $b-c$, $r$, $v_b$, \hspace*{3mm} (\textit{c}) $a$, $b+c$, $r$, + \hspace*{3mm}(\textit{č}) $R$, $r$, $r_a$. + \ezgled + + + +\textbf{\textit{Solution.}} + For each construction we will use the big task - \ref{velikaNaloga}. + We will use + the same labels. + +\begin{figure}[!htb] +\centering +\input{sl.27.1.94_veliki_zadatak_konstr.pic} +\caption{} \label{sl.27.1.94_veliki_zadatak_konstr.pic} +\end{figure} + +\begin{figure}[!htb] +\centering +\input{sl.27.1.94_veliki_zadatak_konstr2.pic} +\caption{} \label{sl.27.1.94_veliki_zadatak_konstr2.pic} +\end{figure} + + + +(\textit{a}) Because $PP_a=b-c$ and point $A_1$ is the common center +of side + $BC$ and + the line $PP_a$, it also holds that $PA_1 = \frac{1}{2}( b - c)$ (Figure +\ref{sl.27.1.94_veliki_zadatak_konstr.pic}). +First, we plan the side $BC$, then its center +$A_1$, point $P$, the inscribed circle of the triangle, tangents from +the vertices $B$ and $C$, and finally the vertices $A$. + + + +(\textit{b}) Similarly to the previous example. First, we plan +the line $PA_1$, then the inscribed circle of the triangle $ABC$ (Figure +\ref{sl.27.1.94_veliki_zadatak_konstr.pic}). +We also need to use the condition of the height from the vertex $B$. With $L$ +we mark the orthogonal projection from the point $A_1$ onto the line $AC$. +The line $A_1L$ is the median of the triangle $CBB'$, so $A_1L +=\frac{1}{2}BB'=\frac{1}{2} v_b$. Therefore, the line $AC$ can +be constructed as the common tangent of the inscribed circle and the circle +$k(A_1, \frac{1}{2}v_b)$. Thus we get the vertex $C$, then the vertices $B$ and $A$. + +(\textit{c}) We know that $RR_a\cong a$ and $AN'=\frac{1}{2}(b+c)$, and that $N'$ is the center of the line $RR_a$ (Figure \ref{sl.27.1.94_veliki_zadatak_konstr2.pic}). So, from the given +data, + we first construct the points $A$, $N'$, $R$ and $R_a$, and then also $S$, $N$ + and +$S_a$. In the end, we draw the dotted and the dashed circle - +the sides of the triangle lie on their common tangents. + + +(\textit{č}) Because $A_1N =\frac{1}{2}(r_a -r)$ and $MN = 2R$, we can +first plan the points $N$, $A_1$ and $M$, and then also the dotted +circle of the triangle $ABC$ and the side $BC$ (Figure \ref{sl.27.1.94_veliki_zadatak_konstr2.pic}). The construction can +be finished in two ways. In the first case, we translate the task into a construction of a triangle that we already know: $a$, $R$, $r$ (example \ref{konstr_Rra}), in +the second case, we use the equality $RR_a\cong a$. + \kdokaz + + + + +%________________________________________________________________________________ +\naloge{Exercises} + +\begin{enumerate} + +\item The sides of a triangle are $6$, $7$ and $9$. Let $k_1$, $k_2$ and $k_3$ be the circles with centers +in the vertices of this triangle. The circles touch each other so that + the circle with the center in the vertex of the smallest angle of the triangle touches the other two circles from the inside, +while the remaining two circles touch from the outside. Calculate the lengths of the radii of these three circles. + +\item Prove that the angle formed by the secants of a circle that intersect each other outside the circle is equal to half the difference of the central angles corresponding to the arcs that lie between the arms of this angle. + +\item The apex of the angle $\alpha$ is an external point of the circle $k$. Between the arms of this angle, on the circle, there are two +arcs, which are in the ratio $3:10$. The larger of these arcs corresponds to the central angle $40^0$. Determine +the measure of the angle $\alpha$. + +\item Prove that the angle formed by the tangents of a circle is equal to half the difference of the central angles corresponding to the arcs that lie between the arms of this angle. + +\item Let $L$ be the orthogonal projection of an arbitrary point $K$ of the circle $k$ +on its tangent $t$ through the point $T\in k$ and $X$ the point that is +symmetric to the point $L$ with respect to the line $KT$. Determine the geometric +position of the points $X$. + +\item Let $BB'$ and $CC'$ be the altitudes of the triangle $ABC$ and $t$ +the tangent of this triangle at the point $A$. Prove that + $B'C'\parallel t$. + +\item In the right triangle $ABC$ is above the cathetus $AC$ as the diameter +drawn circle that intersects the hypotenuse $AB$ at the point $E$. The tangent of this +circle at the point $E$ intersects the other cathetus $BC$ at the point $D$. Prove that $BDE$ +is an isosceles triangle. + +\item In the right angle with the vertex $A$ is drawn a circle that touches the sides of this angle +at the points $B$ and $C$. Any tangent of this circle intersects the lines $AB$ and $AC$, in order +in the points $M$ and $N$ (so that $\mathcal{B}(A,M,B)$). Prove that: +$$\frac{1}{3}\left(|AB|+|AC|\right) < |MB|+|NC| < +\frac{1}{2}\left(|AB|+|AC|\right).$$ + +\item Prove that in the right triangle the sum of the sides is equal to the sum +of the diameters of the inscribed and drawn circle. + +\item Let the similitudes of the internal angles of the convex quadrilateral intersect in six different points. +Prove that four of these points are the vertices of the pedal quadrilateral. + +\item Let: $c$ be the length of the hypotenuse, $a$ and $b$ the lengths +of the catheti and $r$ the radius of the inscribed circle of the right triangle. Prove that: +\begin{enumerate} + \item $2r + c \geq 2 \sqrt{ab}$, \item $a + b + c > 8r$. +\end{enumerate} + +\item Let $P$ and $Q$ be the centers of the shorter arcs $AB$ and $AC$ +of the regular triangle $ABC$ of the drawn circle. Prove that the sides $AB$ and $AC$ of this triangle divide +the chord $PQ$ into three proportional segments. + +\item Let $k_1$, $k_2$, $k_3$, $k_4$ be four circles, each of which from the outside touches one side and two sides of an arbitrary convex +quadrilateral. Prove that the centers of these circles are concircular points. + +\item Circles $k$ and $l$ touch each other from the outside in point $A$. Points $B$ and $C$ are the points of contact of the common external tangent of these two circles. Prove that $\angle BAC$ is a right angle. + +\item Let $ABCD$ be a deltoid ($AB\cong AD$ and $CB\cong CD$). Prove: +\begin{enumerate} + \item $ABCD$ is a tangent quadrilateral, + \item $ABCD$ is a parallelogram exactly when $AB\perp BC$. +\end{enumerate} + +\item Circles $k$ and $k_1$ touch each other from the outside in point $T$, where they intersect lines $p$ and $q$. Line $p$ has two more intersections with the circles, $P$ and $P_1$, line $q$ has $Q$ and $Q_1$. Prove that $PQ\parallel P_1Q_1$. + +\item Let $MN$ be the common tangent of circles $k$ and $l$ ($M$ and $N$ are the points of contact), which intersect in points $A$ and $B$. Calculate the measure of the sum $\angle MAN+\angle MBN$. + +\item Let $t$ be the tangent of triangle $ABC$ of the circumscribed circle in point $A$. A line parallel to the tangent $t$ intersects sides $AB$ and $AC$ in points $D$ and $E$. +Prove that points $B$, $C$, $D$ and $E$ are concyclic. + +\item Let $D$ and $E$ be any points of the semicircle drawn over diameter $AB$. Let $AD\cap BE= \{F\}$ and $AE\cap BD= \{G\}$. +Prove that $FG\perp AB$. + +\item Let $M$ be a point of circle $k(O,r)$. Determine the geometric location of the centers of all the tangents of this circle that have one endpoint in point $M$. + +\item Let $M$ and $N$ be points that are symmetric to the vertex $A'$ of altitude $AA'$ of triangle $ABC$ with respect to side $AB$ and $AC$, and let $K$ be the intersection of lines $AB$ and $MN$. Prove that points $A$, $K$, $A'$, $C$ and $N$ are concyclic. + +\item Let $ABCD$ be a parallelogram, $E$ the altitude point of triangle $ABD$, and $F$ the altitude point of triangle $ABC$. Prove that quadrilateral $CDEF$ +is a parallelogram. + +\item Circles with centers $O_1$ and $O_2$ intersect in points $A$ and $B$. The line $p$, +which goes through point $A$, intersects these two circles in points $M_1$ and $M_2$. Prove that +$\angle O_1M_1B\cong\angle O_2M_2B$. + +\item The circle with center $O$ is drawn over the diameter $AB$. +Let $C$ and $D$ be such points on the line $AB$, that $CO\cong OD$. Parallel lines through points $C$ and $D$ intersect the circle in points $E$ and $F$. +Prove that lines $CE$ and $DF$ are perpendicular to the line $EF$. + +\item On the string $AB$ of the circle $k$ with center $O$ lies the point $C$, point $D$ is the other intersection of the circle $k$ with the drawn circle of the triangle $ACO$. Prove that +$CD\cong CB$. + +\item Let $AB$ be the transversal of the circle $k$. Lines $AC$ and $BD$ are tangents +to the circle $k$ in points $C$ and $D$. +Prove that: + $$||AC|-|BD||< |AB| < |AC|+|BD|.$$ + +\item Let $S$ be the intersection of the sides $AD$ and $BC$ +of the trapezoid $ABCD$ with the base $AB$. Prove that the drawn circles of the triangles $SAB$ and +$SCD$ touch in point $S$. + +\item Lines $PB$ and $PD$ touch the circle $k(O,r)$ in points $B$ and $D$. +Line $PO$ intersects the circle $k$ in points $A$ and $C$ ($\mathcal{B}(P,A,C)$). Prove that the +line $BA$ is the angle bisector of the angle $PBD$. + +\item Quadrilateral $ABCD$ is inscribed in the circle with center $O$. Diagonals $AC$ and +$BD$ are perpendicular. Let $M$ be the perpendicular projection of the center $O$ +on the line $AD$. Prove that + $$|OM|=\frac{1}{2}|BC|.$$ + +\item Lines $AB$ and $BC$ are adjacent sides of a regular nonagon, which is inscribed in the circle $k$ with center $O$. +Point $M$ is the center of the side $AB$, point $N$ is the center +of the radius $OX$ of the circle $k$, which is perpendicular to the line $BC$. Prove that +$\angle OMN=30^0$. + +\item Circles $k_1$ and $k_2$ intersect in points $A$ and $B$. Let $p$ be a line that goes through point $A$, circle $k_1$ intersects also in point $C$, circle $k_2$ intersects also in point $D$, and $q$ be a line that goes through point $B$, circle $k_1$ intersects also in point $E$, circle $k_2$ intersects also in point $F$. Prove that $\angle CBD\cong\angle EAF$. + +\item Circles $k_1$ and $k_2$ intersect in points $A$ and $B$. Draw a line $p$, that goes through point $A$, so that the length of the line $MN$, where $M$ and $N$ are the intersections of line $p$ with circles $k_1$ and $k_2$, is maximal. + +\item Let $L$ be the orthogonal projection of an arbitrary point $K$ of the circle $k$ on its tangent through the point $T\in k$ and $X$ be the point that is symmetric to the point $L$ with respect to the line $KT$. Determine the geometric position of the points $X$. + +\item Prove that the string polygon with an even number of vertices, that has all the internal angles congruent, is a regular polygon. + +\item Two circles touch each other from the inside in the point $A$. The line $AB$ is the diameter of the larger circle, the string $BK$ of the larger circle touches the smaller circle in the point $C$. Prove that the line $AC$ is the bisector of the angle $BAK$. + +\item Let $BC$ be the string of the circle $k$. Determine the geometric position of the altitude points of all the triangles $ABC$, where $A$ is an arbitrary point that lies on the circle $k$. + +\item We have a quadrilateral with three acute internal angles. Prove that the longer diagonal goes through the vertex that belongs to the acute angle. + +\item Let $ABCDEF$ be a string hexagon, $AB\cong DE$ and $BC\cong EF$. Prove that $CD\parallel AF$. + +\item Let $ABCD$ be a convex quadrilateral, where $\angle ABD=50^0$, $\angle ADB=80^0$, $\angle ACB=40^0$ and $\angle DBC=\angle BDC +30^0$. Calculate the measure of the angle $\angle DBC$. + +\item Let $M$ be an arbitrary internal point of the angle with the vertex $A$, points $P$ and $Q$ the orthogonal projections of the point $M$ on the sides of this angle, and point $K$ the orthogonal projection of the vertex $A$ on the line $PQ$. Prove that $\angle MAP\cong \angle QAK$. + +\item In the archery octagon $A_1A_2\ldots A_8$ it holds that $A_1A_2\parallel A_5A_6$, $A_2A_3\parallel A_6A_7$, +$A_3A_4\parallel A_7A_8$. Prove that $A_8A_1\cong A_4A_5$. + +\item A circle intersects each side of a quadrilateral in two points and thus on all sides of the quadrilateral it determines the consistent tautologies. +Prove that this quadrilateral is tangent. + +\item The lengths of the sides of the tangent pentagon $ABCDE$ are natural numbers and at the same time $|AB|=|CD|=1$. +The inscribed circle of the pentagon touches the side $BC$ in the point $K$. +Calculate the length of the line $BK$. + +\item Prove that the circle that passes through the adjacent vertices $A$ and $B$ of the regular +pentagon $ABCDE$ and its center $O$, also passes through the intersection +of its diagonals $AD$ and $BE$. + +\item Let $H$ be the altitude point of the triangle $ABC$, $l$ the circle above the diameter $AH$ +and $P$ and $Q$ the intersections of this circle with the sides $AB$ and $AC$. Prove that the +tangents of the circle $k$ through the points $P$ and $Q$ intersect on the side $BC$. + +\item The circle $l$ touches the circle $k$ from the inside in the point $C$. Let $M$ be any point +of the circle $l$ (different from $C$). The tangent of the circle $l$ in the point $M$ intersects the circle $k$ in +the points $A$ and $B$. Prove that $\angle ACM \cong \angle MCB$. + +\item Let $k$ be the inscribed circle of the triangle $ABC$ and $R$ the center of that arc $AB$ of this circle, +which does not contain the point $C$. The lines $RP$ and $RQ$ are the tautologies of this circle. The first one +is parallel, the second one is perpendicular to the internal angle $\angle BAC$. Prove: +\begin{enumerate} +\item The line $BQ$ is the internal angle $\angle CBA$, +\item The triangle, which is determined by the lines $AB$, $AC$ and $PR$, is a right triangle. + \end{enumerate} + +\item Let $X$ be such an internal point of the triangle $ABC$, that it holds: + $\angle BXC =\angle BAC+60^0$, $\angle AXC =\angle ABC+60^0$ and $\angle AXB =\angle AC B+60^0$. Let +$P$, $Q$ and $R$ be the other intersections of the lines $AX$, $BX$ and $CX$ with the inscribed circle of the triangle $ABC$. Prove that the triangle $PQR$ is a right triangle. + +\item Prove that the tangent points of an inscribed circle of a triangle $ABC$ divide its sides into segments of lengths $s-a$, $s-b$ and $s-c$ ($a$, $b$ and $c$ are the lengths of the sides, $s$ is the semi-perimeter of the triangle). + + + \item Circles $k$, $l$ and $j$ touch each other from the outside in non-linear points $A$, $B$ and $C$. Prove that the circumscribed circle of the triangle $ABC$ is perpendicular to the circles $k$, $l$ and $j$. + + + \item Let $ABCD$ be a square with the center of the circumscribed circle in the point $O$. With $E$ we denote the intersection of its diagonals $AC$ and $BD$ and with $F$, $M$ and $N$ the centers of the lines $OE$, $AD$ and $BC$. If $F$, $M$ and $N$ are collinear points, then $AC\perp BD$ or $AB\cong CD$. Prove. + + + +\item Draw a triangle $ABC$ (see the labels in section \ref{odd3Stirik}): + + (\textit{a}) $a$, $\alpha$, $r$, \hspace*{2mm} + (\textit{b}) $a$, $\alpha$, $r_a$, \hspace*{2mm} + (\textit{c}) $a$, $v_b$, $v_c$, \hspace*{2mm} + + (\textit{d}) $\alpha$, $v_a$, $s$, \hspace*{2mm} + (\textit{e}) $v_a$, $l_a$, $r$, \hspace*{2mm} +(\textit{f}) $\alpha$, $v_a$, $l_a$, \hspace*{2mm} + + (\textit{g}) $\alpha$, $\beta$, $R$, \hspace*{2mm} + (\textit{h}) $c$, $r$, $R$, \hspace*{2mm} + (\textit{i}) $a$, $v_b$, $R$, \hspace*{2mm} + + \item Draw a circle $k$ so that: + + \begin{enumerate} + \item it touches two given non-parallel lines $p$ and $q$, and the tangent that determines the touch points is consistent with the given distance $t$, + \item its center is the given point $S$, and the given line $p$ determines on it the tangent that is consistent with the given distance $t$, + \item it passes through the given points $A$ and $B$, and its center lies on the given circle $l$, + \item it has the given radius $r$ and it touches the two given circles $l$ and $j$, + \item it touches the line $p$ in the point $P$ and passes through the given point $A$. + \end{enumerate} + + \item Draw a square $ABCD$, if the given vertex $B$ and two points $E$ and $F$ that lie on the sides $AD$ and $CD$ are given. + +\item Given is a line $CD$ and points $A$ and $B$ ($A,B\notin CD$). On the line $CD$ draw a point $M$, such that $\angle AMC\cong2\angle BMD$. + + + \item Draw a triangle $ABC$ with the following data: + + (\textit{a}) $v_a$, $t_a$, $\beta-\gamma$, \hspace*{2mm} + (\textit{b}) $v_a$, $l_a$, $R$, \hspace*{2mm} + (\textit{c}) $R$, $\beta-\gamma$, $t_a$, \hspace*{2mm} + + (\textit{d}) $R$, $\beta-\gamma$, $v_a$, \hspace*{2mm} + (\textit{e}) $R$, $\beta-\gamma$, $a$. \hspace*{2mm} + + +\item On the beam of the side $AB$ of the rectangle $ABCD$ draw a point $E$, from which the sides $AD$ and $DC$ are seen under the same angle. When does the task have a solution? + + \item In the convex quadrilateral $ABCD$ it holds that $BC\cong CD$. Draw this quadrilateral, if the sides $AB$ and $AD$ and the internal angle at the vertices $B$ and $D$ are given. + + \item In the given circle $k$ draw a triangle $ABC$, if the vertex $A$, the line $p$, which is parallel to the altitude $AA'$, and the intersection point $B_2$ of the altitude beam $BB'$ and this circle are given. + + \item Draw a regular triangle $ABC$, if its side $BC$ is congruent to the distance $a$, the altitude beams of the sides $AB$ and $AC$ and the line of symmetry of the internal angle $BAC$ go through the given points $M$, $N$ and $P$ one after another. + +\item Draw a triangle $ABC$, if the following are given: + \begin{enumerate} + \item the vertex $A$, the center of the circumscribed circle $O$ and the center of the inscribed circle $S$, + \item the center of the circumscribed circle $O$, the center of the inscribed circle $S$ and the center of the escribed circle $S_a$, + \item the vertex $A$, the center of the circumscribed circle $O$ and the altitude point $V$, + \item the vertices $B$ and $C$ and the internal angle bisector of the internal angle $BAC$, + \item the vertex $A$, the center of the circumscribed circle $O$ and the intersection point $E$ of the side $BC$ with the internal angle bisector of the internal angle $BAC$, + \item the points $M$, $P$ and $N$, in which the altitude and the centroid from the vertex $A$ and the internal angle bisector of the internal angle $BAC$ intersect the circumscribed circle of the triangle, + \item the vertex $A$, the center of the circumscribed circle $O$, the point $N$, in which the internal angle bisector of the internal angle $BAC$ intersects the circumscribed circle of the triangle, and the distance $a$, which is parallel to the side $BC$. + \end{enumerate} + + \item Draw a triangle $ABC$ with the following data: + + (\textit{a}) $a$, $b$, $\alpha=3\beta$, \hspace*{3mm} + (\textit{b}) $t_a$, $t_c$, $v_b$.\hspace*{3mm} + +\item Through the point $M$, which lies inside the given circle $k$, draw such a cord that the difference of its segments (from the point $M$) is equal to the given distance $a$. + +\item Draw a triangle $ABC$, if you know: + + (\textit{a}) $b-c$, $r$, $r_a$, \hspace*{1.8mm} + (\textit{b}) $a$, $r$, $r_a$, \hspace*{1.8mm} + (\textit{c}) $a$, $r_b+r_c$, $v_a$, \hspace*{1.8mm} + (\textit{d}) $b+c$, $r_b$, $r_c$, + + (\textit{e}) $R$, $r_b$, $r_c$, \hspace*{1.8mm} + (\textit{f}) $b$, $R$, $r+r_a$, \hspace*{1.8mm} +(\textit{g}) $a$, $v_a$, $r_a-r$, \hspace*{1.8mm} + (\textit{h}) $\alpha$, $r$, $b+c$. + + + \item The following are given: the circle $k$, its diameter $AB$ and the point $M\notin k$. With only a straightedge, draw a rectangle from the point $M$ to the line $AB$. + +\item The given are: square $ABCD$ and such points $M$ and $N$ on sides $BC$ and $CD$, that $\angle MAN=45^0$. + With only a straightedge draw a rectangle from point $A$ to line $MN$. + +\end{enumerate} + + + + + +% DEL 5 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% VEKTORJI +%________________________________________________________________________________ + + \del{Vectors} \label{pogVEKT} + + +%________________________________________________________________________________ +\poglavje{Vector Definition. The Sum of Vectors} \label{odd5DefVekt} + +Intuitively, a vector is a directed line segment that can be moved parallel\footnote{The concept of a vector was known to the ancient Greeks. The modern concept of vectors, associated with linear algebra and analytic geometry, began to develop in the 19th century as a generalization of complex numbers. In this sense, the English mathematician \index{Hamilton, W. R.}\textit{W. R. Hamilton} (1805--1865) defined the so-called \index{quaternions}\textit{quaternions} $q = w + ix + jy + kz$, $i^2 = j^2 = k^2 = -ijk = -1$ as a generalization of complex numbers in four-dimensional space}. In this sense, the vector $\overrightarrow{AB}$ would represent the entire set of line segments that are consistent, parallel and have the same direction as a given line segment $AB$. We could also write $\overrightarrow{CD}=\overrightarrow{AB}$ for each line segment $CD$ from this set of line segments (Figure \ref{sl.vek.5.1.1.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.1.pic} +\caption{} \label{sl.vek.5.1.1.pic} +\end{figure} + +In this way, we get an idea for a formal definition of vectors. First, we introduce the relation $\varrho$ on the set of pairs of points. Let $A$, $B$, $C$ and $D$ be points in the same plane +(Figure \ref{sl.vek.5.1.2.pic}). We say that $(A,B)\varrho (C,D)$, if one of the three conditions\footnote{In this way - in \index{geometry!affine}affine geometry (without the axioms of congruence) - the vector was defined by the Serbian mathematician \index{Veljković, M.}\textit{M. Veljković} (1954--2008), professor at the Mathematical Gymnasium in Belgrade.}\index{relation!$\varrho$} is fulfilled: + +\begin{enumerate} + \item The quadrilateral $ABDC$ is a parallelogram, + \item There exist points $P$ and $Q$, such that the quadrilaterals $ABQP$ and $CDQP$ are parallelograms, + \item $A=B$ and $C=D$. +\end{enumerate} + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.2.pic} +\caption{} \label{sl.vek.5.1.2.pic} +\end{figure} + +It is intuitively clear that the second condition needs to be added due to the example when points $A$, $B$, $C$ and $D$ are collinear. The third condition will apply to the so-called vector of zero. + +From the definition of the relation $\varrho$ itself, we get the following proposition. + + + + \bizrek \label{vektRelRo} + If $(A,B)\varrho (C,D)$ and $A\neq B$, then the line segments $AB$ and $CD$ + are congruent, parallel, and have the same direction. + \eizrek + +\textbf{\textit{Proof.}} Because $A\neq B$, only the first two conditions from the definition remain. In this case, the relations $AB\parallel CD$ and $AB\cong CD$ are direct consequences of the definition of a parallelogram and proposition \ref{paralelogram}. + +Regarding the orientation of the line segments, we would use the definition: parallel line segments $XY$ and $UV$ are \pojem{equally oriented}, if one of the conditions is fulfilled: +\begin{itemize} + \item $Y,V\ddot{-} XU$; + \item There exist points $S$ and $T$, such that $ST\parallel XY$, $Y,T\ddot{-} XS$ and $V,T\ddot{-} US$. +\end{itemize} +\kdokaz + +The next proposition is needed, so that we can define vectors. + + + \bizrek + The relation $\varrho$ on the set of pairs of points in the plane is an equivalence relation. + \eizrek + + +\textbf{\textit{Proof.}} The reflexivity and symmetry are direct consequences of the definition of the relation $\varrho$. For transitivity, however, it is necessary to check all possible combinations of conditions 1--3 from the definition. +\kdokaz + +\pojem{Vector} \index{vector} is now defined as the class of the equivalence relation $\varrho$: + +$$\overrightarrow{AB}=[AB]_{\varrho}=\{(X,Y);\hspace*{1mm}(X,Y)\varrho(A,B)\}.$$ + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.3.pic} +\caption{} \label{sl.vek.5.1.3.pic} +\end{figure} + + +We will call point $A$ the \index{starting point of a vector}\pojem{starting point}, point $B$ the \index{end point of a vector}\pojem{end point} of the vector $\overrightarrow{AB}$. + +It is clear that in the case of $(A,B)\varrho (C,D)$, the pairs $(A,B)$ and $(C,D)$ are from the same class, which means that then $\overrightarrow{AB}=\overrightarrow{CD}$. The converse is also true: the relation $\overrightarrow{AB}=\overrightarrow{CD}$ means that $(A,B)\varrho (C,D)$ +(Figure \ref{sl.vek.5.1.3.pic}). + +If it is not necessary to emphasize which representative of the class it is, we will also denote vectors with $\overrightarrow{v}$, $\overrightarrow{u}$, $\overrightarrow{w}$, ... + +The vector $\overrightarrow{AA}$, which represents the set of coinciding points, we will call the \index{vector!zero}\pojem{zero vector}. The zero vector will also be denoted by $\overrightarrow{0}$ +(Figure \ref{sl.vek.5.1.4.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.4.pic} +\caption{} \label{sl.vek.5.1.4.pic} +\end{figure} + +We will say that the vector $\overrightarrow{BA}$ is the \index{vector!opposite}\pojem{opposite vector} of the vector $\overrightarrow{AB}$ and denote it by $-\overrightarrow{AB}$. It is clear that the definition of the opposite vector is not dependent on the choice of the representative of the class; the opposite vector is obtained if we exchange the starting and end points of the vector (or of each pair of points of the corresponding class). So if we denote $\overrightarrow{v}=\overrightarrow{AB}$ then +$-\overrightarrow{v}=-\overrightarrow{AB}=\overrightarrow{BA}$ +(Figure \ref{sl.vek.5.1.4.pic}). + + +We denote the set of all vectors of the plane with $\mathcal{V}$. + +A direct consequence of \ref{vektRelRo} is the following assertion. + + + \bizrek \label{vektVzpSkl} + If $\overrightarrow{AB}=\overrightarrow{CD}$, then the line segments $AB$ and $CD$ + are congruent, parallel, and have the same direction. + \eizrek + +Based on the previous statement, we can now correctly define the following concepts: +\begin{itemize} + \item \index{smer vektorja}\pojem{smer} of vector $\overrightarrow{AB}\neq\overrightarrow{0}$ is determined by any line $p\parallel AB$, + \item \index{dolžina!vektorja}\pojem{length} or \index{intenziteta vektorja}\pojem{intensity} of vector $\overrightarrow{AB}$ is $|\overrightarrow{AB}|=|AB|$ ($|\overrightarrow{0}|=0$), + \item \index{usmerjenost vektorja}\pojem{orientation} or \index{orientacija!vektorja}\pojem{orientation} of vector $\overrightarrow{AB}\neq\overrightarrow{0}$ is determined by the ordered pair $(A,B)$, where $A$ is the starting point, and $B$ is the end point of this vector. +\end{itemize} + +If vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ have the same direction, we also say that they are \index{vektorja!vzporedna}\pojem{parallel} or \index{vektorja!kolinearna}\pojem{colinear} and denote it with $\overrightarrow{v}\parallel\overrightarrow{u}$ (in this case we also say that $\overrightarrow{0}$ is colinear with every vector), otherwise the vectors are \index{vektorja!nekolinearna}\pojem{non-colinear}. + +If colinear vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ satisfy $\overrightarrow{v}=\overrightarrow{SA}$ and $\overrightarrow{u}=\overrightarrow{SB}$, we say that vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ are \pojem{equally oriented} (notation $\overrightarrow{u}\rightrightarrows \overrightarrow{v}$), if $A,B\ddot{-} S$, or \pojem{oppositely oriented} (notation $\overrightarrow{u}\rightleftarrows \overrightarrow{v}$), if $A,B\div S$. + +It is clear that colinear vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are equally oriented exactly when lines $AB$ and $CD$ are equally oriented. + + +From these definitions it directly follows that $|-\overrightarrow{v}|=|\overrightarrow{v}|$ and $-\overrightarrow{v}\parallel\overrightarrow{v}$; vectors $\overrightarrow{v}$ and $-\overrightarrow{v}$ are oppositely oriented (if $\overrightarrow{v}=\overrightarrow{AB}$, then $-\overrightarrow{v}=\overrightarrow{BA}$). + +The following statement is very important. + +\bizrek \label{vektABCObst1TockaD} + For every three points there $A$, $B$ and $C$ there exists exactly one point $D$, + such that $\overrightarrow{AB}=\overrightarrow{CD}$, i.e. + $$(\forall A, B, C)(\exists_1 D)\hspace*{1mm}\overrightarrow{AB}=\overrightarrow{CD}.$$ + \eizrek + + +\textbf{\textit{Proof.}} We will consider three different options (Figure \ref{sl.vek.5.1.5.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.5.pic} +\caption{} \label{sl.vek.5.1.5.pic} +\end{figure} + +\textit{1)} If $A=B$, then $D=C$ is the only point for which $(A,B)\varrho (C,D)$ or $\overrightarrow{CD}=\overrightarrow{AB}=\overrightarrow{0}$. + +\textit{2)} Let $A\neq B$ and $C\notin AB$. Then $D$ is the fourth vertex of the parallelogram $ABDC$ (which exists and is only one due to +Playfair's axiom \ref{Playfair1}) and the only point for which $(A,B)\varrho (C,D)$ or $\overrightarrow{CD}=\overrightarrow{AB}$. + +\textit{3)} Let $A\neq B$ and $C\in AB$. Let $P$ be any point that does not lie on the line $AB$. By the proven part \textit{2)} there is exactly one point $Q$, for which $\overrightarrow{PQ}=\overrightarrow{AB}$, and then also exactly one point $D$, for which $\overrightarrow{CD}=\overrightarrow{PQ}=\overrightarrow{AB}$. +\kdokaz + +The previous statement can also be written in another way (Figure \ref{sl.vek.5.1.6.pic}). + + + + \bizrek \label{vektAvObst1TockaB} + For every point $X$ and every vector $\overrightarrow{v}$ there exists exactly one point $Y$, + such that $\overrightarrow{XY}=\overrightarrow{v}$, i.e. + $$(\forall X)(\forall \overrightarrow{v})(\exists_1 Y)\hspace*{1mm}\overrightarrow{XY}=\overrightarrow{v}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.6.pic} +\caption{} \label{sl.vek.5.1.6.pic} +\end{figure} + +As it is evident from the proof, for every vector $\overrightarrow{v}$ we can choose an arbitrary representative from its class with an arbitrary starting point. In a similar way, we could have proven that this is also possible regarding the end point. Intuitively, this means that we can move the vector parallel in only one way to its starting or end point. + +We can therefore put two arbitrary vectors so that they have the same starting point or so that the starting point of the second vector is at the same time the end point of the first vector. +This fact allows us to define the sum of two vectors. + +If $\overrightarrow{v}=\overrightarrow{AB}$ and $\overrightarrow{u}=\overrightarrow{BC}$, the \index{vsota!vektorjev}\pojem{sum of vectors} $\overrightarrow{v}$ and $\overrightarrow{u}$ is the vector $\overrightarrow{v}+\overrightarrow{u}=\overrightarrow{AC}$ (Figure \ref{asl.vek.5.1.7.pic}). So $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ is true. + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.7.pic} +\caption{} \label{asl.vek.5.1.7.pic} +\end{figure} + + +It is necessary to prove the correctness of the definition - that the sum of vectors is not dependent on the choice of representatives of two classes ($\overrightarrow{v}$ and $\overrightarrow{u}$) or on the choice of point $A$. + + \bizrek \label{vektKorektDefSest} + If $A$, $B$, $C$, $A'$, $B'$ and $C'$ are points such that $\overrightarrow{AB}=\overrightarrow{A'B'}$ and + $\overrightarrow{BC}=\overrightarrow{B'C'}$, then there is also $\overrightarrow{AC}=\overrightarrow{A'C'}$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.7a.pic} +\caption{} \label{sl.vek.5.1.7a.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.1.7.pic}). + +In the formal proof, we would consider different possibilities regarding the relation $\varrho$. + \kdokaz + + The rule of adding vectors, for which we use the aforementioned equality: + +\begin{itemize} + \item $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$, +\end{itemize} + +We call it the \index{pravilo!trikotniško}\pojem{triangle rule} or the \index{Chaslesova identiteta}\pojem{Chaslesova\footnote{\index{Chasles, M.} \textit{M. Chasles} (1793–-1880), French mathematician.} identity}. +The rule is named this way even though it is also valid when $A$, $B$ and $C$ are collinear points (or $\overrightarrow{v}$ and $\overrightarrow{u}$ are collinear vectors); in this case, it is not a triangle (Figure \ref{sl.vek.5.1.8.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.8.pic} +\caption{} \label{sl.vek.5.1.8.pic} +\end{figure} + + + +It turns out that the set of all vectors $\mathcal{V}$ with respect to the operation +of adding vectors represents the structure of a so-called \index{group!commutative}\pojem{commutative group} (or \index{group!Abelian}\pojem{Abelian\footnote{\index{Abel, N. H.}\textit{N. H. Abel} (1802--1829), Norwegian mathematician.} group}). The concept of a group (which is not always commutative) was already mentioned in section \ref{odd2AKSSKL} in the context of isometries; we will deal with this structure in more detail in section \ref{odd6Grupe}. The properties of this structure (commutative groups) are given in the statement +of the following theorem. + +\bizrek \label{vektKomGrupa} The ordered pair $(\mathcal{V},+)$ form a commutative group, which means that: + \begin{enumerate} + \item $(\forall \overrightarrow{v}\in\mathcal{V})(\forall \overrightarrow{v}\in\mathcal{V})\hspace*{1mm} + \overrightarrow{v}+\overrightarrow{v}\in\mathcal{V}$ + (\index{grupoidnost}\textit{closure}), + \item $(\forall \overrightarrow{u}\in\mathcal{V}) + (\forall \overrightarrow{v}\in\mathcal{V}) + (\forall \overrightarrow{w}\in\mathcal{V}) + \hspace*{1mm} + \left(\overrightarrow{u}+\overrightarrow{v}\right)+\overrightarrow{w} + = \overrightarrow{u}+\left(\overrightarrow{v}+\overrightarrow{w}\right) + $ + (\index{asociativnost}\textit{associativity}), + \item $(\exists \overrightarrow{e}\in\mathcal{V}) + (\forall \overrightarrow{v}\in\mathcal{V}) + \hspace*{1mm} + \overrightarrow{v}+\overrightarrow{e} + = \overrightarrow{e}+\overrightarrow{v} + $ + (\index{nevtralni element}\textit{identity element}), + \item $(\forall \overrightarrow{v}\in\mathcal{V}) + (\exists \overrightarrow{u}\in\mathcal{V}) + \hspace*{1mm} + \overrightarrow{v}+\overrightarrow{u} + = \overrightarrow{u}+\overrightarrow{v}=\overrightarrow{e} + $ + (\index{inverzni element}\textit{inverse element}), + \item $(\forall \overrightarrow{v}\in\mathcal{V}) + (\forall \overrightarrow{u}\in\mathcal{V}) + \hspace*{1mm} + \overrightarrow{v}+\overrightarrow{u} + = \overrightarrow{u}+\overrightarrow{v} + $ + (\index{komutativnost}\textit{commutativity}). + \end{enumerate} + \eizrek + +\textbf{\textit{Proof.}} + + \textit{1)} The sum of two arbitrary vectors is a vector, which follows from the definition of the sum of two vectors. + + \textit{2)} Let $\overrightarrow{u}$, $\overrightarrow{v}$ and $\overrightarrow{w}$ be arbitrary vectors and $A$, $B$, $C$ and $D$ such points that $\overrightarrow{u}=\overrightarrow{AB}$, $\overrightarrow{v}=\overrightarrow{BC}$ and $\overrightarrow{w}=\overrightarrow{CD}$ (statement \ref{vektAvObst1TockaB}) (Figure \ref{sl.vek.5.1.9a.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.9a.pic} +\caption{} \label{sl.vek.5.1.9a.pic} +\end{figure} + + + By the definition of vector addition (and the correctness of this definition - statement \ref{vektKorektDefSest}) it is: + \begin{eqnarray*} + & & \left(\overrightarrow{u}+\overrightarrow{v}\right)+\overrightarrow{w}= + \left(\overrightarrow{AB}+\overrightarrow{BC}\right)+\overrightarrow{CD}= + \overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AD}\\ + & & \overrightarrow{u}+\left(\overrightarrow{v}+\overrightarrow{w}\right) =\overrightarrow{AB}+\left(\overrightarrow{BC}+\overrightarrow{CD}\right)= + \overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AD} + \end{eqnarray*} + From this it follows that the addition of vectors is associative. + + \textit{3)} In the case of vector addition, the neutral element is the vector of zero or $\overrightarrow{e}=\overrightarrow{0}$, because for every vector $\overrightarrow{v}=\overrightarrow{AB}$ it is true: + + \begin{eqnarray*} + & & \overrightarrow{v}+\overrightarrow{0}= + \overrightarrow{AB}+\overrightarrow{BB}= + \overrightarrow{AB}=\overrightarrow{v}\\ + & & \overrightarrow{0}+\overrightarrow{v}= + \overrightarrow{AA}+\overrightarrow{AB}= + \overrightarrow{AB}=\overrightarrow{v}. + \end{eqnarray*} + + + \textit{4)} In the case of vector addition, for an arbitrary vector $\overrightarrow{v}=\overrightarrow{AB}$ the inverse element is the opposite vector $\overrightarrow{u}=-\overrightarrow{v}=\overrightarrow{BA}$: + +\begin{eqnarray*} + & & \overrightarrow{v}+\left(-\overrightarrow{v}\right)= + \overrightarrow{AB}+\overrightarrow{BA}= + \overrightarrow{AA}=\overrightarrow{0}\\ + & & -\overrightarrow{v}+\overrightarrow{v}= + \overrightarrow{BA}+\overrightarrow{AB}= + \overrightarrow{BB}=\overrightarrow{0}. + \end{eqnarray*} + + \textit{5)} Let $v$ and $u$ be any vectors and $A$, $B$ and $C$ such points that $\overrightarrow{v}=\overrightarrow{AB}$ and $\overrightarrow{u}=\overrightarrow{BC}$ (statement \ref{vektAvObst1TockaB}). We will consider two examples (Figure \ref{sl.vek.5.1.9b.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.9b.pic} +\caption{} \label{sl.vek.5.1.9b.pic} +\end{figure} + + \textit{a)} We assume that points $A$, $B$ and $C$ are nonlinear or $\overrightarrow{v}$ and $\overrightarrow{u}$ are nonlinear vectors. We mark with $D$ the fourth vertex of the parallelogram $ABCD$. By definition of the relation $\varrho$ it is $(A,B)\varrho (D,C)$ and $(B,C)\varrho (A,D)$, by definition of vectors it then follows that $\overrightarrow{AB}=\overrightarrow{DC}$ and $\overrightarrow{BC}=\overrightarrow{AD}$. So: + $$\overrightarrow{v}+\overrightarrow{u}= + \overrightarrow{AB}+\overrightarrow{BC}= + \overrightarrow{AC}= + \overrightarrow{AD}+\overrightarrow{DC}= + \overrightarrow{BC}+\overrightarrow{AB} + = \overrightarrow{u}+\overrightarrow{v}.$$ + +\textit{a)} Let $A$, $B$ and $C$ be collinear points or $\overrightarrow{v}$ and $\overrightarrow{u}$ be collinear vectors. Express the vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ as a sum of $\overrightarrow{v}=\overrightarrow{v}_1+\overrightarrow{v}_2$ and $\overrightarrow{u}=\overrightarrow{u}_1+\overrightarrow{u}_2$, so that none of the vectors $\overrightarrow{v}_1$, $\overrightarrow{v}_2$, $\overrightarrow{u}_1$ and $\overrightarrow{u}_2$ are collinear. If we now use what was proven in case \textit{a)}, we get: + $$\overrightarrow{v}+\overrightarrow{u}= + \overrightarrow{v}_1+\overrightarrow{v}_2+ + \overrightarrow{u}_1+\overrightarrow{u}_2= + \overrightarrow{u}_1+\overrightarrow{u}_2+ + \overrightarrow{v}_1+\overrightarrow{v}_2= + \overrightarrow{u}+\overrightarrow{v},$$ which is what needed to be proven. \kdokaz + +The next statement is a consequence of the previous expression. + + \bzgled \label{vektABCD_ACBD} + For arbitrary points $A$, $B$, $C$ and $D$ is + $$\overrightarrow{AB}=\overrightarrow{CD}\Rightarrow \overrightarrow{AC}=\overrightarrow{BD}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.10.pic} +\caption{} \label{sl.vek.5.1.10.pic} +\end{figure} + +\textbf{\textit{Proof.}} By the definition of vector addition and the expression \ref{vektKomGrupa} (commutativity) is (Figure \ref{sl.vek.5.1.10.pic}): +$\overrightarrow{AC}= +\overrightarrow{AB}+ \overrightarrow{BC}= +\overrightarrow{CD}+ \overrightarrow{BC}= +\overrightarrow{BC}+\overrightarrow{CD}= +\overrightarrow{BD}.$ + \kdokaz + + Another consequence of the commutativity of vector addition (expression \ref{vektKomGrupa}) is the following rule of addition for non-collinear vectors. + + \begin{itemize} + \item \textit{For every three non-collinear points $A$, $B$ and $C$ is $\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AD}$ exactly when $ABDC$ is a parallelogram.} + \end{itemize} + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.11.pic} +\caption{} \label{sl.vek.5.1.11.pic} +\end{figure} + +This rule is called the \index{pravilo!paralelogramsko}\pojem{paralelogramsko pravilo}\footnote{Paralelogramsko pravilo was probably known to the ancient Greeks. It is assumed that the ancient Greek mathematician and philosopher \index{Aristotel} \textit{Aristotel} (384--322 BC) mentioned it.} (Figure \ref{sl.vek.5.1.11.pic}). + +The consequence of the associativity from the izrek \ref{vektKomGrupa} is the following rule for the addition of vectors: + \begin{itemize} + \item $\overrightarrow{A_1A_2}+\overrightarrow{A_2A_3}+\cdots +\overrightarrow{A_{n-1}A_n}=\overrightarrow{A_1A_n}$, + \end{itemize} + +which is called the \index{pravilo!poligonsko}\pojem{poligonsko pravilo} (Figure \ref{asl.vek.5.1.12.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.12.pic} +\caption{} \label{asl.vek.5.1.12.pic} +\end{figure} + +The direct consequence of this rule is the following assertion (Figure \ref{sl.vek.5.1.13.pic}). + + \bzgled + For arbitrary points $A_1$, $A_2$,..., $A_n$ in a plane is + $$\overrightarrow{A_1A_2}+\overrightarrow{A_2A_3}+\cdots +\overrightarrow{A_{n-1}A_n}+\overrightarrow{A_nA_1}=\overrightarrow{0}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.13.pic} +\caption{} \label{sl.vek.5.1.13.pic} +\end{figure} + +With the following izrek we will give an equivalent definition of the center of a line. + + \bzgled \label{vektSredDalj} + A point $S$ is the midpoint of a line segment $AB$ if and only if $\overrightarrow{SA}=-\overrightarrow{SB}$ i.e. $\overrightarrow{SA}+\overrightarrow{SB}=\overrightarrow{0}$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.14.pic} +\caption{} \label{sl.vek.5.1.14.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.1.14.pic}) + +($\Leftarrow$) Let's assume that $\overrightarrow{SA}=-\overrightarrow{SB}$ or $\overrightarrow{AS}=\overrightarrow{SB}$. This means that $AS$ and $SB$ are parallel, have the same direction and are consistent lines (statement \ref{vektVzpSkl}) or $SA\cong BS$ and $\mathcal{B}(A,S,B)$. So $S$ is the center of the line $AB$. + +($\Rightarrow$) Now let $S$ be the center of the line $AB$. It is enough to prove that $\overrightarrow{SA}=\overrightarrow{BS}$ or $\overrightarrow{BS}=\overrightarrow{SA}$. + Let's assume that $\overrightarrow{BS}=\overrightarrow{SA'}$. + But in this case $SA'\cong BS$ and $\mathcal{B}(A',S,B)$. By statement \ref{ABnaPoltrakCX} we have $A=A'$, so $\overrightarrow{BS}=\overrightarrow{SA'}=\overrightarrow{SA}$. +\kdokaz + +Let's also define the operation of subtracting two vectors. \pojem{The difference of vectors} \index{razlika!vektorjev} $\overrightarrow{v}$ and $\overrightarrow{u}$ is the sum of vectors $\overrightarrow{v}$ and $-\overrightarrow{u}$ or +$$\overrightarrow{v}-\overrightarrow{u}=\overrightarrow{v}+(-\overrightarrow{u}).$$ + + \bzgled \label{vektOdsev} + For arbitrary three points $O$, $A$ and $B$ is + $$\overrightarrow{OB}-\overrightarrow{OA}=\overrightarrow{AB}.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.15.pic} +\caption{} \label{sl.vek.5.1.15.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.1.15.pic}) + + From the definition of subtraction and addition of vectors and commutativity of addition + (statement \ref{vektKomGrupa}) it follows: + $$\overrightarrow{OB}-\overrightarrow{OA}= + \overrightarrow{OB}+(-\overrightarrow{OA})= + \overrightarrow{OB}+\overrightarrow{AO}= +\overrightarrow{AO}+\overrightarrow{OB}= + \overrightarrow{AB},$$ which had to be proven. \kdokaz + +\bzgled \label{vektOPi} + Suppose that a point $O$ lies on a line $p$ and let $P_1, P_2, \cdots, P_n$ be points + lying in the same half-plane $\alpha$ with the edge $p$. + If: + $$\overrightarrow{OS_n}=\overrightarrow{OP_1}+\overrightarrow{OP_2}+ + \cdots+\overrightarrow{OP_n},$$ + then the point $S_n$ also lies in the half-plane $\alpha$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.1a.pic} +\caption{} \label{sl.vek.5.2.1a.pic} +\end{figure} + + \textbf{\textit{Solution.}} We will prove the statement by induction + according to $n$ (Figure \ref{sl.vek.5.2.1a.pic}). + + (\textit{i}) For $n=1$ it is clear that $\overrightarrow{OS_1}=\overrightarrow{OP_1}$, + and $P_1\in\alpha\Rightarrow S_1=P_1\in\alpha$. + +(\textit{ii}) We assume that the statement is true for $n=k$. We will prove + that it is then also true for $n=k+1$. Let $P_1, P_2, \cdots, P_k, + P_{k+1}\in\alpha$. We will prove that then also $S_{k+1}\in + \alpha$. It holds: + $$\overrightarrow{OS_{k+1}}=\overrightarrow{OP_1}+\overrightarrow{OP_2}+ + \cdots+\overrightarrow{OP_k}+\overrightarrow{OP_{k+1}}= + \overrightarrow{OS_k}+\overrightarrow{OP_{k+1}}.$$ +The point $S_k$ by the induction assumption lies in the half-plane +$\alpha$. Because $S_k,P_{k+1}\in \alpha$ and $O\in p$, the angle $\angle +S_kOP_{k+1}$ is convex and the whole angle lies in the half-plane $\alpha$. This +means that also the point $S_{n+1}$ (as the fourth vertex of the parallelogram $P_{k+1}OP_kS_{k+1}$) lies in the interior of the angle $\angle +S_kOP_{k+1}$ or in the half-plane $\alpha$. + + \kdokaz + +With the concept of vector addition, we can define the concept of multiplying a vector by an arbitrary natural number: +$$1\cdot \overrightarrow{v} =\overrightarrow{v},\hspace*{2mm} (n+1)\cdot \overrightarrow{v}=n\cdot\overrightarrow{v}+\overrightarrow{v},\hspace*{2mm} (n\in \mathbb{N}).$$ +We can extend this definition to whole numbers: $0\cdot \overrightarrow{v}=\overrightarrow{0}$ and $-n\cdot \overrightarrow{v}=n\cdot(-\overrightarrow{v})$. +It is clear that in this case, the vectors $\overrightarrow{v}$ and $l\cdot \overrightarrow{v}$ ($l\in \mathbb{Z}$) are always collinear and $|l\cdot \overrightarrow{v}|=|l|\cdot |\overrightarrow{v}|$. +This gives us an idea for the definition of multiplying a vector by an arbitrary real number $\lambda\in \mathbb{R}$. + +First, for $\lambda=0$ let us define $0\cdot \overrightarrow{v}=\overrightarrow{0}$. If +$\lambda\neq 0$ and $\overrightarrow{v}=\overrightarrow{AB}$, then $\lambda\cdot\overrightarrow{v}=\overrightarrow{AC}$, where $C$ is such a point on the line $AB$ that (Figure \ref{sl.vek.5.2.1.pic}): + + \begin{itemize} + \item $|AC|=|\lambda| \cdot|AB|$, + \item $C,B\ddot{-} A$, if $\lambda>0$, + \item $C,B\div A$, if $\lambda<0$. + \end{itemize} + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.1.pic} +\caption{} \label{sl.vek.5.2.1.pic} +\end{figure} + + + It is clear that the point $C$ is uniquely determined for each $\lambda\neq 0$, and the vector $\lambda\cdot \overrightarrow{v}$ does not depend on the choice of point $A$. This means that the definition of multiplying a vector by a real number is correct. + +From the definition itself it follows that for every vector $\overrightarrow{v}$ and every real number $\lambda\in \mathbb{R}$ it holds that $|\lambda \cdot \overrightarrow{v}|=|\lambda| \cdot |\overrightarrow{v}|$. + +Multiplying the vector $\overrightarrow{v}$ by the real number $\lambda$ will also be called \index{množenje vektorja s skalarjem}\pojem{multiplying the vector $\overrightarrow{v}$ by the scalar $\lambda$}. + +Similarly to how we omit the multiplication sign $\cdot$ when multiplying algebraic expressions, we will also omit the multiplication sign $\cdot$ when multiplying a vector by a real number, or rather than writing $\lambda\cdot\overrightarrow{v}$ we will just write $\lambda\overrightarrow{v}$. + +The next theorem gives us the necessary and sufficient condition for the collinearity of two vectors. + + + \bizrek \label{vektKriterijKolin} + Vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ ($\overrightarrow{v},\overrightarrow{u}\neq \overrightarrow{0}$) are collinear if and only if there is such $\lambda\in \mathbb{R}$, that is $\overrightarrow{u}=\lambda\cdot\overrightarrow{v}$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.2.pic} +\caption{} \label{sl.vek.5.2.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.2.pic}) + + Let $\overrightarrow{v}$ and $\overrightarrow{u}$ be any vectors ($\overrightarrow{v},\overrightarrow{u}\neq \overrightarrow{0}$), $A$ any point, and $B$ and $C$ such points that $\overrightarrow{AB}=\overrightarrow{v}$ and $\overrightarrow{AC}=\overrightarrow{u}$ (statement \ref{vektAvObst1TockaB}). + + + ($\Leftarrow$) First, let us assume that $\overrightarrow{u}=\lambda\cdot\overrightarrow{v}$ for some $\lambda\in \mathbb{R}$. By definition, point $C$ lies on the line $AB$, which means that vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$, or $\overrightarrow{v}$ and $\overrightarrow{u}$, are collinear. + + ($\Rightarrow$) Now let us assume that vectors $\overrightarrow{v}$ and $\overrightarrow{u}$, or $\overrightarrow{AB}$ and $\overrightarrow{AC}$, are collinear. + In this case: + \begin{eqnarray*} + \overrightarrow{u}&=&\overrightarrow{AC}=\frac{|AC|}{|AB|}\cdot \overrightarrow{AB}= + \frac{|\overrightarrow{u}|}{|\overrightarrow{v}|}\cdot \overrightarrow{v}, + \hspace*{2mm} \textrm{ if } C,B\ddot{-} A;\\ + \overrightarrow{u}&=&\overrightarrow{AC}=-\frac{|AC|}{|AB|}\cdot \overrightarrow{AB}= + -\frac{|\overrightarrow{u}|}{|\overrightarrow{v}|}\cdot \overrightarrow{v}, + \hspace*{2mm} \textrm{ if } C,B\div A, + \end{eqnarray*} + which is what needed to be proven. \kdokaz + +If we combine the operation of adding vectors and multiplying a vector by a scalar, we get what is called a linear combination of vectors. More precisely - for any $n$-tuple of vectors $(\overrightarrow{a_1},\overrightarrow{a_2},\ldots,\overrightarrow{a_n})$ and any $n$-tuple of real numbers $(\alpha_1,\alpha_2,\ldots,\alpha_n)\in \mathbb{R}^n$ the vector + $$\overrightarrow{v}=\alpha_1\cdot \overrightarrow{a_1}+ + \alpha_2\cdot \overrightarrow{a_2} + +\cdots +\alpha_n\cdot \overrightarrow{a_n}$$ + \index{linearna kombinacija vektorjev}\pojem{linearna kombinacija vektorjev} $\overrightarrow{a_1}$, $\overrightarrow{a_2}$, $\ldots$, $\overrightarrow{a_n}$. + + + + \bizrek \label{vektLinKombNicLema} + Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two non-collinear non-zero vectors + and $(\alpha,\beta)\in \mathbb{R}^2$ two real numbers. + If $\alpha\cdot\overrightarrow{a}= + \beta\cdot\overrightarrow{b}$, then $\alpha=\beta=0$. + \eizrek + + \textbf{\textit{Proof.}} We assume the opposite - without loss of generality, that $\alpha\neq0$. Then $\overrightarrow{a}= + \frac{\beta}{\alpha}\cdot\overrightarrow{b}$, which means (statement \ref{vektKriterijKolin}), that the vector $\overrightarrow{b}$ is collinear with the vector $\overrightarrow{a}$. This is in contradiction with the assumption, so $\alpha=\beta=0$. +\kdokaz + +A direct consequence of the previous statement is the following claim. + + + + \bizrek \label{vektLinKombNic} + If for the linear combination of two non-collinear non-zero vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ holds + $$\alpha\cdot\overrightarrow{a}+ + \beta\cdot\overrightarrow{b}=\overrightarrow{0},$$ + then $\alpha=\beta=0$. + \eizrek + +\textbf{\textit{Proof.}} If we write the relation $\alpha\cdot\overrightarrow{a}+ + \beta\cdot\overrightarrow{b}=\overrightarrow{0}$ in the form $\alpha\cdot\overrightarrow{a}= + -\beta\cdot\overrightarrow{b}$, we see that the statement is a direct consequence of the previous expression. +\kdokaz + + The following expression is very important, which refers to the representation of any vector as a linear combination of two nonlinear non-zero vectors. + + + \bizrek \label{vektLinKomb1Razcep} + Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two non-collinear non-zero vectors in the same plane. + Each vector $\overrightarrow{v}$ in this plane can be express in a single way as a linear combination + of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, i.e. there is exactly + one pair of real numbers $(\alpha,\beta)\in \mathbb{R}^2$, such that + $$\overrightarrow{v}=\alpha\cdot\overrightarrow{a}+ + \beta\cdot\overrightarrow{b}.$$ + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.3.pic} +\caption{} \label{sl.vek.5.2.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.3.pic}) + + Let $O$ be an arbitrary point and $A$, $B$ and $V$ such points that $\overrightarrow{OA}=\overrightarrow{a}$, $\overrightarrow{OB}=\overrightarrow{b}$ and $\overrightarrow{OV}=\overrightarrow{v}$ (expression \ref{vektAvObst1TockaB}). We mark with $A'$ the intersection of the line $OA$ and the parallel of the line $OB$ through the point $V$ and with $B'$ the intersection of the line $OB$ and the parallel of the line $OA$ through the point $V$. So the quadrilateral $OA'VB'$ is a parallelogram. By the expression \ref{vektKriterijKolin} there exist such $\alpha,\beta\in \mathbb{R}$, that $\overrightarrow{OA'}= + \alpha\cdot\overrightarrow{OA}= + \alpha\cdot\overrightarrow{a}$ and + $\overrightarrow{OB'}= + \beta\cdot\overrightarrow{OB}= + \beta\cdot\overrightarrow{b}$. By the parallelogram rule it is: + + $$\overrightarrow{v}= + \overrightarrow{OV}=\overrightarrow{OA'}+\overrightarrow{OB'}= + \alpha\cdot\overrightarrow{a}+\beta\cdot\overrightarrow{b}.$$ + +We will prove that $(\alpha,\beta)\in \mathbb{R}^2$ is the only such pair. We assume that for +$(\alpha',\beta')\in \mathbb{R}^2$ it is true that $\overrightarrow{v}= + \alpha'\cdot\overrightarrow{a}+\beta'\cdot\overrightarrow{b}$. From + $\overrightarrow{v}= + \alpha'\cdot\overrightarrow{a}+\beta'\cdot\overrightarrow{b}= + \alpha\cdot\overrightarrow{a}+\beta\cdot\overrightarrow{b}$ it follows that + $(\alpha'-\alpha)\cdot\overrightarrow{a}+(\beta'-\beta)\cdot\overrightarrow{b}= +\overrightarrow{0}$. According to the previous statement \ref{vektLinKombNic} we have that + $\alpha'-\alpha=\beta'-\beta=0$ which means $\alpha'=\alpha$ and $\beta'=\beta$. +\kdokaz + +\bizrek \label{vektVektorskiProstor} + The set $\mathcal{V}$ of all vectors in the plane form so-called + \index{vector space} vector space \footnote{ The first one to, in a modern way, define the concept of a vector space in 1888 was the Italian mathematician \index{Peano, G.}\textit{G. Peano} (1858–-1932).} over the field $\mathbb{R}$, which means that: + \begin{enumerate} + \item the ordered pair $(\mathcal{V},+)$ form a commutative group, + \item $(\forall \alpha \in \mathbb{R})(\forall \overrightarrow{v}\in \mathcal{V})\hspace*{1mm} \alpha\cdot \overrightarrow{v} \in \mathcal{V}$, + \item $(\forall \alpha,\beta \in \mathbb{R})(\forall \overrightarrow{v}\in \mathcal{V})\hspace*{1mm} \alpha\cdot (\beta\cdot\overrightarrow{v})=(\alpha\beta)\cdot \overrightarrow{v}$, + \item $(\forall \overrightarrow{v}\in \mathcal{V})\hspace*{1mm} 1\cdot \overrightarrow{v}=\overrightarrow{v}$, + \item $(\forall \alpha,\beta \in \mathbb{R})(\forall \overrightarrow{v}\in \mathcal{V})\hspace*{1mm} (\alpha+ \beta)\cdot\overrightarrow{v}=\alpha\cdot \overrightarrow{v}+\beta\cdot \overrightarrow{v}$, + \item $(\forall \alpha \in \mathbb{R})(\forall \overrightarrow{v},\overrightarrow{u}\in \mathcal{V})\hspace*{1mm} \alpha\cdot(\overrightarrow{v}+\overrightarrow{u})= + \alpha\cdot \overrightarrow{v}+\alpha\cdot \overrightarrow{u}$. + \end{enumerate} + \eizrek + + + + \textbf{\textit{Proof.}} The claim $\textit{1}$ is, in a sense, already proven in \ref{vektKomGrupa}. The claims $\textit{2}-\textit{5}$ are a direct consequence of the definition of multiplication of a vector with a scalar. The proof of claim $\textit{6}$ is not as simple - in the proof we would have to use Dedekind's axiom of continuity \ref{aksDed}. + \kdokaz + +We will use the notation $\mathcal{\overrightarrow{V}}^2$ for the vector space structure on the set of all vectors of the plane $\mathcal{V}$ over the field $\mathbb{R}$. If we were to consider vectors in Euclidean space, we would also get a vector space. But we can also consider a vector space more abstractly as an arbitrary ordered pair $(\mathcal{\overrightarrow{V}},\mathcal{F})$, because $(\mathcal{\overrightarrow{V}},+)$ represents a commutative group, $(\mathcal{F},+,\cdot)$ is a field, and all conditions $\textit{2}-\textit{6}$ from the previous theorem are satisfied. + +Because every vector in the plane can be expressed as a linear combination of two non-zero vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ of that plane, we say that such two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are called the \index{baza vektorskega prostora}\pojem{baza} (base) of the vector space $\mathcal{\overrightarrow{V}}^2$. If for some vector $\overrightarrow{v}\in \mathcal{V}$ it holds that $\overrightarrow{v}=\alpha\cdot\overrightarrow{a}+ + \beta\cdot\overrightarrow{b}$, we say that the pair $(\alpha,\beta)$ represents the \index{koordinate vektorja}\pojem{koordinate} (coordinates) of the vector $\overrightarrow{v}$ in the base $(\overrightarrow{a},\overrightarrow{b})$. From theorem \ref{vektLinKomb1Razcep} it follows that coordinates of every vector are uniquely determined in any base. + + + +Therefore, there are infinitely many bases in the vector space $\mathcal{\overrightarrow{V}}^2$, the number of vectors in any base is always 2, so we say that the vector space $\mathcal{\overrightarrow{V}}^2$ has \pojem{dimension} 2 or is \pojem{two-dimensional}. + +It is clear that the dimension of the vector space $\mathcal{\overrightarrow{V}}^3$, which is determined by the vectors in Euclidean space, is equal to 3. Euclidean space is therefore \pojem{three-dimensional}, each vector in some base is determined by a triplet of real numbers, which represent the coordinates of that vector. + +In any vector space we determine the base and dimension in a similar way. Because in general the dimension of space is not limited to 3, this allows us to explore \pojem{$n$-dimensional} Euclidean spaces (for any $n\in \mathbb{N}$). + +The following examples are very useful. + +\bzgled \label{vektSredOSOAOB} + Let $O$, $A$ and $B$ be arbitrary points and $S$ the midpoint of the line segment $AB$. Then + $$\overrightarrow{OS}=\frac{1}{2}\cdot\left(\overrightarrow{OA}+ + \overrightarrow{OB}\right).$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.4.pic} +\caption{} \label{sl.vek.5.2.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.4.pic}) + + By \ref{vektSredDalj} we have $\overrightarrow{SA}=-\overrightarrow{SB}$ or $\overrightarrow{AS}=-\overrightarrow{BS}$. By the triangle rule for vector addition we have + $\overrightarrow{OS}=\overrightarrow{OA}+\overrightarrow{AS}$ and + $\overrightarrow{OS}=\overrightarrow{OB}+\overrightarrow{BS}$. If we add the equalities and take into account $\overrightarrow{AS}=-\overrightarrow{BS}$, we get: $2\cdot\overrightarrow{OS}=\overrightarrow{OA}+\overrightarrow{AS} + +\overrightarrow{OB}+\overrightarrow{BS}=\overrightarrow{OA} + +\overrightarrow{OB}$ + or +$\overrightarrow{OS}=\frac{1}{2}\cdot\left(\overrightarrow{OA}+ + \overrightarrow{OB}\right)$. +\kdokaz + + \bzgled \label{vektDelitDaljice} + Let $O$, $A$ and $B$ be arbitrary points and $P$ a point in the line segment $AB$ such that $|AP|:|PB|=n:m$. Then + $$\overrightarrow{OP}=\frac{1}{n+m}\left(m\cdot\overrightarrow{OA}+ + n\cdot\overrightarrow{OB}\right).$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.5.pic} +\caption{} \label{sl.vek.5.2.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.5.pic}) + +First, from $|AP|:|PB|=n:m$ it follows that $|AP|=\frac{n}{n+m}\cdot|AB|$ and $|BP|=\frac{m}{n+m}\cdot|AB|$ or $\overrightarrow{AP}=\frac{n}{n+m}\cdot\overrightarrow{AB}$ and $\overrightarrow{BP}=\frac{m}{n+m}\cdot\overrightarrow{BA}$ (because $\mathcal{B}(A,P,B)$). +Then: +\begin{eqnarray*} + \overrightarrow{OP}&=&\overrightarrow{OA}+\overrightarrow{AP}=\overrightarrow{OA} + +\frac{n}{n+m}\cdot\overrightarrow{AB}\hspace*{3mm} \textrm{ in}\\ + \overrightarrow{OP}&=&\overrightarrow{OB}+\overrightarrow{BP}=\overrightarrow{OB} + +\frac{m}{n+m}\cdot\overrightarrow{BA}. +\end{eqnarray*} + If we multiply the first equality by $m$, the second by $n$ and then add them, we get: +\begin{eqnarray*} +(n+m)\cdot\overrightarrow{OP}&=&m\cdot\overrightarrow{OA} + +n\cdot\overrightarrow{OB}+\frac{nm}{n+m}\cdot\left(\overrightarrow{AB}+ + \overrightarrow{BA}\right)=\\ + &=&m\cdot\overrightarrow{OA} + +n\cdot\overrightarrow{OB}. +\end{eqnarray*} + If we divide the obtained equality by $n+m$, we get the desired relation. +\kdokaz + + + \bzgled \label{vektParamDaljica} + Let $O$, $A$ and $B$ be arbitrary points. A point $X$ lies on the line segment $AB$ if and only if for some scalar $0\leq\lambda\leq1$ is + $$\overrightarrow{OX}=(1-\lambda)\cdot\overrightarrow{OA}+ + \lambda\cdot\overrightarrow{OB}.$$ + \ezgled + + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.6.pic}) + + We first assume that the point $X$ lies on the line $AB$. Then for some $0\leq\lambda\leq1$ it holds that $\overrightarrow{AX}=\lambda\cdot \overrightarrow{AB}$. + + \begin{eqnarray*} + \overrightarrow{OX}&=&\overrightarrow{OA}+\overrightarrow{AX}=\\ + &=& \overrightarrow{OA}+\lambda\cdot \overrightarrow{AB}=\\ + &=& \overrightarrow{OA}+\lambda\cdot \left(\overrightarrow{OB}-\overrightarrow{OA}\right)=\\ + &=& (1-\lambda)\cdot\overrightarrow{OA}+ + \lambda\cdot\overrightarrow{OB}. +\end{eqnarray*} + +Assume now that for some $0\leq\lambda\leq1$ we have + $\overrightarrow{OX}=(1-\lambda)\cdot\overrightarrow{OA}+ + \lambda\cdot\overrightarrow{OB}$. Then: + + \begin{eqnarray*} + \overrightarrow{AX}&=&\overrightarrow{AO}+\overrightarrow{OX}=\\ + &=&\overrightarrow{AO}+(1-\lambda)\cdot\overrightarrow{OA}+ + \lambda\cdot\overrightarrow{OB}=\\ + &=&-\lambda\cdot\overrightarrow{OA}+ + \lambda\cdot\overrightarrow{OB}=\\ + &=&\lambda\cdot\left(\overrightarrow{AO}+ + \overrightarrow{OB}\right)=\\ + &=&\lambda\cdot\overrightarrow{AB}. +\end{eqnarray*} +Therefore, for some $0\leq\lambda\leq1$ we have $\overrightarrow{AX}=\lambda\cdot \overrightarrow{AB}$, + which means that the point $X$ lies on the line $AB$. +\kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.6.pic} +\caption{} \label{sl.vek.5.2.6.pic} +\end{figure} + +A direct consequence is the following theorem. + + $O$, $A$ in $B$ . A point $X$ lies on the line segment $AB$ if and only if for some + + \bzgled + Let $O$, $A$ and $B$ be arbitrary points. A point $X$ lies on the line segment $AB$ if and only if for some scalars $\alpha, \beta\in [0,1]$ ($\alpha+\beta=1$) is + $$\overrightarrow{OX}=\alpha\cdot\overrightarrow{OA}+ + \beta\cdot\overrightarrow{OB}.$$ + \ezgled + +In the next theorem we will give the \index{vector equation of a line} \pojem{vector equation of a line}. + + \bzgled \label{vektParamPremica} + Let $O$, $A$ in $B$ be arbitrary points. A point $X$ lies on the line $AB$ if and only if for some scalar $\lambda\in \mathbb{R}$ is + $$\overrightarrow{OX}=(1-\lambda)\cdot\overrightarrow{OA}+ + \lambda\cdot\overrightarrow{OB}.$$ + \ezgled + + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.6.pic}) + +The proof is the same as in the previous statement, only that we use the fact that the point $X$ lies on the line $AB$ exactly when the vectors $\overrightarrow{AX}$ and $\overrightarrow{AB}$ are collinear, or for some +$\lambda\in \mathbb{R}$ it holds that $\overrightarrow{AX}=\lambda\cdot \overrightarrow{AB}$ (statement \ref{vektKriterijKolin}). +\kdokaz + +\index{količnik kolinearnih vektorjev}\index{razmerje!kolinearnih vektorjev} +If for collinear vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ it holds that $\overrightarrow{v}=\lambda \overrightarrow{u}$ ($\lambda\in\mathbb{R}$) and $\overrightarrow{u}\neq \overrightarrow{0}$, then we can define the \pojem{količnik kolinearnih vektorjev} or the \pojem{razmerje kolinearnih vektorjev} (Figure \ref{sl.vek.5.2.7.pic}): +$$\overrightarrow{v}:\overrightarrow{u} +=\frac{\overrightarrow{v}}{\overrightarrow{u}}=\lambda.$$ + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.7.pic} +\caption{} \label{sl.vek.5.2.7.pic} +\end{figure} + +From the definition of multiplication of a vector with a real number we get the following statement. + + \bizrek \label{vektKolicnDolz} + Let $\overrightarrow{v}$ and $\overrightarrow{u}$ be collinear vectors and $\overrightarrow{u}\neq \overrightarrow{0}$. Then + $$\frac{\overrightarrow{v}}{\overrightarrow{u}}= + \left\{ \begin{array}{lll} + \frac{|\overrightarrow{v}|}{|\overrightarrow{u}|}, & + \textrm{ if } \overrightarrow{u}\rightrightarrows \overrightarrow{v}\\ + -\frac{|\overrightarrow{v}|}{|\overrightarrow{u}|}, & + \textrm{ if } \overrightarrow{u}\rightleftarrows\overrightarrow{v}\\ + 0, & \textrm{ if } \overrightarrow{v}=\overrightarrow{0} + \end{array}\right.$$ + \eizrek + + +The following statement also refers to the newly defined concept. + +\bizrek + \label{izrekEnaDelitevDaljiceVekt} + For any line segment $AB$ and any $\lambda\in\mathbb{R}\setminus\{-1\}$ there exists + exactly one such point $P$ on the line $AB$, that is + $$\frac{\overrightarrow{AP}}{\overrightarrow{PB}}=\lambda.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.8.pic} +\caption{} \label{sl.vek.5.2.8.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.8.pic}) + + Let $P$ be such a point that $$\overrightarrow{AP}=\frac{\lambda}{1+\lambda}\cdot\overrightarrow{AB}.$$ +Since $\lambda\neq-1$, such a point always exists. Because $$\overrightarrow{PB}=\overrightarrow{PA}+\overrightarrow{AB}= +-\frac{\lambda}{1+\lambda}\cdot\overrightarrow{AB}+\overrightarrow{AB}= +\frac{1}{1+\lambda}\overrightarrow{AB},$$ +it also holds that $\frac{\overrightarrow{AP}}{\overrightarrow{PB}}=\lambda$. + +If for another point $P'$ it holds that $\frac{\overrightarrow{AP'}}{\overrightarrow{P'B}}=\lambda$, from +$\overrightarrow{AP'}=\lambda\cdot\overrightarrow{P'B}$ and +$\overrightarrow{P'B}=\overrightarrow{P'A}+\overrightarrow{AB}$ we get $\overrightarrow{AP'}=\frac{\lambda}{1+\lambda}\cdot\overrightarrow{AB}$. Therefore + $\overrightarrow{AP'}=\overrightarrow{AP}$ or + $\overrightarrow{P'P}=\overrightarrow{P'A}+\overrightarrow{AP}= + -\overrightarrow{AP}+\overrightarrow{AP}=\overrightarrow{0}$. So $P'=P$, which means that there is only one point $P$, for which $\frac{\overrightarrow{AP}}{\overrightarrow{PB}}=\lambda$. + \kdokaz + + We say that the point $P$ from the previous statement divides the line segment $AB$ in the \pojem{ratio} $\lambda$. + + + + \bzgled + Let $ABCD$ be a trapezium with the base $AB$. Calculate the ratio in which the line + segment $PD$ divides the diagonal $AC$, if $|AB| = 3\cdot |CD|$ and $P$ is the midpoint of the line segment $AB$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.9.pic} +\caption{} \label{sl.vek.5.2.9.pic} +\end{figure} + + \textbf{\textit{Solution.}} The intersection of the line $DP$ and the diagonal $AC$ is denoted by $S$ (Figure \ref{sl.vek.5.2.9.pic}) +and let + $\overrightarrow{u}=\overrightarrow{AP}$ and $\overrightarrow{v}=\overrightarrow{AD}$. +The vector $\overrightarrow{AS}$ and $\overrightarrow{AC}$ are collinear, so by \ref{vektKriterijKolin} $\overrightarrow{AS}=\lambda\overrightarrow{AC}$, for some $\lambda\in \mathbb{R}$. The vectors $\overrightarrow{PS}$ and $\overrightarrow{PD}$ are also collinear, i.e. for some $\mu\in \mathbb{R}$ we have $\overrightarrow{PS}=\mu\overrightarrow{PD}$. The vector $\overrightarrow{AC}$ and $\overrightarrow{AS}$ are written as a linear combination of the vectors $\overrightarrow{u}$ and $\overrightarrow{v}$: + \begin{eqnarray*} +\hspace*{-1.8mm} \overrightarrow{AC}&=&\overrightarrow{AD}+\overrightarrow{DC} + =\overrightarrow{v}+\frac{2}{3}\overrightarrow{u} + =\frac{2}{3}\overrightarrow{u}+\overrightarrow{v},\\ + \hspace*{-1.8mm}\overrightarrow{AS}&=&\overrightarrow{AP}+\overrightarrow{PS} + =\overrightarrow{u}+\mu\overrightarrow{PD} + =\overrightarrow{u}+\mu(-\overrightarrow{u}+\overrightarrow{v}) + =(1-\mu)\overrightarrow{u}+\mu\overrightarrow{v}. + \end{eqnarray*} +Since $\overrightarrow{AS}=\lambda\overrightarrow{AC}$, we get: +\begin{eqnarray*} + \overrightarrow{AS}&=& + \frac{2}{3}\lambda\overrightarrow{u}+\lambda\overrightarrow{v};\\ + \overrightarrow{AS}&=& + (1-\mu)\overrightarrow{u}+\mu\overrightarrow{v}. + \end{eqnarray*} + Because $\overrightarrow{u}$ and $\overrightarrow{v}$ are collinear vectors, by \ref{vektLinKomb1Razcep} $\frac{2}{3}\lambda=1-\mu$ and $\lambda=\mu$. If we solve this simple system of equations, we get $\lambda=\mu=\frac{3}{5}$. Therefore $\overrightarrow{AS}=\frac{3}{5}\overrightarrow{AC}$, i.e. $AS:SC=3:2$. + \kdokaz + +%________________________________________________________________________________ +\poglavje{Vector Length} \label{odd5DolzVekt} + +We have already defined the length of a vector in section \ref{odd5DefVekt}. From the definition of multiplication of a vector with a scalar in the previous section \ref{odd5LinKombVekt} we saw that for every vector $\overrightarrow{v}$ and every real number $\lambda$ it holds that $|\lambda \cdot \overrightarrow{v}|=|\lambda| \cdot |\overrightarrow{v}|$. In this section we will discuss some more properties of the length of a vector. + +First, from the definition of addition of vectors and the triangle inequality we get the following claim. + + \bizrek \label{neenakTrikVekt} + For any two vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ it holds that + $$|\overrightarrow{v}+\overrightarrow{u}|\leq |\overrightarrow{v}|+|\overrightarrow{u}|.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.7.pic} +\caption{} \label{sl.vek.5.1.7.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.1.7.pic}) + + Let $A$ be an arbitrary point and let $B$ and $C$ be such points that $\overrightarrow{AB}=\overrightarrow{v}$ and $\overrightarrow{BC}=\overrightarrow{u}$ (claim \ref{vektAvObst1TockaB}). From the definition of addition of vectors and the length of a vector and the triangle inequality (claim \ref{neenaktrik}) we get: + \begin{eqnarray*} + |\overrightarrow{v}+\overrightarrow{u}|&=&|\overrightarrow{AB}+\overrightarrow{BC}|=\\ + &=&|\overrightarrow{AC}|=|AC|\leq + |AB|+|AC|= + |\overrightarrow{AB}|+|\overrightarrow{AC}|=\\ + &=&|\overrightarrow{v}|+|\overrightarrow{u}|, + \end{eqnarray*} + which was to be proven. \kdokaz + +In a similar way as we can define a fraction, we can also define the product of collinear vectors. If $\overrightarrow{v}$ and $\overrightarrow{u}$ are two collinear vectors, we define the operation \pojem{multiplication of two collinear vectors}\footnote{This is a special case of the so-called \textit{dot product}, which in linear algebra is defined for any two vectors. In Euclidean space, $\overrightarrow{v}\cdot\overrightarrow{u}= +|\overrightarrow{v}|\cdot|\overrightarrow{u}|\cdot\cos +\angle \overrightarrow{v},\overrightarrow{u}$.}: + \begin{eqnarray*} + \overrightarrow{v}\cdot \overrightarrow{u}= +\left\{ + \begin{array}{ll} + |\overrightarrow{v}|\cdot|\overrightarrow{u}|, & +\overrightarrow{v},\overrightarrow{u}\rightrightarrows ; \\ + -|\overrightarrow{v}|\cdot|\overrightarrow{u}|, & \overrightarrow{v},\overrightarrow{u}\rightleftarrows. + \end{array} +\right. +\end{eqnarray*} + The result of this operation, which is a real number, is called \index{product of collinear vectors} \pojem{product of collinear vectors}. +It is clear from the definition that: + \begin{eqnarray} \label{eqnMnozVektDolzina} + \overrightarrow{v}\cdot \overrightarrow{v}=|\overrightarrow{v}|^2 + \end{eqnarray} + It is also clear that for three collinear points $A$, $B$ and $L$ the following equivalence holds (Figure \ref{sl.vek.5.3.1.pic}): + + \begin{eqnarray} \label{eqnMnozVektRelacijaB} + \overrightarrow{LA}\cdot \overrightarrow{LB}<0 +\hspace*{1mm}\Leftrightarrow\hspace*{1mm} \mathcal{B}(A,L,B) +\hspace*{3mm} \textrm{(}A,B,L \textrm{ are collinear)}. + \end{eqnarray} + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.3.1.pic} +\caption{} \label{sl.vek.5.3.1.pic} +\end{figure} + +\bnaloga\footnote{15. IMO USSR - 1973, Problem 1.} +Point $O$ lies on line $g$; + $\overrightarrow{OP_1},\overrightarrow{OP_2}, + \cdots,\overrightarrow{OP_n}$ +are unit vectors such that points $P_1, P_2, \cdots, P_n$ +all lie in a plane containing $g$ and on one side of $g$. Prove that +if $n$ is odd, + $$|\overrightarrow{OP_1}+\overrightarrow{OP_2}+ + \cdots+\overrightarrow{OP_n}|\geq 1.$$ \label{OlimpVekt15} + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.3.IMO1.pic} +\caption{} \label{sl.vek.5.3.IMO1.pic} +\end{figure} + + \textbf{\textit{Solution.}} Let $k$ be a unit circle with center + at point $O$, which intersects the line $g$ at points $A$ and $B$. All + points $P_1, P_2, \cdots, P_n$ lie on the corresponding + arc, which on the circle $k$ is determined by points $A$ and $B$. + Without loss of generality + we can assume that the vectors are labeled so that it holds: + $\angle BOP_1\leq \angle BOP_2\leq\cdots\leq \angle BOP_n$ + (Figure \ref{sl.vek.5.3.IMO1.pic}). + Let $s$ be the bisector of angle $\angle P_1OP_n$ and $g'$ + the perpendicular of the line $s$ at point $O$. Because $\angle P_1OP_n$ is a convex + angle, the whole lies on the same side of the line $g'$. So are also all + points $P_i$ ($i\in \{1,2,\cdots,n\}$ on the same side of the line $g'$, + because they lie in this angle. So we can sharpen our assumption from the task, that all points $P_1, P_2, \cdots, P_n$ are on the same side of some line + $g$ or $g'$, + with the requirement (which follows from this assumption), + that the first and last vector of the sequence + are symmetric with respect to the perpendicular $s$ of the line $g'$ through point $O$. We will use this fact in the proof. + + By assumption $n$ is an odd number. Let $n=2k-1$, $k\in \mathbb{N}$. + We will prove by induction on $k$. + + (\textit{i}) If $k=1$ or $n=1$, then $|\overrightarrow{OP_1}|=1$ + or $|\overrightarrow{OP_1}|\geq1$ and the statement + is fulfilled. + +(\textit{ii}) Let's assume that the statement is true for $k=l$ or for every sequence + $n=2l-1$ of vectors. We will prove that the statement is also true for $k=l+1$ or in + the case of a sequence $n=2l+1$ of vectors. Let + $\overrightarrow{OP_1},\overrightarrow{OP_2}, + \cdots,\overrightarrow{OP_{2l+1}}$ + be unit vectors, where $P_1, P_2, \cdots, P_{2l+1}$ + are points that + lie in the same plane with the edge $g'$ (in the plane $\alpha$) + and the vectors $\overrightarrow{OP_1}$ + and $\overrightarrow{OP_{2l+1}}$ are symmetrical with respect + to the line $s$. From this it follows that the vector + $\overrightarrow{OU}=\overrightarrow{OP_1}+\overrightarrow{OP_{2l+1}}$ + lies on the line $s$ or $U\in s$. Let + $\overrightarrow{OV}=\overrightarrow{OP_2}+\overrightarrow{OP_3}+ + \cdots+\overrightarrow{OP_{2l}}$. By the induction assumption + $|\overrightarrow{OV}|=|\overrightarrow{OP_2}+\overrightarrow{OP_3}+ + \cdots+\overrightarrow{OP_{2l}}|\geq 1$. Therefore: + \begin{eqnarray*} + |\overrightarrow{OP_1}+\overbrace{\overrightarrow{OP_2}+ + \cdots+\overrightarrow{OP_{2l}}}+\overrightarrow{OP_{2l+1}}|= + |\overrightarrow{OU}+\overrightarrow{OV}|. + \end{eqnarray*} + Let + $\overrightarrow{OW}=\overrightarrow{OU}+\overrightarrow{OV}$. The quadrilateral $VOUW$ is a parallelogram (or the points $O$, $U$, $V$ and $W$ + are collinear). + By the example \ref{vektOPi} the points $U$ and $V$ are in the plane $\alpha$. The point + $U$ lies on the line $s$, which is the line of symmetry of the obtuse angle, which + is determined by the line $g'$ and the point $O$. Therefore $0\leq\angle + UOV\leq 90^0$. From the parallelogram $VOUW$ it then follows that $\angle OVW > + 90^0$. From the triangle inequality \ref{neenaktrik} (for the triangle + $OVW$) it follows that $|\overrightarrow{OW}|>|\overrightarrow{OV}|$. + The inequality is also valid in the case when the points $O$, $U$, $V$ and $W$ + are collinear. Now we have: + \begin{eqnarray*} + |\overrightarrow{OP_1}+\overrightarrow{OP_2}+ + \cdots+\overrightarrow{OP_{2l}}+\overrightarrow{OP_{2l+1}}|= + |\overrightarrow{OU}+\overrightarrow{OV}|= + |\overrightarrow{OW}|>|\overrightarrow{OV}|\geq 1, + \end{eqnarray*} + which had to be proven. \kdokaz + +%________________________________________________________________________________ + \poglavje{Further Use of Vectors} \label{odd5UporabVekt} + + First, let's look at a statement that + is a direct consequence of the definition of vectors (section \ref{odd5DefVekt}). + + \bizrek \label{vektParalelogram} + A quadrilateral $ABCD$ is a parallelogram if and only if + $\overrightarrow{AB}=\overrightarrow{DC}$. + \eizrek + + + We should also mention that we can already +prove the property of the median of a triangle - statement \ref{srednjicaTrik} - +now in the following form. + + \bizrek + \label{srednjicaTrikVekt} + Let $PQ$ be the midsegment of a triangle $ABC$, corresponding + to the side + $BC$. Then + $$ \overrightarrow{PQ} = \frac{1}{2} \overrightarrow{AB}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.4.1c.pic} +\caption{} \label{sl.vek.5.4.1c.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.4.1c.pic}) + The statement is a direct consequence of statement \ref{srednjicaTrik} + \kdokaz + +We have seen that we can add vectors according to the parallelogram rule +(when they have a common starting point) or according to the triangle rule (when the +starting point of the second one is at the end of the first one). Now we will introduce another +rule for adding two vectors that are in any position. This is called the \index{pravilo!splošno za seštevanje vektorjev}\pojem{general vector addition rule}. + + \bizrek \label{vektSestSplosno} + If $S_1$ and $S_2$ + are the midpoints of line segments $A_1B_1$ and $A_2B_2$, then + $$\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}= + 2\overrightarrow{S_1S_2}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.4.1b.pic} +\caption{} \label{sl.vek.5.4.1b.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.4.1b.pic}) + +Since $S_1$ and $S_2$ + are the midpoints of the lines $A_1B_1$ and $A_2B_2$, + $\overrightarrow{S_1A_1}=-\overrightarrow{S_1B_1}$ and $\overrightarrow{S_2A_2}=-\overrightarrow{S_2B_2}$ or $\overrightarrow{S_1A_1}+\overrightarrow{S_1B_1}=\overrightarrow{0}$ and $\overrightarrow{S_2A_2}+\overrightarrow{S_2B_2}=\overrightarrow{0}$. + + If we decompose the vector $\overrightarrow{S_1S_2}$ in two ways using the polygonal rule for adding vectors, we first get: + \begin{eqnarray*} + \overrightarrow{S_1S_2}&=&\overrightarrow{S_1A_1}+\overrightarrow{A_1A_2}+ + \overrightarrow{A_2S_2}\\ + \overrightarrow{S_1S_2}&=&\overrightarrow{S_1B_1}+\overrightarrow{B_1B_2}+ + \overrightarrow{B_2S_2}, + \end{eqnarray*} + then by adding these: + \begin{eqnarray*} + 2\cdot\overrightarrow{S_1S_2}&=&\overrightarrow{S_1A_1}+\overrightarrow{A_1A_2}+ + \overrightarrow{A_2S_2}+\\ + &+&\overrightarrow{S_1B_1}+\overrightarrow{B_1B_2}+ + \overrightarrow{B_2S_2}=\\ + &=& \overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}, + \end{eqnarray*} + which had to be proven. \kdokaz + +We will often use the relation from the previous statement in +the form: + $$\overrightarrow{S_1S_2}= + \frac{1}{2}(\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}),$$ + which is a generalization of the property of the median of a trapezoid and a triangle. + + + + At this point, we will repeat all three rules for adding two vectors. In the first case, we therefore choose representatives of the vectors so that the beginning of the second is at the end of the first, in the second case the vectors have a common starting point, in the third case the representatives are in general position (Figure \ref{sl.vek.5.4.1a.pic}): + +\begin{itemize} +\item \textit{for every three points $A$, $B$ and $C$ it holds that $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ (the triangle rule),} +\item \textit{for every three non-collinear points $A$, $B$ and $C$ it holds that $\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AD}$ exactly when $ABDC$ is a parallelogram (the parallelogram rule),} +\item \textit{if $S_1$ and $S_2$ are the midpoints of the line segments $A_1B_1$ and $A_2B_2$, then +$\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}= +2\overrightarrow{S_1S_2}$ (the general rule).} +\end{itemize} + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.4.1a.pic} +\caption{} \label{sl.vek.5.4.1a.pic} +\end{figure} + +Except for these three, we also have the polygon rule, which is related to the addition of multiple vectors (Figure \ref{sl.vek.5.1.12.pic}): +\begin{itemize} +\item \textit{$\overrightarrow{A_1A_2}+\overrightarrow{A_2A_3}+\cdots +\overrightarrow{A_{n-1}A_n}=\overrightarrow{A_1A_n}$ (the polygon rule).} +\end{itemize} + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.12.pic} +\caption{} \label{sl.vek.5.1.12.pic} +\end{figure} + +\bzgled \label{vektPetkoinikZgled} +Points $M$, $N$, $P$ and $Q$ are the midpoints of the sides $AB$, $BC$, $CD$ and $DE$ of a pentagon $ABCDE$. Prove +that the line segment $XY$, determined by the midpoints of the line segments $MP$ and $NQ$, is parallel to the line $AE$ +and calculate $\frac{\overrightarrow{XY}}{\overrightarrow{AE}}$. +\ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.4.2.pic} +\caption{} \label{sl.vek.5.4.2.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.4.2.pic}) + If we use the expressions \ref{vektSestSplosno} and \ref{srednjicaTrikVekt} we get: + \begin{eqnarray*} + \overrightarrow{XY}&=&\frac{1}{2}\left(\overrightarrow{MQ}+\overrightarrow{PN} \right)=\\ + &=&\frac{1}{2}\left(\frac{1}{2}\left(\overrightarrow{AE}+\overrightarrow{BD}\right) + +\frac{1}{2}\overrightarrow{DB} \right)=\\ + &=&\frac{1}{4}\overrightarrow{AE} + \end{eqnarray*} + So the vectors $\overrightarrow{XY}$ and $\overrightarrow{AE}$ are collinear and $\frac{\overrightarrow{XY}}{\overrightarrow{AE}}=\frac{1}{4}$. + \kdokaz + + + \bzgled + Let $O$ be an arbitrary point in the plane of a triangle $ABC$ and + $D$ and $E$ points of the sides $AB$ and $BC$ such that + $$\frac{\overrightarrow{AD}}{\overrightarrow{DB}}= + \frac{\overrightarrow{BE}}{\overrightarrow{EC}}=\frac{m}{n}.$$ + Let $F$ be the intersection of the line segments $AE$ and $CD$. Express $\overrightarrow{OF}$ as + a function of $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$, $m$ and $n$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.4.3.pic} +\caption{} \label{sl.vek.5.4.3.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.4.3.pic}) + +It is enough to express the vector $\overrightarrow{BF}$ as a linear combination of the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$, because: +\begin{eqnarray*} + \overrightarrow{OF}&=&\overrightarrow{OB}+\overrightarrow{BF}\\ + \overrightarrow{BA}&=&\overrightarrow{OA}-\overrightarrow{OB}\\ + \overrightarrow{BC}&=&\overrightarrow{OC}-\overrightarrow{OB}. + \end{eqnarray*} +By \ref{vektParamPremica} we have: +\begin{eqnarray*} + \overrightarrow{BF}&=&\lambda\overrightarrow{BD}+(1-\lambda)\overrightarrow{BC}= + \lambda\frac{n}{n+m}\overrightarrow{BA}+(1-\lambda)\overrightarrow{BC};\\ + \overrightarrow{BF}&=&\mu\overrightarrow{BA}+(1-\mu)\overrightarrow{BE}= + \mu\overrightarrow{BA}+(1-\mu)\frac{m}{n+m}\overrightarrow{BC} + \end{eqnarray*} +for some numbers $\lambda,\mu\in \mathbb{R}$. Because the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$ are nonlinear, by \ref{vektLinKomb1Razcep} we obtain the system: +\begin{eqnarray*} + & & \lambda\frac{n}{n+m}=\mu\\ + & & 1-\lambda=(1-\mu)\frac{m}{n+m}, + \end{eqnarray*} + which we solve for $\lambda$ and $\mu$ in terms of $m$ and $n$. It is enough to calculate only $\lambda$ and plug it in $\overrightarrow{BF}= + \lambda\frac{n}{n+m}\overrightarrow{BA}+(1-\lambda)\overrightarrow{BC}$. + \kdokaz + + + \bzgled + Let $A_1$, $A_2$, ..., $A_n$ be points on a line $p$ and $B_1$, $B_2$, ..., $B_n$ points on a line $q$ ($n\geq 3$), such that + $$\overrightarrow{A_1A_2}: + \overrightarrow{A_2A_3}:\cdots: + \overrightarrow{A_{n-1}A_n}= + \overrightarrow{B_1B_2}: + \overrightarrow{B_2B_3}:\cdots: + \overrightarrow{B_{n-1}B_n}.$$ + Prove that the midpoints of line segments + $A_1B_1$, $A_2B_2$, ..., $A_nB_n$ lie on one line. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.4.4.pic} +\caption{} \label{sl.vek.5.4.4.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.4.4.pic}) + + It is enough to prove that three arbitrary centers are collinear points. +Without loss of generality, we prove only that points $S_1$, $S_2$, $S_3$ are collinear. Let: +$$\frac{\overrightarrow{A_1A_2}}{\overrightarrow{A_2A_3}}= +\frac{\overrightarrow{B_1B_2}}{\overrightarrow{B_2B_3}}=\lambda.$$ + By \ref{vektSestSplosno} is: + \begin{eqnarray*} + \overrightarrow{S_1S_2}&=&\frac{1}{2}\left(\overrightarrow{A_1A_2}+ + \overrightarrow{B_1B_2}\right)=\\ + &=&\frac{1}{2}\left(\lambda\overrightarrow{A_2A_3}+ + \lambda\overrightarrow{B_2B_3}\right)=\\ + &=&\frac{\lambda}{2}\overrightarrow{S_2S_3}, + \end{eqnarray*} + which means that $\overrightarrow{S_1S_2}$ and $\overrightarrow{S_2S_3}$ are collinear vectors (\ref{vektKriterijKolin}), so $S_1$, $S_2$ and $S_3$ are collinear points. + \kdokaz + + + +%________________________________________________________________________________ + \poglavje{Centroid of a Polygon With Respect to Its Vertices} \label{odd5TezVeck} + +Now we will generalize the concept of the centroid of a triangle to an arbitrary polygon. In section \ref{odd3ZnamTock} we defined the concept of the centroid of a triangle. The next theorem relates to one additional property of this concept, which is related to the concept of a vector. + + + \bizrek \label{tezTrikVekt} + If $T$ is the centroid of a triangle $ABC$, then + $$\overrightarrow{TA}+\overrightarrow{TB}+ + \overrightarrow{TC}=\overrightarrow{0}.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.1.pic} +\caption{} \label{sl.vek.5.5.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} We mark with $A_1$ the center of the side $BC$ (Figure \ref{sl.vek.5.5.1.pic}). By izreku \ref{vektSredOSOAOB} is $\overrightarrow{TA_1}= + \frac{1}{2}\left(\overrightarrow{TB}+\overrightarrow{TC}\right)$, by izreku \ref{tezisce} for the centroid $T$ of the triangle $ABC$ is valid $|AT|:|TA_1|=2:1$ or $\overrightarrow{TA}=-2\cdot\overrightarrow{TA_1}$. So: +$$\overrightarrow{TA}+\overrightarrow{TB}+ + \overrightarrow{TC}=\overrightarrow{TA}+2\cdot\overrightarrow{TA_1}= + \overrightarrow{TA}-\overrightarrow{TA}= + \overrightarrow{0},$$ which had to be proven. \kdokaz + +It is also valid the converse statement: + + + + \bizrek \label{tezTrikVektObr} + Let $A$, $B$ and $C$ be three non-collinear points. If $X$ is a point such that + $$\overrightarrow{XA}+\overrightarrow{XB}+ + \overrightarrow{XC}=\overrightarrow{0},$$ + then + $X$ is the centroid of the triangle $ABC$. + \eizrek + +\textbf{\textit{Proof.}} Let $T$ be the centroid of the triangle. By +the previous izreku \ref{tezTrikVekt} is +$\overrightarrow{TA}+\overrightarrow{TB}+ +\overrightarrow{TC}=\overrightarrow{0}$. By the assumption +is also $\overrightarrow{XA}+\overrightarrow{XB}+ +\overrightarrow{XC}=\overrightarrow{0}$. If we subtract +the two equalities, we get +$3 \cdot \overrightarrow{TX}=\overrightarrow{0}$ or $X=T$. + \kdokaz + +The proven property of the centroid of a triangle gives us an idea for the definition of the centroid of an arbitrary polygon (Figure \ref{sl.vek.5.5.2.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.2.pic} +\caption{} \label{sl.vek.5.5.2.pic} +\end{figure} + +A point $T$ is the \index{center of mass!polygon}\pojem{center of mass of a polygon $A_1A_2\ldots A_n$ with respect to its vertices}, if the following is true: +$$\overrightarrow{TA_1}+\overrightarrow{TA_2}+\cdots +\overrightarrow{TA_n}=\overrightarrow{0}.$$ +The previous relation can also be written as: +$$\sum_{k=1}^n\overrightarrow{TA_k}=\overrightarrow{0}.$$ + +We also mention that the center of mass of any figure $\Phi$ in a plane is defined as a point $T$, for which the following is true: +$$\sum_{X\in \Phi}\overrightarrow{TX}=\overrightarrow{0}.$$ + +In the case of our polygon $A_1A_2\ldots A_n$, we actually found the center of mass of a figure that represents the union of all the vertices of this polygon $\{A_1,A_2,\ldots, A_n\}$. That is why we emphasized that this is the center of mass of a polygon with respect to its vertices. In addition, we could talk about the \pojem{center of mass of a polygon with respect to all its points}, which would more accurately match the general definition of the center of mass of any figure. + +If we consider the center of mass of a figure in a physical sense as - the center of mass - the first variant of the center of mass represents the center of mass, where all the mass is in the vertices and each vertex has the same mass. In the second case, this is the center of mass of the polygon, where the mass is evenly distributed throughout its interior\footnote{\index{Archimedes} + \textit{Archimedes + of Syracuse} (3rd century BC) was a Greek mathematician + who first created the concept of the center of mass, +which he used in many of his writings on mechanics, but +we can only guess what he had in mind when +he considered the center of mass, because none of his surviving writings contain +a clear definition of the concept. The center of mass as the center of mass in a physical sense played an important role in Newton's (\index{Newton, +I.}\textit{I. Newton} (1643-1727), English physicist and mathematician) mechanics, where larger bodies are often considered as points with a certain mass.}. + +If we talk about a general polygon, the aforementioned centroids are only equal in any triangle, already in any quadrilateral the centroids differ. +In the following we will only consider the centroid of the polygon with respect to its vertices, so we will just call it the \pojem{centroid of the polygon}. + +First we will consider the centroid of a quadrilateral. + + \bizrek + The centroid of a parallelogram $ABCD$ is its circumcentre $S$, i.e. + the intersection of its diagonals. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.3.pic} +\caption{} \label{sl.vek.5.5.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.3.pic}) + + By izreku \ref{paralelogram} is the point $S$ the common centre of its diagonals $AC$ and $BD$, so from izreku \ref{vektSredDalj} it follows that $\overrightarrow{SA}+\overrightarrow{SC}=\overrightarrow{0}$ and $\overrightarrow{SB}+\overrightarrow{SD}=\overrightarrow{0}$. + Therefore it holds: + $$\overrightarrow{SA}+\overrightarrow{SB}+ + \overrightarrow{SC}+\overrightarrow{SD}=\overrightarrow{0},$$ +which means that $S$ is the centroid of the parallelogram $ABCD$. +\kdokaz + + + \bizrek \label{vektVarignon} + The centroid of a quadrilateral $ABCD$ is the centroid of its Varignon + parallelogram (see theorem \ref{Varignon}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.1a.pic} +\caption{} \label{sl.vek.5.5.1a.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.1a.pic}) + +Let $P$, $K$, $Q$ and $L$ be the midpoints of sides $AB$, $BC$, $CD$ and $DA$, or $PKQL$ the Varignon parallelogram of the quadrilateral $ABCD$ (the quadrilateral $PKQL$ is a parallelogram by the statement \ref{Varignon}). According to the previous statement \ref{vektVarignon}, the intersection of diagonals $PQ$ and $LK$ (denoted by $T$) is at the same time the centroid of the parallelogram $PKQL$. Then it is (statements \ref{vektSredOSOAOB}) and \ref{paralelogram}): + \begin{eqnarray*} + \overrightarrow{TA}+\overrightarrow{TB}+ + \overrightarrow{TC}+\overrightarrow{TD}&=& + 2\cdot\overrightarrow{TP}+ + 2\cdot\overrightarrow{TQ}=\\ + &=& + 2\cdot\left(\overrightarrow{TP}+ +\overrightarrow{TQ}\right)=\\ + &=&\overrightarrow{0}, + \end{eqnarray*} +which means that $T$ is the centroid of the quadrilateral $ABCD$. +\kdokaz + + + + \bizrek \label{vektTezVeckXT} + If $X$ is an arbitrary point and $T$ the centroid of a polygon $A_1A_2\ldots A_n$, then + $$\overrightarrow{XT}=\frac{1}{n}\left(\overrightarrow{XA_1} + +\overrightarrow{XA_2}+\cdots +\overrightarrow{XA_n}\right).$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.4.pic} +\caption{} \label{sl.vek.5.5.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.4.pic}) + +First, $\overrightarrow{XT}=\overrightarrow{XA_k}+\overrightarrow{A_kT}$ (for every $k\in\{1,2,\ldots,n\}$). If we add all $n$ relations, we get: + +\begin{eqnarray*} + n\cdot\overrightarrow{XT} + &=&\sum_{k=1}^{n}\left(\overrightarrow{XA_k}+\overrightarrow{A_kT}\right)=\\ + &=&\sum_{k=1}^{n}\overrightarrow{XA_k}+\sum_{k=1}^{n}\overrightarrow{A_kT}=\\ + &=& + \sum_{k=1}^{n}\overrightarrow{XA_k}-\sum_{k=1}^{n}\overrightarrow{TA_k}=\\ + &=& + \sum_{k=1}^{n}\overrightarrow{XA_k}-\overrightarrow{0}=\\ + &=&\sum_{k=1}^{n}\overrightarrow{XA_k}. + \end{eqnarray*} + +Therefore: +$$\overrightarrow{XT}=\frac{1}{n}\cdot\sum_{k=1}^{n}\overrightarrow{XA_k},$$ which was to be proven. \kdokaz + +As a result of the previous equation, a particularly useful relation holds for the centroid of an arbitrary triangle. + + + + \bizrek \label{vektTezTrikXT} + If $X$ is an arbitrary point and $T$ the centroid of a triangle $ABC$, then + $$\overrightarrow{XT}=\frac{1}{3}\left(\overrightarrow{XA} + +\overrightarrow{XB}+\overrightarrow{XC}\right).$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.5.pic} +\caption{} \label{sl.vek.5.5.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.5.pic}) + +A direct consequence of the previous equation for $n=3$ +\kdokaz + + + +% The process for an n-gon (n-1)-gon ... + + So far we have had an effective process for determining the centroids of triangles and quadrilaterals. For an arbitrary quadrilateral, we actually do not know whether the centroid exists at all. We will address this issue in the following. + + We will look for the idea in the following facts. If we consider the distance $AB$ as a degenerate $2$-gon, its centroid is the center of the distance $AB$; we denote it with $T_2$. For the centroid $T_3$ of the triangle $ABC$, it then holds that + $\overrightarrow{CT_3}=\frac{2}{3}\cdot \overrightarrow{CT_2}$ (equation \ref{tezisce}). We will generalize this idea in the following equation. + + + \bizrek \label{vektTezVeck} + Every polygon $A_1A_2\ldots A_n$ has exactly one centroid. If + $T_{n-1}$ is the centroid of the polygon $A_1A_2\ldots A_{n-1}$ and $T_n$ a point such that $$\overrightarrow{A_nT_n}=\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}},$$ + then $T_n$ is the centroid of the polygon $A_1A_2\ldots A_n$. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.2a.pic} +\caption{} \label{sl.vek.5.5.2a.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.2a.pic}) + +First, from the relation $\overrightarrow{A_nT_n}=\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}}$ it follows that + $\overrightarrow{T_nA_n}=-\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}}$ and $\overrightarrow{T_nT_{n-1}}=\frac{1}{n}\cdot\overrightarrow{A_nT_{n-1}}$. + Because $T_{n-1}$ is the centroid of the polygon $A_1A_2\ldots A_{n-1}$, by \ref{vektTezVeckXT} it holds that $\overrightarrow{T_nA_1}+\overrightarrow{T_nA_2}+\cdots +\overrightarrow{T_nA_{n-1}}=\left(n-1\right)\cdot \overrightarrow{T_nT_{n-1}}$. + + So: + \begin{eqnarray*} + & & \overrightarrow{T_nA_1}+\overrightarrow{T_nA_2}+\cdots + +\overrightarrow{T_nA_{n-1}}+\overrightarrow{T_nA_n}=\\ + &=&\left(n-1\right)\cdot \overrightarrow{T_nT_{n-1}} +\overrightarrow{T_nA_n}=\\ + &=&\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}} -\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}}=\\ + &=&\overrightarrow{0}, + \end{eqnarray*} + + which means that $T_n$ is the centroid of the polygon $A_1A_2\ldots A_n$. + + Assume that the polygon $A_1A_2\ldots A_n$ has another centroid $T'$. But in this case it holds (\ref{vektTezVeckXT}): + \begin{eqnarray*} + \overrightarrow{0}= + \overrightarrow{T'A_1}+\overrightarrow{T'A_2}+\cdots + \overrightarrow{T'A_n}= + n\cdot\overrightarrow{T'T_n}. + \end{eqnarray*} + So $\overrightarrow{T'T_n}=\overrightarrow{0}$ or $T'=T_n$, which means that the polygon $A_1A_2\ldots A_n$ has only one centroid. + \kdokaz + +The previous statement \ref{vektTezVeck} allows us to effectively construct the centroid of a polygon $A_1A_2\ldots A_n$, so that we first construct the centroids of polygons $A_1A_2$, $A_1A_2A_3$,$A_1A_2A_3A_4$, ... and finally $A_1A_2\ldots A_n$ in order (Figure \ref{sl.vek.5.5.2a.pic}): + \begin{itemize} + \item point $T_2$ is the center of the line segment $A_1A_2$, + \item $T_3$ is such a point that it holds: $\overrightarrow{A_3T_3}=\frac{2}{3}\cdot \overrightarrow{A_3T_2}$, + \item $T_4$ is such a point that it holds: $\overrightarrow{A_4T_4}=\frac{3}{4}\cdot \overrightarrow{A_4T_3}$,\\ + $\vdots$ + \item $T_n$ is such a point that it holds $\overrightarrow{A_nT_n}=\frac{n-1}{n}\cdot \overrightarrow{A_nT_{n-1}}$. + \end{itemize} + + + We get an even easier procedure for determining the centroid of a polygon if we use the relation + $$\overrightarrow{XT}=\frac{1}{n}\left(\overrightarrow{XA_1} + +\overrightarrow{XA_2}+\cdots +\overrightarrow{XA_n}\right)$$ + from statement \ref{vektTezVeckXT}. So for any point $X$ we simply plot the vector $\overrightarrow{XT}$ and get the point $T$. + + We emphasize that in both cases of the centroid construction of a polygon we need the process of planning a point that divides a given line segment in a certain ratio. We will discuss this process in section \ref{odd5TalesVekt} (see \ref{izrekEnaDelitevDaljice} and \ref{izrekEnaDelitevDaljiceNan}). + + + + In the proof of statement \ref{vektTezVeck} we did not use +the fact that the points $A_1$, $A_2$, ..., $A_n$ are in the same plane. +The statement is also true in the case where $ABCD$ ($n=4$) is a so-called \pojem{tetrahedron} +\index{tetrahedron}. The aforementioned point is then called +\index{težišče!tetraedra} \pojem{the centroid of the tetrahedron}. It is +possible to +prove an analogous statement for a tetrahedron; the line segments determined by the vertices of the tetrahedron and the centroids of the opposite faces pass through the centroid of this tetrahedron, which it divides in the ratio $3 :1$. + +In the general case, if $n \in \mathbb{N}$, we can talk about the so-called \index{simplex}\pojem{simplex} (generalization: point, line, triangle, tetrahedron, ...), which lies in the $(n-1)$-dimensional Euclidean space. + + \bzgled + Let $A$, $B$ and $C$ be the centroids of a triangles $OMN$, + $ONP$ and $OPM$, then the centroid $T$ of the triangle $MNP$, + the centroid $T_1$ of the triangle $ABC$ and the point $O$ are + three collinear points. Furthermore, it is $OT_1:T_1T=2:1$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.6.pic} +\caption{} \label{sl.vek.5.5.6.pic} +\end{figure} + + \textbf{\textit{Proof.}} We mark with $P_1$, $M_1$ and $N_1$ the centroids of the sides $MN$, $NP$ and $PM$ of the triangle $PMN$ (Figure \ref{sl.vek.5.5.6.pic}). + If we use the formulas \ref{vektTezTrikXT}, \ref{tezisce} and \ref{vektSredOSOAOB}, we get: + + \begin{eqnarray*} + \overrightarrow{OT_1}&=&\frac{1}{3}\left( + \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC} \right)=\\ + &=&\frac{1}{3}\left( + \frac{2}{3}\overrightarrow{OP_1}+\frac{2}{3}\overrightarrow{OM_1}+ + \frac{2}{3}\overrightarrow{ON_1} \right)=\\ + &=&\frac{1}{3}\left( + \frac{1}{3}\left(\overrightarrow{OM}+\overrightarrow{ON}\right) + +\frac{1}{3}\left(\overrightarrow{ON}+\overrightarrow{OP}\right)+ + \frac{1}{3}\left(\overrightarrow{OP}+\overrightarrow{OM}\right) \right)=\\ + &=&\frac{2}{9}\left( + \overrightarrow{OM}+\overrightarrow{ON}+ + \overrightarrow{OP} \right)=\\ + &=&\frac{2}{3}\overrightarrow{OT} + \end{eqnarray*} + +From $\overrightarrow{OT_1}=\frac{2}{3}\overrightarrow{OT}$ it follows that $\overrightarrow{OT_1}=2\overrightarrow{T_1T}$, which means that the vectors $\overrightarrow{OT_1}$ and $\overrightarrow{T_1T}$ are collinear (also the points $O$, $T_1$ and $T$) and $OT_1:T_1T=2:1$. +\kdokaz + +\bzgled + The centroid of a regular $n$-gon $A_1A_2...A_n$ is its centre (i.e. incentre and circumcentre). + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.7.pic} +\caption{} \label{sl.vek.5.5.7.pic} +\end{figure} + + \textbf{\textit{Proof.}} Let $S$ be the centroid of a regular $n$-gon $A_1A_2...A_n$ (Figure \ref{sl.vek.5.5.7.pic}). +It is enough to prove that: +$$\overrightarrow{SA_1}+\overrightarrow{SA_2}+ +\cdots+\overrightarrow{SA_n}=\overrightarrow{0}.$$ +Although in the case when $n$ is an even number, the statement is trivial, we will prove it for a general value of $n$ (even and odd). +Assume that $\overrightarrow{SA_1}+\overrightarrow{SA_2}+\cdots+\overrightarrow{SA_n}=\overrightarrow{SX}$, where $X\neq S$. +But if we rotate the polygon around the centroid $S$ by an angle $\theta=\frac{360}{n}$ (see the section on rotation \ref{odd6Rotac}), the sum of vectors on the left side of the equality does not change, the result on the right side becomes the vector $\overrightarrow{SX'}$, where $X'$ is the point we get from $X$ with the same rotation. Because the right side of the equality must remain unchanged, we get $\overrightarrow{SX'}=\overrightarrow{SX}$ or $X'=X$. This is possible only in the case when $X=S$ or $\overrightarrow{SX}=\overrightarrow{0}$. +\kdokaz + + + + + + +%________________________________________________________________________________ + \poglavje{Hamilton's Theorem} \label{odd5Hamilton} + +Sedaj bomo nadaljevali z lastnostmi, ki se nanašajo na značilne točke trikotnika (razdelek \ref{odd3ZnamTock}). + + \bizrek \label{HamiltonLema} + If $O$, $V$ and $A_1$ are the circumcentre, the orthocentre and the midpoint + of the side $BC$ of a triangle $ABC$, respectively, then + $$\overrightarrow{AV}=2\cdot \overrightarrow{OA_1}.$$ + \eizrek + + + \textbf{\textit{Proof.}} +Let $B_1$ be the midpoint of the side $AC$ (Figure \ref{sl.vek.5.6.2.pic}). + +The vectors $\overrightarrow{OA_1}$ and $\overrightarrow{AV}$ are collinear, so $\overrightarrow{OA_1}=\alpha \cdot \overrightarrow{AV}$ for some $\alpha \in \mathbb{R}$ (statement \ref{vektKriterijKolin}). Similarly, for the same reasons $\overrightarrow{OB_1}=\beta \cdot \overrightarrow{BV}$ (or $\overrightarrow{B_1O}=\beta \cdot \overrightarrow{VB}$) for some $\beta \in \mathbb{R}$. By statement \ref{srednjicaTrikVekt} (the median of a triangle) we have: + $$\overrightarrow{B_1A_1}=\frac{1}{2}\overrightarrow{AB}= + \frac{1}{2}\left(\overrightarrow{AV}+\overrightarrow{VB}\right)= + \frac{1}{2}\overrightarrow{AV}+\frac{1}{2}\overrightarrow{VB}.$$ + At the same time we also have: + $$\overrightarrow{B_1A_1}=\overrightarrow{B_1O}+\overrightarrow{OA_1}= +\beta\overrightarrow{VB}+\alpha\overrightarrow{AV}= + \alpha\overrightarrow{AV}+\beta\overrightarrow{VB}.$$ + +Since the vectors $\overrightarrow{AV}$ and $\overrightarrow{VB}$ are non-collinear, from statement \ref{vektLinKomb1Razcep} it follows that $\alpha=\frac{1}{2}$ and $\beta=\frac{1}{2}$. Therefore $\overrightarrow{OA_1}=\alpha \overrightarrow{AV}=\frac{1}{2}\overrightarrow{AV}$ or $\overrightarrow{AV}=2\cdot \overrightarrow{OA_1}$. +\kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.6.2.pic} +\caption{} \label{sl.vek.5.6.2.pic} +\end{figure} + + The following statement is very useful. + + \bizrek \label{Hamilton}\index{izrek!Hamiltonov} + (Hamilton's\footnote{\index{Hamilton, W. R.}\textit{W. R. Hamilton} (1805--1865), angleški matematik.} theorem) + If $O$ and $V$ are circumcentre and orthocentre of a triangle $ABC$, respectively then + $$\overrightarrow{OA}+\overrightarrow{OB} + +\overrightarrow{OC}=\overrightarrow{OV}.$$ + + \eizrek + + + + \textbf{\textit{Proof.}} We mark with $A_1$ the centre of the side $BC$ (Figure \ref{sl.vek.5.6.2.pic}). If we use statement \ref{vektSredOSOAOB} and \ref{HamiltonLema}, we get: + +$$\overrightarrow{OA}+\overrightarrow{OB} + +\overrightarrow{OC} + =\overrightarrow{OA}+2\cdot \overrightarrow{OA_1}= + \overrightarrow{OA}+\overrightarrow{AV}= + \overrightarrow{OV},$$ which had to be proven. \kdokaz + + We will continue with the consequences of the previous two equations. + + + \bzgled \label{HamiltonPoslTetiv} + A quadrilateral $ABCD$ is inscribed in a circle with a centre $O$. + The diagonals $AC$ and $BD$ are perpendicular. + If $M$ is the foot of the perpendicular from the centre $O$ on the line $CD$, then + $$|OM|=\frac{1}{2}|AB|.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.6.3a.pic} +\caption{} \label{sl.vek.5.6.3a.pic} +\end{figure} + + \textbf{\textit{Proof.}} Let $V$ be the altitude point of the triangle $BCD$ + (Figure \ref{sl.vek.5.6.3a.pic}). + Because $AC\perp BD$, the point $V$ lies on the diagonal $AC$. By + \ref{HamiltonLema}, $\overrightarrow{OM}=\frac{1}{2} \overrightarrow{BV}$, + so $|OM|=\frac{1}{2}|BV|$. Because there is also (\ref{ObodObodKot} and + \ref{KotaPravokKraki}): + $$\angle BAV=\angle BAC\cong\angle BDC\cong\angle AVB,$$ +it follows that $BV\cong AB$ (\ref{enakokraki}) or +$|OM|=\frac{1}{2}|AB|$. + \kdokaz + + + \bzgled Let $V$ be the orthocentre and $O$ the circumcentre of a triangle $ABC$ and $AV\cong AO$. + Prove that $\angle BAC=60^0$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.6.1a.pic} +\caption{} \label{sl.vek.5.6.1a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $A_1$ be the center of the side $BC$ of the triangle $ABC$ (Figure \ref{sl.vek.5.6.1a.pic}). By the statement \ref{HamiltonLema} it follows that +$|AV|=2\cdot|OA_1|$. Since $OA\cong OC$, it follows that in the right triangle $OA_1C$ it holds that $|OC|=2\cdot|OA_1|$. With $O'$ we mark the point which is symmetrical to the point $O$ with respect to the point $A_1$. From $\triangle OA_1C\cong \triangle O'A_1C$ (the \textit{SAS} statement \ref{SKS}) it follows that $OC\cong O'C\cong OO'$, which means that the $\triangle OO'C$ is an isosceles triangle or $\angle A_1OC=60^0$. From the statement \ref{SredObodKot} and the congruence of the triangles $BOA_1$ and $COA_1$ (the \textit{SSS} statement \ref{SSS}) at the end it follows that +$\angle BAC=\frac{1}{2}\angle BOC=\angle A_1OC=60^0$. + \kdokaz + + \bzgled \label{TetivniVisinska} + Let $ABCD$ be a cyclic quadrilateral and: + $V_A$ the orthocentre of the triangle $BCD$, + $V_B$ the orthocentre of the triangle $ACD$, + $V_C$ the orthocentre of the triangle $ABD$ and + $V_D$ the orthocentre of the triangle $ABC$. + Prove that:\\ + a) the line segments $AV_A$, + $BV_B$, $CV_C$ and $DV_D$ + has a common midpoint,\\ + b) the quadrilateral $V_AV_BV_CV_D$ is congruent + to the quadrilateral $ABCD$. + \ezgled + + +\begin{figure}[htp] +\centering +\input{sl.vek.5.6.3b.pic} +\caption{} \label{sl.vek.5.6.3b.pic} +\end{figure} + + +\begin{figure}[htp] +\centering +\input{sl.vek.5.6.3.pic} +\caption{} \label{sl.vek.5.6.3.pic} +\end{figure} + + + \textbf{\textit{Solution.}} Let $O$ be the center of the circumscribed circle of the cyclic quadrilateral $ABCD$ (Figure \ref{sl.vek.5.6.3.pic}). It is clear that the point $O$ is at the same time the center of the circumscribed circle of the triangles $BCD$, $ACD$, $ABD$ and $ABC$. + By Hamilton's statement \ref{Hamilton} it follows: + +\begin{eqnarray*} + \overrightarrow{OV_A}&=&\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}\\ +\overrightarrow{OV_B}&=&\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OD} + \end{eqnarray*} + We then get: +\begin{eqnarray*} + \overrightarrow{V_BV_A}&=&\overrightarrow{V_BO}+\overrightarrow{OV_A}=\\ +&=&\overrightarrow{OV_A}-\overrightarrow{OV_B}=\\ + &=&\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD} +-(\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OD})=\\ +&=&\overrightarrow{OB}-\overrightarrow{OA}=\\ +&=&\overrightarrow{AB}. + \end{eqnarray*} + + +Therefore + $\overrightarrow{V_BV_A}=\overrightarrow{AB}$. By the statement \ref{vektParalelogram} the quadrilateral $ABV_AV_B$ is a parallelogram, by the statement \ref{paralelogram} its diagonals $AV_A$ and $BV_B$ have a common center - we mark it with $S$ + (Figure \ref{sl.vek.5.6.3b.pic}). In a similar way each of the pairs of lines $AV_A$ and $CV_C$ or $AV_A$ and $DV_D$ have a common center. Because it is the center of the line $AV_A$, it follows that all four lines $AV_A$, + $BV_B$, $CV_C$ and $DV_D$ + have a common center - the point $S$. + + In a similar way as $\overrightarrow{V_BV_A}=\overrightarrow{AB}$ it also follows that $\overrightarrow{V_CV_B}=\overrightarrow{BC}$, $\overrightarrow{V_DV_C}=\overrightarrow{CD}$ and $\overrightarrow{V_DV_A}=\overrightarrow{DA}$. + This means that the quadrilateral $V_AV_BV_CV_D$ and $ABCD$ (statement \ref{vektVzpSkl} and \ref{KotaVzporKraki}) have all the congruent sides and interior angles. Therefore $V_AV_BV_CV_D\cong ABCD$. For a formal proof of this we can use the isometry $\mathcal{I}:A,B,C\mapsto V_A,V_B,V_C$ and prove $\mathcal{I}(D)=V_D$. + \kdokaz + +\bzgled \label{HamiltonSimson}\index{premica!Simsonova} +Let $ABCD$ be a cyclic quadrilateral and: $a$ is the Simson +line with respect to the triangle $BCD$ and the point $A$, $b$ is the Simson +line with respect to the triangle $ACD$ and the point $B$, $c$ is the Simson +line with respect to the triangle $ABD$ and the point $C$ ter $d$ S is the Simson +line with respect to the triangle $ABC$ and the point $D$. +Prove that the lines $a$, $b$, $c$ and +$d$ intersect at a single point. +\ezgled + +\begin{figure}[htp] +\centering +\input{sl.vek.5.6.4.pic} +\caption{} \label{sl.vek.5.6.4.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.6.4.pic}). + +The claim is a direct consequence of theorems +\ref{SimsZgled3} and \ref{TetivniVisinska} - the lines $a$, $b$, $c$ and +$d$ intersect at the point $S$ (from theorem \ref{TetivniVisinska}). +\kdokaz + + + +%________________________________________________________________________________ +\poglavje{Euler Line} \label{odd5EulPrem} + +We will now prove an important property related to the three characteristic points of a triangle. + + \bizrek \label{EulerjevaPremica} + The circumcentre $O$, the centroid $T$ and the orthocentre $V$ + of an arbitrary triangle lies on the same line. Besides that it is + $$|OT|:|TV|=1:2.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.7.1.pic} +\caption{} \label{sl.vek.5.7.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.7.1.pic}) + + If we use theorem \ref{vektTezTrikXT} and Hamilton's theorem \ref{Hamilton}, we get: + $$\overrightarrow{OT}=\frac{1}{3}\left(\overrightarrow{OA} + +\overrightarrow{OB}+\overrightarrow{OC}\right)= + \frac{1}{3}\overrightarrow{OV}.$$ + +The vectors $\overrightarrow{OT}$ and $\overrightarrow{OV}$ are therefore collinear and it holds that $\overrightarrow{OT}:\overrightarrow{OV}=1:3$. This means that the points $O$, $T$ and $V$ are collinear and it holds that $|OT|:|TV|=1:2$. + \kdokaz + +The line from the previous theorem, on which three characteristic points lie, is called the \index{line!Euler's} \pojem{Euler line}. + + In the next theorem we will see the connection between the Euler line and the Euler circle, which we discussed in section \ref{odd3EulKroz}. + + + + + \bizrek \label{EulerKrozPrem1}\index{circle!Euler's} + The centre of Euler of an arbitrary triangle lies on + The Euler line of this triangle. + Furthermore, it is the midpoint of the line segment determined by + the orthocentre and the circumcentre of this triangle. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.7.2.pic} +\caption{} \label{sl.vek.5.7.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Let $AA'$, $BB'$ and $CC'$ be altitudes, and $A_1$, $B_1$ and $C_1$ be the centres of sides $BC$, $AC$ and $AB$ of triangle $ABC$. We mark the centre of the circumscribed circle with $O$, the point of altitude of this triangle with $V$, and the centres of lines $VA$, $VB$ and $VC$ with $V_A$, $V_B$ and $V_C$ (Figure \ref{sl.vek.5.7.2.pic}). + + By theorem \ref{EulerKroznica}, points $A'$, $B'$, $C'$, $A_1$, $B_1$, $C_1$, $V_A$, $V_B$ and $V_C$ lie on one circle - i.e. the Euler circle. We mark the centre of this circle with $E$. Because $\angle V_AA'A_1\cong \angle AA'C=90^0$, by theorem \ref{TalesovIzrKroz2} line $V_AA_1$ is the diameter of this circle, or point $E$ is the centre of line $V_AA_1$. + +According to the statement \ref{HamiltonLema} it holds: +$$\overrightarrow{OA_1}=\frac{1}{2}\cdot \overrightarrow{AV}=\overrightarrow{V_AV}.$$ +Therefore $\overrightarrow{OA_1}=\overrightarrow{V_AV}$, which means that the quadrilateral $A_1OV_AV$ is a parallelogram (\ref{vektParalelogram}). Its diagonals $VO$ and $V_AA_1$ intersect (\ref{paralelogram}), so the point $E$ is the center of the line $OV$ and lies on Euler's line of the triangle $ABC$ (\ref{EulerjevaPremica}). + \kdokaz + +In section \ref{odd7SredRazteg} (\ref{EulerKroznicaHomot}) we will see the continuation of the previous statement, which refers to Euler's circle. + + +%________________________________________________________________________________ + \poglavje{Thales' Theorem - Basic Proportionality Theorem} \label{odd5TalesVekt} + + We have already seen in section \ref{odd5LinKombVekt} that we can talk about the ratio of two collinear vectors. + For collinear vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ ($\overrightarrow{u}\neq \overrightarrow{0}$) we defined their ratio or quotient + $$\overrightarrow{v}:\overrightarrow{u} +=\frac{\overrightarrow{v}}{\overrightarrow{u}}=\lambda,$$ + if for some $\lambda\in\mathbb{R}$ it holds $\overrightarrow{v}=\lambda \overrightarrow{u}$. + +In a similar way as with numbers, we can define the ratio of two pairs of collinear vectors. If $\overrightarrow{a}$ and $\overrightarrow{b}$ ($\overrightarrow{b}\neq \overrightarrow{0}$) are a pair of collinear vectors or $\overrightarrow{c}$ and $\overrightarrow{d}$ ($\overrightarrow{d}\neq \overrightarrow{0}$) are a pair of collinear vectors, we say that the pairs of vectors \index{sorazmerje kolinearnih vektorjev}\pojem{sorazmerna}, if it holds: + + $$\frac{\overrightarrow{a}}{\overrightarrow{b}} + =\frac{\overrightarrow{c}}{\overrightarrow{d}}.$$ + + The next very important statement refers to the defined concept of ratio. + +\bizrek + \label{TalesovIzrek}(Thales'\footnote{Starogrški filozof in matematik \textit{Tales} + \index{Tales} iz Mileta (640--546 pr. n. š.) je obravnaval sorazmerje ustreznih daljic, + ki jih dobimo, če dve premici presekamo z dvema vzporednicama, pri tem pa ni omenjal + vektorske oblike.} theorem - Basic Proportionality Theorem)\\ + Let $a$, $b$, $p$ and $p'$ be lines in the same plane, and $O=a\cap b$, $A=a\cap p$, $A'=a\cap p'$, $B=b\cap p$ and $B'=b\cap p'$.\\ + If $p\parallel p'$, then + $$\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= + \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}= + \frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}.$$ + \index{izrek!Talesov o sorazmerju} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.1.pic} +\caption{} \label{sl.vek.5.8.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.1.pic}) + +Since by assumption $p\parallel p'$, the vectors $\overrightarrow{A'B'}$ and $\overrightarrow{AB}$ are collinear. By \ref{vektKriterijKolin} $\overrightarrow{A'B'}=\lambda\overrightarrow{AB}$ for some $\lambda\in \mathbb{R}$. In a similar way, from the collinearity of the vectors $\overrightarrow{OA'}$ and $\overrightarrow{OA}$ or $\overrightarrow{OB'}$ and $\overrightarrow{OB}$ it follows that $\overrightarrow{OA'}=\alpha\overrightarrow{OA}$ for some $\alpha\in \mathbb{R}$ or $\overrightarrow{OB'}=\beta\overrightarrow{OB}$ for some $\beta\in \mathbb{R}$. From this we obtain: + $$\frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}=\lambda,\hspace*{2mm} + \frac{\overrightarrow{OA'}}{\overrightarrow{OA}}=\alpha,\hspace*{2mm} + \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}=\beta.$$ + It is enough to prove that $\alpha=\beta=\lambda$. + If we use the rule for subtracting vectors \ref{vektOdsev} and \ref{vektVektorskiProstor} (point $\textit{6}$), we get: + \begin{eqnarray*} + \overrightarrow{A'B'}&=&\overrightarrow{OB'}-\overrightarrow{OA'} + =\beta\overrightarrow{OB}-\alpha \overrightarrow{OA};\\ + \overrightarrow{A'B'}&=&\lambda\overrightarrow{AB}= + \lambda\left(\overrightarrow{OB}-\overrightarrow{OA}\right) + =\lambda\overrightarrow{OB}-\lambda\overrightarrow{OA}. + \end{eqnarray*} + Since $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are non-collinear vectors, by \ref{vektLinKomb1Razcep} + $\alpha=\beta=\lambda$ or $\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= + \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}= + \frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}$. +\kdokaz + + + A direct consequence is the Tales theorem in the form of a ratio of distances, which is not in vector form (Figure \ref{sl.vek.5.8.2.pic}). + +\bizrek \label{TalesovIzrekDolzine} +Let $a$, $b$, $p$ and $p'$ be lines in the same plane, and +$O=a\cap b$, $A=a\cap p$, $A'=a\cap p'$, $B=b\cap p$ and $B'=b\cap p'$.\\ +If $p\parallel p'$, then +$$\frac{OA'}{OA}= +\frac{OB'}{OB}= +\frac{A'B'}{AB}$$ +and also +$$\frac{OA'}{OB'}= +\frac{OA}{OB}.$$ +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.2.pic} +\caption{} \label{sl.vek.5.8.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} The claim directly follows from theorems \ref{TalesovIzrek} and \ref{vektKolicnDolz}. +\kdokaz + +We will prove that the converse is also true, i.e. that from the appropriate proportion follows the parallelism of the corresponding lines. + +(Converse Thales' proportionality theorem)\\ +Let $a$, $b$, $p$ and $p'$ be lines in the same plane, and $O=a\cap b$, $A=a\cap p$, $A'=a\cap p'$, $B=b\cap p$ and $B'=b\cap p'$.\\ If +$$\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= +\frac{\overrightarrow{OB'}}{\overrightarrow{OB}},$$ +then $p\parallel p'$ and also +$$\frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}= +\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= +\frac{\overrightarrow{OB'}}{\overrightarrow{OB}}.$$ +\index{izrek!Talesov obratni o sorazmerju} + +\textbf{\textit{Proof.}} We mark $\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= +\frac{\overrightarrow{OB'}}{\overrightarrow{OB}}=\lambda$. In this case, first $\overrightarrow{OA'}=\lambda\overrightarrow{OA}$ and $\overrightarrow{OB'}=\lambda\overrightarrow{OB}$. Therefore, (statement \ref{vektOdsev} and \ref{vektVektorskiProstor}): +$$\overrightarrow{A'B'}=\overrightarrow{OB'}-\overrightarrow{OA'} +=\lambda\overrightarrow{OB}-\lambda\overrightarrow{OA} +=\lambda\left(\overrightarrow{OB}-\overrightarrow{OA}\right) +=\lambda\overrightarrow{AB}.$$ +Since $\overrightarrow{A'B'}=\lambda\overrightarrow{AB}$, according to statement \ref{vektKriterijKolin} vectors $\overrightarrow{A'B'}$ and $\overrightarrow{AB}$ are collinear. This means that $AB\parallel A'B'$ or $p\parallel p'$. + +Finally, from statement \ref{TalesovIzrek} follows the relation +$\frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}= +\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= +\frac{\overrightarrow{OB'}}{\overrightarrow{OB}}$. +\kdokaz + +We also mention some consequences of Tales' statement \ref{TalesovIzrek}. + +\bizrek \label{TalesPosl1} + If parallel lines $p_1$, $p_2$, $p_3$ intersect a line $a$ at points $A_1$, $A_2$, + $A_3$ and a line $b$ at points $B_1$, $B_2$, + $B_3$, then + $$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3} + \hspace*{1mm} + \textrm{ and } \hspace*{1mm} + \frac{A_1A_2}{A_2A_3}=\frac{B_1B_2}{B_2B_3}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.4.pic} +\caption{} \label{sl.vek.5.8.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.4.pic}) + + Without loss of generality, let $\mathcal{B}(A_1,A_2,A_3)$ and $\mathcal{B}(B_1,B_2,B_3)$. + +We mark with $c$ the parallel to the line $b$ through the point $A_1$ and with $C_2$ and $C_3$ the intersections of the line $c$ with the lines $p_2$ and $p_3$. The quadrilateral $B_1B_3C_3A_1$ and $B_2B_3C_3C_2$ are parallelograms, therefore $\overrightarrow{B_1B_3}=\overrightarrow{A_1C_3}$ and $\overrightarrow{B_2B_3}=\overrightarrow{C_2C_3}$. + +If we use the statement \ref{TalesovIzrekDolzine}, we get: + \begin{eqnarray*} + \frac{|A_1A_2|}{|A_2A_3|}&=& + \frac{|A_1A_3|-|A_2A_3|}{|A_2A_3|}= +\frac{|A_1A_3|}{|A_2A_3|}-1=\\ + &=&\frac{|A_1C_3|}{|C_2C_3|}-1= + \frac{|B_1B_3|}{|B_2B_3|}-1=\\ + &=&\frac{|B_1B_3|-|B_2B_3|}{|B_2A_3|}= +\frac{|B_1B_2|}{|B_2B_3|}, + \end{eqnarray*} +or $\frac{A_1A_2}{A_2A_3}=\frac{B_1B_2}{B_2B_3}$ and +$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3}$. + \kdokaz + + + \bizrek \label{TalesPosl2} + If parallel lines $p_1$, $p_2$,..., $p_n$ intersect a line $a$ at points $A_1$, $A_2$,..., + $A_n$ and a line $b$ at points $B_1$, $B_2$,..., + $B_n$, then + $$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3}=\cdots= + \frac{A_{n-1}A_n}{B_{n-1}B_n}\hspace*{1mm} + \textrm{ and } \hspace*{1mm}$$ + $$A_1A_2:A_2A_3:\cdots :A_{n-1}A_n= B_1B_2:B_2B_3:\cdots :B_{n-1}B_n.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.5.pic} +\caption{} \label{sl.vek.5.8.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.5.pic}) + +The statement is a direct consequence of the statement \ref{TalesPosl1}. + \kdokaz + +\bizrek \label{TalesPosl3} +Let $p_1$, $p_2$ and $p_3$ be lines that intersect at a point $O$. If $a$ and $b$ are +parallel lines that intersect the line $p_1$ at points $A_1$ and $B_1$, +the line $p_2$ at points $A_2$ and $B_2$, and +the line $p_3$ at points $A_3$ and $B_3$, then +$$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3} +\hspace*{1mm} +\textrm{ and } \hspace*{1mm} +\frac{A_1A_2}{A_2A_3}=\frac{B_1B_2}{B_2B_3}.$$ +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.6.pic} +\caption{} \label{sl.vek.5.8.6.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.6.pic}) + + By izreku \ref{TalesovIzrekDolzine} we get: + $$\frac{A_1A_2}{B_1B_2}= + \frac{OA_2}{OB_2}=\frac{A_2A_3}{B_2B_3},$$ which was to be proven. \kdokaz + +\bizrek \label{TalesPosl4} + Let $p_1$, $p_2$,..., $p_n$ be lines that intersect at a point $O$. If $a$ and $b$ are + parallel lines that intersect the line $p_1$ at points $A_1$ and $B_1$, + the line $p_2$ at points $A_2$ and $B_2$,..., + the line $p_n$ at points $A_n$ and $B_n$, then + $$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3}=\cdots= + \frac{A_{n-1}A_n}{B_{n-1}B_n}\hspace*{1mm} + \textrm{ and } \hspace*{1mm}$$ + $$A_1A_2:A_2A_3:\cdots :A_{n-1}A_n= B_1B_2:B_2B_3:\cdots :B_{n-1}B_n.$$ + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.8.pic} +\caption{} \label{sl.vek.5.8.8.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.8.pic}) + +The statement is a direct consequence of \ref{TalesPosl3}. + \kdokaz + + The following very well-known and useful planning tasks are next. + + + + \bzgled + \label{izrekEnaDelitevDaljiceNan} + \index{delitev daljice!na enake dele} + Construct points that divide a line segment $AB$ + into $n$ congruent line segments. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.9.pic} +\caption{} \label{sl.vek.5.8.9.pic} +\end{figure} + + +\textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.8.9.pic}) + +Let $X$ be an arbitrary point that does not lie on the line $AB$ and $Q_1$, $Q_2$, ..., $Q_n$ such points on the line segment $AX$, so that +$\overrightarrow{AQ_1}=\overrightarrow{Q_1Q_2}=\cdots =\overrightarrow{Q_{n-1}Q_n}$. +With $P_1$, $P_2$, ..., $P_{n-1}$ we denote the intersection of the line $AB$ with the parallels of the line $BQ_n$ through the points $Q_1$, $Q_2$, ..., $Q_{n-1}$. + +We prove that $P_1$, $P_2$, ..., $P_{n-1}$ are the desired points. By \ref{TalesPosl2} we have: +$$\frac{AP_1}{AQ_1}=\frac{P_1P_2}{Q_1Q_2}=\dots=\frac{P_{n-1}B}{Q_{n-1}Q_n}.$$ +Since, by assumption, $|\overrightarrow{AQ_1}|=|\overrightarrow{Q_1Q_2}|=\cdots =|\overrightarrow{Q_{n-1}Q_n}|$, also +$|AP_1|=|P_1P_2|=\cdots =|P_{n-1}B|$. + \kdokaz + + + \bzgled + \label{izrekEnaDelitevDaljice} + \index{delitev daljice!v razmerju} + Divide a given line segment $AB$ in the ratio $n:m$ ($n,m\in \mathbb{N}$), + i.e. determine such point $T$ on the line $AB$ that + $$\frac{\overrightarrow{AT}}{\overrightarrow{TB}}=\frac{n}{m}.$$ + Prove that there is the only one solution for such point $T$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.10.pic} +\caption{} \label{sl.vek.5.8.10.pic} +\end{figure} + + +\textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.8.10.pic}) + + + By \ref{izrekEnaDelitevDaljiceVekt} there is only one point $T$, for which $\frac{\overrightarrow{AT}}{\overrightarrow{TB}}=\frac{n}{m}$. + +Now we will describe the process of constructing point $T$. Let $\overrightarrow{v}$ be an arbitrary vector that is not collinear with vector $\overrightarrow{AB}$, and let $P$ and $Q$ be such points that $\overrightarrow{AP}=n\overrightarrow{v}$ and $\overrightarrow{BQ}=-m\overrightarrow{v}$. We denote the intersection of lines $AB$ and $PQ$ by $T$ (the intersection exists because $P,Q\div AB$). Then it holds: + \begin{eqnarray*} + \frac{\overrightarrow{AT}}{\overrightarrow{TB}}= + -\frac{\overrightarrow{TA}}{\overrightarrow{TB}}= + -\frac{\overrightarrow{AP}}{\overrightarrow{BQ}}= + -\frac{n\overrightarrow{v}}{-m\overrightarrow{v}}=\frac{n}{m}, + \end{eqnarray*} + which had to be proven. \kdokaz + + In section \ref{odd7Harm} we will further investigate the question of dividing a line segment into a given ratio. + + + + \bzgled \label{vektTrapezZgled} + A line parallel to bases of a trapezium intersects its legs and diagonals in four points + and determine three line segments. Prove that two of them are congruent. \\ + After that, construct a line parallel to the bases of that trapezium such that all + three mentioned line segments are congruent. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.11.pic} +\caption{} \label{sl.vek.5.8.11.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Let $l$ be a line parallel to the bases +$AB$ of trapezium $ABCD$, that intersects +sides $AD$ and $BC$ as well as diagonals $AC$ and $BD$ +in points $M$, $Q$, $N$ and $P$ (Figure \ref{sl.vek.5.8.11.pic}). By Tales' theorem \ref{TalesovIzrekDolzine} it holds: +$MN:DC=MA:DA$ or +$PQ:DC=BQ:BC$. Because by +the consequence \ref{TalesPosl1} of Tales' theorem $MA:AD=BQ:BC$, it also holds +$MN:DC=PQ:DC$ or $MN\cong PQ$. + +If $E$ is the center of the base $AB$ and $N_0$ is the intersection of the lines $DE$ and $AC$, the desired line $l_0$ passes through the point $N_0$ and is parallel to the line $AB$. If $M_0$, $P_0$ and $Q_0$ are the intersections of the line $l_0$ with the lines $AD$, $BD$ and $CB$, then by \ref{TalesPosl3} of Tales' theorem it follows that $M_0N_0\cong N_0P_0$. +\kdokaz + + + \bzgled If $r$ is the inradius and $r_a$, $r_b$ and $r_c$ exradii + of an arbitrary triangle, then $$\frac{1}{r_a} + +\frac{1}{r_b} +\frac{1}{r_c}= \frac{1}{r}.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.12.pic} +\caption{} \label{sl.vek.5.8.12.pic} +\end{figure} + +\textbf{\textit{Proof.}} If we use the labels from the big task +\ref{velikaNaloga} (Figure \ref{sl.vek.5.8.12.pic}), from Tales' theorem \ref{TalesovIzrek} it follows that: +$$ \frac{r}{r_a} = \frac{SQ}{S_aQ_a} = \frac{AQ}{AQ_a} = \frac{s-a}{s}.$$ + Similarly, it is also: + $$ \frac{r}{r_b} = \frac{s-b}{s}\hspace*{2mm} \textrm{ and }\hspace*{2mm} + \frac{r}{r_c} = \frac{s-c}{s}.$$ + By adding the three equalities we get the desired relation. + \kdokaz + + + + \bizrek \label{velNalTockP'} + Suppose that the incircle and the excircle of a triangle $ABC$ touch its side $BC$ in points $P$ and $P_a$. + If $PP'$ is a diameter of the incircle $k(S,r)$ ($P'\in k$), then $Pa$, $P'$ and $A$ are collinear + points. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.13.pic} +\caption{} \label{sl.vek.5.8.13.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Use the labels from the big task +\ref{velikaNaloga} (Figure \ref{sl.vek.5.8.13.pic}). Because the distance $PP'$ + is a diameter of the inscribed circle, the point $S$ is its center. + +Let $\widehat{P'}$ be the intersection of the line segment $PS$ with the line $AP_a$. + We will prove that + $\widehat{P'}=P'$, or that +$S\widehat{P'}=r$. From Tales' theorem it follows that: + $$\frac{S\widehat{P'}}{S_aP_a} = \frac{AS}{AS_a} =\frac{SQ}{S_aQ_a}.$$ + Because $S_a P_a = S_aQ_a = r_a$, + it is also true that $S\widehat{P'}=SQ=r$. + \kdokaz + +In a similar way to \ref{velNalTockP'} we also prove the following theorem. + + + + + \bizrek \label{velNalTockP'1} + Suppose that the incircle and the excircle of a triangle $ABC$ touch its side $BC$ in points $P$ and $P_a$. + If $P_aP_a'$ is a diameter of the excircle $k_a(S_a,r_a)$ ($P_a'\in k_a$), then $P_a'$, $P$ and $A$ are collinear + points + (Figure \ref{sl.vek.5.8.14.pic}). + \eizrek + + + + +\begin{figure}[htp] +\centering +\input{sl.vek.5.8.14.pic} +\caption{} \label{sl.vek.5.8.14.pic} +\end{figure} + +%\vspace*{10mm} + + + + \bizrek \label{velNalTockP'2} + Suppose that the excircles $k_b(S_b,r_b)$ and $k_c(S_c,r_c)$ of a triangle $ABC$ + touch the line $BC$ in points $P_b$ and $P_c$. + If $P_bP_b'$ and $P_cP_c'$ are diameters of the excircles + $k_b$ ($P_b'\in k_b$) and $k_c$ ($P_c'\in k_c$), then $P_c'$, $P_b$ and $A$ are collinear + points and also $P_b'$, $P_c$ and $A$ are collinear + points (Figure \ref{sl.vek.5.8.15.pic}). + \eizrek + + +\begin{figure}[htp] +\centering +\input{sl.vek.5.8.15.pic} +\caption{} \label{sl.vek.5.8.15.pic} +\end{figure} + + + + + + +The last three statements can be used in the construction of triangles, +which we will illustrate in the following task. + +\bzgled + Construct a triangle $ABC$, with given $r$, $b-c$, $t_a$. + \ezgled + + +\begin{figure}[htp] +\centering +\input{sl.vek.5.8.16.pic} +\caption{} \label{sl.vek.5.8.16.pic} +\end{figure} + + \textbf{\textit{Solution.}} + Let $ABC$ be the desired triangle or the triangle, in which the +radius of the inscribed circle is $r$, the centroid $AA_1$ is consistent with +$t_a$ and the difference of sides $AC$ and $AB$ is equal to $b-c$. Use +the notation from the big task \ref{velikaNaloga} and the formula +\ref{velNalTockP'} (Figure \ref{sl.vek.5.8.16.pic}). We know that $PP_a=b-c$, point $A_1$ is +the common center of side $BC$ and distance $PP_a$. From the formula +\ref{velNalTockP'} it follows that the points $P_a$, $P'$ and $A$ +are collinear. + +So first we construct the right triangle $P'PP_a$, then the +circle $k$ with diameter $PP'$ or the inscribed circle + of the triangle $ABC$ and the point $A_1$ as +the center of the line segment $PP_a$. The point $A$ is the intersection of the line segment +$P_aP'$ with the circle $k_1(A_1,t_a)$, points $B$ and $C$ are +the intersections of the line $BC$ with the tangents of the circle $k$ from the point +$A$. + \kdokaz + + \bzgled + Let $A_1$ be the midpoint of the line segment $BC$, $k(S,r)$ + the incircle and $AA'$ the altitude of a triangle $ABC$. + Suppose that $L$ is the intersection of the lines $A_1S$ and $AA'$. + Prove that $AL\cong + r$. + \ezgled + + +\begin{figure}[htp] +\centering +\input{sl.vek.5.8.3.pic} +\caption{} \label{sl.vek.5.8.3.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $P$, $Q$ and $R$ be the points of intersection of the circle $k$ with the sides $BC$, $AC$ and $AB$. Let $k(S_a,r_a)$ be the drawn circle that touches the side $BC$ in the point $P_a$, the lines $AB$ and $AC$ in the points $R_a$ and $Q_a$. Let $P'$ be the intersection of the lines $AP_a$ and $PS$ (Figure \ref{sl.vek.5.8.3.pic}). From the statement \ref{velNalTockP'} and the big exercise \ref{velikaNaloga} it follows: +\begin{itemize} + \item the point $P'$ lies on the circle $k$, + \item the point $A_1$ is the center of the segment $PP_a$. +\end{itemize} +We prove that $AL\cong + r$. The segment $SA_1$ is the median of the triangle +$PP'P_a$, therefore $SA_1\parallel P'P_a$ or $LS\parallel AP'$. +Since $AL\parallel P'S$, the quadrilateral $AP'SL$ is a parallelogram +and $AL\cong P'S\cong r$. + \kdokaz + + + +%________________________________________________________________________________ +\naloge{Exercises} + +\begin{enumerate} + + %Vsota in razlika vektorjev + %_____________________________________ + + \item Draw any vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ and $\overrightarrow{d}$ so that their sum is equal to: + + (\textit{a}) one of these four vectors,\\ + (\textit{b}) the difference of two of these four vectors. + + \item Let $ABCDE$ be a pentagon in some plane. Prove that in this plane there exists +a pentagon with sides that determine the same vectors as the + diagonals of the pentagon $ABCDE$. + + \item Let $A$, $B$, $C$ and $D$ be any points in a plane. Is it generally true that: + + (\textit{a}) $\overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AD}+\overrightarrow{BC}$?\\ + (\textit{b}) $\overrightarrow{AB}=\overrightarrow{DC}\hspace*{1mm}\Rightarrow \hspace*{1mm} \overrightarrow{AC}+\overrightarrow{BD}=2\overrightarrow{BC}$? + + \item Given is a segment $AB$. Using only a ruler (constructions in affine geometry) draw a point $C$ so that: + + (\textit{a}) $\overrightarrow{AC}=-\overrightarrow{AB}$, \hspace*{3mm} + (\textit{b}) $\overrightarrow{AC}=5\overrightarrow{AB}$, \hspace*{3mm} + (\textit{c}) $\overrightarrow{AC}=-3\overrightarrow{AB}$. + +\item Let $ABCD$ be a quadrilateral and $O$ any point in the plane of this quadrilateral. Express the vectors of sides and +diagonals of this quadrilateral with vectors $\overrightarrow{a}=\overrightarrow{OA}$, $\overrightarrow{b}=\overrightarrow{OB}$, $\overrightarrow{c}=\overrightarrow{OC}$ and $\overrightarrow{d}=\overrightarrow{OD}$. + + \item Let $ABCD$ be a quadrilateral and $O$ any point in the plane of this quadrilateral. Is the equivalence true that the quadrilateral $ABCD$ is a parallelogram exactly when $\overrightarrow{OA}+\overrightarrow{OC}= + \overrightarrow{OB}+\overrightarrow{OD}$? + + \item Let $ABCD$ be a parallelogram, $S$ +the intersection of its diagonals and $M$ any point in the plane of this parallelogram. Prove that: + $$\overrightarrow{MS} = \frac{1}{4}\cdot + \left( \overrightarrow{MA}+\overrightarrow{MB} + +\overrightarrow{MC} + \overrightarrow{MD} \right).$$ + + \item Let $ABB_1A_2$, +$BCC_1B_2$ and $CAA_1C_2$ be parallelograms, which are drawn over the sides of triangle $ABC$. Prove that: +$$\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}+ +\overrightarrow{C_1C_2}=\overrightarrow{0}.$$ + + \item The perpendicular lines $p$ and $q$, which intersect in point $M$, intersect the circle $k$ with center $O$ in +points $A$, $B$, $C$ and $D$. Prove that: +$$\overrightarrow{OA}+ \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = 2\overrightarrow{OM}.$$ + +\item Let $A$, $B$, $C$ and $D$ be any points in some plane. Can all six lines, which are determined by these points, be oriented so that the sum of the corresponding six vectors is equal to the vector of nothing? + +\item Let $A_1$, $B_1$ and $C_1$ be the centers of sides $BC$, $AC$ and $AB$ of triangle $ABC$ and $M$ any point. Prove: + +(\textit{a}) $\overrightarrow{AA_1}+\overrightarrow{BB_1}+ +\overrightarrow{CC_1}=\overrightarrow{0}$,\\ + (\textit{b}) $\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}= + \overrightarrow{MA_1}+\overrightarrow{MB_1}+\overrightarrow{MC_1}$,\\ + (\textit{c}) There exists a triangle $PQR$, such that for its vertices it holds:\\ \hspace*{7mm} $\overrightarrow{PQ}=\overrightarrow{CC_1}$, $\overrightarrow{PR}=\overrightarrow{BB_1}$ and $\overrightarrow{RQ}=\overrightarrow{AA_1}$. + + \item Let $M$, $N$, $P$, $Q$, $R$ and $S$ be in order the centers of the sides of an arbitrary hexagon. +Prove that it holds: +$$\overrightarrow{MN}+\overrightarrow{PQ}+ +\overrightarrow{RS}=\overrightarrow{0}.$$ + + + %Linearna kombinacija vektorjev + %_____________________________________ + + \item Let $ABCDEF$ be a convex hexagon, for which $AB\parallel DE$, and points $K$ and $L$ are the centers of the lines determined by the centers of the remaining pairs of opposite sides. Prove that $K=L$ exactly when $AB\cong DE$. + + \item Let $P$ and $Q$ be such points of sides $BC$ and $CD$ of a parallelogram $ABCD$, that $BP:PC=2:3$ and +$CQ:QD=2:5$. Point $X$ is the intersection of lines $AP$ and $BQ$. Calculate the ratios in which point $X$ divides +line $AP$ and $BQ$. + +\item Let $A$, $B$, $C$ and $D$ be arbitrary points of a plane. Point $E$ is the center of line $AB$, $F$ and +$G$ are such points, that $\overrightarrow{EF} = \overrightarrow{BC}$ and $\overrightarrow{EG} = \overrightarrow{AD}$, and $S$ is the center of line $CD$. Prove that $G$, $S$ and $F$ +are collinear points. + +\item Let $K$ and $L$ be such points of side $AD$ and diagonal $AC$ of a parallelogram $ABCD$, that $\frac{\overrightarrow{AK}}{\overrightarrow{KD}}=\frac{1}{3}$ and + $\frac{\overrightarrow{AL}}{\overrightarrow{LC}}=\frac{1}{4}$. Prove that $K$, $L$ and $B$ are collinear points. + +\item Let $X_n$ and $Y_n$ ($n\in \mathbb{N}$) be such points of sides $AB$ and $AC$ of triangle $ABC$, that $\overrightarrow{AX_n}=\frac{1}{n+1}\cdot \overrightarrow{AB}$ and + $\overrightarrow{AY_n}=\frac{1}{n}\cdot \overrightarrow{AC}$. Prove that there exists a point, which lies on all lines +$X_nY_n$ ($n\in \mathbb{N}$). + + \item Let $M$, $N$, $P$ and $Q$ be the centers of sides $AB$, $BC$, $CD$ and $DA$ + of quadrilateral + $ABCD$. Is the following equivalence true, that the quadrilateral $ABCD$ is a parallelogram exactly when: + + (\textit{a}) $2\overrightarrow{MP}=\overrightarrow{BC}+\overrightarrow{AD}$ and + $2\overrightarrow{NQ}=\overrightarrow{BA}+\overrightarrow{CD}$?\\ + (\textit{b}) $2\overrightarrow{MP}+2\overrightarrow{NQ}= + \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}$? + + \item Let $E$, $F$ and $G$ be the centers of sides $AB$, $BC$ and $CD$ of parallelogram $ABCD $, lines $BG$ and $DE$ intersect line $AF$ in points $N$ and $M$. Express $\overrightarrow{AF}$, $\overrightarrow{AM}$ and $\overrightarrow{AN}$ as a linear combination of vectors $\overrightarrow{AB}$ and $\overrightarrow{AD}$. Prove that points $M$ and $N$ divide the distance $AF$ in the ratio $2:2:1$. + + \item Points $K$, $L$, $M$ and $N$ lie on sides $AB$, $BC$, $CD$ and $DA$ +of quadrilateral $ABCD$. If the quadrilateral $KLMN$ is a parallelogram and it is true that +$$\frac{\overrightarrow{AK}}{\overrightarrow{KB}}= +\frac{\overrightarrow{BL}}{\overrightarrow{LC}} +=\frac{\overrightarrow{CM}}{\overrightarrow{MD}}= +\frac{\overrightarrow{DN}}{\overrightarrow{NA}}=\lambda$$ for some $\lambda\neq\pm 1$, +then the quadrilateral $ABCD$ +is a parallelogram. Prove it. + +\item Let $M$ be the center of side $DE$ of regular hexagon $ABCDEF$. Point +$N$ is the center of line $AM$, point $P$ is the center of side $BC$. Express $\overrightarrow{NP}$ as a linear combination of vectors +$\overrightarrow{AB}$ and $\overrightarrow{AF}$. + + + %The length of a vector + %_____________________________________ + + \item Prove that for any points $A$, $B$ and $C$ it is true that: + +(\textit{a}) $|\overrightarrow{AC}|\leq|\overrightarrow{AB}|+|\overrightarrow{BC}|$ \hspace*{6mm} + (\textit{b}) $|\overrightarrow{AC}|\geq|\overrightarrow{AB}|-|\overrightarrow{BC}|$\\ + Under which conditions is the equality true? + + \item Let $M$ and $N$ be points that lie on the lines $AD$ and $BC$, respectively, such that $\frac{\overrightarrow{AM}}{\overrightarrow{MD}}\cdot \frac{\overrightarrow{CN}}{\overrightarrow{NB}}=1$. Prove that: + $$|MN|\leq\max\{|AB|, |CD|\}.$$ + + %Težišče + %_____________________________________ + + \item Points $T$ and $T'$ are the centroids of $n$-sided polygons $A_1A_2...A_n$ and $A'_1A'_2...A'_n$. Calculate: +$$\overrightarrow{A_1A'_1}+\overrightarrow{A_2A'_2}+\cdots+\overrightarrow{A_nA'_n}.$$ + + \item Prove that quadrilaterals $ABCD$ and $A'B'C'D'$ have a common centroid exactly when: +$$\overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}+ +\overrightarrow{DD'}=\overrightarrow{0}.$$ + + \item Let $P$, $Q$, $R$ and $S$ be the centroids of triangles $ABD$, $BCA$, $CDB$ and $DAC$. Prove that quadrilaterals $PQRS$ and $ABCD$ have a common centroid. + + \item Let $A_1A_2A_3A_4A_5A_6$ be an arbitrary hexagon and $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ and $B_6$ are the centroids of triangles $A_1A_2A_3$, $A_2A_3A_4$, $A_3A_4A_5$, $A_4A_5A_6$, $A_5A_6A_1$ and $A_6A_1A_2$, respectively. +Prove that these centroids determine a hexagon with three pairs of parallel sides. + + \item Let $A$, $B$, $C$ and $D$ be four different points. Points $T_A$, $T_B$, $T_C$ and $T_D$ + are the centroids of triangles $BCD$, $ACD$, $ABD$ and $ABC$, respectively. Prove that the lines $AT_A$, $BT_B$, $CT_C$ and $DT_D$ intersect at one point $T$. In what ratio does point $T$ divide these lines? + +\item Let $CC_1$ be the centroid of the triangle $ABC$ and $P$ any point on the side +$AB$ of this triangle. The parallel $l$ of the line $CC_1$ through the point $P$ intersects the line $AC$ and $BC$ in points $M$ and $N$. Prove that: +$$\overrightarrow{PM} + \overrightarrow{PN}= \overrightarrow{AC} + \overrightarrow{BC}.$$ + + + %Hamilton and Euler + %______________________________________________________ + + + \item Let $A$, $B$, $C$ be points of a plane that lie on the same side of the line $p$, and $O$ a point on +the line $p$, for which $|\overrightarrow{OA}| = |\overrightarrow{OB}| = |\overrightarrow{OC}| =1$. Prove that then it also holds: $$|\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}| \geq 1.$$ + +\item Calculate the angles determined by the vectors $\overrightarrow{OA}$, $\overrightarrow{OB}$ and $\overrightarrow{OC}$, if the points $A$, $B$ and $C$ lie +on a circle with center +$O$ and in addition it holds: +$$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{0}.$$ + + \item Let $A$, $B$, $C$ and $D$ be points that lie +on a circle with center +$O$, and it holds +$$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = \overrightarrow{0}.$$ +Prove that $ABCD$ is a rectangle. + +\item Let $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ be vectors of a plane, for which $|\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{c}| =x$. Investigate, in which case it also holds $|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}| = x$. + + + %Tales + %______________________________________________________ + + + \item Divide the distance $AB$: + + (\textit{a}) into five equal distances,\\ + (\textit{b}) in the ratio $2:5$,\\ + (\textit{c}) into three distances, which are in the ratio $2:\frac{1}{2}:1$. + + + \item The distance $AB$ is given. Only by using the rule with the possibility of drawing parallels (constructions in affine geometry) plot the point $C$, if it is: + +(\textit{a}) $\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}$ \hspace*{3mm} + (\textit{b}) $\overrightarrow{AC}=\frac{3}{5}\overrightarrow{AB}$ \hspace*{3mm} + (\textit{c}) $\overrightarrow{AC}=-\frac{4}{7}\overrightarrow{AB}$ + + \item The distances $a$, $b$ and $c$ are given. Plot the distance $x$, so that it will: + +(\textit{a}) $a:b=c:x$ \hspace*{3mm} + (\textit{b}) $x=\frac{a\cdot b}{c}$ \hspace*{3mm} + (\textit{c}) $x=\frac{a^2}{c}$\hspace*{3mm}\\ + (\textit{č}) $x=\frac{2ab}{3c}$\hspace*{3mm} + (\textit{č}) $(x+c):(x-c)=7:2$ + + \item Let $M$ and $N$ be points on the arm $OX$, $P$ point on the arm $OY$ angle +$XOY$ and $NQ\parallel MP$ and $PN\parallel QS$ ($Q\in OY$, $S\in OX$). Prove that +$|ON|^2=|OM|\cdot |OS|$ (for the distance $ON$ in this case we say that it is \index{geometrijska sredina daljic}\pojem{geometrijska sredina} of the distances $OM$ +and $OS$). + + +\item Let $ABC$ be a triangle and $Q$, $K$, $L$, $M$, $N$ and $P$ such points on the sides $AB$, $AC$, $BC$, +$BA$, $CA$ and $CB$, that $AQ\cong CP\cong AC$, $AK\cong BL\cong AB$ and $BM\cong CN\cong BC$. +Prove that $MN$, $PQ$ and $LK$ are three parallel. + +\item Let $P$ be the center of the centroid $AA_1$ of the triangle $ABC$. The point $Q$ is the intersection of the line $BP$ +with the side $AC$. Calculate the ratios $AQ:QC$ and $BP:PQ$. + +\item The points $P$ and $Q$ lie on the sides $AB$ and $AC$ of the triangle $ABC$, and $\frac{|\overrightarrow{PB}|}{|\overrightarrow{AP}|} + +\frac{|\overrightarrow{QC}|}{|\overrightarrow{AQ}|}=1$. Prove that the centroid of this triangle lies on the line $PQ$. + + \item Let $a$, $b$ and $c$ be three sides with a common starting point $S$ and $M$ +point on the side $a$. If the point $M$ "moves" along the side $a$, + the ratio of the distance of this point from the lines $b$ and $c$ is constant. Prove it. + + \item Let $D$ be a point that lies on the side $BC$ of the triangle $ABC$ and + $F$ and $G$ points in which the line passing through the point $D$ and parallel to the centroid $AA_1$, intersects the lines $AB$ and $AC$. +Prove that the sum $|DF|+|DG|$ is constant if the point $D$ "moves" along the side +$BC$. + + + \item + Draw a triangle with the following data: + + (\textit{a}) $v_a$, $r$, $b-c$ \hspace*{3mm} + (\textit{b}) $\beta$, $r$, $b-c$ + +\end{enumerate} + + + + + + +% DEL 6 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% IZOMETRIJE +%________________________________________________________________________________ + +\del{Isometries} \label{pogIZO} + +%________________________________________________________________________________ +\poglavje{Isometries. Identity Map} \label{odd6Ident} + +We have formally defined isometries or isometric transformations of a plane $\mathcal{I}:\mathbb{E}^2\rightarrow \mathbb{E}^2$ in section \ref{odd2AKSSKL} as transformations of a plane that preserve the relation of congruence of pairs of points. We later used them to define the relation of congruence of figures. We intuitively identify them as movements of a plane. Some of them we already know from before (we have not formally introduced them in this book) - rotation and translation (Figure \ref{sl.izo.6.1.1.pic}). Even reflection over a line is an isometry. But this differs from the aforementioned two isometries because it is not a true movement (in a plane). In order to see it as a movement, we need to go to a space where we can rotate the plane by $180^0$. This changes the orientation of the plane. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.1.1.pic} +\caption{} \label{sl.izo.6.1.1.pic} +\end{figure} + +We call such isometries, which change the orientation of the plane, \index{izometrija!indirektna} \pojem{indirect}. +Reflection over a line is therefore an indirect isometry. For those isometries that preserve the orientation of the plane, we say that they are \index{izometrija!direktna}\pojem{direct}. Rotation and translation are examples of direct isometries. At this point, we will not prove the fact that every isometry of a plane is either direct or indirect, or that if one isometry preserves the orientation of one figure, it also preserves the orientation of all other figures. + +It is clear that the composition of two direct or two indirect isometries represents a direct isometry. Similarly, the composition of one direct and one indirect isometry is an indirect isometry. + +Intuitively, with direct isometries we can bring the figure and its image with movement in the plane to the point of overlap. For indirect isometries this is not possible with free movement in the plane - it is necessary to use movement in three-dimensional space. + +Except for the conditions of directness and indirectness, another very important property of isometries is the number of fixed points. We recall that we call a point fixed under an isometry if it is mapped to itself by that isometry (section \ref{odd2AKSSKL}). Intuitively, a rotation has exactly one fixed point – its center. A translation has no fixed points. A reflection over a line has infinitely many, but they are all fixed points on the axis of that reflection. +Is it possible for an isometry to have three non-collinear fixed points? Intuitively, it is clear (and we have also formally proven it in Theorem \ref{IizrekABCident}) that all other points in the plane are also fixed under such an isometry. We have called such an isometry an \index{isometry!identical} identical isometry or the identity, and we have denoted it with $\mathcal{E}$. Obviously, the identity is also a direct isometry, since it preserves all figures. +Because of its importance, we will write Theorem \ref{IizrekABCident} once again, but in a slightly different form. + +\bizrek \label{IizrekABC2} +The identity map $\mathcal{E}$ is the only isometry of a plane having three non-collinear fixed points. + \eizrek + +With the following theorem, we will provide another criterion for the identity. + +\bizrek \label{izo2ftIdent} +A direct isometry of a plane that has at least two fixed points is the identity map. +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.1.2.pic} +\caption{} \label{sl.izo.6.1.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} + We assume that the direct isometry +plane $\mathcal{I}$ has at least two fixed points $A$ and $B$ or +$A'=\mathcal{I}(A)=A$ and $B'=\mathcal{I}(B)=B$ (Figure +\ref{sl.izo.6.1.2.pic}). Let $X$ be an arbitrary point that does not lie on the line $AB$. Let $X'=\mathcal{I}(X)$. Because $\mathcal{I}$ +is a direct isometry, the triangles $AXB$ and $AX'B$ are oriented the same, so the point $X'$ lies in the plane with the edge $AB$ and +with the point $X$. From $\mathcal{I}:A,B,X\mapsto A,B,X'$ it follows + $XA\cong X'A$, $XB\cong X'B$ and $AB\cong AB$. Because +also $XA\cong XA$ and $XB\cong XB$, by + \ref{izomEnaC'} $X=X'$. Therefore, $X$ is also a fixed point. Because $A$, $B$ and $X$ are three + non-linear + points of the isometry $\mathcal{I}$, $\mathcal{I}=\mathcal{E}$ + (statement \ref{IizrekABC2}). + \kdokaz + +In the following we will formally introduce and consider different +types of isometries. + +%________________________________________________________________________________ + \poglavje{Reflections} \label{odd6OsnZrc} + +Although the basic reflection is an intuitively already known transformation, we will +first give a formal definition. + +Let $s$ be a line of some plane. The transformation of this plane, in which each point of the line $s$ is fixed and in which each point $X$, which does not lie on the line $s$, is mapped to such a point $X'$, + that $s$ is the perpendicular bisector of the segment $XX'$, is called + \index{reflection!over a line} \pojem{basic reflection} + or \index{reflection!basic}\pojem{reflection over a line} + $s$ and is denoted by $\mathcal{S}_s$ (Figure \ref{sl.izo.6.2.1.pic}). + The line $s$ is + \index{axis!of reflections} \pojem{the axis of reflections}. + + Because from the notation $\mathcal{S}_s$ + it is already clear that it is a reflection over the line $s$, we will call it + shorter: reflection $\mathcal{S}_s$. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.1.pic} +\caption{} \label{sl.izo.6.2.1.pic} +\end{figure} + +If the figure $\phi$ is mapped onto the figure $\phi'$ by the reflection $\mathcal{S}_s$, or $\mathcal{S}_s: \phi \rightarrow \phi'$, we will say that the figures $\phi$ and $\phi'$ are \pojem{symmetric} with respect to the axis $s$ (Figure \ref{sl.izo.6.2.2.pic}). The axis $s$ is the \pojem{axis of symmetry} of the figures $\phi$ and $\phi'$. + + If $\phi=\phi'$ or $\mathcal{S}_s: \phi \rightarrow \phi$, we will say that the figure $\phi$ is \index{figure!axis-symmetric} \pojem{axis-symmetric} or \pojem{axis-centered}. The line $s$ is the \index{axis!symmetry of a figure}\pojem{axis of symmetry} or \index{centered line} \pojem{centered line} of this figure (Figure \ref{sl.izo.6.2.2.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.2.pic} +\caption{} \label{sl.izo.6.2.2.pic} +\end{figure} + + From the definition it is already clear that all the fixed points of the reflection $\mathcal{S}_s$ lie on the axis $s$ of this reflection. + + Although it is intuitively clear, we need to prove the following property of the defined mapping. + + + + \bizrek \label{izozrIndIzo} + A reflection is an opposite isometry. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.3.pic} +\caption{} \label{sl.izo.6.2.3.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Let $\mathcal{S}_s$ be a reflection over the line $s$ (Figure +\ref{sl.izo.6.2.3.pic}). From the definition it is clear that it represents a bijective mapping. It remains to + prove that for any points + $X$ and $Y$, which are mapped by $\mathcal{S}_s$ into the points $X'$ and $Y'$, it holds that + $XY\cong X'Y'$. We will consider several cases. + + If $X=X'$ and $Y=Y'$, this relation + is automatically fulfilled (statement \ref{sklRelEkv}). + + Let $X\neq + X'$. We will mark with $X_s$ the center of the line segment $XX'$. Because $s$ is the symmetry of + the line segment $XX'$, by definition $X_s\in s$ and $XX'\perp s$. + If $Y=Y'$, the triangles $XX_sY$ and $X'X_sY$ (or $X'X_sY'$) are + congruent (statement + \textit{SAS} \ref{SKS}), therefore $XY\cong X'Y'$. + +We will now look at an example of $X\neq X'$ and $Y\neq Y'$. Similarly, for + the center + $Y_s$ of the line $YY'$ it holds that $Y_s\in s$ and $YY'\perp s$. + The triangles $XX_sY_s$ and $X'X_sY_s$ + are + congruent (by the + theorem + \textit{SAS}, so there exists an isometry $\mathcal{I}$, such that + $\mathcal{I}: X, X_s,Y_s\mapsto X', X_s,Y_s$. This + maps the line segment $X_sY_sX$ onto the line segment $X_sY_sX'$ + (by axiom \ref{aksIII2}). Because the isometry $\mathcal{I}$ preserves angles, the line segment $Y_sY$ is mapped onto the line segment $Y_sY'$, + the point $Y$ is mapped onto the point $Y'$. From + $\mathcal{I}: X, Y\mapsto X', Y'$ it follows at the end that $XY\cong X'Y'$. + + Let $A,B\in s$ and $C\notin s$. Then $\mathcal{S}_s(A)=A$, $\mathcal{S}_s(B)=A$ and $\mathcal{S}_s(C)=C'\neq C$. This means that the triangle $ABC$ is mapped onto the triangle $ABC'$ by the basic reflection $\mathcal{S}_s$. Because $C,C' \div s$ or $C,C' \div AB$, the two triangles are differently oriented, so the basic reflection $\mathcal{S}_s$ is an indirect isometry. + \kdokaz + +It is clear that in the proof of the previous theorem it also holds that +$\mathcal{I}=\mathcal{S}_s$. We can only determine this at the end, when +we prove that $\mathcal{S}_s$ is an isometry and use theorem +\ref{IizrekABC}. + +We will now prove some simple properties of reflections across a line. + + + \bizrek \label{izoZrcPrInvol} + For an arbitrary line $p$ it holds that $$\mathcal{S}_p^2=\mathcal{E}\hspace*{1mm}\textrm{ i.e. } + \hspace*{1mm}\mathcal{S}_p^{-1}=\mathcal{S}_p.$$ + \eizrek + + +\textbf{\textit{Proof.}} It is enough to prove that $\mathcal{S}_p^2(X)=X$ for every point $X$ of the plane. Let +$\mathcal{S}_p(X)=X'$. If $X\in p$ or $X=X'$, the relation +$\mathcal{S}_p^2(X)=X$ is automatically fulfilled. If $X\notin p$, by definition the line $p$ is the perpendicular bisector of the line segments $XX'$ (and also $X'X$), +so $\mathcal{S}_p(X')=X$ or $\mathcal{S}_p^2(X)=X$. + \kdokaz + +Every isometry (as well as every mapping) $f:\mathbb{E}^2\rightarrow \mathbb{E}^2$, + for which $f^2=\mathcal{E}$, is called + \pojem{involution}\index{involution}. + The basic reflection is therefore an involution. + + + + \bizrek \label{zrcFiksKroz} + Let $l$ be an arbitrary circle + with the centre $S$ and $p$ a line in the plane. + If $S\in p$, then $\mathcal{S}_p(l)=l$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.4.pic} +\caption{} \label{sl.izo.6.2.4.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Let $X\in l$ be an arbitrary point of the +circle $l$ and $\mathcal{S}_p(X)=X'$ (Figure \ref{sl.izo.6.2.4.pic}). +Because $\mathcal{S}_p:S,X\mapsto S, X'$, we have $SX\cong SX'$ or +$X'\in l$. Therefore $\mathcal{S}_p(l)\subseteq l$. Similarly, an +arbitrary point $Y$ of the circle $l$ is the image of the point +$Y'=\mathcal{S}_p^{-1}(Y)=\mathcal{S}_p(Y)$, which lies on this +circle. Therefore $\mathcal{S}_p(l)\supseteq l$. Therefore +$\mathcal{S}_p(l)=l$. + \kdokaz + + \bzgled \label{izoSimVekt} + If + $\mathcal{S}_s:A, B\mapsto A', B'$, the vector + $\overrightarrow{v}=\overrightarrow{AB}+\overrightarrow{A'B'}$ + is parallel to the line $s$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.5.pic} +\caption{} \label{sl.izo.6.2.5.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Let $A_s$ and $B_s$ be the centres of the +lines $AA'$ and $BB'$ (Figure \ref{sl.izo.6.2.5.pic}). Because $p$ +is the symmetry of these lines, $A_s, B_s\in p$. Therefore the +vector + \begin{eqnarray*} + \overrightarrow{v}&=&\overrightarrow{AB}+\overrightarrow{A'B'}=\\ + &=&(\overrightarrow{AA_s}+\overrightarrow{A_sB_s}+\overrightarrow{B_sB})+ + (\overrightarrow{A'A_s}+\overrightarrow{A_sB_s}+\overrightarrow{B_sB'})=\\ + &=& 2\overrightarrow{A_sB_s} + \end{eqnarray*} +is parallel to the line $s$. + \kdokaz + +We will continue with the properties of the basic reflection in the next section, + but now let's look at the use of this isometry. + + + + \bzgled \label{HeronProbl} + (Heron's\footnote{This problem was + posed by \index{Heron}\textit{Heron of Alexandria} (20--100). In + his work 'Catoprica' he formulated a law that says that a beam that goes from + point $A$ and is reflected from line $p$ through point $B$, passes + the shortest possible path.} problem) \index{problem!Heron's} + Two points $A$ and $B$ are given on the same side of a line $p$. + Find a point $X$ on the line $p$ such that the sum $|AX|+|XB|$ is minimal. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.6.pic} +\caption{} \label{sl.izo.6.2.6.pic} +\end{figure} + + +\textbf{\textit{Solution.}} + Let $A'$ be the image of point $A$ under +reflection $\mathcal{S}_p$ (Figure \ref{sl.izo.6.2.6.pic}). With $X$ +we mark the intersection of line $p$ and $A'B$ (points $A'$ and $B$ are on +different sides of line $p$). We will prove that $X$ is the desired point. + +Let $Y\neq X$ be an arbitrary point on line $p$. Because reflection +$\mathcal{S}_p$ is an isometry that maps line segment $AX$ to line segment $A'X$ and line segment $AY$ to line segment $A'Y$, we have $AX\cong A'X$ and $AY\cong A'Y$. If +we also use the triangle inequality - \ref{neenaktrik} (for triangle $A'YB$), we get: $$|AX| + |XB| = |A'X| + |XB| = |A'B| < +|A'Y| + |YB| = |AY| + |YB|,$$ which was to be proven. \kdokaz + + + + \bzgled + Let $k$ and $l$ be circles on the same side of a line $p$ in the same plane. + Construct a point $S$ on the line $p$ such that + the tangents from this point to the circles $k$ and $l$ + determine the congruent angles with the line $p$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.7.pic} +\caption{} \label{sl.izo.6.2.7.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $q$ and $r$ be the tangents in that order of the circles $k$ and $l$, which intersect on the line $p$ in the point $S$ and with it determine the corresponding angle (Figure \ref{sl.izo.6.2.7.pic}). The line $p$ is the angle bisector of the angle determined by the tangents $q$ and $r$, therefore $\mathcal{S}_p(r)=q$. The line $q$ is therefore also a tangent of the circle $l'=\mathcal{S}_p(l)$. This means that we can plan the line $q$ as the common tangent of the circles $k$ and $l'$ (see example \ref{tang2ehkroz}). Then $S=q\cap p$ and $r=\mathcal{S}_p(q)$. + \kdokaz + + + \bzgled \label{FagnanLema} + Let $AP$, $BQ$ and $CR$ be the altitudes of a triangle $ABC$. + If $P'=\mathcal{S}_{AB}(P)$ and + $P''=\mathcal{S}_{AC}(P)$, prove that $P'$, $R$, $Q$ and $P''$ + are collinear points. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.9a.pic} +\caption{} \label{sl.izo.6.2.9a.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.2.9a.pic}) First, +from $\angle BRC=90^0$ and $\angle BQC=90^0$ by Tales' theorem +\ref{TalesovIzrKroz} it follows that points $R$ and $Q$ lie on a +circle with diameter $BC$. Therefore $BRQC$ is a chordal quadrilateral +and by theorem \ref{TetivniPogojZunanji} it holds that $\angle +ARQ\cong \angle BCQ = \gamma$. Similarly (from the chordality of +$CARP$ quadrilateral) it also holds that $\angle BRP\cong\gamma$, so +$\angle ARQ\cong \angle BRP$. Because $\mathcal{S}_{AB}:P, R, B\mapsto +P', R, B$, it holds that $\angle BRP\cong \angle BRP'$. From the +previous relations it follows that $\angle ARQ\cong \angle BRP'$. +Triangle $ABC$ is acute, which means that points $P$ and $Q$ are on +the same side of line $AB$. From this it follows that points $Q$ and +$P'$ are on different sides of line $AB$. From the proven relation +$\angle ARQ\cong \angle BRP'$ it now follows that $P'$, $R$ and $Q$ +are collinear points. Similarly, points $P''$, $R$ and $Q$ are +collinear, which means that all four points $P'$, $R$, $Q$ and $P''$ +lie on the same line. + \kdokaz + + + + \bizrek \label{FagnanLema1} + Let $A$ be a point not lying on lines $p$ and $q$ in the same plane. + Construct points $B\in p$ and $C\in q$ such that + the perimeter of the triangle $ABC$ is minimal. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.8.pic} +\caption{} \label{sl.izo.6.2.8.pic} +\end{figure} + + +\textbf{\textit{Solution.}} + Let $A'=\mathcal{S}_p(A)$, $A''=\mathcal{S}_q(A)$ and $B$ and $C$ + be the intersection points of line $A'A''$ with lines $p$ and $q$ + (Figure \ref{sl.izo.6.2.8.pic}). + +We will prove that $ABC$ is the desired triangle. If $B_1$ and $C_1$ are any points (where $B_1\neq B$ or $C_1\neq C$) that lie on the lines $p$ and $q$, the perimeter of the triangle $AB_1C_1$ is equal to the length of the broken line $A'B_1C_1A''$ ($AB_1\cong A'B_1$ and $AC_1\cong A'' C_1$), and the perimeter of the triangle $ABC$ is equal to the length of the line segment $A'A''$ ($AB\cong A'B$ and $AC\cong A'' C$). From Theorem \ref{neenakIzlLin} it follows that the first length is greater than the second, so the perimeter of the triangle $ABC$ is less than the perimeter of the triangle $AB_1C_1$. + \kdokaz + + + \bzgled + For a given acute triangle $ABC$ determine the inscribed triangle of minimal perimeter + \index{problem!Fagnano}(Fagnano's problem\footnote{\textit{G. F. Fagnano} + \index{Fagnano, G. F.} (1717--1797), Italian mathematician, + posed this problem in 1775 and solved + it + with the methods + of differential calculus. A more + elementary solution to this problem was later given by the Hungarian + mathematician \index{Fejer, L.} \textit{L. Fejer} (1880--1959).}). + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.9.pic} +\caption{} \label{sl.izo.6.2.9.pic} +\end{figure} + + +\textbf{\textit{Solution.}} We mark for $AP$, $BQ$ and $CR$ the altitudes of the triangle $ABC$ (Figure \ref{sl.izo.6.2.9.pic}). + If $X$ +is any point on the side $BC$, then the minimal perimeter of the triangle +$XYZ$, where the vertices $Y$ and $Z$ lie on the sides $AC$ and $AB$, +is obtained in the way described in the previous Theorem +\ref{FagnanLema1}. The points $Y$ and $Z$ are therefore the intersections of the sides +$AC$ and $AB$ with the line $X'X''$ ($X'=\mathcal{S}_{AB}(X)$, +$X''=\mathcal{S}_{AC}(X)$). In this case, the perimeter of the triangle $XYZ$ is equal +to the length of the line segment $X'X''$. + +So it remains for us to determine for which point $X$ of side +$BC$ is the perimeter of triangle $XYZ$ the smallest or the distance $X'X''$ +the shortest. From the properties of reflection it follows that $AX'\cong AX\cong AX''$, +$\angle X'AB\cong \angle BAX$ and $\angle X''AC\cong \angle CAX$ +or $\angle X'AX''=2\angle BAC$. As $X'AX''$ is an isosceles triangle $X'AX''$ is constant (depending on the point $X$), therefore +its base $X'X''$ is the shortest when its shortest leg $AX'$ (by \ref{SkladTrikLema}) or the distance +$AX\cong AX'\cong AX''$. This is fulfilled when $X$ is the foot of the altitude of triangle $ABC$ from point $A$ or $X=P$. From \ref{FagnanLema} it follows that $Y=Q$ and $Z=R$. So the inscribed triangle with the smallest perimeter is actually the pedal triangle of triangle +$ABC$. + \kdokaz + + + + \bzgled + Let $P$ and $Q$ be interior points of an angle $aOb$. + Construct a point $X$ on the side $a$ of this angle such that + the rays $XP$ and $XQ$ intersect the side $b$ at points $Y$ and $Z$ such that $XY\cong XZ$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.10.pic} +\caption{} \label{sl.izo.6.2.10.pic} +\end{figure} + +\textbf{\textit{Solution.}} + Let $X$, $Y$ and $Z$ be the sought points (Figure \ref{sl.izo.6.2.10.pic}). + Let $S$ be the center of the segment $YZ$ and $P'=\mathcal{S}_a(P)$. + The measure of the angle +$aOb$ we denote by $\omega$. The triangle $XYZ$ is equilateral, therefore the +circumcenter $XS$ is at the same time the altitude and the symmetry of the internal angle $XYZ$ + of this +triangle. So it holds that $\angle OSX=90^0$ and $\angle YXS\cong +\angle ZXS$. From $\mathcal{S}_a:P, X, O\mapsto P', X, O$ it follows +that $\angle PXO\cong \angle P'XO$. Now we can calculate the measure of the angle +$P'XQ$: + \begin{eqnarray*} +\angle P' XQ &=& \angle P' XP + \angle PXQ = 2\angle OXP + 2\angle +PXS\\ &=& 2\angle OXS = 2(90° - \omega) =180° - 2\omega. + \end{eqnarray*} +Since the points $P'$ and $Q$ are known to us, the point $X$ is the +intersection of the segment $a$ and the arc with the chord $P'Q$ and the central angle $180° +- 2\omega$ (the theorem \ref{ObodKotGMT}). + \kdokaz + + + + \bzgled + Let $ABCD$ be a rectangle such that $|AB|=3|BC|$. Suppose that $E$ and $F$ are + points on the side $AB$ such that + $AE\cong EF\cong FB$. + Prove $$\angle AED+\angle AFD+\angle + ABD=90^0.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.11.pic} +\caption{} \label{sl.izo.6.2.11.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $F'=\mathcal{S}_{CD}(F)$ and +$B'=\mathcal{S}_{CD}(B)$ + (Figure \ref{sl.izo.6.2.11.pic}). + Then $DF'\cong DF$ and $\angle DF'F\cong \angle DFF'$. From the similarity of +triangles $DAF$ and $F'B'B$ (the \textit{SAS} theorem \ref{SKS}) it follows that +$DF\cong F'B$ and $\angle DFA\cong \angle B'BF'$. It is also true that +$\angle FF'B\cong \angle B'BF'$ (the \ref{KotiTransverzala} theorem). +Therefore: + $$\angle DF'B=\angle DF'F+\angle FF'B=\angle DFF'+\angle DFA=\angle AFF'=90^0.$$ + Because $DF'\cong F'B$, $DF'B$ is an isosceles right triangle with the hypotenuse $BD$, therefore (the \ref{enakokraki} theorem) $\angle DBF'=\angle BDF'=45^0$. Because $DAE$ is also an isosceles right triangle with the hypotenuse $DE$, $\angle AED=45^0$ or $\angle AED=\angle DBF'$. Therefore: +$$\angle AED+\angle AFD+\angle ABD=\angle DBF'+\angle F'BB'+ +\angle ABD=\angle ABC=90^0,$$ which was to be proven. \kdokaz + + At the next example we will describe the standard procedure of using + isometries in design tasks. We assume that it is necessary to construct points $X\in k$ and $Y\in l$, where $k$ and $l$ are given circles + (it can also be lines or a circle and a line), but we know that for some + isometry $\mathcal{I}$ it is true that $\mathcal{I}(X)=Y$. In this case + from $X\in k$ it follows that $Y=\mathcal{I}(X)\in \mathcal{I}(k)$. Because + $Y$ is also in $l$, we can construct the point $Y$ from the condition $Y\in + l\cap \mathcal{I}(k)$. + + + + \bzgled + Two circles $k$ and $l$ on different sides of a line $p$ are given. + Construct a square $ABCD$ such that $A\in k$, $C\in l$ and $B,D\in p$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.12.pic} +\caption{} \label{sl.izo.6.2.12.pic} +\end{figure} + + +\textbf{\textit{Solution.}} + (Figure \ref{sl.izo.6.2.12.pic}) + +Let $ABCD$ be the sought square. Then $\mathcal{S}_p(C)=A$. The point $C$ lies on the circle $l$, therefore its image -- point $A$ -- after reflection $\mathcal{S}_p$ lies on the image of the circle $l$ -- circle $l'$, which we can plot. But the point $A$ also lies on the circle $k$, therefore $A\in l'\cap k$. In this way, we first find the vertex $A$, and then $C=\mathcal{S}_p(A)$. The center of the square $S$ is found as the center of the diagonal $AC$. In the end, the vertices $B$ and $D$ are the intersections of the line $p$ and the circle with center $S$ and radius $SA$. + +The number of solutions depends on the number of intersections of the circles $l'$ and $k$. \kdokaz + + + + \bnaloga\footnote{38. IMO Argentina - 1997, Problem 2.} + The angle at $A$ is the smallest angle of triangle $ABC$. The points $B$ and $C$ + divide the circumcircle of the triangle into two arcs. Let $U$ be an interior point + of the arc between $B$ and $C$ which does not contain $A$. The perpendicular + bisectors of $AB$ and $AC$ meet the line $AU$ at $V$ and $W$, respectively. The lines + $BV$ and $CW$ meet at $T$. Show that $$|AU| =|TB| + |TC|.$$ + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.IMO1.pic} +\caption{} \label{sl.izo.6.2.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let's mark with $q$ and $p$ the + sides $AB$ and $AC$ of the triangle $ABC$ + (Figure \ref{sl.izo.6.2.IMO1.pic}). Their intersection -- point + $O$ -- is the center of the circumscribed circle $l$ of this triangle. According to + the assumption, $V=AU\cap q$ and $W=AU\cap p$. Let $D$ be the other + intersection of the line segment $CW$ with the circle $l$. With $\mathcal{S}_p$ + we mark the reflection over + the line $p$. $\mathcal{S}_p$ maps the points $A$, $C$, $W$ and $O$ + in order, into the points $C$, $A$, $W$ and $O$ and the line segment $AW$ + into the line segment $CW$. Because $O\in p$, by the theorem \ref{zrcFiksKroz} + $\mathcal{S}_p(l) =l$. From this follows: + $$\mathcal{S}_p(U) =\mathcal{S}_p(AW\cap l)= + \mathcal{S}_p(AW)\cap \mathcal{S}_p(l)= + CW\cap l=D.$$ + Therefore $\mathcal{S}_p:\hspace*{1mm} A,U\mapsto C,D$, so $AU\cong CD$ and $AD\cong CU$. From the similarity of the line segments $AD$ + and $CU$ follows the similarity of the corresponding angles + (theorem \ref{SklTetSklObKot}) or equivalently + $\angle ABD\cong\angle UAC$. But the angles $BDC$ and $BAC$ are also similar over the line segment $BC$ (theorem\ref{ObodObodKot}). + + Because the point $V$ lies on the line $q$, which is the symmetry of the line segment $AB$, + $\mathcal{S}_q:\hspace*{1mm} A,B,V\mapsto B,A,V$. This means that + $\angle BAV\cong\angle ABV$. + + From the similarity of the corresponding angles it follows: + \begin{eqnarray*} \angle BDT&=&\angle BDC\cong\angle BAC=\\ + &=&\angle BAU+ + \angle UAC=\\ + &=&\angle ABV+\angle ABD=\\ + &=&\angle DBT. + \end{eqnarray*} + Therefore the triangle $DTB$ is isosceles and by the theorem \ref{enakokraki} + it holds that $TD\cong TB$. In the end we get: + $$|AU|= |CD|=|CT|+|TD|=|CT|+|TB|,$$ which was to be proven. \kdokaz + +\bnaloga \footnote{50. IMO Germany - 2009, Problem 4.} +Let $ABC$ be a triangle with $|AB|=|AC|$. The angle bisectors of $\angle CAB$ and $\angle ABC$ +meet the sides $BC$ and $CA$ at $D$ and $E$, respectively. Let $K$ be the incentre of triangle $ADC$. +Suppose that $\angle BEK=45^0$. Find all possible values of $\angle CAB$. +\enaloga + + + + \textbf{\textit{Solution.}} We denote by $S$ the centre of the inscribed circle + of triangle $ABC$. + This means that $S$ is the intersection of the altitudes $AD$ and $BE$ + of angles $\alpha=\angle CAB$ and $\beta=\angle ABC$. The point $S$ also lies on + the altitude +$CF$ ($F\in AB$) of angle $\gamma=\angle ACB$. From $|AB|=|AC|$ it follows +$\beta\cong\gamma$, therefore $\angle SBC\cong\angle SCB$. + +By assumption, the point $K$ is the centre of the inscribed circle + of triangle $ADC$, therefore the point $K$ lies on the altitude $CS$ of + angle $\angle ACD=\angle ACB$. The segment $DK$ is the altitude of + angle $ADC$. From the similarity of triangles $ADB$ and $ADC$ (the \textit{SAS} theorem \ref{SKS}) + it follows that $\angle + ADC\cong\angle ADB=90^0$. From this it follows that $\angle SDK=45^0$. + +Let $E'=\mathcal{S}_{SC}(E)$. Because $CS$ is the altitude of angle $ACB$, +the segment $CA$ is mapped onto the segment $CB$ by the reflection $\mathcal{S}_{SC}$. Therefore $E'\in CB$. Because $S,K,C\in SC$, we have +$\mathcal{S}_{SC}:\hspace*{1mm}S,K,C\mapsto S,K,C$. + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.IMO2a.pic} +\caption{} \label{sl.izo.6.2.IMO2a.pic} +\end{figure} + +We will consider two cases. If $E'=D$ (Figure +\ref{sl.izo.6.2.IMO2a.pic}), then $\angle SEC\cong \angle SDC=90^0$. +Triangles $ABE$ and $CBE$ are similar in this case (the \textit{ASA} +theorem \ref{KSK}), therefore $AB\cong BC$. Triangle $ABC$ is +isosceles or $\angle BAC=60^0$. + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.IMO2.pic} +\caption{} \label{sl.izo.6.2.IMO2.pic} +\end{figure} + +Let $E'\neq D$ (Figure \ref{sl.izo.6.2.IMO2.pic}). From +$\mathcal{S}_{SC}:\hspace*{1mm}S,K\mapsto S,K$ it follows that $\angle SE'K +\cong \angle SEK=45^0$. Because $\angle SDK \cong \angle +SE'K=45^0$, by izreku \ref{ObodKotGMT} the points $S$, $K$, $D$ and +$E'$ are concircle. From izreku \ref{TalesovIzrKroz} it then follows +that $\angle SKE'\cong \angle SDE'=90^0$. Because the reflection +$\mathcal{S}_{SC}$ maps the angle $SKE$ into the angle $SKE'$, $\angle +SKE\cong \angle SKE'=90^0$ (the points $E$, $K$ and $E'$ are +therefore collinear). From this and izreku \ref{VsotKotTrik} for the triangle +$SKE$ it follows that $\angle KSE=45^0$. It is the outer angle of the triangle +$SBC$, so by izreku \ref{zunanjiNotrNotr} $\angle SBC +\angle +SCB=45^0$ or $\beta+\gamma=90^0$. From this relation it follows that +$\angle BAC=90^0$. + + The possible values of the angle + $\angle CAB$ are therefore $60^0$ and $90^0$. +\kdokaz + + +%________________________________________________________________________________ + \poglavje{More on Reflections} \label{odd6Sopi} + +In the initial research on isometries, we found that every direct isometry with at least two fixed points represents the identity. It is intuitively clear that an opposite isometry of the plane with at least one fixed point is a reflection. We will prove an even stronger statement. + + + + \bizrek \label{izo1ftIndZrc} + An opposite isometry of the plane with at least one fixed point is a reflection. + The axis of this reflection passes through this point. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.1.pic} +\caption{} \label{sl.izo.6.3.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} + (Figure \ref{sl.izo.6.3.1.pic}) + +Let $A$ be a fixed point of an indirect isometry $\mathcal{I}$ of +some plane. Because it is indirect, the isometry $\mathcal{I}$ is +not the identity, so there is such a point $B$ of this plane that +$\mathcal{I}(B)=B'\neq B$. Let $p$ be the perpendicular bisector of +the segment $BB'$. Therefore, $\mathcal{I}$ is an isometry that +maps the point $A$ and $B$ to the point $A$ and $B'$, so $AB\cong +AB'$. This means that the point $A$ lies on the perpendicular +bisector $p$ of the segment $BB'$. We prove that the isometry +$\mathcal{I}$ represents the reflection over the line $p$ or +$\mathcal{I} = \mathcal{S}_p$. The composition $\mathcal{S}_p \circ +\mathcal{I}$ (of two indirect isometries) is a direct isometry +with two fixed points $A$ and $B$, so based on the aforementioned +formula \ref{izo2ftIdent} it represents the identity. Therefore, +$\mathcal{S}_p \circ \mathcal{I}=\mathcal{E}$. From this it follows +that $ \mathcal{I}=\mathcal{S}_p^{-1} \circ\mathcal{E}$ or +$\mathcal{I} = \mathcal{S}_p$. + +The following formula is very useful. + + \bizrek \label{izoTransmutacija} + Let $\mathcal{I}$ be an isometry and $p$ an arbitrary line in the plane. + Suppose that this isometry maps the line $p$ to a line $p'$. If $\mathcal{S}_p$ is + a reflection with axis $p$, then + $$\mathcal{I}\circ \mathcal{S}_p\circ \mathcal{I}^{-1} = \mathcal{S}_{p'}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.2.pic} +\caption{} \label{sl.izo.6.3.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Let $Y$ be an arbitrary point of the line $p'$ and $X = \mathcal{I}^{-1}(Y)$ + (Figure \ref{sl.izo.6.3.2.pic}). + +Since $\mathcal{I}(p)=p'$ or $\mathcal{I}^{-1}(p')=p$, the point $X$ lies on the line $p$ and is $\mathcal{S}_p(X)=X$. Therefore, it holds: +$$\mathcal{I}\circ\mathcal{S}_p\circ\mathcal{I}^{-1}(Y)= + \mathcal{I}\circ\mathcal{S}_p(X)= + \mathcal{I}(X)=Y.$$ + Thus, the composition $\mathcal{I}\circ\mathcal{S}_p\circ\mathcal{I}^{-1}$ is an indirect isometry with a fixed point $Y$, therefore, according to the previous statement \ref{izo1ftIndZrc}, it represents the central reflection. But $Y$ is an arbitrary point of the line $p'$, which means that the axis of this reflection is just the line $p'$ or $\mathcal{I}\circ \mathcal{S}_p\circ \mathcal{I}^{-1} = \mathcal{S}_{p'}$. + \kdokaz + +The transformation +$\mathcal{I}\circ\mathcal{S}_p\circ\mathcal{I}^{-1}$ from the previous +statement is called the \index{transmutation!reflection across a line} +\pojem{transmutation} of the reflection $\mathcal{S}_p$ with the isometry $\mathcal{I}$. + +We prove a simple consequence of the previous statement. + + + + \bzgled \label{izoZrcKomut} + Two reflections in the plane commute if and only if + their axis are perpendicular or coincident, i.e. + $$\mathcal{S}_p \circ \mathcal{S}_q + = \mathcal{S}_q \circ \mathcal{S}_p \Leftrightarrow + (p\perp q \vee p=q).$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.3.pic} +\caption{} \label{sl.izo.6.3.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} + (Figure \ref{sl.izo.6.3.3.pic}) + +First, $\mathcal{S}_p \circ \mathcal{S}_q += \mathcal{S}_q \circ \mathcal{S}_p +\Leftrightarrow \mathcal{S}_p= \mathcal{S}_q \circ + \mathcal{S}_p \circ \mathcal{S}_q^{-1}$. If we use the previous +statement + \ref{izoTransmutacija} for $\mathcal{I} = \mathcal{S}_q$, +we get that the last equality is equivalent to $\mathcal{S}_p = +\mathcal{S}_{p'}$, where $p' = \mathcal{S}_q(p)$. This is exactly the case +when $p=p'=\mathcal{S}_q(p)$ i.e. when $p\perp q$ or $p=q$. + \kdokaz + +The composite of two basic reflections $\mathcal{I}=\mathcal{S}_p \circ +\mathcal{S}_q$ will be very important in further research +isometries. We immediately notice that the composite of two +indirect isometries represents a direct isometry. +If the axes of these reflections coincide, we have +already determined that it is the identity (statement \ref{izoZrcPrInvol}). +It is interesting to ask what $\mathcal{I}$ represents in the general case, +when the axes of $p$ and $q$ of these reflections are different and +coplanar; especially in the case when they intersect or are parallel. +We will find the answer to this question in the following sections. In +the next statement we will find out something about the fixed points of the composite +of two basic reflections. + + \bizrek \label{izoKomppqX} + Let $p$ and $q$ be two distinct coplanar lines. + Then + $$\mathcal{S}_q\circ\mathcal{S}_p(X)=X \Leftrightarrow p\cap q=\{X\}.$$ + \eizrek + +\textbf{\textit{Proof.}} + +($\Leftarrow$) Trivial, because from $X\in p$ and $X\in q$ it follows +$\mathcal{S}_p(X)= \mathcal{S}_q(X)=X$ i.e. +$\mathcal{S}_q\circ\mathcal{S}_p(X)=X$. + +($\Rightarrow$) Let $\mathcal{S}_q\circ\mathcal{S}_p(X)=X$ in +$\mathcal{S}_p(X)=X'$. From this it follows that $\mathcal{S}_q(X')=X$. +Assume that $X\neq X'$. In this case, according to the definition of +the axis of reflection $p$ and $q$ are the symmetrals of the line +segment $XX'$. But this is not possible, because the line segment has +one symmetral, while $p$ and $q$ are different by assumption. +Therefore, $X=X'$. From this it follows that +$\mathcal{S}_p(X)=\mathcal{S}_q(X)=X$, which means that $X\in p$ and +$X\in q$ or $X\in p\cap q$. Because $p$ and $q$ are different lines, +by Axiom \ref{AksI1} $p\cap q=\{X\}$. + \kdokaz + + If the lines $p$ and $q$ intersect, from the previous statement it follows that the composite + $\mathcal{S}_q\circ\mathcal{S}_p$ + has only one fixed point, which is their intersection point. If $p$ and +$q$ are parallel, the composite $\mathcal{S}_q\circ\mathcal{S}_p$ +has no fixed points. + +For further research of isometries, the following concept is very +characteristic. + +The set of all lines of a plane that pass through one point $S$ of +this plane, is called the \index{šop!konkurentnih premic} + \pojem{set of concurrent lines} (or \index{šop!eliptični}\pojem{elliptic set}) with center $S$ and is denoted by + $\mathcal{X}_S$. + +The set of all lines of a plane that are parallel to one line $s$ of +this plane, is called the \index{šop!vzporednic} \pojem{set of +parallel lines} (or \index{šop!parabolični}\pojem{parabolic set} or \index{snop premic}\pojem{set of lines}) and is denoted by + $\mathcal{X}_s$. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.4.pic} +\caption{} \label{sl.izo.6.3.4.pic} +\end{figure} + + + + In both cases we will talk about the \pojem{set of lines} + (Figure \ref{sl.izo.6.3.4.pic}). If we will not specifically emphasize whether it is a set + of concurrent lines or a set of parallel lines, we will denote this set + only by $\mathcal{X}$. + +In connection with the introduced concepts, the following statement +is valid. + +\bizrek \label{izoSop} +Let $p$, $q$ and $r$ be lines in the plane. +The product $\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p$ +is a reflection if and only if +the lines $p$, $q$ and $r$ belong to the same family of lines. +The axis of the new reflection also belongs to this family. +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.5.pic} +\caption{} \label{sl.izo.6.3.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} + (Figure \ref{sl.izo.6.3.5.pic}) + + ($\Leftarrow$) First, let us assume that the lines $p$, $q$ and $r$ + belong to the same family of lines $\mathcal{X}$. We will consider two + cases. + +\textit{1)} Let $\mathcal{X}$ be a family of concurrent lines with +a center $S$ or $p, q, r\in \mathcal{X}_S$. In this case, +the composition $\mathcal{I}=\mathcal{S}_r \circ \mathcal{S}_q \circ +\mathcal{S}_p$ is an indirect isometry with a fixed point $S$, so according to +the izorek \ref{izo1ftIndZrc} it represents a reflection $\mathcal{S}_t$ with an axis that goes through the point $S$ or $t\in +\mathcal{X}_S$. + +\textit{2)} Let $\mathcal{X}$ be a set of parallel lines and $n$ any common + rectangle of lines $p$, $q$ and $r$. In this case, the composite + $\mathcal{I}=\mathcal{S}_r \circ \mathcal{S}_q \circ +\mathcal{S}_p$ is an indirect isometry and it holds that $\mathcal{I}(n)=n$. +We will now prove that there exists + a fixed point on line $n$ of + this isometry. Let $A$ be any point + on line + $n$ and $\mathcal{I}(A)=A'$. If $A$ is not a fixed point of isometry + $\mathcal{I}$, with $O$ + we denote the center of line $AA'$ and with $O'=\mathcal{I}(O)$. + $\mathcal{I}$ is an indirect isometry, therefore vectors + $AO$ and $A'O'$ are congruent and have different orientation on line $n$. + From this and from the fact that $O$ is the center of line $AA'$, it follows that + $\overrightarrow{A'O}=\overrightarrow{OA}=\overrightarrow{A'O'}$. + Therefore $O=O'$ or $O$ is a fixed + point of isometry $\mathcal{I}$, which represents some basic reflection $\mathcal{S}_t$ (statement \ref{izo1ftIndZrc}). In this case + $\mathcal{S}_t(n)=\mathcal{I}(n)=n$. This means that $t=n$ or + $t\perp n$. The first option is not possible, because from $\mathcal{S}_r + \circ \mathcal{S}_q \circ +\mathcal{S}_p=\mathcal{S}_n$ it follows that $\mathcal{S}_q \circ +\mathcal{S}_p=\mathcal{S}_r\circ \mathcal{S}_n$. The last relation +is not possible, because lines $p$ and $q$ are parallel, lines $r$ and $n$ +are perpendicular and they intersect (statement \ref{izoKomppqX}). Therefore $t\perp n$, which means that $t$ is also a line from the set of parallel lines $\mathcal{X}$. + + + ($\Rightarrow$) We will now assume that + $\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p=\mathcal{S}_t$ + for some line $t$. In this case it holds that + $\mathcal{S}_q \circ \mathcal{S}_p=\mathcal{S}_r + \circ\mathcal{S}_t$. We will again consider two cases. + + \textit{1)} We will assume that $p$ and $q$ intersect in some point $S$. + In this case, it holds that $\mathcal{S}_r + \circ\mathcal{S}_t(S) = \mathcal{S}_q \circ \mathcal{S}_p(S)=S$. According to + statement \ref{izoKomppqX}, lines $r$ and $t$ also intersect in point $S$, therefore $p, q, r, t\in \mathcal{X}_S$. + +\textit{2)} Let's assume that the lines $p$ and $q$ are parallel. + From the aforementioned equation \ref{izoKomppqX} it follows that + the composite $\mathcal{S}_q \circ \mathcal{S}_p$ has no fixed points. + Then neither does the composite $\mathcal{S}_r \circ \mathcal{S}_t$, + so $r\parallel t$. Let $n$ + be the common perpendicular of the lines $p$ and $q$. Then $\mathcal{S}_r + \circ\mathcal{S}_t(n) = \mathcal{S}_q \circ \mathcal{S}_p(n)=n$. + From this it follows that $\mathcal{S}_t(n)=\mathcal{S}_r(n)=n'$. + If $n\neq n'$, then both parallels $p$ and $q$ are axes + of symmetry of the lines $n$ and $n'$, which is not possible. Therefore $n=n'$ and + $\mathcal{S}_t(n)=\mathcal{S}_r(n)=n$. + So $r,t\perp n$, which means that the lines $p$, $q$, $r$ and + $t$ + belong to the family of lines that are all perpendicular to the line $n$. + \kdokaz + +The consequence of the proven statement is the following equation. + + + + \bizrek \label{izoSop2n+1} + The product of an odd number of reflections with the axes + from the same family of lines is also a reflection. + \eizrek + + \textbf{\textit{Proof.}} We will carry out the proof by induction on + the number $n\geq 1$, where $m=2n+1$ (odd) is the number of lines. + + For $n=1$ or $m=3$, the statement is a direct consequence of the previous equation + \ref{izoSop}. + + We assume that for every $k\in \mathbb{N}$ ($k\geq 1$) + and every $(2 k+1)$-tuple + of lines + $p_1,p_2,\ldots ,p_{2k+1}$, which are all from the same family, the composite + $\mathcal{S}_{p_{2k+1}}\circ + \cdots \circ\mathcal{S}_{p_2}\circ\mathcal{S}_{p_1}$ represents + some basic reflection $\mathcal{S}_p$ (inductive assumption). + +It is necessary to prove that then for $k+1$ +and every $(2(k+1)+1)$-terica +of lines +$p_1,p_2,\ldots ,p_{2(k+1)+1}$, which are all from the same family of lines, the composition +$\mathcal{I}=\mathcal{S}_{p_{2(k+1)+1}}\circ + \cdots \circ\mathcal{S}_{p_2}\circ\mathcal{S}_{p_1}$ represents + a certain basic reflection $\mathcal{S}_{p'}$. But: + \begin{eqnarray*} + \mathcal{I}&=&\mathcal{S}_{p_{2(k+1)+1}}\circ + \cdots \circ\mathcal{S}_{p_2}\circ\mathcal{S}_{p_1}=\\ + &=&\mathcal{S}_{p_{2k+3}}\circ\mathcal{S}_{p_{2k+2}}\circ + \mathcal{S}_{p_{2k+1}}\circ + \cdots \circ\mathcal{S}_{p_2}\circ\mathcal{S}_{p_1}=\\ + &=& \mathcal{S}_{p_{2k+3}}\circ\mathcal{S}_{p_{2k+2}}\circ + \mathcal{S}_p= \hspace*{14mm} \textrm{ (by the induction hypothesis)}\\ + &=& \mathcal{S}_{p'} \hspace*{48mm} \textrm{ (by \ref{izoSop}),} + \end{eqnarray*} + which had to be proven. \kdokaz + + + We shall also prove some direct consequences of \ref{izoSop}. + + \bzgled \label{iropqrrqp} + If $p$, $q$ and $r$ are lines from the same family of lines $\mathcal{X}$, then + $$\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p = + \mathcal{S}_p \circ \mathcal{S}_q \circ \mathcal{S}_r.$$ + \ezgled + +\textbf{\textit{Proof.}} Because $p$, $q$ and $r$ are lines from the same +family of lines $\mathcal{X}$, by the previous \ref{izoSop} it follows that +$\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p +=\mathcal{S}_t$ for some line $t\in\mathcal{X}$. Therefore: + $$(\mathcal{S}_r \circ +\mathcal{S}_q \circ \mathcal{S}_p)^2 +=\mathcal{S}_t^2=\mathcal{E},\hspace*{3mm}\textrm{or,}$$ + $$\mathcal{S}_r \circ +\mathcal{S}_q \circ \mathcal{S}_p \circ\mathcal{S}_r \circ +\mathcal{S}_q \circ \mathcal{S}_p=\mathcal{E}.$$ + If we multiply the last +relation from the left by $\mathcal{S}_r$, +$\mathcal{S}_q$ and $\mathcal{S}_p$, we get the desired equality. + \kdokaz + +\bzgled \label{izoSopabc} + If $a$, $b$ and $c$ are lines such that $\mathcal{S}_a(b)=c$, + then those lines are from the same family of lines. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.6a.pic} +\caption{} \label{sl.izo.6.3.6a.pic} +\end{figure} + + \textbf{\textit{Proof.}} + (Figure \ref{sl.izo.6.3.6a.pic}) + +From the transmutation theorem \ref{izoTransmutacija} it follows that +$\mathcal{S}_a\circ\mathcal{S}_b\circ\mathcal{S}_a= +\mathcal{S}_{\mathcal{S}_a(b)}= \mathcal{S}_c$. By multiplying +the obtained relation +$\mathcal{S}_a\circ\mathcal{S}_b\circ\mathcal{S}_a= \mathcal{S}_c$ +first from the right with $\mathcal{S}_a$, and then from the left with +$\mathcal{S}_c$, we get +$\mathcal{S}_c\circ\mathcal{S}_a\circ\mathcal{S}_b=\mathcal{S}_a$. +By theorem \ref{izoSop} the lines $a$, $b$ and $c$ belong to the same +family of lines. + \kdokaz + + It is clear that the line $a$ from the previous theorem is the line of symmetry of the lines $b$ and $c$. + + With the following theorem we will determine the line $t$ from the relation + $\mathcal{S}_r \circ +\mathcal{S}_q \circ \mathcal{S}_p =\mathcal{S}_t$ for + lines $p$, $q$ and $r$ from the same family of lines. + + + \bizrek \label{izoSoppqrt} + If $p$, $q$, $r$ and $t$ are lines such that + $\mathcal{S}_r \circ + \mathcal{S}_q \circ \mathcal{S}_p =\mathcal{S}_t$, then the axis of symmetry of the lines + $p$ and $r$ is also + the axis of symmetry of the lines $q$ and $t$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.6.pic} +\caption{} \label{sl.izo.6.3.6.pic} +\end{figure} + +\textbf{\textit{Proof.}} + From $\mathcal{S}_r \circ +\mathcal{S}_q \circ \mathcal{S}_p =\mathcal{S}_t$ it follows that +the lines $p$, $q$, $r$ and $t$ + belong to the same set $\mathcal{X}$ (statement \ref{izoSop}). + Let $s$ be the line of symmetry of the lines $p$ and $r$ (Figure \ref{sl.izo.6.3.6.pic}). + This means that $\mathcal{S}_s(p)=r$, + so according to the transmutation statement \ref{izoTransmutacija} also + $\mathcal{S}_s\circ\mathcal{S}_p\circ\mathcal{S}_s + =\mathcal{S}_{\mathcal{S}_s(p)} + =\mathcal{S}_r$. From this it further follows: + $$\mathcal{S}_t = \mathcal{S}_r \circ\mathcal{S}_q \circ \mathcal{S}_p + = (\mathcal{S}_s\circ\mathcal{S}_p\circ\mathcal{S}_s)\circ + \mathcal{S}_q\circ\mathcal{S}_p = +\mathcal{S}_s\circ(\mathcal{S}_p\circ \mathcal{S}_s\circ +\mathcal{S}_q)\circ \mathcal{S}_p.$$ + Because $\mathcal{S}_s(p)=r$, from the previous example \ref{izoSopabc} it follows that + the lines $s$, $p$ and $r$ +belong to the same set, so $s\in \mathcal{X}$. From statement +\ref{iropqrrqp} it follows $\mathcal{S}_p\circ \mathcal{S}_s\circ +\mathcal{S}_q = \mathcal{S}_q\circ \mathcal{S}_s\circ +\mathcal{S}_p$. If we use the transmutation statement +\ref{izoTransmutacija} again, we get: +$$\mathcal{S}_t = +\mathcal{S}_s\circ\mathcal{S}_p\circ \mathcal{S}_s\circ +\mathcal{S}_q\circ \mathcal{S}_p= +\mathcal{S}_s\circ\mathcal{S}_q\circ \mathcal{S}_s\circ +\mathcal{S}_p\circ \mathcal{S}_p= +\mathcal{S}_s\circ\mathcal{S}_q\circ \mathcal{S}_s= +\mathcal{S}_{\mathcal{S}_s(q)}.$$ + From $\mathcal{S}_t =\mathcal{S}_{\mathcal{S}_s(q)}$ it follows + $\mathcal{S}_s(q)=t$, which means that the line $s$ is the line of symmetry of the lines $q$ and $t$. + \kdokaz + + In the next planning task we will illustrate two methods + of solving. + +\bizrek +Let $p$, $q$ and $r$ be lines in the plane. +Construct a triangle $ABC$ such that its vertices $B$ and $C$ +lie on the line $p$ and the lines $q$ and $r$ are perpendicular +bisectors of the sides $AB$ and $AC$. +\eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.7.pic} +\caption{} \label{sl.izo.6.3.7.pic} +\end{figure} + + \textbf{\textit{Solution.}} (first way) + + Let $q$ and $r$ be the perpendicular bisectors of the sides $AB$ and $AC$ + (Figure \ref{sl.izo.6.3.7.pic}\textit{a}). Then + $\mathcal{S}_q(B)=A$ and $\mathcal{S}_r(C)=A$. +The points $B$ and $C$ lie on the line $p$, so the point $A$ lies on +the images of this line with respect to the reflections $\mathcal{S}_q$ and +$\mathcal{S}_r$. Therefore, we can construct the point $A$ as the +intersection of the lines $p'_q=\mathcal{S}_q(p)$ and +$p'_r=\mathcal{S}_r(p)$. + + \textbf{\textit{Solution.}} (second way) + + + The intersection of the lines $q$ and $r$ is the center of the circumscribed circle of this triangle + (see \ref{SredOcrtaneKrozn}); +we denote it by $O$ (Figure \ref{sl.izo.6.3.7.pic}\textit{b}). Therefore, we can also construct the third perpendicular bisector $s$ of the side $BC$, which is +perpendicular to the line $p$ at the point $O$. The composition $\mathcal{S}_q +\circ \mathcal{S}_r \circ \mathcal{S}_s$ by \ref{izoSop} +is a certain basic reflection $\mathcal{S}_l$ (because $q, r, s\in +\mathcal{X}_O$). The fixed points of this composition are $O$ and $B$, +so $\mathcal{S}_q \circ \mathcal{S}_r \circ \mathcal{S}_s = +\mathcal{S}_{OB}$. Therefore, we can construct the line $OB=l$ as the +perpendicular bisector of the segment $XX'$, where $X$ is an arbitrary point and +$X'=\mathcal{S}_q \circ \mathcal{S}_r \circ \mathcal{S}_s(X)$. In +the case when $X=X'$, we have $l=OX$. Finally, we obtain the point $B$ as the +intersection of the lines $p$ and $l$. + \kdokaz + +\bzgled + Let $ABCDE$ be a pentagon with a right angle + at the vertex $A$. + The perpendicular bisectors of the sides $AE$, $BC$, + and $CD$ (lines $p$, $q$, and $r$) intersect + at a point $O$ and the perpendicular bisectors of the sides $AB$ and $DE$ + (lines $x$ and $y$) intersect at a point $S$ + ($S\neq O$). Prove that one of the vertices of a triangle $KLM$, + such that $p$, $q$ + and $r$ are the perpendicular bisectors of its sides, + belongs to the line $OS$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.8.pic} +\caption{} \label{sl.izo.6.3.8.pic} +\end{figure} + + \textbf{\textit{Solution.}} + (Figure \ref{sl.izo.6.3.8.pic}) + +Let $\mathcal{I} = \mathcal{S}_y \circ\mathcal{S}_r +\circ\mathcal{S}_q \circ\mathcal{S}_x \circ\mathcal{S}_p$. +$\mathcal{I}$ is an indirect isometry, with a fixed point $E$, +so by \ref{izo1ftIndZrc} it represents some reflection $S_l$. Because $\angle EAB$ is a right angle, simetrali $p$ and $x$ +are perpendicular, so the reflections $\mathcal{S}_p$ and $\mathcal{S}_x$ +commute (\ref{izoZrcKomut}). So we have: + $$\mathcal{I} =\mathcal{S}_l= + \mathcal{S}_y \circ\mathcal{S}_r +\circ\mathcal{S}_q \circ\mathcal{S}_x \circ\mathcal{S}_p= + \mathcal{S}_y \circ (\mathcal{S}_r +\circ\mathcal{S}_q \circ\mathcal{S}_p) \circ\mathcal{S}_x.$$ + Lines $p$, $q$ and $r$ belong to the same +$\mathcal{X}_O$, so the composite $\mathcal{S}_r \circ\mathcal{S}_q +\circ\mathcal{S}_p$ represents some reflection $\mathcal{S}_t$, +where it also holds that $t\in\mathcal{X}_O$ or the $t$-axis goes through the point +$O$ (\ref{izoSop}). + So we have $\mathcal{S}_l= + \mathcal{S}_y +\circ\mathcal{S}_t \circ\mathcal{S}_x$, so lines $x$, $t$ and +$y$ belong to the same, or the $t$-line goes through the point $S=x\cap +y$. Because the $t$-line goes through different points $O$ and $S$, $t=OS$. So $\mathcal{S}_{OS}=\mathcal{S}_t=\mathcal{S}_r +\circ\mathcal{S}_q \circ\mathcal{S}_p$. Let $p$, $q$ and $r$ +be the simetrali of sides $KL$, $LM$ and $MK$ of triangle $KLM$. In this +case $\mathcal{S}_{OS}(K)=\mathcal{S}_t(K)=\mathcal{S}_r +\circ\mathcal{S}_q \circ\mathcal{S}_p(K)=K$, so point $K$ +lies on the line $OS$. + \kdokaz + + + +%________________________________________________________________________________ + \poglavje{Rotations} \label{odd6Rotac} + +So far we have learned about isometries that have more than one fixed point. +We have proven that there are two such types of isometries, namely identity as a direct isometry with at least two +fixed points (Theorem \ref{izo2ftIdent}) and reflection as an indirect isometry with at least one +fixed point (Theorem \ref{izo1ftIndZrc}). In this section we will +learn about a type of direct isometries that have exactly one +fixed point. + +Let $S$ be an arbitrary point and $\omega\neq 0$ be an oriented plane. The transformation of this plane, in which the point +$S$ is fixed, each other point $X\neq S$ of this plane is mapped to +such a point $X'$, that $\measuredangle XSX'\cong \omega$ and +$SX'\cong SX$, is called a rotation with center $S$ for angle $\omega$; +we denote it with $\mathcal{R}_{S,\omega}$ (Figure +\ref{sl.izo.6.4.1.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.1.pic} +\caption{} \label{sl.izo.6.4.1.pic} +\end{figure} + +From the definition itself, the following theorems directly follow. + + + + \bizrek \label{RotacFiksT} + The only fixed point of a rotation $\mathcal{R}_{S,\omega}$ + is its center $S$ i.e. + $$\mathcal{R}_{S,\omega}(X)=X + \hspace*{1mm} \Leftrightarrow \hspace*{1mm} X=S.$$ + \eizrek + + + \bizrek \label{rotacEnaki} + Rotations $\mathcal{R}_{S,\alpha}$ and $\mathcal{R}_{B,\beta}$ + are equal if and only if $A=B$ and $\alpha=\beta$ i.e. + $$\mathcal{R}_{S,\alpha}=\mathcal{R}_{B,\beta} + \hspace*{1mm} \Leftrightarrow \hspace*{1mm} A=B \hspace*{1mm} + \wedge\hspace*{1mm} \alpha=\beta.$$ + \eizrek + + + + \bizrek + The inverse transformation of a rotation is a rotation with the same center + and congruent angle with opposite orientation, i.e. + $$\mathcal{R}_{S,\omega}^{-1}=\mathcal{R}_{S,-\omega}.$$ + \eizrek + + + + Although it is intuitively clear, we need to prove that rotation + is an isometry. + +\bizrek + Rotations are isometries of the plane. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.2.pic} +\caption{} \label{sl.izo.6.4.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} Let $\mathcal{R}_{S,\omega}$ be an arbitrary + rotation. From the definition it is clear that it represents a bijective mapping. It remains to be proven that for any points $X$ and $Y$ + and their images $X'$ and $Y'$ ($\mathcal{R}_{S,\omega}:X,Y\mapsto X', Y'$) it holds + $X'Y'\cong XY$ (Figure \ref{sl.izo.6.4.2.pic}). + +If one of the points $X$ or $Y$ is equal to the center of the rotation $S$ (for example +$X=S$), the relation $X'Y'\cong XY$ is automatically fulfilled, since in +this case, according to the definition of the rotation, it holds $SY'\cong SY$. + +Assume that none of the points $X$ and $Y$ is the center of the rotation +$S$. First, according to the definition of the rotation $SX'\cong SX$ and $SY'\cong +SY$. Then (from relation \ref{orientKotVsota}): + \begin{eqnarray*} + \measuredangle Y'SX'&=& \measuredangle Y'SX+\measuredangle XSX'= + \measuredangle Y'SX + \omega=\\ + &=& \measuredangle Y'SX + \measuredangle YSY'= + \measuredangle YSY'+ \measuredangle Y'SX=\\ &=&\measuredangle YSX + \end{eqnarray*} + This means that the triangle $X'SY'$ and $XSY$ are congruent (\textit{SAS} theorem \ref{SKS}), + therefore + $X'Y'\cong XY$. + \kdokaz + + As every isometry, also the rotation maps a line into a line. + In the next theorem we will see the relation between a line and + its image under the rotation. + + + + \bizrek \label{rotacPremPremKot} + A line and its rotated image determine an angle congruent + to the angle of this rotation. + If the angle of the rotation measures $180^0$, + the line and its image are parallel. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.3.pic} +\caption{} \label{sl.izo.6.4.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} Let $p'$ be the image of the line $p$ under the rotation + $\mathcal{R}_{S,\omega}$ (Figure \ref{sl.izo.6.4.3.pic}). + +If $S\in p$, the proof is trivial. Assume that $S\notin p$. + We denote with $P$ the orthogonal projection of the center $S$ onto + the line $p$ and its image $P'=\mathcal{R}_{S,\omega}(P)$. Because $S\notin + p$, + it follows that $P\neq S$ or $P'\neq P$. From + $P\in p$ it follows that $P'\in p'$. Because the isometry preserves the angles, from + $SP\perp p$ it follows that $SP'\perp p'$. + + If $\omega=180^0$, the points $P$, $S$ and $P'$ are collinear and + the lines $p$ and $p'$ have a common perpendicular + $PP'$, which means that $p\parallel p'$. + + In the case $\omega\neq 180^0$, the lines intersect (in the opposite + case, from $p\parallel p'$ it would follow that $SP\parallel SP'$, which is not possible). + Let $V$ be their intersection point. The angle, determined by + the lines + $p$ and $p'$, is equal to the angle determined by the lines $SP$ and + $SP'$ (the angle with perpendicular legs - statement \ref{KotaPravokKraki}), + which is just the angle of rotation $\omega$. + \kdokaz + + + In the proof of the previous statement, the process of designing the image + $p'$ + of the line $p$ under the rotation $\mathcal{R}_{S,\omega}$ is also described. + First, we draw the orthogonal projection $P$ of the center $S$ onto + the line $p$, and then its image $P=\mathcal{R}_{S,\omega}(P')$. + The line $p'$ is obtained as the perpendicular of the line $SP'$ in the point + $P'$. + Of course, another way would be to rotate two arbitrary points + of the line $p$. + + In the next important statement, we will see how to express + the rotation using the basic reflections. + +\bizrek \label{rotacKom2Zrc} +Any rotation +$\mathcal{R}_{S,\omega}$ can be expressed as the product of two reflections +$\mathcal{S}_p$ and $\mathcal{S}_q$ where +$p$ and $q$ are arbitrary lines, such that $S=p\cap q$ +and $\measuredangle pq=\frac{1}{2}\omega$.\\ The reverse is also true - +the product of two reflections $\mathcal{S}_p$ and $\mathcal{S}_q$ ($S=p\cap q$) +is a rotation with the centre $S$ and the angle +$\omega=2\cdot\measuredangle pq$, i.e.\\ +$$\mathcal{R}_{S,\omega}=\mathcal{S}_q\circ\mathcal{S}_p +\hspace*{1mm} \Leftrightarrow \hspace*{1mm} S=p\cap q \hspace*{1mm} +\wedge\hspace*{1mm} \measuredangle pq=\frac{1}{2}\omega.$$ +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.4.pic} +\caption{} \label{sl.izo.6.4.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} + + ($\Leftarrow$) Assume that $p$ + and $q$ are lines that intersect in the point $S$. Let $\omega + =2\cdot\measuredangle pq$ + (Figure \ref{sl.izo.6.4.4.pic}). We will prove that + $\mathcal{R}_{S,\omega}=\mathcal{S}_q\circ\mathcal{S}_p$, + or $\mathcal{R}_{S,\omega}(X)=\mathcal{S}_q\circ\mathcal{S}_p(X)$ for every +point $X$ of the plane. We will consider two cases. + + \textit{1)} If $X=S$, then $\mathcal{S}_p(X)=\mathcal{S}_q(X)=X$ (because +the lines $p$ and $q$ intersect in the point $S$). So +$\mathcal{R}_{S,\omega}(S)=S=\mathcal{S}_q\circ\mathcal{S}_p(S)$. + +\textit{2)} Let $X\neq S$ and $\mathcal{S}_q\circ\mathcal{S}_p(X)=X'$. + Prove that also $\mathcal{R}_{S,\omega}(X)=X'$. Let +$\mathcal{S}_p(X)=X_1$. Then $\mathcal{S}_q(X_1)=X'$. Because +of +this, first $SX\cong SX_1\cong SX'$, and then: + \begin{eqnarray*} + \measuredangle XSX'&=& \measuredangle XSX_1+ \measuredangle + X_1SX'=\\ + &=& 2\cdot\measuredangle p,SX_1+2\cdot\measuredangle SX_1,q=\\ + &=& 2\cdot(\measuredangle p,SX_1+ \measuredangle SX_1,q) =\\ + &=& 2\cdot\measuredangle pq=\omega. + \end{eqnarray*} + By the definition of rotation, $\mathcal{R}_{S,\omega}(X)=X'$. + + ($\Rightarrow$) Let + $\mathcal{R}_{S,\omega}=\mathcal{S}_q\circ\mathcal{S}_p$. Then + it + is + $\mathcal{S}_q\circ\mathcal{S}_p(S)=\mathcal{R}_{S,\omega}(S)=S$. + By izrek \ref{izoKomppqX} $S=p\cap q$. It remains to prove + $\measuredangle pq=\frac{1}{2}\omega$ or $\omega + =2\cdot\measuredangle pq$. We denote $\widehat{\omega} + =2\cdot\measuredangle pq$. By the first part of the proof ($\Leftarrow$) + it is + $\mathcal{S}_q\circ\mathcal{S}_p=\mathcal{R}_{S,\widehat{\omega}}$. + Therefore $\mathcal{R}_{S,\omega}=\mathcal{S}_q\circ\mathcal{S}_p= + \mathcal{R}_{S,\widehat{\omega}}$. From izrek \ref{rotacEnaki} + it follows $\omega =\widehat{\omega} + =2\cdot\measuredangle pq$. + \kdokaz + + + + \bizrek \label{RotacDirekt} + A rotation is a direct isometry. + \eizrek + + \textbf{\textit{Proof.}} A direct consequence of the previous izrek + \ref{rotacKom2Zrc}, because the composition of two basic reflections is an indirect isometry. + \kdokaz + +Similarly to the basic reflection, also for the rotation, there is an izrek about +transmutation\index{transmutation!rotations}. + +\bizrek \label{izoTransmRotac} +For an arbitrary rotation $\mathcal{R}_{O,\alpha}$ +and an arbitrary isometry $\mathcal{I}$ is +$$\mathcal{I}\circ +\mathcal{R}_{O,\alpha}\circ\mathcal{I}^{-1}= +\mathcal{R}_{\mathcal{I}(O),\alpha'},$$ + where: $\alpha'=\alpha$, if $\mathcal{I}$ a direct isometry, or +$\alpha'=-\alpha$, if $\mathcal{I}$ is an opposite isometry. +\eizrek + +\textbf{\textit{Proof.}} By izrek \ref{rotacKom2Zrc} we can write the rotation + $\mathcal{R}_{O,\alpha}$ + as a composite of two basic +reflections, namely +$\mathcal{R}_{O,\omega}=\mathcal{S}_q\circ\mathcal{S}_p$, where + $O=p\cap q$ and $\measuredangle pq=\frac{1}{2}\alpha$. +If we use the transmutation theorem for basic reflections, we get: + \begin{eqnarray*} + \mathcal{I}\circ + \mathcal{R}_{O,\alpha}\circ\mathcal{I}^{-1}&=& + \mathcal{I}\circ\mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{I}=\\ + &=& + \mathcal{I}\circ\mathcal{S}_q\circ\mathcal{I}^{-1} + \circ\mathcal{I}\circ\mathcal{S}_p\circ\mathcal{I}^{-1}=\\ + &=&\mathcal{S}_{q'}\circ\mathcal{S}_{p'}=\\ + &=&\mathcal{R}_{O',\alpha'}, + \end{eqnarray*} +where $p'$ and $q'$ are the images of lines $p$ and $q$ under the isometry +$\mathcal{I}$, $O'=p'\cap q'$ and $\alpha'=2\cdot\measuredangle +p'q'$. Because $O=p\cap q$, we have $\mathcal{I}(O)= +\mathcal{I}(p\cap q)=\mathcal{I}(p)\cap \mathcal{I}(q)=p'\cap +q'=O'$. Because isometries preserve the relation of congruence of angles, we have + $\alpha'=2\cdot\measuredangle p'q'=2\cdot\measuredangle pq=\alpha$, if + $\mathcal{I}$ is direct, or + $\alpha'=2\cdot\measuredangle p'q'=-2\cdot\measuredangle pq=-\alpha$, if + $\mathcal{I}$ is indirect. + \kdokaz + + In the next theorem we will prove the important fact that for two non-parallel lines there exists a rotation that transforms the first line into the second. + +\bizrek + Let $AB$ and $A'B'$ be congruent line segments that are not parallel. + There is exactly one rotation that maps the line segment $AB$ to the line segment $A'B'$, + such that the point $A$ maps to the point $A'$ and the point $B$ to the point $B'$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.10Rotac.pic} +\caption{} \label{sl.skl.3.1.10Rotac.pic} +\end{figure} + + \textbf{\textit{Proof.}} Let $S$ be the intersection of the line segments $AA'$ and $BB'$, and $\omega=\measuredangle AB, A'B'$. We will prove that $\mathcal{R}_{S,\omega}$ is the desired rotation. + Because the point $S$ lies on the line segments $AA'$ and $BB'$, we have $SA\cong SA'$ and $SB\cong SB'$ (Figure \ref{sl.skl.3.1.10Rotac.pic}). Then from $AB\cong A'B'$ by the \textit{SSS} \ref{SSS} it follows that $\triangle SAB\cong \triangle SA'B'$ (see also example \ref{načrt1odd3}). Therefore, we have $\measuredangle ASB\cong \measuredangle A'SB'$, or: + $$\measuredangle ASA'=\measuredangle ASB+\measuredangle BSA'= + \measuredangle A'SB'+\measuredangle BSA'= \measuredangle BSB'.$$ + By the definition of rotation, we have $\mathcal{R}_{S,\measuredangle ASA'}:A,B\mapsto A',B'$. By izreku \ref{rotacPremPremKot} we have $\measuredangle ASA'=\measuredangle AB, A'B'=\omega$, so $\mathcal{R}_{S,\omega}:A,B\mapsto A',B'$. + Because rotation as an isometry preserves the relation $\mathcal{B}$, the line segment $AB$ is mapped by this rotation to the line segment $A'B'$. + +Let $\mathcal{R}_{\widehat{S},\widehat{\omega}}$ be another rotation, for which we have $\mathcal{R}_{\widehat{S},\widehat{\omega}}:A,B\mapsto A',B'$. Then +$\mathcal{R}^{-1}_{S,\omega} \circ \mathcal{R}_{\widehat{S},\widehat{\omega}}$ is a direct isometry with two fixed points $A$ and $B$, so by izreku \ref{izo2ftIdent} it represents the identity $\mathcal{E}$, or +$\mathcal{R}_{\widehat{S},\widehat{\omega}} = \mathcal{R}_{S,\omega}$. + \kdokaz + + + In the following examples we will use the fact that a triangle + $ABC$ is equilateral if and only if + $\mathcal{R}_{A,60^0}(B)=C$. + +\bzgled + Let $P$ be an interior point of an equilateral triangle + $ABC$.\\ + a) Prove that + $|PA|+|PB|\geq |PC|.$\\ + b) Suppose that $\angle BPA=\mu$, + $\angle CPA= \nu$ and $\angle BPC= \xi$. Calculate the interior angles of a triangle, + with sides that are congruent to the line segments $PA$, $PB$ in $PC$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.5.pic} +\caption{} \label{sl.izo.6.4.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} If $P'=\mathcal{R}_{A,60^0}(P)$, + the triangle $APP'$ is equilateral (Figure \ref{sl.izo.6.4.5.pic}). + Because $\mathcal{R}_{A,60^0}(B)=C$, + the line segment $BP$ is mapped to +the corresponding line segment $CP'$ by this rotation. Therefore, the triangle $PP'C$ has sides that are +congruent to the line segments $PA$, $PB$ and $PC$. By the triangle inequality +- statement \ref{neenaktrik} - we have $|PA|+|PB|=|PP'|+|P'C|\geq PC.$ + +Let us calculate the angles of the triangle $PP'C$ as well. The rotation +$\mathcal{R}_{A,60^0}$ maps the triangle $ABP$ to the triangle +$ACP'$, so the triangles are congruent and $\angle BPA=\angle CP'A$. From +this and from statement \ref{VsotKotTrik} it follows that: + \begin{eqnarray*} + \angle CP'P &=& \angle CP'A-\angle PP'A=\angle + BPA-60^0=\mu-60^0,\\ + \angle CPP' &=& \angle CPA-60^0=\nu-60^0,\\ + \angle PCP' &=& 180^0 + -(\mu-60^0)-(\nu-60^0)=300^0-(\mu+\nu)=\\ + &=& 300^0-(360^0-\xi)=\xi-60^0, + \end{eqnarray*} + which had to be calculated. \kdokaz + + + + \bzgled + Let $ABCD$ be a rhombus, such that the interior angle at the vertex $A$ + is equal to $60^0$. + If a line $l$ intersects the sides $AB$ and $BC$ of this rhombus at points $P$ and + $Q$ such that $|BP|+|BQ|=|AB|$, then $PQD$ is a regular triangle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.6.pic} +\caption{} \label{sl.izo.6.4.6.pic} +\end{figure} + +\textbf{\textit{Proof.}} From $AB\cong AD$ and $\angle DAB=60^0$ + it follows that the triangle $ABD$ is regular (Figure \ref{sl.izo.6.4.6.pic}). + Similarly, the + triangle $BCD$ is also regular. This means that $\mathcal{R}_{D,60^0}:A,B\mapsto + B,C$. Because $|BP|+|BQ|=|AB|$ and $|BP|+|AP|=|AB|$, we have $AP\cong BQ$. + But the rotation $\mathcal{R}_{D,60^0}$ maps the segment $AB$ +to the segment $BC$, so because of the condition $AP\cong BQ$ it also +maps the point $P$ to the point $Q$. Therefore, $\mathcal{R}_{D,60^0}(P)=Q$, which +means that the triangle $DPQ$ is regular. + \kdokaz + +At the next example we will illustrate the use of rotation in +design tasks. + + + + \bzgled + Let $A$ be an interior point of an angle $pOq$. Construct points + $B$ and $C$ on the sides $p$ and $q$ + such that $ABC$ is a regular triangle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.7.pic} +\caption{} \label{sl.izo.6.4.7.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.4.7.pic}). +Let $ABC$ be such a regular triangle that its vertices $B$ and +$C$ lie on the sides $p$ and $q$. Then +$\mathcal{R}_{A,60^0}(B)=C$. The point $B$ lies on the segment $p$, so +its image - the point $C$ - lies on the segment +$p'=\mathcal{R}_{A,60^0}(p)$. The vertices $C$ can be obtained as +the intersection of the segments $p'$ and $q$, and then the vertex $B$ as +$B=\mathcal{R}_{A,-60^0}(C)$. If the segments $p'$ and $q$ do not +intersect, there is no solution. But if they lie on the same line, the task + has infinitely many solutions. + \kdokaz + +If in the previous task the condition was that the triangle $ABC$ +was isosceles and right-angled at the vertex $A$, we would +use the rotation with center $A$ for the angle $45^0$. Similarly, if we needed to plan the square $PQRS$ with center at the point +$A$ under the condition $P\in p$, $Q\in q$ (Figure \ref{sl.izo.6.4.8.pic}). + +\bnaloga\footnote{1. IMO Romania - 1959, Problem 5.} + An arbitrary point $M$ is selected in the interior of the segment $AB$. The + squares $AMCD$ and $MBEF$ are constructed on the same side of $AB$, with + the segments $AM$ and $MB$ as their respective bases. The circles circumscribed + about these squares, with centres $P$ and $Q$, intersect at $M$ and also + at another point $N$. Let $N'$ denote the point of intersection of the straight + lines $AF$ and $BC$. + + (a) Prove that the points $N$ and $N'$ coincide. + + (b) Prove that the straight lines $MN$ pass through a fixed point $S$ independent + of the choice of $M$. + + (c) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between + $A$ and $B$ + \enaloga + + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.IMO1.pic} +\caption{} \label{sl.izo.6.4.IMO1.pic} +\end{figure} + + \textbf{\textit{Solution.}} Let $k$ and $l$ + denote the circumscribed circles + of the squares + $AMCD$ and $MBEF$ + (Figure \ref{sl.izo.6.4.IMO1.pic}). + + (\textit{i}) The rotation $\mathcal{R}_{M,-90^0}$ maps the points + $A$ and $F$ in order to the points $C$ and $B$ or the line $AF$ to the line $CB$. By the statement + of \ref{rotacPremPremKot} + these two lines determine the angle of rotation, + so $AF\perp CB$. Therefore $\angle AN'C\cong\angle BN'F =90^0$, which means + (by the statement + \ref{TalesovIzrKroz}) that the point $N'$ lies on both circles + $k$ and $l$ + with diameters $AC$ and $BF$. Because of this, + $N'=N$. + + (\textit{ii}) Let $j$ be a circle with diameter $AB$ and $S$ the center + of that arc (semicircle) $AB$, which is on the opposite side of the line + $AB$ with respect to the squares $AMCD$ and $MBEF$. The point $S$ is not dependent + on the choice of point $M$. We prove that all lines $MN$ go through the point $S$. + Because $\angle ANB=90^0$, by the statement + \ref{TalesovIzrKroz} also the point $N$ lies + on the circle $j$. From the statement + \ref{ObodObodKot} + it follows that $\angle ANM =\angle ADM=45^0$ and $\angle MNB =\angle + MEB=45^0$, so $NM$ is the perpendicular bisector of the angle $ANB$. By the statement + \ref{TockaN} the perpendicular bisector $NM$ goes through the point $S$. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.IMO1a.pic} +\caption{} \label{sl.izo.6.4.IMO1a.pic} +\end{figure} + + (\textit{iii}) Let $O$ be the center of the line segment $PQ$. + We denote with $P'$, $Q'$ and $O'$ the orthogonal projections of points + $P$, $Q$ and $O$ on the line $AB$ (Figure + \ref{sl.izo.6.4.IMO1a.pic}). The line $OO'$ is the median of the rectangle + $P'Q'QP$, so by Theorem \ref{srednjTrapez}: + $$|OO'|=\frac{1}{2}\left(|PP'|+|QQ'| \right)= + \frac{1}{2}\left(\frac{1}{2}|AM|+\frac{1}{2}|MB| \right)=\frac{1}{4}|AB|.$$ + Therefore, the distance of the point $O$ from the line $AB$ is constant or + independent of the choice of the point $M$. + + Let $L$ be the center of the square $ABGH$, which is on the same side + of the line $AB$ as the square $AMCD$ and $MBEF$. With $O_A$ and $O_B$ + we denote the centers of the line segments $LA$ and $LB$. Because $O_AO_B$ + is the median of the triangle $ALB$ with the base $AB$, $O_AO_B \parallel + AB$. + From $|AO_A|=\frac{1}{2}|AL|=\frac{1}{4}|AG|$ it follows that + the distance of the parallels $O_AO_B$ and $AB$ is equal to $\frac{1}{4}|AB|$. From + the already proven fact $d(O,AB)=\frac{1}{4}|AB|$ it follows that the point + $O$ lies on the line $O_AO_B$. But, when the point $M$ moves inside the line segment $AB$, the points + $P$ and $Q$ move inside the line segments $AL$ and $BL$. This means that the point $O$ is + inside the triangle $ALB$ or $O$ lies on the open line + segment $O_AO_B$. + + We will also prove that any point $O$ of the line segment $O_AO_B$ + is the center of a line segment $PQ$ for a certain choice of the squares + $AMCD$ and $MBEF$ or the points $M$. In this case, we get the point $M$ + from the condition $|AO'|=\frac{1}{2}|AM|+\frac{1}{4}|AB|$ or + $|AM|=2|AO'|-\frac{1}{2}|AB|$. Such a point $M$ always + exists, if $\frac{1}{4}|AB|<|AO'|<\frac{3}{4}|AB|$, or + when the point $O$ lies on the open line segment $O_AO_B$. + +The desired geometric location of points $M$ is therefore the line $O_AO_B$. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Half-Turn} \label{odd6SredZrc} + +We will consider the rotation by the angle $180^0$ as +a special type of isometry. + +The rotation of the plane $\mathcal{R}_{S,\omega}$ by the angle $\omega=180^0$ +we will call the \index{zrcaljenje!središčno}\pojem{central reflection} +or \index{zrcaljenje!čez točko}\pojem{reflection over a point} with +\index{središče!središčnega zrcaljenja}\pojem{center} $S$ and denote it by $\mathcal{S}_S$ (Figure \ref{sl.izo.6.5.1.pic}). So +$\mathcal{S}_S=\mathcal{R}_{S,180^0}$. + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.1.pic} +\caption{} \label{sl.izo.6.5.1.pic} +\end{figure} + +From the definition it is clear that for any point $X\neq S$ it holds +$\mathcal{S}_S(X)=X'$ exactly when $S$ is the center of the line +$XX'$. + +Directly from the definition it also follows the next statement. + + \bizrek \label{izoSredZrcInv} + A half-turn is an involution, i.e. + $\mathcal{S}^2_S=\mathcal{E}$. \eizrek + + Because the central reflection is a type of rotation, it also has all + the properties of rotation. The only fixed point + of the central reflection is therefore the center of this reflection. + + Just like the rotation, the central reflection + $\mathcal{S}_S$ + can be represented as the composition of two basic reflections. The angle, which is determined by the axes, is equal to half the angle of the rotation (statement \ref{rotacKom2Zrc}) - + in our case + this is $\frac{180^0}{2}=90^0$. This means that the axes of the two + reflections are perpendicular in the case of central reflection. In this case, according to + \ref{izoZrcKomut}, the basic reflections commute + (Figure \ref{sl.izo.6.5.2.pic}). + So the following statement holds. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.2.pic} +\caption{} \label{sl.izo.6.5.2.pic} +\end{figure} + +\bizrek \label{izoSrZrcKom2Zrc} + Any half-turn around a point $S$ + can be expressed as the product of two reflections + $\mathcal{S}_p$ and $\mathcal{S}_q$ where + $p$ and $q$ are arbitrary perpendicular lines intersecting at the point $S$.\\ + The reverse is also true - + the product of two reflections + $\mathcal{S}_p$ and $\mathcal{S}_q$ ($S=p\cap q$ and + $p\perp q$) + is the half-turn around the point $S$, i.e.\\ + $$\mathcal{S}_S=\mathcal{S}_q\circ\mathcal{S}_p= + \mathcal{S}_p\circ\mathcal{S}_q + \hspace*{1mm} \Leftrightarrow \hspace*{1mm} S=p\cap q \hspace*{1mm} + \wedge\hspace*{1mm} p\perp q.$$ + \eizrek + + + + \bizrek \label{izoKomp3SredZrc} + The product of three half-turns is a half-turn. + If the centres of these half-turns are three non-collinear points, + then the centre of the new half-turn is the fourth vertex of a parallelogram. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.3.pic} +\caption{} \label{sl.izo.6.5.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Let $\mathcal{S}_A$, $\mathcal{S}_B$ and $\mathcal{S}_C$ be three central +symmetries with centers $A$, $B$ and $C$ (Figure +\ref{sl.izo.6.5.3.pic}). We denote by $p$ the line $AB$, by $a$, $b$ +and $c$ the rectangles of the line $p$ through the points $A$, $B$ and +$C$, and by $c'$ the rectangle of the line $c$ through the point $C$. Then by izreku \ref{izoSrZrcKom2Zrc}: + $$\mathcal{S}_C \circ \mathcal{S}_B \circ \mathcal{S}_A= + \mathcal{S}_{c'} \circ \mathcal{S}_c \circ \mathcal{S}_b + \circ \mathcal{S}_p \circ \mathcal{S}_p \circ \mathcal{S}_a + = \mathcal{S}_{c'} \circ \mathcal{S}_c \circ \mathcal{S}_b \circ \mathcal{S}_a.$$ + The axes $a$, $b$ and $c$ belong to the same set of parallels $\mathcal{X}_a$, +because they are all perpendicular to the line $p$. The composite +$\mathcal{S}_c \circ \mathcal{S}_b \circ \mathcal{S}_a$ therefore +represents the basic reflection $S_d$, where the axis $d$ belongs to the same set +$\mathcal{X}_a$ (izrek \ref{izoSop}). Then both lines $c'$ +and $d$ are perpendicular, so: + $$\mathcal{S}_C \circ \mathcal{S}_B \circ \mathcal{S}_A= + \mathcal{S}_{c'} \circ \mathcal{S}_d=\mathcal{S}_D,$$ +where $D=d\cap c'$. By izreku \ref{izoSoppqrt} the pairs of lines $a$, +$c$ and $b$, $d$ have a common somernica. This means that if +$A$, $B$ and $C$ are three non-collinear points, the quadrilateral $ABCD$ +is a parallelogram. + \kdokaz + +The consequence of the proven statement is the following izrek. + + + + + \bizrek \label{izoKomp2n+1SredZrc} + The product of an odd number of half-turns is a half-turn. + \eizrek + + \textbf{\textit{Proof.}} We will prove the statement by induction on + the number $n\geq 1$, where $m=2n+1$ (odd) is the number of points. + + For $n=1$, i.e. $m=3$, the statement is a direct consequence of the previous izrek + \ref{izoKomp3SredZrc}. + +Let's assume that for every $k\in \mathbb{N}$ ($k\geq 1$) + and every $(2k+1)$-tuple + of coplanar points + $A_1,A_2,\ldots ,A_{2k+1}$ the composite + $\mathcal{S}_{A_{2k+1}}\circ + \cdots \circ\mathcal{S}_{A_2}\circ\mathcal{S}_{A_1}$ represents + some central reflection $\mathcal{S}_A$ (inductive assumption). + + It is necessary to prove that then also for $k+1$ + and every $(2(k+1)+1)$-tuple + of coplanar points + $A_1,A_2,\ldots ,A_{(2(k+1)+1)}$ the composite + $\mathcal{I}=\mathcal{S}_{A_{2(k+1)+1}}\circ + \cdots \circ\mathcal{S}_{A_2}\circ\mathcal{S}_{A_1}$ represents + some central reflection $\mathcal{S}_{A'}$. So: + + \begin{eqnarray*} + \mathcal{I}&=&\mathcal{S}_{A_{2(k+1)+1}}\circ + \cdots \circ\mathcal{S}_{A_2}\circ\mathcal{S}_{A_1}=\\ + &=&\mathcal{S}_{A_{2k+3}}\circ\mathcal{S}_{A_{2k+2}}\circ + \mathcal{S}_{A_{2k+1}}\circ + \cdots \circ\mathcal{S}_{A_2}\circ\mathcal{S}_{A_1}=\\ + &=& \mathcal{S}_{A_{2k+3}}\circ\mathcal{S}_{A_{2k+2}}\circ + \mathcal{S}_A= \hspace*{14mm} \textrm{ (by the inductive assumption)}\\ + &=& \mathcal{S}_{A'} \hspace*{48mm} \textrm{ (by \ref{izoKomp3SredZrc}),} + \end{eqnarray*} + which was to be proven. \kdokaz + +A direct consequence of \ref{izoTransmRotac}, which concerns rotation, is the following theorem or izrek about transmutation of central reflection\index{transmutation!of central reflection}. + + + \bizrek \label{izoTransmSredZrc} + For an arbitrary half-turn + $\mathcal{S}_{O}$ + and an arbitrary isometry $\mathcal{I}$ is + $$\mathcal{I}\circ + \mathcal{S}_{O}\circ\mathcal{I}^{-1}= + \mathcal{R}_{\mathcal{I}(O)}.$$ + \eizrek + + + Just like the basic reflection, also the central + reflection is often used in planning + tasks. + +\bzgled + Let $S$ be an interior point of an angle $pOq$. Construct a square + $ABCD$ with the centre $S$ such that the vertices + $A$ and $C$ lie on the sides $p$ and $q$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.4.pic} +\caption{} \label{sl.izo.6.5.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} Let $ABCD$ be a square that satisfies the given conditions - + $S$ is its center, $A\in p$ and $C\in q$ (Figure \ref{sl.izo.6.5.4.pic}). + Because the point $S$ is the center of the diagonal + $AC$, $\mathcal{S}_S(A)=C$. From $A\in p$ it follows that $C\in + \mathcal{S}_S(p)$. + We therefore obtain the point $C$ from the condition $C\in q\cap\mathcal{S}_S(p)$. + Finally, $A=\mathcal{S}_S(C)$, $D=\mathcal{R}_{S,90^0}(C)$ and + $B=\mathcal{R}_{S,-90^0}(C)$. + \kdokaz + + \bzgled + Let $A$ be one of the two intersections of circles $k$ and $l$. Construct + a common secant of these two circles passing through the point $A$ and determine + a two congruent chord with these circles. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.5.pic} +\caption{} \label{sl.izo.6.5.5.pic} +\end{figure} + + \textbf{\textit{Solution.}} + Let $B$ be the other intersection of circles $k$ and $l$ + (Figure \ref{sl.izo.6.5.5.pic}). Then one solution to the task + is the line $AB$, since the distance $AB$ + is the common secant of both circles. + + Let $s\ni A$ be a common secant of circles $k$ and $l$, which also intersects + in points $K$ and $L$, so that $AK\cong AL$. In the case $K=L$ + we have $K=L=B$, which is the already mentioned solution. If $K\neq L$, + from the condition $A, K, L\in s$ it follows that $A$ is the center of the line $KL$. + Therefore $\mathcal{S}_A(K)=L$, which means that we obtain the point $L$ + from the condition $L\in l\cap \mathcal{S}_A(k)$. Finally, + $K=\mathcal{S}_A(L)$. + \kdokaz + +\bzgled +Construct a quadrilateral $ABCD$ with the sides that are congruent to +the four given line segments $a$, $b$, $c$, and $d$ and +the line segment defined by the midpoints of +the two opposite sides $AD$ and $BC$ that is congruent to the given line segment $l$. +\ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.6.pic} +\caption{} \label{sl.izo.6.5.6.pic} +\end{figure} + + \textbf{\textit{Solution.}} +Let $P$ and $Q$ be the midpoints of sides $AD\cong d$ and $BC\cong b$ +of quadrilateral $ABCD$, so that $PQ\cong l$, $AB\cong a$ and +$CD\cong c$ still hold (Figure \ref{sl.izo.6.5.6.pic}). Let $S$ +be the midpoint of diagonal $AC$. We can construct triangle $PQS$ since $PQ\cong l$, $QS=\frac{1}{2}a$ and $PS =\frac{1}{2}c$ (by Theorem \ref{srednjicaTrik}). Points $C$ and $A$ lie on +circles $k_d=k(P,\frac{1}{2}d)$ and $k_b=k(Q,\frac{1}{2}b)$, respectively. Since +$C=\mathcal{S}_S(A)$, point $C$ also lies on the image $k'_d$ +of circle $k_d$ under central reflection $\mathcal{S}_S$. +Point $C$ is therefore one of the intersections of circles $k'_d$ and $k_b$. Then $A=\mathcal{S}_S(C)$, $D=\mathcal{S}_P(A)$ and +$B=\mathcal{S}_Q(C)$. + \kdokaz + + + \bzgled + Let $P$, $Q$, $R$, $S$ and $T$ be points in the plane. Construct + a pentagon $ABCDE$ such that the points $P$, $Q$, $R$, $S$ and $T$ are + the midpoints of its sides $AB$, $BC$, $CD$, $DE$ and $EA$, respectively. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.7.pic} +\caption{} \label{sl.izo.6.5.7.pic} +\end{figure} + +\textbf{\textit{Solution.}} +Let $ABCDE$ be the desired pentagon + (Figure \ref{sl.izo.6.5.7.pic}) and: + $$\mathcal{I} = \mathcal{S}_T \circ \mathcal{S}_S + \circ \mathcal{S}_R \circ \mathcal{S}_Q \circ \mathcal{S}_P.$$ + From \ref{izoKomp2n+1SredZrc} it follows that the isometry + $\mathcal{I}$ is a central +reflection $\mathcal{S}_O$. Since +$\mathcal{S}_O(A)=\mathcal{I}(A)=A$, we have $O=A$ or $\mathcal{I}=\mathcal{S}_A$. The point +$A$ can therefore be constructed as the center of the line $XX'$, where +$X$ is any point on the given plane and $X'= \mathcal{I}(X)$. + \kdokaz + + + \bzgled + Let $A_iB_i$ ($i\in \{1,2,3\}$) + be parallel chords of a circle $k$. Suppose that $S_1$, $S_2$ and $S_3$ + are the midpoints of the line segments $A_2A_3$, $A_1A_3$ and $A_1A_2$, + respectively. Prove that + $C_i=\mathcal{S}_{S_i}(B_i)$ are three collinear points. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.4a.pic} +\caption{} \label{sl.izo.6.5.4a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $S$ be the center of the circle $k$ +(Figure \ref{sl.izo.6.5.4a.pic}). Because $SA_i\cong SB_i$ ($i\in +\{1,2,3\}$), all the simetrals $s_i$ of the chords $A_iB_i$ ($i\in +\{1,2,3\}$) pass through the point $S$. From $s_i\perp A_iB_i$ and the fact that +the chords are parallel to each other, it follows that all the simetrals $s_i$ +are equal to each other - we denote them with $s$. Therefore, $s$ is the common simetral of +the chords $A_iB_i$, so $\mathcal{S}_s(A_i)=B_i$ ($i\in +\{1,2,3\}$). + +Since $S_1$, $S_2$, and $S_3$ are the centers of the lines $C_1B_1$, $C_2B_2$, and $C_3B_3$, and the sides $A_2A_3$, $A_1A_3$, and $A_1A_2$ of the triangle $A_1A_2A_3$, according to the theorem \ref{vektSestSplosno}, +$\overrightarrow{S_1S_2}= +\frac{1}{2}(\overrightarrow{C_1C_2}+\overrightarrow{B_1B_2})$ or +$\overrightarrow{C_1C_2}= +2\overrightarrow{S_1S_2}-\overrightarrow{B_1B_2} +=\overrightarrow{A_2A_1}-\overrightarrow{B_1B_2}=\overrightarrow{A_2A_1} ++\overrightarrow{B_2B_1}$. +Since the points $A_2$ and $A_1$ are reflected across the line $s$ at the points $B_2$ and $B_1$, the vector +$\overrightarrow{C_1C_2}=\overrightarrow{A_2A_1} ++\overrightarrow{B_2B_1}$ is collinear with the line $s$ (see +\ref{izoSimVekt}). Similarly, the vector $\overrightarrow{C_2C_3}$ +is collinear with the line $s$, so the points $C_1$, $C_2$, and $C_3$ +lie on the same line parallel to the line $s$. + \kdokaz + + + +%________________________________________________________________________________ + \poglavje{Translations} \label{odd6Transl} + + So far we have considered isometries that have at least one fixed point. Of the direct isometries we had the identity and rotation, and of the indirect ones, the basic reflection. Now we will introduce a new type of isometries that have no fixed points. + + Let $\overrightarrow{v}$ + be an arbitrary vector of the Euclidean plane ($\overrightarrow{v}\neq \overrightarrow{0}$). The transformation of this plane, in which the +point $X$ is mapped to such a point $X'$, that $\overrightarrow{XX'}=\overrightarrow{v}$, is called the \index{translacija} \pojem{translation} or the \index{vzporedni premik} \pojem{parallel shift} for +the vector $\overrightarrow{v}$, and is denoted by $\mathcal{T}_{\overrightarrow{v}}$ (Figure \ref{sl.izo.6.6.1.pic}). The vector $\overrightarrow{v}$ is\index{vektor!translacije} \pojem{the vector of translation}. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.1.pic} +\caption{} \label{sl.izo.6.6.1.pic} +\end{figure} + +We will now prove the first basic properties of translation. + +\bizrek \label{translEnaki} + Two translations are equal if and only + the vectors of these translations are equal, i.e. + $$\mathcal{T}_{\overrightarrow{v}}= + \mathcal{T}_{\overrightarrow{u}}\Leftrightarrow \overrightarrow{v}=\overrightarrow{u}.$$ + \eizrek + + \textbf{\textit{Proof.}} The part ($\Leftarrow$) is trivial. We prove part ($\Rightarrow$). For an arbitrary point $X$ we have + $\mathcal{T}_{\overrightarrow{v}}(X)= + \mathcal{T}_{\overrightarrow{u}}(X)=X'$. In this case + $\overrightarrow{v}=\overrightarrow{u}=\overrightarrow{XX'}$. + \kdokaz + + + \bizrek + A translation has no fixed points. + \eizrek + +\textbf{\textit{Proof.}} If $X$ is a fixed point of the translation $\mathcal{T}_{\overrightarrow{v}}$ or $\mathcal{T}_{\overrightarrow{v}}(X)=X$, then $\overrightarrow{v}=\overrightarrow{XX}=\overrightarrow{0}$, which according to the definition of translation is not possible. +\kdokaz + + + \bizrek + The inverse transformation of a translation is a translation + with the opposite vector, i.e. + $\mathcal{T}^{-1}_{\overrightarrow{v}}= + \mathcal{T}_{-\overrightarrow{v}}$ + \eizrek + +\textbf{\textit{Proof.}} Let $Y$ be an arbitrary point and $X$ such a point that $\overrightarrow{XY}=\overrightarrow{v}$. From this it follows that $\mathcal{T}_{\overrightarrow{v}}(X)=Y$. From $\overrightarrow{YX}=-\overrightarrow{v}$ it follows that $\mathcal{T}_{-\overrightarrow{v}}(Y)=X$. Because $\mathcal{T}^{-1}_{\overrightarrow{v}}(Y)=X$, it follows that $\mathcal{T}^{-1}_{\overrightarrow{v}}(Y)= + \mathcal{T}_{-\overrightarrow{v}}(Y)$. Because this is true for an arbitrary point $Y$, we have $\mathcal{T}^{-1}_{\overrightarrow{v}}= + \mathcal{T}_{-\overrightarrow{v}}$. + \kdokaz + + \bizrek + Translations are isometries of the plane. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.2.pic} +\caption{} \label{sl.izo.6.6.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $\mathcal{T}_{\overrightarrow{v}}$ be an arbitrary translation. From the definition it is clear that it represents a bijective mapping. It remains to be proven that for any points $X$ and $Y$ + and their images $\mathcal{T}_{\overrightarrow{v}}:X,Y\mapsto X', Y'$ it holds that + $X'Y'\cong XY$ (Figure \ref{sl.izo.6.6.2.pic}). + From $\overrightarrow{XX'}=\overrightarrow{v}$ and $\overrightarrow{YY'}=\overrightarrow{v}$ it follows that $\overrightarrow{XX'}=\overrightarrow{YY'}$. By \ref{vektABCD_ACBD} it is then also true that $\overrightarrow{XY}=\overrightarrow{X'Y'}$ or $X'Y'\cong XY$. + \kdokaz + +We will now show that, just like a rotation, any translation can be expressed as +the composition of two basic reflections. + + + + \bizrek \label{translKom2Zrc} + Any translation $\mathcal{T}_{\overrightarrow{v}}$ + can be expressed as the product of two reflections + $\mathcal{S}_p$ and $\mathcal{S}_q$ where + $p$ and $q$ are arbitrary parallel lines with a common + perpendicular line ($P\in p$ and $Q\in q$) + such that $\overrightarrow{v} + =2\overrightarrow{PQ}$, i.e. + $$\mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_q\circ\mathcal{S}_p + \hspace*{1mm} \Leftrightarrow \hspace*{1mm} p\parallel q \hspace*{1mm} + \wedge\hspace*{1mm} \overrightarrow{v} + =2\overrightarrow{PQ}\hspace*{1mm} (PQ\perp p,\hspace*{1mm} + P\in p,\hspace*{1mm} Q\in q).$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.3.pic} +\caption{} \label{sl.izo.6.6.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.6.3.pic}) + +($\Leftarrow$) Let us assume that $p\parallel q$, $\overrightarrow{v}=2\overrightarrow{PQ}$, $PQ\perp p$, + $P\in p$ and $Q\in q$. + We need to prove that + $\mathcal{T}_{\overrightarrow{v}}(X)= + \mathcal{S}_q\circ\mathcal{S}_p(X)$ for any point $X$ of this plane. + Let $\mathcal{S}_q\circ\mathcal{S}_p(X)=X'$. We will prove that also $\mathcal{T}_{\overrightarrow{v}}(X)=X'$ or $\overrightarrow{XX'}= \overrightarrow{v}$. + We will mark +$\mathcal{S}_p(X)=X_1$. In this case it is clear that $\mathcal{S}_q(X_1)=X'$. Because of this, first the line $XX'$ is the perpendicular +bisector of the parallels $p$ and $q$, which means that the vector $\overrightarrow{XX'}$ is parallel to the vector $\overrightarrow{v}$. No matter the position of the point $X$, in +every case the length of the line $XX'$ is twice as long as the distance between +the parallels $p$ and $q$, which is equal to the length of the line $PQ$. So $\overrightarrow{XX'} =2\overrightarrow{PQ}= \overrightarrow{v}$. With this we have proven that +$\mathcal{T}_{\overrightarrow{v}}(X)=X'$. + +($\Rightarrow$) Let now $\mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_q\circ\mathcal{S}_p$. +The parallels $p$ and $q$ are in this case parallel, because otherwise the intersection point of these two lines would be a fixed point of the composition $\mathcal{S}_q\circ\mathcal{S}_p$, but the translation $\mathcal{T}_{\overrightarrow{v}}$ has no fixed points. +From the first part of the proof ($\Leftarrow$) it follows that the composition $\mathcal{S}_q\circ\mathcal{S}_p$ is equal to the translation $\mathcal{T}_{\overrightarrow{v_1}}$ for the vector $\overrightarrow{v_1}$, where $\overrightarrow{v_1}=2\overrightarrow{PQ}$, $PQ\perp p$, + $P\in p$ and $Q\in q$. But from +$\mathcal{T}_{\overrightarrow{v}}= +\mathcal{S}_q\circ\mathcal{S}_p=\mathcal{T}_{\overrightarrow{v_1}}$ + it follows that $\overrightarrow{v}=\overrightarrow{v_1}$ (from statement \ref{translEnaki}) or $\overrightarrow{v}=2\overrightarrow{PQ}$. + \kdokaz + +From the previous statement it follows that the translation as the composition +of two indirect isometries is a direct isometry. + + + \bizrek + A translation is a direct isometry. + \eizrek + +We will prove that we can represent a translation as +the composition of two central reflections. + +\bizrek \label{transl2sred} +The product of two half-turns is a translation, such that +$$\mathcal{S}_Q\circ\mathcal{S}_P= +\mathcal{T}_{2\overrightarrow{PQ}}.$$ +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.4.pic} +\caption{} \label{sl.izo.6.6.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $X$ be an arbitrary point, $\mathcal{S}_Q\circ\mathcal{S}_P(X)=X'$ and $\mathcal{S}_P(X)=X_1$ (Figure \ref{sl.izo.6.6.4.pic}). Then $\mathcal{S}_Q(X_1)=X'$. By the definition of central reflections, points $P$ and $Q$ are the centers of the lines $XX_1$ and $X_1X'$, which means that the line $PQ$ is the median of the triangle $XX_1X'$ for the base $XX'$. By Theorem \ref{srednjicaTrikVekt}, +$\overrightarrow{XX'}=2\overrightarrow{PQ}$, so $\mathcal{T}_{2\overrightarrow{PQ}}(X)=X'= +\mathcal{S}_Q\circ\mathcal{S}_P(X)$. Since this is true for every point $X$, $\mathcal{S}_Q\circ\mathcal{S}_P= +\mathcal{T}_{2\overrightarrow{PQ}}$. +\kdokaz + +\bizrek \label{translKomp} +The product of two translations is a translation for the vector +that is the sum of the vectors of these two translations, i.e. +$$\mathcal{T}_{\overrightarrow{u}}\circ +\mathcal{T}_{\overrightarrow{v}}= +\mathcal{T}_{\overrightarrow{v}+\overrightarrow{u}}.$$ +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.5.pic} +\caption{} \label{sl.izo.6.6.5.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.6.5.pic}). + Let $P$ be an arbitrary point, $Q$ such a point that + $\overrightarrow{PQ} = \frac{1}{2}\overrightarrow{v}$, +and $R$ a point for which +$\overrightarrow{QR} = \frac{1}{2}\overrightarrow{u}$. + Because $\overrightarrow{v} + \overrightarrow{u} + = 2\overrightarrow{PQ} + 2\overrightarrow{QR} = 2\overrightarrow{PR}$, + by the previous statement \ref{transl2sred} it follows: + $$\mathcal{T}_{\overrightarrow{u}}\circ + \mathcal{T}_{\overrightarrow{v}}= + \mathcal{T}_{2\overrightarrow{QR}}\circ + \mathcal{T}_{2\overrightarrow{PQ}}= + \mathcal{S}_R\circ\mathcal{S}_Q\circ + \mathcal{S}_Q\circ\mathcal{S}_P= + \mathcal{S}_R\circ\mathcal{S}_P= + \mathcal{T}_{2\overrightarrow{PR}}= + \mathcal{T}_{\overrightarrow{v}+\overrightarrow{u}},$$ which was to be proven. \kdokaz + + The consequence of the proven is the following statement. + + \bizrek + The product of translations is commutative, i.e. + $$\mathcal{T}_{\overrightarrow{u}}\circ + \mathcal{T}_{\overrightarrow{v}}= + \mathcal{T}_{\overrightarrow{v}}\circ + \mathcal{T}_{\overrightarrow{u}}$$ + \eizrek + + \textbf{\textit{Proof.}} By the previous statement \ref{translKomp} it is: + $$\mathcal{T}_{\overrightarrow{u}}\circ + \mathcal{T}_{\overrightarrow{v}}= + \mathcal{T}_{\overrightarrow{v}+\overrightarrow{u}}= + \mathcal{T}_{\overrightarrow{u}+\overrightarrow{v}}= + \mathcal{T}_{\overrightarrow{v}}\circ + \mathcal{T}_{\overrightarrow{u}},$$ which was to be proven. \kdokaz + + +Similarly to the basic reflection, rotation and central reflection, also for translation the statement about +transmutation\index{transmutation!of translation} holds. + +\bizrek \label{izoTransmTrans} +For an arbitrary translation +$\mathcal{T}_{2\overrightarrow{PQ}}$ + and an arbitrary isometry $\mathcal{I}$ is +$$\mathcal{I}\circ +\mathcal{T}_{2\overrightarrow{PQ}}\circ\mathcal{I}^{-1}= +\mathcal{T}_{2\overrightarrow{\mathcal{I}(P)\mathcal{I}(Q)}}.$$ +\eizrek + +\textbf{\textit{Proof.}} By \ref{transl2sred} we can write the translation + $\mathcal{T}_{2\overrightarrow{PQ}}$ + as the composite of two central +reflections, namely +$\mathcal{T}_{2\overrightarrow{PQ}}=\mathcal{S}_Q\circ\mathcal{S}_P$. +If we use the transmutation theorem for central reflections \ref{izoTransmSredZrc}, we get: + \begin{eqnarray*} + \mathcal{I}\circ + \mathcal{T}_{2\overrightarrow{PQ}}\circ\mathcal{I}^{-1}&=& + \mathcal{I}\circ\mathcal{S}_Q\circ\mathcal{S}_P\circ\mathcal{I}=\\ + &=& + \mathcal{I}\circ\mathcal{S}_Q\circ\mathcal{I}^{-1} + \circ\mathcal{I}\circ\mathcal{S}_P\circ\mathcal{I}^{-1}=\\ + &=&\mathcal{S}_{\mathcal{I}(Q)}\circ\mathcal{S}_{\mathcal{I}(P)}=\\ + &=&\mathcal{T}_{2\overrightarrow{\mathcal{I}(P)\mathcal{I}(Q)}}, + \end{eqnarray*} + which is what needed to be proven. \kdokaz + + + + \bizrek \label{izoKompTranslRot} + The product of a translation and a rotation (also a rotation and a translation) + is a rotation with the same angle. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.5a.pic} +\caption{} \label{sl.izo.6.6.5a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $\mathcal{I}=\mathcal{R}_{S,\omega}\circ + \mathcal{T}_{\overrightarrow{v}}$ be the composition of the translation + $\mathcal{T}_{\overrightarrow{v}}$ and the rotation + $\mathcal{R}_{S,\omega}$. Let $b$ be a line passing through the point + $S$ and perpendicular to the vector $\overrightarrow{v}$ + (Figure \ref{sl.izo.6.6.5a.pic}). According to + \ref{translKom2Zrc} for a given line $a$, $a\parallel b$, it holds that + $\mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_b\circ + \mathcal{S}_a$. Let $c$ be a line passing through the point $S$, for which + $\measuredangle b,c=\frac{1}{2}\omega$. According to + \ref{rotacKom2Zrc} is $\mathcal{R}_{S,\omega}=\mathcal{S}_c\circ + \mathcal{S}_b$. Therefore: + $$\mathcal{I}=\mathcal{R}_{S,\omega}\circ + \mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_c\circ + \mathcal{S}_b\circ\mathcal{S}_b\circ + \mathcal{S}_a=\mathcal{S}_c\circ + \mathcal{S}_a.$$ + By Playfair's axiom, the lines $a$ and $c$ intersect in some point + $S_1$, according to \ref{KotiTransverzala} is + $\measuredangle a,c=\measuredangle b,c=\frac{1}{2}\omega$. + If we use \ref{rotacKom2Zrc} once again, we get: +$$\mathcal{R}_{S,\omega}\circ + \mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_c\circ + \mathcal{S}_a=\mathcal{R}_{S_1,\omega}.$$ + In a similar way it can be proven that the composition + of the rotation and + translation $\mathcal{T}_{\overrightarrow{v}} + \circ\mathcal{R}_{S,\omega}$ is a rotation for the same angle. + \kdokaz + +In the following we will consider the use of translation. First, we +will see how we can use translation in planning tasks. + + + \bzgled + Let $A$ and $B$ be interior points of an angle $pOq$. + Construct points $C$ and $D$ on the sides $p$ and $q$ + such that the quadrilateral $ABCD$ is a parallelogram. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.6.pic} +\caption{} \label{sl.izo.6.6.6.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.6.6.pic}) +Since $ABCD$ is a parallelogram, $\overrightarrow{AB}= \overrightarrow{DC} = \overrightarrow{v}$. + Therefore, $\mathcal{T}_{\overrightarrow{v}}(D)= C$. From the given +conditions, the point $D$ lies on the line segment $p$, so its image $C$ under +the translation $\mathcal{T}_{\overrightarrow{v}}$ lies on the image $p'$ of the line segment $p$ under this translation (the image $p'$ can be obtained by drawing the image $O'$ of the point $O$, and then the parallel line segment of the line segment $p$ from the point $O'$). Since the point $C$ also lies on the line segment $q$, we can draw it from the condition $C\in p'\cap q$. The number of solutions to the task depends on whether the line segments $p'$ and $q$ have any common points. Since the line segments $p$ and $q$ are not parallel (it is the angle between them), the line segments $p'$ and $q$ do not have more than one common point. Therefore, the task has one or no solution. + \kdokaz + + + + \bzgled + Let $AB$ be a chord of a circle $k$, $P$ and $Q$ points of this circle lying on the + same side of the line $AB$ and $d$ a line segment in the plane. + Construct a point $L$ on the circle + $k$ such that $XY\cong d$, where $X$ and $Y$ + are intersections of the lines $LP$ and $LQ$ with the chord $AB$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.7.pic} +\caption{} \label{sl.izo.6.6.7.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.6.7.pic}) +Although the points $X$ and $Y$ are unknown, the vector +$\overrightarrow{v}=\overrightarrow{XY}$ is known, which has the same length +as the distance $d$ and is parallel and has the same direction as the vector +$\overrightarrow{AB}$. Also, as $\omega=\angle PLQ$ is known, +because it is the central angle for the chord $PQ$ (statement \ref{ObodKotGMT}). Let +$P'=\mathcal{T}_{\overrightarrow{v}}(P)$. Because +$\overrightarrow{PP'}=\overrightarrow{v}=\overrightarrow{XY}$, the quadrilateral $PP'YX$ is a parallelogram, so the angles $P'YQ$ and $PLQ$ +are complementary (statement \ref{KotaVzporKraki}). We can therefore +construct the point $Y$ as the intersection of the chord $AB$ with the appropriate chord +$P'Q$ and central angle $\omega$. Then +$X=\mathcal{T}^{-1}_{\overrightarrow{v}}(Y)$. + \kdokaz + + + + \bzgled \footnote{Predlog za MMO 1996. (SL 10.)} + Let $H$ be the orthocentre of a triangle $ABC$ and $P$ the point lying on + the circumcircle of this triangle different from its vertices. + $E$ is the foot of the altitude from the vertex $B$ of the triangle $ABC$. + Suppose that the quadrilaterals $PAQB$ and $PARC$ are parallelograms. The lines + $AQ$ and $HR$ intersect at a point $X$. Prove that $EX\parallel AP$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.8.pic} +\caption{} \label{sl.izo.6.6.8.pic} +\end{figure} + + \textbf{\textit{Solution.}} We mark with $O$ the center of the circumscribed + circle of the triangle $ABC$ and with $\mathcal{T}_{\overrightarrow{PA}}$ the translation for the vector $\overrightarrow{PA}$ (Figure \ref{sl.izo.6.6.8.pic}). + +Since $PAQB$ and $PARC$ are parallelograms, $\overrightarrow{BQ}=\overrightarrow{PA}=\overrightarrow{CR}$. +This means that the translation $\mathcal{T}_{\overrightarrow{PA}}$ maps the triangle $BPC$ +into the triangle $QAR$, the altitude point $H_1$ of the triangle $BPC$ into the altitude point $H'_1$ +of the triangle $QAR$. We prove that $H'_1=H$ or $H=\mathcal{T}_{\overrightarrow{PA}}(H_1)$ +or $\overrightarrow{H_1H}=\overrightarrow{PA}$. If we use Hamilton's theorem \ref{Hamilton} +for the triangle $ABC$ and $PBC$, which have a common center of the inscribed circle, we get +$\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC} +=\overrightarrow{OH}$ and $\overrightarrow{OP}+\overrightarrow{OB}+\overrightarrow{OC} +=\overrightarrow{OH_1}$. From this it follows first $\overrightarrow{OH}-\overrightarrow{OA}= +\overrightarrow{OH_1}-\overrightarrow{OP}$, then $\overrightarrow{OH}-\overrightarrow{OH_1}= +\overrightarrow{OA}-\overrightarrow{OP}$ or $\overrightarrow{H_1H}=\overrightarrow{PA}$. + +So $H$ is the altitude point of the triangle $ARQ$, which means that $RH\perp AQ$ or $\angle AXH=90^0$. +Since $\angle AEH=90^0$ as well, the points $X$ and $E$ lie on the circle +with diameter $AH$ (theorem \ref{TalesovIzrKrozObrat}), so $AXEH$ is a trapezoid. So: +\begin{eqnarray*} +\angle AXE&=&180^0-\angle AHE \hspace*{4mm} \textrm{(theorem \ref{TetivniPogoj})}\\ +&=&180^0-\angle ACB \hspace*{4mm} \textrm{(theorem \ref{KotaPravokKraki})}\\ +&=&180^0-\angle APB \hspace*{4mm} \textrm{(theorem \ref{ObodObodKot})}\\ +&=&\angle QAP \hspace*{4mm} \textrm{(by theorem \ref{paralelogram}, because }PAQB\textrm{ is a parallelogram)} +\end{eqnarray*} + +By theorem \ref{KotiTransverzala}, $EX\parallel AP$. +\kdokaz + +%________________________________________________________________________________ + \poglavje{Composition of Two Rotations} \label{odd6KompRotac} + +In the previous sections, we have already considered the composition of some mappings. +We have found that the composition of two basic reflections is a direct isometry, namely the identity or a rotation or a translation, depending on whether the lines are equal or intersect or are parallel. In the previous section, we proved that the composition of two central reflections and also the composition of two translations is a translation. + +In this section, we will investigate the composition of two rotations. We start with the following statement. + + + \bizrek \label{rotacKomp2rotac} + Let $R_{A,\alpha}$ and $R_{B,\beta}$ be rotations of the plane + and $\mathcal{I}=R_{B,\beta}\circ R_{A,\alpha}$. + Then: + \begin{enumerate} + \item if $\alpha+\beta\notin\{k\cdot 360^0;k\in \mathbb{Z}\}$ and $A=B$, + the product $\mathcal{I}$ is a rotation for the angle $\alpha+\beta$ with the centre $A$, + \item if $\alpha+\beta\notin\{k\cdot 360^0;k\in \mathbb{Z}\}$ and $A\neq B$, + the product $\mathcal{I}$ is a rotation for the angle $\alpha+\beta$ + with the centre $C$,\\ where $\measuredangle CAB=\frac{1}{2}\alpha$ + and $\measuredangle ABC=\frac{1}{2}\beta$, + \item if $\alpha+\beta\in\{k\cdot 360^0;k\in \mathbb{Z}\}$ and $A=B$, + the product $\mathcal{I}$ is the identity map, + \item if $\alpha+\beta\in\{k\cdot 360^0;k\in \mathbb{Z}\}$ and $A\neq B$, + the product $\mathcal{I}$ is a translation. + \end{enumerate} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.1.pic} +\caption{} \label{sl.izo.6.7.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.7.1.pic}) + +Let us assume that the centres $A$ and $B$ are different (the case +$A=B$ is trivial). + +Let $p$ and $q$ be such lines and planes, that +$\measuredangle p,AB=\frac{1}{2}\alpha$ and $\measuredangle +AB,q=\frac{1}{2}\beta$. If we decompose each of the rotations with the help of +reflections on the axes (statement \ref{rotacKom2Zrc}), we get: +$$\mathcal{I}=R_{B,\beta}\circ R_{A,\alpha}= +\mathcal{S}_q\circ\mathcal{S}_{AB}\circ\mathcal{S}_{AB}\circ +\mathcal{S}_p=\mathcal{S}_q\circ +\mathcal{S}_p.$$ + +If $\alpha+\beta\in\{0^0,360^0,-360^0\}$, +then $\frac{1}{2}\left(\alpha+\beta\right)\in\{0^0,180^0,-180^0\}$, +which means that the lines $p$ and $q$ are parallel (statement \ref{KotiTransverzala}). + In this case, the composition $\mathcal{I}=\mathcal{S}_q\circ\mathcal{S}_p$ + is a translation (statement \ref{translKom2Zrc}). + +If $\alpha+\beta\notin\{0^0,360^0,-360^0\}$, +then $\frac{1}{2}\left(\alpha+\beta\right)\notin\{0^0,180^0,-180^0\}$ +and the lines $p$ and $q$ intersect in some point $C$ (statement \ref{KotiTransverzala}). +As $\measuredangle pCq$ is the exterior angle of the triangle $ABC$, it is equal to the sum +of the angles $\measuredangle p,AB$ and $\measuredangle AB,q$ (statement \ref{zunanjiNotrNotr}). +Therefore $\measuredangle pCq=\frac{1}{2}\left(\alpha+\beta\right)$. +The composition +$\mathcal{I}=\mathcal{S}_q\circ\mathcal{S}_p$ in this case represents +a rotation (statement \ref{rotacKom2Zrc}) $\mathcal{R}_{C,2\measuredangle pCq}=\mathcal{R}_{C,\alpha+\beta}$. + \kdokaz + + In the following we will see the use of the previous statement in various tasks. + + + \bzgled + Let $P$, $Q$ and $R$ be non-collinear points. + Construct a triangle $ABC$ such that $ARB$, $BPC$ and $CQA$ are regular triangles + with the same orientation. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.2.pic} +\caption{} \label{sl.izo.6.7.2.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.7.2.pic}) + +If $A$, $B$ and $C$ are points that satisfy the given +conditions, it holds: +$$\mathcal{R}_{Q,60^0} \circ \mathcal{R}_{R,60^0} \circ \mathcal{R}_{P,60^0}(C)=C.$$ +But according to the previous statement \ref{rotacKomp2rotac} +the composite $\mathcal{R}_{Q,60^0} \circ \mathcal{R}_{R,60^0} \circ \mathcal{R}_{P,60^0}$ +is a central symmetry (because $60^0+60^0+60^0=180^0$) with a +fixed +point $C$, so: +$$\mathcal{R}_{Q,60^0} \circ \mathcal{R}_{R,60^0} \circ \mathcal{R}_{P,60^0}=\mathcal{S}_C.$$ +The vertex $C$ can therefore be constructed as the center of the line segment $XX'$, +where $X'$ is the image of an arbitrary point $X$ under the composite +$\mathcal{R}_{Q,60^0} \circ \mathcal{R}_{R,60^0} \circ +\mathcal{R}_{P,60^0}$. Then $B=\mathcal{R}_{P,60^0}(C)$ and +$A=\mathcal{R}_{R,60^0}(B)$. + \kdokaz + + + + \bzgled \label{rotKompZlato} + Let $BALK$ and $ACPQ$ be squares with the same orientation and + $Z$ the midpoint of the line segment $PK$. + Prove that $BZC$ is an isosceles right triangle with the hypotenuse $BC$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.3.pic} +\caption{} \label{sl.izo.6.7.3.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.7.3.pic}) + + Let $\mathcal{I}=\mathcal{R}_{B,90^0}\circ \mathcal{R}_{C,90^0}$. Because +$90°+90°= 180^0$, according to statement \ref{rotacKomp2rotac} +the isometry $\mathcal{I}$ represents the central symmetry $\mathcal{S}_Y$, +where $\measuredangle YCB=\frac{1}{2}\cdot 90^0=45^0$ and $\measuredangle CBY=\frac{1}{2}\cdot 90^0=45^0$. +Therefore, the point $Y$ is the vertex of the isosceles right triangle $BYC$ with the hypotenuse $BC$. But since also +$\mathcal{S}_Y(P)=\mathcal{I}(P)=K$, it follows that $Y=Z$. + \kdokaz + +\bzgled \label{rotZgl1} +On each side of a quadrilateral $ABCD$ squares + $BALK$, $CBMN$, $DCSR$ and $ADPQ$ +are externally erected. +Let $E$, $F$, $G$ and $H$ be the midpoints of the line segments +$PK$, $MR$, $LN$ and $SQ$. + Prove that the quadrilaterals $BFDE$ and $AGCH$ are also squares. +\ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.4.pic} +\caption{} \label{sl.izo.6.7.4.pic} +\end{figure} + + \textbf{\textit{Solution.}} By the previous statement \ref{rotKompZlato} + the triangles $DEB$ and $BFD$ are isosceles and right with a common + hypotenuse $BD$ (Figure \ref{sl.izo.6.7.4.pic}). From this it follows that the quadrilateral + $BFDE$ is a square. In the same way we get that the quadrilateral $AGCH$ is a square. + \kdokaz + + + + \bzgled \label{rotZgl2} + Let $A_1$, $B_1$ and $C_1$ be the centres of squares + externally erected on the sides $BC$, $AC$ and $AB$ of an arbitrary + triangle $ABC$ + and $P$ the midpoint of the side $AC$. Prove that: + \begin{enumerate} + \item $C_1PA_1$ is an isosceles right triangle, + \item $C_1B_1$ and $A_1A$ are perpendicular and congruent line segments, + \item the line segments $AA_1$, $BB_1$ and $CC_1$ intersect at + a single point\footnote{To je drugi od treh planimetričnih + izrekov, ki jih je objavil italijanski matematik + \index{Bellavitis, G.} \textit{G. + Bellavitis} (1803--1880) na kongresu v Milanu 1844.}. + \end{enumerate} + \ezgled + + +\textbf{\textit{Solution.}} + +\textit{1)} Let $\mathcal{I}= \mathcal{R}_{C_1,90^0} +\mathcal{R}_{A_1,90^0}$ (Figure \ref{sl.izo.6.7.5.pic}). Because $90^0+90^0= 180^0$, the isometry $\mathcal{I}$ represents the central reflection $\mathcal{S}_Y$, where the point $Y$ is the vertex of an isosceles right triangle $C_1YA_1$ with the hypotenuse $C_1A_1$ (statement \ref{rotacKomp2rotac}). Because $\mathcal{S}_Y(C)=\mathcal{I}(C)=A$, it follows that $Y=P$. + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.5.pic} +\caption{} \label{sl.izo.6.7.5.pic} +\end{figure} + + + +\textit{2)} With the rotation $\mathcal{R}_{P,90^0}$, the line segment $B_1C_1$ is mapped to the line segment $AA_1$ (Figure \ref{sl.izo.6.7.5a.pic}). Therefore, the line segments are congruent and perpendicular (statement \ref{rotacPremPremKot}). + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.5a.pic} +\caption{} \label{sl.izo.6.7.5a.pic} +\end{figure} + +\textit{3)} From the proven part \textit{(2)} it follows that the lines $AA_1$, $BB_1$, $CC_1$ are the altitudes of the triangle $A_1B_1C_1$ (Figure \ref{sl.izo.6.7.5a.pic}), so they intersect in its altitude point (statement \ref{VisinskaTocka}). + \kdokaz + + + + \bzgled \label{rotZgl3} + Let $ABCD$ and $AB_1C_1D_1$ be squares with the same orientation, + $P$ and $Q$ the midpoints of the line segments $BD_1$ and $B_1D$. + Suppose that $O$ and $S$ are the centres of these squares. + Prove that the quadrilateral $POQS$ is also a square. + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.6.pic} +\caption{} \label{sl.izo.6.7.6.pic} +\end{figure} + +\textbf{\textit{Solution.}} + The statement is obvious if we use part \textit{(1)} + from the previous example \ref{rotZgl2} twice (Figure \ref{sl.izo.6.7.6.pic}). + \kdokaz + + + + If we connect the facts from examples \ref{rotZgl1} and \ref{rotZgl3}, + we get the following statement (Figure \ref{sl.izo.6.7.4a.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.4a.pic} +\caption{} \label{sl.izo.6.7.4a.pic} +\end{figure} + + + + \bzgled \label{rotZgl4} + Let $BALK$, $CBMN$, $DCSR$ and $ADPQ$ be the + squares externally erected on the four sides of an arbitrary quadrilateral $ABCD$. + Then all quadrilaterals defined by the following vertices are also squares: + \begin{enumerate} + \item the point $B$, the midpoint of the line segment $MR$, the point $D$ and the midpoint of the line segment $PK$, + \item the point $A$, the midpoint of the line segment $LN$, the point $C$ and the midpoint of the line segment $SQ$, + \item the midpoints of the line segments $QL$, $LB$, $BD$ and $DQ$, + \item the midpoints of the line segments $KM$, $MC$, $CA$ and $AK$, + \item the midpoints of the line segments $NS$, $SD$, $DB$ and $BN$, + \item the midpoints of the line segments $RP$, $PA$, $AC$ and $CR$. + \end{enumerate} + + \ezgled + + In the next example we will use the composition of rotations in a situation where the sum of the angles of these rotations is equal to $360^0$. + + + \bzgled \label{rotZgl5} + Let $ABC$ and $A'B'C'$ be isosceles triangles of the same orientation with the bases $BC$ and + $B'C'$ and $\angle BAC\cong\angle B'A'C'=\alpha$. + Suppose that $A_0$, $B_0$ and $C_0$ are the midpoints of the line segments $AA'$, $BB'$ and $CC'$. + Prove that $A_0B_0C_0$ is also an isosceles triangle + with the base $B_0C_0$ and $\angle B_0A_0C_0\cong\alpha$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.7.pic} +\caption{} \label{sl.izo.6.7.7.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.7.7.pic}). + +Let $\mathcal{I}= \mathcal{S}_{C_0} +\circ \mathcal{R}_{A',\alpha} \circ \mathcal{S}_{B_0} \circ +\mathcal{R}_{A,-\alpha}$. The composite $\mathcal{I}$ is a direct isometry. Because the sum of the corresponding +angles of rotation is equal to $360^0$ and $\mathcal{I}(C)=C$, it must be +$\mathcal{I}=\mathcal{E}$ (statement \ref{rotacKomp2rotac}). If we use the statement about +the transmutation of rotation \ref{izoTransmRotac}, we get + $\mathcal{S}_{A_0} \circ\mathcal{R}_{A,\alpha} \circ \mathcal{S}_{A_0} =\mathcal{R}_{A',\alpha}$. + From what has been proven and from the statement \ref{transl2sred} it follows: + \begin{eqnarray*} + \mathcal{E}&=&\mathcal{I}= \mathcal{S}_{C_0} + \circ \mathcal{R}_{A',\alpha} \circ \mathcal{S}_{B_0} + \circ \mathcal{R}_{A,-\alpha}=\\ + &=& \mathcal{S}_{C_0} \circ (\mathcal{S}_{A_0} \circ\mathcal{R}_{A,\alpha} + \circ \mathcal{S}_{A_0}) \circ \mathcal{S}_{B_0} \circ \mathcal{R}_{A,-\alpha}=\\ + &=& (\mathcal{S}_{C_0} \circ \mathcal{S}_{A_0}) \circ\mathcal{R}_{A,\alpha} + \circ (\mathcal{S}_{A_0} \circ \mathcal{S}_{B_0}) \circ \mathcal{R}_{A,-\alpha}=\\ + &=& \mathcal{T}_{2\overrightarrow{A_0C_0}} \circ\mathcal{R}_{A,\alpha} + \circ \mathcal{T}_{2\overrightarrow{B_0A_0}} \circ \mathcal{R}_{A,-\alpha} + \end{eqnarray*} + Therefore $\mathcal{E}=\mathcal{T}_{2\overrightarrow{A_0C_0}} + \circ\mathcal{R}_{A,\alpha} \circ \mathcal{T}_{2\overrightarrow{B_0A_0}} + \circ \mathcal{R}_{A,-\alpha}$, or $\mathcal{T}_{2\overrightarrow{C_0A_0}}= + \mathcal{R}_{A,\alpha} \circ \mathcal{T}_{2\overrightarrow{B_0A_0}} \circ + \mathcal{R}_{A,-\alpha}$. If we use the statement about the transmutation of translation + \ref{izoTransmTrans}, we get: + $$\mathcal{T}_{2\overrightarrow{C_0A_0}}= \mathcal{R}_{A,\alpha} + \circ \mathcal{T}_{2\overrightarrow{B_0A_0}} \circ + \mathcal{R}_{A,-\alpha}=\mathcal{T}_{2\overrightarrow{B'_0A'_0}},$$ + where $\mathcal{R}_{A,\alpha}:A_0, B_0\mapsto A'_0, B'_0$. Therefore + $\overrightarrow{C_0A_0}=\overrightarrow{B'_0A_0'}$ or + $\overrightarrow{A_0C_0}=\overrightarrow{A'_0B'_0}$. + The vector $\overrightarrow{A_0B_0}$ is transformed into the vector $\overrightarrow{A'_0B'_0}=\overrightarrow{A_0C_0}$ by the rotation $\mathcal{R}_{A,\alpha}$, + so $|A_0B_0|=|A_0C_0|$ and $\measuredangle B_0A_0C_0=\measuredangle\overrightarrow{A_0B_0}, + \overrightarrow{A_0C_0}=\alpha$. + \kdokaz + +\bzgled \label{RotacZglVeck} +Let $A_1A_2...A_n$ and $B_1B_2...B_n$ be regular + $n$-gons with the same orientation. Suppose that + $S_1$, $S_2$, ..., $S_n$ are the midpoints of the line segments +$A_1B_1$, $A_2B_2$, ..., $A_nB_n$. Prove that $S_1S_2...S_n$ is also a regular $n$-gon. +\ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.8.pic} +\caption{} \label{sl.izo.6.7.8.pic} +\end{figure} + + \textbf{\textit{Solution.}} +Since $n$-gons $A_1A_2...A_n$ and $B_1B_2...B_n$ +are regular and have the same orientation, $A_1A_2A_3$ and $B_1B_2B_3$ +are equilateral and have the same orientation +as the triangle with the bases $A_1A_3$ and $B_1B_3$ (Figure \ref{sl.izo.6.7.8.pic}). +In addition, $\angle A_1A_2A_3\cong \angle B_1B_2B_3=\frac{(n - 2)\cdot 180^0}{n}$ (\ref{pravVeckNotrKot}). + +From the previous example \ref{rotZgl5} it follows that +also $S_1S_2S_3$ is an equilateral triangle and +$\angle S_1S_2S_3=\frac{(n - 2)\cdot 180^0}{n}$. +Therefore, the sides $S_1S_2$ and $S_2S_3$ are congruent, the interior angle at +the vertex $S_2$ is congruent to the angle of the regular $n$-gon. +We prove analogously that all +sides of the polygon $S_1S_2...S_n$ are congruent and all its +interior angles are congruent, which means that this polygon is regular. + \kdokaz + +We will consider one interesting consequence of the composition of rotations in section \ref{odd7Napoleon}. + + +%________________________________________________________________________________ + \poglavje{Glide Reflections} \label{odd6ZrcDrs} + +So far we have learned about some types of isometries - three direct isometries +(identity, rotation and translation) and one indirect isometry (reflection in a point). +Now we will define another indirect isometry that has no fixed points. + +Let $\mathcal{T}_{\overrightarrow{v}}$ be a translation + for vector $\overrightarrow{v} = 2\overrightarrow{PQ}$ +and $\mathcal{S}_{PQ}$ be a reflection over the line $PQ$. The composite +$\mathcal{S}_{PQ}\circ \mathcal{T}_{\overrightarrow{v}}$ is called +\index{glide reflection}\pojem{glide reflection} with the axis $PQ$ for vector $\overrightarrow{v} = 2\overrightarrow{PQ}$ + (Figure \ref{sl.izo.6.8.1.pic}). We denote it by +$\mathcal{G}_{2\overrightarrow{PQ}}$. + + + From the definition it is clear that a glide reflection is determined by its axis and vector. + + We prove the basic properties of glide reflection. + + + + \bizrek \label{IzoZrcDrs1} + Any glide reflection can be expressed as the product of three reflections.\\ + The reflection and translation in the product, as the presentation of glide reflection, commute, i.e. + $$\mathcal{G}_{2\overrightarrow{PQ}}= + \mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}= + \mathcal{T}_{2\overrightarrow{PQ}}\circ \mathcal{S}_{PQ}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.8.2.pic} +\caption{} \label{sl.izo.6.8.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} Let $\mathcal{G}_{2\overrightarrow{PQ}}= + \mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}$ with the axis $PQ$ + and vector $2\overrightarrow{PQ}$. Let $p$ and $q$ be the perpendiculars + of the line $PQ$ at points $P$ and $Q$ (Figure \ref{sl.izo.6.8.2.pic}). + Because $p,q\perp PQ$ and $p\parallel q$, we can represent the translation + $\mathcal{T}_{2\overrightarrow{PQ}}$ as + the composite $\mathcal{S}_q\circ\mathcal{S}_p$. In this case, the reflection $\mathcal{S}_{PQ}$ + commutes with reflections $\mathcal{S}_p$ and $\mathcal{S}_q$ (statement \ref{izoZrcKomut}). Therefore: + $$\mathcal{G}_{2\overrightarrow{PQ}}= + \mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}= + \mathcal{S}_{PQ}\circ\mathcal{S}_q\circ\mathcal{S}_p= + \mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_{PQ}= + \mathcal{T}_{2\overrightarrow{PQ}}\circ\mathcal{S}_{PQ},$$ which had to be proven. \kdokaz + +\bizrek +\label{izoZrcdrsZrcdrs} +The product of a glide reflection with itself is a translation (Figure \ref{sl.izo.6.8.3.pic}). +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.8.3.pic} +\caption{} \label{sl.izo.6.8.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} We use the previous statement \ref{IzoZrcDrs1} + and the statement \ref{translKomp}: + $$\mathcal{G}^2_{2\overrightarrow{PQ}}= + \mathcal{T}_{2\overrightarrow{PQ}}\circ \mathcal{S}_{PQ}\circ + \mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}= + \mathcal{T}^2_{2\overrightarrow{PQ}}= + \mathcal{T}_{4\overrightarrow{PQ}},$$ which had to be proven. \kdokaz + + From the proof of statement \ref{IzoZrcDrs1} it follows that every glide reflection can + be represented as the composition of three basic +reflections, where the axis of one reflection is perpendicular to the axes of the other two. +A glide reflection is therefore an indirect +isometry as the composition of three +basic reflections - indirect isometries. The same fact also follows from +the definition of glide reflection, because it is the composition of one direct and one indirect isometry. + + \bizrek + A glide reflection has no fixed points. + \eizrek + +\textbf{\textit{Proof.}} By statement \ref{IzoZrcDrs1} a glide reflection + can be represented as the composition of three basic reflections: +$$\mathcal{G}_{2\overrightarrow{PQ}}= + \mathcal{S}_{PQ}\circ\mathcal{S}_q\circ\mathcal{S}_p,$$ + where $p$ and $q$ are perpendicular to the line $PQ$ in points $P$ and $Q$. + If the glide reflection $\mathcal{G}_{2\overrightarrow{PQ}}$ or + the composition $\mathcal{S}_{PQ}\circ\mathcal{S}_q\circ\mathcal{S}_p$ + had a fixed point, $\mathcal{S}_{PQ}\circ\mathcal{S}_q\circ\mathcal{S}_p$ + as an indirect isometry would represent the basic reflection (statement \ref{izo1ftIndZrc}). + By statement \ref{izoSop} in this case the lines $p$, $q$ and $PQ$ would belong to the same + bundle, which is not possible. The glide reflection $\mathcal{G}_{2\overrightarrow{PQ}}$ therefore has no fixed points. + \kdokaz + +We have already used the fact from the statement \ref{IzoZrcDrs1}, +that we can always represent a glide reflection as a composite of three basic +reflections, where the axes of these reflections are not in the same family. But does the converse hold +- that the composite of three basic reflections, whose axes are not from the same family, is always a glide reflection? +In connection with this, the following statement, which is also very important for the classification +of isometries, which we will carry out in the next section, will hold. + + + \bizrek \label{izoZrcdrsprq} + If lines $p$, $q$ and $r$ in the plane + are not from the same family of lines, then product + $\mathcal{S}_p \circ + \mathcal{S}_q\circ \mathcal{S}_r$ is a glide reflection. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.8.4.pic} +\caption{} \label{sl.izo.6.8.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.8.4.pic}) + The line $q$ intersects either the line $p$ or the line +$r$, because otherwise all three lines would belong to one set of parallel +lines. Without loss of generality, let $q\cap r=\{A\}$. The lines $p$, +$q$ and $r$ are not from the same set, so $A\notin p$. Let $s$ +be a line that goes through the point $A$ and is perpendicular to the line $p$ at +the point $B$. The lines $s$, $q$ and $r$ belong to the same set +$\mathcal{X}_A$, so by izrek \ref{izoSop} $\mathcal{S}_r \circ +\mathcal{S}_q \circ \mathcal{S}_ s = \mathcal{S}_t$, where $t$ is also +$t\in \mathcal{X}_A$. From this it follows that $\mathcal{S}_r \circ +\mathcal{S}_q = \mathcal{S}_t \circ \mathcal{S}_s$ or (if we use +izrek \ref{izoSrZrcKom2Zrc}): +$$\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p = \mathcal{S}_t +\circ \mathcal{S}_s \circ \mathcal{S}_p = \mathcal{S}_t \circ \mathcal{S}_B.$$ +Let $C$ be the perpendicular projection of the point $B$ onto the line $t$ +(in this case $B\neq C$, because otherwise the lines $s$ +and $t$, and consequently $r$ and $q$, would overlap) and $b$ the line that is perpendicular to the line $BC$ at +the point $B$. +Then, by izrek \ref{izoSrZrcKom2Zrc} and \ref{translKom2Zrc}: +$$\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p = \mathcal{S}_t \circ \mathcal{S}_B= +\mathcal{S}_t \circ \mathcal{S}_b \circ \mathcal{S}_{BC}= +\mathcal{T}_{2\overrightarrow{BC}} \circ \mathcal{S}_{BC}= +\mathcal{G}_{2\overrightarrow{BC}},$$ which was to be proven. \kdokaz + + \bzgled \label{izoZrcDrsKompSrOsn} + Any glide reflection can be expressed as the product of a reflection and a half-turn, specifically + $$\mathcal{G}_{2\overrightarrow{PQ}}=\mathcal{S}_q\circ \mathcal{S}_P,$$ + where $q$ is perpendicular to the line $PQ$ in the point $Q$. + \ezgled + +By definition, + $\mathcal{G}_{2\overrightarrow{PQ}}=\mathcal{S}_{PQ} + \circ \mathcal{T}_{2\overrightarrow{PQ}}$. If + we use the expression \ref{transl2sred} and \ref{izoSrZrcKom2Zrc}, it is: + $$\mathcal{G}_{2\overrightarrow{PQ}}=\mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}= + \mathcal{S}_{PQ}\circ\mathcal{S}_Q\circ\mathcal{S}_P= + \mathcal{S}_{PQ}\circ\mathcal{S}_{PQ}\circ + \mathcal{S}_q\circ\mathcal{S}_P= + \mathcal{S}_q\circ\mathcal{S}_P,$$ which had to be proven. \kdokaz + + +%________________________________________________________________________________ + \poglavje{Classification of Plane Isometries. Chasles' Theorem} \label{odd6KlasifIzo} + +In the previous sections we investigated certain types of isometries. +The question arises as to whether these are the only isometries +or whether there is another type of isometry that we have not +encountered yet. In this section we will prove that the answer to this +question is negative and make a final classification of all +isometries of the plane. + +First, we will consider direct isometries. So far we have mentioned +the identity, rotation and translation. We will prove that these are the only +types of direct isometries. First, let us recall that each of +these isometries can be represented as the composition of two reflections over a line. Depending on whether the axes overlap, intersect or are +parallel, we got the identity, rotation and translation. These are also the only possibilities for the mutual position of two lines in +the plane. Can all direct isometries be +represented as the composition of two reflections over a line? In that case, the three +aforementioned direct isometries would really be the only ones. We will use this idea in the following theorem. + + + \bizrek \label{Chaslesov+} + Any direct isometry can be expressed as the product of two reflections. + The only direct isometries are + identity map, rotations and + translations. + \eizrek + + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.10.1.pic}) + + Let $\mathcal{I} : \mathbb{E}^2\rightarrow \mathbb{E}^2$ + be a direct isometry of the plane. We will carry out the proof according to the number +of fixed points. + +\textit{1)} If a direct isometry $\mathcal{I}$ has at least two fixed points, +it is, according to \ref{izo2ftIdent}, the identity. We can represent it as the composite $\mathcal{I} = \mathcal{S}_p \circ +\mathcal{S}_p$ (\ref{izoZrcPrInvol}) for an arbitrary line +$p$. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.10.1.pic} +\caption{} \label{sl.izo.6.10.1.pic} +\end{figure} + + \textit{2)} Let's assume that the isometry +$\mathcal{I}$ has exactly one fixed point $S$. Let $p$ +be an arbitrary line passing + through the point $S$. The composite $\mathcal{S}_p\circ \mathcal{I}$ + is an indirect isometry with a fixed +point $S$, therefore, according to \ref{izo1ftIndZrc}, it represents a central +reflection - for example $\mathcal{S}_q$, where the axis $q$ passes through the point +$S$. So we have $\mathcal{S}_p\circ \mathcal{I} = \mathcal{S}_q$ +or $\mathcal{I} +=\mathcal{S}_p^{-1}\circ\mathcal{S}_q=\mathcal{S}_p\circ\mathcal{S}_q$ +(\ref{izoZrcPrInvol}). The lines $p$ and $q$ intersect in the point +$S$ (from $p=q$ we get $\mathcal{I}=\mathcal{S}_p \circ +\mathcal{S}_p=\mathcal{E}$), therefore, according to \ref{rotacKom2Zrc}, +$\mathcal{I}$ represents a rotation with the center in the point $S$. + +\textit{3)} Let $\mathcal{I}$ be an isometry without fixed points. +Then for any point $A$ of this plane $\mathcal{I}(A) = A'\neq +A$. Let $p$ be the perpendicular bisector of the line segment $AA'$ and $S_p$ the reflection over the line $p$. In this case $\mathcal{S}_p\circ \mathcal{I}(A)= +\mathcal{S}_p(A')=A$. The composition $\mathcal{S} \circ +\mathcal{I}_p$ is therefore an indirect isometry with a fixed point $A$, so +according to theorem \ref{izo1ftIndZrc} it represents the reflection over some line $q$ - $\mathcal{S}_q$, where the axis of $q$ goes through the point $A$. +As in the previous example, +$\mathcal{I}=\mathcal{S}_p\circ\mathcal{S}_q$. The lines $p$ and $q$ +do not intersect and are not equal, because otherwise from the already proven +$\mathcal{I}$ it would represent the identity or a rotation and would have +at least one fixed point. So the only possibility left is that the lines +$p$ and $q$ are parallel. In this case $\mathcal{I}$ is a translation +(theorem \ref{translKom2Zrc}). + \kdokaz + +We are left with indirect isometries. The only ones mentioned so far +are the basic reflection and the glide reflection. +Are they also the only ones? We will answer this + in the next + theorem. + + + + \bizrek \label{Chaslesov-} + Any opposite isometry is either a reflection + either it can be represented as the product of three reflections. + The only opposite isometries are + reflections and + glide reflections. + \eizrek + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.10.2.pic}) + +Let $\mathcal{I} : \mathbb{E}^2\rightarrow \mathbb{E}^2$ +be an indirect isometry of the plane. We will again carry out the proof +depending on the number of fixed points. + +\textit{1)} If the isometry $\mathcal{I}$ has at least one fixed +point, $\mathcal{I}$ according to theorem \ref{izo1ftIndZrc} is the reflection over a line. + +\textit{2)} Let's assume that $\mathcal{I}$ is an isometry without any fixed points. + In this case, for any point $A$ on this plane, we have +$\mathcal{I}(A) = A'\neq A$. Let $p$ be the line of symmetry of the line segment $AA'$ and $\mathcal{S}_p$ be the reflection over the line $p$. Then +$\mathcal{S}_p\circ \mathcal{I}(A)= \mathcal{S}_p(A')=A$. Therefore, the composition $\mathcal{S}_p \circ \mathcal{I}$ is a direct isometry with a +fixed point at $A$, which, according to the previous statement, represents a rotation or the identity (a translation has no fixed points). The other case is not possible, +because otherwise $\mathcal{I} = \mathcal{S}_p$ and the isometry $\mathcal{I}$ would have fixed points, which contradicts +the initial assumption. So the composition $\mathcal{S} \circ +\mathcal{I}_p$ is a rotation with center at $A$, which we can represent +as the composition of two reflections over the lines $q$ and $r$, which intersect +at the point $A$. Therefore, $\mathcal{S}_p\circ \mathcal{I}=\mathcal{S}_q\circ +\mathcal{S}_r$ or $\mathcal{I}=\mathcal{S}_p\circ \mathcal{S}_q\circ +\mathcal{S}_r$. Since $A'\neq A$ and $p$ is the line of symmetry of the line segment $AA'$, +the point $A$ does not lie on the line $p$. This means that the lines $p$, $q$ +and $r$ are not in the same plane. Therefore, according to \ref{izoZrcdrsprq}, +$\mathcal{I}=\mathcal{S}_p\circ \mathcal{S}_q\circ \mathcal{S}_r$ +is a glide reflection. + \kdokaz + + Because of the previous two statements, + we can say + that these isometries are the only type of isometries. +We can also say that we can represent any isometry of the plane as the composition of +the basic isometries, where we can choose the axes so that +there are no more than three in the composition. From the proof of the aforementioned statements, it is clear that we can determine any isometry only by the number +of fixed points and whether the isometry is direct or indirect. +We will formulate these facts in the next two statements. + + + + \bizrek \label{IzoKompZrc} + Any isometry of the plane can be expressed as the product of + one, two or three reflections. + \eizrek + +\bizrek \label{Chaslesov} \index{izrek!Chaslesov} + (Chasles’\footnote{\index{Chasles, M.} + \textit{M. Chasles} (1793--1880), + French geometer, who derived this classification in 1831.}) + The only isometries of the plane + $\mathcal{I} : E^2 \rightarrow E^2$ are: + identity map, reflections, rotations, + translations and glide reflections. Specifically:\\ + \hspace*{3mm}(i) if $\mathcal{I}$ is + a direct isometry and has at least two fixed points, then $\mathcal{I}$ + is the identity map,\\ + \hspace*{3mm}(ii) if $\mathcal{I}$ is a direct isometry and has exactly one fixed point, + then $\mathcal{I}$ is a rotation (or specially a half-turn),\\ + \hspace*{3mm}(iii) if $\mathcal{I}$ is + a direct isometry and has no fixed points, + then $\mathcal{I}$ is a translation,\\ + \hspace*{3mm}(iv) if $\mathcal{I}$ is + an opposite isometry and has at least one fixed point, + then $\mathcal{I}$ is a reflection,\\ + \hspace*{3mm}(v) + if $\mathcal{I}$ is + an opposite isometry and has no fixed points, + then $\mathcal{I}$ is a glide reflection. + \eizrek + + All that we have said in this section about the classification of isometries is +illustrated in the following table (Figure \ref{IzoKlas.eps}): + +\vspace*{-2mm} + +%\begin{figure}[!htb] +%\centering +%\input{sl.izo.6.10.3.pic} +%\caption{} \label{sl.izo.6.10.3.pic} +%\end{figure} + +\begin{figure}[!htb] +\centering + \includegraphics[width=0.85\textwidth]{IzoKlas.eps} +\caption{} \label{IzoKlas.eps} +\end{figure} + +%________________________________________________________________________________ + \poglavje{Hjelmslev's Theorem} \label{odd6Hjelmslev} + +The following theorem, which refers to opposite isometries, is very useful. + +(Hjelmslev's\footnote{ +\index{Hjelmslev, J. T.} +\textit{J. T. Hjelmslev} (1873--1950), Danish +mathematician.}) \index{theorem!Hjelmslev's} +The midpoints of all line segments defined by corresponding pairs +of points of an arbitrary indirect isometry lie on the same line. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.11.1.pic} +\caption{} \label{sl.izo.6.11.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.11.1.pic}) + +Let $\mathcal{I}$ be an indirect isometry and $X'=\mathcal{I}(X)$ for +an arbitrary point $X$ and $X_s$ the center of the line segment $XX'$. From Chasles' +theorem \ref{Chaslesov} it follows that the only types of indirect isometries of a plane +are reflection and glide reflection. We prove that in both cases +the sought line is exactly the line of reflection or glide reflection. In +the first case, where $\mathcal{I}=\mathcal{S}_s$, it is trivial that $X_s\in s$. Let $\mathcal{I}=\mathcal{G}_{2\overrightarrow{PQ}}= +\mathcal{S}_s\circ \mathcal{T}_{2\overrightarrow{PQ}}$, where $s=PQ$. If we denote by $X_1=\mathcal{T}_{2\overrightarrow{PQ}}(X)$ and +$X_2$ the center of the line segment $X_1X'$, then the line segment $X_sX_2$ is the median of the triangle $XX_1X'$ with the base $XX_1$. Therefore, (from theorem \ref{srednjicaTrik}) $X_sX_2\parallel XX_1\parallel s$ or $X_s\in +s$ (Playfair's axiom \ref{Playfair}). + \kdokaz + +The proven theorem can be used if we have congruent figures +or if we find an indirect isometry that maps one set +of points to another. We will illustrate this with the following examples. + + \bzgled + Let $ABC$ and $A'B'C'$ be congruent triangles with the opposite orientation. + Prove that the midpoints of the line segments + $AA$', $BB'$ and $CC'$ lie on the same line. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.11.2.pic} +\caption{} \label{sl.izo.6.11.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.11.2.pic}) +Since $ABC$ and $A'B'C'$ are congruent triangles with different orientations, there exists an indirect isometry $\mathcal{I}$, which maps triangle $ABC$ to triangle $A'B'C'$. In this case, $A$ and $A'$, $B$ and $B'$, and $C$ and $C'$ are pairs of this isometry, so by \ref{Chasles-Hjelmsleva} the midpoints of the line segments $AA'$, $BB'$, and $CC'$ lie on the same line. + \kdokaz + + \bzgled + Let $A$ be a point and $p$ a line in the plane. Suppose that points $X_i$ + lie on the line $p$ and $AX_iY_i$ are regular triangles with the same orientation. + Prove that the midpoints of the line segments + $X_iY_i$ lie on the same line. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.11.3.pic} +\caption{} \label{sl.izo.6.11.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.11.3.pic}) +The isometry $\mathcal{R}_{A,60^0}\circ\mathcal{S}_p$ is an indirect isometry, which maps points $X_i$ to points $Y_i$, + so by \ref{Chasles-Hjelmsleva} the midpoints + of the line segments +$X_iY_i$ lie +on the same line. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Isometry Groups. Symmetries of Figures} \label{odd6Grupe} + +In section \ref{odd2AKSSKL} we found that the set + $\mathfrak{I}$ of all isometries +of a plane together with the operation of the composition of +mappings represents the so-called +structure \index{group}\pojem{group}\footnote{The theory of groups +was discovered by the brilliant young French mathematician \index{Galois, E.} +\textit{E. Galois} (1811--1832).}. + This means that the following properties are fulfilled: +\begin{enumerate} + \item $(\forall f\in \mathfrak{I})(\forall g\in \mathfrak{I}) + \hspace*{1mm}f\circ g\in \mathfrak{I}$, + \item $(\forall f\in \mathfrak{I})(\forall g\in \mathfrak{I}) + (\forall h\in \mathfrak{I}) + \hspace*{1mm}(f\circ g)\circ h=f\circ (g\circ h)$, + \item $(\exists e\in \mathfrak{I})(\forall f\in \mathfrak{I}) + \hspace*{1mm}f\circ e=e\circ f=f$, + \item $(\forall f\in \mathfrak{I})(\exists g\in \mathfrak{I}) + \hspace*{1mm}f\circ g=g\circ f=e$. +\end{enumerate} + + Property (2) is valid in general for +the composition of mappings. Properties (1), (3) and (4) were introduced with +the axiom \ref{aksIII4}. Property (1) means that the composition of two +isometries is again an isometry, (3) and (4) refer to the identity +and the inverse isometry. The group mentioned, which is determined by the set +$\mathfrak{I}$ of all isometries of a plane with respect to the operation of +the composition of mappings, is called the \index{group!isometries} +\pojem{group of all isometries of the plane}. We will also denote it by $\mathfrak{I}$. + +There are also other groups of isometries, which we obtain +if we take an appropriate subset of all isometries of the plane. Property (1) +tells us that this set cannot be arbitrary. For example, the set of all +rotations of the plane is not a group, because the composition of two +rotations is not always a rotation (it can also be a translation)." + +From the properties of translation it follows that the set of all translations +together with the identity $\mathcal{E}$ forms a group - i.e. +\index{group!translation}\pojem{translation group} with the designation +$\mathfrak{T}$. We call it a \index{subgroup} +\pojem{subgroup} of the group $\mathfrak{I}$ of all isometries of the plane (this +fact is denoted by $\mathfrak{T}<\mathfrak{I}$). In fact, every +group of isometries of the plane is a subgroup of the group $\mathfrak{I}$ of all isometries of this plane. +However, the group of translations is not the only subgroup of the group +$\mathfrak{I}$. All direct isometries of the plane in fact form one +such subgroup; we denote it by $\mathfrak{I}^+$. This means that +$\mathfrak{I^+}<\mathfrak{I}$. Since translations are direct +isometries, $\mathfrak{T}<\mathfrak{I^+}$ is also true. It is clear that +the set of all indirect isometries does not determine a group, since the composition +of two indirect isometries is a direct isometry (even the identity +is not an indirect isometry). + +There are also finite subgroups of the group $\mathfrak{I}$ (which have +a finite number of isometries). An example of such a subgroup is the so-called +Klein group +$\mathfrak{K}$ ($\mathfrak{K}<\mathfrak{I}$) (or \textit{Klein's quadrilateral}), which is determined by +the set of isometries $\{\mathcal{E}, \mathcal{S}_p, \mathcal{S}_q, +\mathcal{S}_O\}$, where $p$ and $q$ are rectangles that intersect at +the point $O$. This group can also be represented by the following table: + +\vspace*{5mm} + +\hspace*{22mm}\begin{tabular}{|c||c|c|c|c|} \hline + % after \\ : \hline or \cline{col1-col2} \cline{col3-col4} ... + $\circ$ & $\mathcal{E}$ & $\mathcal{S}_p$ & $\mathcal{S}_q$& $\mathcal{S}_O$ + \\ \hline \hline + $\mathcal{E}$ & $\mathcal{E}$& $\mathcal{S}_p$& $\mathcal{S}_q$& $\mathcal{S}_O$ + \\ \hline + $\mathcal{S}_p$ & $\mathcal{S}_p$ & $\mathcal{E}$ & $\mathcal{S}_O$& $\mathcal{S}_q$ + \\ \hline + $\mathcal{S}_q$ & $\mathcal{S}_q$ & $\mathcal{S}_O$ & $\mathcal{E}$& $\mathcal{S}_p$ + \\ \hline + $\mathcal{S}_O$ & $\mathcal{S}_O$ & $\mathcal{S}_q$ &$\mathcal{S}_p$ & $\mathcal{E}$ + \\ \hline +\end{tabular} + + +\vspace*{5mm} + +Since Klein's group $\mathfrak{K}$ contains the glide reflection, which is an indirect isometry, $\mathfrak{K}$ is not a subgroup of the group $\mathfrak{I}^+$. + + We will call the group of isometries consisting only of + the identity $\{\mathcal{E}\}$ the \index{group!trivial} + \pojem{trivial group} with the notation $\mathfrak{E}$. This group is obviously a subgroup + of every group + of isometries; e.g. $\mathfrak{E}<\mathfrak{T}<\mathfrak{I^+}<\mathfrak{I}$ or + $\mathfrak{E}<\mathfrak{K}<\mathfrak{I}$. + +If we take only the identity $\{\mathcal{E}$ and +one glide reflection $\mathcal{S}_p\}$ we get another finite group of isometries. +We get the same structure if instead of the glide reflection over the line we take +the glide reflection over the point, i.e. $\{\mathcal{E}$ and $\mathcal{S}_O\}$. Although the +sets are different, the structure of the group is the same, as we illustrate with tables. +We say that in this case the groups +\index{group!isomorphic}\pojem{isomorphic}. + +\vspace*{5mm} + +\hspace*{12mm}\begin{tabular}{|c||c|c|} \hline + % after \\ : \hline or \cline{col1-col2} \cline{col3-col4} ... + $\circ$ & $\mathcal{E}$ & $\mathcal{S}_p$ + \\ \hline \hline + $\mathcal{E}$ & $\mathcal{E}$& $\mathcal{S}_p$ + \\ \hline + $\mathcal{S}_p$ & $\mathcal{S}_p$ & $\mathcal{E}$ + \\ \hline +\end{tabular} +\hspace*{22mm} +\begin{tabular}{|c||c|c|} \hline + % after \\ : \hline or \cline{col1-col2} \cline{col3-col4} ... + $\circ$ & $\mathcal{E}$ & $\mathcal{S}_O$ + \\ \hline \hline + $\mathcal{E}$ & $\mathcal{E}$& $\mathcal{S}_O$ + \\ \hline + $\mathcal{S}_O$ & $\mathcal{S}_O$ & $\mathcal{E}$ + \\ \hline +\end{tabular} + +\vspace*{5mm} + +We have already shown that every isometry of the plane can be represented +as a composite of a finite number of reflections over a line (we can always +choose at most three reflections) - the statement +\ref{IzoKompZrc}. We say that reflections over a line +\pojem{generate} the group $\mathfrak{I}$ of all isometries of that plane or that they are \pojem{generators} of the group $\mathcal{I}$. + We recall +that the composite of an even number of reflections over a line represents a direct isometry, +while in the case of an indirect isometry, the number of reflections over a line is odd. + + There is another important type of groups of isometries that we + will present in the following statement. + + + + + \bizrek + Let $\phi$ be a figure in the plane. The set of all isometries + of that plane that map the figure $\phi$ to itself forms a group. + \eizrek + +\textbf{\textit{Proof.}} + Let $\mathfrak{G}$ be the set of all isometries + of the plane that map the figure $\phi$ to itself. We will prove + that this set determines a group. For any isometries + $f,g\in \mathfrak{G}$ we have $f(\phi)=\phi$ and +$g(\phi)=\phi$. But in this case we also have $g \circ f (\phi) = +\phi$ and $f^{-1}(\phi) = \phi$. So conditions +(1) and (4) are satisfied. Regarding condition (2), we have +already said that the operation of composition of mappings is always satisfied. +Condition (3) is also satisfied in our case, since +we have $\mathcal{E}(\phi)=\phi$. + \kdokaz + +The group from the previous statement - the set of all isometries +of the plane that map the figure $\phi$ onto itself - is called the +\index{group!symmetry}\emph{symmetry group} of the figure $\phi$ and is +denoted by $\mathfrak{G}(\phi)$. It is clear that for every figure $\phi$ +it holds that $\mathfrak{G}(\phi)<\mathfrak{I}$. We mention that in the +proof that a subset of a group (in our case the group $\mathfrak{I}$) +represents a group or its subgroup, it is enough to check only the +conditions (\textit{i}) and (\textit{iv}). The condition (\textit{ii}) +is always satisfied, while the condition (\textit{iii}) directly +follows from the conditions (\textit{i}) and (\textit{iv}). + +There is another group that is a subgroup of the symmetry group +$\mathfrak{G}(\phi)$. This group consists of the set of all direct +isometries from $\mathfrak{G}(\phi)$. We call it the +\index{group!rotation}\emph{rotation group} of the figure $\phi$ and +denote it by $\mathfrak{G}^+(\phi)$. It is clear that for every figure +its rotation group is a subgroup of its symmetry group, i.e. +$\mathfrak{G}^+(\phi)<\mathfrak{G}(\phi)$. + +In the next example we will determine the symmetry groups of +different figures. + +\bzgled \label{grupeSimPrimeri} +Determine the symmetry group and the rotation group of: +(i) a square, (ii) a rectangle, (iii) a trapezium, +(iv) a line, (v) a ray, (vi) a circle. +\ezgled + +\textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.12.1.pic}) + +(\textit{i}) A square with the center $O$ has four sides that intersect +in the point $O$ and four rotations (including the identity), therefore: + +$\mathfrak{G}(\phi)=\{\mathcal{S}_p, \mathcal{S}_q, +\mathcal{S}_r, \mathcal{S}_s, \mathcal{E}, \mathcal{R}_{O,90^0} , +\mathcal{R}_{O,180^0} , \mathcal{R}_{O,270^0} \}$ and + +$\mathfrak{G}^+(\phi)=\{\mathcal{E}, \mathcal{R}_{O,90^0} , +\mathcal{R}_{O,180^0} , \mathcal{R}_{O,270^0} \}$. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.1.pic} +\caption{} \label{sl.izo.6.12.1.pic} +\end{figure} + +(\textit{ii}) A rectangle with center $O$ has only two diagonals, +so $\mathfrak{G}(\phi)=\{\mathcal{S}_p, \mathcal{S}_q, +\mathcal{E}, \mathcal{S}_O \}$ and +$\mathfrak{G}^+(\phi)=\{\mathcal{E}, \mathcal{S}_O \}$. It is +interesting to note that the first group of rectangles +$\mathfrak{G}(\phi)$ is actually Klein's group $\mathfrak{K}$. + + (\textit{iii}) If $\phi$ is an arbitrary trapezoid, the identity + is the only one that maps it onto itself. This means that + $\mathfrak{G}(\phi)=\mathfrak{G}^+(\phi) + =\{\mathcal{E}\}$ is a group that we have already called the trivial group. + +(\textit{iv}) The groups of symmetry and rotation of the line $p$ +are infinite groups and specifically: + \begin{eqnarray*} + \mathfrak{G}(p)&=&\{\mathcal{S}_l;l\perp p\}\cup + \{\mathcal{S}_p\}\cup + \{\mathcal{S}_O;O\in p\}\cup + \{ \mathcal{T}_{\overrightarrow{v}};\overrightarrow{v} \parallel p\}\cup + \{\mathcal{E}\}\\ + \mathfrak{G}^+(p)&=& + \{\mathcal{S}_O;O\in p\}\cup + \{ \mathcal{T}_{\overrightarrow{v}};\overrightarrow{v} \parallel p\}\cup + \{\mathcal{E}\} + \end{eqnarray*} + +(\textit{v}) For the line segment $h=OA$ we have +$\mathfrak{G}(h)=\{\mathcal{E}, \mathcal{S}_{h}\}$ and +$\mathfrak{G}^+(h)=\{\mathcal{E}\}$. + +(\textit{vi}) For the circle $k(O,r)$ all reflections through the +point $O$ and all rotations with center $O$ (including the identity) +belong to the group of symmetry. The group of rotations is +composed only of the aforementioned rotations. So: + \begin{eqnarray*} + \mathfrak{G}(k)&=&\{\mathcal{S}_l;l\ni O\}\cup + \{\mathcal{R}_{O,\alpha};\alpha \textrm{ any angle}\}\cup + \{\mathcal{E}\}\\ + \mathfrak{G}^+(k)&=&\{\mathcal{R}_{O,\alpha};\alpha \textrm{ any angle}\}\cup + \{\mathcal{E}\}, + \end{eqnarray*} + which was to be proven. \kdokaz + +We have seen that a rectangle has exactly two diagonals and is also +centrally symmetric. Now we will prove a general statement. + +\bzgled + If a figure in the plane has exactly two axes of symmetry, + then it also has a centre of symmetry. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.2.pic} +\caption{} \label{sl.izo.6.12.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $p$ and $q$ be the only axes of symmetry of the figure $\phi$. We shall prove that $p\perp q$ (Figure \ref{sl.izo.6.12.2.pic}). +Since $\mathcal{S}_p(\phi)=\mathcal{S}_q(\phi)=\phi$, by the +transmutability theorem \ref{izoTransmutacija} it is also +$$\mathcal{S}_{\mathcal{S}_p(q)}(\phi)= +\mathcal{S}_p\circ\mathcal{S}_q\circ\mathcal{S}_p^{-1}(\phi)=\phi.$$ +The lines $p$ and $q$ are the only axes of symmetry of the figure $\phi$, therefore either $\mathcal{S}_p(q)=p$ or $\mathcal{S}_p(q)=q$. In both cases +it follows that $p=q$ or $p\perp q$. However, the first possibility is excluded, since $p$ and $q$ are different by assumption. Therefore the lines $p$ and $q$ are perpendicular and intersect in a point $O$, so: +$$\mathcal{S}_O(\phi)= +\mathcal{S}_q\circ\mathcal{S}_p(\phi)=\mathcal{S}_q(\mathcal{S}_p(\phi))= +\phi,$$ which means that the figure is centrally symmetric. + \kdokaz + +In group terms we can write the previous theorem in the following way. + + + + \bzgled If the symmetry group of some figure contains exactly two reflections + then it also contains a half-turn. + \ezgled + +First of all, we notice that different figures can have the same symmetry group. For example, the symmetry group of a rectangle and a line is the Klein group $\mathfrak{K}$. Also, an isosceles triangle and an isosceles trapezoid have the same symmetry group $\{\mathcal{E}, \mathcal{S}_{p}\}$. + +Also, the first three and the fifth symmetry groups from the example +\ref{grupeSimPrimeri} are finite groups, while the fourth and the sixth are infinite. From this example we can also see that the infinity of the symmetry group of a figure is not related to the boundedness of this figure. The symmetry group of a bounded figure can be finite, but it can also be infinite. The same goes for unbounded figures. For further use, we shall define the concept of boundedness more precisely. + +The figure $\phi$ is \index{figure!bounded} \pojem{bounded}, if there exists such a +distance $AB$, that for any points $X$ and $Y$ of this figure it holds +$XY|XX_n|=|n\overrightarrow{v}|$. +This should hold for every $n\in \mathbb{N}$, which of course is not +the case. The last relation therefore leads us to a contradiction, which +means that in the group $\mathfrak{G}(\phi)$ there are no translations. + \kdokaz + + From the previous izrek it follows, that the group of symmetry + of a bounded figure also does not contain a glide reflection, since +its square is a translation (example \ref{izoZrcdrsZrcdrs}). But how is it with reflections and rotations? + + + + \bizrek \label{GrupaSomer1} If a bounded figure has more axes of symmetry, + then they all intersect at one point. + \eizrek + +\textbf{\textit{Proof.}} Let $p$ and $q$ be the axes of symmetry of a +bounded figure $\phi$. The lines $p$ and $q$ are not parallel, because otherwise the composition of the appropriate reflections $\mathcal{S}_p \circ \mathcal{S}_q$ +would be a translation. So $p$ and $q$ intersect in some point $S$. If +$r$ is a third arbitrary axe, it contains the point $S$, because otherwise (from \ref{izoZrcdrsprq}) the composition $\mathcal{S}_p \circ +\mathcal{S}_q\circ \mathcal{S}_r$ would be a mirror glide. + \kdokaz + + + + \bizrek \label{GrupaSomerRot} If a bounded figure has at least one axe of symmetry + and at least one center of rotation, then + this center lies on the axe of symmetry. + \eizrek + + \textbf{\textit{Proof.}} + We assume the opposite - let $p$ be the axe of symmetry of a bounded figure $\phi$ and + $S$ + the center of rotation +$\mathcal{R}_{S,\alpha}$, which maps this figure into itself, but +$S\notin p$. The rotation $\mathcal{R}_{S,\alpha}$ can +be represented as the composition of two reflections over the lines with the points, which +go through the point $S$. So $\mathcal{S}_p \circ +\mathcal{R}_{S,\alpha} = \mathcal{S}_p \circ \mathcal{S}_q\circ +\mathcal{S}_r$. The lines $p$, $q$ and $r$ are not in the same bundle, so +(from \ref{izoZrcdrsprq}) the composition $\mathcal{S}_p +\circ \mathcal{R}_{S,\alpha}$ is a mirror glide, which cannot +be in the group $\mathfrak{G}(\phi)$, which means that the center $S$ lies +on the axe $p$. + \kdokaz + + + + \bizrek \label{GrupaRot} All rotations of a bounded figure + have the same center. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.4.pic} +\caption{} \label{sl.izo.6.12.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.12.4.pic}). + Let's assume the opposite - that $\mathcal{R}_{O,\alpha}$ and + $\mathcal{R}_{S,\beta}$ +are rotations that map the bounded figure $\phi$ into itself and $O\neq +S$. Then $O'=\mathcal{R}_{S,\beta}(O)\neq S$. By \ref{izoTransmRotac} +$\mathcal{R}_{O',\alpha}=\mathcal{R}_{S,\beta}\circ +\mathcal{R}_{O,\alpha}\circ\mathcal{R}_{S,\beta}^{-1}$, so + $\mathcal{R}_{O',\alpha}\in \mathfrak{G}(\phi)$ and then also + $\mathcal{I}=\mathcal{R}_{O',\alpha}^{-1}\circ + \mathcal{R}_{O,\alpha}\in \mathfrak{G}(\phi)$. + Let $p$ be a line that with the line $OO'$ in the point +$O$ determines the angle $\frac{1}{2}\alpha$, and $q$ its parallel through the point $O'$. By \ref{rotacKom2Zrc} $\mathcal{I} = +\mathcal{S}_q \circ \mathcal{S}_{OO'} \circ \mathcal{S}_{OO'} +\circ \mathcal{S}_p = \mathcal{S}_q \circ \mathcal{S}_p$, so +$\mathcal{I}$ is a translation, which is not possible. This means that +$O\neq S$ is not true, so all rotations of a bounded figure +have the same centre. + \kdokaz + + From the previous equations we get the following statement. + + + + + \bizrek \label{GrupaOmejenLik} If symmetry group of a bounded figure $\phi$ is not trivial, + then there is a point $S$ such that the only possible isometries in this group are: + \\ + (i) the identity map,\\ + (ii) rotations with the centre $S$,\\ + (iii) reflections with axes, + which all contains the point $S$. + \eizrek + + In this way we have determined all possible groups of bounded figures. + These groups are not necessarily +finite. What are then the finite groups? Let's explore +some examples of symmetry groups. + + + \bzgled \label{GrupaDiederska} Determine the symmetry group and the rotation group + of a regular $n$-gon. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.5.pic} +\caption{} \label{sl.izo.6.12.5.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.12.5.pic}) + No matter if $n$ is an even or odd number, a regular $n$-gon has + exactly $n$ +diagonals (see section \ref{odd3PravilniVeck}). The basic angle +of rotation that transforms a $n$-gon into itself is $\theta = \frac{360^0}{n}$. +Therefore, if we denote the group of symmetries and the group of rotations + of a regular $n$-gon with $\mathcal{D}_n$ and $\mathcal{C}_n$, and its center with $O$, we get: + \begin{eqnarray*} + \mathfrak{D}_n&=&\{ \mathcal{S}_{p_1}, \mathcal{S}_{p_2},\ldots, + \mathcal{S}_{p_n}, + \mathcal{E}, \mathcal{R}_{O,\theta}, \mathcal{R}_{O,2\theta},\ldots + \mathcal{R}_{O,(n-1)\theta}\}\\ + \mathfrak{C}_n&=&\{ \mathcal{E}, \mathcal{R}_{O,\theta}, + \mathcal{R}_{O,2\theta},\ldots \mathcal{R}_{O,(n-1)\theta}\}, + \end{eqnarray*} + which is what needed to be determined. \kdokaz + + The group of symmetries $\mathfrak{D}_n$ of a regular $n$-gon from + the previous example + is called + \index{group!diedral} \pojem{diedral group}, the group of rotations + of a regular $n$-gon $\mathfrak{C}_n$ is + \index{group!cyclic} \pojem{cyclic group}. It is clear that + $\mathfrak{C}_n<\mathfrak{D}_n$. + + Groups $\mathfrak{D}_n$ and $\mathfrak{C}_n$ ($n\geq 3$) + can also be generalized for the cases $n < 3$, even though in this case we are no longer + talking about an $n$-gon. $\mathfrak{D}_2$ is thus +actually Klein's group $\mathfrak{K}$ (the group of symmetries of a line +or some sort of a "2-gon"). The group $\mathfrak{C}_2$ +consists of the identity and one central symmetry. This group can be +represented as the group of rotations of a line. The group $\mathfrak{D}_1$ +contains the identity and one reflection, but it is isomorphic to the group +$\mathfrak{C}_2$ (an example we have already mentioned at the beginning of this +section). The group $\mathfrak{C}_1$ is the trivial group $\mathfrak{E}$. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.5a.pic} +\caption{} \label{sl.izo.6.12.5a.pic} +\end{figure} + +We mention that the groups $\mathfrak{D}_n$ and $\mathfrak{C}_n$ ($n\in +\mathbb{N}$) are not symmetry groups only for regular $n$-sided polygons (Figure +\ref{sl.izo.6.12.5a.pic}). They can also be symmetry groups of unlimited +shapes. For example, Klein's group $\mathfrak{D}_2$ is also a symmetry group +of the hyperbola with equation $x\cdot y= 1$ in the Cartesian coordinate system +$O_{xy}$. The asymptotes are then determined by the equations $y = x$ and $y = +-x$, and the center of symmetry is the origin $O$. $\mathfrak{D}_2$ is also a symmetry group of the ellipse, given by +the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. + +It is clear that all groups $\mathfrak{D}_n$ and $\mathfrak{C}_n$ +($n\in \mathbb{N}$) are finite groups of isometries. We will show that these are the only finite groups of isometries! + + + + \bizrek \label{GrupaLeonardo} \index{izrek!Leonarda da Vincija} + (Leonardo da Vinci\footnote{Famous Italian painter, architect and + inventor \index{Leonardo da Vinci} + \textit{Leonardo da Vinci} (1452--1519) studied all possible + symmetries of the central building and possible arrangements of + chapels around + it, which preserve the basic symmetry.}) The only finite groups of isometries are $\mathfrak{D}_n$ and $\mathfrak{C}_n$. + \eizrek + +\textbf{\textit{Proof.}} Let $\mathfrak{G}$ be a finite group of +isometries. If it contained a translation +$\mathcal{T}_{\overrightarrow{v}}$, it would have an infinite +subgroup $\{ \mathcal{E}, \mathcal{T}_{\overrightarrow{v}}, +\mathcal{T}_{\overrightarrow{v}}^2, +\mathcal{T}_{\overrightarrow{v}}^3,\ldots \}$, which is not possible. +Therefore, $\mathfrak{G}$ is a group without translations. In a +similar way as before (statement \ref{GrupaOmejenLik}), we can +prove that the only possible isometries in this group (except for +the identical mapping) are rotations with the same center $S$ and +reflections over a line with eight, which go through the point $S$. +If the group has no rotations, at most one basic reflection is +possible (already two would generate a rotation). In this case, +the only possible groups are $\mathfrak{D}_1$ and $\mathfrak{C}_1$. + +Without loss of generality, we assume that all angles of rotation +are positive. Because the group $\mathfrak{G}$ is finite, there +exists a rotation $\mathcal{R}_{O,\theta}$ with the smallest +angle $\theta$. For the same reason, there exists a natural +number $n$, for which $\mathcal{R}_{O,\theta}^n = \mathcal{E}$ +(otherwise, the finite group $\mathfrak{G}$ would have an infinite +subgroup $\{ \mathcal{E}, \mathcal{R}_{O,\theta}, +\mathcal{R}_{O,\theta}^2, \mathcal{R}_{O,\theta}^3,\ldots \}$). +Therefore, $\theta = \frac{360^0}{n}$. + +Let $\mathcal{R}_{O,\delta}$ be an arbitrary rotation from +$\mathfrak{G}$. We prove that the angle $\delta$ is a multiple of +the angle $\theta$ or $\delta=k\cdot \theta$ for some $k\in +\mathbb{N}$. We assume the opposite, that such $k$ does not exist. +Because $\delta>\theta$, for some $l\in \mathbb{N}$ we have +$l\theta<\delta<(l+1)\theta$. If we rearrange, we get +$0<\delta-l\theta<\theta$. But +$\mathcal{R}_{O,\delta-l\theta}=\mathcal{R}_{O,\delta}\circ +\left(\mathcal{R}_{O,\theta}^{-1}\right)^l\in \mathfrak{G}$, which +is contradictory to how we defined the angle $\theta$. Therefore, +all rotations from $\mathfrak{G}$ are of the form +$\mathcal{R}_{O,\theta}^k$, $k\in \{ 1,2,\ldots, n\}$. + +If there are no basic reflections in the group, then $\mathfrak{G}$ is a cyclic +group or $\mathfrak{G}=\mathfrak{C}_n$ for some $n\in \mathbb{N}$. If there are +also reflections over a line in the group, their composites +(of each two reflections) are rotations, so each pair of axes determines an angle, which is +half of some angle of rotation. In this case, +$\mathfrak{G}$ is a dihedral group or $\mathfrak{G}=\mathfrak{D}_n$ for some $n\in \mathbb{N}$. + \kdokaz + + We emphasize once again that the final group of symmetries and the group of symmetries + of a limited figure are not the same concept. + The group of symmetries of limited figures is not necessarily final + (for example, the group of symmetries of a circle). Similarly, unlimited +figures can have a finite group of symmetries (for example, an equilateral triangle) and +unlimited figures with an infinite group of symmetries (for example, a line). We will illustrate these facts with the following diagram (Figure +\ref{sl.izo.6.12.6.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.6.pic} +\caption{} \label{sl.izo.6.12.6.pic} +\end{figure} + + + We will also mention another type of infinite groups of isometries that preserve + certain tessellations of the plane (see section + \ref{odd3Tlakovanja}). So it's not just about the groups of symmetries + of these tessellations, but also about all subgroups of these groups. + +For example, if we choose + tiling + a plane with an arbitrary parallelogram $ABCD$ (Figure + \ref{sl.izo.6.12.7.pic}), the + infinite group of symmetries of this tiling is generated by + reflections in the vertex $A$ and the centers of the sides $AB$ and $AD$. In this + group are then also reflections in the vertices, centers of sides and + intersections + of diagonals of all + parallelograms of this tiling and all translations that are composites + of two of + these reflections. + But one of its subgroups, which also preserves the same tiling, is the group + of all mentioned translations. It is generated by + two + translations $\mathcal{T}_{\overrightarrow{AB}}$ + and $\mathcal{T}_{\overrightarrow{AD}}$. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.7aa.pic} +\input{sl.izo.6.12.7bb.pic} +\caption{} \label{sl.izo.6.12.7.pic} +\end{figure} + + The group of symmetries in + the case of tiling with a regular triangle $ABC$ (Figure + \ref{sl.izo.6.12.7.pic}) + or tiling $(3,6)$ contains + rotations in the vertices of the grid at angles $k\cdot 60^0$, + reflections over lines, which are + determined by the sides and heights of these triangles, and translations, + generated by translations $\mathcal{T}_{\overrightarrow{AB}}$ + and $\mathcal{T}_{\overrightarrow{AC}}$. This group has several + different subgroups, all preserving tiling $(3,6)$. + + All such groups that preserve certain tilings are called + \index{group!discrete} \pojem{discrete groups of isometries}. + We formally define them as follows: A group of isometries $\mathfrak{G}$ is a discrete group of isometries, + if there exists such + $\varepsilon>0$, that the lengths of vectors of all translations and + measures of angles of all rotations from this group are greater than $\varepsilon$. + +There are two types of discrete groups of isometries: \index{group!frieze} + \pojem{frieze + groups}\footnote{The term \textit{frieze}, which + was used by the Ancient Greeks, + meant a repeating border pattern.} and \index{group!wallpaper}\pojem{wallpaper groups}. In frieze + groups, the subgroup + of all translations is generated by one single translation, in wallpaper + groups by two translations, determined by two non-linear + vectors. It has been proven that there are exactly 7 frieze groups and 17 + wallpaper groups\footnote{All 17 groups or the corresponding types of ornamentation were + known to the Egyptians, + and were often used by Muslim artists. In the Alhambra palace (Spain), the Moors + painted all 17 types of ornamentation in the 14th century. The formal proof that there + are + exactly + 17 wallpaper groups was first given by the Russian mathematician, + crystallographer and mineralogist \index{Fedorov, E.} + \textit{E. Fedorov} + (1853–-1919) in 1891, + and then completed and continued by the Hungarian mathematician + \index{Pólya, G.}\textit{G. Pólya} (1887–-1985) in 1924, who is more famous for his + famous book ‘‘How to Solve Mathematical Problems’’.}. + Frieze groups determine 7 different types of \index{border} + \pojem{borders} - + strips with repeating patterns (Figure \ref{sl.izo.6.12.7}a\footnote{http://mathworld.wolfram.com/WallpaperGroups.html}), wallpaper + groups 17 different types of \pojem{ornamentation} - covering a plane + with identical patterns (Figure \ref{sl.izo.6.12.7}b\footnote{http://www.quadibloc.com/math/tilint.htm}). + +\begin{figure}[!h] +\centering + \includegraphics[bb=0 0 7cm 7cm]{bands.eps}\\ +\vspace*{15mm} +\includegraphics[width=1\textwidth]{wall17_phpbtUJbf.eps} +\caption{} \label{sl.izo.6.12.7} +\end{figure} + +%v bmp +%\begin{figure}[!h] +%\centering + %\includegraphics[bb=0 0 7cm 7cm]{bands.bmp}\\ +%\vspace*{8mm} +%\includegraphics[bb=0 0 12cm 9.85cm]{wall17_phpbtUJbf.bmp} +%\caption{} \label{sl.izo.6.12.7} +%\end{figure} + +\bnaloga\footnote{40. IMO Romania - 1999, Problem 1.} + Determine all finite sets $\mathbf {S}$ of at least three points in the plane which satisfy the + following condition:\\ + for any two distinct points $A$ and $B$ in $\mathbf {S}$, the perpendicular bisector + of the line segment $AB$ is an axis of symmetry for $\mathbf {S}$. + \enaloga + + +\textbf{\textit{Solution.}} + Let $\mathbf {S}=\{A_1,A_2,\ldots +A_n \}$ in $\mathcal{M}=\{s_{A_iA_j};\hspace*{1mm}A_i,A_j\in +\mathcal{S}\}$ be a set of all symmetries. Because $\mathbf {S}$ +is a finite set (or figure), by Theorem \ref{GrupaSomer1} all +perpendiculars of this set or symmetries from the set $\mathcal{M}$ go through +one point - we denote it with $O$. It is clear that $O$ +is the intersection of any two symmetries from $\mathcal{M}$, so $O=s_{A_1A_2}\cap s_{A_2A_3}$. + +Let $k$ be a circle with center $O$ that goes through the point $A_1$. +We first prove that all points of the set $\mathbf {S}$ lie on this +circle. Let $A_i\in \mathbf {S}$ be an arbitrary point. From +the proven $O\in s_{A_1A_i}$ and $A_1 \in k$ it follows +$A_i=\mathcal{S}_{s_{A_1A_i}}(A_1)\in +\mathcal{S}_{s_{A_1A_i}}(k)=k$ (Figure \ref{sl.izo.6.12.IMO1.pic}). + +Without loss of generality, we can assume that the points $\mathbf +{S}$ are ordered in sequence on the circle $k$, so that $\angle +A_1OA_2<\angle A_1OA_3<\cdots \angle A_1OA_n$ (otherwise we can +make a new labeling of these points). This means that $A_1A_2\ldots +A_n$ represents a polygon, which is inscribed in the circle $k$. We +also prove that this polygon is regular. Because this polygon is +concyclic, it is enough to +prove that all sides are congruent. Because $s_{A_1A_3}$ +is the altitude of the set $\mathbf {S}$ (and therefore also of the +polygon $A_1A_2\ldots +A_n$), $A_1A_2\cong A_2A_3$. Similarly, $s_{A_iA_{i+2}}$ +($i\in \{2,3,\ldots n-1\}$) is the altitude of the polygon $A_1A_2\ldots A_n$ +and $A_iA_{i+1}\cong A_{i+1}A_{i+2}$, which means that +$A_1A_2\ldots A_n$ is a regular polygon. + +The set $\mathbf {S}$ therefore represents the vertices of a regular +polygon. + \kdokaz + +%\vspace*{31mm} + +\newpage + +\naloge{Exercises} + +\begin{enumerate} + + \item Given a line $p$ and points $A$ and $B$, which lie on + opposite sides of the line $p$. Construct a point + $X$, which lies on +the line $p$, so that the difference $|AX|-|XB|$ is maximal. + + \item In the plane are given lines $p$, $q$ and $r$. Construct + an equilateral triangle $ABC$, so that + the vertex $B$ lies on the line $p$, $C$ on $q$, +and the altitude from the vertex $A$ lies on the line $r$. + +\item Given a quadrilateral $ABCD$ and a point $S$. Draw a parallelogram +with the center in the point $S$, so +that its vertices lie on the lines of +the sides of the given quadrilateral. + +\item Let $\mathcal{I}$ be an indirect isometry of the plane, which +maps the point $A$ to the point $B$, +$B$ to $A$. Prove that $\mathcal{I}$ is a central symmetry. + +\item Let $K$ and $L$ be points, which are symmetric to the vertex +$A$ of the triangle $ABC$ with respect to +the lines of symmetry of the internal angles at the vertices $B$ and $C$. Let $P$ be the point of tangency of the inscribed circle of this triangle and the sides $BC$. +Prove that $P$ is the center of the segment $KL$. + +\item Let $k$ and $l$ be two circles on different sides of the line +$p$. Draw an isosceles triangle $ABC$, so that its altitude $AA'$ lies on the line $p$, the vertex $B$ lies on the circle $k$, and the vertex $C$ lies on the circle $l$. + +\item Let $k$ be a circle and $a$, $b$, and $c$ be lines in the same plane. Draw a triangle $ABC$ inscribed in the circle $k$, so that its sides $BC$, $AC$, and $AB$ are parallel to the lines $a$, $b$, and $c$, respectively. + +\item Let $ABCDE$ be a quadrilateral with $BC\parallel DE$ and $CD\parallel EA$. Prove that the vertex $D$ lies on the perpendicular bisector of the line segment $AB$. + +\item The lines $p$, $q$, and $r$ lie in the same plane. Prove that the composition $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p + =\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_r$ is equivalent to the condition that the lines $p$, $q$, and $r$ belong to the same bundle. + +\item Let $O$, $P$, and $Q$ be three non-collinear points. Construct a square $ABCD$ (in the plane $OPQ$) with the center at the point $O$, so that the points $P$ and $Q$ lie on the lines $AB$ and $BC$, respectively. + +\item Let $\mathcal{R}_{S,\alpha}$ be a rotation and $\mathcal{S}_p$ be a reflection in the same plane and $S\in p$. Prove that the composition $\mathcal{R}_{S,\alpha}\circ\mathcal{S}_p$ and $\mathcal{S}_p\circ\mathcal{R}_{S,\alpha}$ represent a reflection. + +\item Given a point $A$ and a circle $k$ in the same plane. Draw a square $ABCD$, so that the midpoints of the diagonals $BD$ lie on the circle $k$. + +\item Let $ABC$ be an arbitrary triangle. Prove: + $$\mathcal{R}_{C,2\measuredangle BCA}\circ + \mathcal{R}_{B,2\measuredangle ABC}\circ + \mathcal{R}_{A,2\measuredangle CAB}=\mathcal{E}.$$ + +\item Prove that the composition of a reflection $\mathcal{S}_p$ +and a central reflection $\mathcal{S}_S$ ($S\in p$) +represents a reflection. + +\item Let $O$, $P$, and $Q$ be three non-collinear points. Construct a square $ABCD$ (in the plane $OPQ$) with the center at the point $O$, so that the points $P$ and $Q$ lie on the lines $AB$ and $CD$, respectively. + +\item What does the composition of a translation and a central reflection represent? + +\item Given are a line $p$ and circles $k$ and $l$, which +lie in the same plane. Draw a line that is parallel to +the line $p$, so that it determines the congruent tangents on the circles $k$ and $l$. + +\item Let $c$ be a line that intersects the parallels $a$ and $b$, and $l$ + be a distance. Draw an equilateral triangle $ABC$, so that $A\in a$, $B\in b$, $C\in c$ and $AB\cong l$. + +\item Prove that the composition of a rotation and an axial reflection +of a plane +represents a glide reflection exactly when the center +of rotation does not lie on the axis of the axial reflection. + +\item Let $ABC$ be an equilateral triangle. Prove +that the composition $\mathcal{S}_{AB} + \circ\mathcal{S}_{CA} + \circ\mathcal{S}_{BC}$ +represents a glide reflection. Also determine the vector and the axis of this glide reflection. + +\item Given are points $A$ and $B$ on the same side of a line +$p$. +Draw a line $XY$, which lies on the line $p$ and is congruent +to the given distance $l$, so that the sum +$|AX|+|XY|+|YB|$ is minimal. + +\item Let $ABC$ be an isosceles right triangle with the right angle at the vertex $A$. What does the composition: +$\mathcal{G}_{\overrightarrow{AB}}\circ \mathcal{G}_{\overrightarrow{CA}}$ represent? + +\item In the same plane are given lines $a$, $b$ and $c$. +Draw points $A\in a$ and $B\in b$ +so that $\mathcal{S}_c(A)=B$. + +\item Given are lines $p$ and $q$ and a point $A$ in the same plane. +Draw points $B$ and $C$ so +that the lines $p$ and $q$ will be the internal angle bisectors at +the vertices $B$ and $C$ of the triangle $ABC$. + +\item Let $s$ be the angle bisector of one of the angles determined by the lines $p$ and $q$. Prove that $\mathcal{S}_s\circ\mathcal{S}_p = +\mathcal{S}_q\circ\mathcal{S}_s$. + +\item Let $S$ be the center of the triangle $ABC$ inscribed in the circle and $P$ be the point where this circle touches the side $BC$. Prove: $$\mathcal{S}_{SC} \circ\mathcal{S}_{SA}\circ\mathcal{S}_{SB} =\mathcal{S}_{SP}.$$ + +\item The lines $p$, $q$ and $r$ of a plane go through the center $S$ of the circle $k$. Draw a triangle $ABC$, which is inscribed in this circle, so that the lines $p$, $q$ and $r$ are the altitudes of the vertices $A$, $B$ and $C$ of this triangle. + +\item The lines $p$, $q$, $r$, $s$ and $t$ of a plane intersect in the point $O$, and the point $M$ lies on the line $p$. Draw a pentagon so that $M$ is the center of one of its sides, and the lines $p$, $q$, $r$, $s$ and $t$ are the altitudes. + +\item The point $P$ lies in the plane of the triangle $ABC$. Prove that the lines which are symmetric to the lines $AP$, $BP$ and $CP$ with respect to the altitudes at the vertices $A$, $B$ and $C$ of this triangle, belong to the same family. + +\item Calculate the angle determined by the lines $p$ and $q$, if it holds that $\mathcal{S}_p\circ\mathcal{S}_q\circ\mathcal{S}_p = \mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_q$. + +\item Let $\mathcal{R}_{A,\alpha}$ and $\mathcal{R}_{B,\beta}$ be rotations in the same plane. Determine all points $X$ in this plane, for which it holds that $\mathcal{R}_{A,\alpha}(X)=\mathcal{R}_{B,\beta}(X)$. + +\item The lines $p$ and $q$ intersect at an angle of $60^0$ in the center $O$ of the equilateral triangle $ABC$. Prove that the segments determined by the sides of the triangle $ABC$ on the lines $p$ and $q$ are congruent. + +\item The point $S$ is the center of the regular pentagon $ABCDE$. Prove that it holds: + $$\overrightarrow{SA} + \overrightarrow{SB} + \overrightarrow{SC} + + \overrightarrow{SD} + \overrightarrow{SE} = \overrightarrow{0}.$$ + +\item Prove that the diagonals of the regular pentagon intersect in the points which are also the vertices of the regular pentagon. + +\item Let $ABP$ and $BCQ$ be two triangles with the same orientation and $\mathcal{B}(A,B,C)$. The points $K$ and $L$ are the midpoints of the lines $AQ$ and $PC$. Prove that $BLK$ is a triangle. + +\item There are three concentric circles and a line in the same plane. Draw a triangle so that its vertices are on these circles in order, and one side is parallel to the given line. + +\item The point $P$ is an internal point of the triangle $ABC$, so that $\angle APB=113^0$ and $\angle BPC=123^0$. Calculate the size of the angles of the triangle whose sides are consistent with the distances $PA$, $PB$ and $PC$. + +\item The points $P$, $Q$ and $R$ are given. Draw a triangle $ABC$ so that $P$, $Q$ and $R$ are the centers of the squares constructed over the sides $BC$, $CA$ and $AB$ of this triangle. + +\item Let $A$ and $B$ be points and $p$ a line in the same plane. Prove that the composite $\mathcal{S}_B\circ\mathcal{S}_p\circ\mathcal{S}_A$ is a reflection exactly when $AB\perp p$. + +\item Let $p$, $q$ and $r$ be tangents to the triangle $ABC$ of the inscribed circles that are parallel to its sides $BC$, $AC$ and $AB$. Prove that the lines $p$, $q$, $r$, $BC$, $AC$ and $AB$ determine such a hexagon, in which the pairs of opposite sides are consistent with the distances. + +\item Draw a triangle with the data: $\alpha$, $t_b$, $t_c$. + +\item Let $ALKB$ and $ACPQ$ be two squares outside the triangle $ABC$ drawn over the sides $AB$ and $AC$, and $X$ the center of the side $BC$. Prove that $AX\perp LQ$ and $|AX|=\frac{1}{2}|QL|$. + +\item Let $O$ be the center of the triangle $ABC$ and $D$ and $E$ the points of the sides $CA$ and $CB$, so that $CD\cong CE$. The point $F$ is the fourth vertex of the parallelogram $BODF$. Prove that the triangle $OEF$ is a right triangle. + +\item Let $L$ be the point in which the inscribed circle of the triangle $ABC$ touches its side $BC$. Prove: $$\mathcal{R}_{C,\measuredangle ACB}\circ\mathcal{R}_{A,\measuredangle BAC} \circ\mathcal{R}_{B,\measuredangle CBA} =\mathcal{S}_L.$$ + +\item Points $P$ and $Q$ as well as $M$ and $N$ are the centers of two squares, +which are drawn +outside of the opposite sides of any quadrilateral. Prove that +$PQ\perp MN$ and $PQ\cong MN$. + +\item Let $APB$ and $ACQ$ be two right triangles, which are drawn +outside of the triangle $ABC$ +on the sides $AB$ and $AC$. Point $S$ is the center +of the side $BC$ and $O$ is the center of the triangle $ACQ$. Prove that +$|OP|=2|OS|$. + +\item Prove that the reflection in an axis and the translation of a plane +commute only when the axis of this reflection is parallel to +the vector of the translation. + +\item In the same plane, given are a line $p$, circles $k$ and $l$ and a distance $d$. +Draw a rhombus $ABCD$ with a side, which is congruent to the distance $d$, the side $AB$ lies +on the line $p$, points $C$ and $D$ in turn lie +on the circles $k$ and $l$. + +\item Let $p$ be a line, $A$ and $B$ points which lie +on the same side of the line $p$, and $d$ a distance in the same plane. +Draw points $X$ and $Y$ on the line $p$ so that $AX\cong BY$ and $XY\cong d$. + +\item Let $H$ be the altitude of the triangle $ABC$ and $R$ the radius of the circumscribed circle of this +triangle. Prove that $|AB|^2+|CH|^2=4R^2$. + +\item Let $EAB$ be a triangle, which is drawn on the side $AB$ of the square +$ABCD$. Let also $M=pr_{\perp AE}(C)$ and $N=pr_{\perp BE}(D)$ and point $P$ +be the intersection of the lines $CM$ and $DN$. Prove that $PE\perp AB$. + +\item Draw an equilateral triangle $ABC$ so that its vertices in turn +lie on three parallel lines $a$, $b$ and $c$ in the same plane, +the center of this triangle lies on the line $s$, which intersects +the lines $a$, $b$ and $c$. + +\item If a pentagon has at least two axes of symmetry, it is regular. Prove. + +\item Let $A$, $B$ and $C$ be three collinear points. What does +the composite $\mathcal{G}_{\overrightarrow{BC}}\circ \mathcal{S}_A$ represent? + +\item Let $p$, $q$ and $r$ be lines that are not in the same plane, and let $A$ be a point in the same plane. Draw a line $s$ through the point $A$ such that $\mathcal{S}_r\circ \mathcal{S}_q\circ \mathcal{S}_p(s)=s'$ and $s\parallel s'$. + +%new tasks +%___________________________________ + +\item Let $Z$ and $K$ be inner points of the rectangle $ABCD$. +Draw points $A_1$, $B_1$, $C_1$ and $D_1$, which in turn lie on the sides $AB$, $BC$, $CD$ and $DA$ of this rectangle, so that $\angle ZA_1A\cong\angle B_1A_1B$, $\angle A_1B_1B\cong\angle C_1B_1C$, $\angle B_1C_1C\cong\angle D_1C_1D$ and $\angle C_1D_1D\cong\angle KD_1A$. + +\item The point $A$ lies on the line $a$, and the point $B$ lies on the line $b$. +Determine the rotation that maps the line $a$ to the line $b$ and the point $A$ to the point $B$. + +\item In the center of the square, two rectangles intersect. +Prove that these rectangles intersect the sides of the square at the vertices of a new square. + +\item Given a circle $k$ and lines $a$, $b$, $c$, $d$ and $e$, which lie in the same plane. Draw a pentagon on the circle $k$ with sides that are in turn parallel to the lines $a$, $b$, $c$, $d$ and $e$. + +\item The point $P$ lies inside the angle $aOb$. Draw a line $p$ through the point $P$, which with the sides $a$ and $b$ determines a triangle with the smallest area. + +\item The parallelogram $PQKL$ is drawn in the parallelogram $ABCD$ (the vertices of the first lie on the sides of the second). Prove that the parallelograms have a common center. + +\item The arcs $l_1, l_2,\cdots , l_n$ lie on the circle $k$ and the sum of their lengths is less than the radius of this circle. Prove that there exists such a diameter $PQ$ of the circle $k$, that none of its endpoints lies on any of the arcs $l_1, l_2,\cdots , l_n$. + +\item Given is a circle $k(S,20)$. Players $\mathcal{A}$ and $\mathcal{B}$ take turns drawing circles with radii $x_i$ ($1AB$, +or $b>c$. Since $\angle AMB =\omega$, by the statement +\ref{ObodKotGMT} the point $M$ lies on the arc $l$ with the chord $AB$ and the central angle +$\omega$. The extension $h_{B,2}$ maps the points $B$ and $M$ in a row to +the points $B$ and $C$, and the arc $l$ to the arc $l'$. From $M\in l$ it follows that $C\in +l'$. Since $AC=b$, it follows that the point $C$ lies on +the circle $k(A,b)$ as well. So $C\in l'\cap k(A,b)$. The proven facts +allow the construction. + + First of all, we draw the segment $AB=c$ and the arc $l$ with this chord and the central angle +$\omega$, then the arc $l'=h_{B,2}(l)$, and finally the point $C$ as +one of the intersections of the arc $l'$ with the circle $k(A,b)$. + +We will prove that the constructed triangle $ABC$ satisfies the conditions of the task. By construction, $AB=c$ immediately. From $C\in k(A,b)$ it follows that $AC=b$. Let $M=h^{-1}_{B,2}(C)$. Because $C\in l'=h_{B,2}(l)$, $M \in l$. Because $l$ is a line segment with this string and the obtuse angle $\omega$, $\angle AMB=\omega$. It remains to be proven that the point $M$ is the center of the line segment $BC$, which follows directly from the relation $M=h^{-1}_{B,2}(C)$. + +We will find the conditions for the number of solutions to the task. We have already mentioned that, due to the condition $\omega<90^0$, $b>c$ must be true. In the case of $b\leq c$, there is no solution. The number of solutions to the task is further dependent on the number of intersections of the line segment $l'$ with the circle $k(A,b)$. +\kdokaz + + + + + +%________________________________________________________________________________ + \poglavje{Classification of Similarity Transformations} \label{odd7PrezentTransPod} + +In section \ref{odd6KlasifIzo} we classified isometries, and here we will do a similar classification of similarity transformations. Also in this classification, +the number of fixed points and the fact that the similarity transformation is direct or indirect will be important. + +In the previous two sections we have found that all isometries and central stretches represent similarity transformations. Also, their composition is a similarity transformation (statement \ref{TransPodGrupa}). Now we will prove that the converse is also true. + + + + \bizrek \label{TransPodKompHomIzo} + Each similarity transformations $f$ with coefficient $k$ can be expressed as the product + of one isometry and one homothety with an arbitrary centre: + $$f=h_{S,k}\circ\mathcal{I}_1=\mathcal{I}_2\circ h_{S,k}.$$ + \eizrek + +\textbf{\textit{Proof.}} + +Let $f$ be an arbitrary similarity transformation with coefficient $k$. +If we denote the central stretch with arbitrary center $S$ and coefficient $\frac{1}{k}$ with $h_{S,\frac{1}{k}}$, then the composite $h_{S,\frac{1}{k}}\circ f$ represents +a similarity transformation with coefficient $k\cdot \frac{1}{k}=1$ (statement \ref{TransPodGrupa}) or isometry. So $h_{S,\frac{1}{k}}\circ f=\mathcal{I}_1$, where $\mathcal{I}_1$ is some isometry. According to statement \ref{homotGrupa}, $f=h_{S,\frac{1}{k}}^{-1}\circ\mathcal{I}_1=h_{S,k}\circ\mathcal{I}_1$. +Similarly, $f\circ h_{S,\frac{1}{k}}=\mathcal{I}_2$, where $\mathcal{I}_2$ is some isometry, so $f=\mathcal{I}_2\circ h_{S,k}$. + + + \bizrek \label{TransPodOhranjaKote} + Similarity transformations preserve the measure of angles, i.e. there map an angle to the congruent angle. + \eizrek + + +\textbf{\textit{Proof.}} + The statement is a direct consequence of statements \ref{TransPodKompHomIzo} and \ref{homotOhranjaKote}. +\kdokaz + + \bizrek \label{homotTransm} + For each isometry $\mathcal{I}$ and each homothety $h_{S,k}$ + it is: + $$\mathcal{I}\circ h_{S,k}\circ \mathcal{I}^{-1}=h_{\mathcal{I}(S),k}$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.2.pic} +\caption{} \label{sl.pod.7.1p.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure +\ref{sl.pod.7.1p.2.pic}) + +Let $\mathcal{I}(S)=S_1$. We denote $f=\mathcal{I}\circ h_{S,k}\circ \mathcal{I}^{-1}$. It is necessary to prove that $f=h_{S_1,k}$ or $f(X_1)=h_{S_1,k}(X_1)$ for an arbitrary point $X_1$ of this plane. + +Let us also denote $X=\mathcal{I}^{-1}(X_1)$, $X'=h_{S,k}(X)$ and +$X'_1=\mathcal{I}(X_1)$. Then we have: +\begin{eqnarray*} +f(X_1)&=& \mathcal{I}\circ h_{S,k}\circ \mathcal{I}^{-1}(X_1)\\ + &=& \mathcal{I}\circ h_{S,k}(X)\\ + &=& \mathcal{I}(X')\\ + &=& X'_1 +\end{eqnarray*} + + Therefore $f(X_1)=X'_1$. From $X'=h_{S,k}(X)$ it follows that $\overrightarrow{SX'}=k\cdot\overrightarrow{SX}$. Because $\mathcal{I}$ is an isometry and $\mathcal{I}:\hspace*{1mm}S,X,X'\mapsto S_1,X_1,X'_1$, also +$\overrightarrow{S_1X'_1}=k\cdot\overrightarrow{S_1X_1}$ or $h_{S_1,k}(X_1)=X'_1$. +This means that for any point $X_1$ we have $f(X_1)=X'_1=h_{S_1,k}(X_1)$, so +$f=h_{S_1,k}$. +\kdokaz + + + + \bizrek \label{homotIzomKom} + An Isometry $\mathcal{I}$ and a homothety $h_{S,k}$ commute + if and only if + the centre of this homothety is a fixed point of the isometry $\mathcal{I}$, i.e.: + $$\mathcal{I}\circ h_{S,k}=h_{S,k}\circ\mathcal{I}\hspace*{1mm} + \Leftrightarrow\hspace*{1mm}\mathcal{I}(S)=S.$$ + \eizrek + + +\textbf{\textit{Proof.}} + By the previous statement (\ref{homotTransm}) we have: + \begin{eqnarray*} +\mathcal{I}\circ h_{S,k}=h_{S,k}\circ\mathcal{I} +\hspace*{1mm} + &\Leftrightarrow& \hspace*{1mm} \mathcal{I}\circ h_{S,k}\circ\mathcal{I}^{-1}=h_{S,k}\\ +\hspace*{1mm} + &\Leftrightarrow& \hspace*{1mm} h_{\mathcal{I}(S),k}=h_{S,k}\\ +\hspace*{1mm} + &\Leftrightarrow& \hspace*{1mm} \mathcal{I}(S)=S, +\end{eqnarray*} + which was to be proven. \kdokaz + +If we choose a rotation for the isometry from statement \ref{TransPodKompHomIzo} with the same centre as the central dilation, we get a very useful type of similarity transformation. + +The composition of a rotation and a central stretch with the same center is called a \index{rotational stretch} \pojem{rotational stretch} (Figure +\ref{sl.pod.7.1p.1.pic}): +$$\rho_{S,k,\omega}=h_{S,k}\circ \mathcal{R}_{S,\omega}$$ +with the \index{center!of a rotational stretch}\pojem{center} $S$, the \index{coefficient!of a rotational stretch}\pojem{coefficient} $k$, and the \index{angle!of a rotational stretch}\pojem{angle} $\omega$. + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.1.pic} +\caption{} \label{sl.pod.7.1p.1.pic} +\end{figure} + +By \ref{homotIzomKom} we have: +$$\rho_{S,k,\omega}=h_{S,k}\circ \mathcal{R}_{S,\omega}= +\mathcal{R}_{S,\omega}\circ h_{S,k}.$$ + + +Since a central stretch and a rotation are direct transformations (\ref{homotDirekt} and \ref{RotacDirekt}), a stretch rotation is also a direct transformation of similarity. + +It is clear that a central stretch can also be considered as a type of stretch rotation, if we assume that the identity is a rotation by the angle $0^0$: +$$h_{S,k}=\rho_{S,k,0^0}.$$ + +Similarly, we can also see a rotation as a type of stretch rotation: +$$\mathcal{R}_{S,\omega}=\rho_{S,1,\omega}.$$ + + + \bizrek \label{rotacRaztKot} + An arbitrary line and its image under a stretch rotation + determine an oriented angle which is congruent to the angle of this stretch rotation: + $$\rho_{S,k,\omega}(p)=p'\hspace*{1mm} \Rightarrow + \hspace*{1mm} \angle p,p'=\omega.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.3.pic} +\caption{} \label{sl.pod.7.1p.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Let $p'=\rho_{S,k,\omega}(p)$ be the image of the line $p$ under the stretch rotation $\rho_{S,k,\omega}=h_{S,k}\circ \mathcal{R}_{S,\omega}$ (Figure \ref{sl.pod.7.1p.3.pic}) and $p_1=\mathcal{R}_{S,\omega}(p)$. Then $h_{S,k}(p_1)=p'$. By \ref{rotacPremPremKot} we have $\measuredangle p,p_1=\omega$. By \ref{RaztPremica} we have $p_1\parallel p'$. Therefore $\measuredangle p,p'=\angle p,p_1=\omega$ (\ref{KotiTransverzala1}). +\kdokaz + + + + \bizrek \label{rotacRaztKompSredZrc} + The product of a half-turn and a stretch rotation + with the same centre is a stretch rotation. Furthermore: + $$\rho_{S,k,\omega}\circ \mathcal{S}_S= + \mathcal{S}_S\circ\rho_{S,k,\omega}=\rho_{S,-k,\omega}.$$ + \eizrek + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.3a.pic}) + +As we have already mentioned in section \ref{odd7SredRazteg}, $\mathcal{S}_S=h_{S,-1}$. +By \ref{homotGrupa} we have: +\begin{eqnarray*} +\rho_{S,k,\omega}\circ \mathcal{S}_S= +\mathcal{R}_{S,\omega}\circ h_{S,k}\circ h_{S,-1}= +\mathcal{R}_{S,\omega}\circ h_{S,-k}=\rho_{S,-k,\omega}. +\end{eqnarray*} +Similarly we have: +\begin{eqnarray*} +\mathcal{S}_S\circ\rho_{S,k,\omega}= +h_{S,-1}\circ h_{S,k}\circ \mathcal{R}_{S,\omega}= +h_{S,-k}\circ\mathcal{R}_{S,\omega}=\rho_{S,-k,\omega}, +\end{eqnarray*} + which is what needed to be proven. \kdokaz + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.3a.pic} +\caption{} \label{sl.pod.7.1p.3a.pic} +\end{figure} + + A direct consequence is the following theorem. + +\bizrek \label{rotacRaztNegKoefk} + For each stretch rotation is: + $$\rho_{S,-k,\omega}=\rho_{S,k,180^0+\omega}.$$ + \eizrek + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.3a.pic}) + +By the previous statement \ref{rotacRaztKompSredZrc} and the statement \ref{rotacKomp2rotac} it is: +\begin{eqnarray*} +\rho_{S,-k,\omega}= +\mathcal{S}_S\circ\rho_{S,k,\omega}= +\mathcal{S}_S\circ\mathcal{R}_{S,\omega}\circ h_{S,k} +=\mathcal{R}_{S,180^0+\omega}\circ h_{S,k}= +\rho_{S,k,180^0+\omega}, +\end{eqnarray*} +which had to be proven. \kdokaz + +If for the isometry from the statement \ref{TransPodKompHomIzo} we choose the reflection over a line that goes through the center of the central stretch, we get another type of similarity transformation. + +The composite of the central stretch $s$ and the central stretch $h_{S,k}$ with the center $S\in s$ is called the \index{osni razteg} \pojem{osni razteg} (Figure +\ref{sl.pod.7.1p.1a.pic}): +$$\sigma_{S,k,s}=h_{S,k}\circ \mathcal{S}_s;\hspace*{2mm} (S\in s)$$ +with the \index{središče!osnega raztega}\pojem{center}, the \index{koeficient!rotacijskega raztega}\pojem{coefficient} $k$ and the \index{os!osnega raztega}\pojem{axis} $s$. + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.1a.pic} +\caption{} \label{sl.pod.7.1p.1a.pic} +\end{figure} + +By the statement \ref{homotIzomKom} it is: +$$\sigma_{S,k,s}=h_{S,k}\circ \mathcal{S}_s= +\mathcal{S}_s\circ h_{S,k}.$$ + +Since the central stretch is a direct and the rotation is an indirect transformation (the statement \ref{homotDirekt} and \ref{izozrIndIzo}), the axis stretch is an indirect transformation of similarity. + + + + + \bizrek \label{transPod1FixTocLema} + Let $f$ be a similarity transformation that is not an isometry. + If $f$ maps each line to its parallel line, then $f$ is a homothety. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.1bb.pic} +\caption{} \label{sl.pod.7.1p.1bb.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.1bb.pic}) + +Let $k$ be the coefficient of similarity of the transformation $f$. Because $f$ is not an isometry, $k \neq 1$. + +By assumption, $f$ maps each line into a parallel line. We first prove that there are at least two lines that intersect and are not mapped into themselves. +Let $X$, $Y$ and $Z$ be any three non-collinear points of the plane and let $p$, $q$ and $r$ be lines determined by the points $X$, $Y$ and $Z$: $p = XY$, $q = YZ$ and $r = XZ$. Let $p' = f (p)$, $q' = f (q)$ and $r' = f (r)$. By assumption, $p \parallel p'$, $q \parallel q'$ and $r \parallel r'$. We prove that at least one of the three lines is not mapped into itself. Assume the contrary, that $p = p'$, $q = q'$ and $r = r'$. But in this case, $f (X) = f (p \cap q) = f (p) \cap f (q) = p \cap q = X$ and similarly, for example, $f (Y) = Y$. The distance $XY$ would be mapped into itself in this case, which is not possible, since $k \neq 1$. Without loss of generality, let $p \neq p'$. In the same way, by using a triangle in which no altitude is parallel to the line $p$, we can prove that there is another line that intersects the line $p$ and is not mapped into itself by $f$. + +So there are lines $b$ and $c$ that intersect at point $A$ and for $b'=f(b)$ and $c'=f(c)$ it holds that $b\parallel b'$, $c\parallel c'$, $b\neq b'$ and $c\neq c'$. Let $B\in b$ and $C\in c$ be any points that are different from point $A$. +We mark $A'=b\cap c$, $B'=f(B)$ and $C'=f(C)$. First, $f(A)=f(b\cap c)=f(b)\cap f(c)=b'\cap c'=A'$, $B'\in b'$ and $C'\in c'$. +Because $b\parallel b'$, also $AB\parallel A'B'$. The distance $AB$ is mapped by transformation $f$ into the distance $A'B'$, so $A'B'=k\cdot AB$. +The lines $AA'$ and $BB'$ are not parallel. Otherwise the quadrilateral $ABB'A'$ would be a parallelogram or $AB\cong A'B'$ (by statement \ref{paralelogram}), which is not possible because $k\neq 1$. We mark with $S$ the intersection of the lines $AA'$ and $BB'$. Because $b\parallel b'$, by Tales' statement: +$$\frac{SA'}{SA}=\frac{SB'}{SB}=\frac{A'B'}{AB}=k.$$ + This means that the central stretch $h_{S,k_1}$ with center $S$ and coefficient $k_1=k$ (or $k_1=-k$) maps points $A$ and $B$ into points $A'$ and $B'$. + Let $\widehat{C'}=h_{S,k_1}(C)$. Because by assumption $C'=f(C)$ or $A'C'=k\cdot AC$, by statement \ref{RaztPremica} $\widehat{C'}=C'$ or $h_{S,k_1}(C)=C'$. + + The mapping +$$g=f^{-1}\circ h_{S,k_1}$$ + is by statement \ref{TransPodGrupa} a similarity transformation with similarity coefficient $\frac{1}{k}\cdot |k_1|=\frac{1}{k}\cdot k=1$, so it represents an isometry. +But $g(A)=f^{-1}\circ h_{S,k_1}(A)=f^{-1}(A')=A$, or $A$ is a fixed point of isometry $g$. In a similar way we prove that also $B$ and $C$ are fixed points of isometry $g$, which means that $g=\mathcal{E}$ is the identity (by statement \ref{IizrekABC2}). So $f^{-1}\circ h_{S,k_1}=g=\mathcal{E}$ or $f=h_{S,k_1}$. +\kdokaz + + Now we are ready for the next important statement. + +\bizrek \label{transPod1FixToc} + Each similarity transformation other than isometry has exactly one fixed point. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.1c.pic} +\caption{} \label{sl.pod.7.1p.1c.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.1c.pic}) + +Let $f$ be a similarity transformation with coefficient $k$. By assumption, $k\neq 1$. +Similarly to the proof of the previous izrek \ref{transPod1FixTocLema}, we quickly convince ourselves that $f$ cannot have two fixed points, because $k=1$ would be. + +We will prove that $f$ has a fixed point. +According to the previous statement (\ref{transPod1FixTocLema}), we can assume that there exists at least one line $p$ that is not mapped into its parallel. Indeed, if we assume that each plane line is mapped into its parallel, then according to the statement \ref{transPod1FixTocLema} the mapping $f$ is a central stretch with a fixed point (its center). +Let $p'=f(p)$ and $ p'\nparallel p$ ($\neg p'\parallel p$). We mark the intersection of lines $p$ and $p'$ with $A$. Let $A'=f(A)$. If $A'=A$, then $A$ is a fixed point of transformation $f$ and the proof is finished. So let $A'\neq A$. From $A\in p$ it follows that $A'\in p'$. We mark the parallel to the line $p$ through the point $A'$ with $q$. Because $A'\neq A$, $q\neq p$. Let $q'=f(q)$. Because $p\parallel q$ and $f:\hspace*{1mm}p,q\rightarrow p',q'$, also $p'\parallel q'$ (if $p'$ and $q'$ would intersect in a point $T$, then also lines $p$ and $q$ would intersect in the point $f^{-1}(T)$). In the same way from $q\neq p$ it follows that $q'\neq p'$. This means that lines $p$, $q$, $p'$ and $q'$ determine the parallelogram $AA'BC$, where + $B=q\cap q'$ and $C=p\cap q'$. Let $B'=f(B)$. From $B\in q$ it follows that $B'\in q'$. If $B'=B$, then $B$ is a fixed point and the proof is finished. So we can further assume that $B'\neq B$. From $p'\parallel q'$ it follows that $AA'\parallel BB'$. In the case $AB\parallel A'B'$ the quadrilateral $AA'B'B$ would be a parallelogram or $AB\cong A'B'$, which is not possible, because $f:\hspace*{1mm}A,B\rightarrow A',B'$ and $k\neq 1$. So lines $AB$ and $A'B'$ intersect in a point $S$. We mark $S'=f(S)$. We will prove that $S$ is a fixed point or $S'=S$. +According to Tales' statement: +\begin{eqnarray} \label{eqnTransfPod1Ft1} + \frac{AS}{SB}=\frac{A'S}{SB'} +\end{eqnarray} + Because $f:\hspace*{1mm}A,B,S\rightarrow A',B',S'$, also (statement \ref{TransPodOhranjajoRazm}): +\begin{eqnarray} \label{eqnTransfPod1Ft2} + \frac{AS}{SB}=\frac{A'S'}{S'B'} +\end{eqnarray} +Also from $S\in AB$ it follows that $S'\in A'B'$. +If we connect the last two equalities \ref{eqnTransfPod1Ft1} and + \ref{eqnTransfPod1Ft2}, it is: +\begin{eqnarray} \label{eqnTransfPod1Ft3} + \frac{A'S}{SB'}=\frac{A'S'}{S'B'}, +\end{eqnarray} +where $S$ and $S'$ are points of the line $A'B'$. But similarity transformations preserve the relation $\mathcal{B}$ (statement \ref{TransPodB}), or: +\begin{eqnarray*} + \mathcal{B}(A,S,B)\hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} \mathcal{B}(A',S',B');\\ + \mathcal{B}(S,A,B)\hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} \mathcal{B}(S',A',B');\\ +\mathcal{B}(A,B,S)\hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} \mathcal{B}(A',B',S'). +\end{eqnarray*} + From this and from the relation \ref{eqnTransfPod1Ft3} it follows: +\begin{eqnarray*} + \frac{\overrightarrow{A'S}}{\overrightarrow{SB'}}= +\frac{\overrightarrow{A'S'}}{\overrightarrow{S'B'}}, +\end{eqnarray*} + so according to the statement \ref{izrekEnaDelitevDaljiceVekt} $S'=S$ or $S$ is a fixed point of similarity transformation $f$. +\kdokaz + +Now we can make the predicted classification of similarity transformations. + + + + \bizrek \label{transPodKlasif} + The only similarity transformations of the plane are: + \begin{itemize} + \item isometries, + \item homotheties, + \item stretch rotations, + \item stretch reflections. + \end{itemize} + \eizrek + +\textbf{\textit{Proof.}} +Let $f$ be an arbitrary similarity transformation with coefficient $k$. + +If $k=1$, $f$ is an isometry. + +Assume that $k\neq 1$ or $f$ is not an isometry. According to Theorem \ref{transPod1FixToc}, $f$ has exactly one fixed point - we denote it with $S$. So $f(S)=S$. According to Theorem \ref{TransPodKompHomIzo}, we can represent the transformation of similarity $f$ as the composition of a central stretch with an arbitrary center (we choose the point $S$ as the center) and a coefficient $k$ and one isometry $\mathcal{I}$: +$$f=\mathcal{I}\circ h_{S,k}.$$ + Because $S$ is a fixed point of the transformation of similarity $f$ and the central stretch $h_{S,k}$, +it is: +$$S=f(S)=\mathcal{I}\circ h_{S,k}(S)=\mathcal{I}(S).$$ +So $\mathcal{I}(S)=S$ or $\mathcal{I}$ is an isometry with a fixed point $S$. According to Theorem \ref{Chaslesov}, $\mathcal{I}$ can be: identity, rotation with center $S$ or reflection over a line that goes through the point $S$: +\begin{eqnarray*} +\mathcal{I}=\left\{ + \begin{array}{l} + \mathcal{E}, \\ + \mathcal{R}_{S,\omega}, \\ + \mathcal{S}_s (S\in s) + \end{array} + \right. +\end{eqnarray*} +Therefore: +\begin{eqnarray*} +f=\mathcal{I}\circ h_{S,k}=\left\{ + \begin{array}{l} + h_{S,k}, \\ + \rho_{S,k,\omega}, \\ + \sigma_{S,k,s} + \end{array} + \right. +\end{eqnarray*} + which had to be proven. \kdokaz + + A direct consequence is the following theorem. + +\bizrek \label{transPodKlasifDirInd} + The only direct similarity transformations of the plane other than isometries are: + \begin{itemize} + \item homotheties, + \item stretch rotations. + \end{itemize} + The only opposite similarity transformations of the plane other than isometries are: + \begin{itemize} + \item stretch reflections. + \end{itemize} + \eizrek + + +It is very useful to know the property of the composite of two stretch rotations. + +\bizrek \label{RotRazKomoz} + The product of two stretch rotations $\rho_{S_1,k_1,\omega_1}$ and $\rho_{S_2,k_2,\omega_2}$ + is a direct isometry, a homothety or a stretch rotation. Furthermore (Figure \ref{sl.pod.7.1p.5a.pic}): + \begin{eqnarray*} + \rho_{S_2,k_2,\omega_2}\circ \rho_{S_1,k_1,\omega_1}= + \left\{ + \begin{array}{ll} + \mathcal{R}_{S,\omega}, & k_1k_2=1, \hspace*{2mm} + \omega=\omega_1+\omega_2\neq n\cdot 180^0; \\ + \mathcal{T}_{\overrightarrow{v}}, & k_1k_2=1, \hspace*{2mm} + \omega=\omega_1+\omega_2=n\cdot 180^0; \\ + h_{S,k}, & k=k_1k_2\neq 1, \hspace*{2mm} + \omega=\omega_1+\omega_2=n\cdot 180^0; \\ + \rho_{S,k,\omega}, & k=k_1k_2\neq 1, \hspace*{2mm} + \omega=\omega_1+\omega_2\neq n\cdot 180^0. + \end{array} + \right. + \end{eqnarray*} + for $n\in \mathbb{Z}$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.5.pic} +\caption{} \label{sl.pod.7.1p.5.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.5.pic}) + +Without loss of generality, assume that $k_1>0$ and $k_2>0$. For example, if $k_1<0$, we can by \ref{rotacRaztNegKoefk} write + $\rho_{S,k_1,\omega}=\rho_{S,-k_1,180^0+\omega}$, where $-k_1>0$. + +Let $f=\rho_{S_2,k_2,\omega_2}\circ \rho_{S_1,k_1,\omega_1}$. By the formulas \ref{RaztTransPod} and \ref{TransPodGrupa}, $f$ is a similarity transformation +with coefficient $k=k_1\cdot k_2$. It is a direct transformation as the composite of two direct transformations. By the formula \ref{transPodKlasifDirInd}, $f$ can be an isometry, central dilation or rotational dilation. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.5.pic} +\caption{} \label{sl.pod.7.1p.5.pic} +\end{figure} + + +Let $p$ be an arbitrary line and $f(p)=p'$. Calculate the measure of the oriented angle $\measuredangle p,p'$. Let $\rho_{S_1,k_1,\omega_1}(p)=p_1$ and consequently $\rho_{S_2,k_2,\omega_2}(p_1)=p'$. By the formula \ref{rotacRaztKot}, $\angle p,p_1=\omega_1$ and $\angle p_1,p'=\omega_2$. +If $\omega=\omega_1+\omega_2\neq n\cdot 180^0$ for every $n\in \mathbb{Z}$, then +$\angle p,p'=\omega_1+\omega_2$ (formula \ref{zunanjiNotrNotr}). +If $\omega=\omega_1+\omega_2=n\cdot 180^0$ for some $n\in \mathbb{Z}$, then $p\parallel p'$ (formula \ref{KotiTransverzala}). + + +If $k_1\cdot k_2=1$, the composite $f$ is a similarity transformation with coefficient $k=1$ and it is a direct isometry $f=\mathcal{I}\in \mathfrak{I}^+$. It can be a translation or a rotation (in the special case of identity) (formula \ref{Chaslesov+}). + If $\omega=\omega_1+\omega_2\neq n\cdot 180^0$ or +$\angle p,p'=\omega_1+\omega_2$, it is a rotation by the angle $\omega=\omega_1+\omega_2$ (formula \ref{rotacPremPremKot}). + If $\omega=\omega_1+\omega_2=n\cdot 180^0$ or $p\parallel p'$, it is a translation (or identity). + +Let us now assume that $k_1\cdot k_2\neq 1$. As we have already mentioned, in this case $f$ can be a central extension or a rotational extension with a coefficient $k=k_1\cdot k_2$. + If $\omega=\omega_1+\omega_2\neq n\cdot 180^0$ or +$\angle p,p'=\omega_1+\omega_2$, it is a rotational extension for an angle $\omega=\omega_1+\omega_2$ (statement \ref{rotacRaztKot}). + If $\omega=\omega_1+\omega_2=n\cdot 180^0$ or $p\parallel p'$, it is a central extension according to statement \ref{RaztPremica} (or a central reflection if $k_1\cdot k_2=-1$). +\kdokaz + +A direct consequence is the following theorem. + + + \bizrek + + The product of two homotheties + is a direct isometry or a homothety. Furthermore: + \begin{eqnarray*} + h_{S_2,k_2}\circ h_{S_1,k_1}= + \left\{ + \begin{array}{ll} + \mathcal{T}_{\overrightarrow{v}}, & k_1k_2=1; \\ + h_{S,k}, & k=k_1k_2\neq \pm 1. + \end{array} + \right. + \end{eqnarray*} + \eizrek + +\textbf{\textit{Proof.}} +The statement follows directly from the previous statement, if we write +$h_{S_1,k_1}=\rho_{S_1,k_1,0^0}$ and $h_{S_2,k_2}=\rho_{S_2,k_2,0^0}$. +\kdokaz + + + + \bnaloga\footnote{17. IMO Bulgaria - 1975, Problem 3.} + On the sides of an arbitrary triangle $ABC$, triangles $ABR$, $BCP$, $CAQ$ are + constructed externally with $\angle CBP\cong\angle CAQ=45^0$, $\angle BCP\cong\angle ACQ=35^0$, + $\angle ABR\cong\angle BAR=15^0$. Prove that $\angle QRP=90^0$ and $QR\cong RP$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1.IMO2.pic} +\caption{} \label{sl.pod.7.1.IMO2.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $S$ be the third vertex +of the isosceles triangle $BAS$, which is drawn over the side $BA$ +of the triangle $ABC$ (Figure \ref{sl.pod.7.1.IMO2.pic}). Because $\angle +RBS\cong\angle SAR=45^0$ and $\angle BSR\cong\angle ASR=30^0$, the +triangles $BPC$, $AQC$, $BRS$ and $ARS$ are similar to each other. + +Let $f_1$ and $f_2$ be the stretching rotations: + \begin{eqnarray*} + &&f_1=\rho_{B,k,45^0}=h_{B,k}\circ \mathcal{R}_{B,45^0}\\ + &&f_2=\rho_{A,\frac{1}{k},45^0}=h_{A,\frac{1}{k}}\circ + \mathcal{R}_{A,45^0}, + \end{eqnarray*} +where: +$$k=\frac{|BC|}{|BP|}=\frac{|AC|}{|AQ|}=\frac{|BS|}{|BR|} +=\frac{|AS|}{|AR|}.$$ + Let $f=f_2\circ f_1$. By the statement \ref{RotRazKomoz} it is: + $$f=f_2\circ f_1=\rho_{A,\frac{1}{k},45^0}\circ + \rho_{B,k,45^0}=\rho_{T,1,90^0}=\mathcal{R}_{T,90^0}.$$ + Then it is valid: + \begin{eqnarray*} + &&\mathcal{R}_{T,90^0}(P)=f(P)=f_2\circ f_1(P)=f_2(C)=Q\\ + &&\mathcal{R}_{T,90^0}(R)=f(R)=f_2\circ f_1(R)=f_2(S)=R. + \end{eqnarray*} + From the previous (second) relation we see that $\mathcal{R}_{T,90^0}(R)=R$, which means that $R=T$ (statement \ref{RotacFiksT}) or + $\mathcal{R}_{T,90^0}=\mathcal{R}_{R,90^0}$. + From this and from the first relation now follows: + $$\mathcal{R}_{R,90^0}(P)=Q,$$ + which means that $\angle QRP=90^0$ + and $QR\cong RP$. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Similar Figures. Similarity of Triangles} \label{odd7PodobTrik} + +As we have already announced in the introduction to this chapter, similarity transformations allow us to define the concept of similarity of figures. + +We say that the figure $\Phi$ \index{figure!similar}\pojem{similar} to the figure $\Phi'$ from the same plane (the notation $\Phi\sim \Phi'$), + if there exists a similarity transformation $f$ of this plane, which maps the figure $\Phi$ to the figure $\Phi'$ or $f(\Phi)=\Phi'$ (Figure \ref{sl.pod.7.2.1.pic}). + The coefficient of similarity transformation $f$ is at the same time \index{coefficient!of similarity of figures}\pojem{the coefficient of similarity of figures} $\Phi$ and $\Phi'$. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.1.pic} +\caption{} \label{sl.pod.7.2.1.pic} +\end{figure} + +It is clear that for $k=1$ we get the congruence of figures as a special case of similarity. Congruent figures are therefore also similar, but the converse is not true, that is: +$$\Phi\cong \Phi' \hspace*{1mm}\Rightarrow \hspace*{1mm}\Phi\sim \Phi'.$$ + +We prove the most important property of the similarity of figures. + + \bizrek + The similarity of figures is an equivalence relation. + \eizrek + + \textbf{\textit{Proof.}} It is necessary (and sufficient) to prove that the similarity of figures is reflexive, symmetric and transitive. + + + (\textit{R}) For every figure $\Phi$ it holds that $\Phi\sim \Phi$, because the identity $\mathcal{E}$, which maps the figure $\Phi$ to itself, is a similarity transformation with the coefficient $k=1$ (Figure \ref{sl.pod.7.2.1r.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.1r.pic} +\caption{} \label{sl.pod.7.2.1r.pic} +\end{figure} + + (\textit{S}) We assume that for the figures $\Phi_1$ and $\Phi_2$ it holds that $\Phi_1\sim \Phi_2$ (Figure \ref{sl.pod.7.2.1s.pic}). By definition there exists a similarity transformation $f$, which maps the figure $\Phi_1$ to the figure $\Phi_2$, or $f:\hspace*{1mm}\Phi_1\rightarrow \Phi_2$. + By \ref{TransPodGrupa} the inverse mapping $f^{-1}$, for which $f^{-1}:\hspace*{1mm}\Phi_2\rightarrow \Phi_1$, is also a similarity transformation. Therefore it holds that $\Phi_2\sim \Phi_1$. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.1s.pic} +\caption{} \label{sl.pod.7.2.1s.pic} +\end{figure} + +(\textit{T}) Let's assume that for figures $\Phi_1$, $\Phi_2$ and $\Phi_3$ it holds that $\Phi_1\sim \Phi_2$ and $\Phi_2\sim \Phi_3$ (Figure \ref{sl.pod.7.2.1t.pic}). Prove that then also $\Phi_1\sim \Phi_3$ holds. By definition, there exist similarity transformations $f$ and $g$, such that $f:\hspace*{1mm}\Phi_1\rightarrow \Phi_2$ and $g:\hspace*{1mm}\Phi_2\rightarrow \Phi_3$ hold. But by \ref{TransPodGrupa} the composition of mapping $g\circ f:\hspace*{1mm}\Phi_1\rightarrow \Phi_3$ + is also a similarity transformation, therefore $\Phi_1\sim \Phi_3$ holds. +\kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.1t.pic} +\caption{} \label{sl.pod.7.2.1t.pic} +\end{figure} + + + +Because the relation of similarity between figures is symmetric, in the case of $\Phi\sim \Phi'$ we will say that figures $\Phi$ and $\Phi'$ are similar. + +In the following we will specifically consider the similarity between triangles. Let's assume that triangles $ABC$ and $A'B'C'$ are similar, i.e. $\triangle ABC\sim\triangle A'B'C'$. This means that there exists a similarity transformation that maps triangle $ABC$ to triangle $A'B'C'$. In the case of triangles (and also polygons) we will additionally require that the vertices are mapped in order by the similarity transformation, i.e. $f:\hspace*{1mm}A,B,C\mapsto A',B',C'$. +Since a similarity transformation maps lines to lines and angles to angles (\ref{TransPodKol}), sides $AB$, $BC$ and $CA$ of triangle $ABC$ are mapped to sides $A'B'$, $B'C'$ and $C'A'$, and the internal angles $BAC$, $ABC$ and $ACB$ of triangle $ABC$ are mapped to the internal angles $B'A'C'$, $A'B'C'$ and $A'C'B'$ of triangle $A'B'C'$. For pairs of elements in this mapping we will say that they are \pojem{corresponding} or \pojem{congruent}. + +According to the statement \ref{TransPodOhranjajoRazm} similarity transformations preserve the ratio of distances, which means that the corresponding sides are proportional, that is\footnote{By using the similarity of the isosceles right triangles with the help of the appropriate ratio, \index{Tales}\textit{Tales from Miletus} (7th-6th century BC) calculated the height of the pyramid of Cheops.} (Figure \ref{sl.pod.7.2.2.pic}): + \begin{eqnarray} \label{eqnPodTrik1} + \triangle ABC\sim\triangle A'B'C'\hspace*{1mm} \Rightarrow \hspace*{1mm} \frac{A'B'}{AB}=\frac{A'C'}{AC}=\frac{B'C'}{BC}=k, + \end{eqnarray} +where $k$ is the coefficient of similarity. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.2.pic} +\caption{} \label{sl.pod.7.2.2.pic} +\end{figure} + +According to the statement \ref{TransPodOhranjaKote} similarity transformations preserve angles, that is (Figure \ref{sl.pod.7.2.2.pic}): + \begin{eqnarray} \label{eqnPodTrik2} + \triangle ABC\sim\triangle A'B'C'\hspace*{1mm} \Rightarrow \hspace*{1mm} + \left\{ + \begin{array}{l} + \angle B'A'C'\cong\angle BAC; \\ + \angle A'B'C'\cong\angle ABC; \\ + \angle A'C'B'\cong\angle ACB. + \end{array} + \right. + \end{eqnarray} + + + + From the fact that the similarity transformation preserves angles (statement \ref{TransPodOhranjaKote}) and the ratio of distances (statement \ref{TransPodOhranjajoRazm}), it also follows that the heights, centroids, ... of one triangle are mapped into the heights, centroids, ... of the other triangle, while the ratio of the corresponding elements is preserved. + +In the case of similarity of triangles, we have the same problem as with congruence - it is not always easy to prove the similarity of triangles directly by definition. So, analogously to congruence, we also get the \index{statement!about similarity of triangles}\pojem{statements about similarity of triangles}. +We have already seen that from the similarity of two triangles $ABC$ and $A'B'C'$ we get the ratio of the corresponding sides \ref{eqnPodTrik1} or the congruence of the corresponding angles \ref{eqnPodTrik2}. The following statements speak about what conditions are sufficient for the triangles to be similar. + +\bizrek \label{PodTrikSKS} +Triangles $ABC$ and $A'B'C'$ are similar +if two pairs of the sides of the triangles are proportional and the pair of included angles between the sides is congruent, i.e.: + \begin{eqnarray*} + \left. + \begin{array}{l} + \angle B'A'C'\cong \angle BAC\\ + \frac{A'B'}{AB}=\frac{A'C'}{AC} + \end{array} + \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC\sim\triangle A'B'C' + \end{eqnarray*} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.3.pic} +\caption{} \label{sl.pod.7.2.3.pic} +\end{figure} + + + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.3.pic}) +Let it be: + $$k=\frac{A'B'}{AB}=\frac{A'C'}{AC}.$$ +Let $B''$ and $C''$ be such points on the line segments +$A'B'$ and $A'C'$, that $A'B''\cong AB$ and +$A'C''\cong AC$. Triangles + $A'B''C''$ and $ABC$ are similar (\textit{SAS} \ref{SKS}), therefore there exists an isometry $\mathcal{I}$, which maps triangle $ABC$ to triangle +$A'B''C''$. Because + $$\frac{A'B'}{A'B''}=\frac{A'B'}{AB}=k\hspace*{1mm}\textrm{ and }\hspace*{1mm}\frac{A'C'}{A'C''}=\frac{A'C'}{AC}=k,$$ or (because points $B''$ and $C''$ lie on the line segments +$A'B'$ and $A'C'$) $\overrightarrow{A'B'}=k\cdot \overrightarrow{A'B''}$ and $\overrightarrow{A'C'}=k\cdot \overrightarrow{A'C''}$, central stretch $h_{A',k}$ maps points $A'$, $B''$ and $C''$ to points $A'$, $B'$ and $C'$ or triangle $A'B''C''$ to triangle $A'B'C'$. Therefore the composition $f=h_{A',k}\circ \mathcal{I}$, which is a similarity transformation, maps triangle $ABC$ to triangle $A'B'C'$, which means that $\triangle ABC\sim\triangle A'B'C'$. + \kdokaz + +\bizrek \label{PodTrikKKK} + Triangles $ABC$ and $A'B'C'$ are similar + if two pairs of the angles of the triangles are congruent, i.e.: + \begin{eqnarray*} + \left. + \begin{array}{l} + \angle B'A'C'\cong\angle BAC\\ + \angle A'B'C'\cong\angle ABC + \end{array} + \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC\sim\triangle A'B'C' + \end{eqnarray*} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.4.pic} +\caption{} \label{sl.pod.7.2.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.4.pic}) +Let's mark: + $$k=\frac{A'B'}{AB}.$$ +Let $B''$ and $C''$ be such points on the line segments +$A'B'$ and $A'C'$, that $A'B''\cong AB$ and +$A'C''\cong AC$ holds. Then $\frac{A'B'}{A'B''}=k$ as well. Triangles + $A'B''C''$ and $ABC$ are similar (by the \textit{SAS} theorem \ref{SKS}), so there exists an isometry $\mathcal{I}$, that maps triangle $ABC$ to triangle +$A'B''C''$. From this similarity follows that $\angle A'B''C''\cong\angle ABC$ as well. Because $\angle A'B'C'\cong\angle ABC$ by assumption, it is also true that $\angle A'B''C''\cong\angle A'B'C'$. By \ref{KotiTransverzala} theorem $B''C''\parallel B'C'$. By Tales' \ref{TalesovIzrek} theorem +$$\frac{A'C'}{A'C''}=\frac{A'B'}{A'B''}=k.$$ Because $B''$ and $C''$ are points on the line segments +$A'B'$ and $A'C'$, it is also true that $\overrightarrow{A'B'}=k\cdot \overrightarrow{A'B''}$ and $\overrightarrow{A'C'}=k\cdot \overrightarrow{A'C''}$. This means that the central stretch $h_{A',k}$ maps points $A'$, $B''$ and $C''$ to points $A'$, $B'$ and $C'$ or triangle $A'B''C''$ to triangle $A'B'C'$. The composition $f=h_{A',k}\circ \mathcal{I}$, which is a similarity transformation, therefore maps triangle $ABC$ to triangle $A'B'C'$, so $\triangle ABC\sim\triangle A'B'C'$. + \kdokaz + +\bizrek \label{PodTrikSSS} +Triangles $ABC$ and $A'B'C'$ are similar +if three pairs of the sides of the triangles are proportional, i.e. +\begin{eqnarray*} +\frac{A'B'}{AB}=\frac{A'C'}{AC}=\frac{B'C'}{BC} +\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC\sim\triangle A'B'C' +\end{eqnarray*} +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.5.pic} +\caption{} \label{sl.pod.7.2.5.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.5.pic}) +Let's mark: +$$k=\frac{A'B'}{AB}=\frac{A'C'}{AC}=\frac{B'C'}{BC}.$$ +Let $B''$ and $C''$ be such points of the line segments +$A'B'$ and $A'C'$, that $A'B''\cong AB$ and +$A'C''\cong AC$ hold. Then it is also true that $$\frac{A'B'}{A'B''}=\frac{A'C'}{A'C''}=k.$$ +By the converse of Tales' theorem \ref{TalesovIzrekObr} it follows that $B''C''\parallel B'C'$ and $$\frac{B'C'}{B''C''}=\frac{A'B'}{A'B''}.$$ +Therefore $$\frac{B'C'}{B''C''}=\frac{A'B'}{A'B''}=\frac{A'C'}{A'C''}=k=\frac{B'C'}{BC},$$ +which means that $B''C''\cong BC$ holds. From this it follows that +the triangles + $A'B''C''$ and $ABC$ are congruent (\textit{SSS} theorem \ref{SSS}), so there exists an isometry $\mathcal{I}$, which maps the triangle $ABC$ to the triangle +$A'B''C''$. Similarly (as in the proof of the previous two theorems) the central stretch $h_{A',k}$ maps the points $A'$, $B''$ and $C''$ to the points $A'$, $B'$ and $C'$ or the triangle $A'B''C''$ to the triangle $A'B'C'$. In the end, the composition $f=h_{A',k}\circ \mathcal{I}$, which is a similarity transformation, maps the triangle $ABC$ to the triangle $A'B'C'$, so $\triangle ABC\sim\triangle A'B'C'$. +\kdokaz + +The next theorem will be given without a proof (Figure \ref{sl.pod.7.2.6.pic}). + +\bizrek \label{PodTrikSSK} +Triangles $ABC$ and $A'B'C'$ are similar +if two pairs of the sides of the triangles are proportional and the pair of the angles opposite to the longer sides is congruent. +\eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.6.pic} +\caption{} \label{sl.pod.7.2.6.pic} +\end{figure} + +We will continue by using theorems about similarity of triangles. + + + + + + \bzgled + Let $k$ be the circumcircle of a triangle $ABC$ and $B'$ and $C'$ + the foots of the perpendiculars from the vertices $B$ and $C$ on the tangent of the circle $k$ in the vertex $A$. + Prove that the altitude $AD$ of this triangle is the geometric mean of the line segments $BB'$ and $CC'$, i.e.: + $$|AD|=\sqrt{|BB'|\cdot |CC'|}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.7.pic} +\caption{} \label{sl.pod.7.2.7.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.7.pic}) + +Since $BAB'$ is congruent to +the angle subtended by the chord $AB$ at +the tangent of the circle $k$ in the vertex $A$ or $\angle BAB'\cong\angle ACD$ (theorem \ref{ObodKotTang}). Since also +$\angle AB'B\cong\angle CDA=90^0$, we have +$\triangle AB'B\sim\triangle CDA$ (theorem\ref{PodTrikKKK}). From this it follows: +\begin{eqnarray} \label{eqnPodTrikZgl1a} +AD:BB'=AC:BA. +\end{eqnarray} + +In the same way we prove $\triangle AC'C\sim\triangle BDA$ or +\begin{eqnarray} \label{eqnPodTrikZgl1b} + CC':AD=AC:BA. +\end{eqnarray} + From relations \ref{eqnPodTrikZgl1a} and \ref{eqnPodTrikZgl1b} +it follows $AD:BB'=CC':AD$ or $|AD|=\sqrt{|BB'|\cdot |CC'|}$. + \kdokaz + +\bzgled \label{izrekSinusni} +Let $v_a$ be the length of the altitude $AD$, $k(O,R)$ the circumcircle of a triangle $ABC$ +and $b=|AC|$ and $c=|AB|$. Prove that\footnote{The theorem from this example in trigonometry represents the so-called \index{izrek!Sinusni}\textit{sinusoidal theorem}, which was proved by the Arab mathematician \index{al-Biruni, A. R.}\textit{A. R. al-Biruni} (973--1048). Namely, if we insert $\frac{v_a}{c}=\sin \beta$ into the given equality, we get $\frac{b}{\sin\beta}=2R$ and similarly +$\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2R$.}: +$$bc=2R\cdot v_a.$$ +\ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.8a.pic} +\caption{} \label{sl.pod.7.2.8a.pic} +\end{figure} + + \textbf{\textit{Proof.}} We also mark $A_0=\mathcal{S}_O(A)$ (Figure \ref{sl.pod.7.2.8a.pic}). The distance $AA_0$ is the diameter of the circle $k$, so according to Theorem \ref{TalesovIzrKroz2} $\angle ACA_0=90^0=\angle ADB$. The central angles $CBA$ and $CA_0A$ over the arc $AC$ of the circle $k$ are congruent (Theorem \ref{ObodObodKot}), i.e. $\angle DBA=\angle CBA\cong\angle CA_0A$. According to Theorem \ref{PodTrikKKK}, $\triangle ABD\sim\triangle AA_0C$, so: +$$\frac{AB}{AA_0}=\frac{AD}{AC},$$ +i.e. +$bc=2R\cdot v_a$. +\kdokaz + + + + \bzgled + Suppose that the bisector of the interior angle $BA$C of a triangle $ABC$ intersects + its side $BC$ at the point $E$ and the circumcircle of this triangle at the point $N$ + ($N\neq A$). Let's denote $b=|AC|$, $c=|AB|$ and $l_a=|AE|$. Prove that: + $$|AN|=\frac{bc}{l_a}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.8.pic} +\caption{} \label{sl.pod.7.2.8.pic} +\end{figure} + +Suppose that the bisector of the interior angle $BA$C of a triangle $ABC$ intersects +its side $BC$ at the point $E$ and the circumcircle of this triangle at the point $N$ +($N\neq A$). Let's denote $b=|AC|$, $c=|AB|$ and $l_a=|AE|$. Prove that: +$$|AN|=\frac{bc}{l_a}.$$ + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.8.pic} +\caption{} \label{sl.pod.7.2.8.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let us also mark with $AD$ the altitude ($v_a=|AD|$), with $k(O,R)$ the circumscribed circle of the triangle $ABC$ and $M=\mathcal{S}_O(N)$ (Figure \ref{sl.pod.7.2.8.pic}). According to Theorem \ref{TockaN}, the point $N$ lies on the side $BC$'s simetrical line, which means that $NM$ is the diameter of the circle $k$ and $MN\perp BC$. According to Theorem \ref{TalesovIzrKroz2}, $\angle NAM=90^0=\angle EDA$. From $AD,MN\perp BC$ it follows that $AD\parallel MN$, therefore according to Theorem \ref{KotiTransverzala} $\angle DAE\cong\angle MNA$. The triangle $ADE$ and $NAM$ are therefore similar (Theorem \ref{PodTrikKKK}), so: +$$\frac{AD}{AN}=\frac{AE}{NM}.$$ +If we use the previous Theorem \ref{izrekSinusni}, we get: + $$|AN|=\frac{2R\cdot v_a}{l_a}=\frac{bc}{l_a},$$ which had to be proven. \kdokaz + + + +%________________________________________________________________________________ + \poglavje{The Theorems of Ceva and Menelaus} \label{odd7MenelCeva} + + + In this section we will discuss the so-called \index{theorem!theorem of the double}\emph{theorem of the double}. + + + + + \btheorem \label{izrekCeva}\index{theorem!Ceva's} + (Ceva\footnote{\index{Ceva, G.} \textit{G. Ceva} (1648--1734), Italian mathematician, who + proved this theorem in 1678.}) + Let $P$, $Q$ and $R$ be points lying on the lines containing the sides $BC$, $CA$ and $AB$ + of a triangle $ABC$. Then the lines $AP$, $BQ$ and $CR$ belong + to the same family of lines if and only if: + \begin{eqnarray}\label{formulaCeva} + \frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=1. + \end{eqnarray} + \etheorem + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.5.1.pic} +\caption{} \label{sl.pod.7.5.1.pic} +\end{figure} + +%slikaNova1-3-1 +%\includegraphics[width=100mm]{slikaNova1-3-1.pdf} + +\textbf{\textit{Proof.}} +($\Rightarrow$) We assume that the lines $AP$, $BQ$ and $CR$ +belong to the same elliptic conic, that is, they intersect in a +point $S$ (we leave it to the reader to prove the statement in +the case when the lines are parallel). We denote the line that +contains point $A$ and is parallel to line $BC$ with $p$, and its +intersections with lines $BQ$ and $CR$ with $K$ and $L$ (Figure +\ref{sl.pod.7.5.1.pic}). Now from Tales' theorem \ref{TalesovIzrek} it follows that + $\frac{\overrightarrow{BP}}{\overrightarrow{KA}}= + \frac{\overrightarrow{SP}}{\overrightarrow{SA}}= + \frac{\overrightarrow{PC}}{\overrightarrow{AL}}.$ +From this we obtain: + \begin{eqnarray*} + \frac{\overrightarrow{BP}}{\overrightarrow{PC}}= + \frac{\overrightarrow{KA}}{\overrightarrow{AL}}. + \end{eqnarray*} + Similarly, we also have: + \begin{eqnarray*} + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}= + -\frac{\overrightarrow{QC}}{\overrightarrow{QA}}= + -\frac{\overrightarrow{CB}}{\overrightarrow{AK}}= + \frac{\overrightarrow{CB}}{\overrightarrow{KA}},\\ + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= + -\frac{\overrightarrow{RA}}{\overrightarrow{RB}}= + -\frac{\overrightarrow{AL}}{\overrightarrow{BC}}= + \frac{\overrightarrow{AL}}{\overrightarrow{CB}}, + \end{eqnarray*} + which means that + $$\frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= + \frac{\overrightarrow{KA}}{\overrightarrow{AL}}\cdot + \frac{\overrightarrow{CB}}{\overrightarrow{KA}}\cdot + \frac{\overrightarrow{AL}}{\overrightarrow{CB}}=1.$$ + ($\Leftarrow$) Now we assume that relation + (\ref{formulaCeva}) holds and that + $S$ is the intersection of lines $BQ$ and $CR$ (the case when $BQ$ and $CR$ are parallel, + we again leave it to the reader). If we now denote with $P'$ the intersection of line $AS$ + with line $BC$, from the first part of the proof it follows (Figure \ref{sl.pod.7.5.1.pic}): + $$\frac{\overrightarrow{BP'}}{\overrightarrow{P'C}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=1.$$ + By the assumption (\ref{formulaCeva}) we then have + $\frac{\overrightarrow{BP}}{\overrightarrow{PC}}= + \frac{\overrightarrow{BP'}}{\overrightarrow{P'C}}$ or $P=P'$ (statement \ref{izrekEnaDelitevDaljiceVekt}). + Therefore, lines $AP$, $BQ$, $CR$ intersect in point $S$. + \kdokaz + +We observe that the statement is true in the case when the lines $AP$, $BQ$ and $CR$ +intersect in one point (elliptical cone), as well as in the case when these lines +are parallel (parabolic cone). It seems that we could more easily prove +the statement if we introduced +points at infinity and thus generalized the concept of (elliptical) cone +also to the case of parallel lines. These are the ideas that led to the development of so-called \index{geometrija!projektivna}\pojem{projective geometry}. + + + + + \bizrek \index{izrek!Menelajev}\label{izrekMenelaj} + (Menelaj\footnote{\index{Menelaj} \textit{The ancient Greek mathematician Menelaus of Alexandria} (1st century) + proved this statement in his work \textit{Spharica}, namely + in the case of spherical triangles. He mentions the statement in the plane + as already known. Since the previous documents about this are not preserved, we call + the statement Menelaus's statement. Because of its similarity, + Menelaus's and Ceva's statements are also called \textit{theorems of the twins}. The 1500-year period that + separates these + two discoveries is testimony to how + geometry gradually developed during this + period.}) + Let $P$, $Q$ and $R$ be points lying on the lines containing the sides $BC$, $CA$ and $AB$ + of a triangle $ABC$. Then the points $P$, $Q$ and $R$ are collinear if and only if: + \begin{eqnarray}\label{formulaMenelaj} + \frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=-1. + \end{eqnarray} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.5.2.pic} +\caption{} \label{sl.pod.7.5.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} +($\Rightarrow$) Let $P$, $Q$ and $R$ be points of a line $l$. +We denote with $A'$, $B'$ and $C'$ the intersection points of the perpendiculars from the vertices +$A$, $B$ and $C$ of the triangle on the line $l$ (Figure \ref{sl.pod.7.5.2.pic}). +By using Tales' theorem (\ref{TalesovIzrek}) we get: +\begin{eqnarray*} + \frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= + -\frac{\overrightarrow{PB}}{\overrightarrow{PC}}\cdot + \frac{\overrightarrow{QC}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{RA}}{\overrightarrow{RB}}= + -\frac{\overrightarrow{BB'}}{\overrightarrow{CC'}}\cdot + \frac{\overrightarrow{CC'}}{\overrightarrow{AA'}}\cdot + \frac{\overrightarrow{AA'}}{\overrightarrow{BB'}}= + -1. + \end{eqnarray*} + ($\Leftarrow$) Let now the relation (\ref{formulaMenelaj}) be satisfied. With $P'$ + we denote the intersection of the lines $QR$ and $BC$. If the lines $QR$ and $BC$ + were parallel, from this by Tales' theorem it would follow + $\frac{\overrightarrow{CQ}}{\overrightarrow{QA}}= + \frac{\overrightarrow{RB}}{\overrightarrow{AR}}$ + and then from (\ref{formulaMenelaj}) also + $\frac{\overrightarrow{BP}}{\overrightarrow{PC}}=-1$ or + $\overrightarrow{BP}=\overrightarrow{CP}$, which + is not possible. Let therefore $P'=QR\cap BC$. Then the points + $P'$, $Q$, and $R$ are collinear. By the proven in the first part + of the theorem + we have: + \begin{eqnarray*} + \frac{\overrightarrow{BP'}}{\overrightarrow{P'C}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=-1. + \end{eqnarray*} + From this and the assumed relations (\ref{formulaMenelaj}) it follows + $\frac{\overrightarrow{BP}}{\overrightarrow{PC}}= + \frac{\overrightarrow{BP'}}{\overrightarrow{P'C}}$ or $P=P'$ (theorem \ref{izrekEnaDelitevDaljiceVekt}), + which means that the points $P$, $Q$ and $R$ are collinear. + \kdokaz + + +We continue with the use of two proven theorems. + +\bizrek + The lines, joining the vertices of a triangle to the tangent points of the incircle, intersect at one point (so-called \index{točka!Gergonova}\pojem{Gergonne\footnote{To trditev je dokazal \index{Gergonne, J. D.}\textit{J. D. Gergonne} (1771--1859), francoski matematik. V 19. stoletju je bila posvečena posebna + pozornost metričnim lastnostim trikotnika, tako so bile odkrite tudi druge karakteristične točke trikotnika (glej naslednja zgleda).} point} \color{blue} of this triangle). + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.5.3.pic} +\caption{} \label{sl.pod.7.5.3.pic} +\end{figure} + + + \textbf{\textit{Proof.}} +Let $P$, $Q$ and $R$ be the points in which the inscribed circle of the triangle $ABC$ (see izrek \ref{SredVcrtaneKrozn}) touches its +sides $BC$, $CA$ and $AB$ (Figure \ref{sl.pod.7.5.3.pic}). We prove that the lines $AP$, $BQ$ and $CR$ intersect at one point. By izrek \ref{TangOdsek} the corresponding tangent lines are: $BP\cong BR$, $CP\cong CQ$ and +$AQ\cong AR$. Because $\mathcal{B}(B,P,C)$, $\mathcal{B}(C,Q,A)$ and $\mathcal{B}(A,R,B)$, we have $\frac{\overrightarrow{BP}}{\overrightarrow{PC}} +\cdot \frac{\overrightarrow{CQ}}{\overrightarrow{QA}} +\cdot \frac{\overrightarrow{AR}}{\overrightarrow{RB}} >0$. Therefore: +$$\frac{\overrightarrow{BP}}{\overrightarrow{PC}} +\cdot \frac{\overrightarrow{CQ}}{\overrightarrow{QA}} +\cdot \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= +\frac{|BP|}{|PC|}\cdot\frac{|CQ|}{|QA|}\cdot\frac{|AR|}{|RB|}=1.$$ + By Ceva's izrek \ref{izrekCeva} the lines $AP$, $BQ$ and $CR$ belong to one family. Because of Pash's axiom \ref{PaschIzrek} the lines $AP$ and $BQ$ intersect, it is an elliptic family, which means that the lines $AP$, $BQ$ and $CR$ intersect at one point. + \kdokaz + +\bzgled + Prove that the lines, joining the vertices of a triangle to the tangent points of the excircles to the opposite sides, intersect at one point (so-called \index{točka!Nagelova}\pojem{Nagel\footnote{\index{Nagel, C. H.}\textit{C. H. Nagel} (1803--1882), German mathematician, who published this theorem in 1836. Each of the lines in this theorem divides the area of the triangle into two equal parts, so the theorem is also called \index{izrek!o polobsegu trikotnika}\pojem{the statement about the half-perimeter of the triangle}. + point} \color{green1} of this triangle). + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.5.6.pic} +\caption{} \label{sl.pod.7.5.6.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + We use the notation from the big task (\ref{velikaNaloga}). We prove that the lines $AP_a$, $BQ_b$ and $CR_c$ intersect at one point (Figure \ref{sl.pod.7.5.6.pic}). If we use the fact $\frac{\overrightarrow{BP_a}}{\overrightarrow{P_aC}} +\cdot \frac{\overrightarrow{CQ_b}}{\overrightarrow{Q_bA}} +\cdot \frac{\overrightarrow{AR_c}}{\overrightarrow{R_cB}} >0$ and the relations from the big task (\ref{velikaNaloga}), we get: +\begin{eqnarray*} +\frac{\overrightarrow{BP_a}}{\overrightarrow{P_aC}} +\cdot \frac{\overrightarrow{CQ_b}}{\overrightarrow{Q_bA}} +\cdot \frac{\overrightarrow{AR_c}}{\overrightarrow{R_cB}}&=& +\frac{|BP_a|}{|P_aC|}\cdot\frac{|CQ_b|}{|Q_bA|}\cdot\frac{|AR_c|}{|R_cB|}=\\&=& +\frac{s-c}{s-b}\cdot\frac{s-a}{s-c}\cdot\frac{s-b}{s-a}= +1 +\end{eqnarray*} +By Ceva's theorem \ref{izrekCeva} the lines $AP_a$, $BQ_b$ and $CR_c$ belong to one pencil. Since, as a consequence of Pash's axiom \ref{PaschIzrek}, the lines $AP_a$ and $BQ_b$ intersect, it is an elliptic pencil, which means that the lines $AP_a$, $BQ_b$ and $CR_c$ intersect at one point. + \kdokaz + +\bzgled + Prove that the lines, joining the vertices of a triangle to the + points dividing opposite sides in the ratio of squares of adjacent sides, + intersect at one point (so-called \index{točka!Lemoinova} + \pojem{Lemoine\footnote{\index{Lemoine, E. M. H.}\textit{E. M. H. Lemoine} (1751--1816), French mathematician.} point} \color{green1} of this triangle). + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.5.5.pic} +\caption{} \label{sl.pod.7.5.5.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Let $X$, $Y$ and $Z$ be points on sides $BC$, $AC$ and $AB$ of triangle $ABC$, such that $\frac{|BX|}{|XC|}=\frac{|BA|^2}{|AC|^2}$, $\frac{|CY|}{|YA|}=\frac{|CB|^2}{|BA|^2}$ and $\frac{|AZ|}{|ZB|}=\frac{|AC|^2}{|CB|^2}$ + (Figure \ref{sl.pod.7.5.5.pic}). + Because $\frac{\overrightarrow{BX}}{\overrightarrow{XC}} +\cdot \frac{\overrightarrow{CY}}{\overrightarrow{YA}} +\cdot \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}} >0$, we have: +\begin{eqnarray*} +\frac{\overrightarrow{BX}}{\overrightarrow{XC}} +\cdot \frac{\overrightarrow{CY}}{\overrightarrow{YA}} +\cdot \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}}&=& +\frac{|BX|}{|XC|}\cdot\frac{|CY|}{|YA|}\cdot\frac{|AZ|}{|ZB|}=\\ +&=& +\frac{|BA|^2}{|AC|^2}\cdot\frac{|CB|^2}{|BA|^2}\cdot\frac{|AC|^2}{|CB|^2}= +1. +\end{eqnarray*} + By Ceva's theorem \ref{izrekCeva} lines $AX$, $BY$ and $CZ$ belong to one pencil. Because of Pash's axiom \ref{PaschIzrek} lines $AX$ and $BY$ intersect, it is an elliptic pencil, which means that lines $AX$, $BY$ and $CZ$ intersect in one point. + \kdokaz + +\bzgled + Let $Z$ and $Y$ be points of the sides $AB$ and $AC$ of a triangle + $ABC$ such that $AZ:ZB=1:3$ and $AY:YC=1:2$, and $X$ the point, + in which the line $YZ$ intersects the line containing the side $BC$ of this triangle. Calculate + $\overrightarrow{BX}:\overrightarrow{XC}$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.5.4.pic} +\caption{} \label{sl.pod.7.5.4.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + (Figure \ref{sl.pod.7.5.4.pic}) + +Since the points $X$, $Y$ and $Z$ are collinear, by Menelaus's theorem \ref{izrekMenelaj} it follows that: +$$\frac{\overrightarrow{BX}}{\overrightarrow{XC}} +\cdot\frac{\overrightarrow{CY}}{\overrightarrow{YA}} +\cdot\frac{\overrightarrow{AZ}}{\overrightarrow{ZB}}=-1.$$ +If we use the given conditions, we get: +$$\frac{\overrightarrow{BX}}{\overrightarrow{XC}} +\cdot\frac{2}{1} +\cdot\frac{1}{3}=-1$$ +or $\overrightarrow{BX}:\overrightarrow{XC}=-3:2$. +\kdokaz + + + + \bzgled + Prove that the tangents of the circumcircle of an arbitrary scalene triangle $ABC$ at its + vertices $A$, $B$ and $C$ intersect the lines containing the opposite sides of this triangle + at three collinear points\footnote{This is a special case of Pascal's theorem \ref{izrekPascalEvkl} (section \ref{odd7PappusPascal}). \index{Pascal, B.} \textit{B. Pascal} (1623--1662), French mathematician and philosopher.}. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.2a.pic} +\caption{} \label{sl.pod.7.12.2a.pic} +\end{figure} + +\textbf{\textit{Solution.}} + Let $X$, $Y$ and $Z$ be the points of intersection + of the tangent to the circle $k$ of the triangle $ABC$ in the vertices $A$, $B$ and $C$ + with the sides $BC$, $AC$ and $AB$ of this + triangle (Figure \ref{sl.pod.7.12.2a.pic}). We prove that $X$, $Y$ and $Z$ are collinear points. By \ref{ObodKotTang} we have $\angle XAB\cong\angle ACB$. Because the triangle $AXB$ and $CXA$ have a common internal angle at the vertex $X$, we have $\triangle AXB\sim \triangle CXA$ (\ref{PodTrikKKK}), so: + $$\frac{|AX|}{|CX|}=\frac{|AB|}{|AC|}=\frac{|BX|}{|AX|}.$$ +From $\frac{|AX|}{|CX|}=\frac{|BX|}{|AX|}$ it follows that $|XB|\cdot |XC|=|XA|^2$. If we divide this relation by $|XC|^2$, we get $\frac{|BX|}{|CX|}=\frac{|AX|^2}{|CX|^2}$. Since $\frac{|AX|}{|CX|}=\frac{|AB|}{|AC|}$, we finally get: + \begin{eqnarray*} +&& \frac{|BX|}{|XC|}=\frac{|BA|^2}{|AC|^2} + \end{eqnarray*} + and similarly + \begin{eqnarray*} +&& \frac{|CY|}{|YA|}=\frac{|CB|^2}{|BA|^2},\\ +&& \frac{|AZ|}{|ZB|}=\frac{|AC|^2}{|CB|^2}. + \end{eqnarray*} + Because the points $X$, $Y$ and $Z$ do not lie on the sides $BC$, $AC$ and $AB$, we have $\frac{\overrightarrow{BX}}{\overrightarrow{XC}} +\cdot \frac{\overrightarrow{CY}}{\overrightarrow{YA}} +\cdot \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}} <0$, so: +\begin{eqnarray*} +\frac{\overrightarrow{BX}}{\overrightarrow{XC}} +\cdot \frac{\overrightarrow{CY}}{\overrightarrow{YA}} +\cdot \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}}&=& +-\frac{|BX|}{|XC|}\cdot\frac{|CY|}{|YA|}\cdot\frac{|AZ|}{|ZB|}=\\ +&=& +-\frac{|BA|^2}{|AC|^2}\cdot\frac{|CB|^2}{|BA|^2}\cdot\frac{|AC|^2}{|CB|^2}= +-1. + \end{eqnarray*} + By Menelaus' theorem \ref{izrekMenelaj}, the points $X$, $Y$ and $Z$ are collinear. + \kdokaz + +The next property of quadrilaterals is a continuation + of examples + \ref{TetivniVcrtana} and \ref{TetivniVisinska}. Namely, we are dealing with the following problem. + Given is a quadrilateral $ABCD$ with the characteristic points of the triangles $BCD$, $ACD$, $ABD$ and $ABC$ as vertices. What does the quadrilateral represent, that has the centroid of the triangle $BCD$, $ACD$, $ABD$ and $ABC$ as vertices? The answer is trivial if the centroid of the circumscribed circle is given, since in that case $ABCD$ is a quadrilateral with one point as the centroid of the circumscribed circles and the sought quadrilateral does not exist. The cases of the centroids of the circumscribed circles and the altitude points are treated in the aforementioned examples \ref{TetivniVcrtana} and \ref{TetivniVisinska}. The next example gives the answer for the case of the centroids of the triangles, in the case of an arbitrary quadrilateral. + + + + + \bzgled \label{TetivniTezisce} + Let $ABCD$ be an arbitrary quadrilateral and + $T_A$ the centroid of the triangle $BCD$, + $T_B$ the centroid of the triangle $ACD$, + $T_C$ the centroid of the triangle $ABD$ and + $T_D$ the centroid of the triangle $ABC$. + Prove that:\\ + a) a) the lines $AT_A$, $BT_B$, $CT_C$ and $DT_D$ intersect at one point which is the common centroid + of the quadrilaterals $ABCD$ and $T_AT_BT_CT_D$,\\ + b) the quadrilateral $T_AT_BT_CT_D$ is similar to the quadrilateral $ABCD$ with the coefficient + of similarity equal to $\frac{1}{3}$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.9.pic} +\caption{} \label{sl.pod.7.2.9.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.9.pic}) + +The lines $AT_A$, $BT_B$, $CT_C$ and $DT_D$ intersect at one point which is the common centroid of the quadrilaterals $ABCD$ and $T_AT_BT_CT_D$. The quadrilateral $T_AT_BT_CT_D$ is similar to the quadrilateral $ABCD$ with the coefficient of similarity equal to $\frac{1}{3}$. + +\textit{a)} +We mark with $P$, $K$, $Q$, $L$, $M$ and $N$ the centers of the lines $AB$, $BC$, $CD$, $DA$, $AC$ and $BD$. +Since the point $T_A$ is the center of gravity of the triangle $BCD$, the intersection of its medians $BQ$, $CN$ and $DK$, it also holds that $QT_A:T_AB=1:2$ (\ref{tezisce}) or $\overrightarrow{QT_A}=\frac{1}{3}\overrightarrow{QB}$. Similarly, from the triangle $ACD$ we get that $\overrightarrow{QT_B}=\frac{1}{3}\overrightarrow{QA}$. From the last two relations and \ref{vektVektorskiProstor} we obtain: $$\overrightarrow{T_BT_A}=\overrightarrow{T_BQ}+\overrightarrow{QT_A} +=\frac{1}{3}\overrightarrow{AQ}+\frac{1}{3}\overrightarrow{QB}= +\frac{1}{3}\left(\overrightarrow{AQ}+\overrightarrow{QB} \right)=\frac{1}{3}\overrightarrow{AB}.$$ +Therefore: +\begin{eqnarray} \label{eqnCevaTez1} +\overrightarrow{T_BT_A}=\frac{1}{3}\overrightarrow{AB}, +\end{eqnarray} +so $T_BT_A\parallel AB$. We mark with $T$ the intersection of the lines $AT_A$ and $BT_B$. By Tales' theorem and \ref{eqnCevaTez1} it holds that: +\begin{eqnarray} \label{eqnCevaTez2} +\frac{\overrightarrow{TT_A}}{\overrightarrow{TA}} +=\frac{\overrightarrow{TT_B}}{\overrightarrow{TB}}= +\frac{\overrightarrow{T_AT_B}}{\overrightarrow{AB}}=-\frac{1}{3}. +\end{eqnarray} +The lines $AT_A$ and $BT_B$ therefore intersect in the point $T$, which divides it in the ratio $2:1$. Similarly, the lines $AT_A$ and $CT_C$ or the lines $AT_A$ and $DT_D$ intersect in a point which divides it in the ratio $2:1$, and that is (because of the line $AT_A$) precisely the point $T$. This means that the lines $AT_A$, $BT_B$, $CT_C$ and $DT_D$ intersect in the point $T$. + +We will now prove that point $T$ is the center of mass of the quadrilaterals $ABCD$ and $T_AT_BT_CT_D$. +Since +$$\frac{\overrightarrow{QT_B}}{\overrightarrow{T_BA}}\cdot +\frac{\overrightarrow{AP}}{\overrightarrow{PB}}\cdot +\frac{\overrightarrow{BT_A}}{\overrightarrow{T_AQ}}= +\frac{1}{2}\cdot\frac{1}{1}\cdot\frac{2}{1}=1, +$$ +by Ceva's theorem \ref{izrekCeva} for the triangle $QAB$ the lines $AT_A$, $BT_B$ and $PQ$ intersect in one point - point $T$. Therefore point $T$ lies on the line $PQ$. Similarly, point $T$ lies on the line $KL$, which means that point $T$ is the intersection of the diagonals $PQ$ and $KL$ of the quadrilateral $PKQL$. Since this is a Varignon parallelogram of the quadrilateral $ABCD$, point $T$ is, by theorem \ref{vektVarignon}, the center of mass of the quadrilateral $ABCD$. Since from \ref{eqnCevaTez2} it also follows that: +$$\overrightarrow{TT_A}+\overrightarrow{TT_B}+\overrightarrow{TT_C}+\overrightarrow{TT_D}= +-\frac{1}{3}\cdot\left( +\overrightarrow{TA}+\overrightarrow{TB}+\overrightarrow{TC}+\overrightarrow{TD} +\right)=-\frac{1}{3}\cdot \overrightarrow{0}=\overrightarrow{0},$$ +point $T$ is also the center of mass of the quadrilateral $T_AT_BT_CT_D$. + +\textit{b)} From \ref{eqnCevaTez2} it follows that $\overrightarrow{TT_A}=-\frac{1}{3}\overrightarrow{TA}$, +$\overrightarrow{TT_B}=-\frac{1}{3}\overrightarrow{TB}$, +$\overrightarrow{TT_C}=-\frac{1}{3}\overrightarrow{TC}$ and +$\overrightarrow{TT_D}=-\frac{1}{3}\overrightarrow{TD}$, or +$h_{T,-\frac{1}{3}}:\hspace*{1mm}A,B,C,D\mapsto T_A,T_B,T_C,T_D$, which means that the quadrilateral $T_AT_BT_CT_D$ is similar to the quadrilateral + $ABCD$ with the similarity coefficient $\frac{1}{3}$. +\kdokaz + +%________________________________________________________________________________ + \poglavje{Harmonic Conjugate Points. Apollonius Circle} + \label{odd7Harm} + + +Since it is very important for what follows, we will write theorem \ref{izrekEnaDelitevDaljiceVekt} from section \ref{odd5LinKombVekt} in another form. + +\bizrek \label{HarmCetEnaSamaDelitev} + If $A$ and $B$ are different points on the line $p$ and $\lambda\neq -1$ + an arbitrary real number, then there is exactly one point $L$ on the line $p$ such that: + $$ \frac{\overrightarrow{AL}}{\overrightarrow{LB}}=\lambda.$$ + \eizrek + + +In section \ref{odd5TalesVekt} we found out how to divide the distance $AB$ in the ratio $m:n$ (statement \ref{izrekEnaDelitevDaljice}). We will now use this idea to construct the point $L$ from the previous statement \ref{HarmCetEnaSamaDelitev}. + + + + \bzgled + The points $A$ and $B$ and the real number $\lambda\neq -1$ are given. + Construct a point $L$ on the line $AB$ such that: + $$\frac{\overrightarrow{AL}}{\overrightarrow{LB}}=\lambda.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.1.pic} +\caption{} \label{sl.pod.7.4.1.pic} +\end{figure} + + \textit{\textbf{Solution.}} (Figure \ref{sl.pod.7.4.1.pic}) + +In the case $\lambda=0$ the proof is direct - then $L=A$. Let $\lambda\neq 0$. +Construct +any parallel lines $a$ and $b$ through the points $A$ and $B$. With $P$ and $Q$ we mark such points on the lines $a$ and $b$, +that $P,Q\div AB$ and $|AP|=|\lambda|$ and $|BQ|=1$. We also mark $Q'=\mathcal{S}_B(Q)$. +We will consider two cases. + +\textit{1)} Let $\lambda>0$. It is clear that for the sought point $L$ it must hold that $\mathcal{B}(A,L,B)$. With $L_1$ we mark +the intersection of the lines $AB$ and $PQ$. By Tales' statement \ref{TalesovIzrek} it holds: + $$\frac{\overrightarrow{AL_1}}{\overrightarrow{L_1B}}= +\frac{\overrightarrow{AP}}{\overrightarrow{QB}}=\lambda.$$ + + \textit{2)} If $\lambda<0$, the sought point does not lie on the line $AB$. With $L_1$ we mark +the intersection of the lines $AB$ and $PQ'$. Because $\lambda\neq -1$, this intersection exists. By Tales' statement \ref{TalesovIzrek} it holds: + $$\frac{\overrightarrow{AL_2}}{\overrightarrow{L_2B}}= +\frac{\overrightarrow{AP}}{\overrightarrow{Q'B}} +=-\frac{\overrightarrow{AP}}{\overrightarrow{BQ'}}=-\left(-\lambda\right)=\lambda,$$ which was to be proven. \kdokaz + +From the previous example it is therefore clear that for each of the conditions $\frac{\overrightarrow{AL}}{\overrightarrow{LB}}=\lambda$ ($\lambda>0$) or $\frac{\overrightarrow{AL}}{\overrightarrow{LB}}=\lambda$ ($\lambda<0$ and $\lambda\neq -1$) there is exactly one solution for the point $L$ on the line $AB$. In the first case, the point $L$ lies on the line $AB$, so we say that this is the \index{delitev daljice!notranja}\pojem{inner division} of the line $AB$ in the ratio $\lambda$ ($\lambda>0$). In the second case, the point $L$ does not lie on the line $AB$, so we say that this is the \index{delitev daljice!zunanja}\pojem{outer division} of the line $AB$ in the ratio $|\lambda|$ ($\lambda<0$ and $\lambda\neq -1$). + +We can also say that for the condition $\frac{AL}{LB}=\lambda$ ($\lambda>0$, $\lambda\neq 1$) there are two solutions for the point $L$ on the line - the inner or the outer division. + +If $L_1$ and $L_2$ are the inner and outer division of the line (for the same $\lambda$), it holds: +$$\frac{\overrightarrow{AL_1}}{\overrightarrow{L_1B}}: +\frac{\overrightarrow{AL_2}}{\overrightarrow{L_2B}}=-1.$$ + If we define the expression $\frac{\overrightarrow{AL_1}}{\overrightarrow{L_1B}}: +\frac{\overrightarrow{AL_2}}{\overrightarrow{L_2B}}$ as the \index{dvorazmerje parov točk}\pojem{ratio of pairs of points} $(A,B)$ and $(L_1,L_2)$ and denote +$$d(A,B;L_1,L_2)=\frac{\overrightarrow{AL_1}}{\overrightarrow{L_1B}}: +\frac{\overrightarrow{AL_2}}{\overrightarrow{L_2B}},$$ + we see that for the inner and outer division of the line $AB$ with points $L_1$ and $L_2$ the corresponding ratio is $-1$, or $d(A,B;L_1,L_2)=-1$ +This gives us the idea for a new definition. + +We say that different collinear points $A$, $B$, $C$ and $D$ determine + \index{harmonična četverica točk} + \pojem{harmonično četverico točk} oz. that the pair $(A, B)$, + \pojem{harmonično konjugiran} s parom $(C, D)$, oznaka + $\mathcal{H}(A,B;C,D)$, if: + \begin{eqnarray}\label{formulaHarmEvkl} + \frac{\overrightarrow{AC}}{\overrightarrow{CB}}= + -\frac{\overrightarrow{AD}}{\overrightarrow{DB}}, + \end{eqnarray} or: + \begin{eqnarray*} +d(A,B;C,D)=\frac{\overrightarrow{AC}}{\overrightarrow{CB}}: +\frac{\overrightarrow{AD}}{\overrightarrow{DB}}=-1. + \end{eqnarray*} + +We prove the basic properties of the defined relation. + As we have already mentioned, the points $C$ and $D$, for which $\mathcal{H}(A,B;C,D)$ is true, represent +the internal and external division of the distance $AB$ in some ratio $\lambda$ ($\lambda>0$ and $\lambda\neq 1$). There are therefore an infinite number of such pairs $C$, $D$. But if one of the points $C$ or $D$ is given, the other is uniquely determined. We express this in other words in the following proposition. + + + + \bizrek \label{HarmCetEnaSamaTockaD} + Let $C$ be a point on a line $AB$ different from the points $A$ and $B$ + and also from the midpoint of the line segment $AB$, + then there is exactly one point $D$ such that + $\mathcal{H}(A,B;C,D)$. + \eizrek + + + \textit{\textbf{Proof.}} Let $\frac{\overrightarrow{AC}}{\overrightarrow{CB}}=\lambda$. Because $C$ is different from $A$, $B$ and + the midpoint of the line segment $AB$, $\lambda\neq 0$ and $\lambda\neq 1$. We are looking for a point $D$, for which $\mathcal{H}(A,B;C,D)$ is true, or $d(A,B;C,D)=-1$ or equivalently $\frac{\overrightarrow{AD}}{\overrightarrow{DB}}=-\lambda$. Because $-\lambda\neq -1$, by \ref{HarmCetEnaSamaDelitev} there is only one point $D$, for which this is fulfilled. +\kdokaz + +\bizrek \label{HarmCetEF} + Let $A$, $B$, $C$ and $D$ be four different collinear points + on a line $p$ and $O$ a point not lying on this line. Suppose that + a line that is parallel to the line $OA$ through the point $B$ + intersects the lines $OC$ and $OD$ at the points $E$ and $F$. + Then: + $$\mathcal{H}(A,B;C,D) \hspace*{1mm} \Leftrightarrow \hspace*{1mm} \mathcal{S}_B(E)=F.$$ + \eizrek + + + \textit{\textbf{Proof.}} + (Figure \ref{sl.pod.7.4.2.pic}) + +($\Rightarrow$) If $\mathcal{H}(A,B;C,D)$ is true, by Tales' theorem \ref{TalesovIzrek}: +$$\frac{\overrightarrow{AO}}{\overrightarrow{EB}}= +\frac{\overrightarrow{AC}}{\overrightarrow{CB}}= +-\frac{\overrightarrow{AD}}{\overrightarrow{DB}}= +-\frac{\overrightarrow{AO}}{\overrightarrow{FB}}= +\frac{\overrightarrow{AO}}{\overrightarrow{BF}},$$ + it follows that +$\overrightarrow{EB}=\overrightarrow{BF}$ or $\mathcal{S}_B(E)=F$. + +($\Leftarrow$) Now suppose that $\mathcal{S}_B(E)=F$. From this it follows that $\overrightarrow{EB}=\overrightarrow{BF}$, so (again by Tales' theorem \ref{TalesovIzrek}): +$$\frac{\overrightarrow{AC}}{\overrightarrow{CB}}= +\frac{\overrightarrow{AO}}{\overrightarrow{EB}}= +\frac{\overrightarrow{AO}}{\overrightarrow{BF}}= +-\frac{\overrightarrow{AO}}{\overrightarrow{FB}}= +-\frac{\overrightarrow{AD}}{\overrightarrow{DB}},$$ +so by definition $\mathcal{H}(A,B;C,D)$. +\kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.2.pic} +\caption{} \label{sl.pod.7.4.2.pic} +\end{figure} + +The previous theorem allows us to effectively construct the fourth point in a harmonic set of points. + + + + \bzgled \label{HarmCetEnaSamaTockaDKonstr} + Let $C$ be a point that lies on a line $AB$. Suppose that $C$ is different from the points $A$, + $B$ and the midpoint of the line segment $AB$. Construct a point $D$ such that + $\mathcal{H}(A,B;C,D)$. + \ezgled + +\textit{\textbf{Solution.}} According to the theorem \ref{HarmCetEnaSamaTockaD} there is only one such point $D$, for which $\mathcal{H}(A,B;C,D)$. Now we will construct it. Let $O$ be an arbitrary point, which does not lie on the line $AB$ (Figure \ref{sl.pod.7.4.2.pic}), and $l$ is a parallel to the line $AO$ in the point $B$. With $E$ we mark the intersection of the lines $OC$ and $l$ and $F=\mathcal{S}_B(E)$. The point $D$ is the intersection of the lines $OF$ and $AB$. +By the previous theorem \ref{HarmCetEF} it is $\mathcal{H}(A,B;C,D)$. +\kdokaz + +In a similar way as in the previous example, for given collinear points $A$, $B$ and $D$ (the point $D$ does not lie on the line $AB$) we can plan such a point $C$, that $\mathcal{H}(A,B;C,D)$ is valid. + +It is intuitively clear that from $\mathcal{H}(A,B;C,D)$ it follows $\mathcal{H}(A,B;D,C)$, because if the points $C$ and $D$ divide the line $AB$ in the same ratio, then the same is true for the points $D$ and $C$. It is interesting that from $\mathcal{H}(A,B;C,D)$ it also follows $\mathcal{H}(C,D;A,B)$, which means that if the points $C$ and $D$ divide the line $AB$ in the same ratio, then the points $A$ and $B$ divide the line $CD$ in the same ratio. We prove both properties formally. + +\bizrek +a) $\mathcal{H}(A,B;C,D) \hspace*{1mm} \Rightarrow +\hspace*{1mm} \mathcal{H}(A,B;D,C)$; \\ +\hspace*{22mm}b) $\mathcal{H}(A,B;C,D) +\hspace*{1mm} \Rightarrow +\hspace*{1mm} \mathcal{H}(C,D;A,B)$; +\eizrek + + +\textit{\textbf{Proof.}} + +$$a)\hspace*{1mm}\mathcal{H}(A,B;C,D) +\hspace*{1mm} \Rightarrow + \hspace*{1mm} + \frac{\overrightarrow{AC}}{\overrightarrow{CB}}= -\frac{\overrightarrow{AD}}{\overrightarrow{DB}} +\hspace*{1mm} \Rightarrow + \hspace*{1mm} +\frac{\overrightarrow{AD}}{\overrightarrow{DB}}= -\frac{\overrightarrow{AC}}{\overrightarrow{CB}} +\hspace*{1mm} \Rightarrow + \hspace*{1mm} +\mathcal{H}(A,B;D,C).$$ + + + $$b)\hspace*{1mm}\mathcal{H}(A,B;C,D) +\hspace*{1mm} \Rightarrow + \hspace*{1mm} + \frac{\overrightarrow{AC}}{\overrightarrow{CB}}= -\frac{\overrightarrow{AD}}{\overrightarrow{DB}} +\hspace*{1mm} \Rightarrow + \hspace*{1mm} +\frac{\overrightarrow{CA}}{\overrightarrow{AD}}= -\frac{\overrightarrow{CB}}{\overrightarrow{BD}} +\hspace*{1mm} \Rightarrow + \hspace*{1mm} +\mathcal{H}(C,D;A,B),$$ which had to be proven. \kdokaz + + + + \bizrek \label{izrek 1.2.1} + Let $A$, $B$, $C$ and $D$ be different collinear points. + Then $\mathcal{H}(A,B;C,D)$ if and only if there exists such a quadrilateral $PQRS$, + that: + $$A=PQ\cap RS, \hspace*{2mm}B=QR\cap PS, \hspace*{2mm} C\in PR \hspace*{2mm} + \textrm{and} + \hspace*{2mm} D\in QS.$$ + \eizrek + +\textit{\textbf{Proof.}} ($\Rightarrow$) Let $PQRS$ be a quadrilateral, +such that $A=PQ\cap RS$, $B=QR\cap PS$, $C\in PR$ and $D\in QS$ +(Figure \ref{sl.pod.7.4.7a.pic}). By Menelaus' theorem +\ref{izrekMenelaj} for the triangle $ABP$ and the line $QS$ we get: +$$\frac{\overrightarrow{AD}}{\overrightarrow{DB}}\cdot + \frac{\overrightarrow{BS}}{\overrightarrow{SP}}\cdot + \frac{\overrightarrow{PQ}}{\overrightarrow{QA}}=-1.$$ + Similarly, by Ceva's theorem \ref{izrekCeva} for the same + triangle and the point $R$ we get: + $$\frac{\overrightarrow{AC}}{\overrightarrow{CB}}\cdot + \frac{\overrightarrow{BS}}{\overrightarrow{SP}}\cdot + \frac{\overrightarrow{PQ}}{\overrightarrow{QA}}=1.$$ +From these two relations it follows +$\frac{\overrightarrow{AC}}{\overrightarrow{CB}}= +-\frac{\overrightarrow{AD}}{\overrightarrow{DB}}$ or +$\mathcal{H}(A,B;C,D)$. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.7a.pic} +\caption{} \label{sl.pod.7.4.7a.pic} +\end{figure} + + ($\Leftarrow$) Now, let's assume that + $\mathcal{H}(A,B;C,D)$ holds. Without loss of generality, let the point + $C$ be between the points $A$ and $B$; in the other two cases the proof + proceeds in the same way. Let $P$ be an arbitrary point outside the line $AB$ and $Q$ + an arbitrary point between the points $A$ and $P$. By Pasch's\footnote{\index{Pasch, M.} + \textit{M. Pasch} + (1843--1930), German mathematician, who introduced the relation of order of points in his work + \textit{Predavanja o novejši geometriji} (Lectures on Modern Geometry) from 1882.} \ref{PaschIzrek} + axiom, the line $PC$ + and the line $QB$ intersect in some point $R$ and the line $AR$ and the line $PB$ intersect in some + point $S$, therefore the line $AB$ and the line $QS$ intersect in some point $D_1$ (from $AB\parallel QS$ + it follows that + $C$ is the center of the line segment $AB$, which is not possible due to + $\mathcal{H}(A,B;C,D)$). Now, from the first part of the proof ($\Rightarrow$) + it follows $\mathcal{H}(A,B;C,D_1)$. Since by the assumption we also have + $\mathcal{H}(A,B;C,D)$, by the uniqueness of the fourth point of a harmonic + quadrilateral of points (theorem \ref{HarmCetEnaSamaTockaD}) + it follows $D=D_1$. Therefore $PQRS$ is the desired + quadrilateral. + \kdokaz + +We observe that the previous statement is also valid in the case when $C$ is the center of the line $AB$ and $D$ is a point at infinity (Figure \ref{sl.pod.7.4.7a.pic}). + + The statement on the right side of the equivalence from the previous statement (the existence of the appropriate quadrilateral) could be accepted as the definition of the harmonic quadruple of points in Euclidean geometry. In projective geometry, this definition is even more natural, since in such a definition we do not use metric. + + From the previous statement it also follows that the relation of harmonic quadruple is preserved in the so-called \pojem{central projection} in space. Indeed, if $A$, $B$, $C$ and $D$ are points for which $\mathcal{H}(A,B;C,D)$ holds, and $A'$, $B'$, $C'$ and $D'$ are the central projections of these points, then from the existence of the appropriate quadrilateral for the quadruple $A$, $B$, $C$, $D$ it follows that the appropriate quadrilateral exists also for the quadruple $A'$, $B'$, $C'$, $D'$. The latter is the central projection of the first quadrilateral, since the central projection preserves collinearity. Of course, we should also include in the previous consideration the case when some of the central projections are points at infinity. + + \index{geometrija!projektivna}In this sense, projective geometry can be described as the geometry that deals with objects and properties that are preserved in the central projection. Because "being a harmonic quadruple of points" is one of such properties, it is the subject of study in projective geometry. Even more - in this geometry, this relation is one of the basic concepts (see \cite{Mitrovic})." + +We have already seen that on the line $AB$ there are exactly two points that divide the distance $AB$ in +the ratio $\lambda>0$, $\lambda\neq 1$ (in the case $\lambda= 1$ this is only one point - the center of the line). These points represent +the inner and outer division of the line in this ratio and together with points $A$ and $B$ determine a harmonic quadruple of points. The question arises: What does the set of all +such points $X$ in the plane mean, so that $\frac{AX}{XB}=\lambda$ ($\lambda\neq 1$)? The answer is given by the following theorem. + + + + + + + + + \bizrek \label{ApolonijevaKroznica} + Suppose that $A$ and $B$ are points in the plane and $\lambda>0$, $\lambda\neq 1$ + an arbitrary real number. If $C$ and $D$ are points of the line $AB$ such that: + $$\frac{\overrightarrow{AC}}{\overrightarrow{CB}}= + -\frac{\overrightarrow{AD}}{\overrightarrow{DB}}=\lambda,$$ + i.e. $\mathcal{H}(A,B;C,D)$, then the set of all points $X$ of this plane such that: + $$\frac{AX}{XB}=\lambda,$$ + is a circle with the diameter $CD$ + (so-called Apollonius Circle\footnote{\index{Apolonij} + \textit{Apolonij iz Perge} (3.-- 2. st. pr. n. + š.), starogrški matematik.}). + \index{krožnica!Apolonijeva} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.3.pic} +\caption{} \label{sl.pod.7.4.3.pic} +\end{figure} + + \textit{\textbf{Proof.}} +Let $k$ be a circle with diameter $CD$. Let $X$ be an arbitrary point of this plane and $p$ a parallel line through point $B$ to line $AX$. With $E$ and $F$ we mark the intersections of lines $XC$ and $XD$ +with line $p$ (Figure \ref{sl.pod.7.4.3.pic}). By Theorem \ref{HarmCetEF}, point $B$ is the center +of the line $EF$, because $\mathcal{H}(A,B;C,D)$. It needs to be proven that: +$$\frac{AX}{XB}=\lambda \hspace*{1mm} \Leftrightarrow \hspace*{1mm} X\in k.$$ + +($\Leftarrow$) Let $X\in k$. Then $\angle EXF=\angle CXD=90^0$ and $BE\cong BF\cong BX$ (statement \ref{TalesovIzrKroz2}). From this and from Tales's statement \ref{TalesovIzrek} it follows: +$$\frac{AX}{XB}=\frac{AX}{EB}=\frac{AC}{CB}=\lambda.$$ + +($\Rightarrow$) Let $\frac{AX}{XB}=\lambda$. From this and from Tales's statement \ref{TalesovIzrek} it follows: +$$\frac{AX}{XB}=\lambda=\frac{AC}{CB}=\frac{AX}{EB},$$ +so $XB\cong EB\cong BF$. Therefore $EXF$ +is a right angled triangle (statement \ref{TalesovIzrKroz2}). Then $\angle CXD=\angle EXF=90^0$, which means that the point $X$ lies on the circle $k$ above +the diameter $CD$ (statement \ref{TalesovIzrKroz2}). +\kdokaz + + +For given points $A$ and $B$ and for different values $\lambda\in \mathbb{R}$ ($\lambda>0$, $\lambda\neq 1$) we have different Apollonius's circles (Figure \ref{sl.pod.7.4.4a.pic}). We will mark them with $\mathcal{A}_{AB,\lambda}$, but if we know which distance it is, we can also write it shorter $k_{\lambda}$. In the case $\lambda=1$ we are actually looking for a set of all such points $X$ of this plane, for which: + $\frac{AX}{XB}=1$, i.e. $AX\cong BX$. Then the sought set does not represent a circle, but the symmetry $s_{AB}$ of the distance $AB$. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.4a.pic} +\caption{} \label{sl.pod.7.4.4a.pic} +\end{figure} + + + + \bzgled + Points $A$ and $B$ and a line $l$ in the plane are given. + Construct the point $L$ on the line $l$ such that $LA:LB=5:2$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.4.pic} +\caption{} \label{sl.pod.7.4.4.pic} +\end{figure} + + \textit{\textbf{Solution.}} + (Figure \ref{sl.pod.7.4.4.pic}) + +Let's first draw a point $C$ on the line $AB$, so that $AC:CB=5:2$ + (see example \ref{izrekEnaDelitevDaljice}), and then the fourth point of the harmonic quartet $\mathcal{H}(A,B;C,D)$ (see example \ref{HarmCetEnaSamaTockaDKonstr}). Construct the Apollonius circle $k_{\frac{5}{2}}$ over the diameter $CD$. The point $L$ is then the one for which $L\in k\cap l$. + +By the previous theorem (\ref{ApolonijevaKroznica}) we have: +$$\frac{AL}{LB}=\frac{AC}{CB}=\frac{5}{2}.$$ + +The task has 0, 1 or 2 solutions, depending on the number of intersections of the circle $k$ and the line $l$. +\kdokaz + + + + \bizrek \label{HarmOhranjaVzporProj} + A parallel projection preserves the relation of a harmonic conjugate points, + i.e. if $A$, $B$, $C$, $D$ and $A'$, $B'$, $C'$, $D'$ are two quartets + of collinear points such that $AA'\parallel BB'\parallel CC'\parallel DD'$, then: + $$\mathcal{H}(A,B;C,D)\hspace*{1mm}\Rightarrow\hspace*{1mm} + \mathcal{H}(A',B';C',D').$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.5.pic} +\caption{} \label{sl.pod.7.4.5.pic} +\end{figure} + + \textit{\textbf{Solution.}} + (Figure \ref{sl.pod.7.4.5.pic}) + +Let $\mathcal{H}(A,B;C,D)$ or $\frac{\overrightarrow{AC}}{\overrightarrow{CB}}: +\frac{\overrightarrow{AD}}{\overrightarrow{DB}}=-1$. By Tales' theorem \ref{TalesovIzrek} we have: +$$\frac{\overrightarrow{A'C'}}{\overrightarrow{C'B'}}: +\frac{\overrightarrow{A'D'}}{\overrightarrow{D'B'}}= +\frac{\overrightarrow{AC}}{\overrightarrow{CB}}: +\frac{\overrightarrow{AD}}{\overrightarrow{DB}}=-1,$$ +which means that $\mathcal{H}(A',B';C',D')$ is also true. +\kdokaz + +It is interesting that the central projection also preserves the relation of harmonic points. Therefore, the mentioned relation is the subject of research in projective geometry (see \cite{Mitrovic}). + +\bizrek \label{HarmCetSimKota} + Suppose that a line $BC$ intersects the bisectors of the interior and + the exterior angle at the vertex $A$ of a triangle $ABC$ at points $E$ and $F$. Then: + + a) $BE:CE=BF:CF=AB:AC$, + + b) $\mathcal{H}(B,C;E,F)$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.6.pic} +\caption{} \label{sl.pod.7.4.6.pic} +\end{figure} + + \textit{\textbf{Solution.}} Let $L$ be an arbitrary point for which $\mathcal{B}(C,A,L)$ holds. +We mark with $M$ and $N$ the points in which the parallel to the line $AC$ through the point $B$ intersects in order the simetrals $AE$ and $AF$ of the internal and the external angle (Figure \ref{sl.pod.7.4.6.pic}). + +\textit{a)} By izrek \ref{KotiTransverzala} we have: + \begin{eqnarray*} +\angle BMA &\cong& \angle CAM=\angle CAE\cong\angle BAE=\angle BAM\\ +\angle BNA &\cong& \angle LAN=\angle LAF\cong\angle BAF=\angle BAN +\end{eqnarray*} + +Therefore, $AMB$ and $ANB$ are isosceles triangles with the bases $AM$ and $AN$ (izrek \ref{enakokraki} and it holds: + $$BM\cong BA\cong BN.$$ +From this and from the consequence of Tales' theorem \ref{TalesovIzrekDolzine} (because $AC\parallel NM$) we get: +\begin{eqnarray*} +\frac{BE}{EC}&=& \frac{BM}{AC}=\frac{BA}{AC},\\ +\frac{BF}{FC}&=& \frac{BN}{AC}=\frac{BA}{AC}. +\end{eqnarray*} + +\textit{b)} Because $\mathcal{B}(B,E,C)$ and $\neg\mathcal{B}(B,F,C)$, we have: +\begin{eqnarray*} +\frac{\overrightarrow{BE}}{\overrightarrow{EC}}&=& \frac{BA}{AC},\\ +\frac{\overrightarrow{BF}}{\overrightarrow{FC}}&=& -\frac{BA}{AC}. +\end{eqnarray*} +Therefore: +\begin{eqnarray*} +\frac{\overrightarrow{BE}}{\overrightarrow{EC}}= +-\frac{\overrightarrow{BF}}{\overrightarrow{FC}}, +\end{eqnarray*} +which means that $\mathcal{H}(B,C;E,F)$ holds. +\kdokaz + +We will continue by using the previous theorem. + + \bzgled \label{HarmTrikZgl1} + Construct a triangle $ABC$, with given: $a$, $t_a$, $b:c=2:3$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.8.pic} +\caption{} \label{sl.pod.7.4.8.pic} +\end{figure} + + \textit{\textbf{Solution.}} +Let $ABC$ be a triangle, in which the side $BC\cong a$, the median $AA_1\cong t_a$ and $AC:AB=2:3$ (Figure \ref{sl.pod.7.4.8.pic}). So we can +construct the side $BC$ and its center $A_1$. +From the given conditions, the vertex $A$ lies on the circle +$k(A_1,t_a)$ and the Apollonius circle $\mathcal{A}_{BC,\frac{3}{2}}$, because $AB:AC=3:2$ (statement \ref{ApolonijevaKroznica}). So we get the vertex $A$ as one of the intersections of the circles +$k$ and $\mathcal{A}_{BC,\frac{3}{2}}$. + +Although we do not need this fact in the construction, we mention that +the Apollonius circle $\mathcal{A}_{BC,\frac{3}{2}}$ represents the circle with the diameter $EF$, where $E$ and $F$ are defined as in the statement \ref{HarmCetSimKota}. +\kdokaz + + + + \bzgled + Construct a parallelogram, with the sides congruent to + given line segments $a$ and $b$, and the diagonals in the ratio $3:7$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.9.pic} +\caption{} \label{sl.pod.7.4.9.pic} +\end{figure} + + \textit{\textbf{Solution.}} (Figure \ref{sl.pod.7.4.9.pic}) + +Let $S$ be the intersection of the diagonals of the sought parallelogram $ABCD$, in which $AB\cong a$, $BC\cong b$ and $AC:BD=3:7$. We also mark with $P$ the center of the side $AB$. +Since the diagonals of a +parallelogram are divided in half, the line segments $SA$ and $SB$ are in the ratio $2:3$. This means that we can first construct +the triangle $ASB$, similar to the previous example \ref{HarmTrikZgl1}: +$AB\cong a$, $SP=\frac{1}{2}b$ and $SA:SB=3:7$. + \kdokaz + +\bizrek \label{harmVelNal} + Suppose that $k(S, r)$ is the incircle and $k_a (S_a, r_a)$ the excircle of a triangle $ABC$ + and $P$ and $P_a$ the touching points of these circles with the aide $BC$. + Let $A'$ be the foot of the altitude $AA'$ and $E$ + intersection of the bisectors of the + interior angle at the vertex $A$ and the side $BC$. + If $L$ and $L_a$ are foots of the perpendiculars from the points $S$ and $S_a$ on the line $AA'$, then: + + \hspace*{2mm} (i) $\mathcal{H}(A,E;S,Sa)$ \hspace*{2mm} (ii) + $\mathcal{H}(A,A';L,La)$ \hspace*{2mm} (iii) + $\mathcal{H}(A',E;P,Pa)$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.3.7.pic} +\caption{} \label{sl.pod.7.3.7.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.3.7.pic}) + +\textit{(i)} The lines $CS$ and $CS_a$ are the altitudes of the +triangle $ACE$, therefore by \ref{HarmCetSimKota} we have $\mathcal{H}(A,E;S,S_a)$. + +\textit{(ii)} The points $A$, $A'$, $L$ and $L_a$ +are the orthogonal projections of the points $A$, $E$, $S$ and $S_a$ on the line $AA'$. By \ref{HarmOhranjaVzporProj} we have $\mathcal{H}(A,A';L,La)$. + +\textit{(iii)} Similarly to the previous statement, since +the points $A'$, $E$, $P$ and $P_a$ + are the orthogonal projections of the points $A$, $E$, $S$ and $S_a$ on the line $BC$. +\kdokaz + + + \bzgled Construct a triangle $ABC$, with given: + + \hspace*{4mm} (i) $r$, $a$, $v_a$ \hspace*{5mm} (ii) $v_a$, + $r$, $b-c$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.10.pic} +\caption{} \label{sl.pod.7.4.10.pic} +\end{figure} + +\textbf{\textit{Solution.}} +We use the notation from the big exercise \ref{velikaNaloga} and the statement \ref{harmVelNal} (Figure \ref{sl.pod.7.4.10.pic}). In both cases \textit{(i)} and \textit{(ii)} we can use the fact $\mathcal{H}(A,A';L,La)$ from the statement \ref{harmVelNal}, because from $AA'\cong v_a$ and $LA'\cong r_a$ we can plot the fourth point in the harmonic quadruple $\mathcal{H}(A,A';L,La)$. So we get $r_a\cong A'L_a$. + +\textit{(i)} We use the relation $RR_a\cong a$ (big exercise \ref{velikaNaloga}), we plot the circles $k(S,SR)$ and $k_a(S_a,R_a)$, and then their three common tangents (example \ref{tang2ehkroz}). + + +\textit{(ii)} We use the relation $PP_a\cong b-c$ (big exercise \ref{velikaNaloga}), we plot the circles $k(S,SP)$ and $k_a(S_a,P_a)$, and then their three common tangents. +\kdokaz + +In the previous example we saw how, with the help of the proven relation $\mathcal{H}(A,A';L,La)$, we can get $r_a$ from the elements of the triangle $v_a$ and $r$. In a similar way, from each known pair of a triple $(v_a,r,r_a)$ we get the third element. We will write this fact in the form $\langle v_a,r,r_a\rangle$. + + + + + + \bzgled + Construct a cyclic quadrilateral $ABCD$, with the sides congruent to + given line segments $a$, $b$, $c$ in $d$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.11.pic} +\caption{} \label{sl.pod.7.4.11.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.4.11.pic}) + +Let $ABCD$ be a cyclic quadrilateral with +sides congruent to +given line segments $a$, $b$, $c$ and $d$, $h_{A,k}$ central stretch with center $A$ +and coefficient $k=\frac{AD}{AB}=\frac{d}{a}$, $\mathcal{R}_{A,\alpha}$ rotation with +center $A$ for the oriented angle $\alpha=\angle BAD$ and the composite: + $$\rho_{A,k,\alpha}=\mathcal{R}_{A,\alpha}\circ h_{A,k}$$ +rotational stretch with center $A$, coefficient $k$ and angle $\alpha$. + +Let $B'=h_{A,k}(B)$ and $C'=h_{A,k}(C)$. Because $|AB'|=k\cdot |AB|=\frac{|AD|}{|AB|}\cdot |AB|= |AD|$ and $\angle B'AD=\angle BAD= \alpha$, it follows that $R_{A,\alpha}(B')=D$, so $\rho_{A,k,\alpha}(B)=D$. Let $E=R_{A,\alpha}(C')$ +or $E=\rho_{A,k,\alpha}(C)$. Then the triangle $ADE$ is the image of the triangle $ABC$ under the rotational stretch $\rho_{A,k,\alpha}$ (which is a similarity transformation), so these +two triangles are similar with a similarity coefficient of $k$. From this and from the statement \ref{TetivniPogoj} it follows that: +$$\angle EDC=\angle EDA+\angle ADC=\angle CBA+\angle ADC=180^0,$$ +which means that the points $E$, $D$ and $C$ are collinear. The distance $ED$ can be constructed (using Tales' theorem \ref{TalesovIzrek}), because: +$$ED\cong B'C'=k\cdot BC=\frac{AD}{AB}\cdot BC=\frac{b\cdot d}{a}.$$ +After the construction of the points $C$, $D$ and $E$, we can also construct the point $A$, because: +$$\frac{AE}{AC}=\frac{AC'}{AC}=k,$$ +so the point $A$ lies on the Apollonius circle $\mathcal{A}_{EC,k}$, as well as on the circle $k(D,d)$. +\kdokaz + + + \bnaloga\footnote{44. IMO Japan - 2003, Problem 4.} + $ABCD$ is cyclic. The foot of the perpendicular from $D$ to the + lines $AB$, $BC$, $CA$ are $P$, $Q$, $R$, respectively. Show that the angle bisectors of + $\angle ABC$ and $\angle CDA$ meet on the line $AC$ if and only if $RP \cong RQ$. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.3.IMO1.pic} +\caption{} \label{sl.pod.7.3.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Without loss of generality, we assume that +$\mathcal{B}(B,C,Q)$ and $\mathcal{B}(B,P,A)$ hold. According to +\ref{SimpsPrem}, the points $P$, $Q$ and $R$ are collinear and lie on +\index{premica!Simsonova} Simson's line (Figure +\ref{sl.pod.7.3.IMO1.pic}). From the proof of this theorem it also follows that + $\angle CDQ\cong \angle ADP$, + which means that $CDQ$ and $ADP$ are similar right triangles, + so we have: $$CD:CQ=AD:AP \textrm{ or } CD:AD=CQ:AP.$$ + Let us denote $P_1=\mathcal{S}_A(P)$, from the previous relation we get: + \begin{eqnarray} \label{eqn.pod.7.3.IMO1} + CD:AD=CQ:AP_1. + \end{eqnarray} +($\Rightarrow$) First, let us assume that the altitudes of the angles $ABC$ +and $CDA$ intersect at the point $E$, which lies on the line + $AC$. According to \ref{HarmCetSimKota}, we then have: + $BC:BA=CE:EA=DC:DA$. From this and from the proven relation + \ref{eqn.pod.7.3.IMO1} it follows that $BC:BA=CQ:AP_1$ or $BC:CQ=BA:AP_1$. + Because $\mathcal{B}(B,C,Q)$ + and $\mathcal{B}(B,A,P_1)$ hold, from the previous relation we get + $BC:BQ=BA:BP_1$. By the converse of Tales' theorem + \ref{TalesovIzrekObr}, $CA\parallel QP_1$ or $AR\parallel QP_1$, + so by Tales' + theorem + \ref{TalesovIzrek} $PR:RQ=PA:AP_1=1:1$ or $PR\cong RQ$. + + ($\Leftarrow$) Now, let $PR\cong RQ$ hold. Because $A$ is the center of the line segment $PP_1$, + $RA$ is the median + of the triangle $PQP_1$. From this it follows that $AR\parallel QP_1$ or + $CA\parallel QP_1$. By Tales' theorem, we have $BC:CQ=BA:AP_1$ + or $BC:BA=CQ:AP_1$. From the proven relation + \ref{eqn.pod.7.3.IMO1} we now get $BC:BA=CD:AD$. Let $E_1$ + and $E_2$ be the intersections of the altitudes of the angles $ABC$ and $CDA$ with the line $AC$. If we use + the previous relation and \ref{HarmCetSimKota}, we get:\\ + $CE_1:AE_1=BC:BA=CD:AD=CE_2:AE_2$. Because the points $E_1$ and $E_2$ both lie + on the line segment $AC$, $E_1=E_2$ (\ref{HarmCetEnaSamaDelitev}), so the altitudes of the angles + $ABC$ and $CDA$ intersect on the line $AC$. + \kdokaz + +%________________________________________________________________________________ + \poglavje{The Right Triangle Altitude Theorem. Euclid's Theorems} \label{odd7VisinEvkl} + +In algebra and mathematical analysis, the concepts of arithmetic or geometric mean of two (or more) numbers are known. +At this point, we will first define the so-called arithmetic and geometric mean of two line segments. +We say that $x$ \index{aritmetična sredina daljic} \pojem{arithmetic mean} of line segments $a$ and $b$, if it holds: + +\begin{eqnarray} \label{eqnVisEvkl1} +|x|=\frac{1}{2}\left( |a|+|b| \right). +\end{eqnarray} +Similarly, $y$ \index{geometrijska sredina daljic}\pojem{geometric (also geometrical) mean} of line segments $a$ and $b$, if it holds: + +\begin{eqnarray} \label{eqnVisEvkl2} +|y|= \sqrt{|a|\cdot |b|}. +\end{eqnarray} + +Therefore, arithmetic or geometric mean of two line segments is a line segment with a length that is equal to the arithmetic or geometric mean of the lengths of these two line segments. Relations \ref{eqnVisEvkl1} and \ref{eqnVisEvkl2} will often be written in a shorter form: +\begin{eqnarray*} +x=\frac{1}{2}\left( a+b \right)\hspace*{1mm} \textrm{ or }\hspace*{1mm} +y=\sqrt{ab}. +\end{eqnarray*} + +It is clear that we can construct the arithmetic mean of two given line segments in a very simple way. In this section we will derive the construction of the geometric mean of two given line segments. We will first prove the main statement that relates to right-angled triangles. + + + + + \bizrek \index{izrek!višinski}\label{izrekVisinski} + The altitude on the hypotenuse of a right-angled triangle + is the geometric mean + of the line segments into which it divides the hypotenuse.\\ + (The right triangle altitude theorem) + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6E.1.pic} +\caption{} \label{sl.pod.7.6E.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Let $ABC$ be a right-angled triangle with +hypotenuse $AB$ and altitude $CC'$. We denote $v_c=|CC'|$, $b_1=|AC'|$ and $a_1=|BC'|$ (Figure \ref{sl.pod.7.6E.1.pic}). + +First, $\angle ACC'=90^0-\angle CAC'=\angle CBC'$ and + $\angle BCC'=90^0-\angle CBC'=\angle CAC'$. +From the similarity of the right-angled triangles $CC'A$ and $BC'C$ (statement \ref{PodTrikKKK}) it follows that: +$$\frac{CC'}{BC'}=\frac{C'A}{C'C},$$ +hence: +\begin{eqnarray} \label{eqnVisinski} +v_c^2=a_1\cdot b_1. +\end{eqnarray} +\kdokaz + + + + \bizrek \index{izrek!Evklidov}\label{izrekEvklidov} + Each leg of a right-angled triangle + is the geometric mean + of the hypotenuse and the line segment of the hypotenuse adjacent to the leg.\\ + (Euclid's\footnote{Starogrški filozof in matematik \index{Evklid}\textit{Evklid iz Aleksandrije} (3. st. pr. n. š.).} theorems) + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6E.2.pic} +\caption{} \label{sl.pod.7.6E.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Let $ABC$ be a right-angled triangle with +hypotenuse $AB$ and altitude $CC'$. We denote $a=|BC|$, $b=|AC|$, $c=|AB|$, $b_1=|AC'|$ and $a_1=|BC'|$ (Figure \ref{sl.pod.7.6E.2.pic}). + +As in the proof of statement \ref{izrekVisinski}, $\angle ACC'=90^0-\angle CAC'=\angle CBA$ and + $\angle CAC'=\angle CAB$. +From the similarity of the right-angled triangles $CC'A$ and $BCA$ (statement \ref{PodTrikKKK}) it follows that: +$$\frac{CA}{BA}=\frac{C'A}{CA},$$ +or: +\begin{eqnarray} \label{eqnEvklidov1} +b^2=b_1\cdot c. +\end{eqnarray} +Similarly, we have: +\begin{eqnarray} \label{eqnEvklidov2} +a^2=a_1\cdot c. +\end{eqnarray} +\kdokaz + + + + \bzgled \label{EvklVisPosl} + If $c$ is the length of the hypotenuse, $a$ and $b$ the lengths of the legs and $v_c$ + the length of the altitude on the hypotenuse of a right-angled triangle $ABC$, then + $$c\cdot v_c=a\cdot b.$$ + \ezgled + +\textbf{\textit{Proof.}} + By multiplying relations \ref{eqnEvklidov1} and \ref{eqnEvklidov2} from Euclid's theorem \ref{izrekEvklidov} and inserting the relation \ref{eqnVisinski} from the altitude theorem \ref{izrekVisinski} we get: + $$a^2b^2=a_1b_1 c^2=v^2c^2,$$ + from which it follows that $ab=cv_c$. + \kdokaz + + Now we will derive the predicted construction. + + + + \bzgled + Construct a line segment $x$ that is the geometric mean of given line segments + $a$ and $b$, i.e. $x=\sqrt{ab}$. + \ezgled + + +\textbf{\textit{Proof.}} We will derive the construction in two ways. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6E.3.pic} +\caption{} \label{sl.pod.7.6E.3.pic} +\end{figure} + +\textit{1)} + First, let's draw points $P$, $Q$ and $R$ so that $PQ\cong a$, $QR\cong b$ and $\mathcal{B}(P,Q,R)$ (Figure \ref{sl.pod.7.6E.3.pic}), then the circle $k$ over the diameter $PR$ and finally the point $X$ as the intersection of the circle $k$ with the rectangle line $PR$ at point $Q$. + + We prove that $x=QX$ is the desired line segment. By Theorem \ref{TalesovIzrKroz2}, $PXR$ is a right triangle with hypotenuse $PR$, $XQ$ is its altitude on this hypotenuse. By the altitude theorem \ref{izrekVisinski}, $x$ is the geometric mean of line segments $a$ and $b$. + +\textit{2)} Without loss of generality, let $a>b$ (if $a=b$ then also $x=a$). + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6E.4.pic} +\caption{} \label{sl.pod.7.6E.4.pic} +\end{figure} + + First, let's draw points $P$, $Q$ and $R$ so that $PQ\cong a$, $QR\cong b$ and $\mathcal{B}(P,R,Q)$ (Figure \ref{sl.pod.7.6E.4.pic}), then the circle $k$ over the diameter $PQ$ and finally the point $X$ as the intersection of the circle $k$ with the rectangle line $PQ$ at point $R$. + +We will now prove that $x=QX$ is the desired distance. According to the \ref{TalesovIzrKroz2} theorem, $PXQ$ is a right angled triangle with hypotenuse $PQ$, and $XR$ is its height on that hypotenuse. According to Euclid's \ref{izrekEvklidov} theorem, $x$ is the geometric mean of the distances $a$ and $b$. +\kdokaz + +In the next example, we will consider another important construction. + + \bzgled \label{konstrKoren} + Construct a line segment $x=e\sqrt{6}$, for a given line segment $e$. + \ezgled + +\textbf{\textit{Proof.}} Similarly to the previous example, we will carry out the construction in two ways. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6E.5.pic} +\caption{} \label{sl.pod.7.6E.5.pic} +\end{figure} + +\textit{1)} + First, let's draw points $P$, $Q$ and $R$, such that: $PQ=3e$, $QR=2e$ and $\mathcal{B}(P,Q,R)$ (Figure \ref{sl.pod.7.6E.5.pic}), then the circle $k$ over the diameter $PR$ and finally the point $X$ as the intersection of the circle $k$ with the perpendicular line $PR$ at point $Q$. + + We will now prove that $x=QX$ is the desired distance. According to the \ref{TalesovIzrKroz2} theorem, $PXR$ is a right angled triangle with hypotenuse $PR$, and $XQ$ is its height on that hypotenuse. According to the \ref{izrekVisinski} theorem, $x$ is the geometric mean of the distances $PQ=3e$ and $QR=2e$, i.e. $x=\sqrt{3e\cdot 2e}=e\sqrt{6}$. + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6E.6.pic} +\caption{} \label{sl.pod.7.6E.6.pic} +\end{figure} + +\textit{2)} + First, let's draw points $P$, $Q$ and $R$, such that: $PQ=3e$, $QR=2e$ and $\mathcal{B}(P,R,Q)$ (Figure \ref{sl.pod.7.6E.6.pic}), then the circle $k$ over the diameter $PQ$ and finally the point $X$ as the intersection of the circle $k$ with the perpendicular line $PQ$ at point $R$. + +We will prove that $x=QX$ is the desired distance. By the \ref{TalesovIzrKroz2} theorem, $PXQ$ is a right angled triangle with hypotenuse $PQ$, $XR$ is its height on that hypotenuse. By \ref{izrekEvklidov} theorem, $x$ is the geometric mean of the distances $PQ=3e$ and $QR=2e$, i.e. $x=\sqrt{3e\cdot 2e}=e\sqrt{6}$. +\kdokaz + + In the previous example, we have described the process of constructing the distance $x=e\sqrt{n}$, where $e$ is a given distance, and $n\in \mathbb{N}$ is a given natural number. We always have two options for construction - using the altitude theorem or the Pythagorean theorem. But if $n$ is a composite number, for example $n=12$, we can choose $PQ=3e$ and $QR=4e$, or $PQ=6e$ and $QR=2e$. If $n$ is a prime number, for example $n=7$, it is most advantageous to choose $PQ=7e$ and $QR=e$. + + +%________________________________________________________________________________ + \poglavje{Pythagoras' Theorem} \label{odd7Pitagora} + +We will prove the famous Pythagorean theorem. + + + + \bizrek \label{PitagorovIzrek} + \index{izrek!Pitagoras} + The square of the length of the hypotenuse of a right-angled triangle is + equal to the sum of the squares of the lengths of both legs, i.e. for a right-angled triangle + $ABC$ with the hypotenuse of length $c$ and the legs of lengths $a$ and $b$ is + $$a^2+b^2=c^2$$ + (Pythagoras'\footnote{Predpostavlja se, da je bil ta izrek znan že Egipčanom (pribl. 3000 let pr. n. š.) in Babiloncem (pribl. 2000 + let pr. n. š.), toda starogrški filozof in matematik \index{Pitagora}\textit{Pitagora z otoka Samosa} (582--497 pr. n. š.) ga je verjetno prvi dokazal. Prvi pisni dokument dokaza Pitagorovega izreka je dal \index{Evklid}\textit{Evklid iz Aleksandrije} (3. st. pr. n. š.) v svojem delu ‘‘Elementi’’. + Pri Starih Grkih se je Pitagorov izrek običajno nanašal na zvezo med ploščinami kvadratov nad stranicami pravokotnega trikotnika.} theorem) + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.0.pic} +\caption{} \label{sl.pod.7.6.0.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.6.0.pic}) + +If we use the statement \ref{izrekEvklidov} and the labels from the proof of this statement, we get: +$$a^2 + b^2 = c\cdot a_1 + c\cdot b_1 = c\cdot (a_1 + b_1) = c^2,$$ which was to be proven. \kdokaz + +In the previous form, Pythagorean Theorem applies to the squares of the lengths of the sides of a right-angled triangle. The second form, which applies to the relationship between the areas of the squares over the sides of a right-angled triangle, will be considered in section \ref{odd8PloTrik}. + +Pythagorean Theorem allows us to calculate the third side if the first two sides of a right-angled triangle are given. If we denote the length of the hypotenuse with $c$ and the length of the catheti with $a$ and $b$, from Pythagorean Theorem we get: +\begin{eqnarray*} +c&=&\sqrt{a^2+b^2}\\ +a&=&\sqrt{c^2-b^2}\\ +b&=&\sqrt{c^2-a^2} +\end{eqnarray*} + + + + \bizrek \label{PitagorovIzrekObrat} + \index{izrek!Obratni Pitagorov} + Let $ABC$ be an arbitrary triangle. If + $$|AC|^2+|BC|^2=|AB|^2,$$ + then $ABC$ is a right-angled triangle with the right angle at the vertex $C$.\\ + (Converse of Pythagorean Theorem) + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.0a.pic} +\caption{} \label{sl.pod.7.6.0a.pic} +\end{figure} + +\textbf{\textit{Proof.}} We mark: $c=|AB|$, $b=|AC|$ and $a=|BC|$ (Figure \ref{sl.pod.7.6.0a.pic}). So $a^2+b^2=c^2$ is true, which means: +\begin{eqnarray} \label{eqnPitagObrat1} +|AB|=c=\sqrt{a^2+b^2}. +\end{eqnarray} +Let $A'B'C'$ be such a right-angled triangle with a right angle at the vertex $C'$, that +$A'C'\cong AC$ and $B'C'\cong BC$. By Pythagoras' theorem +$|A'B'|^2= |A'C'|^2+|B'C'|^2=b^2+a^2$ +or: +\begin{eqnarray} \label{eqnPitagObrat2} +|A'B'|=\sqrt{a^2+b^2}, +\end{eqnarray} + therefore from \ref{eqnPitagObrat1} and \ref{eqnPitagObrat2} it follows that $A'B'\cong AB$. +The triangles $ABC$ and $A'B'C'$ are therefore congruent (by the \textit{SSS} \ref{SSS} theorem), from which it follows that $\angle ACB\cong \angle A'C'B'=90^0$, which means that $ABC$ is a right-angled triangle with a right angle at the vertex $C$. +\kdokaz + + + +If the lengths of the sides $a$, $b$ and $c$ of the right-angled triangle $ABC$ (where $c$ is the length of the hypotenuse) are natural numbers, we say that the triple $(a,b,c)$ represents the \index{pitagorejska trojica}\pojem{Pythagorean triplet}. According to Pythagoras' \ref{PitagorovIzrek} theorem, for the Pythagorean triplet $(a, b, c)$ it is true that $a^2+b^2=c^2$. According to the converse Pythagoras' theorem, if for natural numbers $a$, $b$ and $c$ it is true that $a^2+b^2=c^2$, then $(a,b,c)$ is a Pythagorean triplet. The most famous Pythagorean triplet\footnote{This triplet was known to the Ancient Egyptians and Babylonians. It is also called the \index{trikotnik!egipčanski}\pojem{Egyptian triangle}, because the Ancient Egyptians used it to determine the right angle on the ground.} is $(3, 4, 5)$ (Figure \ref{sl.pod.7.6.0b.pic}), because $3^2+4^2=5^2$. We often encounter the Pythagorean triplets $(5, 12, 13)$ and $(7, 24, 25)$ as well. + +In reality, there are an infinite number of Pythagorean triples. From just one such triple $(a,b,c)$, we can get an infinite number of them: $(ka, kb, kc)$ (for any $k\in \mathbb{N}$). Obviously, in this case, we are dealing with similar right-angled triangles. +We say that a \index{pitagorejska trojica!primitivna}\pojem{Pythagorean triple} is \pojem{primitive}, if $a$, $b$ and $c$ do not have a common divisor. It turns out that there are also an infinite number of primitive Pythagorean triples. + +We can calculate new Pythagorean triples using the following method. +If $m$ and $n$ are any natural numbers ($m > n$), then: + \begin{eqnarray*} + a &=& m^2 - n^2,\\ + b &=& 2mn,\\ + c &=& m^2 + n^2. + \end{eqnarray*} + With a simple calculation, we can convince ourselves that $(a, b, c)$ really is a Pythagorean triple. + + We will use Pythagoras' theorem to calculate other shapes in the following sections. + + + + + \bzgled \label{PitagorovPravokotnik} + If $a$ and $b$ are the lengths of the sides and $d$ the lengths of the diagonal of + a rectangle, then: + $$d=\sqrt{a^2+b^2}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.0c.pic} +\caption{} \label{sl.pod.7.6.0c.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $ABCD$ be any rectangle. We label: $a=|AB|$, $b=|BC|$ and $d=|AC|$ (Figure \ref{sl.pod.7.6.0c.pic}). Because $ABC$ is a right-angled triangle with hypotenuse $AC$, by Pythagoras' theorem \ref{PitagorovIzrek}, $d^2=a^2+b^2$ or $d=\sqrt{a^2+b^2}$. +\kdokaz + + A direct consequence (for $a=b$) is the following statement. + + + + + \bzgled \label{PitagorovKvadrat} + If $a$ is the length of the side and $d$ the length of the diagonal of a square, + then (Figure \ref{sl.pod.7.6.0d.pic}): + $$d=a\sqrt{2}.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.0d.pic} +\caption{} \label{sl.pod.7.6.0d.pic} +\end{figure} + +The fact that $\sqrt{2}$ is an irrational number ($\sqrt{2}\notin \mathbb{Q}$) or that it cannot be written in the form of a fraction, results in the fact that the diagonal $d$ of a square and the side $a$ are not \index{proportional lines}\pojem{proportional} or \index{comparable lines}\pojem{comparable} lines\footnote{The ancient Greeks did not yet know of irrational numbers. The Pythagoreans - a philosophical school founded by the Greek mathematician \index{Pythagoras} + \textit{Pythagoras of Samos} + (582--497 BC) - were of the belief that everything is a number. They meant rational numbers and believed that any two lines are proportional. The first to discover the opposite was the Greek mathematician and philosopher from the Pythagorean school \index{Hipas}\textit{Hipas of Metapontum} (5th century BC). Several legends are associated with this discovery, which was therefore in complete contrast to Pythagorean philosophy - from his suicide to the fact that the Pythagoreans drowned him or that they simply excluded him from their circle.}. This means that there is no line $e$ as a unit such that for some natural numbers $n,m\in \mathbb{N}$ it would hold that $d=n\cdot e$ and $a=m\cdot e$. + + + + \bzgled \label{PitagorovEnakostr} + Suppose that $a$ is the length of the side of an equilateral triangle $ABC$. + If $v$ is the length of the altitude, $R$ the circumradius and $r$ the inradius + of that triangle, then + $$v=\frac{a\sqrt{3}}{2},\hspace{2mm} R=\frac{a\sqrt{3}}{3}, + \hspace{2mm} r=\frac{a\sqrt{3}}{6}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.0e.pic} +\caption{} \label{sl.pod.7.6.0e.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $AA'$ be the altitude of the triangle $ABC$ (Figure \ref{sl.pod.7.6.0e.pic}). + +In section \ref{odd3ZnamTock} we have proven that in an equilateral triangle all four characteristic points are equal ($O=S=T=V$). This means that the center $T$ of the triangle $ABC$ is at the same time the center of the circumscribed and inscribed circle of this triangle. Also, $T$ is at the same time the altitude point of this triangle, which means that $AA'$ is also the median, therefore the point $A'$ is the center of the side $BC$, or $|BA'|=\frac{a}{2}$. By the theorem \ref{tezisce} it follows that $AT:TA'=2:1$. Therefore: + \begin{eqnarray} \label{eqnPitagEnakostr1} + R=|TA|=\frac{2}{3}|AA'|=\frac{2}{3}v \hspace*{1mm} \textrm{ and } \hspace*{1mm} + r=|TA'|=\frac{1}{3}|AA'|=\frac{1}{3}v. + \end{eqnarray} + +Since $ABA'$ is a right triangle with hypotenuse $AB$, by the Pythagorean theorem \ref{PitagorovIzrek} +\begin{eqnarray*} + v=\sqrt{a^2-\left(\frac{a}{2} \right)^2}=\sqrt{\frac{3a^2}{4}}=\frac{a\sqrt{3}}{2}. + \end{eqnarray*} + If we use the relation from \ref{eqnPitagEnakostr1}, we get $R=\frac{a\sqrt{3}}{3}$ + and $r=\frac{a\sqrt{3}}{6}$. +\kdokaz + + + + \bzgled \label{PitagorovRomb} + If $a$ is the length of the side and $e$ and $f$ the lengths of the diagonals of a rhombus, then + $$\left(\frac{e}{2}\right)^2+\left(\frac{f}{2}\right)^2=a^2.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.0f.pic} +\caption{} \label{sl.pod.7.6.0f.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $S$ be the intersection of the diagonals of the rhombus $ABCD$ (Figure \ref{sl.pod.7.6.0f.pic}). By the theorem \ref{paralelogram} $S$ is the center of the diagonals $AC$ and $BD$. Therefore $|SA|=\frac{e}{2}$ and $|SB|=\frac{f}{2}$. By the theorem \ref{RombPravKvadr} the diagonals $AC$ and $BD$ are perpendicular, therefore $ASB$ is a right triangle with hypotenuse of length $|AB|=c$ and the cathets of lengths $|SA|=\frac{e}{2}$ and $|SB|=\frac{f}{2}$. The desired relation is now a direct consequence of the Pythagorean theorem \ref{PitagorovIzrek}. +\kdokaz + +\bzgled + If $c$ is the length of the hypotenuse, $a$ and $b$ the lengths of the legs + and $v_c$ the length of the altitude on the hypotenuse of a right-angled + triangle, then + $$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{v_c^2}.$$ + \ezgled + + +\textbf{\textit{Proof.}} + If we use the Pythagorean Theorem \ref{PitagorovIzrek} and the statement from Example \ref{EvklVisPosl}, we get: + $$\frac{1}{a^2}+\frac{1}{b^2}=\frac{a^2+b^2}{a^2b^2} + =\frac{c^2}{a^2b^2}=\frac{c^2}{c^2v_c^2}=\frac{1}{v_c^2},$$ which was to be proven. \kdokaz + +In Example \ref{konstrKoren} we considered two ways of constructing the line segment $x=e\sqrt{n}$ ($n\in \mathbb{N}$) using the altitude and Euclid's theorem. At this point we will solve this task with the help of the Pythagorean theorem - again in two ways. + + + \bzgled \label{konstrKorenPit} + Construct a line segment $e\sqrt{7}$, for a given line segment $e$. + \ezgled + +\textbf{\textit{Solution.}} We will carry out the construction in two ways. + + +\textit{1)} (Figure \ref{sl.pod.7.6.3.pic}). +The idea is to design a right-angled triangle $ABC$ with the cathetus $a=e\sqrt{7}$, where the other cathetus $b=n\cdot e$ and the hypotenuse $c=m\cdot e$ ($n$ and $m$ are natural or at least rational numbers). +By the Pythagorean Theorem \ref{PitagorovIzrek} we have $\left(e\sqrt{7}\right)^2+b^2=c^2$ or $m^2 e^2-n^2e^2=7e^2$ or equivalently $(m+n)(m-n)=7$. We get one solution of this equation by $m$ and $n$ if we solve the system: +\begin{eqnarray*} +&& m+n=7\\ +&& m-n=1 +\end{eqnarray*} +By adding and subtracting the equations from this system, we therefore get one possibility $m=4$ and $n=3$; from this we get $c=4e$ and $b=3e$. This idea allows us to construct it. + +Let's first draw a right angled triangle $ABC$ with hypotenuse $AB=4e$ and cathetus $AC=3e$. According to Pythagoras' theorem \ref{PitagorovIzrek} $BC^2=\left(4e\right)^2-\left(3e\right)^2=7e^2$ or $BC=e\sqrt{7}$. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.3.pic} +\caption{} \label{sl.pod.7.6.3.pic} +\end{figure} + +\textit{2)} (Figure \ref{sl.pod.7.6.3.pic}). +Let's first construct a right angled triangle $A_0A_1A_2$ with cathetuses $A_0A_1=A_1A_2=e$, then a right angled triangle $A_0A_2A_3$ with cathetuses $A_0A_2$ and $A_2A_3=e$. We continue the process and construct a sequence of right angled triangles $A_0A_{n-1}A_n$ with cathetuses $A_0A_{n-1}$ and $A_{n-1}A_n=e$. According to Pythagoras' theorem: + \begin{eqnarray*} +A_0A_2&=&\sqrt{A_0A_1^2+A_1A_2^2}=\sqrt{e^2+e^2}=e\sqrt{2}\\ +A_0A_3&=&\sqrt{A_0A_2^2+A_2A_3^2}=\sqrt{2e^2+e^2}=e\sqrt{3}\\ +&\vdots&\\ +A_0A_n&=&\sqrt{A_0A_{n-1}^2+A_{n-1}A_n^2}=\sqrt{(n-1)e^2+e^2}=e\sqrt{n} +\end{eqnarray*} +We therefore get the desired distance from the sixth right angled triangle: $A_0A_7=e\sqrt{7}$. +\kdokaz + +Some of the following examples relate to the use of Pythagoras' theorem in circles. + + + + \bzgled \label{PitagoraCofman} + (Example from the book \cite{Cofman}) + Let $a$ and $b$ be two circles of the same + radius $r=1$, touching each other externally at the point $P$, + and $t$ their common external tangent. + If $c_1$, $c_2$, $\ldots$, $c_n$,... is a sequence of circles touching + circles $a$ and $b$, the first of them touching the line $t$ and each circle from the sequence + touches the previous one. Calculate the diameter of the circle $c_n$.\\ + + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.4.pic} +\caption{} \label{sl.pod.7.6.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Let us denote: with $A$ and $B$ the centers of the circles $a$ and $b$, with $A'$ and $B'$ the points of tangency of these circles with +the tangent $t$, with $S_n$ the centers of the circles $c_n$ ($n =1,2,\ldots$), with $r_n$ their radii, with $d_n=2r_n$ their +diameters, with $P_n$ the points of tangency of the circles $c_n$ and $c_{n-1}$ (or the circles $c_1$ and the line $t$) and with $S_n'$ and $X_n$ the perpendicular projections of the points $S_n$ +or $P_n$ on the line $AA'$ (Figure \ref{sl.pod.7.6.4.pic}). Let $x_n=|AX_n|$. If we use the Pythagorean Theorem \ref{PitagoraovIzrek} for the triangle +$AS_nS_n'$, we get $|AS_n|^2=|S_nS_n'|^2+|S_n'A|^2$ or +$\left(1+r_n \right)^2=1+\left(x_n-r_n \right)^2$. From this, solving for $r_n$, we get: + \begin{eqnarray} \label{eqnPitagoraCofman1} + r_n=\frac{x_n^2}{2\left(1+x_n \right)}. + \end{eqnarray} + On the other hand, it is clear that $x_n-x_{n+1}=d_n=2r_n$. If we connect this with the relation \ref{eqnPitagoraCofman1}, we get: + \begin{eqnarray*} + x_{n+1}=x_n-2r_n=x_n-\frac{x_n^2}{1+x_n}=\frac{x_n}{1+x_n}. + \end{eqnarray*} + Because in this case $x_1=1$, by direct calculation we get $x_2=\frac{1}{2}$, $x_3=\frac{1}{3}$, $\ldots$ From this we intuitively conclude that: + \begin{eqnarray} \label{eqnPitagoraCofman2} + x_n=\frac{1}{n}. + \end{eqnarray} +We will prove the relation \ref{eqnPitagoraCofman2} formally - by using mathematical induction. The relation \ref{eqnPitagoraCofman2} is clearly true for $n=1$. We assume that the relation \ref{eqnPitagoraCofman2} is true for $n=k$ ($x_k=\frac{1}{k}$) and prove that from this it follows that it is true for $n=k+1$ ($x_{k+1}=\frac{1}{k+1}$): + \begin{eqnarray*} + x_{k+1}=\frac{x_k}{1+x_k}=\frac{\frac{1}{k}}{1+\frac{1}{k}}=\frac{1}{k+1}. + \end{eqnarray*} + With this we have proven that the relation \ref{eqnPitagoraCofman2} is true for every $n\in \mathbb{N}$. + +At the end, from relations \ref{eqnPitagoraCofman1} and \ref{eqnPitagoraCofman2} it follows: +\begin{eqnarray*} + d_n=2r_n=\frac{x_n^2}{1+x_n}= + \frac{\left(\frac{1}{n}\right)^2}{1+\frac{1}{n}}=\frac{1}{n(n+1)} + \end{eqnarray*} +or +\begin{eqnarray} \label{eqnPitagoraCofman3} + d_n=\frac{1}{n(n+1)}, + \end{eqnarray} + which was needed to calculate. +\kdokaz + +With the help of the statement from the previous example or relation \ref{eqnPitagoraCofman3} we can come to an interesting infinite sequence. Namely, $c_1, c_2, \ldots +c_n,\ldots$ is an infinite sequence of circles and in this case it holds: +\begin{eqnarray*} + d_1+d_2+\cdots + d_n+\cdots=|PP_1|=1. + \end{eqnarray*} +Now, from relation \ref{eqnPitagoraCofman3} we get: +\begin{eqnarray*} + \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{n\cdot (n+1)}+\cdots=1, + \end{eqnarray*} + or +\begin{eqnarray*} + \sum_{n=1}^{\infty}\frac{1}{n (n+1)}=1. + \end{eqnarray*} + + + + \bizrek \label{Jung} + Let $\mathcal{P}=\{A_1,A_2,\ldots A_n\}$ be a finite set of + points in the plane and $d=\max \{|A_iA_j|;\hspace*{1mm}1\leq i,j\leq n\}$. + Prove that there exists a circle with radius $\frac{d}{\sqrt{3}}$, + which contains all the points of this set (\index{izrek!Jungov}Jung's theorem\footnote{Nemški matematik + \index{Jung, H. W. E.}\textit{H. W. E. Jung} (1876--1953) + je leta 1901 dokazal + splošno trditev za $n$-razsežni primer.}). + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.1a.pic} +\caption{} \label{sl.pod.7.6.1a.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.6.1a.pic}) + +According to the statement \ref{lemaJung} it is enough to prove that for every three of these $n$ points there exists an appropriate circle. Let these points be $A$, $B$ and $C$. According to the assumption none of the sides of the triangle $ABC$ exceeds $d$. Without loss of generality let $BC$ be the longest side of this triangle. Then $|AC|, |AB| \leq |BC| \leq d$. We will consider two cases: + +\textit{1)} In the case when the triangle $ABC$ is rectangular or top-rectangular, the circle $\mathcal{K}'(S, \frac{|BC|}{2})$ (where $S$ is the center of the side $BC$) contains all three points $A$, $B$ and $C$. Because $\frac{|BC|}{2}\leq \frac{d}{2}<\frac{d}{\sqrt{3}}$, points $A$, $B$ and $C$ are also in the circle $\mathcal{K}(S, \frac{d}{\sqrt{3}})$. + +\textit{2)} If the triangle $ABC$ is acute, then $\angle BAC \geq 60^0$, because $BC$ is the longest side of this triangle. Let $BA'C$ be a right triangle, so that points $A$ and $A'$ are on the same side of the line $BC$, and $\mathcal{K}'(S,R)$ is a circle defined by the circumscribed circle of this triangle. Because of the condition $\angle BAC \geq 60^0$, point $A$ is either on the edge of the circle $\mathcal{K}'$ or its inner point. The radius of this circle is, according to the \ref{PitagorovEnakostr} equality, equal to $R=\frac{|BC|\sqrt{3}}{3}=\frac{|BC|}{\sqrt{3}}$. Because $|BC|\leq d$, points $A$, $B$ are also in the circle $\mathcal{K}(S, \frac{d}{\sqrt{3}})$. +\kdokaz + + + + \bizrek + Let $M$ and $N$ be common points of two congruent circles + $k$ and $l$ with a radius $r$. Points $P$ and $Q$ are the intersections of these two circles with + a line defined by their centres such that $P$ and $Q$ are on the same side of the line $MN$. Prove that + $$|MN|^2 + |PQ|^2 = 4r^2.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.1.pic} +\caption{} \label{sl.pod.7.6.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Let $O$ and $S$ be the centers of these two circles and $\overrightarrow{v} += \overrightarrow{OS}$ (Figure \ref{sl.pod.7.6.1.pic}). The translation +$\mathcal{T}_{\overrightarrow{v}}$ +maps the circle $k$ to the circle $l$ and the point $M$ to some point $M'$. +Since the point $M$ lies +on the circle +$k$, its image $M'$ lies on the circle $l$. For the same reasons, + $\mathcal{T}_{\overrightarrow{v}}(P )=Q$. Therefore, +$\overrightarrow{MM'} = \overrightarrow{v} = \overrightarrow{OS} = \overrightarrow{PQ}$, which means + that the lines $MM'$ and $OS$ are parallel and $MM'\cong OS\cong PQ$. Because the line +$MN$ is perpendicular to the line $OS$ ($OS$ is the perpendicular bisector of the segment $MN$), +the line $MN$ is also perpendicular to the parallel +$MM'$ of the line $OS$. +Therefore, $\angle NMM'$ is a right angle, so $NM'$ is the diameter of the circle $l$. + If we use the Pythagorean Theorem \ref{PitagorovIzrek}, we get: +$$|MN|^2 + |PQ|^2 = |MN|^2 + |MM'|^2 = |NM'|^2 = 4r^2,$$ which was to be proven. \kdokaz + + + + \bzgled + Lines $b$ and $c$ and a point $A$ in the same plane are given. Construct + a square $ABCD$ such that the vertices $B$ and $C$ lie on the lines $b$ + and $c$, respectively. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.2.pic} +\caption{} \label{sl.pod.7.6.2.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.6.2.pic}) + +We use the rotational dilation $\rho_{A,\sqrt{2},45^0}$. Since $\rho_{A,\sqrt{2},45^0}(B)=C$, we can plan the point $C$ from the condition +$C\in c\cap\rho_{A,\sqrt{2},45^0}(b)$. +\kdokaz + + + + \bnaloga\footnote{1. IMO, Romania - 1959, Problem 4.} + Construct a right triangle with given hypotenuse $c$ such that the median + drawn to the hypotenuse is the geometric mean of the two legs of the triangle. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.IMO1.pic} +\caption{} \label{sl.pod.7.6.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} + Let's assume that $ABC$ is a right angled triangle with hypotenuse + $AB\cong c$ and its center of gravity is on the hypotenuse $CM=t_c$ + geometrical middle of the cathetus $AC=b$ and $BC=a$ or + $t_c=\sqrt{a b}$ (Figure \ref{sl.pod.7.6.IMO1.pic}). + + From the \ref{TalesovIzrKroz} follows $t_c=\frac{1}{2}\cdot c$. If + we use the Pythagorean theorem (\ref{PitagorovIzrek}), we get: + $$(a+b)^2=a^2+b^2+2ab=c^2+2t^2_c=c^2+2\left( \frac{1}{2}\cdot c\right)^2= + \frac{6}{4}\cdot c^2$$ + or + $$a+b=\frac{\sqrt{6}}{2}\cdot c.$$ + The latter relation allows the construction. + + First, we describe the auxiliary construction of the distance, which has the length + $\frac{\sqrt{6}}{2}\cdot c$. First, we draw the line $UV=c$ + and its center $W$, then the right angled triangle $TUW$ + ($TU=c$ and $\angle TUW=90^0$) and the right angled triangle + $TWZ$ ($WZ=\frac{1}{2}\cdot c$ + and $\angle TWZ=90^0$). + + Now we will construct the triangle $ABC$. Draw the line $DB\cong + TZ$ and $\angle BDX=45^0$. The point $A$ will be one of + the intersections of the circle $k(B,c)$ and the arc $DX$. In the end, draw the point + $C$ as the intersection of the line $BD$ with the line + $s_{DA}$ of the line $DA$. + + We prove that the triangle $ABC$ satisfies all the conditions of the task. + First, from the construction it follows that $A\in k(B,c)$, which means that + $AB=c$. The point $C$ by construction lies on the line + $DA$, so $DC\cong AC$ or (\ref{enakokraki}) $\angle + DAC\cong \angle ADC =45^0$. From the triangle $ACD$ is by \ref{VsotKotTrik} + $\angle ACD=90^0$. Because by construction $\mathcal{B}(D,C,B)$, is + also $\angle ACB=90^0$ and $ABC$ is a right angled triangle with + hypotenuse $AB=c$. + +We will now prove that its + centroid $CM=t_c$ + is the geometrical mean of its cathets $AC=b$ and $BC=a$ or $t_c=\sqrt{a b}$. + By construction, the triangles $TUW$ and $TWZ$ are right angled and + $UW=WZ=\frac{1}{2}\cdot c$ and $TW=c$, + so from Pythagoras' theorem it follows: + \begin{eqnarray*} + |TZ|^2&=&|TW|^2+|WZ|^2=\\ + &=& |TU|^2+|UW|^2+|WZ|^2=\\ + &=& + c^2+\left(\frac{c}{2}\right)^2+\left(\frac{c}{2}\right)^2=\\ + &=& \frac{6}{4}\cdot c^2 + \end{eqnarray*} + By construction, $BD\cong TZ$, so $|BD|^2=\frac{6}{4}\cdot + c^2$. Therefore: + \begin{eqnarray*} + \frac{6}{4}\cdot c^2&=&|BD|^2=\left(|BC|+|CD|\right)^2=\\ + &=&\left(|BC|+|CA|\right)^2=(a+b)^2=\\ + &=&a^2+b^2+2ab=\\ + &=&c^2+2ab\\ + \end{eqnarray*} + From this it follows that $\left(\frac{c}{2} \right)^2=ab$. Because according to the theorem + \ref{TalesovIzrKroz} $t_c=\frac{c}{2}$, in the end we get + $t_c^2=ab$. + + The number of solutions to the task + depends on the number of intersections of the circle $k(B,c)$ and the line + $DX$. + \kdokaz + + + + \bnaloga\footnote{23. IMO, Hungary - 1982, Problem 5.} + The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are + divided by the inner points $M$ and $N$, respectively, so that + $$\frac{AM}{AC}=\frac{CN}{CE}=r.$$ + Determine $r$ if $B$, $M$ and $N$ are collinear. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.IMO2.pic} +\caption{} \label{sl.skk.4.2.IMO2.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $O$ be the centre of the regular +hexagon $ABCDEF$ (Figure \ref{sl.skk.4.2.IMO2.pic}) and $a$ the +length of its side. + From the assumption $\frac{AM}{AC}=\frac{CN}{CE}=r$ and from the fact $AC\cong CE$ + it follows that $CM\cong EN$ and $AM\cong CN$. +The triangles $ACB$ and $CED$ are congruent (the \textit{SSS} +theorem \ref{SSS}), therefore $\angle ACB\cong\angle CED$. This means that the triangles $BCM$ and $DEN$ are also congruent (the \textit{ASA} theorem \ref{KSK} or $\angle BMC\cong \angle DNE$. + The triangle $ACE$ is regular, therefore $\angle ACE=60^0$. So: + \begin{eqnarray*} + \angle BND&=&\angle BNC+\angle CND\\ + &=&\angle BMC- \angle ECA+180^0-\angle DNE=120^0 + \end{eqnarray*} + Because also $\angle BOD=120^0$ and $CO\cong CD\cong CB$, + the point $N$ lies on the circle $k(C,CO)$. Therefore $AM\cong CN\cong + CB=a$. Let $v$ be the length of the altitude of the regular triangle + $OCD$ and use the Pythagorean theorem: + $$r=\frac{CN}{CE}=\frac{a}{2v}=\frac{\sqrt{3}}{3},$$ which had to be calculated. \kdokaz + +Let $ABC$ be an acute-angled triangle with circumcentre $O$. Let $P$ on $BC$ be the foot of the altitude from $A$. +Suppose that $\angle BCA>\angle ABC+30^0$. +Prove that $\angle CAB+\angle COP<90^0$. + + \bnaloga\footnote{42. IMO, USA - 2001, Problem 1.} + Let $ABC$ be an acute-angled triangle with circumcentre $O$. Let $P$ on $BC$ be the foot of the altitude from $A$. + Suppose that $\angle BCA>\angle ABC+30^0$. + Prove that $\angle CAB+\angle COP<90^0$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.IMO3.pic} +\caption{} \label{sl.pod.7.6.IMO3.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $A_1$ be the centre of the side +$BC$, $P'$ and $A'$ the other intersections of the lines $AP$ and $AO$ with +the circumscribed circle $k(O,R)$ of the triangle $ABC$, and $Q$ the orthogonal projection of the point $A'$ on the line $BC$ (Figure +\ref{sl.pod.7.6.IMO3.pic}). Let $\alpha$, $\beta$ and +$\gamma$ be the internal angles of the triangle $ABC$ at the vertices $A$, $B$ and +$C$. + +Since $AA'$ is the diameter of the circle $k$, $\angle AP'A'=90^0$ (by the statement +\ref{TalesovIzrKroz}). This means that $PP'A'Q$ is a rectangle and +it holds: + \begin{eqnarray} +PQ\cong A'P'. \label{pod.7.6.IMO31} + \end{eqnarray} + The point $O$ is the center of the diameter $AA'$, $A_1$ is by Tales' statement +\ref{TalesovIzrek} the center of the line $PQ$. The point $A_1$ is +thus the common center of the lines $BC$ and $PQ$. + +By the statement \ref{TockaNbetagama} it is also $\angle A'AP=\angle +OAP=\gamma-\beta$. From the given condition $\angle BCA>\angle ABC+30^0$ +or $\gamma-\beta>30^0$ it thus follows: + \begin{eqnarray} +\angle A'AP>30^0. \label{pod.7.6.IMO32} +\end{eqnarray} +By the statement \ref{SredObodKot} it is $\angle BOC=2\alpha$. Since $BOC$ +is an isosceles triangle with the base $BC$, $\angle +OCB=\frac{1}{2}\left( 180^0-2\alpha\right)=90^0-\alpha$ (by the statement +\ref{enakokraki} and \ref{VsotKotTrik}). The condition $\angle CAB+\angle +COP<90^0$ or $\angle COP<90^0-\alpha$, which we want to prove, is +thus equivalent to the condition $\angle COP<\angle OCB=\angle OCP$ +(it holds $\angle OCB=\angle OCP$, because $ABC$ is an acute-angled triangle and $\mathcal{B}(B,P,C)$ is +obtuse-angled). The latter is by the statement +\ref{vecstrveckot} equivalent to the condition $CP|BC|^2. \label{pod.7.6.IMO3b} + \end{eqnarray} + +From inequality \ref{pod.7.6.IMO32} and from \ref{SredObodKot} it follows that $\angle A'OP'>60^0$, so in the equilateral triangle $OA'P'$ $\angle A'OP'$ is the largest angle (\ref{enakokraki} and \ref{VsotKotTrik}). This means that $A'P'>OA'=R$ is true. From \ref{pod.7.6.IMO31} it follows that $PQ\cong A'P'>R$. In the end: +\begin{eqnarray*} +\hspace*{-1.5mm} 4|OA_1|^2+2|BC|\cdot|PQ|>2|BC|\cdot|PQ|>2|BC|\cdot R=|BC|\cdot +2R>|BC|^2, + \end{eqnarray*} +which had to be proven. + \kdokaz + + + +%_______________________________________________________________________________ + \poglavje{Napoleon Triangles} \label{odd7Napoleon} + +We will first define two triangles, which we will adapt to any triangle $ABC$. +We will construct equilateral triangles on the outside of its sides. The centers of these triangles are the vertices of the so-called \index{trikotnik!Napoleonov zunanji}\pojem{outer Napoleon triangle} (Figure \ref{sl.pod.7.7n.1.pic}). If we construct equilateral triangles on the inside in the same way, we get the so-called \index{trikotnik!Napoleonov notranji}\pojem{inner Napoleon triangle}. + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7n.1.pic} +\caption{} \label{sl.pod.7.7n.1.pic} +\end{figure} + + + + \bizrek + The outer and inner Napoleon triangles are equilateral. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7n.2.pic} +\caption{} \label{sl.pod.7.7n.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Let $P$,$ Q$ and $R$ be the centers of the regular triangles, which are constructed on the outside of the sides of any triangle $ABC$, so $PQR$ is the outer Napoleon triangle of triangle $ABC$ (Figure \ref{sl.pod.7.7n.2.pic}). We will prove that $PQR$ is an equilateral triangle. + +Let: +$$\mathcal{I}=\mathcal{R}_{R,120^0}\circ +\mathcal{R}_{P,120^0}\circ\mathcal{R}_{Q,120^0}.$$ +Since $120^0+120^0+120^0=360^0$, the composite $\mathcal{I}$ is either the identity or a translation (statement \ref{rotacKomp2rotac}). However, the latter option is ruled out, since $\mathcal{I}(A)=A$. Therefore $\mathcal{I}=\mathcal{E}$ or: + $$\mathcal{R}_{R,120^0}\circ +\mathcal{R}_{P,120^0}\circ\mathcal{R}_{Q,120^0}=\mathcal{E}.$$ +It follows that: + $$\mathcal{R}_{R,120^0}\circ\mathcal{R}_{P,120^0}= + \mathcal{R}^{-1}_{Q,120^0}=\mathcal{R}_{Q,-120^0}=\mathcal{R}_{Q,240^0}.$$ + Let $Q'$ be the third vertex of the equilateral triangle $PQ'R$. By statement \ref{rotacKomp2rotac}, we have: + $$\mathcal{R}_{R,120^0}\circ\mathcal{R}_{P,120^0}=\mathcal{R}_{Q',240^0}.$$ + Therefore $\mathcal{R}_{Q,240^0}=\mathcal{R}_{Q',240^0}$ or $Q=Q'$, which means that $PQR$ is an equilateral triangle. + In a similar way, we prove (by using rotations for an angle of $-120^0$) that the inner Napoleon triangle is also equilateral. + \kdokaz + + +We shall now prove an additional property of the two Napoleon triangles. + + + + \bizrek + The outer and inner Napoleon triangles of an arbitrary + triangle $ABC$ have the same centre, which is at the same time the centroid of the triangle $ABC$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7n.3.pic} +\caption{} \label{sl.pod.7.7n.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Let $PQR$ and $P'Q'R'$ be the outer and inner Napoleon triangles of the triangle $ABC$ (Figure \ref{sl.pod.7.7n.3.pic}). We denote by $Y$ and $A_1$ the centres of the lines $PR$ and $BC$. We shall first prove: + +\begin{enumerate} + \item The triangles $QP'C$ and $RBP'$ are congruent and similar to the triangle $ABC$. + \item The quadrilaterals $ARP'Q$ and $CP'RQ'$ are parallelograms. + \item $\overrightarrow{YA_1}=\frac{1}{2}\overrightarrow{AQ}$. +\end{enumerate} + +\textit{1}) First, from $\angle BCP'\cong \angle ACQ=30^0$ it follows that $\angle P'CQ\cong\angle BCA$ (with the rotation $\mathcal{R}_{C.30^0}$, the angle $\angle P'CQ$ is mapped into the angle $\angle BCA$). It is also true that $\frac{CQ}{CA}=\frac{CP'}{CB}=\frac{\sqrt{3}}{3}$ (by the Pythagorean equality \ref{PitagorovEnakostr}, because $CQ$ and $CP'$ are the radii of the circumscribed circles of two isosceles triangles with sides $CA$ and $CB$), so the triangles $ABC$ and $QP'C$ are similar (by the similarity theorem \ref{PodTrikSKS}) with the similarity coefficient $k=\frac{\sqrt{3}}{3}$. + Similarly, the triangles $ABC$ and $RBP'$ are also similar with the same coefficient of similarity, which means that the triangles $QP'C$ and $RBP'$ are congruent. + +\textit{2}) From the congruence of the triangles $QP'C$ and $RBP'$ and from the fact that the triangles $RAB$ and $QAC$ are of the same size, it follows: $QP'\cong RB\cong RA$ and $P'R\cong CQ\cong QA$, so the quadrilateral $ARP'Q$ is a parallelogram. Similarly, we prove that the quadrilateral $CP'RQ'$ is a parallelogram. + + +\textit{3}) The line $YA_1$ is the median of the triangle $RPP'$ with the base $RP'$, so by the theorems \ref{srednjicaTrikVekt} and \ref{vektParalelogram}: + +$$\overrightarrow{YA_1}=\frac{1}{2}\overrightarrow{RP'}=\frac{1}{2}\overrightarrow{AQ}$$ + +We only need to prove the initial claim. Because $\overrightarrow{YA_1}=\frac{1}{2}\overrightarrow{AQ}$, + the centroid $AA_1$ of the triangle $ABC$ and +the centroid $QY$ of the triangle $PQR$ intersect in a point which they divide in the ratio $2:1$. So this is the common centroid of the triangles $ABC$ and $PQR$ (by the theorem \ref{tezisce}). Similarly, it can be proven that the triangles +$ABC$ and $P'Q'R'$ also have a common centroid. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Ptolemy's Theorem} \label{odd7Ptolomej} + +We will now prove a very important theorem which is related to the cord quadrilaterals. + +\bizrek \label{izrekPtolomej} + \index{izrek!Ptolomejev}(Ptolemy’s\footnote{\index{Ptolomej Aleksandrijski}\textit{Ptolomej Aleksandrijski} (2. st.) je dokazal ta izrek v svojemu delu ‘‘Veliki zbornik’’.} theorem) + If $ABCD$ is a cyclic quadrilateral then + the product of the lengths of its diagonals + is equal to the sum of the products of the lengths of the pairs of opposite sides, + i.e. + $$|AC|\cdot |BD|=|AB|\cdot |CD|+|BC|\cdot |AD|.$$ + + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.1.pic} +\caption{} \label{sl.pod.7.7.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Brez škode za splošnost predpostavimo, da je $\angle CBD\leq\angle ABD$. +Naj bo $L$ takšna točka diagonale $AC$, da velja +$\angle ABL\cong \angle CBD$ (Figure \ref{sl.pod.7.7.2.pic}). Ker je še $\angle BAL=\angle BAC\cong\angle BDC$ (izrek \ref{ObodObodKot}), sta trikotnika +$ABL$ in $DBC$ podobna (izrek \ref{PodTrikKKK}), zato je $AB:DB=AL:DC$ oziroma: +\begin{eqnarray} \label{eqnPtolomej1} +|AB|\cdot |CD|=|AL|\cdot |BD|. +\end{eqnarray} +Ker je $\angle BCL=\angle BCA\cong\angle BDA$ (izrek \ref{ObodObodKot}) in $\angle LBC\cong\angle ABD$ (iz predpostavke $\angle CBD\cong\angle ABL$), velja +tudi $\triangle BCL\sim \triangle BDA$ (izrek \ref{PodTrikKKK}), zato je $BC:BD=CL:DA$ oziroma: +\begin{eqnarray} \label{eqnPtolomej2} +|BC|\cdot |AD|=|CL|\cdot |BD|. +\end{eqnarray} + Po seštevanju relacij \ref{eqnPtolomej1} in \ref{eqnPtolomej2} dobimo iskano enakost. +\kdokaz + + +Trditev iz zgleda \ref{zgledTrikABCocrkrozP} bomo na tem mestu dokazali na bolj enostaven način - z uporabo Ptolomejevega izreka. + +\bzgled \label{zgledTrikABCocrkrozPPtol} +Let $k$ be the circumcircle of a regular triangle $ABC$. + If $P$ is an arbitrary point lying +on the shorter arc $BC$ of the circle $k$, then + $$|PA|=|PB|+|PC|.$$ +\ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.2a.pic} +\caption{} \label{sl.pod.7.7.2a.pic} +\end{figure} + +\textbf{\textit{Solution.}} +The quadrilateral +$ABPC$ is a cyclic quadrilateral (Figure \ref{sl.pod.7.7.2a.pic}), therefore from Ptolomey's theorem \ref{izrekPtolomej} it follows that +$|PA|\cdot |BC|=|PB|\cdot |AC|+|PC|\cdot |AB|$. Because $ABC$ is an equilateral triangle, i.e. $|BC|=|AC|= |AB|$, from the previous relation it follows that $|PA|=|PB|+|PC|$. + \kdokaz + + Another application of Ptolomey's theorem relates to an isosceles trapezium. + + + + + \bzgled + Let $a$ and $c$ be the bases, $b$ the leg and $d$ the diagonal + isosceles trapezium. Prove that + $$ac=d^2-b^2.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.4.pic} +\caption{} \label{sl.pod.7.7.4.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.7.4.pic}) + +By theorem \ref{trapezTetivEnakokr} the isosceles trapezium is cyclic, therefore we can use Ptolomey's theorem \ref{izrekPtolomej} and obtain $d^2=ac+b^2$ or $ac=d^2-b^2$. +\kdokaz + +It is interesting that if we take the special case $a=c$ in the previous theorem (Figure \ref{sl.pod.7.7.4.pic}), when the isosceles trapezium is a rectangle, we obtain the formula for the diagonal of a rectangle $d^2=a^2+b^2$, which is otherwise a direct consequence of Pythagoras' theorem \ref{PitagorovIzrek}. In this sense, we can also say that Pythagoras' theorem is a consequence of Ptolomey's theorem, if we complete the right-angled triangle to a rectangle. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.5.pic} +\caption{} \label{sl.pod.7.7.5.pic} +\end{figure} + +Of course, the use of Ptolemy's theorem in a square leads to the well-known relationship (theorem \ref{PitagorovKvadrat}) between the diagonal $d$ and the side $a$ of the square: $d^2=2a^2$ or $d=a\sqrt{2}$ (Figure \ref{sl.pod.7.7.5.pic}). + +We know that for every regular $n$-gon there is a circumscribed circle \ref{sredOcrtaneKrozVeck}. This allows us to use Ptolemy's theorem, if in a regular $n$-gon (in the case $n>4$) we choose the appropriate four vertices. +Now we will use this idea in the cases $n=5$ and $n=7$. + + + + \bzgled \label{PtolomejPetkotnik} + Let $a$ be the side and $d$ the diagonal of a regular pentagon ($5$-gon). + Prove that + $$d=\frac{1+\sqrt{5}}{2}a.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.3.pic} +\caption{} \label{sl.pod.7.7.3.pic} +\end{figure} + +\textbf{\textit{Solution.}} + Let $A_1A_2A_3A_4A_5$ be a regular pentagon with side $a$ and diagonal $d$ + (Figure \ref{sl.pod.7.7.3.pic}). If we use Ptolemy's theorem \ref{izrekPtolomej} for the quadrilateral $A_1A_2A_3A_5$, we get: $d^2=ad+a^2$ or $d^2-ad-a^2=0$. The positive solution of this quadratic equation in $d$ is precisely $d=\frac{1+\sqrt{5}}{2}a$. + \kdokaz + +The previous example allows us to construct a regular pentagon with a side that is congruent to the line $a$. +Indeed, if $ABCDE$ is the desired regular pentagon, we can first +plan the triangle $ABC$, where $AB\cong BC\cong a$ and $AC =\frac{1+\sqrt{5}}{2}a$. To construct $\frac{1+\sqrt{5}}{2}a$, we use the Pythagorean Theorem (\ref{PitagorovIzrek}) for a right triangle with sides $a$ and $2a$ - its hypotenuse measures $a\sqrt{5}$. We add this hypotenuse to the line $a$ and divide the resulting line into two lines. The center of the drawn circle of the regular pentagon $ABCDE$ is also the center of the drawn circle of the triangle $ABC$ (Figure \ref{sl.pod.7.7.3a.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.3a.pic} +\caption{} \label{sl.pod.7.7.3a.pic} +\end{figure} + + +In a similar way, we can construct a regular pentagon that is inscribed in a given circle. We do this by first constructing a regular pentagon with any side, and then using the central extension (Figure \ref{sl.pod.7.7.3a.pic}). + +We will talk more about the construction of regular $n$-gons in section \ref{odd9LeSestilo}. + + + + \bzgled + Let $a$ be the side, $d$ the shorter and $D$ the longer diagonal + of a regular heptagon ($7$-gon). Prove that + $$\frac{1}{a}=\frac{1}{D}+\frac{1}{d}$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.2.pic} +\caption{} \label{sl.pod.7.7.2.pic} +\end{figure} + +\textbf{\textit{Solution.}} + +Let $A_1A_2A_3A_4A_5A_6A_7$ be a regular heptagon and $k$ the drawn +circle (Figure \ref{sl.pod.7.7.2.pic}). The quadrilateral +$A_1A_2A_3A_5$ is a tetragon, so from Ptolomey's theorem \ref{izrekPtolomej} it follows +that $ad+aD=dD$ or $\frac{1}{a}=\frac{1}{D}+\frac{1}{d}$. + \kdokaz + +If we use the Ptolomey Theorem multiple times, we get one generalization of task \ref{zgledTrikABCocrkrozPPtol}: + +\bzgled + Let $P$ be an arbitrary point of the shorter arc $A_1A_{2n+1}$ + of the circumcircle of a regular polygon $A_1A_2\cdots A_{2n+1}$. + If we denote $d_i=PA_i$ ($i\in \{1,2,\cdots , 2n+1 \}$), + prove that (Figure \ref{sl.pod.7.7.6.pic}) + $$d_1+d_3+\cdots +d_{2n+1}=d_2+d_4+\cdots +d_{2n}.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.6.pic} +\caption{} \label{sl.pod.7.7.6.pic} +\end{figure} + +%________________________________________________________________________________ + \poglavje{Stewart's Theorem} \label{odd7Stewart} + +The next theorem refers to a very interesting and important metric property of a triangle. + + + + \bizrek \label{StewartIzrek} + \index{izrek!Stewartov} (Stewart's\footnote{\index{Stewart, M.}\textit{M. Stewart} (1717--1785), + English mathematician, who in 1746 proved and published this theorem. His teacher - English mathematician \index{Simson, R.}\textit{R. Simson} (1687--1768) - introduced him to the theorem, which he published only in 1749. It is assumed + that the theorem was known to \index{Arhimed} \textit{Archimedes of Syracuse} (3rd century BC) already.} theorem) + If $X$ is an arbitrary point of the side + $BC$ of a triangle $ABC$, then + $$|AX|^2=\frac{|BX|}{|BC|}|AC|^2+\frac{|CX|}{|BC|}|AB|^2-|BX|\cdot |CX|.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.1.pic} +\caption{} \label{sl.pod.7.8.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Let $a$, $b$ and $c$ denote the lengths of the sides $BC$, $AC$ and $AB$, and $p$, $q$ and $x$ the lengths of the lines $BX$, $CX$ and $AX$, in order. With $A'$ we denote the +intersection point of the altitude $v_a$ from the vertex $A$ of the triangle $ABC$ (Figure \ref{sl.pod.7.8.1.pic}). We assume that $\mathcal{B}(B,A',C)$. + +We first prove the claim for the case when $X=A'$. From Pythagoras' theorem \ref{PitagorovIzrek} it follows that $v_a^2 =b^2- q^2$ +and $v_a^2 =c^2- p^2$. By multiplying the first equality by $p$ and the second by $q$ and adding the resulting relations and taking into account that $p+q=a$, we obtain $av_a^2=pb^2+qc^2-apq$ or: +$$v_a^2=\frac{p}{a}b^2+\frac{q}{a}c^2-pq,$$ +which means that in the case of altitudes the claim is true. + +Let now $X\neq A'$. The line $v_a$ is the altitude of the triangles $ABX$ and $ABC$ from the vertex $A$. Without loss of generality, we assume that $\mathcal{B}(B,A',X)$. If we denote the length of the line $BA'$ with $y$ and use the already proven part of the claim for altitudes, +we obtain: +\begin{eqnarray*} +v_a^2&=&\frac{y}{p}x^2+\frac{p-y}{p}c^2-y(p-y),\\ +v_a^2&=&\frac{y}{a}b^2+\frac{a-y}{a}c^2-y(a-y). +\end{eqnarray*} +If we equate the right sides of these two equalities and simplify, we obtain: +$$x^2=\frac{p}{a}b^2+\frac{q}{a}c^2-pq,$$ +which means that the claim is also true in the case when $X\neq A'$. + +The proof of the altitude $v_a$ is similar also in the case when $\mathcal{B}(B,A',C)$ is not true, only then we obtain $v_a^2=\frac{p}{a}b^2+\frac{q}{a}c^2+pq$. +\kdokaz + + +Stewart's theorem \ref{StewartIzrek} can also be written in another form: + +\bizrek \label{StewartIzrek2} +Let $a=|BC|$, $b=|AC|$ and $c=|AB|$ be the length of +the sides of a triangle $ABC$. If $X$ is the point that divides the side $BC$ of this triangle +in the ratio $n:m$ and $x=|AX|$ then +$$x^2=\frac{n}{n+m}b^2+\frac{m}{n+m}c^2-\frac{mn}{(m+n)^2}bc.$$ +\eizrek + +The most well-known use of Stewart's theorem is for the centroids of a triangle. + + + + \bizrek \label{StwartTezisc} + If $a$, $b$ and $c$ are the sides and $t_a$ the triangle median + on the side $a$, then + $$t_a^2=\frac{1}{2}b^2+\frac{1}{2}c^2-\frac{1}{4}a^2.$$ + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.2.pic} +\caption{} \label{sl.pod.7.8.2.pic} +\end{figure} + + +\textbf{\textit{Proof.}} The theorem is a direct consequence of Stewart's theorem \ref{StewartIzrek2}, since in the case of the centroid $n=m=1$ (Figure \ref{sl.pod.7.8.2.pic}). +\kdokaz + + +A direct consequence is the following theorem. + + + + \bzgled \label{StwartTezisc2} + If $a$, $b$ and $c$ are the sides and $t_a$, $t_b$ and $t_c$ the triangle medians + on those sides, respectively, then + $$t_a^2+t_b^2+t_c^2=\frac{3}{4}\left(a^2+ b^2+c^2\right).$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.3.pic} +\caption{} \label{sl.pod.7.8.3.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.8.3.pic}) + +If we use the proven relation from the example \ref{StwartTezisc} three times, we get: + \begin{eqnarray*} + t_a^2&=&\frac{1}{2}b^2+\frac{1}{2}c^2-\frac{1}{4}a^2\\ + t_b^2&=&\frac{1}{2}a^2+\frac{1}{2}c^2-\frac{1}{4}b^2\\ + t_c^2&=&\frac{1}{2}a^2+\frac{1}{2}b^2-\frac{1}{4}c^2 + \end{eqnarray*} + By adding all three equations we get the desired relation. +\kdokaz + +In the case of an isosceles triangle, the equality on the right side of Stewart's theorem is simplified. + + + + \bzgled \label{StewartEnakokraki} + If $ABC$ is an isosceles triangle with the base $BC$ and + $X$ an arbitrary point of this base, then + $$|AX|^2=|AB|^2-|BX|\cdot |CX|.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.8.pic} +\caption{} \label{sl.pod.7.8.8.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.8.8.pic}) + + Therefore, if $AB\cong AC$, directly from Stewart's theorem (\ref{StewartIzrek}) it follows: +\begin{eqnarray*} +|AX|^2 &=& \frac{|BX|}{|BC|}|AC|^2+\frac{|CX|}{|BC|}|AB|^2-|BX|\cdot |CX|=\\ +&=&\left(\frac{|BX|}{|BC|}+\frac{|CX|}{|BC|}\right)|AB|^2-|BX|\cdot |CX|=\\ +&=&\frac{|BX|+|CX|}{|BC|}|AB|^2-|BX|\cdot |CX|=\\ +&=&|AB|^2-|BX|\cdot |CX|, +\end{eqnarray*} + which had to be proven. \kdokaz + + +We will continue a little bit more using Stewart's theorem. + + + + \bzgled + Let $E$ be the intersection of the side $BC$ with the bisector of the interior angle + $BAC$ of a triangle $ABC$. If we denote + $a=|BC|$, $b=|AC|$, $c=|AB|$, $l_a=|AE|$ and $s=\frac{a+b+c}{2}$, then + $$l_a=\frac{2\sqrt{bc}}{b+c}\sqrt{s(s-a)}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.4.pic} +\caption{} \label{sl.pod.7.8.4.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.8.4.pic}) + + According to the statement \ref{HarmCetSimKota} it is: $BE:CE=c:b$. If we use Stewart's statement \ref{StewartIzrek2} for the triangle $ABC$ and the distance $AE$, we get: +$$l_a^2=\frac{c}{b+c}b^2+\frac{b}{b+c}c^2-\frac{bc}{(b+c)^2}a^2.$$ +After a simple rearrangement and simplification of the expression on the right side of the equality, we get the desired relation. +\kdokaz + + + \bizrek \label{izrekEulerStirik}\index{izrek!Eulerjev za štirikotnike} + (Euler's\footnote{Švicarski matematik \index{Euler, L.}\textit{L. Euler} (1707--1783).} theorem for quadrilaterals) + If $P$ and $Q$ are the midpoints of + the diagonals $e$ and $f$ of an arbitrary quadrilateral with sides $a$, $b$, $c$ and $d$, then + $$|PQ|^2=\frac{1}{4}\left(a^2+b^2+c^2+d^2-e^2-f^2 \right).$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.5.pic} +\caption{} \label{sl.pod.7.8.5.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Let $P$ and $Q$ be the centers of the diagonals $AC$ and $BD$ of an arbitrary quadrilateral $ABCD$ (Figure \ref{sl.pod.7.8.5.pic}). +We also mark $a=|AB|$, $b=|BC|$, $c=|CD|$, $d=|DA|$, $e=|AC|$ and $f=|BD|$. The distance $PQ$ is the median of the triangle $AQC$, so according to the statement \ref{StwartTezisc} it holds: + \begin{eqnarray} \label{eqnEulStirik} + |PQ|^2=\frac{1}{2}|AQ|^2+\frac{1}{2}|CQ|^2-\frac{1}{4}e^2. + \end{eqnarray} + Similarly, the distances $QA$ and $QC$ are the medians of the triangles $ABD$ and +$CBD$, so (from the statement \ref{StwartTezisc}): + \begin{eqnarray*} + |AQ|^2&=&\frac{1}{2}d^2+\frac{1}{2}a^2-\frac{1}{4}f^2,\\ + |CQ|^2&=&\frac{1}{2}c^2+\frac{1}{2}b^2-\frac{1}{4}f^2. + \end{eqnarray*} + If we insert the last two equalities into \ref{eqnEulStirik}, we get the desired relation. + \kdokaz + + A direct consequence of the previous statement is obtained if we choose a parallelogram for the quadrilateral. + + \bizrek + A quadrilateral is a parallelogram if and only if the sum of + the squares of all of its sides is equal to the sum of the squares of its diagonals. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.6.pic} +\caption{} \label{sl.pod.7.8.6.pic} +\end{figure} + + +\textbf{\textit{Proof.}} +Let $P$ and $Q$ be the centers of the diagonals $AC$ and $BD$ of an arbitrary quadrilateral $ABCD$. We also mark $a=|AB|$, $b=|BC|$, $c=|CD|$, $d=|DA|$, $e=|AC|$ and $f=|BD|$. According to the statement \ref{paralelogram} the quadrilateral $ABCD$ is a parallelogram exactly when $P=Q$ (Figure \ref{sl.pod.7.8.6.pic}) or $|PQ|=0$. The latter, according to the previous statement \ref{izrekEulerStirik}, is true exactly when +$\frac{1}{4}\left(a^2+b^2+c^2+d^2-e^2-f^2 \right)=0$ or $a^2+b^2+c^2+d^2=e^2+f^2$. +\kdokaz + +\bizrek \label{GMTmnl} + Let $A$ and $B$ be given points in the plane and + $m, n, l\in R^+\setminus \{0\}$. + Determine a set of all points of this plane such that + $$m|AX|^2 + n|BX|^2 = l^2.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.7.pic} +\caption{} \label{sl.pod.7.8.7.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.8.7.pic}) + + According to the statement from Example \ref{izrekEnaDelitevDaljice} there exists only one such point $S$ on the line $AB$, that satisfies +$\overrightarrow{AS}:\overrightarrow{SB}=n:m$. Let $X$ be an arbitrary point. If we +use Stewart's Theorem \ref{StewartIzrek2} for the triangle $AXB$ and the line $XS$, we get: +\begin{eqnarray*} +|XS|^2=\frac{m}{n+m}|AX|^2+\frac{n}{n+m}|BX|^2-\frac{nm}{(n+m)^2}|AB|^2, \textrm{ i.e.} +\end{eqnarray*} + \begin{eqnarray} \label{eqnStewMnozTock} +|XS|^2=\frac{1}{n+m}\left(m|AX|^2+n|BX|^2\right)-\frac{nm}{(n+m)^2}|AB|^2. +\end{eqnarray} +The point $X$ lies on the desired set of points exactly when $m|AX|^2 + n|BX|^2 = l^2$. Because of \ref{eqnStewMnozTock} this is exactly when: + \begin{eqnarray*} +|XS|^2=\frac{1}{n+m}l^2-\frac{nm}{(n+m)^2}|AB|^2=c, +\end{eqnarray*} +where $c$ is a constant, that is not dependent on the point $X$. So if $c>0$, the desired +set of points is a circle $k(S, c)$. If $c=0$, the set of points is only $\{S\}$, but if $c<0$, the set +of points is an empty set. +\kdokaz + +\bnaloga\footnote{50. IMO, Germany - 2009, Problem 2.} + Let $ABC$ be a triangle with circumcentre $O$. The points $P$ and $Q$ are interior points + of the sides $CA$ and $AB$, respectively. Let $K$, $L$ and $M$ be the midpoints of the segments $BP$, $CQ$ + and $PQ$, respectively, and let $l$ be the circle passing through $K$, $L$ and $M$. Suppose that the line + $PQ$ is tangent to the circle $l$. Prove that $|OP|=|OQ|$. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.IMO2.pic} +\caption{} \label{sl.pod.7.8.IMO2.pic} +\end{figure} + +\textbf{\textit{Solution.}} We denote with $k(O,R)$ the circumscribed +circle of the triangle $ABC$ (Figure \ref{sl.pod.7.8.IMO2.pic}). + The lines $MK$ and $ML$ are the medians of the triangles $QPB$ and $PQC$, + so (from the statement \ref{srednjicaTrikVekt}): + \begin{eqnarray} \label{eqn72} + \overrightarrow{MK}=\frac{1}{2}\overrightarrow{QB} \hspace*{3mm} + \textrm{and} \hspace*{3mm} \overrightarrow{ML}=\frac{1}{2}\overrightarrow{PC} + \end{eqnarray} + From the statements \ref{KotiTransverzala} and \ref{KotaVzporKraki} it follows: + \begin{eqnarray} \label{eqn73} + \angle AQP \cong \angle QMK,\hspace*{3mm} + \angle APQ \cong \angle PML\hspace*{3mm} \textrm{and} \hspace*{3mm} + \angle BAC \cong \angle KML + \end{eqnarray} + Because, by assumption, the line $PQ$ is tangent to the circle $l$, by + the statement \ref{ObodKotTang}: + $$\angle MLK\cong\angle QMK \hspace*{3mm} \textrm{and} \hspace*{3mm} + \angle MKL\cong\angle PML.$$ + From this and \ref{eqn73} it follows: + \begin{eqnarray*} + \angle AQP \cong \angle MLK,\hspace*{3mm} + \angle APQ \cong \angle MKL\hspace*{3mm} \textrm{and} \hspace*{3mm} + \angle BAC \cong\angle KML, + \end{eqnarray*} + which means that the triangles $AQP$ and $MLK$ are similar (it is enough to prove + the congruence of two pairs of corresponding angles, by + the statement \ref{PodTrikKKK}). So, from the definition of similarity + of figures it follows: + $\frac{AQ}{ML}=\frac{AP}{MK}$. If we combine this relation with the relation + \ref{eqn72}, we get $\frac{AQ}{AP}=\frac{ML}{MK}= + \frac{\frac{1}{2}\cdot CP}{\frac{1}{2}\cdot + BQ}=\frac{CP}{BQ}$. So it holds: + \begin{eqnarray} \label{eqn74} + |AQ|\cdot |BQ| = |AP|\cdot |CP| + \end{eqnarray} + From Stewart's statement \ref{StewartIzrek} for the isosceles triangle $AOB$ + ($|OA|=|OB|=R$) it follows: + $$|OQ|^2=|OA|^2\cdot \frac{QB}{AB}+|OB|^2\cdot \frac{QA}{AB} + -|AQ|\cdot |BQ|=R^2-|AQ|\cdot |BQ|.$$ + Similarly, from the triangle $AOC$ by the same statement we get: + $$|OP|^2=R^2-|AP|\cdot |CP|.$$ +From the proven relation \ref{eqn74} at the end it follows $|OQ|^2=|OP|^2$ +or $|OQ|=|OP|$. + \kdokaz + +%________________________________________________________________________________ +\poglavje{Desargues' Theorem} \label{odd7Desargues} + +The next theorem is historically connected to the development of projective geometry. + + + + \bizrek \label{izrekDesarguesEvkl} \index{izrek!Desarguesov} + (Desargues’\footnote{ + \index{Desargues, G.} \textit{G. Desargues} (1591--1661), French architect, who was one of + the founders of projective geometry.} theorem) + Let $ABC$ and $A'B'C'$ be two triangles in the plane such that the lines $AA'$, $BB'$ and $CC'$ intersect + at a point $S$ + (i.e the triangles are \index{perspective triangles}\pojem{perspective with respect to the centre} + \color{blue} $S$). + If $P=BC\cap B'C'$, $Q=AC\cap A'C'$ and $R=AB\cap A'B'$, + then the points $P$, $Q$ and + $R$ are collinear + (i.e. triangles are \pojem{perspective with respect to the axis} \color{blue} $PQ$). + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10D.1.pic} +\caption{} \label{sl.pod.7.10D.1.pic} +\end{figure} + + +\textit{\textbf{Proof.}} (Figure +\ref{sl.pod.7.10D.1.pic}) + +If we use Menelaus' theorem \ref{izrekMenelaj} for +the triangles $SA'B'$, $SA'C'$ and $SB'C'$, we get: + \begin{eqnarray*} +\hspace*{-2mm} \frac{\overrightarrow{SA}}{\overrightarrow{AA'}}\cdot + \frac{\overrightarrow{A'R}}{\overrightarrow{RB'}}\cdot + \frac{\overrightarrow{B'B}}{\overrightarrow{BS}}=-1, \hspace*{1mm} +\frac{\overrightarrow{SA}}{\overrightarrow{AA'}}\cdot + \frac{\overrightarrow{A'Q}}{\overrightarrow{QC'}}\cdot + \frac{\overrightarrow{C'C}}{\overrightarrow{CS}}=-1, \hspace*{1mm} + \frac{\overrightarrow{SC}}{\overrightarrow{CC'}}\cdot + \frac{\overrightarrow{C'P}}{\overrightarrow{PB'}}\cdot + \frac{\overrightarrow{B'B}}{\overrightarrow{BS}}=-1. + \end{eqnarray*} + From these three relations it follows: + \begin{eqnarray*} + \frac{\overrightarrow{A'Q}}{\overrightarrow{QC'}}\cdot + \frac{\overrightarrow{C'P}}{\overrightarrow{PB'}}\cdot + \frac{\overrightarrow{B'R}}{\overrightarrow{RA'}}=-1. + \end{eqnarray*} +Therefore, according to Menelaus' theorem \ref{izrekMenelaj} (in the opposite direction) for the triangle $A'B'C'$ +the points $P$, $Q$ and $R$ are collinear. + \kdokaz + +We prove the converse statement in a similar way. + + + + \bizrek \label{izrekDesarguesObr} \index{izrek!Desarguesov obratni} + Let $ABC$ and $A'B'C'$ be two triangles in the + plane such that the lines $AA'$ and $BB'$ intersect at the point $S$. + If $P=BC\cap B'C'$, + $Q=AC\cap A'C'$ and $R=AB\cap A'B'$ are collinear points, then also + $S\in CC'$.\\ + (Converse of Desargues’ theorem) + \eizrek + + +The following theorems are in a certain way similar to Desargues' theorem \ref{izrekDesarguesEvkl}. + + + + \bizrek \label{izrekDesarguesZarkVzp} + Let $ABC$ and $A'B'C'$ be two triangles in the plane such that the lines $AA'$, $BB'$ and $CC'$ + are parallel to each other. + If $P=BC\cap B'C'$, $Q=AC\cap A'C'$ and $R=AB\cap A'B'$, + then the points $P$, $Q$ and + $R$ are collinear. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10D.2.pic} +\caption{} \label{sl.pod.7.10D.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} + (Figure \ref{sl.pod.7.10D.2.pic}) + + After using Tales' Theorem \ref{TalesovIzrek} three times, we get: + \begin{eqnarray*} + \frac{\overrightarrow{BP}}{\overrightarrow{PC}}= + \frac{\overrightarrow{BB'}}{\overrightarrow{C'C}},\hspace*{4mm} + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}= + \frac{\overrightarrow{CC'}}{\overrightarrow{A'A}},\hspace*{4mm} + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= + \frac{\overrightarrow{AA'}}{\overrightarrow{B'B}}. + \end{eqnarray*} + After multiplying these three relations, we get: + \begin{eqnarray*} + \frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=-1, + \end{eqnarray*} +and therefore, according to Menelaus' Theorem \ref{izrekMenelaj} (in the opposite direction), for the triangle $ABC$ +the points $P$, $Q$ and $R$ are collinear. + \kdokaz + + + If $P=BC\cap B'C'$, $Q=AC\cap A'C'$ and $R=AB\cap A'B'$, + then the points $P$, $Q$ and + $R$ are collinear. + + \bizrek \label{izrekDesarguesOsNesk} + Let $ABC$ and $A'B'C'$ be two triangles in the plane such that + the lines $AA'$, $BB'$ and $CC'$ intersect + at a point $S$. + If + $BC\parallel B'C'$ and + $AC\parallel A'C'$, then also $AB\parallel A'B'$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10D.3.pic} +\caption{} \label{sl.pod.7.10D.3.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.10D.3.pic}) + +Because $BC\parallel B'C'$ and + $AC\parallel A'C'$, from Tales' theorem \ref{TalesovIzrek} it follows + $\frac{\overrightarrow{SB}}{\overrightarrow{SB'}}= + \frac{\overrightarrow{SC}}{\overrightarrow{SC'}}$ and + $\frac{\overrightarrow{SA}}{\overrightarrow{SA'}}= + \frac{\overrightarrow{SC}}{\overrightarrow{SC'}}$. + From the previous two relations it follows first + $\frac{\overrightarrow{SB}}{\overrightarrow{SB'}}= + \frac{\overrightarrow{SA}}{\overrightarrow{SA'}}$, so by + Tales' theorem (in the opposite direction) \ref{TalesovIzrekObr} it is also true that $AB\parallel A'B'$. + \kdokaz + + The formulations of the previous three theorems are very similar, even though the + proofs of these + theorems are essentially different. The formulations would not differ at all, if we + had assumed that all parallels (in one direction) of a plane intersect at + the same point at infinity, and that all points at infinity of a plane + determine exactly one line at infinity. This was actually the main motivation for the development of projective geometry, + in which every two lines in a plane intersect (see \cite{Mitrovic}). + + In the following we will see some consequences of Desargues' theorem. + + + + \bzgled \label{zgled 3.2} + Let $p$, $q$ and $r$ be three lines in the plane which + intersect at the same point, and points $A$, $B$ and $C$ of this plane which do not belong + to these lines. Construct a triangle whose vertices belong to the given + lines, and the sides contain the given points. + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10D.4.pic} +\caption{} \label{sl.pod.7.10D.4.pic} +\end{figure} + + + \textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.10D.4.pic}) + + Let $PQR$ be such a triangle, that + its vertices $P$, $Q$ and $R$ belong to the lines $p$, $q$ and + $r$, and sides $QR$, $PR$ and $PQ$ contain points $A$, $B$ and + $C$. Let $S$ be the common point of lines $p$, $q$ and $r$. + +If $P'Q'R'$ is an arbitrary triangle that is perspective to the triangle $PQR$ with respect to the center $S$, whose sides $R'Q'$ and $R'P'$ contain the points $A$ and $B$ (the condition regarding the point $C$ is omitted), then by Desargues' Theorem \ref{izrekDesarguesEvkl} the triangle $PQR$ and $P'Q'R'$ are perspective with respect to some line $s$. Therefore, the line $s$ contains the points $A$, $B$ and $Z=PQ\cap P'Q'$. + +We construct the triangle $PQR$ by first constructing the auxiliary triangle $P'Q'R'$, where the point $R'\in r$ is arbitrary. Then we construct the point $Z$ as the intersection of the lines $AB$ and $P'Q'$. With the points $Z$ and $C$ the side $PQ$ is determined. + +In the proof that $PQR$ is the desired triangle, we use the converse of Desargues' Theorem \ref{izrekDesarguesObr}. + \kdokaz + + + +%________________________________________________________________________________ +\poglavje{Power of a Point} \label{odd7Potenca} + +One of the most interesting characteristics of a circle that highlights some of its metric properties is the power of a point\footnote{The term power was first used in this sense by the Swiss mathematician \index{Steiner, J.}\textit{J. Steiner} (1769--1863).}. Before moving on to the definition, we prove the following theorem. + + + + \bizrek \label{izrekPotenca} + Suppose that $P$ is an arbitrary point in the plane of a circle $k(S,r)$. + For any line of this plane containing the point $P$ and intersecting the circle $k$ at points + $A$ and $B$, the expression $\overrightarrow{PA}\cdot \overrightarrow{PB}$ + (Figure \ref{sl.pod.7.12.1b.pic}) + is constant, furthermore: + $$\overrightarrow{PA}\cdot \overrightarrow{PB} = |PS|^2 - r^2.$$ + If $P$ is an exterior point of the circle $k$ and $PT$ its tangent at a point $T$, then: + $$\overrightarrow{PA}\cdot \overrightarrow{PB} = |PT|^2.$$ + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.1b.pic} +\caption{} \label{sl.pod.7.12.1b.pic} +\end{figure} + + +\textbf{\textit{Proof.}} We will consider three possible cases. + + +\textit{1)} (Figure \ref{sl.pod.7.12.1.pic}) + +Let $P$ be an external point of the circle $k$. In this case, there is no $\mathcal{B}(A,P,B)$, so (the equivalence of \ref{eqnMnozVektRelacijaB} from section \ref{odd5DolzVekt}): +\begin{eqnarray} \label{eqnPotenIzr1} +\overrightarrow{PA}\cdot \overrightarrow{PB}>0. +\end{eqnarray} +Without loss of generality, assume that $\mathcal{B}(P,A,B)$ is true. Because $\angle PTA\cong\angle TBA=\angle TBP$ (the statement of \ref{ObodKotTang}) and $\angle TPA=\angle BPT$, +the triangles $PAT$ and $PTB$ are similar (the statement of \ref{PodTrikKKK}), so +$PA:PT=PT:PB$. If we use the Pythagorean theorem, we get: + $$|PA|\cdot |PB| = |PT|^2 = |PS|^2 - r^2.$$ + From this, due to relation \ref{eqnPotenIzr1}, it follows: +$$\overrightarrow{PA}\cdot \overrightarrow{PB} = |PS|^2 - r^2.$$ + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.1.pic} +\caption{} \label{sl.pod.7.12.1.pic} +\end{figure} + +\textit{2)} (Figure \ref{sl.pod.7.12.1a.pic}) + +If the point $P$ lies on the circle $k$, then $P=A$ or $P=B$, therefore: +$$\overrightarrow{PA}\cdot \overrightarrow{PB} = 0 = |PS|^2 - r^2.$$ + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.1a.pic} +\caption{} \label{sl.pod.7.12.1a.pic} +\end{figure} + + +\textit{3)} (Figure \ref{sl.pod.7.12.1a.pic}) + +Let $P$ be an inner point of the circle $k$. In this case, $\mathcal{B}(A,P,B)$, so (the equivalence of \ref{eqnMnozVektRelacijaB} from section \ref{odd5DolzVekt}): +\begin{eqnarray} \label{eqnPotenIzr2} +\overrightarrow{PA}\cdot \overrightarrow{PB}<0. +\end{eqnarray} +Let $A_1$ and $B_1$ be the intersection points of the line $SP$ with the circle $k$ (without loss of generality, let $\mathcal{B}(A_1,S,P)$). Because of the compatibility of the corresponding +circumferential angles (the statement of \ref{ObodObodKot}), $\triangle APA_1\sim \triangle B_1PB$ (the statement of \ref{PodTrikKKK}), so $AP:B_1P = PA_1:PB$, therefore +$$|PA|\cdot |PB|=|PA_1|\cdot |PB_1|= +\left(r+|PS|\right)\cdot\left(r-|PS|\right)=r^2-|PS|^2.$$ + From this, due to relation \ref{eqnPotenIzr1}, it follows +$$\overrightarrow{PA}\cdot \overrightarrow{PB} = |PS|^2 - r^2,$$ which had to be proven. \kdokaz + +The constant product $\overrightarrow{PA}\cdot \overrightarrow{PB}$ from the previous statement of \ref{izrekPotenca} is called +\index{potenca točke}\pojem{potenca točke} $P$ with respect to the circle $k$ and is denoted $p(P,k)$. + +According to the previous statement of \ref{izrekPotenca}, the power of the point $P$ with respect to the circle $k(S,r)$ is the number $|PS|^2 - r^2$. +This number is positive, negative or zero, depending on whether $P$ is an outer or inner or point on the circle $k$. So: + \begin{eqnarray*} + p(P,k)\hspace*{1mm}\left\{ + \begin{array}{ll} + >0, & \textrm{if } OP>r; \\ + =0, & \textrm{if } OP=r; \\ + <0, & \textrm{if } OP0$, by the previous statement (\ref{PotencOsLema}) the desired set +is a rectangle of the line $S_1S_2$. In the case $r_1=r_2$ it is clear that $|S_1P|=|S_2P|$, thus the desired set is the line $S_1S_2$'s symmetry. +We denote this line with $p$. + +In the special case when the circles intersect in points $A$ and $B$, we have $p(A,k_1)=p(A,k_2)=0$ and $p(B,k_1)=p(B,k_2)=0$. This means that points $A$ and $B$ lie on the desired line $p$, thus this is the line $AB$. + +If the circles touch in point $T$, we have $p(T,k_1)=p(T,k_2)=0$. This means that the point $T$ lies on the line $p$, which is the rectangle of the line $S_1S_2$. Thus the line $p$ is the common tangent of two given circles through their point of contact $T$. +\kdokaz + + The line $p$ from the previous statement is called the \index{potent!line} + \pojem{potent line} of two + circles. We will denote the potent line of circles $k_1$ and $k_2$ with $p(k_1,k_2)$. + +It is interesting to find out how to construct the power line of two given circles $k_1$ and $k_2$ effectively. +In the cases where the circles intersect or touch, we have already given the answer in Theorem \ref{PotencnaOs} (Figure \ref{sl.pod.7.12.4.pic}). +It remains to construct the power line in the case where the circles have no common points. One option is to use Theorem \ref{PotencOsLema} directly. A slightly faster procedure is related to the construction +of the auxiliary circle $l$, which intersects the given circle at points $A$ and $B$ or $C$ and $D$. Then the intersection +of the lines $AB$ and $CD$ - point $X$ - lies on the desired power line $p(k_1,k_2)$. Indeed, from $p(k_1,l)=AB$ and $p(k_2,l)=CD$ it follows that $X\in p(k_1,l)$ and $X\in p(k_1,l)$ or $p(X,p_1)=p(X,l)=p(X,p_2)$. + +Theorem \ref{PotencnaOs} does not consider one case - when +$k_1(S,r_1)$ and $k_2(s,r_2)$ are concentric circles. In this case, the mentioned set of points is an empty set. We get this from the condition +$|S_1P|^2-|S_2P|^2=r_1^2-r_2^2$. Namely, if $S_1=S_2$ and $r_1\neq r_2$, we get the condition $0=r_1^2-r_2^2\neq 0$, which is not satisfied for any point $P$. + + +Let us also define some concepts related to the power line of two circles. + +Let the set of all such circles of a plane be such that each has two power lines $p$ that are equal for each pair of circles of this set. We call this set \index{set of circles}\pojem{set of circles}. The line $p$ is the \index{power!line}\pojem{power line} of this set of circles (Figure \ref{sl.pod.7.12.5.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.5.pic} +\caption{} \label{sl.pod.7.12.5.pic} +\end{figure} + +Let $p$ be the power line of a set of circles. From Theorem \ref{PotencnaOs} it is clear that all the centers of the circles of this set lie on the same line $s$, which is perpendicular to the line $p$. +We will consider three cases. + +\textit{1)} If at least two circles of a family intersect in points $A$ and $B$, then $p=AB$, so all circles of this family go through points $A$ and $B$. In this case we say that it is a \index{family of circles!elliptic}\pojem{elliptic family of circles}. + +\textit{2)} If at least two circles of a family touch in point $T$, then $p$ is a perpendicular of line $s$ in point $T$, so all circles have a common tangent $p$ in point $T$. In this case we say that it is a \index{family of circles!parabolic}\pojem{parabolic family of circles}. + +\textit{3)} If no two circles of a family have common points, we say that it is a \index{family of circles!hyperbolic}\pojem{hyperbolic family of circles}. In this case the following property is valid: if an arbitrary circle intersects circles of this family (it is not necessary that it intersects all of them) in points $A_i$ and $B_i$, then all lines $A_iB_i$ go through one point that lies on line $p$. + + + + + \bizrek \label{PotencnoSr} + Let $k$, $l$ and $j$ be three non-concentric circles with non-collinear centres. + Then there is exactly one point that has the same power with respect to all three circles. + This point is the intersection of their three radical axes $p(k,l)$, $p(l,j)$ and $p(k,j)$. + \eizrek + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.6.pic}) Because the centres of circles $k$, $l$ and $j$ are three non-linear points, no two of lines $p(k,l)$, $p(l,j)$ and $p(k,j)$ are parallel. Let $P=p(k,l)\cap p(l,j)$. Then +$$p(P,k)=p(P,l)=p(P,j),$$ + i.e. $P\in p(k,j)$, which means that the power lines $p(k,l)$, $p(l,j)$ and $p(k,j)$ intersect in point $P$. + +If for another point $\widehat{P}$ of this plane it is valid that $p(\widehat{P},k)=p(\widehat{P},l)=p(\widehat{P},j)$, then $\widehat{P}\in p(k,l),\hspace*{1mm}p(l,j),\hspace*{1mm}p(k,j)$, so $\widehat{P}=P$. +\kdokaz + +The point from the previous statement (\ref{PotencnoSr}) is called the \index{središče!potenčno} + \pojem{potential center} of three circles. We will denote the potential center of circles $k$, $l$ and $j$ with $p(k,l,j)$. + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.6.pic} +\caption{} \label{sl.pod.7.12.6.pic} +\end{figure} + + In the case when the centers of three circles are three collinear points, and the circles are not from the same pencil, all three radical axes are parallel, because they are all perpendicular to the common center of these circles (statement \ref{PotencnaOs}). + From this and statement \ref{PotencnoSr} we directly get the following statement (Figure \ref{sl.pod.7.12.6.pic}). + + +Let $k$, $l$ and $j$ be three non-concentric circles with non-collinear centres. + Then there is exactly one point that has the same power with respect to all three circles. + This point is the intersection of their three radical axes $p(k,l)$, $p(l,j)$ and $p(k,j)$. + + Radical axes of three circles in the plane that are not + from the same pencil and no two of them are concentric, + belong to the same family of lines. + + + \bizrek \label{PotencnoSrSop} + Radical axes of three circles in the plane that are not + from the same pencil and no two of them are concentric, + belong to the same family of lines. + \eizrek + +A special case of statement \ref{PotencnoSr}, when each two lines intersect, is the following theorem. + + + \bzgled + Let $k$, $l$, and $j$ be three circles of some plane with nonlinear + centres and: + \begin{itemize} + \item $A$ and $B$ intersections of the circles $k$ and $l$, + \item $C$ and $D$ intersections of the circles $l$ and $j$, + \item $E$ and $F$ intersections of the circles $j$ and $k$. + \end{itemize} + Prove that the lines $AB$, $CD$ and $EF$ intersect at a single point. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.7.pic} +\caption{} \label{sl.pod.7.12.7.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.7.pic}) + +According to the \ref{PotencnaOs} theorem, $p(k,l)=AB$, $p(l,j)=CD$ and $p(j,k)=EF$ are the corresponding power axes. According to the \ref{PotencnoSr} theorem, they intersect at one point - the power center $P=p(k,l,j)$ of these three circles. +\kdokaz + +The next statement is very similar. + + + + + \bzgled + Let $k$, $l$, and $j$ be three circles of some plane with nonlinear + centres and: + \begin{itemize} + \item $t_1$ the common tangent of the circles $k$ in $l$, + \item $t_2$ the common tangent of the circles $l$ in $j$, + \item $t_3$ the common tangent of the circles $j$ in $k$. + \end{itemize} + Prove that the lines $t_1$, $t_2$ and $t_3$ intersect at a single point. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.8.pic} +\caption{} \label{sl.pod.7.12.8.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.8.pic}) + +According to the \ref{PotencnaOs} theorem, $p(k,l)=t_1$, $p(l,j)=t_2$ and $p(j,k)=t_3$ are the corresponding power axes. According to the \ref{PotencnoSr} theorem, they intersect at one point - the power center $P=p(k,l,j)$ of these three circles. +\kdokaz + +\bzgled + a) Suppose that circles $k$ and $l$ are touching each other externally, and a line $t$ is + is the common tangent of these circles at their common point. Let $AB$ be a second + common tangent of these circles at touching points $A$ and $B$. Prove that the midpoint + the line segment $AB$ lies on the tangent $t$.\\ + b) Suppose that circles $k$ and $l$ intersect at points $P$ and $Q$. Let $AB$ be a common tangent + of these circles at touching points $A$ and $B$. Prove that the midpoint of the line segment $AB$ lies on + line $PQ$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.9.pic} +\caption{} \label{sl.pod.7.12.9.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.9.pic}) + +\textit{a)} Let $S$ be the midpoint of the line segment $AB$. Then we have: +$p(S,k)= |SA|^2 = |SB|^2 = p(S,l)$, +so the point $S$ lies on the power curve of circles $k$ and $l$ or on the line $t$ (statement \ref{PotencnaOs}). + +\textit{b)} Just as in the previous example, only that the power curve of circles $k$ and $l$ is the line $AB$ in this case. +\kdokaz + + + + \bizrek \label{EulerjevaFormula} + \index{formula!Eulerjeva} + (Euler's\footnote{\index{Euler, L.} + \textit{L. Euler} + (1707--1783), švicarski matematik.} formula) If $k(S,r)$ is the incircle and $l(O,R)$ the circumcircle +of an arbitrary triangle, then + $$|OS|^2=R^2- 2Rr.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.10.pic} +\caption{} \label{sl.pod.7.12.10.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.10.pic}) + +We denote by $A$, $B$ and $C$ the vertices of a triangle, $NM$ the diameter of the circumscribed circle $l$, which is perpendicular to +the side $BC$ (and also $A,N\div BC$). By \ref{TockaN} the point $N$ lies on the bisector of the internal angle at the vertex $A$ or on the altitude +$AS$ (\ref{SredVcrtaneKrozn}). By \ref{TockaN.NBNC} we have $NS\cong NC$. If we use the power of the point $S$ with respect to the circle $l$ (\ref{PotencnoSr}) and the fact $\mathcal{B}(A,S,N)$, we get $p(S,l)= |SO|^2 - R^2 = \overrightarrow{SA}\cdot \overrightarrow{SN} + = -|SA|\cdot |SN| += -|SA|\cdot |CN|$, thus: + \begin{eqnarray} \label{eqnEulFormOS} + |SO|^2 - R^2 = -|SA|\cdot |CN|. +\end{eqnarray} +We denote by $Q$ the point of contact of the inscribed circle $k$ with the side $AC$ of the triangle $ABC$. +By \ref{TangPogoj} and \ref{TalesovIzrKroz2} we have $\angle AQS\cong\angle MCN=90^0$, from \ref{ObodObodKot} we also get $\angle SAQ=\angle NAC\cong\angle NMC$. +Therefore, the triangles $AQS$ and $MCN$ are similar (\ref{PodTrikKKK}), so: +$$\frac{AS}{MN}=\frac{SQ}{NC}$$ or $|AS|\cdot |NC|=|MN|\cdot |SQ|=2Rr$. If we insert this into \ref{eqnEulFormOS}, we get: +$$|SO|^2 = R^2-|SA|\cdot |CN|=R^2-2Rr,$$ +which had to be proven. \kdokaz + + +The next task is a special case of the previous formula and is therefore +its consequence (Figure \ref{sl.pod.7.12.10a.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.10a.pic} +\caption{} \label{sl.pod.7.12.10a.pic} +\end{figure} + + + \bnaloga\footnote{4. IMO, Czechoslovakia - 1962, Problem 6.} + Consider an isosceles triangle. Let $r$ be the radius of its circumscribed circle + and $\rho$ the radius of its inscribed circle. Prove that the distance $d$ between + the centres of these two circles is $$d = \sqrt{r(r-2\rho)}.$$ + \enaloga + +The next statement is also a direct consequence of \ref{EulerjevaFormula}. + +\bizrek +If $k(S,r)$ is the incircle and $l(O,R)$ the circumcircle +of an arbitrary triangle, then +$$R\geq 2r.$$ +Equality is achieved for an equilateral triangle. +\eizrek + +The next design task is one of the ten Apollonius' problems on the tangency of circles, which we will investigate in more detail in section \ref{odd9ApolDotik}. + + +\bzgled +Construct a circle through two given points $A$ and $B$ and tangent to a given line $t$. +\ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.11.pic} +\caption{} \label{sl.pod.7.12.11.pic} +\end{figure} + + +\textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.12.11.pic}) + +Let $k$ be the desired circle, which passes through points $A$ and $B$ and is tangent to the line $t$ in point $T$. + +If the lines $AB$ and $t$ are parallel, +the third point $T$ of the circle $k$ is obtained as the intersection of the line +$t$ with the perpendicular of the line $AB$. + +Let $P$ be the intersection of the lines $AB$ and $t$. + +We will use the power of point $P$ with respect to + the circle $k$. With $l$ we denote any +circle, which passes through points $A$ and $B$. By Theorem \ref{PotencnaOs} the line $AB$ +is the power line of the circles $k$ and $l$, therefore $p(P,k)=p(P,l)$. We denote with $PT$ and +$PT_1$ the tangents of the circles $k$ and $l$ in their points $P$ and $P_1$. Then it holds (Theorem \ref{izrekPotenca}): + $$|PT|^2=p(P,k)=p(P,l)=|PT_1|^2$$ +or $|PT|=|PT_1|$. + +The last relation allows us to construct the third point $T$ of the circle $k$. + \kdokaz + +\bzgled + Let $E$ be the intersection of the bisector of the interior angle at the vertex + $A$ with the side $BC$ of a triangle $ABC$ and $A_1$ the midpoint of this side. + Let $P$ and $Q$ + be intersections of the circumcircle of the triangle $AEA_1$ with the sides $AB$ + and $AC$ of the triangle $ABC$. Prove that: + $$BP\cong CQ.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.12.pic} +\caption{} \label{sl.pod.7.12.12.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.12.pic}) + +We'll mark the circumcircle of the triangle $AEA_1$ with $k$. If we use the power of points $B$ and $C$ according to the circle $k$, the relation $\mathcal{B}(B,P,A)$ and $\mathcal{B}(C,Q,A)$ (because of the assumption that points $P$ and $Q$ lie on the sides $AB$ and $AC$ of the triangle $ABC$) and +the equivalence \ref{eqnMnozVektRelacijaB} from section \ref{odd5DolzVekt}, we get: + \begin{eqnarray*} +p(B,k)&=&|BP|\cdot |BA|=|BE|\cdot |BA_1|,\\ +p(C,k)&=&|CP|\cdot |CA|=|CE|\cdot |CA_1|. +\end{eqnarray*} + From this and the relation $BA_1\cong CA_1$ and the theorem \ref{HarmCetSimKota} we get: +\begin{eqnarray*} +\frac{|BP|\cdot |BA|}{|CP|\cdot |CA|} +=\frac{|BE|\cdot |BA_1|}{|CE|\cdot |CA_1|} +=\frac{|BE|}{|CE|} +=\frac{|BA|}{|CA|}. +\end{eqnarray*} +Therefore $\frac{|BP|\cdot |BA|}{|CP|\cdot |CA|}=\frac{|BA|}{|CA|}$ or $|BP|=|CP|$. + \kdokaz + + \bzgled + Construct a circle that is perpendicular to three given circles + $k$, $l$ in $j$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.13.pic} +\caption{} \label{sl.pod.7.12.13.pic} +\end{figure} + + +\textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.12.13.pic}) + +We will first assume that the centers of these circles $k$, $l$ and $j$ are nonlinear points. Let $x$ be the desired circle with center $P$, which is perpendicular to the circles $k$, $l$ and $j$, and $A\in x\cap k$, $B\in x\cap l$ and $C\in x\cap j$. Because $x\perp k,j,l$, $PA$, $PB$ and $PC$ are tangents of these circles from point $P$ (by \ref{pravokotniKroznici}). By \ref{izrekPotenca} we have $p(P,k)=|PA|^2$, $p(P,l)=|PB|^2$ and $p(P,j)=|PC|^2$. Because points $A$, $B$ and $C$ lie on the circle $x$ with center $P$, we have $|PA|^2=|PB|^2=|PC|^2$, i.e. $p(P,k)=p(P,l)= p(P,j)$. This means that $P=p(k,l,j)$ is the power center of the circles $k$, $l$ and $j$. + +Therefore, the desired circle $x$ can be constructed by first drawing its center $P=p(k,l,j)$, then the radius $PA$, where the line $PA$ is tangent to the circle $k$ at point $A$. +It is clear that in the case where $P$ is an internal point of one of the circles $k$, $l$, $j$, the task has no solution. + +Even in the case where the centers of the +circles are collinear points, the desired circle +does not exist (it is a "degenerate circle" or a line representing their common center). + \kdokaz + + + + \bzgled + Circles $k(O,r)$ and $l(S,\rho)$ and a point $P$ in the same + plane are given. + Construct a line passing through the point $P$, which determine congruent chords + on the circles $k$ and $l$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.14.pic} +\caption{} \label{sl.pod.7.12.14.pic} +\end{figure} + + +\textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.12.14.pic}) + +Let $p$ be a line that goes through the point $P$, +the circles $k$ and $l$ intersect in such points $A$ and $B$ or $C$ and $D$, that $AB\cong CD$. Let +$\overrightarrow{v}$ be a vector that is determined by +the centers of the lines $AB$ and $CD$. +Then +$\mathcal{T}_{\overrightarrow{v}}:\hspace*{1mm}A,B\mapsto C,D$, +the circle $k$ is transformed by this translation into +the circle $k'$, that goes through the points $C$ and $D$. So +the circles $k'$ and $l$ intersect in the points $C$ and $D$. +Thus the problem of constructing the line $p$ is translated into the problem +of constructing the vector + $\overrightarrow{v}$ or the point +$O'= \mathcal{T}_{\overrightarrow{v}}(O)$, which represents the center of the circle $k'$. + +The line, that +is determined by the centers $S$ and $O'$ of the circles $l$ and $k'$, is perpendicular to their common chord $CD$. +Because $\overrightarrow{OO'}= \overrightarrow{v} \parallel p$, it follows + $$\angle OO'S=90^0.$$ +By \ref{TalesovIzrKroz2} the point $O'$ lies on the circle above the diameter $OS$. + +The point $P$ lies on the power line $p$ of the circles $k'$ and $l$ (\ref{PotencnaOs}). From this it follows $PL\cong PK$, where $PL$ and +$PK$ are tangents to the circles $l$ and $k'$ in the points $L$ and $K$. +This means that we can construct a triangle that is similar to the right triangle $PKO'$ ($PK\cong PL$, $\angle PO'K=90^0$ and $KO'\cong r$), and thus also + $d$, which is similar to the distance $PO'$. So the point $O'$ belongs +to the intersection of the circle with center $P$ and radius $d$ and the circle above the diameter $OS$. + \kdokaz + + + + \bnaloga\footnote{36. IMO Canada - 1995, Problem 1.} + Let $A$, $B$, $C$, $D$ be four distinct points on a line, in that order $\mathcal{B}(A,B,C,D)$. The + circles with diameters $AC$ and $BD$ intersect at $X$ and $Y$. The line $XY$ + meets $BC$ at $Z$. Let $P$ be a point on the line $XY$ other than $Z$. The + line $CP$ intersects the circle with diameter $AC$ at $C$ and $M$, and the + line $BP$ intersects the circle with diameter $BD$ at $B$ and $N$. Prove + that the lines $AM$, $DN$, $XY$ are concurrent. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.IMO1.pic} +\caption{} \label{sl.pod.7.12.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $k_1$ and $k_2$ be the circles +over the segments $AC$ and $BD$ (Figure \ref{sl.pod.7.12.IMO1.pic}). By +the theorem \ref{KroznPresABpravokOS} the line $XY$ is perpendicular to +the central line $BC$ of these two circles and is also their power line +(theorem \ref{PotencnaOs}). Let $S_1=AM\cap XY$ and $S_2=DN\cap +XY$. It is enough to prove $S_1=S_2$. + + Because $M\in k_1$, by the theorem \ref{TalesovIzrek} $\angle + AMC=90^0$ or $AS_1\perp MC$. Because of this the + $\angle MCA\cong\angle AS_1Z$ (angle with perpendicular legs + - theorem \ref{KotaPravokKraki}). From this similarity of angles + it follows that $\frac{AZ}{PZ}=\frac{ZS_1}{ZC}$ or + $|ZS_1|=\frac{|ZC|\cdot |ZA|}{|PZ|}$. But $|ZC|\cdot + |ZA|=p(Z,k_1)=|ZX|\cdot + |ZY|=|ZX|^2$. From this it follows: + $$|ZS_1|=\frac{|ZX|^2}{|PZ|}.$$ + In the same way it can be proven that: + $|ZS_2|=\frac{|ZX|^2}{|PZ|}$. Because $S_1$ and $S_2$ are on + the same line segment $ZP$, $S_1=S_2$. + \kdokaz + + + + \bzgled + Let $P$ be an arbitrary point in the plane of a triangle $ABC$ which + does not lie on any of lines containing altitudes of this triangle. Suppose $A_1$ is a point, + in which a perpendicular line of the line $AP$ at the point $P$ intersects the line $BC$. Analogously + we can also define points $B_1$ and $C_1$. + Prove that $A_1$, $B_1$ and $C_1$ are three collinear points. + \ezgled + +\textbf{\textit{Solution.}} We mark with $A_C$ and $B_C$ the +orthogonal projections of points $A$ and $B$ on the line $CP$. +Similarly, $A_B$ and $C_B$ are the orthogonal projections of +points $A$ and $C$ on the line $BP$, and $B_A$ and $C_A$ are the +orthogonal projections of points $B$ and $C$ on the line $AP$ +(Figure \ref{sl.pd.7.4.6.pic}). By Tales' theorem \ref{TalesovIzrek} +it is: + \begin{eqnarray} \label{4.1}\frac{AC_1}{C_1B} + \cdot \frac{BA_1}{A_1C} + \cdot \frac{CB_1}{B_1A}= +\frac{A_CP}{PB_C} + \cdot +\frac{B_AP}{PC_A} + \cdot +\frac{C_BP}{PA_B} + \end{eqnarray} + +\begin{figure}[!htb] +\centering +\input{sl.pd.7.4.6.pic} +\caption{} \label{sl.pd.7.4.6.pic} +\end{figure} + + + From $\angle AA_CC\cong\angle AC_AC=90^0$ it follows that + points $A_C$ and $C_A$ lie on the circle with diameter $AC$. Therefore + the power of point $P$ on this circle is equal (by theorem \ref{izrekPotenca}) + $\overrightarrow{PC}\cdot \overrightarrow{PA_C}= + \overrightarrow{PA}\cdot \overrightarrow{PC_A}$. + Similarly, $\overrightarrow{PB}\cdot \overrightarrow{PC_B}= + \overrightarrow{PC}\cdot \overrightarrow{PB_C}$ and + $\overrightarrow{PB}\cdot \overrightarrow{PA_B}= + \overrightarrow{PA}\cdot \overrightarrow{PB_A}$. From these relations + we get: $\frac{PA_C}{PC_A}=\frac{PA}{PC}$, + $\frac{PC_B}{PB_C}=\frac{PC}{PB}$ and + $\frac{PB_A}{PA_B}=\frac{PB}{PA}$. If we insert these relations in + \ref{4.1}, we get $\frac{AC_1}{C_1B} + \cdot \frac{BA_1}{A_1C} + \cdot \frac{CB_1}{B_1A}=1$. + Because $\frac{\overrightarrow{AC_1}}{\overrightarrow{C_1B}} + \cdot \frac{\overrightarrow{BA_1}}{\overrightarrow{A_1C}} + \cdot \frac{\overrightarrow{CB_1}}{\overrightarrow{B_1A}}<0$, + it is + $\frac{\overrightarrow{AC_1}}{\overrightarrow{C_1B}} + \cdot \frac{\overrightarrow{BA_1}}{\overrightarrow{A_1C}} + \cdot \frac{\overrightarrow{CB_1}}{\overrightarrow{B_1A}}=-1$. By + Menelaus' theorem points $A_1$, $B_1$ and $C_1$ are collinear. +\kdokaz + +\bnaloga\footnote{41. IMO, S. Korea - 2000, Problem 1.} + $AB$ is tangent to the circles $CAMN$ and $NMBD$. $M$ lies + between $C$ and $D$ on the line $CD$, and $CD$ is parallel to $AB$. The chords + $NA$ and $CM$ meet at $P$; the chords $NB$ and $MD$ meet at $Q$. The rays $CA$ + and $DB$ meet at $E$. Prove that $PE\cong QE$. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.IMO3.pic} +\caption{} \label{sl.pod.7.12.IMO3.pic} +\end{figure} + +\textbf{\textit{Solution.}} We mark with $L$ the intersection of the lines $MN$ +and $AB$ and with $k$ and $l$ the circumscribed circles + $CAMN$ and $NMBD$ (Figure \ref{sl.pod.7.12.IMO3.pic}). + +By Theorem \ref{PotencnaOs} the line $MN$ is the power line of the circles $k$ +and $l$, therefore for its point $L\in MN$ we have +$|LA|^2=p(L,k)=p(L,l)=|LB|^2$, which means that $L$ is the center +of the line segment $AB$. Because, by assumption, $AB\parallel CD$, i.e. +$AB\parallel PQ$, by Tales' theorem $MP:MQ=LA:LB=1$. Therefore +the point $M$ is the center of the line segment $PQ$ or $MP\cong MQ$. + +If we use Theorem \ref{ObodKotTang} and \ref{KotiTransverzala}, +we get: + \begin{eqnarray*} + \angle EAB &\cong& \angle AMC \cong\angle MAB\\ + \angle EBA &\cong& \angle BMQ \cong\angle MBA + \end{eqnarray*} +This means that the triangles $AEB$ and $AMB$ are similar (by the \textit{ASA} Theorem \ref{KSK}), +namely, they are symmetric with respect to the line $AB$. This +means that $EM\perp AB$. Because $AB\parallel PQ$, $EM\perp PQ$ or $\angle PME\cong\angle QME =90^0$. +We have already proven that $MP\cong MQ$, therefore the triangles $PME$ and $QME$ are similar +(by the \textit{SAS} Theorem \ref{SKS}), from which it follows that $PE\cong QE$. + \kdokaz + +\bnaloga\footnote{40. IMO, Romania - 1999, Problem 5.} + Two circles $k_1$ and $k_2$ are contained inside the circle $k$, and are tangent to $k$ + at the distinct points $M$ and $N$, respectively. $k_1$ passes through the center of + $k_2$. The line passing through the two points of intersection of $k_1$ and $k_2$ meets + $k$ at $A$ and $B$. The lines $MA$ and $MB$ meet $k_1$ at $C$ and $D$, respectively. + Prove that $CD$ is tangent to $k_2$. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.IMO4.pic} +\caption{} \label{sl.pod.7.12.IMO4.pic} +\end{figure} + +\textbf{\textit{Solution.}} We mark with $O_1$ and $O_2$ the centers +of the circles $k_1$ and $k_2$, with $r_1$ and $r_2$ their radii and +with $E$ the other point of intersection of the line $AN$ with the +circle $k_2$ (Figure \ref{sl.pod.7.12.IMO4.pic}). Without loss of +generality we assume $r_1\geq r_2$. + +We will first prove that the line $CE$ is the common tangent of the circles $k_1$ and $k_2$. Let $\widehat{E}$ be the other intersection of the drawn circle $k'$ of the triangle $CMN$ and the circle $k_2$. The point $A$ lies on the power axes of the circles $k_1$ and $k_2$ or $k_1$ and $k'$, so according to the \ref{PotencnoSr} theorem, the point $A$ is the power center of the circles $k_1$, $k_2$ and $k'$. This means that the point $A$ lies on the power line of the circles $k_2$ and $k'$ - the line $N\widehat{E}$. From this it follows that $\widehat{E}\in AN\cap k_2$ or $\widehat{E}=E$. So the points $M$, $C$, $E$ and $N$ are conciliatory and according to the \ref{ObodKotTang} theorem, $\angle ACE \cong\angle ANM$. We mark with $L$ any point of the common tangent of the circles $k$ and $k_1$ in the point $M$, which lies in the plane with the edge $AC$, in which the points $B$ and $D$ are not. According to the same theorem \ref{ObodKotTang} (with respect to the circles $k$ and $k_1$), we have $\angle LMA \cong\angle MBA$ and $\angle LMC \cong\angle MDC$. From the \ref{ObodObodKot} theorem (for the circle $k$ and the cord $AM$), we also get $\angle ANM \cong\angle ABM$. If we connect the proven relations, we get: +$$\angle ACE \cong \angle ANM +\cong \angle ABM\cong \angle LMA \cong\ \angle CDM. $$ + From $\angle ACE \cong \angle CDM$ and according to the \ref{ObodKotTang} theorem, it follows that $EC$ is the tangent of the circle $k_1$. Because in the proof we have not yet used the fact that $O_2\in k_1$, we analogously prove that $CE$ is the tangent of the circle $k_2$. + +Let $T$ be the intersection of line $O_2O_1$ and circle $k_2$. It is enough to +prove that $T\in CD$ and $\angle CTO_2=90^0$. Let $O'_2$ +be the orthogonal projection of point $O_2$ on line $O_1C$. Because $CE$ +is the common tangent of circles $k_1$ and $k_2$, the radii $O_1C$ and +$O_2E$ are perpendicular to this tangent. Therefore, $CEO_2O'_2$ +is a rectangle and $O'_2C\cong O_2E=r_2$ holds. From this it follows +that $O_1O'_2=r_1-r_2=O_1T$. This means that the triangle +$O_1O'_2O_2$ is similar (by the \textit{SAS} theorem \ref{SKS}), to triangle +$O_1TC$, so $\angle CTO_1\cong\angle O_2O'_2O_1=90^0$ or $\angle +CTO_2=90^0$. Line $CT$ and line $AB$ are perpendicular to +the center $O_1O_2$ of the two circles. Therefore, $CT\parallel AB$. Because +(due to the already proven relation $\angle ABM\cong \angle CDM$), also $CD\parallel AB$. By Playfair's axiom \ref{Playfair}, +line $CT$ and line $CD$ are the same line, or $T\in CD$. This means that line +$CD$ is tangent to circle $k_2$ at point $T$. +\kdokaz + + + + +%________________________________________________________________________________ +\poglavje{The Theorems of Pappus and Pascal} \label{odd7PappusPascal} + +Theorems in this section are historically connected to the development of \index{geometry!projective}projective geometry. + + + + \bizrek \label{izrek Pappus} \index{theorem!Pappus'}(Pappus'\footnote{\index{Pappus} \textit{Pappus} from Alexandria (3rd century), one of the last + great ancient Greek geometers. He proved this theorem in the Euclidean case, + using metric. But the fundamental role of Pappus' theorem + in projective geometry was discovered only sixteen centuries + later.} theorem) + Let $A$, $B$ and $C$ be three different + points of a line $p$ and $A'$, $B'$ and $C'$ three different points of another line + $p'$ in the same plane. Then the points + $X=BC'\cap B'C$, $Y=AC'\cap A'C$ and $Z=AB'\cap A'B$ are collinear. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10.1.pic} +\caption{} \label{sl.pod.7.10.1.pic} +\end{figure} + + + \textbf{\textit{Proof.}} +Let $L=AB'\cap BC'$, $M=AB'\cap CA'$ and $N=CA'\cap BC'$ (Figure \ref{sl.pod.7.10.1.pic}). We use Menelaus' theorem (\ref{izrekMenelaj}) five times +with respect to the triangle $LMN$ and the lines $BA'$, $AC'$, $CB'$, $AB$, $A'B'$: + +\begin{eqnarray*} +& & \frac{\overrightarrow{LZ}}{\overrightarrow{ZM}}\cdot \frac{\overrightarrow{MA'}}{\overrightarrow{A'N}}\cdot \frac{\overrightarrow{NB}}{\overrightarrow{BL}}=-1,\\ +& & \frac{\overrightarrow{LA}}{\overrightarrow{AM}}\cdot \frac{\overrightarrow{MY}}{\overrightarrow{YN}}\cdot \frac{\overrightarrow{NC'}}{\overrightarrow{C'L}}=-1,\\ +& & \frac{\overrightarrow{LB'}}{\overrightarrow{B'M}}\cdot \frac{\overrightarrow{MC}}{\overrightarrow{CN}}\cdot \frac{\overrightarrow{NX}}{\overrightarrow{XL}}=-1,\\ +& & \frac{\overrightarrow{LA}}{\overrightarrow{AM}}\cdot \frac{\overrightarrow{MC}}{\overrightarrow{CN}}\cdot \frac{\overrightarrow{NB}}{\overrightarrow{BL}}=-1,\\ +& & \frac{\overrightarrow{LB'}}{\overrightarrow{B'M}}\cdot \frac{\overrightarrow{MA'}}{\overrightarrow{A'N}}\cdot \frac{\overrightarrow{NC'}}{\overrightarrow{C'L}}=-1. +\end{eqnarray*} +From these five relations (if we multiply the first three, then insert the fourth and fifth into the resulting relation) we get: + $$\frac{\overrightarrow{LZ}}{\overrightarrow{ZM}}\cdot \frac{\overrightarrow{MY}}{\overrightarrow{YN}}\cdot \frac{\overrightarrow{NX}}{\overrightarrow{XL}}=-1.$$ + By the converse of Menelaus' theorem (\ref{izrekMenelaj}) the points $X$, $Y$ and $Z$ are collinear. + \kdokaz + + We now prove Pascal's\footnote{It is not known how the French mathematician and philosopher \index{Pascal, B.} + \textit{B. Pascal} (1623--1662) proved this statement for a circle, because + the original proof is lost. However, we can assume that he used + the results + and methods of that time, which means that he probably used Menelaus' theorem.} + theorem for + a circle. + +\bizrek \index{izrek!Pascalov} \label{izrekPascalEvkl} + Let $A$, $B$, $C$, $D$, $E$ and $F$ be arbitrary points on + some circle $k$. Then the points $X=AE\cap BD$, $Y=AF\cap CD$ and + $Z=BF\cap CE$ are collinear. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10.2.pic} +\caption{} \label{sl.pod.7.10.2.pic} +\end{figure} + + + \textbf{\textit{Proof.}} +We mark $L=AE\cap BF$, $M=AE\cap CD$ and $N=CD\cap BF$. + If we use Menelaus' theorem (\ref{izrekMenelaj}) three times for the triangle $LMN$ and the lines $BD$, +$AF$ and $CE$, we get (Figure \ref{sl.pod.7.10.2.pic}): + \begin{eqnarray*} + \hspace*{-2mm} \frac{\overrightarrow{LX}}{\overrightarrow{XM}}\cdot + \frac{\overrightarrow{MD}}{\overrightarrow{DN}}\cdot + \frac{\overrightarrow{NB}}{\overrightarrow{BL}}=-1,\hspace*{1mm} + \frac{\overrightarrow{LA}}{\overrightarrow{AM}}\cdot + \frac{\overrightarrow{MY}}{\overrightarrow{YN}}\cdot + \frac{\overrightarrow{NF}}{\overrightarrow{FL}}=-1,\hspace*{1mm} + \frac{\overrightarrow{LE}}{\overrightarrow{EM}}\cdot + \frac{\overrightarrow{MC}}{\overrightarrow{CN}}\cdot + \frac{\overrightarrow{NZ}}{\overrightarrow{ZL}}=-1. + \end{eqnarray*} + If we use the powers of the points $M$, $N$ and $L$ with respect to the circle + $k$ (theorem \ref{izrekPotenca}) or consider the similarity of the corresponding triangles, we get: + \begin{eqnarray*} + \overrightarrow{MC}\cdot\overrightarrow{MD}= + \overrightarrow{MA}\cdot\overrightarrow{ME},\hspace*{3mm} + \overrightarrow{NC}\cdot\overrightarrow{ND}= + \overrightarrow{NF}\cdot\overrightarrow{NB},\hspace*{3mm} + \overrightarrow{LA}\cdot\overrightarrow{LE}= + \overrightarrow{LB}\cdot\overrightarrow{LF}. + \end{eqnarray*} + From the previous six relations it follows: + \begin{eqnarray*} + \frac{\overrightarrow{LX}}{\overrightarrow{XM}}\cdot + \frac{\overrightarrow{MY}}{\overrightarrow{YN}}\cdot + \frac{\overrightarrow{NZ}}{\overrightarrow{ZL}}=-1. + \end{eqnarray*} +Therefore, according to Menelaus' theorem (\ref{izrekMenelaj}, the inverse direction), the points $X$, $Y$ and +$Z$ are collinear. + \kdokaz + +The previous statement, which refers to a circle, + can be generalized to any conic section\footnote{The study of conic sections began +with the Ancient Greeks. The terms ellipse, parabola, hyperbola +were first used by the Greek mathematician \index{Apolonij} \textit{Apolonij} from Perga (262--200 BC) in + his +famous work \textit{Razprava o presekih stožca} (Discussion of the Intersections of a Cone), which +consists + of +eight books, in which he defines a conic section as the intersection of a plane with +a circular cone. A commentary on this work by Apolonij was written by the Greek philosopher and mathematician, and the last representative of ancient science, \index{Hipatija} \textit{Hipatija} from Alexandria (370--415). The interest in conic sections was revived + by the German astronomer \index{Kepler, J.} \textit{J. Kepler} (1571--1630) and +the French mathematician and philosopher \index{Pascal, B.} \textit{B. +Pascal} (1623--1662) in the 17th century.} + or a curve of the second class. + In the Euclidean case, these are: ellipse, parabola and hyperbola. + We can determine this by defining + a conic section + as the intersection of the generators of all + sides of the cone and some plane. + Because of this, we generally look at a conic section $\mathcal{K}$ as + the central + projection of some circle $k$ onto some plane. The center of projection is + the apex of the cone $S$. We have already mentioned that + a central projection preserves collinearity. If $A'$, $B'$, $C'$, + $D'$, $E'$ and $F'$ are points on the conic section $\mathcal{K}$ and $X'$, $Y'$ and $Z'$ + are the corresponding points, defined as in statement \ref{izrekPascalEvkl}, + then these points are the images of some points $A$, $B$, $C$, $D$, $E$, $F$, $X$, $Y$ and + $Z$. + In this case, the first six points are on the circle $k$, and the last three are + collinear by statement \ref{izrekPascalEvkl}. It follows that the points $X'$, $Y'$ and $Z'$ are also collinear (Figure \ref{sl.pod.7.10.3.pic}). This means that statement \ref{izrekPascalEvkl} + really holds in the general case for any conic section. This general statement is + known as Pascal's statement for conic sections\footnote{\index{Pascal, B.} \textit{B. Pascal} +(1623--1662), French mathematician and philosopher, who + as a sixteen-year-old proved this important statement about + conic sections, and published it +in 1640, but at that time the statement did not directly apply to +projective geometry.}. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10.3.pic} +\caption{} \label{sl.pod.7.10.3.pic} +\end{figure} + + The ideas we have discussed in this section lead us to the conclusion that we can also derive Pascal's theorem in projective geometry. Even more - in projective geometry we can define and study also the conics, but it is not possible to distinguish between ellipse, hyperbola or parabola (see \cite{Mitrovic}). + + +%________________________________________________________________________________ + \poglavje{The Golden Ratio} \label{odd7ZlatiRez} + + We say that point $Z$ of the line $AB$ divides this line in the ratio of the \index{golden!ratio}\pojem{golden ratio}\footnote{Such a division was already considered by \index{Pythagoras}\textit{Pythagoras from the island of Samos} (582--497 BC), an ancient Greek philosopher and mathematician. The first known records of the golden ratio were created by the ancient Greek mathematician \index{Euclid} + \textit{Euclid of Alexandria} (3rd century BC). In his famous work \textit{Elements} he posed the problem: ‘‘\textit{Given a line, divide it into two unequal parts so that the area of the rectangle whose length is equal to the length of the whole line and whose height is equal to the length of the shorter part of the line is equal to the area of the square drawn on the longer part of the line.}’’ +The term golden ratio, which we use today, was introduced by \index{Leonardo da Vinci}\textit{Leonardo da Vinci} (1452--1519), +an Italian painter, architect and inventor. The golden ratio has been used by people for thousands of years in painting and architecture.}, if the ratio of the length of the whole line to the longer part is equal to the ratio of the longer part to the shorter part (Figure \ref{sl.pod.7.15.1.pic}) or: + + \begin{eqnarray} \label{eqnZlatiRez} + AB:AZ=AZ:ZB. + \end{eqnarray} + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.1.pic} +\caption{} \label{sl.pod.7.15.1.pic} +\end{figure} + +If +we now insert the shorter part "into" the longer part, we get the same +ratio. Indeed, because +$AZ:ZB=AB:AZ=(AZ+ZB):AZ$, it also holds that +$$ZB:(AZ-ZB)=AZ:ZB.$$ +We can continue this process (Figure \ref{sl.pod.7.15.2.pic}) + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.2.pic} +\caption{} \label{sl.pod.7.15.2.pic} +\end{figure} + +Of course, the question arises as to how to construct such a point $Z$. We will describe this construction in the next example. + + + +\bzgled + For a given line $AB$, construct a point $Z$ that divides the line segment into + the golden ratio. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.3.pic} +\caption{} \label{sl.pod.7.15.3.pic} +\end{figure} + + + \textbf{\textit{Solution.}} +First, let's draw a circle $k(S,SB)$ with radius +$|SB|=\frac{1}{2}|AB|$, which touches the line $AB$ at point $B$ (Figure \ref{sl.pod.7.15.3.pic}). Let's also construct the intersections +of this circle with the line $AS$ - we'll mark them with $X$ and $Y$ (so that $\mathcal{B}(A,X,S)$ is true). We now get the point $Z$ as +the intersection of the line $AB$ and the circle +$l(A,AX)$. + +We will prove that $Z$ is the desired point. If we use the power of point $A$ with respect to the circle $k$ (statement \ref{izrekPotenca}), we get: +$$\hspace*{-1.5mm} |AB|^2=p(A,k)=|AX|\cdot |AY|=|AX|\cdot(|AX|+|XY|)=|AZ|\cdot(|AZ|+|AB|).$$ + Therefore: +\begin{eqnarray} \label{eqnZlatiRez2} + |AB|^2=|AZ|\cdot(|AZ|+|AB|) + \end{eqnarray} + Next: + \begin{eqnarray*} + |AB|^2=|AZ|\cdot(|AZ|+|AB|)\hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} + \frac{|AB|}{|AZ|}=\frac{|AZ|+|AB|}{|AB|}\\ + \hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} + \frac{|AB|}{|AZ|}=\frac{|AZ|}{|AB|}+1\\ + \hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} + \frac{|AB|}{|AZ|}-1=\frac{|AZ|}{|AB|}\\ + \hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} + \frac{|AB|-|AZ|}{|AZ|}=\frac{|AZ|}{|AB|}\\ + \hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} + \frac{|BZ|}{|AZ|}=\frac{|AZ|}{|AB|} + \end{eqnarray*} + which is equivalent to relation \ref{eqnZlatiRez}. This means that point $Z$ divides the line segment $AB$ in the golden ratio. + \kdokaz + + In the next example we will calculate the value of the ratio determined by the golden ratio. + + + + \bizrek \label{zlatiRezStevilo} + If a point $Z$ divides a line segment $AB$ into the golden ratio + ($AZ$ is the longer part), then + \begin{eqnarray*} + && AZ:ZB=AB:AZ=\frac{\sqrt{5}+1}{2}, \hspace*{1mm}\textrm{ i.e.}\\ + && AZ=\frac{\sqrt{5}-1}{2}AB. + \end{eqnarray*} + \eizrek + + \textbf{\textit{Proof.}} + Relation \ref{eqnZlatiRez2} from the previous statement is equivalent to the relation: + $$|AB|^2-|AB|\cdot |AZ|-|AZ|^2=0.$$ + If we solve this quadratic equation for $|AZ|$, we get: + $$|AZ|=\frac{\sqrt{5}+1}{2}|AB|,$$ + from which the desired equalities follow. + \kdokaz + +The number $$\Phi=\frac{\sqrt{5}+1}{2}$$ from the previous equation, which therefore represents the value of the ratio determined by the golden cut, is called the \index{number!golden}\pojem{golden number}. Of course, it is an irrational number ($\Phi\notin \mathbb{Q}$). Its approximate value is: +$$\Phi=\frac{\sqrt{5}+1}{2}\doteq 1.62.$$ + +Similarly, it is: +$$\frac{\sqrt{5}-1}{2}\doteq 0.62,$$ +which means that the longer part $AZ$ of the golden cut is approximately $62\%$ of the entire distance $AB$. + + . + + \bzgled \label{zlatiRezKonstr} + Construct a golden rectangle $ABCD$ with two sides + $a$ and $b$ in the golden ratio - so-called \index{golden!rectangle}\pojem{golden rectangle}\color{green1}. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.4.pic} +\caption{} \label{sl.pod.7.15.4.pic} +\end{figure} + + \textbf{\textit{Solution.}} We draw any line segment $AB$, then the point $Z$, which divides the line segment $AB$ in the golden cut so that $AZ$ is the longer part (see the previous example \ref{zlatiRezKonstr}). In the end, we get $D=\mathcal{R}_{A,90^0}(Z)$ and $C=\mathcal{T}_{\overrightarrow{AB}}(D)$ + (Figure \ref{sl.pod.7.15.4.pic}). + \kdokaz + + If we divide the golden rectangle along its longer side in the golden ratio, we get a square and another golden rectangle. We divide the new golden rectangle in the same way and continue the process. If we connect the corresponding vertices (the places of the diagonals of the squares that determine the cusp), so that we draw circular arcs with a central angle of $90^0$, we get an approximate construction of the so-called \index{golden!spiral}\pojem{golden spiral} or \index{logarithmic!spiral}\pojem{logarithmic spiral} (Figure \ref{sl.pod.7.15.5.pic}). + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.5.pic} +\caption{} \label{sl.pod.7.15.5.pic} +\end{figure} + +Logarithmic spiral\footnote{The proposal for the name logarithmic spiral was made by the French mathematician \index{Varignon, P.}\textit{P. Varignon} (1645--1722). The logarithmic spiral has the property that each straight line from the center of the spiral intersects it at the same angle. It appears in various forms in nature: from spirals in sunflower petals, snail shells and spider webs to distant galaxies. We also call it the spiral mirabilis (miraculous spiral) - this name was proposed by the Swiss mathematician \index{Bernoulli, J.}\textit{J. Bernoulli} (1667–-1748) because of its wonderful properties.} is a curve defined by equations in parametric form: +\begin{eqnarray*} +x&=&a e^{bt}\cdot\cos t \\ +y&=&a e^{bt}\cdot\sin t, +\end{eqnarray*} +where $t\in \mathbb{R}$ is a parameter, $a$ and $b$ are arbitrary real constants and $e\doteq 2,72$ is the \index{number!Euler's }\pojem{Euler's\footnote{Number $e$ is an irrational number and represents the value of the limit $\lim_{n\rightarrow \infty}\left(1+\frac{1}{n} \right)^n$ =e. It was named after the Swiss mathematician \index{Euler, L.}\textit{L. Euler} (1707--1783).} number.}. +The previous construction (with circular arcs) is, as we have already mentioned, approximate, but it represents a very good approximation of this curve (Figure \ref{sl.pod.7.15.5a.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.5a.pic} +\caption{} \label{sl.pod.7.15.5a.pic} +\end{figure} + +We will continue with a regular pentagon. + + + + \bzgled \label{zlatiRezPravPetk1} + A diagonal and a side of a regular pentagon are in the golden ratio. + \ezgled + + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.15.6.pic}) + + Let's mark with $a$ the side and with $d$ the diagonal of the regular pentagon $ABCDE$. From \ref{PtolomejPetkotnik} it follows: + $$d=\frac{1+\sqrt{5}}{2}a.$$ +Therefore, according to \ref{zlatiRezStevilo}, the segments $d$ and $a$ are in the golden ratio. +\kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.6.pic} +\caption{} \label{sl.pod.7.15.6.pic} +\end{figure} + +\bzgled + Two diagonals of a regular pentagon intersect at a point that + divides them in the golden ratio\footnote{The Pythagoreans knew this property of the pentagon. The Pythagorean school was founded by the Greek philosopher and mathematician \index{Pitagora}\textit{Pitagora from the island of Samos} (582--497 BC) in Croton in southern + Italy. His students were engaged in philosophy, mathematics and natural sciences. For their distinctive sign they + chose the pentagram, which is composed of the pentagon's diagonals. The influence of the Pythagorean school on the mathematics of the Ancient + Greeks lasted for several centuries after Pythagora's death.}. + \ezgled + + + \textbf{\textit{Proof.}} +We denote by $P$ the intersection of the diagonals $AC$ and $BD$ of a regular pentagon $ABCDE$ and by $k$ +its circumscribed circle (Figure \ref{sl.pod.7.15.6.pic}). The angles $EAD$, $DAC$, $CAB$ and $DBC$, which are adjacent to the +chords $ED$, $CD$ and $CB$, are congruent (statement \ref{SklTetSklObKot}). Because according to statement \ref{pravVeckNotrKot} the interior angle $EAB$ of a regular pentagon is equal to +$108^0$, the angles $EAD$, $DAC$, $CAB$ (or $PAB$) and +$DBC$ are equal to $36^0$. From this we obtain $\angle PBA=\angle DBA=\angle CBA-\angle CBD=108^0-36^0= 72^0$. From the sum of the angles of triangle $ABP$ according to statement \ref{VsotKotTrik} it follows +$\angle APB=180^0-\angle PBA-\angle PAB=72^0$. Therefore, the triangle $PAB$ is isosceles with the base $PB$ (statement \ref{enakokraki}), +or $AP\cong AB$. From this it follows: +$$AC:AP=AC:AB.$$ +If we use the previous statement \ref{zlatiRezPravPetk1}, we conclude that the point $P$ divides the diagonal $AC$ in the golden ratio. +\kdokaz + +If in a regular pentagon $ABCDE$ we choose the vertices $A$, $B$ and $D$, we get an isosceles triangle $ABD$ with a base that is equal to the side $a$ of the pentagon $ABCDE$, and the legs are equal to the diagonals $d$ of this pentagon. The leg and the base of this triangle are therefore in the golden ratio (Figure \ref{sl.pod.7.15.7.pic}). Therefore, the triangle $ABD$ is called the \index{golden!triangle}\pojem{golden triangle}. The angles of the golden triangle measure $72^0$, $72^0$ and $36^0$. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.7.pic} +\caption{} \label{sl.pod.7.15.7.pic} +\end{figure} + +Similarly to the golden rectangle, if we use the similitudes of the corresponding internal angles at the base of the golden triangle, we get a sequence of golden triangles, all of which are similar. With the help of the corresponding circular arcs, we can also derive another approximate construction of the golden (or logarithmic) spiral (Figure \ref{sl.pod.7.15.7.pic}). + +%________________________________________________________________________________ + \poglavje{Morley's Theorem and Some More Theorems} \label{odd7Morly} + +We say that the segments $SP$ and $SQ$ \index{trisektrisa}\pojem{trisektrisi} of the angle $ASB$, if the points $P$ and $Q$ lie in this angle and it holds $$\angle ASP\cong\angle PSQ\cong\angle QSB,$$ +or if it is a segment that divides the angle into three congruent angles +(Figure \ref{sl.pod.7.16.0.pic}). + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.0.pic} +\caption{} \label{sl.pod.7.16.0.pic} +\end{figure} + + We first prove an auxiliary assertion - lemma. + +\bizrek \label{izrekMorleyLema} + Let $Y'$, $Z$, $Y$ and $Z'$ be points in the plane such that + $Y'Y\cong ZY\cong ZZ'$ and + $$\angle Z'ZY\cong \angle ZYY'=180^0-2\alpha>60^0.$$ + If $A$ is a point in this plane that is on the different side + of the line $Z'Y'$ with respect to + the point $Z$ and also $Y'AZ'=3\alpha$, then the points $A$, $Y'$, $Z$, $Y$ and $Z'$ are + concyclic and also + $$\angle Z'AZ\cong \angle ZAY\cong \angle YAY'=\alpha.$$ +\eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.1a.pic} +\caption{} \label{sl.pod.7.16.1a.pic} +\end{figure} + + + \textbf{\textit{Proof.}} The given condition $180^0-2\alpha>60^0$ is equivalent to the condition $\alpha<60^0$. + +Let $s$ be the perpendicular bisector of the line segment $ZY$, and $\mathcal{S}_s$ be the reflection over the line $s$ (Figure \ref{sl.pod.7.16.1a.pic}). Thus $\mathcal{S}_s(Z)=Y$. Because $\mathcal{S}_s$ is an isometry that preserves angles and lengths of line segments, $\mathcal{S}_s(Z')=Y'$ as well. From $ZY,Z'Y'\perp s$ it follows that $ZY\parallel Z'Y'$, which means that the trapezoid $ZYY'Z'$ is isosceles. + By the \ref{trapezTetivEnakokr} theorem, this trapezoid is tangent - we denote its inscribed circle with $k$. By the \ref{KotiTransverzala} theorem, the angles $YZZ'$ and $Y'Z'Z$ are supplementary, so we have: + \begin{eqnarray} \label{eqnMorleyLema1} + \angle ZZ'Y'=180^0-\angle Z'ZY=2\alpha. + \end{eqnarray} + From $Z'Z\cong ZY$ by the \ref{SklTetSklObKot} theorem it follows that $\angle ZZ'Y\cong \angle YZ'Y'$, so by the relation \ref{eqnMorleyLema1} $\angle ZZ'Y\cong \angle YZ'Y'=\alpha$. The triangle $Z'ZY$ is isosceles with the base $Z'Y$, so by the \ref{enakokraki} theorem it is also $\angle ZYZ'\cong \angle ZZ'Y=\alpha$. Analogously, $\angle YY'Z\cong \angle ZY'Z'=\alpha$ and $\angle YZY'\cong \angle YY'Z=\alpha$. If we put everything together, we get $\angle ZZ'Y\cong \angle YZ'Y'\cong \angle ZYZ'\cong + \angle YY'Z\cong \angle ZY'Z'\cong\angle YZY'=\alpha$, so: + \begin{eqnarray} \label{eqnMorleyLema2} + \angle Z'Y'Z \cong \angle Y'Z'Y= \alpha. + \end{eqnarray} + Because $\angle Z'ZY'=\angle Z'ZY-\angle Y'ZY=180^0-2\alpha-\alpha=180^0-3\alpha$ (this angle exists because $\alpha<60^0$) or $\angle Z'ZY'+\angle Z'AY'=180^0-3\alpha+3\alpha=180^0$, by the \ref{TetivniPogoj} theorem the quadrilateral $Z'ZY'A$ is tangent, which means that the point $A$ lies on the circle $k$. + + From the relation \ref{eqnMorleyLema2} and the \ref{ObodObodKot} theorem it follows: + \begin{eqnarray*} + &&\angle Z'AZ\cong \angle Z'Y'Z=\alpha,\\ + &&\angle Y'AY\cong \angle Y'Z'Y=\alpha. + \end{eqnarray*} + Finally, we have: + \begin{eqnarray*} + \angle ZAY=\angle Z'AY'-\angle Z'AZ-\angle Y'AY =3\alpha-\alpha-\alpha=\alpha, + \end{eqnarray*} + which was to be proven. \kdokaz + +Now we are ready for the basic theorem. + + + + + + \bizrek \label{izrekMorley}\index{izrek!Morleyev} + If $X$, $Y$ and $Z$ are three points of intersection of the adjacent angle trisectors + of a triangle $ABC$, then $XYZ$ is an equilateral triangle.\\ + (Morley's\footnote{\index{Morley, F.}\textit{F. Morley} (1860--1937), English mathematician, discovered this property of a triangle in 1904, but published it only 20 years + later. At that time, the theorem was published as a task in the journal \textit{Educational Times}. Here we will give one of the solutions proposed at that time.} theorem) + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.1.pic} +\caption{} \label{sl.pod.7.16.1.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + (Figure \ref{sl.pod.7.16.1.pic}) + + Let $X$ and $V$ be the points of intersection of the respective trisectors of the internal angles $ABC$ and $ACB$ of the triangle $ABC$, so that $X$ is the internal point of the triangle $BVC$. We also denote by $Z_1$ and $Y_1$ the points of the trisector $BV$ and $CV$, for which $\angle Z_1XV \cong \angle VXY_1=30^0$. We first prove that $XY_1Z_1$ is an equilateral triangle. +Since the lines $BX$ and $CX$ are the angle bisectors of the angles $VBC$ and $VCB$, the point $X$ is the center of the inscribed circle of the triangle $BVC$, so the line $VX$ is the angle bisector of the angle $BVC$ (theorem \ref{SredVcrtaneKrozn}). From the similarity of the triangles $VXZ_1$ and +$VXY_1$ (theorem \textit{ASA} \ref{KSK}) it follows that $XZ_1\cong XY_1$. Since $\angle Z_1XY_1=60^0$, $XY_1Z_1$ is an equilateral triangle (theorem \ref{enakokraki}). It is enough to prove that $Y_1=Y$ and $Z_1=Z$ or $\angle BAZ_1\cong\angle Z_1AY_1\cong\angle Z_1AC$. + +From the congruence of the triangles $VXZ_1$ and +$VXY_1$ it follows that $Z_1V \cong Y_1V$. Therefore, $VZ_1Y_1$ is an isosceles triangle with a base $Z_1Y_1$, so according to the theorem \ref{enakokraki}: +\begin{eqnarray} \label{eqnMorley1} +\angle VZ_1Y_1\cong\angle VY_1Z_1. + \end{eqnarray} +Let $Z'$ and $Y'$ be such points on the sides $AB$ and $AC$ of the triangle $ABC$, that $BZ'\cong BX$ and $CY'\cong CX$. +The triangles $BZ'Z_1$ and $BXZ_1$ are congruent (theorem \textit{SAS} \ref{SKS}), so $Z'Z_1\cong XZ_1$. Also, $\angle Z'Z_1B\cong\angle XZ_1B$ or the line $BV$ is the perpendicular bisector of the angle $Z'Z_1X$. +Similarly, $Y'Y_1\cong XY_1$. From this and from the fact that $XY_1Z_1$ is an equilateral triangle, it follows: +\begin{eqnarray} \label{eqnMorley2} +Z'Z_1\cong Z_1Y_1\cong Y_1Y'. + \end{eqnarray} +We denote with $3\alpha$, $3\beta$, $3\gamma$ the measures +of the internal angles of the triangle $ABC$, at the vertices $A$, $B$ and $C$. +It is clear that $3\alpha+3\beta+3\gamma=180^0$, therefore: +\begin{eqnarray} \label{eqnMorley3} +2\alpha+2\beta+2\gamma=120^0. + \end{eqnarray} +If we use the fact that the line $BV$ is the perpendicular bisector of the convex or non-convex angle $Z'Z_1X$ and the relations \ref{eqnMorley1} and \ref{eqnMorley3}, after simple calculations we get: +\begin{eqnarray*} + \angle Z'Z_1Y_1&=&\angle Z'Z_1V+\angle VZ_1Y_1=\\ + &=&\angle VZ_1X+\angle VZ_1Y_1=\\ + &=&60^0+2\angle VZ_1Y_1=\\ + &=&60^0+\angle VZ_1Y_1+\angle VY_1Z_1=\\ + &=&60^0+180^0-\angle Z_1VY_1=\\ + &=&60^0+2\beta+2\gamma=\\ + &=&60^0+120^0-2\alpha=\\ + &=&180^0-2\alpha. + \end{eqnarray*} + Similarly, $\angle Y'Y_1Z_1=180^0-2\alpha$, so (if we take into account the relation \ref{eqnMorley2}) by theorem \ref{izrekMorleyLema} + $\angle Z'AZ_1\cong\angle Z_1AY_1\cong\angle Y_1AY'= \alpha$, which was to be proven. + \kdokaz + +\bizrek (Leibniz's\footnote{\index{Leibniz, G. W.}\textit{G. W. Leibniz} (1646--1716), German mathematician.} theorem) + \label{izrekLeibniz}\index{theorem!Leibniz's} + If $T$ is the centroid and $X$ an arbitrary + point in the plane of a triangle $ABC$, then + $$|XA|^2 + |XB|^2 + |XC|^2 = \frac{1}{3}\left(|AB|^2 +|BC|^2 +|CA|^2\right) + 3|XT|^2 + ,\textrm{ i.e.}$$ + $$|XA|^2 + |XB|^2 + |XC|^2 = |TA|^2 +|TB|^2 +|TC|^2 + 3|XT|^2.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.2.pic} +\caption{} \label{sl.pod.7.16.2.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.16.2.pic}) + Let $A_1$ be the center +of the side $BC$ or $|AA_1|=t_a$ its +centroid. Because $AT:TA_1=2:1$ (statement \ref{tezisce}), by +Stewart's theorem \ref{StewartIzrek2} +with respect to the triangle +$AXA_1$ it follows: +\begin{eqnarray} \label{eqnLeibniz1} +|XT|^2=\frac{1}{3}|XA|^2+\frac{2}{3}|XA_1|^2-\frac{2}{9}t_a^2 +\end{eqnarray} +If we use Stewart's theorem \ref{StewartIzrek2} again (or its consequence for the centroid \ref{StwartTezisc}) with respect to the triangle $BXC$, we get: + +\begin{eqnarray} \label{eqnLeibniz2} +|XA_1|^2=\frac{1}{2}|XB|^2+\frac{1}{2}|XC|^2-\frac{1}{4}|BC|^2 +\end{eqnarray} +if we insert \ref{eqnLeibniz2} in \ref{eqnLeibniz1}: +\begin{eqnarray} \label{eqnLeibniz3} +|XT|^2=\frac{1}{3}\left(|XA|^2+|XB|^2+|XC|^2\right)-\frac{1}{6}|BC|^2-\frac{2}{9}t_a^2 +\end{eqnarray} +and similarly: +\begin{eqnarray} \label{eqnLeibniz4a} +\hspace*{-4mm} |XT|^2&=&\frac{1}{3}\left(|XA|^2+|XB|^2+|XC|^2\right)-\frac{1}{6}|AC|^2-\frac{2}{9}t_b^2\\ +\hspace*{-4mm} |XT|^2&=&\frac{1}{3}\left(|XA|^2+|XB|^2+|XC|^2\right)-\frac{1}{6}|AB|^2-\frac{2}{9}t_c^2\label{eqnLeibniz4} +\end{eqnarray} +By adding the equality from \ref{eqnLeibniz3} - \ref{eqnLeibniz4} we get: +\begin{eqnarray*} +3|XT|^2&=&|XA|^2+|XB|^2+|XC|^2-\frac{1}{6}\left(|BC|^2+|AC|^2+|AB|^2\right)-\\ +&-&\frac{2}{9}\left(t_a^2+t_b^2+t_c^2\right) +\end{eqnarray*} +Finally, from the statement \ref{StwartTezisc2} it follows: +\begin{eqnarray*} +3|XT|^2=|XA|^2+|XB|^2+|XC|^2-\frac{1}{3}\left(|BC|^2+|AC|^2+|AB|^2\right) +\end{eqnarray*} +or both relations from the statement. +\kdokaz + +A direct consequence of Leibniz's formula is the following statement. + + + + + \bizrek + A point in the plane of a triangle for which the squared distances + from its vertices has a minimum value is its centroid. + \eizrek + + \textbf{\textit{Proof.}} By Leibnitz's formula for any point $X$ in the plane of the triangle $ABC$ with centroid $T$ it holds: +$$|XA|^2 + |XB|^2 + |XC|^2 = \frac{1}{3}\left(|AB|^2 +|BC|^2 +|CA|^2\right) + 3|XT|^2.$$ +The minimum of the sum $|XA|^2 + |XB|^2 + |XC|^2$ with respect to $X$ is therefore achieved when $|XT|$ is the smallest, which is when $X=T$. +\kdokaz + +\bizrek (Carnot's\footnote{\index{Carnot, L. N. M.}\textit{L. N. M. Carnot} (1753--1823), + French mathematician.} theorem)\index{theorem!Carnot's} + Let $P$, $Q$ and $R$ be points on the lines containing the + sides $BC$, $CA$ and $AB$ of a triangle $ABC$. Perpendicular lines on the lines $BC$, $CA$. + and $AB$ at the points $P$, $Q$ and $R$ intersect at one point if and only if + \begin{eqnarray} \label{eqnCarnotIzrek1} + |BP|^2 - |PC|^2 + |CQ|^2 - |QA|^2 + |AR|^2 - |RB|^2 = 0. + \end{eqnarray} + + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.3.pic} +\caption{} \label{sl.pod.7.16.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} We mark with $p$, $q$ and $r$ the perpendiculars of the lines $BC$, $CA$ and $AB$ through the points $P$, $Q$ and $R$ (Figure \ref{sl.pod.7.16.3.pic}). + +($\Rightarrow$) We first assume that the lines $p$, $q$ and $r$ intersect at some point $L$. If we use the Pythagorean theorem \ref{PitagorovIzrek} six times, we get: + \begin{eqnarray*} +\begin{array}{cc} + |AR|^2+|RL|^2=|AL|^2 & \hspace{6mm} -|AQ|^2-|QL|^2=-|AL|^2 \\ + |BP|^2+|PL|^2=|BL|^2 & \hspace{6mm} -|BR|^2-|RL|^2=-|BL|^2\\ + |CQ|^2+|QL|^2=|CL|^2 & \hspace{6mm} -|CP|^2-|PL|^2=-|CL|^2 +\end{array} +\end{eqnarray*} + If we add all six equalities, we get relation \ref{eqnCarnotIzrek1}. + +($\Leftarrow$) Now let's assume that the +relation \ref{eqnCarnotIzrek1} is true. +The rectangles $q$ and $r$ of the sides $AC$ and $AB$ are not parallel (because otherwise, as a result of Playfair's axiom \ref{Playfair1}, the points $A$, $B$ and $C$ would be collinear). We denote by $\widehat{L}$ the intersection of the lines $q$ and $r$. We denote by $\widehat{P}$ the orthogonal projection of the point $\widehat{L}$ onto the line $BC$. Because the rectangles of the sides of the triangle $ABC$ intersect at the points $\widehat{P}$, $Q$ and $R$ in the point $\widehat{L}$, from the first part of the proof ($\Rightarrow$) it follows: + \begin{eqnarray} \label{eqnCarnotIzrek2} + |B\widehat{P}|^2 - |\widehat{P}C|^2 + |CQ|^2 - |QA|^2 + |AR|^2 - |RB|^2 = 0. + \end{eqnarray} + From \ref{eqnCarnotIzrek1} and \ref{eqnCarnotIzrek2} it follows: +$$|B\widehat{P}|^2 - |\widehat{P}C|^2=|BP|^2 - |PC|^2.$$ + In the same way as at the end of the second part of the proof of the \ref{PotencOsLema} theorem, we get $P=\widehat{P}$, i.e. the lines $p$, $q$ and $r$ intersect in one point. +\kdokaz + + + + + \bizrek (Butterfly theorem\footnote{An English mathematician \index{Horner, W. J.}\textit{W. J. Horner} (1786--1837) published a proof of this theorem in 1815.}) + \index{theorem!butterfly} + Let $S$ be the midpoint of chord $PQ$ of a circle $k$. Suppose that $AB$ and $CD$ + are arbitrary chords of this circle passing through the point $S$. If $X$ and $Y$ + are the points of intersection of the chords $AD$ and $BC$ with the chord $PQ$, + then $S$ is the midpoint of the line segment $XY$. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.4.pic} +\caption{} \label{sl.pod.7.16.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let's mark $x=|SX|$ and $y=|SY|$ +(Figure \ref{sl.pod.7.16.4.pic}). + Let +$X_1$ and $X_2$ or $Y_1$ and $Y_2$ +be the perpendicular projections of points $X$ or $Y$ on the lines $AB$ +and $CD$. +By Tales' theorem (\ref{TalesovIzrekDolzine}) it is: +\begin{eqnarray*} + \frac{x}{y}=\frac{XX_1}{YY_1}=\frac{XX_2}{YY_2} +\end{eqnarray*} + If we use the latter equality, then +similarity of triangles $AXX_1$ and $CYY_2$ or +triangles $DXX_2$ and $BYY_1$ (statement \ref{PodTrikKKK}) and finally the power +of points $X$ and $Y$ with respect to the circle $k$ (statement \ref{izrekPotenca}), we get: +\begin{eqnarray*} + \frac{x^2}{y^2}&=& \frac{|XX_1|}{|YY_1|}\cdot\frac{|XX_2|}{|YY_2|}= + \frac{|XX_1|}{|YY_2|}\cdot\frac{|XX_2|}{|YY_1|}=\\ + &=& \frac{|AX|}{|CY|}\cdot\frac{|DX|}{|BY|}=\frac{|AX|\cdot |DX|}{|CY|\cdot |BY|}=\\ + &=& \frac{|XP|\cdot |XQ|}{|YP|\cdot |YQ|}= + \frac{\left(|PS|-x\right)\cdot\left(|PS|+x\right)} + {\left(|PS|+y\right)\cdot\left(|PS|-y\right)}=\\ + &=& \frac{|PS|^2-x^2}{|PS|^2-y^2}. +\end{eqnarray*} + From $$\frac{x^2}{y^2}=\frac{|PS|^2-x^2}{|PS|^2-y^2}$$ it finally follows that $x=y$. +\kdokaz + + + \bizrek \index{krožnica!Taylorjeva} \label{izrekTaylor} + Let $A'$, $B'$ and $C'$ be the foots of the altitudes of a triangle $ABC$. + The foot of the perpendiculars from the points $A'$, + $B'$ and $C'$ on the lines containing the adjacent sides + of that triangle lie on a circle, so-called \pojem{Taylor\footnote{\index{Taylor, B.} \textit{B. Taylor} (1685--1731), angleški matematik.} circle} \color{blue} of that triangle. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.5.pic} +\caption{} \label{sl.pod.7.16.5.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Let $A_c$ and $A_b$ be the orthogonal projections of points $C'$ and $B'$ on the line $BC$, + $B_a$ and $B_c$ be the orthogonal projections of points $A'$ and $C'$ on the line $AC$ and + $C_a$ and $C_b$ be the orthogonal projections of points $A'$ and $B'$ on the line $AB$ (Figure \ref{sl.pod.7.16.5.pic}). + + Triangle $A'B'C'$ is + the pedal triangle of triangle $ABC$, so from the proof of Theorem \ref{PedalniVS} it follows +$\angle B'A'C\cong\angle BAC=\alpha$ and $\angle C'B'A\cong\angle CBA=\beta$. Because $\angle B'B_aA'=\angle B'A_bA'=90^0$, by Theorem \ref{TalesovIzrKroz2} $A'A_bB_aB'$ is a trapezoid, so $\angle A_bB_aC\cong B'A'C=\alpha$ (Theorem \ref{TetivniPogojZunanji}). From this, by Theorem \ref{KotiTransverzala} it follows $A_bB_a \parallel BA$. + +Triangle $C'A_cB_c$ and $VA'B'$ are perspective with respect to point $C$ (this means that the lines $C'V$, $A_cA'$ and $B_cB'$ intersect at point $C$). Because $C'A_c\parallel VA'$ and $C'B_c\parallel VB'$, by the generalization of Desargues' Theorem \ref{izrekDesarguesOsNesk} it follows that $A_cB_c\parallel A'B'$. By Theorem \ref{KotiTransverzala1} it follows that $\angle B_cA_cA'\cong B'A'C=\alpha$. Because therefore $\angle B_cA_cA'=\alpha=A_bB_aC$, $A_cA_bB_aB_c$ is a trapezoid (Theorem \ref{TetivniPogojZunanji}); we denote its circumscribed circle with $k$. + +It remains to prove that the points $C_a$ and $C_b$ also lie on the circle $k$. Analogously to the proven $A_cB_c\parallel A'B'$, it is also $C_aB_a\parallel C'B'$ and analogously to the proven, that $\angle A_bB_aC\cong B'A'C=\alpha$, it is also $\angle C_aA_cB\cong\angle A'C'B=\gamma$. +From the parallelism $C_aB_a\parallel C'B'$ it follows (from the statement \ref{KotiTransverzala1}) +$\angle C_aB_aA \cong\angle C'B'A = \beta$, therefore: + $$\angle C_aB_aA_b=180^0-\angle A_bB_aC-\angle C_aB_aA= 180^0-\alpha-\beta=\gamma.$$ + Therefore $\angle C_aA_cB\cong\angle C_aB_aA_b=\gamma$, which means (from the statement \ref{TetivniPogojZunanji}), that $C_aA_cA_bB_a$ is a cyclic quadrilateral, therefore the point $C_a$ lies on the circle $k$. Analogously we prove that also the point $C_b$ lies on the circle $k$. +\kdokaz + + A direct consequence is the following theorem. + + + + \bizrek + Let $A'$, $B'$ and $C'$ be the foots of the altitudes of a triangle $ABC$. + The foot of the perpendiculars from the points $A'$, + $B'$ and $C'$ on the lines containing the adjacent sides + of that triangle determine three congruent line segments. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.6.pic} +\caption{} \label{sl.pod.7.16.6.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.16.6.pic}) + +In the proof of the previous statement \ref{izrekTaylor} we have determined that $A_bB_a\parallel BA$, therefore the quadrilateral $C_bC_aA_bB_a$ is a trapezoid. This +trapezoid is also a cyclic (the previous statement \ref{izrekTaylor}) and therefore it is also an isosceles trapezoid (\ref{trapezTetivEnakokr}). Its diagonals are congruent (the statement \ref{trapezEnakokraki}), or. it is +$A_bC_b\cong C_aB_a$. Analogously it is also $A_bC_b\cong A_cB_c$. +\kdokaz + + + + \vspace*{12mm} + + +%Tales + +\poglavje{The basics of Geometry} \label{osn9Geom} + +It remains to prove that the points $C_a$ and $C_b$ also lie on the circle $k$. Analogously to the proven $A_cB_c\parallel A'B'$, it is also $C_aB_a\parallel C'B'$ and analogously to the proven, that $\angle A_bB_aC\cong B'A'C=\alpha$, it is also $\angle C_aA_cB\cong\angle A'C'B=\gamma$. +From the parallelism $C_aB_a\parallel C'B'$ it follows (from the statement \ref{KotiTransverzala1}) +$\angle C_aB_aA \cong\angle C'B'A = \beta$, therefore: + $$\angle C_aB_aA_b=180^0-\angle A_bB_aC-\angle C_aB_aA= 180^0-\alpha-\beta=\gamma.$$ + Therefore $\angle C_aA_cB\cong\angle C_aB_aA_b=\gamma$, which means (from the statement \ref{TetivniPogojZunanji}), that $C_aA_cA_bB_a$ is a cyclic quadrilateral, therefore the point $C_a$ lies on the circle $k$. Analogously we prove that also the point $C_b$ lies on the circle $k$. +\kdokaz + + A direct consequence is the following theorem. + + + + \bizrek + Let $A'$, $B'$ and $C'$ be the foots of the altitudes of a triangle $ABC$. + The foot of the perpendiculars from the points $A'$, + $B'$ and $C'$ on the lines containing the adjacent sides + of that triangle determine three congruent line segments. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.6.pic} +\caption{} \label{sl.pod.7.16.6.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.16.6.pic}) + +In the proof of the previous statement \ref{izrekTaylor} we have determined that $A_bB_a\parallel BA$, therefore the quadrilateral $C_bC_aA_bB_a$ is a trapezoid. This +trapezoid is also a cyclic (the previous statement \ref{izrekTaylor}) and therefore it is also an isosceles trapezoid (\ref{trapezTetivEnakokr}). Its diagonals are congruent (the statement \ref{trapezEnakokraki}), or. it is +$A_bC_b\cong C_aB_a$. Analogously it is also $A_bC_b\cong A_cB_c$. +\kdokaz + + + + \vspace*{12mm} + + +%Tales + +\poglavje{The basics of Ge + +\item +Let $S$ be the intersection of the diagonal $AC$ and $BD$ of the trapezoid $ABCD$. Let $P$ and $Q$ be the intersections of the parallels to the bases $AB$ and $CD$ through the point $S$ with the sides of this trapezoid. Prove that $S$ is the center of the line $PQ$. + +\item +Let $ABCD$ be a trapezoid with the base $AB$, the point $S$ be the intersection of its diagonals and $E$ be the intersection of the altitudes of this trapezoid. Prove that the line $SE$ goes through the centers of the bases $AB$ and $CD$. + +\item +Let $P$, $Q$ and $R$ be points in which an arbitrary line through the point $A$ intersects the altitudes of the sides $BC$ and $CD$ and the diagonal $BD$ of the parallelogram $ABCD$. Prove that $|AR|^2=|PR|\cdot |QR|$. + +\item +The points $D$ and $K$ lie on the sides $BC$ and $AC$ of the triangle $ABC$ so that $BD:DC=2:5$ and $AK:KC=3:2$. Calculate the ratio in which the line $BK$ divides the line $AD$. + +\item +Let $P$ be a point on the side $AD$ of the parallelogram $ABCD$ such that $\overrightarrow{AP}= +\frac{1}{n}\overrightarrow{AD}$, and $Q$ be the intersection of the lines $AC$ and $BP$. Prove that: + $$\overrightarrow{AQ}=\frac{1}{n + 1}\overrightarrow{AC}.$$ + + + +%Homotetija + +\item +Given are: the point $A$, the lines $p$ and $q$ and the lines $m$ and $n$. Draw a line $s$ through the point $A$, +which intersects the lines $p$ and $q$ in such points $X$ and $Y$, that $XA:AY= m:n$. + +\item Given are: the point $S$, the lines $p$, $q$ and $r$ and the lines $m$ and $n$. Draw a line $s$ through the point $S$, +which intersects the lines $p$, $q$ and $r$ in such points $X$, $Y$ and $Z$, that $XY:YZ=m:n$. + + \item In the given triangle $ABC$ draw such a rectangle $PQRS$, that the side $PQ$ lies on the side $BC$, +the points $R$ and $S$ lie on the sides $AB$ and $AC$, and also $PQ=2QR$. + +\item +Draw:\\ +(\textit{a}) a rhombus, if given are the side $a$ and the ratio of the diagonals $e:f$;\\ + (\textit{b}) a trapezoid, if given are: the angles $\alpha$ and $\beta$ at one of the bases, the ratio of this base and the height $a:v$ and the other base $c$. + +\item +Let $A$ be a point inside the angle $pSq$, $l$ a line and $\alpha$ an angle in some plane. Draw a triangle +$APQ$, such that: $P\in p$, $Q\in q$, $\angle PAQ\cong \alpha$ and +$PQ\parallel l$. + +\item +Draw a circle $l$, that touches the given circle $k$ and the given line $p$, if the point of contact is given:\\ +(\textit{a}) $l$ and $p$;\hspace*{3mm} +(\textit{b}) $l$ and $k$. + +\item +Draw a circle, that touches the arms of the angle $pSq$ and:\\ +(\textit{a}) goes through the given point,\\ +(\textit{b}) touches the given circle. + +\item Let $p$ and $q$ be lines, $S$ a point ($S\notin p$ and $S\notin q$) and the lines $m$ and $n$. +Draw circles $k$ and $l$, that touch each other from the outside in the point $S$, the first one touches the line $p$, +the second line $q$, and the ratio of the radii is $m:n$. + + +\item +In a plane are given: a line $p$ and points $B$ and $C$, that lie on the circle $k$. Draw a point $A$ +on the circle $k$, so that the centroid of the triangle $ABC$ lies on the line $p$. + +\item +Let $k$ be a circle with the diameter $PQ$. Draw a square $ABCD$, so that $A,B\in PQ$ and $C,D\in k$. + +\item +In the same plane are given: lines $p$ and $q$, a point $A$ and lines $m$ and $n$. Draw a rectangle $ABCD$, so that $B\in p$, $D\in q$ +and $AB:AD=m:n$. + +\item In the given triangle $ABC$ draw a triangle, so that its sides are parallel to the given lines $p$, $q$ and $r$. + +\item Let $p$, $q$ and $r$ be three lines of some plane. Draw a line $t$, that +is perpendicular to the line $p$ and intersects the lines $p$, $q$ and $r$ in such points $P$, $Q$ and $R$, that $PQ\cong QR$. + + + + +%Similarity of triangles + +\item Let $P$ be an inner point of the triangle $ABC$ and $A_1$, $B_1$ and $C_1$ be the orthogonal projections of +the point $P$ on the sides of the triangle $BC$, $AC$ and $AB$. Similarly, the points $A_2$, $B_2$ and $C_2$ are determined by the point $P$ +and the triangle $A_1B_1C_1$,..., the points $A_{n+1}$, $B_{n+1}$ and $C_{n+1}$ with the point $P$ and the triangle $A_nB_nC_n$... Which of the triangles $A_1B_1C_1$, $A_2B_2C_2$, ... are similar to the triangle $ABC$? + +\item +Let $k$ be the inscribed circle of the quadrilateral $ABCD$, $E$ be the intersection of its diagonals and $CB\cong CD$. Prove that $\triangle ABC \sim\triangle BEC$. + +\item +Let $ABCD$ be a parallelogram. In the points $E$ and $F$ of the triangle $ABC$ the inscribed circle intersects the line $AD$ and $CD$. Prove that $\triangle EBC\sim\triangle EFD$. + +\item +Let $AA'$ and $BB'$ be the altitudes of the acute angled triangle $ABC$. Prove that +$\triangle ABC\sim\triangle A'B'C$. + +\item +Let the altitude $AD$ of the triangle $ABC$ be tangent to the inscribed circle of this triangle. Prove that $|AD|^2=|BD|\cdot |CD|$. + +\item +In the triangle $ABC$ let the internal angle at the vertex $A$ be twice +as large as the internal angle at the vertex $B$. Prove that $|BC|^2= |AC|^2+|AC|\cdot |AB|$. + +\item Prove that the radii of the inscribed circles of two similar triangles are proportional to the corresponding sides of these two triangles. + + + \item The quadrilateral $ABCD$ is inscribed in a circle with center $S$. The diagonals of this quadrilateral are perpendicular and intersect at the point $E$. The line through the point $E$ and perpendicular to the side $AD$ intersects the side $BC$ at the point $M$. \\ +(\textit{a}) Prove that the point $M$ is the center of the segment $BC$.\\ +(\textit{b}) Determine the set of all points $M$, if the diagonal $BD$ changes and is always perpendicular to the diagonal $AC$. + +\item Let $t$ be the tangent to the inscribed circle $l$ of the triangle $ABC$ at the vertex $A$. +Let $D$ be such a point of the line $AC$, that $BD\parallel t$. Prove that +$|AB|^2=|AC|\cdot |AD|$. + +\item The altitude point of an acute angled triangle should divide its altitude in an equal ratio (from the vertex to the knife point of the altitude). Prove that it is an isosceles triangle. + +\item In the triangle $ABC$, the altitude $BD$ touches the circumscribed circle of this triangle. +Prove:\\ +(\textit{a}) that the difference of the angles at the base $AC$ is equal to $90^0$,\\ +(\textit{b}) that $|BD|^2=|AD|\cdot |CD|$. + + \item The circle with the center on the base $BC$ of the isosceles triangle $ABC$ touches +the sides $AB$ and $AC$. The points $P$ and $Q$ are the intersections of these sides with any +tangent of this circle. Prove that $4\cdot |PB|\cdot |CQ|=|BC|^2$. + +\item Let $V$ be the altitude point of the acute angled triangle $ABC$, the point $V$ the center of the altitude +$AD$, and the altitude $BE$ the point $V$ divides in the ratio $3:2$. Calculate the ratio in which $V$ divides the altitude $CF$. + +\item Let $S$ be an external point of the circle $k$. $P$ and $Q$ +are the points in which the circle $k$ touches its tangent from the point $S$, $X$ and $Y$ are the intersections of these circles with any line passing through the point $S$. Prove that $XP:YP=XQ:YQ$. + +\item Let $D$ be a point lying on the side $BC$ of the triangle $ABC$. The points $S_1$ and $S_2$ shall be the centers of the circumscribed circles of the triangles $ABD$ and $ACD$. Prove that $ \triangle ABC\sim\triangle AS_1S_2$. + +\item The point $P$ lies on the hypotenuse $BC$ of the triangle $ABC$. The perpendicular of the line $BC$ at the point $P$ intersects the line $AC$ and $AB$ at the points $Q$ and $R$ and the circumscribed circle of the triangle $ABC$ at the point $S$. Prove that $|PS|^2=|PQ|\cdot |PR|$. + +\item The point $A$ lies on the leg $OP$ of the right angle $POQ$. Let +$B$, $C$ and $D$ be such points of the leg $OQ$ that $\mathcal{B}(O,B,C)$, $\mathcal{B}(B,C,D)$ and +$OA\cong OB\cong BC\cong CD$. Prove that also $\triangle ABC\sim\triangle DBA$. + +\item Draw a triangle if the following data is known:\\ +(\textit{a}) $\alpha$, $\beta$, $R+r$, \hspace*{3mm} + (\textit{b}) $a$, $b:c$, $t_c-v_c$,\hspace*{3mm} + (\textit{c}) $v_a$, $v_b$, $v_c$. + +\item Let $AB$ and $CD$ be the bases of an isosceles trapezoid $ABCD$, and let $r$ be the radius of the inscribed circle. Prove that $|AB|\cdot |CD|=4r^2$. + + + + +%Harmonic cetverica + +\item Given is a circle $k$ and points $A$ and $B$. Draw a point $X$ on the circle $k$ such that $AX:XB=2:5$. + +\item Draw a triangle with the given data:\\ +(\textit{a}) $a$, $v_a$, $b:c$, \hspace*{3mm} + (\textit{b}) $a$, $t_a$, $b:c$,\hspace*{3mm} + (\textit{c}) $a$, $b$, $b:c$,\\ + (\textit{d}) $a$, $\alpha$, $b:c$,\hspace*{3mm} + (\textit{e}) $a$, $l_a$, $b:c$. + +\item Draw a triangle if the following data is given:\\ +(\textit{a}) $v_a$, $r$, $\alpha$, \hspace*{3mm} +(\textit{b}) $v_a$, $r_a$, $a$, \hspace*{3mm} +(\textit{c}) $v_a$, $t_a$, $b-c$. + +\item Draw a parallelogram, where one side and the corresponding altitude are consistent with the given distances $a$ and $v_a$, and the diagonals are in the ratio $3:5$. + +\item Let $E$ be the intersection of the internal angle $BAC$ of the triangle $ABC$ with its side $BC$. Prove that + $$\overrightarrow{AE}=\frac{|AC|}{|AB|+|AC|}\cdot\overrightarrow{AB}+ + \frac{|AB|}{|AB|+|AC|}\cdot\overrightarrow{AC}.$$ + +\item Given are four collinear points for which $\mathcal{H}(A,B;C,D)$ holds. Draw a point $L$, from which the distances $AC$, $CB$ and $BD$ are seen at the same angle. + +\item Let $AE$ ($E\in BC$) be the internal angle of the triangle $ABC$ and let $a=|BC|$, $b=|AC|$ and $c=|AB|$. Prove that +$$|BE|=\frac{ac}{b+c} \hspace*{1mm} \textrm{ and } \hspace*{1mm} |CE|=\frac{ab}{b+c}.$$ + +\item Let $AE$ ($E\in BC$) and $BF$ ($F\in AC$) be the internal angle bisectors and $S$ the center of the inscribed circle of triangle $ABC$. Prove that $ABC$ is an isosceles triangle (with the base $AB$) exactly when $AS:SE=BS:SF$. + + + + +%Menelaus' Theorem + + +\item Prove that the external angle bisectors of any triangle intersect the opposite sides in three collinear points. + + + +%Pythagoras' Theorem + +\item Given $a$, $b$ and $c$ ($a>b$), construct such a distance $x$, that:\\ +(\textit{a}) $x=\sqrt{a^2+b^2}$, \hspace*{3mm} +(\textit{b}) $x=\sqrt{a^2-b^2}$, \hspace*{3mm} +(\textit{c}) $x=\sqrt{3ab}$,\\ +(\textit{d}) $x=\sqrt{a^2+bc}$, \hspace*{3mm} +(\textit{e}) $x=\sqrt{3ab-c^2}$, \hspace*{3mm} +(\textit{f}) $x=\frac{a\sqrt{ab+c^2}}{b+c}$. + + + +%Stewart's Theorem + +\item Let $a$, $b$ and $c$ be the sides of a triangle and $a^2+b^2=5c^2$. Prove that the centroids corresponding to sides $a$ and $b$ are perpendicular to each other. + +\item Let $a$, $b$, $c$ and $d$ be the sides, $e$ and $f$ the diagonals and $x$ the distance determined by the midpoints of sides $b$ and $d$ of a quadrilateral. Prove: +$$x^2 = \frac{1}{4} \left(a^2 +c^2 -b^2 -d^2 +e^2 +f^2 \right).$$ + +\item Let $a$, $b$ and $c$ be the sides of triangle $ABC$. Prove that the distance of the centroid $A_1$ of side $a$ from the foot $A'$ of the altitude on that side is equal to: +$$|A_1A'|=\frac{|b^2-c^2|}{2a}.$$ + + + +%Pappus and Pascal + +\item Let ($A$, $B$, $C$) and ($A_1$, $B_1$, $C_1$) be two triples of collinear points of a plane that are not on the same line. If $AB_1\parallel A_1B$ and $AC_1\parallel A_1C$, then also $CB_1\parallel C_1B$. (\textit{Pappus' Theorem}\footnote{Pappus of Alexandria\index{Pappus} (3rd century BC), Greek mathematician. This is a generalization of Pappus' Theorem (see Theorem \ref{izrek Pappus}), if we choose points $X$, $Y$ and $Z$ at infinity.}) + + + +%Desargues' Theorem + +\item Let $P$, $Q$ and $R$ be points of sides $BC$, $AC$ and $AB$ of triangle $ABC$, such that lines $AP$, $BQ$ and $CR$ are from the same pencil. Prove: If $X=BC\cap QR$, $Y=AC\cap PR$ and $Z=AB\cap PQ$, then points $X$, $Y$ and $Z$ are collinear. + + +\item Let $AA'$, $BB'$ and $CC'$ be altitudes of triangle $ABC$ and $X=B'C'\cap BC$, $Y=A'C'\cap AC$ and $Z=A'B'\cap AB$. Prove that points $X$, $Y$ and $Z$ are collinear points. + +\item Let $A$ and $B$ be points outside line $p$. Draw the intersection of lines $p$ and $AB$ without drawing line $AB$ directly. + +\item Let $p$ and $q$ be lines of a plane that intersect in point $S$, which is "outside the paper", and $A$ is a point of that plane. Draw the line that goes through points $A$ and $S$. + +\item Draw a triangle so that its vertices lie on three given parallel lines and the carriers of its sides go through three given points. + + + + +%Ppotenca + +\item Given is a circle $k(S,r)$.\\ +(\textit{a}) Which values can the power of a point have with respect to circle $k$?\\ + (\textit{b}) What is the smallest value of this power and for which point is this minimal value achieved?\\ +(\textit{c}) Determine the set of all points for which the power with respect to circle is equal to $\lambda\in \mathbb{R}$. + +\item Let $k_a(S_a,r_a)$ and $l(O,R)$ be drawn and dashed circles of some triangle. Prove the equality\footnote{The statement is a generalization of Euler's formula for a circle (see Theorem \ref{EulerjevaFormula}). \index{Euler, L.} + \textit{L. Euler} + (1707--1783), Swiss mathematician.}: + $$S_aO^2=R^2+2r_aR.$$ + + +\item Draw a circle that goes through given points $A$ and $B$ and touches a given circle $k$. + +\item Prove that the centers of lines that are determined by common tangents of two circles are collinear points. + +\item Draw a circle that is perpendicular to two given circles and intersects the third given circle in points that determine the diameter of that third circle. + +\item Let $M$ and $N$ be the intersection points of sides $AB$ and $AC$ of triangle $ABC$ with a line, +which passes through the center of the inscribed circle of this triangle and is parallel to its +side $BC$. Express the length of the line $MN$ as a function of the lengths of the sides of triangle $ABC$. + +\item Let $AA_1$ be the median of triangle $ABC$. Points $P$ and $Q$ shall be the intersection points +of the altitudes $AA_1B$ and $AA_1C$ with sides $AB$ and $AC$. Prove that $PQ\parallel BC$. + +\item In triangle $ABC$ let the sum (or difference) of the internal angles $ABC$ and $ACB$ be equal to a right +angle. Prove that $|AB|^2+|AC|^2=4r^2$, where $r$ is the radius of the circumscribed circle of this triangle. + +\item Let $AD$ be the altitude of triangle $ABC$. Prove that +the sum (or difference) of the internal angles $ABC$ and $ACB$ is equal to a right angle exactly when: +$$\frac{1}{|AB|^2}+\frac{1}{|AC|^2}=\frac{1}{|AD|^2}.$$ + +\item Express the distance between the centroid and the center of the circumscribed circle of a triangle as +a function of the lengths of its sides and the radius of the circumscribed circle. + +\item Prove that in triangle $ABC$ the altitude of the external angle at the vertex $A$ and the altitudes of +the internal angles at the vertices $B$ and $C$ intersect the opposite sides in three collinear points. + +\item Prove that in triangle $ABC$ with altitude $AD$, the center of the circumscribed circle and the point, in which the side $BC$ touches the circumscribed circle of this triangle, are three +collinear points. + +\item Prove Simson's theorem \ref{SimpsPrem} by using Menelaus' theorem \ref{izrekMenelaj}. + +\item Through point $M$ of side $AB$ of triangle $ABC$ a line is constructed, which intersects +the line $AC$ in point $K$. Calculate the ratio, in which the line $MK$ divides the side $BC$, +if $AM:MB=1:2$ and $AK:AC=3:2$. + +\item Let $A_1$ be the center of side $BC$ of triangle $ABC$ and let $P$ and $Q$ be such points of sides +$AB$ and $AC$, that $BP:PA=2:5$ and $AQ:QC=6:1$. Calculate the ratio, in which the line $PQ$ divides the median $AA_1$. + +\item Prove that in any triangle, the lines determined by the vertices and the points of contact of one of the inscribed circles of the opposite sides intersect in a common point. + +\item What does the set of all points representing the intersection of two lines tangent to two given circles with respect to the centers and the points of contact of the circles represent? + +\item Let $PP_1$ and $QQ_1$ be the external tangents of the circles $k(O,r)$ and $k_1(O_1,r_1)$ (the points $P$, $P_1$, $Q$ and $Q_1$ are the corresponding points of contact). Let $S$ be the intersection of these two tangents, $A$ one of the intersections of the circles $k$ and $k_1$, and $L$ and $L_1$ the intersections of the line $SO$ with the lines $PQ$ and $P_1Q_1$. Prove that $\angle LAO\cong\angle L_1AO_1$. + +\item Prove that the side of a regular pentagon is equal to the larger part of the division of the radius of the inscribed circle of the pentagon in the golden ratio. + +\item Let $a_5$, $a_6$ and $a_{10}$ be the sides of a regular pentagon, hexagon and decagon, which are inscribed in the same circle. Prove that: + $$a_5^2=a_6^2+a_{10}^2.$$ + + +\item Let $t_a$, $t_b$ and $t_c$ be the centroids and $s$ the semiperimeter of a triangle. Prove that: + $$t_a^2+t_b^2+t_c^2\geq s^2.$$ % zvezek - dodatni MG + + +\end{enumerate} + + + + +% DEL 8 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% PLOŠČINA +%________________________________________________________________________________ + + \del{Area of Figures} \label{pogPLO} + + +%________________________________________________________________________________ +\poglavje{Area of Figures. Definition} \label{odd8PloLik} + + In this section we will define the concept of the area of a certain class of figures. The area is based on the concept of the length of a line in a certain system of measurement and on the theory of infinitesimal calculus\footnote{The first steps in this direction were made by the Greek mathematician \index{Euclid}\textit{Euclid} (408--355 BC) and \index{Archimedes}\textit{Archimedes} (287--212 BC) in the calculation of the volume of bodies.}. + +Let $\Omega_0$ be a square with a side length of $1$ in a given measuring system. The aforementioned square represents the \index{unit of measurement of area}\pojem{unit of measurement of area} - we also call it the \index{unit square}\pojem{unit square}. Intuitively, the area of a figure $\Phi$ represents the number of unit squares that are consistent with square $\Omega_0$ and with which we can ''cover'' figure $\Phi$. + +Now we will approach the definition of area more formally. Let $\Phi$ be an arbitrary figure in a plane. The unit square $\Omega_0$ defines the tiling $(4,4)$ (see section \ref{odd3Tlakovanja}) - we denote it with $\mathcal{T}_0$. With $\underline{S}_0$ we denote the number of squares of tiling $\mathcal{T}_0$, which are in figure $\Phi$, or are its subset. +With $\overline{S}_0$ we denote the number of squares of tiling $\mathcal{T}_0$, which have at least one common point with figure $\Phi$ (Figure \ref{sl.plo.8.1.1.pic}). It is clear that it holds: + $$0\leq\underline{S}_0\leq\overline{S}_0.$$ + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.1.1.pic} +\caption{} \label{sl.plo.8.1.1.pic} +\end{figure} + + +If we divide each side of the square into $10$ lines, we can divide the unit square $\Omega_0$ into $10^2$ consistent squares, which are all consistent with one of these squares $\Omega_1$, which has a side length of $\frac{1}{10}$. Square $\Omega_1$ defines a new tiling $\mathcal{T}_1$. Similarly to the previous example, let $\underline{S}_1$ be the number of squares of tiling $\mathcal{T}_1$, which are in figure $\Phi$, and $\overline{S}_1$ be the number of squares of tiling $\mathcal{T}_1$, which have at least one common point with figure $\Phi$. It is clear that $\underline{S}_1\leq\overline{S}_1$ and also: + $$0\leq\underline{S}_0\leq\frac{\underline{S}_1}{10^2} +\leq\frac{\overline{S}_1}{10^2}\leq\overline{S}_0.$$ + +If we continue the process, we get a sequence of squares $\Omega_n$, which have a side length of $\frac{1}{10^n}$, tiling $\mathcal{T}_n$ and pairs of numbers $\underline{S}_n$ and $\overline{S}_n$, for which the following is true: +$$0\leq\underline{S}_0\leq\frac{\underline{S}_1}{10^2}\leq \cdots \leq \frac{\underline{S}_n}{10^{2n}}\leq\cdots\leq \frac{\overline{S}_n}{10^{2n}}\leq\cdots +\leq\frac{\overline{S}_1}{10^2}\leq\overline{S}_0.$$ + +The sequence $\frac{\underline{S}_n}{10^{2n}}$ is increasing and bounded from above, so it is convergent by a known theorem of mathematical analysis and has the following limit: +$$\underline{S}=\lim_{n\rightarrow\infty}\frac{\underline{S}_n}{10^{2n}}.$$ + +Similarly, the sequence $\frac{\overline{S}_n}{10^{2n}}$ is decreasing and bounded from below, so it is convergent and has the following limit: +$$\overline{S}=\lim_{n\rightarrow\infty}\frac{\overline{S}_n}{10^{2n}}.$$ + +If $\underline{S}=\overline{S}$, we say that the figure $\Phi$ is measurable. The number $S=\underline{S}=\overline{S}$ is its \index{area of a figure}\pojem{area}. We denote it by $p(\Phi)$ or $p_{\Phi}$. + +It is intuitively clear that the area of a figure $\Phi$ is not dependent on the location of the unit square - that is, it is only dependent on the system of measuring lengths, in which the side of the unit square has a length of 1. We will not formally prove this fact here. + +We will now prove the first important property of area. + + + + \bizrek \label{ploscDaljice} + The area of an arbitrary point is 0. + The area of an arbitrary line segment is 0. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.1.2.pic} +\caption{} \label{sl.plo.8.1.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} + +\textit{1)} Let $A$ be an arbitrary point. It is clear that for every $n\in \mathbb{N}$ it holds: $\underline{S}_n=0$ and $0\leq\overline{S}_n\leq4$ (the point lies in at most four paving tiles $T_n$). So it holds: +$$\underline{S}=\lim_{n\rightarrow\infty}\frac{\underline{S}_n}{10^{2n}} +=\lim_{n\rightarrow\infty}\frac{0}{10^{2n}}=0$$ +and +$$0\leq\overline{S}=\lim_{n\rightarrow\infty}\frac{\overline{S}_n}{10^{2n}}\leq +\lim_{n\rightarrow\infty}\frac{4}{10^{2n}}=0,$$ +thus $p_A=S=\underline{S}=\overline{S}=0$ + +\textit{2)} Let $AB$ be an arbitrary distance (Figure \ref{sl.plo.8.1.2.pic}) and $|AB|=d$. With $k$ we denote the integer part of the number $d$, i.e. $k=[d]$. We choose a unit square $\Omega_0$, so that one of its vertices is the point $A$, one of its sides lies on the distance $AB$. In this case $\underline{S}_n=0$ and $\overline{S}_n\leq (k+1)\cdot 10^n$. Then it holds: +$$\underline{S}=\lim_{n\rightarrow\infty}\frac{\underline{S}_n}{10^{2n}} +=\lim_{n\rightarrow\infty}\frac{0}{10^{2n}}=0$$ +and +$$0\leq\overline{S}=\lim_{n\rightarrow\infty}\frac{\overline{S}_n}{10^{2n}}\leq +\lim_{n\rightarrow\infty}\frac{(k+1)10^n}{10^{2n}}= +\lim_{n\rightarrow\infty}\frac{k+1}{10^n}=0,$$ +thus $p_{AB}=S=\underline{S}=\overline{S}=0.$ +\kdokaz + +The next theorem, which refers to the basic properties of the area, we will state without a proof (see \cite{Lucic}). + +\bizrek \label{ploscGlavniIzrek} + Let $p$ be an area, defined on a set $\mu$ of measurable figures + of a plane. + Then it holds: + \begin{enumerate} + \item $p(\Omega_0)=1$; + \item $(\forall \Phi\in\mu)\hspace*{1mm}p(\Phi)\geq 0$; + \item $(\forall \Phi_1,\Phi_2\in\mu)\hspace*{1mm} + \left(\Phi_1\cong\Phi_2 + \hspace*{1mm}\Rightarrow\hspace*{1mm}p(\Phi_1)=p(\Phi_2)\right)$; + \item $(\forall \Phi_1,\Phi_2\in\mu)\hspace*{1mm} + \left(p(\Phi_1\cap\Phi_2)=0 + \hspace*{1mm}\Rightarrow\hspace*{1mm}p(\Phi_1\cup\Phi_2)= + p(\Phi_1)+p(\Phi_2)\right)$. + \end{enumerate} + \eizrek + +The following theorem is very useful. + + + \bizrek \label{ploscLomljenke} + The area of any break is equal to 0. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.1.3.pic} +\caption{} \label{sl.plo.8.1.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.1.3.pic}) + +The claim is a direct consequence of theorems \ref{ploscDaljice} and \ref{ploscGlavniIzrek} (\textit{4}). +\kdokaz + + + +%________________________________________________________________________________ + \poglavje{Area of Rectangles, Parallelogram, and Trapezoids} \label{odd8PloParalel} + + + In the following we will derive formulas for the area of certain quadrilaterals. + + \bizrek \label{ploscPravok} + If $p$ is the area of a rectangle with sides of lengths $a$ and $b$, then + it is: + $$p=ab.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.1a.pic} +\caption{} \label{sl.plo.8.2.1a.pic} +\end{figure} + + + \textbf{\textit{Proof.}} +Let $ABCD$ be a rectangle with sides $AB$ and $AD$ of lengths $|AB|=a$ and $|AD|=b$ (Figure \ref{sl.plo.8.2.1a.pic}). +Choose a unit square $\Omega_0$, so that one of its vertices is point $A$, one of its sides lies on the line $AB$. Let $\mathcal{T}_n$ ($n\in \mathbb{N}\cup\{0\}$) be the corresponding tiling that is determined by the unit square $\Omega_0$. + +We denote with $a_n=[a\cdot 10^n]$ and $b_n=[b\cdot 10^n]$ (where $[x]$ represents the integer part of number $x$). First, we have: + \begin{eqnarray} \label{eqnPloscPrav1a} + \frac{a_n}{10^n}\leq a\leq \frac{a_n+1}{10^n} \hspace*{2mm} \textrm{ and } \hspace*{2mm} +\frac{b_n}{10^n}\leq b\leq \frac{b_n+1}{10^n}, + \end{eqnarray} +then: + \begin{eqnarray} \label{eqnPloscPrav2a} + \underline{S}_n=\frac{a_n\cdot b_n}{10^{2n}} \hspace*{2mm} \textrm{ and } \hspace*{2mm} +\overline{S}_n=\frac{\left(a_n+1\right)\cdot \left(b_n+1\right)}{10^{2n}}. + \end{eqnarray} +From \ref{eqnPloscPrav2a} and \ref{eqnPloscPrav1a} we get: +\begin{eqnarray*} +0&\leq&\lim_{n\rightarrow\infty}\left(\overline{S}_n-\underline{S}_n\right)=\\ +&=&\lim_{n\rightarrow\infty}\left(\frac{\left(a_n+1\right)\cdot \left(b_n+1\right)}{10^{2n}}-\frac{a_n\cdot b_n}{10^{2n}}\right)=\\ +&=& \lim_{n\rightarrow\infty}\frac{a_n+ b_n+1}{10^{2n}}\leq\\ +&\leq& \lim_{n\rightarrow\infty}\frac{a+b+1}{10^n}=\\ +&=& 0. + \end{eqnarray*} +Therefore: +\begin{eqnarray*} +\lim_{n\rightarrow\infty}\left(\overline{S}_n-\underline{S}_n\right)=0, + \end{eqnarray*} +which means that $\underline{S}=\lim_{n\rightarrow\infty}\underline{S}_n +=\lim_{n\rightarrow\infty}\overline{S}_n=\overline{S}$, which means that the rectangle $ABCD$ is a measurable figure with the area $p=S=\underline{S}=\overline{S}$. + +We will now prove $p=ab$. From \ref{eqnPloscPrav1a} we get: + \begin{eqnarray*} + \frac{a_n\cdot b_n}{10^{2n}} \leq ab\leq \frac{\left(a_n+1\right)\cdot \left(b_n+1\right)}{10^{2n}}, + \end{eqnarray*} +Since this is true for every $n\in \mathbb{N}$, it follows: + \begin{eqnarray*} + \hspace*{-2mm} \underline{S}=\lim_{n\rightarrow\infty}\underline{S}_n= +\lim_{n\rightarrow\infty}\frac{a_n\cdot b_n}{10^{2n}} \leq ab\leq \lim_{n\rightarrow\infty}\frac{\left(a_n+1\right)\cdot \left(b_n+1\right)}{10^{2n}}=\lim_{n\rightarrow\infty}\overline{S}_n=\overline{S}. + \end{eqnarray*} +From $p=S=\underline{S}=\overline{S}$ at the end it follows $p=ab$. +\kdokaz + +A direct consequence of the previous \ref{ploscPravok} is the following claim (Figure \ref{sl.plo.8.2.2.pic}). + + \bizrek \label{ploscKvadr} + If $p_{\square}$ is the area of a square with a side of length $a$, then: + $$p_{\square}=a^2.$$ + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.2.pic} +\caption{} \label{sl.plo.8.2.2.pic} +\end{figure} + +We will now derive formulas for the area for some other quadrilaterals. + + + \bizrek \label{ploscParal} + If $p$ is the area of a parallelogram with sides of length $a$ and $b$ and + the corresponding heights of length $v_a$ and $v_b$, then + it is: + $$p=av_a=bv_b.$$ + \eizrek + +\textbf{\textit{Proof.}} +Let $ABCD$ be a parallelogram with side $AB$ of length $|AB|=a$ and the corresponding height $v_a$. We mark with $E$ and $F$ the orthogonal projections of the vertices $D$ and $C$ on the line $AB$. The quadrilateral $EFCD$ is a rectangle with sides $|CD|=a$ and $|FC|=v_a$, so according to the theorem \ref{ploscPravok}: +\begin{eqnarray} +p_{EFCD}=av_a. \label{eqnPloscPrav2} +\end{eqnarray} +The right-angled triangle $AED$ and $BFC$ are congruent (theorem \textit{ASA} \ref{KSK}) so according to the theorem \ref{ploscGlavniIzrek} \textit{3)}: +\begin{eqnarray} +p_{AED}=p_{BFC}. \label{eqnPloscPrav1} +\end{eqnarray} +Without loss of generality, we assume that $\angle BAD\leq 90^0$. In this case, the points $B$ and $E$ are on the same side of the point $A$ (in the case $E=A$ it is a rectangle $ABCD$ and the statement follows directly from the theorem \ref{ploscPravok}). We will consider several possible cases: $\mathcal{B}(A,E,B)$, $E=B$ and $\mathcal{B}(A,B,E)$. + +\textit{1)} Let $\mathcal{B}(A,E,B)$ (Figure \ref{sl.plo.8.2.3.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.3.pic} +\caption{} \label{sl.plo.8.2.3.pic} +\end{figure} + +If we use the relations \ref{eqnPloscPrav2} and \ref{eqnPloscPrav1} and theorems \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscDaljice}, we get: + \begin{eqnarray*} + p_{ABCD}=p_{AED}+p_{EBCD}=p_{BFC}+p_{EBCD}=p_{EFCD}=av_a. +\end{eqnarray*} + +\textit{2)} In the case $E=B$ (Figure \ref{sl.plo.8.2.3a.pic}) similarly from the relations \ref{eqnPloscPrav1} and \ref{eqnPloscPrav1} and theorems \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscDaljice} we get: + \begin{eqnarray*} + p_{ABCD}=p_{AED}+p_{BCD}=p_{BFC}+p_{BCD}=p_{EFCD}=av_a. +\end{eqnarray*} + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.3a.pic} +\caption{} \label{sl.plo.8.2.3a.pic} +\end{figure} + + + +\textit{3)} We assume that $\mathcal{B}(A,B,E)$ (Figure \ref{sl.plo.8.2.3b.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.3b.pic} +\caption{} \label{sl.plo.8.2.3b.pic} +\end{figure} + +The lines $BC$ and $DE$ intersect in some point $L$. By the formulas \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscDaljice} it is: +$p_{AED}=p_{ABLD}+p_{BEL}$ and $p_{BFC}=p_{EFCL}+p_{BEL}$. Because by the relation \ref{eqnPloscPrav1} $p_{AED}=p_{BFC}$, it is also: + \begin{eqnarray*} + p_{ABLD}=p_{EFCL}. +\end{eqnarray*} +From this and from the formulas \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscDaljice} it follows: + \begin{eqnarray*} + p_{ABCD}=p_{ABLD}+p_{DLC}=p_{EFCL}+p_{DLC}=p_{EFCD}=av_a, +\end{eqnarray*} +which was to be proven. \kdokaz + + \bizrek \label{ploscTrapez} + If $p$ is the area of a trapezoid with bases of lengths $a$ and $c$ and + height of length $v$, then + it is: + $$p=\frac{a+c}{2}\cdot v.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.4.pic} +\caption{} \label{sl.plo.8.2.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $ABCD$ be a trapezoid with bases $AB$ and $CD$ of lengths $|AB|=a$ and $|CD|=c$ and a height of length $v$ (Figure \ref{sl.plo.8.2.4.pic}). +With $S$ we mark the center of the leg $BC$ and $\mathcal{S}_S:\hspace*{1mm}A,D\mapsto A', D'$. Because $S$ is the common center of the lines $AA'$ and $DD'$, by \ref{paralelogram} $AD'A'D$ is a parallelogram. Because $\mathcal{S}_S(B)=C$ and $\mathcal{B}(A,B,D')$, $\mathcal{B}(A',C,D)$ is also true. The isometry $\mathcal{S}$ maps the trapezoid $ABCD$ into a similar trapezoid $A'CBD'$, therefore $|BD'|=|CD|=c$ or $|AD'|=a+c$, and by \ref{ploscGlavniIzrek} \textit{3)} also $p_{ABCD}=p_{A'CBD'}$. So the parallelogram $AD'A'D$ with a base $AD'$ of length $|AD'|=a+c$ and a height which is equal to the height of the trapezoid $ABCD$ of length $v$, is divided into two similar trapezoids $ABCD$ and $A'CBD'$ with equal areas. + +From this and from \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscDaljice} and \ref{ploscParal} it follows: + \begin{eqnarray*} + 2\cdot p_{ABCD}=p_{ABCD}+p_{A'CBD'}=p_{AD'A'D}=\left(a+c\right)v, +\end{eqnarray*} +or the desired relation. +\kdokaz + + +If we use \ref{srednjTrapez}, we see that the expression $\frac{a+c}{2}$ represents the length of the median of the trapezoid, so for the area of the trapezoid the formula: + $$p=mv,$$ +where $m$ is the length of the median of the trapezoid and $v$ is its height + (Figure \ref{sl.plo.8.2.4a.pic}) is also true. + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.4a.pic} +\caption{} \label{sl.plo.8.2.4a.pic} +\end{figure} + + + + + \bzgled \label{ploscStirikPravok} + If $p$ is the area of a quadrilateral with perpendicular diagonals of lengths $e$ and + $f$, then: + $$p=\frac{ef}{2}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.5.pic} +\caption{} \label{sl.plo.8.2.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} Let $ABCD$ be a quadrilateral with perpendicular diagonals $AC$ and $BD$ of lengths $|AC|=e$ and $|BD|=f$ (Figure \ref{sl.plo.8.2.5.pic}). The parallels of these diagonals through the vertices of the quadrilateral $ABCD$ determine the rectangle $A'B'C'D'$ with sides of lengths $|A'B'|=f$ and $|B'C'|=e$. From the similarity of triangles $ASD$ and $DA'A$ (the \textit{SAS} theorem \ref{SKS}) it follows that $p_{ASD}=p_{DA'A}$ (the \ref{ploscGlavniIzrek} \textit{3)}). Similarly, $p_{ASB}=p_{BB'A}$, $p_{CSB}=p_{BC'C}$ and $p_{CSD}=p_{DD'C}$. From this and from theorems \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscDaljice} and \ref{ploscPravok} it follows: + \begin{eqnarray*} + 2\cdot p_{ABCD}&=&2\cdot\left(p_{ASD}+p_{ASB}+p_{CSB}+p_{CSD}\right)=\\ +&=&2\cdot p_{ASD}+2\cdot p_{ASB}+2\cdot p_{CSB}+2\cdot p_{CSD}=\\ +&=& p_{ASD}+ p_{DA'A}+ p_{ASB}+p_{BB'A}+\\ + && + p_{CSB}+p_{BC'C} +p_{CSD}+p_{DD'C}=\\ +&=& p_{A'B'C'D'}= ef, +\end{eqnarray*} + +or the desired relation. +\kdokaz + + +As a consequence, we have the following theorem (Figure \ref{sl.plo.8.2.5a.pic}). + + + \bzgled \label{ploscDeltoid} + If $p$ is the area of a deltoid with diagonals of lengths $e$ and + $f$, then: + $$p=\frac{ef}{2}.$$ + \ezgled + + \textbf{\textit{Proof.}} + + The theorem is a direct consequence of theorem \ref{ploscStirikPravok} and the definition of a deltoid. +\kdokaz + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.5a.pic} +\caption{} \label{sl.plo.8.2.5a.pic} +\end{figure} + +\bzgled \label{ploscRomb} +If $p$ is the area of a rhombus with diagonals of lengths $e$ and +$f$, then: +$$p=\frac{ef}{2}.$$ +\ezgled + +\textbf{\textit{Proof.}} +The statement is a direct consequence of Theorems \ref{ploscStirikPravok} and \ref{RombPravKvadr}. +\kdokaz + + + +\bzgled \label{ploscRomb1} +If: $a$ is the length of a side, $v$ is the height, and $e$ and +$f$ are the lengths of the diagonals of the rhombus, then: +$$av=\frac{ef}{2}.$$ +\ezgled + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.6.pic} +\caption{} \label{sl.plo.8.2.6.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.2.5a.pic}) + + The statement is a direct consequence of Theorems \ref{ploscRomb1} and \ref{ploscParal}. +\kdokaz + + +%________________________________________________________________________________ + \poglavje{Area of Triangles} \label{odd8PloTrik} + +In this section we will derive several formulas for the area of a triangle. + + \bizrek \label{PloscTrik} If $p_\triangle$ is the + area of the triangle $ABC$, $v_a$, $v_b$ and $v_c$ are the lengths of the heights corresponding to the sides $BC$, $AC$ and $AB$ with lengths $a$, $b$ and $c$, then: + $$p_\triangle=\frac{a\cdot v_a}{2}=\frac{b\cdot v_b}{2} + =\frac{c\cdot v_c}{2}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.1a.pic} +\caption{} \label{sl.plo.8.3.1a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $D$ be the fourth vertex of the parallelogram $ABCD$ (Figure \ref{sl.plo.8.3.1a.pic}). The triangles $ABC$ and $ADC$ are congruent (by \ref{paralelogram} and \ref{SSS}), therefore $p_{ABC}=p_{ADC}$ (by \ref{ploscGlavniIzrek} \textit{3)}). The parallelogram $ABCD$ has the side $BC$ and the corresponding altitude of lengths $a$ and $v_a$, therefore by \ref{ploscParal} $p_{ABCD}=av_a$. If we use \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscDaljice} as well, we get: + \begin{eqnarray*} + 2\cdot p_{ABC}=p_{ABC}+p_{ADC}=p_{ABCD}=av_a, +\end{eqnarray*} +or the desired relation $p_\triangle=\frac{a\cdot v_a}{2}$. Similarly, $p_\triangle=\frac{b\cdot v_b}{2}$ or $p_\triangle=\frac{c\cdot v_c}{2}$. +\kdokaz + + \bizrek \label{PloscTrikVcrt} If $p_\triangle$ is the area + of the triangle $ABC$ with + the semi-perimeter $s=\frac{a+b+c}{2}$ and the radius of the inscribed circle + $r$, then + $$p_\triangle=sr.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.1.pic} +\caption{} \label{sl.plo.8.3.1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $S$ be the center of the inscribed circle of the triangle $ABC$ (Figure \ref{sl.plo.8.3.1.pic}). + +By \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscDaljice} and \ref{PloscTrik} we have: +\begin{eqnarray*} + p_{ABC}=p_{SBC}+p_{ASC}+p_{ABS}&=& + \frac{a\cdot r}{2}+\frac{b\cdot r}{2}+\frac{c\cdot r}{2}=\\ +&=& + \frac{a+b+c}{2}\cdot r=sr, +\end{eqnarray*} + which was to be proven. \kdokaz + + + \bizrek \label{PloscTrikPricrt} If $p_\triangle$ is the + area of the triangle $ABC$ with + the semi-perimeter $s=\frac{a+b+c}{2}$ and the radius of the circumscribed circle + $r_a$, then + $$p_\triangle=(s-a)r_a.$$ + \eizrek + +\textbf{\textit{Proof.}} We use the notation from the big task \ref{velikaNaloga} (Figure \ref{sl.plo.8.3.2.pic}). From the same task we have: $AR=s-a$ and $AR_a=s$. + +The right angled triangle $ARS$ and $AR_aS_a$ are similar (statement \ref{PodTrikKKK}), so $\frac{SR}{S_aR_a}=\frac{AR}{AR_a}$, i.e. $\frac{r}{r_a}=\frac{s-a}{s}$. If we use the previous statement \ref{PloscTrikVcrt}, we get: +\begin{eqnarray*} + p_{ABC}=sr=(s-a)r_a, +\end{eqnarray*} + which was to be proven. \kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.2.pic} +\caption{} \label{sl.plo.8.3.2.pic} +\end{figure} + + + \bizrek \label{PloscTrikHeron} If $p_\triangle$ + is the area of the triangle $ABC$ with + the semiperimeter $s=\frac{a+b+c}{2}$, then it holds + \index{formula!Heronova} + (Heron's\footnote{\index{Heron} + \textit{Heron from Alexandria}(20--100), + ancient Greek matemetician.} formula): + $$p_\triangle=\sqrt{s(s-a)(s-b)(s-c)}.$$ + \eizrek + + + \textbf{\textit{Proof.}} Again, we use the notation from the big task \ref{velikaNaloga} (Figure \ref{sl.plo.8.3.2.pic}). From the same task we have: $BP=s-b$ and $BP_a=s-c$. + +The angle $SBP$ and $BS_aP_a$ are complementary, because they have a pair of perpendicular sides (statement \ref{KotaPravokKraki}). Therefore, the right angled triangle $SBP$ and $BS_aP_a$ are similar (statement \ref{PodTrikKKK}), so $\frac{SP}{BP_a}=\frac{BP}{S_aP_a}$ i.e. $\frac{r}{s-c}=\frac{s-b}{r_a}$ i.e. $rr_a=(s-b)(s-c)$. + If we use the statement \ref{PloscTrikVcrt} and \ref{PloscTrikPricrt}, we get: +\begin{eqnarray*} +p_{ABC}^2=sr(s-a)r_a=s(s-a)rr_a=s(s-a)(s-b)(s-c), +\end{eqnarray*} + which was to be proven. \kdokaz + +\bizrek \label{PloscTrikOcrt} If $p_\triangle$ is the area of the triangle $ABC$ with the side lengths $a$, $b$ and $c$ and $R$ is the radius of the circumscribed circle, then the following is true: +$$p_\triangle=\frac{abc}{4R}.$$ +\eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.3.pic} +\caption{} \label{sl.plo.8.3.3.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.3.3.pic}) + + The statement is a direct consequence of the formulas \ref{PloscTrik} and \ref{izrekSinusni}: +\begin{eqnarray*} +p_{ABC}=\frac{av_a}{2}=\frac{a}{2}\cdot v_a=\frac{a}{2}\cdot \frac{bc}{2R}=\frac{abc}{4R}, +\end{eqnarray*} + which was to be proven. \kdokaz + + +The previous formulas for the area of a triangle can be used to calculate the radius of the circumscribed, inscribed and excribed circles of a triangle as functions of its sides. + + \bzgled \label{PloscTrikOcrtVcrt} If $R$ and $r$ are the radius of the circumscribed and inscribed circles, $r_a$, $r_b$ and $r_c$ are the radius of the excribed circles and + $s=\frac{a+b+c}{2}$ is the perimeter + of the triangle $ABC$, + then the following is true: + + (i) $R=\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$,\hspace*{7mm} + (ii) $r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$,\\ + (iii) $r_a=\sqrt{\frac{s(s-b)(s-c)}{s-a}}$,\hspace*{2.7mm} + (iv) $r_b=\sqrt{\frac{(s-a)s(s-c)}{s-b}}$,\hspace*{2.7mm} + (v) $r_c=\sqrt{\frac{(s-a)(s-b)s}{s-c}}$. + + \ezgled + + + + \textbf{\textit{Proof.}} + The statement is a direct consequence of the formulas \ref{PloscTrikOcrt}, \ref{PloscTrikVcrt}, \ref{PloscTrikPricrt} and \ref{PloscTrikHeron}. +\kdokaz + +\bzgled \label{PloscTrikVcrtPricrt} If $r$ is the radius of the inscribed circle, $r_a$, $r_b$ and $r_c$ are the radii of the circumscribed circles, and $p_{\triangle}$ is the area of the triangle $ABC$, then: +$$p_{\triangle}=\sqrt{rr_ar_br_c}.$$ +\ezgled + + \textbf{\textit{Proof.}} + If we use the statements from the previous example \ref{PloscTrikOcrtVcrt}, we get: + \begin{eqnarray*} +rr_ar_br_c=s(s-a)(s-b)(s-c)=p_{\triangle}^2, +\end{eqnarray*} +which was to be proven. \kdokaz + + \bizrek \label{ploscTrikPedalni} + Let $A'B'C'$ be the pedal triangle of the obtuse triangle $ABC$, $s'$ be the perimeter + of the triangle $A'B'C'$, $R$ be the radius of the circumscribed circle, and $p_{\triangle}$ be the area + of the triangle $ABC$, then: + $$p_{\triangle} = s'R.$$ + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.3a.pic} +\caption{} \label{sl.plo.8.3.3a.pic} +\end{figure} + + + \textbf{\textit{Proof.}} We mark with $O$ the center of the circumscribed circle of the triangle $ABC$ (Figure \ref{sl.plo.8.3.3a.pic}). By the statement \ref{PedalniLemaOcrtana}, $OA\perp B'C'$, so from the statement \ref{ploscStirikPravok} it follows that $p_{AC'OB'}=\frac{|AO|\cdot |B'C'|}{2}=\frac{R\cdot |B'C'|}{2}$. Analogously, $p_{BA'OC'}=\frac{R\cdot |A'C'|}{2}$ and $p_{CB'OA'}=\frac{R\cdot |A'B'|}{2}$. Because $ABC$ is an obtuse triangle, the point $O$ is in its interior (see section \ref{odd3ZnamTock}), so by the statements \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscDaljice}, it holds: +\begin{eqnarray*} +p_{\triangle}&=&p_{AC'OB'}+p_{BA'OC'}+p_{CB'OA'}=\\ +&=&\frac{R}{2}\left(|B'C'|+|A'C'|+|A'B'| \right)=s'R, +\end{eqnarray*} + which was to be proven. \kdokaz + +\bzgled \label{ploscTrikPedalni1} + Let $R$ and $r$ be the radii of the circumscribed and inscribed circles of the acute angled triangle $ABC$ and $s$ and $s'$ be the semi-perimeter of this triangle and its pedal triangle. + Prove that: + $$\frac{R}{r}=\frac{s}{s'}.$$ + \ezgled + +\textbf{\textit{Proof.}} By the formulas \ref{PloscTrikVcrt} and \ref{ploscTrikPedalni} we have: +\begin{eqnarray*} +p_{ABC}=sr=s'R. +\end{eqnarray*} +From this we obtain the desired relation. +\kdokaz + +In a special case we obtain formulas for the area of a right angled and an equilateral triangle. + + \bzgled \label{PloscTrikPravokotni} If $p_\triangle$ is the area + of the right angled triangle $ABC$ with the sides of lengths $a$ and $b$, then: + $$p_\triangle=\frac{ab}{2}.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.4a.pic} +\caption{} \label{sl.plo.8.3.4a.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.3.4a.pic}) + The statement is a direct consequence of the formula \ref{PloscTrik}, since in this case $v_a=b$. +\kdokaz + + \bizrek \label{PloscTrikEnakostr} If $p_\triangle$ is the area + of the equilateral triangle $ABC$ with the side of length $a$, then: + $$p_\triangle=\frac{a^2\sqrt{3}}{4}.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.4.pic} +\caption{} \label{sl.plo.8.3.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.3.4.pic}) +The statement is a direct consequence of theorems \ref{PloscTrik} and \ref{PitagorovEnakostr}: +\begin{eqnarray*} +p_{ABC}=\frac{av_a}{2}=\frac{a}{2}\cdot v_a=\frac{a}{2}\cdot \frac{a\sqrt{3}}{2}=\frac{a^2\sqrt{3}}{4}, +\end{eqnarray*} +which had to be proven. \kdokaz + +We will continue using the formulas for the area of a triangle. + + + + + + \bzgled \label{CarnotOcrtLema} + Let $P$ be an inner point of the triangle $ABC$ and $x$, $y$ and $z$ the distances of this point from the sides $BC$, $AC$ and $AB$ with lengths $a$, $b$ and $c$. If $r$ is the radius of the inscribed circle of this triangle, then: + $$xa+yb+zc=2\cdot p_{\triangle ABC}=r(a+b+c).$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.6.pic} +\caption{} \label{sl.plo.8.3.6.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.3.6.pic}) + +$x$, $y$ and $z$ are the lengths of the altitudes of the triangles $PBC$, $APC$ and $ABP$, so +according to theorems \ref{PloscTrik} and \ref{PloscTrikVcrt} it follows: + \begin{eqnarray*} + xa+yb+zc &=& 2\cdot p_{\triangle PBC}+2\cdot p_{\triangle APC}+2\cdot p_{\triangle ABP}\\ + &=& 2\cdot p_{\triangle ABC}\\ + &=& r(a+b+c), + \end{eqnarray*} + which had to be proven. \kdokaz + + + \bzgled \index{izrek!Vivianijev} (Vivianijev\footnote{\index{Viviani, V.}\textit{V. Viviani} (1622--1703), + Italian mathematician and physicist.} izrek) + Let $P$ be an inner point of the equilateral triangle $ABC$ and $x$, $y$ and $z$ the distances of this point from the sides $BC$, $AC$ and $AB$. If $v$ is the height of this triangle, then: + $$x+y+z=v.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.6a.pic} +\caption{} \label{sl.plo.8.3.6a.pic} +\end{figure} + +\textbf{\textit{Proof.}} +By the statement from the previous example \ref{CarnotOcrtLema} and the formula \ref{PloscTrik}: +$xa+ya+za=2\cdot p_{\triangle ABC}=v_aa$. +\kdokaz + + \bizrek \label{CarnotOcrt}\index{izrek!Carnotov o očrtani krožnici}(Carnotov\footnote{\index{Carnot, L. N. M.}\textit{L. N. M. Carnot} (1753--1823), francoski matematik.} izrek o očrtani krožnici.) + The sum of the distances from the center of the circumscribed circle to the sides of the triangle is equal to the sum of the radii of the circumscribed and inscribed circle of that triangle. So if $l(O,R)$ is the circumscribed and $k(S,r)$ the inscribed circle and $A_1$, $B_1$ and $C_1$ are the centers of the sides of the triangle $ABC$, it holds: + $$|OA_1|+|OB_1|+|OC_1|=R+r.$$ + + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.7.pic} +\caption{} \label{sl.plo.8.3.7.pic} +\end{figure} + +\textbf{\textit{Proof.}} We mark $x=|OA_1|$, $y=|OB_1|$ and $z=|OC_1|$ (Figure \ref{sl.plo.8.3.7.pic}). According to the statement from the example \ref{CarnotOcrtLema} it is: +\begin{eqnarray} \label{eqnCarnotOcrt1} +xa+yb+zc=r(a+b+c) +\end{eqnarray} + From the similarity of triangles $BA_1O$ and $CA_1O$ (\textit{SAS} \ref{SKS}) and \ref{SredObodKot} it follows that $\angle BAC=\frac{1}{2}\angle BOC=\angle BOA_1$. This means that $\triangle OA_1B\sim \triangle AB'B \sim \triangle AC'C$ (\ref{PodTrikKKK}) is valid, therefore: + \begin{eqnarray*} + \frac{OA_1}{AB'}=\frac{OB}{AB} \hspace*{2mm} \textrm{ and } \hspace*{2mm} + \frac{OA_1}{AC'}=\frac{OB}{AC} +\end{eqnarray*} +or: +\begin{eqnarray*} + cx=R\cdot |AB'| \hspace*{2mm} \textrm{ and } \hspace*{2mm} + bx=R\cdot |AC'| +\end{eqnarray*} +After adding the last two equations we get: +\begin{eqnarray*} + (b+c)x=R(|AB'|+|AC'|) +\end{eqnarray*} +and similarly: +\begin{eqnarray*} + (a+c)y=R\cdot (|BA'|+|BC'|)\\ + (a+b)z=R\cdot (|CA'|+|CB'|). +\end{eqnarray*} + +After adding the last three relations we get: +\begin{eqnarray} \label{eqnCarnotOcrt2} +(b+c)x+(a+c)y+(a+b)z=R(a+b+c). +\end{eqnarray} +If we finally add the equality \ref{eqnCarnotOcrt1} and \ref{eqnCarnotOcrt2} and divide the obtained equality by $a+b+c$, we get: +\begin{eqnarray*} +x+y+z=r+R, +\end{eqnarray*} + which had to be proven. \kdokaz + + + \bnaloga\footnote{12. IMO, Hungary - 1970, Problem 1.} + Let $M$ be a point on the side $AB$ of triangle $ABC$. Let $r_1$, $r_2$ and $r$ be the radii + of the inscribed circles of triangles $AMC$, $BMC$ and $ABC$. Let $q_1$, $q_2$ and $q$ + be the radii of the escribed circles of the same triangles that lie in the angle + $ACB$. Prove that + $$\frac{r_1}{q_1}\cdot\frac{r_2}{q_2}=\frac{r}{q}.$$ + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.IMO1.pic} +\caption{} \label{sl.plo.8.3.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} First, we mark $a=|BC|$, $b=|AC|$, +$c=|AB|$, $x=|AM|$, $y=|BM|$ and $m=|CM|$ (Figure +\ref{sl.plo.8.3.IMO1.pic}). + + If we use the expression \ref{PloscTrikVcrt} and + \ref{PloscTrikPricrt}, we get: + \begin{eqnarray*} + && p_{\triangle ABC}=\frac{a+b+x+y}{2}\cdot r=\frac{a+b-x-y}{2}\cdot q\\ + && p_{\triangle AMC}=\frac{b+m+x}{2}\cdot r_1=\frac{b+m-x}{2}\cdot q_1\\ + && p_{\triangle ABC}=\frac{a+m+y}{2}\cdot r_2=\frac{a+m-y}{2}\cdot q_2 + \end{eqnarray*} +It follows that: + \begin{eqnarray*} + \frac{r}{q}=\frac{a+b+x+y}{a+b-x-y},\hspace*{6mm} + \frac{r_1}{q_1}=\frac{b+m+x}{b+m-x},\hspace*{6mm} + \frac{r_2}{q_2}=\frac{a+m+y}{a+m-y} + \end{eqnarray*} +Therefore, the relation +$\frac{r_1}{q_1}\cdot\frac{r_2}{q_2}=\frac{r}{q}$, which we want to +prove, is equivalent to: +$$(a+b+x+y)(a+m-y)(b+m-x)=(a+b-x-y)(a+m+y)(b+m+x),$$ +which, after rearranging, is equivalent to: +$$(a+b)(a+m)x+(a+b)(b+m)y=c(a+m)(b+m)+(x+y)xy,$$ +or, if we take into account $x+y=c$ and rearrange further, to: +$$m^2=a^2\frac{x}{c}+b^2\frac{y}{c}-xy.$$ + The last relation is true, because it represents Stewart's theorem + \ref{StewartIzrek} for the triangle $ABC$ and the distance $CM$. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Area of Polygons} \label{odd8PloVeck} + +The area of any polygon is obtained by dividing it with its diagonals into a union of triangles (Figure \ref{sl.plo.8.4.1.pic}). According to the expressions \ref{ploscGlavniIzrek} and \ref{ploscDaljice}, the area of this polygon is equal to the sum of the areas of these triangles. + + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.4.1.pic} +\caption{} \label{sl.plo.8.4.1.pic} +\end{figure} + +The following theorem applies specifically to a regular hexagon. + +\bizrek +If $p$ is the area and $a$ is the length of a side of a regular hexagon, then: +$$p=\frac{3\sqrt{3}\cdot a^2}{2}.$$ +\eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.4.2.pic} +\caption{} \label{sl.plo.8.4.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.4.2.pic}) + +The theorem is a direct consequence of formulas \ref{ploscGlavniIzrek}, \ref{ploscDaljice} \ref{PloscTrikEnakostr} and the fact that a regular hexagon can be divided into six regular triangles (section \ref{odd3PravilniVeck}). +\kdokaz + +A similar formula to that in the case of triangles applies for the area of tangent polygons (formula \ref{PloscTrikVcrt}). + + \bizrek \label{ploscTetVec} + If: $s$ is the circumference, $k(S,r)$ is the inscribed circle of the tangent polygon + $A_1A_2\ldots A_n$ and $p$ + is its area, then $$p = sr.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.4.3.pic} +\caption{} \label{sl.plo.8.4.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.4.3.pic}) + +The triangles $A_1SA_2$, $A_2SA_3$, ... , $A_{n-1}SA_n$ and $A_nSA_1$ all have the same length of altitude from the vertex $S$, which is equal to $r$. If we use formulas \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscDaljice} and \ref{PloscTrik}, we get: +\begin{eqnarray*} + p &=& p_{A_1SA_2}+p_{A_2SA_3}+\cdots +p_{A_{n-1}SA_n} +p_{A_nSA_1}=\\ + &=& \frac{|A_1A_2|\cdot r}{2}+\frac{|A_2A_3|\cdot r}{2}+\cdots+ +\frac{|A_{n-1}A_n|\cdot r}{2}+\frac{|A_n A_1|\cdot r}{2}=\\ +&=&sr, +\end{eqnarray*} +which is what needed to be proven. \kdokaz + +\bnaloga\footnote{30. IMO, Germany - 1989, Problem 2.} + Let $S$, $S_a$, $S_b$ and $S_c$ be incentre and excentres of an acute-angled triangle $ABC$. + $N_a$, $N_b$ and $N_c$ are midpoints of line segments $SS_a$, $SS_b$ and $SS_c$, respectively. + Let $\mathcal{S}_{X_1X_2\ldots X_n}$ be the area of a polygon $X_1X_2\ldots X_n$. + Prove that: + $$\mathcal{S}_{S_aS_bS_c} = + 2\mathcal{S}_{AN_cBN_aCN_b} \geq 4\mathcal{S}_{ABC}.$$ + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.IMO2.pic} +\caption{} \label{sl.plo.8.3.IMO2.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $l$ be the circle circumscribed +about the triangle $ABC$ (Figure \ref{sl.plo.8.3.IMO2.pic}). By the +great theorem (\ref{velikaNaloga}) the point $N_a$ is the centre +of that arc $BC$ of the circle $l$, on which the point $A$ does not +lie - we denote this arc with $l_{BC}$. Similarly, $N_b$ and $N_c$ +are the centres of the corresponding arcs $AC$ and $AB$ on the +circle $l$. + + First, from $SN_a\cong N_aS_a$ it follows that + $\mathcal{S}_{SCN_a}=\mathcal{S}_{N_aCS_a}$ or + $\mathcal{S}_{SCS_a}=2\cdot\mathcal{S}_{SCN_a}$. If we use + analogous equalities and take into account that $S$ is an inner point of the triangle + $S_aS_bS_c$, we get: + \begin{eqnarray*} +\mathcal{S}_{S_aS_bS_c} &=& +\mathcal{S}_{SCS_a}+\mathcal{S}_{SBS_a}+ +\mathcal{S}_{SCS_b}+\mathcal{S}_{SAS_b}+ +\mathcal{S}_{SAS_c}+\mathcal{S}_{SBS_c}=\\ + &=& +2\cdot\left(\mathcal{S}_{SCN_a}+\mathcal{S}_{SBN_a}+ +\mathcal{S}_{SCN_b}+\mathcal{S}_{SAN_b}+ +\mathcal{S}_{SAN_c}+\mathcal{S}_{SBN_c}\right)=\\ +&=& 2\mathcal{S}_{AN_cBN_aCN_b}. + \end{eqnarray*} + +We will now prove that $2\mathcal{S}_{AN_cBN_aCN_b} \geq 4\mathcal{S}_{ABC}$, or $\mathcal{S}_{AN_cBN_aCN_b} \geq 2\mathcal{S}_{ABC}$. Let $V$ be the altitude point of triangle $ABC$ and $V_a=S_{BC}(V)$. Triangle $ABC$ is acute by assumption, so point $V$ lies within it. By \ref{TockaV'} point $V_a$ lies on the arc $l_{BC}$. We have already mentioned that $N_a$ is the center of this arc, so the altitude of triangle $BNC$ is longer than the altitude of triangle $BVC$ by the same base $BC$. It follows that $\mathcal{S}_{BNC}\geq \mathcal{S}_{BVC}$. Similarly, we have $\mathcal{S}_{ANC}\geq \mathcal{S}_{AVC}$ and $\mathcal{S}_{ANB}\geq \mathcal{S}_{AVB}$. Since $V$ is an internal point of triangle $ABC$, we have: + \begin{eqnarray*} +\mathcal{S}_{AN_cBN_aCN_b}&=&\mathcal{S}_{ABC}+ +\mathcal{S}_{BNC}+\mathcal{S}_{ANC}+\mathcal{S}_{ANB}\geq\\ + &\geq& \mathcal{S}_{ABC}+ +\mathcal{S}_{BVC}+\mathcal{S}_{AVC}+\mathcal{S}_{AVB}=\\ +&=& 2\mathcal{S}_{ABC}, + \end{eqnarray*} + which was to be proven. \kdokaz + + + +%________________________________________________________________________________ +\poglavje{Koch Snowflake} \label{odd8PloKoch} + +In this section we will consider a figure which is bounded (a subset of a circle), has a finite area, but its perimeter is infinite. + +First, we will define a special type of polygon. An equilateral triangle $ABC$ with a side length $a=a_0$ is denoted by $\mathcal{K}_0(a)$. The polygon $\mathcal{K}_1(a)$ is obtained if each side of the triangle $\mathcal{K}_0(a)$ is divided into three equal parts and an equilateral triangle with a side length $a_1=\frac{1}{3}\cdot a_0$ is drawn over the middle. The process is continued and the polygon $\mathcal{K}_n(a)$ is obtained if each side of the triangle $\mathcal{K}_{n-1}(a)$ is divided into three equal parts and an equilateral triangle with a side length $a_n=\frac{1}{3}\cdot a_{n-1}$ is drawn over the middle (Figure \ref{sl.plo.8.5.Koch1.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.Koch1.pic} +\caption{} \label{sl.plo.8.5.Koch1.pic} +\end{figure} + + If the described process is continued to infinity, we get the so-called +\index{Kochova snežinka} \pojem{Koch's\footnote{This figure was defined by the Swedish mathematician \index{Koch, H.}\textit{Helge von Koch} (1870--1924) in 1904. Today, Koch's snowflake is classified +as a \textit{fractal}.} snowflake} $\mathcal{K}(a)$. + +First, let's calculate the perimeter of the polygon $\mathcal{K}_n(a)$. + +From the process by which we obtained the polygon $\mathcal{K}_n(a)$, it is clear that the polygon $\mathcal{K}_n(a)$ has four times as many sides as $\mathcal{K}_{n-1}(a)$ (each side of the polygon $\mathcal{K}_{n-1}(a)$ is replaced by four sides of the polygon $\mathcal{K}_n(a)$). We denote by $s_n$ the number of sides of the polygon $\mathcal{K}_n(a)$. Obviously, $s_n$ is a geometric sequence with a quotient of $q=4$ and an initial term of $s_0=3$, so: +\begin{eqnarray} \label{eqnKoch1} +s_n=3\cdot 4^n +\end{eqnarray} +Each side $a_n$ of the polygon $\mathcal{K}_n(a)$ is, by construction, three times shorter than the side $a_{n-1}$ of the polygon $\mathcal{K}_{n-1}(a)$. So we have a geometric sequence $a_n$ again with a quotient of $q=\frac{1}{3}$ and an initial term of $a_0=a$, so: +\begin{eqnarray} \label{eqnKoch2} +a_n=\left( \frac{1}{3} \right)^n\cdot a +\end{eqnarray} +Because for the perimeter $o_n$ of the polygon $\mathcal{K}_n(a)$ it holds that $o_n=s_n\cdot a_n$, from \ref{eqnKoch1} and \ref{eqnKoch2} we get: +\begin{eqnarray} \label{eqnKoch3} +o_n=3\cdot \left( \frac{4}{3} \right)^n\cdot a +\end{eqnarray} +If we want to obtain the perimeter $o$ of the Koch snowflake $\mathcal{K}(a)$, we calculate it as $o=\lim_{n\rightarrow\infty}o_n$, but from \ref{eqnKoch3} we get (because $\frac{4}{3}>1$): +\begin{eqnarray} \label{eqnKoch4} +o=\lim_{n\rightarrow\infty}o_n=\lim_{n\rightarrow\infty}3\cdot \left( \frac{4}{3} \right)^n\cdot a=3a\cdot\lim_{n\rightarrow\infty}\left( \frac{4}{3} \right)^n=\infty +\end{eqnarray} +which means that the Koch snowflake $\mathcal{K}(a)$ has no finite perimeter, that is, the perimeter $o_n$ of the polygon $\mathcal{K}_n(a)$ increases without bound as the number of steps $n$ increases. + +Let's now calculate the area $p$ of the Koch snowflake $\mathcal{K}(a)$. We'll denote the area of the polygon $\mathcal{K}_n(a)$ with $p_n$. +The area $p_n$ is obtained by adding a certain number of smaller triangle areas to the area $p_{n-1}$. How many of these triangles are there? By the construction of the polygon $\mathcal{K}_n(a)$, we get it from the polygon $\mathcal{K}_{n-1}(a)$ by drawing a smaller triangle at each side of the polygon $\mathcal{K}_{n-1}(a)$. So the number of these triangles is equal to $s_{n-1}$. Because of this and the \ref{PloscTrikEnakostr} equation, we have: +\begin{eqnarray*} +p_n=p_{n-1}+s_{n-1}\cdot \frac{a_n^2\cdot \sqrt{3}}{4}=p_{n-1} ++\frac{3}{16}\cdot \left(\frac{4}{9} \right)^n\cdot a^2\sqrt{3}. +\end{eqnarray*} +The area $p$ of the Koch snowflake $\mathcal{K}(a)$ is therefore the infinite sum: +\begin{eqnarray*} +p&=&p_0+ \sum_{n=1}^{\infty}\frac{3}{16}\cdot \left(\frac{4}{9} \right)^n\cdot a^2\sqrt{3}=\\ +&=&\frac{a^2\cdot \sqrt{3}}{4}+ \frac{3}{16} a^2\sqrt{3}\cdot\sum_{n=1}^{\infty} \left(\frac{4}{9}\right)^n=\\ +&=&\frac{a^2\cdot \sqrt{3}}{4}+ \frac{3}{16} a^2\sqrt{3}\cdot\left(\frac{4}{9}+\left(\frac{4}{9}\right)^2+\cdots + \left(\frac{4}{9}\right)^n+\cdots \right). +\end{eqnarray*} +If we calculate the sum of an infinite geometric sequence and simplify it, we get: +\begin{eqnarray} \label{eqnKoch5} +p=\frac{2a^2\sqrt{3}}{5}. +\end{eqnarray} + + +%________________________________________________________________________________ +\poglavje{Circumference and Area of a Circle} \label{odd8PloKrog} + +%OBSEG + +In section \ref{odd3NeenTrik} we have already defined the circumference of a polygon as the sum of all its sides. In this sense, the circumference represented a distance. In the following, we will understand the circumference also as the length of this distance, or the sum of the lengths of all the sides of the polygon. Here we will deal with the circumference of a circle. Although it is intuitively clear, this new concept needs to be defined first. + +Let $k(S,r)$ be an arbitrary circle. The corresponding circle is denoted by $\mathcal{K}(S,r)$. The circumference of this circle intuitively represents the length of the circle $k$. Let $A_1A_2\ldots A_n$ ($n\in \mathbb{N}$ and $n\geq 3$) be a regular polygon inscribed in the circle $k$. The tangents of this circle at the vertices of the polygon $A_1A_2\ldots A_n$ determine the sides of the regular polygon $B_1B_2\ldots B_n$, which is circumscribed to the circle $k$ (Figure \ref{sl.plo.8.5.1.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.1.pic} +\caption{} \label{sl.plo.8.5.1.pic} +\end{figure} + +We denote by $\underline{o}_n$ the circumference of the polygon $A_1A_2\ldots A_n$ and $\overline{o}_n$ the circumference of the polygon $B_1B_2\ldots B_n$. By the triangle inequality (statement \ref{neenaktrik}) we have $|A_1B_1|+|B_1A_2|>|A_1A_2|$, $|A_2B_2|+|B_2A_3|>|A_2A_3|$,..., $|A_nB_n|+|B_nA_1|>|A_nA_1|$ (Figure \ref{sl.plo.8.5.1a.pic}). If we add all these inequalities we get $\overline{o}_n>\underline{o}_n$. Similarly, $\underline{o}_n<\underline{o}_{n+1}$ and $\overline{o}_n>\overline{o}_{n+1}$. Therefore, for every $n\in \mathbb{N}$, $n\geq 3$ we have: +\begin{eqnarray*} +\underline{o}_1<\underline{o}_2<\cdots< +\underline{o}_n<\overline{o}_n<\cdots<\overline{o}_2<\overline{o}_1. +\end{eqnarray*} + This means that $\underline{o}_n$ is an increasing sequence bounded from above by $\overline{o}_1$, so it is a known statement of mathematical analysis that it is convergent and has its limit: +\begin{eqnarray*} +\underline{o}=\lim_{n\rightarrow\infty}\underline{o}_n. +\end{eqnarray*} + Similarly, $\overline{o}_n$ is a decreasing sequence bounded from below by $\underline{o}_1$, so it has its limit: +\begin{eqnarray*} +\overline{o}=\lim_{n\rightarrow\infty}\overline{o}_n. +\end{eqnarray*} + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.1a.pic} +\caption{} \label{sl.plo.8.5.1a.pic} +\end{figure} + +We will not formally prove the intuitively clear fact that for a sufficiently large natural number $n$, the difference $\overline{o}_n-\underline{o}_n$ is arbitrarily small, i.e. $\underline{o}=\overline{o}$. \index{obseg!kroga}\pojem{Circumference of a circle} $\mathcal{K}$ with the designation $o$ then represents $o=\underline{o}=\overline{o}$. + +We prove the following important statement. + + \bizrek + If $o$ is the circumference of a circle with radius $r$, then the ratio + $o:2r$ + is constant and thus independent of the choice of the circle. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.1b.pic} +\caption{} \label{sl.plo.8.5.1b.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.1b.pic}) + +Let $\mathcal{K}(S,r)$ and $\mathcal{K}'(S',r')$ be two arbitrary circles with circumferences $o$ and $o'$, and $k(S,r)$ and $k'(S',r')$ the corresponding circles. According to Theorem \ref{RaztKroznKrozn1}, there exists a central stretch that maps the circle $k$ to the circle $k'$. According to Theorem \ref{RaztTransPod}, this central stretch is a similarity transformation (denoted by $f$) with some coefficient $\lambda$. First, $r':r=\lambda$ or: +\begin{eqnarray} + r'=\lambda r. \label{eqnPloscKrogrr'oo'} +\end{eqnarray} +Let $A_1A_2\ldots A_n$ and $B_1B_2\ldots B_n$ ($n\in \mathbb{N}$ and $n\geq 3$) be inscribed and circumscribed regular polygons of the circle $k$ with circumferences $\underline{o}_n$ and $\overline{o}_n$. We denote: +$$f:S,A_1,A_2,\ldots,A_n,B_1,B_2,\ldots,B_n\mapsto S',A_1',A_2',\ldots,A_n',B_1',B_2',\ldots,B_n'.$$ +The polygons $A_1'A_2'\ldots A_n'$ and $B_1'B_2'\ldots B_n'$ ($n\in \mathbb{N}$ and $n\geq 3$) are inscribed and circumscribed regular polygons of the circle $k$ and for their circumferences $\underline{o'}_n$ and $\overline{o'}_n$ it holds that $\underline{o'}_n=\lambda\underline{o}_n$ and $\overline{o'}_n=\lambda\overline{o}_n$. From this and from the definition of the circumference of a circle it follows: +\begin{eqnarray}\label{eqnPloscKrogrr'oo'1} + o'=\underline{o'}=\lim_{n\rightarrow\infty}\underline{o'}_n= +\lim_{n\rightarrow\infty}\lambda\underline{o}_n= +\lambda\lim_{n\rightarrow\infty}\underline{o}_n=\lambda \underline{o}=\lambda o. +\end{eqnarray} +From \ref{eqnPloscKrogrr'oo'} and \ref{eqnPloscKrogrr'oo'1} it follows: +$$\frac{o'}{2r'}=\frac{\lambda o}{2\lambda r}=\frac{o}{2r},$$ which was to be proven. \kdokaz + +The constant from the previous theorem is called +\index{number!$\pi$}\pojem{number $\pi$} (\index{Archimedes' constant}\pojem{Archimedes' constant} or \index{number!Ludolf's}\pojem{Ludolf's number}) and we also denote it by $\pi$. + +From the previous theorem and from the definition of the number $\pi$ we get the following theorem. + +\bizrek \label{obsegKtoznice} +If $o$ is the circumference of a circle with radius $r$, then: +$$o=2r\pi.$$ +\eizrek + +First, we will give a rough estimate of the number $\pi$. + +\bizrek \label{stevPiOcena} +For the number $\pi$ it holds: +$$3<\pi<4.$$ +\eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.2.pic} +\caption{} \label{sl.plo.8.5.2.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.2.pic}) + +Let $\mathcal{K}$ be an arbitrary circle with radius $r$ and circumference $o$, and $k$ the corresponding circle. By \ref{obsegKtoznice}, $o=2r\pi$. +We choose the regular inscribed polygon $n=6$ and the regular circumscribed polygon $n=4$. From the definition of the circumference of a circle it follows: +$$\underline{o}_6r_1$) be two circles with the same center (the corresponding circles $k_1$ and $k_2$ are concentric). The set $\mathcal{K}_2\setminus \mathcal{K}_1$ is called +\index{krožni!kolobar}\pojem{circular sector}. + + + \bizrek \label{ploscKrozKolob} + If $p_{kl}$ is the area of the circular sector determined by the circles $k_1(S,r_1)$ + and $k_2(S,r_2)$ ($r_2>r_1$), then: + $$p_{kl}=\left( r_2^2-r_1^2 \right)\cdot\pi.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.4.pic} +\caption{} \label{sl.plo.8.5.4.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.4.pic}) + +We mark with $\mathcal{K}_1(S,r_1)$ and $\mathcal{K}_2(S,r_2)$ the corresponding circles of the circles $k_1(S,r_1)$ and $k_2(S,r_2)$. + +From $\mathcal{K}_2=\mathcal{K}_2\setminus \mathcal{K}_1\cup \mathcal{K}_1 $ and $\left( \mathcal{K}_2\setminus \mathcal{K}_1 \right) \cap \mathcal{K}_1=\emptyset$ according to the formulas \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscKrog} we get + $p_{\mathcal{K}_2}=p_{\mathcal{K}_2\setminus \mathcal{K}_1} + p_{\mathcal{K}_1}$ i.e.: +$$p_{kl}=p_{\mathcal{K}_2\setminus \mathcal{K}_1}= p_{\mathcal{K}_2}- p_{\mathcal{K}_1}=r_2^2\pi-r_1^2\pi +=\left( r_2^2-r_1^2 \right)\cdot\pi,$$ which had to be proven. \kdokaz + +In a similar way as for the length of the circular arc (formula \ref{obsegKrozLok}) we get the following statement. + +\bizrek \label{ploscKrozIzsek} +If $p_i$ is the area of the circular section with the corresponding central angle with a measure (in degrees) $\alpha$ and with a radius $r$, then: + $$p_i=\frac{r^2\pi\cdot\alpha}{360^0}.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.5.pic} +\caption{} \label{sl.plo.8.5.5.pic} +\end{figure} + + + + \bzgled + If in the formula for the area of the circle we replace the number $\pi$ with its approximate value $3$, we get the formula for the area of the inscribed regular dodecagon\footnote{This task was solved by the Chinese mathematician + \index{Liu, H.}\textit{H. Liu} (3rd century)}. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.6.pic} +\caption{} \label{sl.plo.8.5.6.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.6.pic}) + +Let $A_1A_2\ldots A_{12}$ be a regular dodecagon, which is inscribed in the circle $k(S,r)$. Then $A_1A_3\ldots A_{11}$ is a regular hexagon, which is inscribed in the same circle. We denote by $p_{12}$ the area of the mentioned dodecagon. By \ref{ploscGlavniIzrek} \textit{4)} we have $p_{12}=12\cdot p_{SA_1A_2}$. + +Since the triangle $A_1SA_3$ is a right triangle, the height $A_1P$ + of the triangle $SA_1A_2$ is equal to half of the side $A_1A_3$ +of the hexagon $A_1A_3\ldots A_{11}$, that is: + $$|A_1P|=\frac{1}{2}\cdot |A_1A_3|=\frac{r}{2}.$$ +Then, by \ref{PloscTrik} we have: + $$p_{12}=12\cdot p_{SA_1A_2}=12\cdot\frac{|SA_2|\cdot |A_1P|}{2}=12\cdot\frac{r^2}{4}=3\cdot r^2,$$ which had to be proven. \kdokaz + + + + \bzgled \label{HipokratoviLuni} + Above the catheti of a right triangle we construct circles + and from the thus enlarged triangle we cut out a circle above the hypotenuse + (Figure \ref{sl.plo.8.5.7.pic}). Show that the area of the remainder + is equal to the area of the original + right triangle\footnote{A special claim (the next example in this section), if the given triangle is isosceles, was proven by \index{Hipokrat}\textit{Hipokrat from Kios} (5th century BC), an ancient +Greek mathematician. The aforementioned figure is called \index{Hipokratovi luni}\pojem{Hipokratovi luni}. Hipokrat was the first to discover that some figures with a curved edge can be converted into a square with a straight edge and a hexagon.}. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.7.pic} +\caption{} \label{sl.plo.8.5.7.pic} +\end{figure} + +\textbf{\textit{Solution.}} +Let $ABC$ be the aforementioned triangle with hypotenuse $c$ and sides $a$ and $b$. +Let $p$ be the desired area, $p_{\triangle}$ the area of the triangle $ABC$, and $p_a$, $p_b$, and $p_c$ the areas of the corresponding semicircles. By the formulas \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscKrozIzsek} and \ref{PloscTrik} and the Pythagorean theorem \ref{PitagorovIzrek}, we have: + \begin{eqnarray*} + p&=& p_a+p_b-p_c+p_{\triangle}=\\ + &=& \left(\frac{a}{2} \right)^2\cdot \pi ++\left(\frac{a}{2} \right)^2\cdot \pi-\left(\frac{a}{2} \right)^2\cdot \pi+p_{\triangle}=\\ + &=& \frac{\pi}{4}\cdot \left(a^2+b^2-c^2 \right)+p_{\triangle}=\\ + &=& \frac{\pi}{4}\cdot 0+p_{\triangle}=\\ + &=& p_{\triangle}, + \end{eqnarray*} + which had to be proven. \kdokaz + + \bzgled \label{HipokratoviLuni2} + The day is the square $PQRS$. + Draw a square that has the same area as the figure that results when + the circle drawn around the square $PQRS$ + is cut by + a circle with + radius $PQ$ + (Figure \ref{sl.plo.8.5.8.pic}). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.8.pic} +\caption{} \label{sl.plo.8.5.8.pic} +\end{figure} + +\textbf{\textit{Solution.}} +If we choose the isosceles right triangle $QSQ'$ (where $Q'=\mathcal{S}_p(Q)$), by the previous formula (\ref{HipokratoviLuni}) the area of the desired figure is equal to the area of the isosceles right triangle $SPQ$ (Figure \ref{sl.plo.8.5.8.pic}), and its area + is equal to the area of the square $PLSN$, where $L$ is the center of the line $SQ$ and $N=\mathcal{S}_{SP}(L)$ + (Figure \ref{sl.plo.8.5.8a.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.8a.pic} +\caption{} \label{sl.plo.8.5.8a.pic} +\end{figure} +\kdokaz + + + \bzgled + Let $ABCD$ be a square which we increase by four semicircles above the sides, and from the resulting figure we cut out the circle circumscribed to the square (Figure \ref{sl.plo.8.5.9.pic}). Prove that + the area of the remainder is equal to the area of the original square. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.9.pic} +\caption{} \label{sl.plo.8.5.9.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $S$ be the center of the square $ABCD$. +The statement is a direct consequence of the statement of the previous theorem (\ref{HipokratoviLuni2}) - we use the appropriate relation for the triangles $ASB$, $BSC$, $CSD$ and $DSA$ and add everything up. +\kdokaz + + + \bzgled \label{arbelos} + Let $C$ be an arbitrary point on the diameter $AB$ of the semicircle $k$ and let $D$ be the point of intersection of this + semicircle with the rectangle with diameter $AB$ at point $C$. Let $j$ and $l$ + be semicircles with diameters $AC$ and $BC$. Prove that the area of the figure bounded by semicircles $k$, $l$ and $j$ is equal to the area of the circle with diameter + $CD$\footnote{This task was published in his 'Book of Lemmas' by the Greek mathematician \index{Arhimed}\textit{Archimedes of Syracuse} (3rd century BC). +He named the aforementioned figure \index{arbelos}\textit{arbelos} (which in Greek means 'shoemaker's knife').}. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.10.pic} +\caption{} \label{sl.plo.8.5.10.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.10.pic}) + +Let's mark $a=|AC|$ and $b=|BC|$ the appropriate diameters, $p$ the area of the aforementioned figure and $p_0$ the area of the circle with diameter $|CD|$. According to \ref{izrekVisinski} +it is $|CD|^2 = ab$. If we use \ref{ploscKrozIzsek} and \ref{ploscKrog} as well, +we get: +\begin{eqnarray*} +p&=&\frac{1}{2}\left(\frac{a+b}{2}\right)^2\cdot\pi- +\frac{1}{2}\left(\frac{a}{2}\right)^2\cdot\pi- +\frac{1}{2}\left(\frac{b}{2}\right)^2\cdot\pi=\\ +&=& \frac{ab}{4}\cdot\pi=\\ +&=& \left(\frac{|CD|}{2}\right)^2\cdot\pi=p_0, +\end{eqnarray*} +which was to be proven. \kdokaz + + + \bzgled + Let $k$ be a drawn circle of a right triangle and $\mathcal{K}$ the corresponding circle. Then draw three + smaller circles, which touch the circle $k$ and two sides of + the triangle, then three smaller circles, which touch the previous three + circles and two sides of the triangle, and so on until infinity. Is the area of all the corresponding circles, except for the circle $\mathcal{K}$, + smaller + than half the area of the circle $\mathcal{K}$? + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.11.pic} +\caption{} \label{sl.plo.8.5.11.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.plo.8.5.11.pic}) + +Let $ABC$ be a given triangle and $k(S,r)$ the inscribed circle of this triangle. With $k_i(S_i ,r_i)$ ($i\in\mathbb{N}$) we denote the circles from the given sequence of circles at the vertex $A$ and with $p_i$ ($i \in \mathbb{N}$) the areas of the corresponding circles. We also denote the circle $k$ with $k_0(S_0 ,r_0)$ and the area of the corresponding circle $\mathcal{K}$ with $p=p_0$. Let $k_n(S_n ,r_n)$ and $k_{n+1}(S_{n+1} ,r_{n+1})$ be the consecutive circles from the mentioned sequence. We denote their tangent points with the side $AB$ of the triangle $ABC$ with $T_n$ and $T_{n+1}$, and the orthogonal projection of the center $S_{n+1}$ onto the line $S_nT_n$ with $L_n$. In the right triangle $S_{n+1}L_nS_n$ the length of the hypotenuse is $|S_nL_n| = r_n- r_{n+1}$ and the hypotenuse $|S_{n+1}S_n| = r_n+ r_{n+1}$. As $L_nS_{n+1}S_n$ is equal to half of the internal angle at the vertex $A$, namely $30°$. From this it follows that $\angle S_{n+1}S_nL_n=60^0$. Let $S_n'=\mathcal{S}_{S_{n+1}L_n}(S_n)$. The triangles $S_{n+1}L_nS_n$ and $S_{n+1}L_nS_n'$ are congruent (by the \textit{SAS} \ref{SKS}), so also $\angle S_{n+1}S_n'L_n=60^0$. Therefore, $S_{n+1}S_n'S_n$ is an isosceles triangle, which means that $|S_{n+1}S_n| =|S_n'S_n| =2\cdot|S_nL_n|$ or +$$r_n+ r_{n+1}=2\cdot(r_n-r_{n+1}).$$ +If we express $r_{n+1}$ from the last equality, we get: + \begin{eqnarray*} + r_{n+1}=\frac{1}{3}\cdot r_n. +\end{eqnarray*} +From this, by \ref{ploscKrog}, it follows: + \begin{eqnarray*} + p_{n+1}=\frac{1}{9}\cdot p_n. +\end{eqnarray*} +The sequence $p_n$ is therefore a geometric sequence with the coefficient $q=\frac{1}{9}$ and the initial value $p_0=p$, so: + \begin{eqnarray*} + p_n=\left( \frac{1}{9} \right)^n \cdot p. +\end{eqnarray*} +This means that the total area $p_A$ of the circles from the sequence at the vertex $A$ represents the sum of an infinite geometric sequence $p_n$ ($n\in \mathbb{N}$). Therefore: +\begin{eqnarray*} + p_A&=&p_1+p_2+p_3+\cdots=\\ +&=&\frac{1}{9}\cdot p+\left(\frac{1}{9}\right)^2\cdot p+ +\left(\frac{1}{9}\right)^3\cdot p+\cdots=\\ +&=&\frac{1}{9}p\cdot \frac{1}{1-\frac{1}{9}}=\\ +&=& \frac{1}{8}\cdot p. +\end{eqnarray*} +The sum of the areas of all circles at the vertices $A$, $B$ and $C$ is then equal to $\frac{3}{8}\cdot p$ and is not greater than half of $p$. +\kdokaz + +\bnaloga\footnote{6. IMO, USSR - 1964, Problem 3.} + A circle is inscribed in triangle $ABC$ with sides $a$, $b$, $c$. Tangents to the circle + parallel to the sides of the triangle are constructed. Each of these tangents + cuts off a triangle from triangle $ABC$. In each of these triangles, a circle is inscribed. + Find the sum of the areas of all four inscribed circles (in terms of $a$, $b$, $c$). + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.IMO1.pic} +\caption{} \label{sl.plo.8.5.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $k(S,\rho)$ be the inscribed circle +of triangle $ABC$ with altitudes $v_a$, $v_b$ and $v_c$ and $p$ the +area of the corresponding circle. We denote by $AB_aC_a$, $A_bBC_b$ +and $A_cB_cC$ the new triangles, $k_a$, $k_b$ and $k_c$ the inscribed +circles of these triangles with radii $\rho_a$, $\rho_b$ and $\rho_c$ +and $p_a$, $p_b$ and $p_c$ the areas of the corresponding circles. +(Figure \ref{sl.plo.8.5.IMO1.pic}). + +The triangles $ABC$ and $AB_aC_a$ are similar with the coefficient +$\frac{v_a-2\rho}{v_a}$. Therefore: + $$\rho_a=\frac{v_a-2\rho}{v_a}\cdot \rho.$$ + Because according to the statement \ref{PloscTrikVcrt} for the area of the triangle $ABC$ + it holds $p_\triangle=s\cdot \varrho$ + (where $s=\frac{a+b+c}{2}$ is the semi-perimeter of the triangle $ABC$), it follows: +$$\rho_a^2=\left(1-\frac{2\rho}{v_a}\right)^2\cdot \rho^2= + \left(1-\frac{2\frac{P_\triangle}{s}}{2\frac{P_\triangle}{a}}\right)^2\cdot \rho^2= + \frac{(s-a)^2}{s^2}\cdot \rho^2.$$ +Similarly we obtain $\rho_b^2= + \frac{(s-b)^2}{s^2}\cdot \rho^2$ and $\rho_c^2= + \frac{(s-c)^2}{s^2}\cdot \rho^2$. From the statement \ref{PloscTrikOcrtVcrt} + it follows + $\rho^2=\frac{(s-a)(s-b)(s-c)}{s}$, therefore: +\begin{eqnarray*} + && p+p_a+p_b+p_c= \pi (\rho^2+\rho_a^2+\rho_b^2+\rho_c^2)=\\ + &&=\pi + \rho^2(1+\frac{(s-a)^2}{s^2}+\frac{(s-b)^2}{s^2}+\frac{(s-c)^2}{s^2})=\\ + &&= + \frac{(s-a)(s-b)(s-c)(s^2+(s-a)^2+(s-b)^2+(s-c)^2)}{s^3}\cdot \pi, +\end{eqnarray*} + which had to be expressed. \kdokaz + + +%________________________________________________________________________________ +\poglavje{Pythagoras' Theorem and Area} \label{odd8PloPitagora} + + In the section \ref{odd7Pitagora} we have proven Pythagoras' statement \ref{PitagorasIzrek} and predicted that we will express it in the form that relates to the areas. + + \bizrek \index{statement!Pythagoras for areas} \label{PitagorasIzrekPlosc}(Pythagoras' statement in the form of areas)\\ + The area of the square over the hypotenuse of a right-angled triangle is equal to the sum of the areas of the squares over the two cathets of this triangle. + + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.6.1.pic} +\caption{} \label{sl.plo.8.6.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} At this point we will provide two proofs and one idea of the proof of this statement. + +\textit{1)} The statement is a direct consequence of the Pythagorean Theorem \ref{PitagorovIzrek} and the formula for the area of a square \ref{ploscKvadr} (Figure +\ref{sl.plo.8.6.1.pic}). + + +\textit{2)\footnote{This proof was published by \index{Euclid}\textit{Euclid of Alexandria} (3rd century BC) in his famous work 'Elements', which consists of 13 books.}} Let $AMNB$, $BPQC$ and $CKLA$ be squares constructed on the hypotenuse and +the legs of the right triangle $ABC$. Let $D$ and $E$ be the projections of point $C$ on the lines $AB$ and $MN$ (Figure +\ref{sl.plo.8.6.1.pic}). + +Since the area of the square $AMNB$ is equal to the sum of the areas of the rectangles $NBDE$ and $EDAM$, it is enough to prove that the areas of the squares $BPQC$ and +$CKLA$ are equal to the areas of the rectangles $NBDE$ and $EDAM$. We will prove only the first equality $p_{BPQC}=p_{NBDE}$ (the +second equality $p_{CKLA}=p_{EDAM}$ is then analogous). + +For this relation $p_{BPQC}=p_{NBDE}$ it is enough to prove the equality of the areas of the triangles $BPQ$ and +$NBD$, since $p_{BPQC}=2\cdot p_{BPQ}$ and $p_{NBDE}=2\cdot p_{NBD}$. + Then: + \begin{eqnarray*} + p_{BPQ}&=&p_{BPA}=\hspace*{3mm} \textrm{(the triangles have the same base and height)}\\ + &=&p_{NBC}=\hspace*{3mm} \textrm{(the triangles are congruent due to } \mathcal{R}_{B,90^0+\angle CBA}\textrm{)}\\ + &=&p_{NBD}\hspace*{4mm} \textrm{(the triangles have the same base and height)} + \end{eqnarray*} +Therefore, the triangles $BPQ$ and $NBD$ have the same area, which was enough to prove. + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.6.2.pic} +\caption{} \label{sl.plo.8.6.2.pic} +\end{figure} + +\textit{3)\footnote{This proof (based on the illustration) is attributed to the Ancient Indians. We can say that it only needs the comment: "Look!" - so it is simple and elegant.}} The idea of the proof is given by the +picture \ref{sl.plo.8.6.2.pic}. We will leave the formal proof to the reader. +\kdokaz + +\bzgled + The area of a circular sector, determined by the inscribed and circumscribed circle of a regular $n$-gon with a side length $a$, is not dependent on the number $n$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.6.3.pic} +\caption{} \label{sl.plo.8.6.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $AB$ be one side, $S$ the center of this side and $O$ the center of the inscribed or circumscribed circle of any regular $n$-gon $\mathcal{V}_n(a)$ (Figure \ref{sl.plo.8.6.3.pic}). According to the assumption, $|AB|=a$, therefore $|AS|=\frac{a}{2}$. We mark with $R$ and $r$ the radius of the inscribed and circumscribed circle of the $n$-gon $\mathcal{V}_n(a)$. From the congruence of triangles $OSA$ and $OSB$ (\textit{SSS} \ref{SSS}) it follows that $\angle OSA\cong \angle OSB=90^0$. Therefore $OSA$ is a right triangle with hypotenuse $OA$, so by the Pythagorean Theorem \ref{PitagorovIzrek} $|OA|^2-|OS|^2=|AS|^2$ or $R^2-r^2=\left(\frac{a}{2}\right)^2$. From this relation and the formula for the area of a circular sector \ref{ploscKrozKolob} we get: + $$p_k=\left(R^2-r^2\right)\cdot \pi=\left(\frac{a}{2}\right)^2\cdot \pi=\frac{a^2\pi}{4},$$ +which means that the area of the sector is only dependent on the length of the side of the regular $n$-gon $\mathcal{V}_n(a)$. +\kdokaz + + \bzgled + Using the statement from Example \ref{PitagoraCofman}, prove the inequality: + \begin{eqnarray*} + \frac{1}{1^2\cdot 2^2}+\frac{1}{2^2\cdot 3^2}+\cdots +\frac{1}{n^2 \left(n+1\right)^2}+\cdots<\frac{2}{3} + \end{eqnarray*} + or + \begin{eqnarray*} + \sum_{n=1}^{\infty}\frac{1}{n^2 \left(n+1\right)^2}<\frac{2}{3}. + \end{eqnarray*} + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.6.4.pic} +\caption{} \label{sl.plo.8.6.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.6.4.pic}) + +We use the notation from the example \ref{PitagoraCofman}. According to the relation \ref{eqnPitagoraCofman3} from this example, $d_n=\frac{1}{n(n+1)}$, therefore $r_n=\frac{1}{2n(n+1)}$. If we denote the area of the corresponding circle of the circle $c_n$ with $p_n$, it is (formula \ref{ploscKrog}): +\begin{eqnarray*} +p_n=\pi r_n^2=\frac{\pi}{4}\frac{1}{n^2\left(n+1\right)^2}. +\end{eqnarray*} +Therefore, the sum of the areas of all corresponding circles is equal to: +\begin{eqnarray} \label{eqnPitagoraCofmanPlo2} +\sum_{n=1}^{\infty} p_n=\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{1}{n^2\left(n+1\right)^2}. +\end{eqnarray} +But this sum is smaller than the area $p$ of the figure bounded by the line +$A'B'$ and the circular arc $A'P$ and $B'P$ of the circles $a$ and $b$ with the central angle $90^0$. The area $p$ is equal to the difference of the area of the rectangle $A'ABB'$ and the area of two corresponding circular sectors with the central angle $90^0$. Therefore (formula \ref{ploscPravok} and \ref{ploscKrozIzsek}): + \begin{eqnarray} \label{eqnPitagoraCofmanPlo3} +p=2-2\cdot\frac{\pi}{4}=2-\frac{\pi}{2}. +\end{eqnarray} +Because $\sum_{n=1}^{\infty} p_n3$ (formula \ref{stevPiOcena}), we get: +\begin{eqnarray*} +\sum_{n=1}^{\infty}\frac{1}{n^2\left(n+1\right)^2}<\frac{8}{\pi}-2<\frac{8}{3}-2=\frac{2}{3}, +\end{eqnarray*} +which was to be proven. \kdokaz + + + +% Ttrikotniki + +\item Let $T$ be the centroid of the triangle $ABC$. Prove: + $$p_{TBC}=p_{ATC}=p_{ABT}.$$ + +\item Let $c$ be the hypotenuse, $v_c$ be the corresponding altitude, and $a$ and $b$ be the catheti of a right triangle. Prove that $c+v_c>a+b$. +% zvezek - dodatni MG + + + \item Draw a square that has the same area as a rectangular triangle with catheti that are consistent with the given distances $a$ and $b$. % (Hipokratovi luni) + +\item Let $ABC$ be an isosceles right triangle with the length of the catheti $|CB|=|CA|=a$. We mark with $A_1$, $B_1$, $C_1$ and $C_2$ the points for which it holds: $\overrightarrow{CA_1}=\frac{n-1}{n}\cdot \overrightarrow{CA}$, $\overrightarrow{CB_1}=\frac{n-1}{n}\cdot \overrightarrow{CB}$, $\overrightarrow{AC_1}=\frac{n-1}{n}\cdot \overrightarrow{AB}$ and $\overrightarrow{AC_2}=\frac{n-2}{n}\cdot \overrightarrow{AB}$ for some $n\in \mathbb{N}$. Express the area of the quadrilateral determined by the lines $AB$, $A_1B_1$, $CC_1$ and $CC_2$, as a function of $a$ and $n$. + +\item Given is a right triangle $ABC$ with hypotenuse $AB$ and area $p$. Let $C'=\mathcal{S}_{AB}(C)$, $B'=\mathcal{S}_{AC}(B)$ and $A'=\mathcal{S}_{BC}(A)$. Express the area of the triangle $A'B'C'$ as a function of $p$. + +\item Let $R$ and $Q$ be points in which the inscribed circle of the triangle $ABC$ intersects its sides $AB$ and $AC$. Let the internal angle bisector of $ABC$ intersect the line $QR$ in the point $L$. Determine the ratio of the areas of the triangles $ABC$ and $ABL$. + % pripremni zadaci - naloga 193 + + +\item Determine a point inside the triangle $ABC$, for which the product of its distances from the sides of this triangle is maximal. % zvezek - dodatni MG + +\item The triangle + $ABC$ with sides of lengths $a$, $b$ and $c$ has an inscribed circle. We draw + the tangents of this circle that are parallel + to the sides + of the triangle. + Each of the tangents inside the triangle determines the appropriate distances of lengths $a_1$, $b_1$ and $c_1$. Prove that: + $$\frac{a_1}{a}+\frac{b_1}{b}+\frac{b_1}{b}=1.$$ % zvezek - dodatni MG + +\item Let $T$ and $S$ be the centroid and the center of the inscribed circle of the triangle $ABC$. Let also $|AB|+|AC|=2\cdot |BC|$. Prove that $ST\parallel BC$. % zvezek - dodatni MG + +\item Let $p$ be the area of the triangle $ABC$, $R$ the radius of the circumscribed circle and $s'$ the perimeter of the pedal triangle. Prove that $p=Rs'$. %Lopandic - nal 918 + +% Sstirikotniki + + +\item Let $L$ be an arbitrary point inside the parallelogram $ABCD$. Prove that: + $$p_{LAB}+p_{LCD}=p_{LBC}+p_{LAD}.$$ %Lopandic - nal 890 + +\item Let $P$ be the center of the leg $BC$ of the trapezoid $ABCD$. Prove: + $$p_{APD}=\frac{1}{2}\cdot p_{ABCD}.$$ + +\item Let $o$ be the perimeter, $v$ the height and $p$ the area of the tangent trapezoid. Prove that: $p=\frac{o\cdot v}{4}$. + +\item Draw a line $p$, which goes through the vertex $D$ of the trapezoid $ABCD$ ($AB>CD$), so that it divides this trapezoid into two area-equal figures. + + +\item Draw lines $p$ and $q$, which go through the vertex $D$ of the square $ABCD$ and divide it into area-equal figures. + +\item Let $ABCD$ be a square, $E$ the center of its side $BC$ and $F$ a point, for which $\overrightarrow{AF}=\frac{1}{3}\cdot \overrightarrow{AB}$. The point $G$ is the fourth vertex of the rectangle $FBEG$. What part of the area of the square $ABCD$ does the area of the triangle $BDG$ represent? + +\item Let $\mathcal{V}$ and $\mathcal{V}'$ be similar polygons with the similarity coefficient $k$. Then: + $$p_{\mathcal{V}'}=k^2\cdot p_{\mathcal{V}}.$$ Prove. + +\item Let $a$, $b$, $c$ and $d$ be the lengths of the sides, $s$ the semiperimeter and $p$ the area of an arbitrary quadrilateral. Prove that: + $$p=\sqrt{(s-a)(s-b)(s-c)(s-d)}.$$ %Lopandic - nal 924 + +\item Let $a$, $b$, $c$ and $d$ be the lengths of the sides and $p$ the area of the string-tangent quadrilateral. Prove that: + $$p=\sqrt{abcd}.$$ %Lopandic - nal 925 + +% Veckotniki + +\item There is a rectangle $ABCD$ with sides of lengths $a=|AB|$ and $b=|BC|$. Calculate the area of the figure that represents the union of the rectangle $ABCD$ and its image under reflection across the line $AC$. + +\item Let $ABCDEF$ be a regular hexagon, and let $P$ and $Q$ be the centers of its sides $BC$ and $FA$. What part of the area of this hexagon is the area of the triangle $PQD$? + + +% KKrog + +\item In a square, four congruent circles are drawn so that each circle touches two sides and two circles. Prove that the sum of the areas of these circles is equal to the area of the square inscribed circle. + +\item Calculate the area of the circle that is inscribed in the triangle with sides of lengths 9, 12 and 15. %resitev 54 + +\item Let $P$ be the center of the base $AB$ of the trapezoid $ABCD$, for which $|BC|=|CD|=|AD|=\frac{1}{2}\cdot |AB|=a$. Express the area of the figure determined by the base $CD$ and the shorter circular arcs $PD$ and $PC$ of the circles with centers $A$ and $B$, as a function of the base $a$. + +\item The cord $PQ$ ($|PQ|=d$) of the circle $k$ touches its conchoid $k'$. Express the area of the lune determined by the circles $k$ and $k'$, as a function of the cord $d$. + +\item Let $r$ be the radius of the inscribed circle of the polygon $\mathcal{V}$, which is divided into triangles $\triangle_1,\triangle_2,\ldots,\triangle_n$, so that no two triangles have common interior points. Let $r_1,r_2,\ldots , r_n$ be the radii of the inscribed circles of these triangles. Prove that: + $$\sum_{i=1}^n r_i\geq r.$$ + +\end{enumerate} + + +% DEL 3 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% INVERZIJA +%________________________________________________________________________________ + + + \del{Inversion} \label{pogINV} +\poglavje{Definition and Basic Properties of Inversion} +\label{odd9DefInv} + +In this chapter we will define inversion - a new mapping, which in a certain way represents the "reflection" over a circle. +We expect this new mapping to have similar properties to the reflection over a line. It is clear that inversion cannot be an isometry, but despite this, we will list the following desired common properties of the "two reflections" (Figure \ref{sl.inv.9.1.1.pic}): + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.1.1.pic} +\caption{} \label{sl.inv.9.1.1.pic} +\end{figure} + +\begin{itemize} + \item the reflection is a bijective mapping, + \item the reflection is an involution, + \item the reflection maps one of the areas of the plane, which the (circle) axis of the reflection determines, into the other area, + \item all +fixed points of the reflection lie on the (circle) axis of this +reflection, + \item if $X'$ is the image of the point $X$ under the reflection, then the line + $XX'$ +is perpendicular to the (circle) axis of this reflection. +\end{itemize} + + The mapping, which we will now +define, should therefore satisfy all these properties. Instead of the name + "reflection over a circle" we will rather use + the term "inversion"\index{inversion}\footnote{Inversion was first introduced by the German mathematician +\index{Magnus, L. I.} \textit{L. I. Magnus} (1790--1861) in 1831. +The first ideas of inversion appeared already in the ancient Greek mathematician +\index{Apolonij} \textit{Apolonij iz Perge} (3.--2. st. pr. n. +š.) and the Swiss mathematician \index{Steiner, J.} \textit{J. +Steiner} (1796--1863)}. + + + In the Euclidean plane we will define inversion in the following way. +Let $k(O,r)$ be a circle in the Euclidean plane $\mathbb{E}^2$. The point $X'$ is the image of some point $X\in \mathbb{E}^2\setminus +\{O\}$ in the \index{inversion} \pojem{inversion} $\psi_k$ with respect to +the circle $k$ (its \index{inverse image} \pojem{inverse image}), if it belongs to the segment $OX$ and $|OX|\cdot +|OX'|=r^2$, $k$ is the \index{circle!inversion}\pojem{circle +of inversion}, $O$ is the \index{center!inversion} \pojem{center +of inversion}. + +We will often denote the inversion $\psi_k$ with respect to the circle $k(O,r)$ by $\psi_{O,r}$. + +From the definition itself it immediately follows that the only fixed points of the inversion are precisely the points on the inversion circle. And all points of the inversion circle are fixed. Therefore, the equivalence holds: +$$\psi_k(X)=X\hspace*{1mm}\Leftrightarrow\hspace*{1mm}X\in k.$$ + +The inversion $\psi_k$ is a bijective mapping of the set $\mathbb{E}^2\setminus \{O\}$. It also holds that $\psi_k^{-1}=\psi_k$, therefore $\psi_k^2$ is the identical mapping. Both properties follow directly from the definition. It also directly follows that the external points of the circle $k$ with the inversion $\psi_k$ are mapped to internal points and vice versa (without the point O). + +From the definition it follows that the line $XX'$ is perpendicular to the inversion circle, which means that the last desired property is also fulfilled. + +Now we will construct the image $X'=\psi_k(X)$ of an arbitrary point $X$ with the inversion. From everything said above (especially from the fact that $\psi_k^{-1}=\psi_k$) it follows that it is enough to describe the construction when $X$ is outside the circle $k$. Let in this case $OX$ be the tangent of the circle $k$ in its point $T$, then from $|OX|\cdot |OX'|=r^2=|OT|^2$ it follows that the triangles $OTX$ and $OX'T$ are similar and therefore $\angle TX'O=90^0$ (Figure \ref{sl.inv.9.1.2.pic}). Since the point $X'$ is also on the segment $OX$, we get this point as the perpendicular projection of the point $T$ on the segment $OX$. + +From the relation in the definition of inversion $|OX|\cdot |OX'|=r^2$ it follows that if the point $X$ approaches the center $O$, its image $X'$ "moves away to infinity". It also holds - if the point $X$ is close to the circle $k$, its image is also close to this circle, of course on the other side. We can write all these findings in a more formal way. + +\bizrek \label{invUrejenost} If $\psi_k:A,B \mapsto A',B'$ and +$\mathcal{B}(O,A,B)$, + then $\mathcal{B}(O,B',A')$. + \eizrek + + The point $O$ has no image. Intuitively, its image is a point at infinity. We will say more about this in chapter \ref{odd9InvRavn}. + +We prove the following important property of inversion. + +\bizrek \label{invPodTrik} Let $O$, $A$ and $B$ be three +non-collinear points and $\psi_k$ be the inversion with respect to the circle +$k(O,r)$. If $A'$ and $B'$ are the images of points $A$ and $B$ in this +inversion, then the triangles $OAB$ and $OB'A'$ are similar, that is: +$$\psi_{O,r}:A,B\mapsto A',B' \Rightarrow \triangle OAB \sim \triangle +OB'A'.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.1.3.pic} +\caption{} \label{sl.inv.9.1.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} Because $\psi_{O,r}:A,B\mapsto A',B'$, + it is also +$OA\cdot OA'=OB\cdot OB'=r^2$ (Figure \ref{sl.inv.9.1.3.pic}). +The similarity of the triangles now follows from $OA:OB'=OB:OA'$ and $\angle AOB +\cong \angle B'OA'$. + \kdokaz + + The next statement is a direct consequence of the previous proposition. + +\bizrek Let $O$, $A$ and $B$ be three non-collinear points and +$\psi_k$ be the inversion with respect to the circle $k(O,r)$. If $A'$ and +$B'$ are the images of points $A$ and $B$ in this inversion, then: + + (i) $\angle +OAB \cong \angle OB'A'$ and $\angle OBA \cong \angle OA'B'$, + +(ii) points $A$, $B$, $B'$ and $A'$ are concircular, + +(iii) the circle passing through points $A$, $B$, $B'$ and $A'$ is +orthogonal to the inversion circle. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.1.4.pic} +\caption{} \label{sl.inv.9.1.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} + \textit{(i)} The angle congruence follows from the similarity of triangles $OAB$ and + $OB'A'$(the previous proposition \ref{invPodTrik}). + +\textit{(ii)} Let $l$ be a circle drawn through the triangle +$ABA'$ (Figure \ref{sl.inv.9.1.4.pic}). From the definition of +inversion it follows that $OA\cdot OA'=OB\cdot +OB'=r^2$. This means that the power of the point $O$ on the circle $l$ +is equal to $OA\cdot OA'=OB\cdot +OB'$, thus the point $B'$ lies on this circle. + +We mention that the tautness of the quadrilateral $AA'B'B$ also follows directly +from $\angle +OAB \cong \angle OB'A'$, because then $\angle A'AB + \angle BB'A' += 180^0 $. + +\textit{(iii)} From $\psi_{O,r}(A)=A'$ it follows that one of the points +$A$ and $A'$ is in the interior, the other in the exterior of the inversion circle $k$. By the consequence of Dedekind's axiom +(\ref{DedPoslKrozKroz}) the circles $k$ and $l$ have two common +points; one of them we denote by $T$. Because $OA\cdot OA'=OB\cdot +OB'=r^2=OT^2$, it follows that $OT$ is a tangent to the circle $k$, +which means that the circles $k$ and $l$ are perpendicular (by the statement +\ref{pravokotniKroznici}). +\kdokaz + +In the same way as in part (\textit{iii}) of the previous statement, +we also prove the following claim. + +\bizrek \label{invPravKrozn} Let +$X'$ be the image of the point $X$ ($X\neq X'$) under the inversion $\psi_k$ +with respect to the circle $k(O,r)$. Then every circle that goes +through the points $X$ and $X'$ (also the line $XX'$) is perpendicular to +the inversion circle $k$. + \eizrek + + In the next example we will see how we can translate an inversion with the help of + certain stretches into an inversion with a concentric + inversion circle. + +\bzgled \label{invRazteg} The composition of the inversion $\psi_{S,r}$ and the dilation $h_{S,k}$ with the same center $S$ and positive coefficient ($k>0$) is an inversion with center $S$ and coefficient $r\sqrt{k}$. + \ezgled +\textbf{\textit{Proof.}} Let $f=h_{S,k}\circ \psi_{S,r}$. For any point $X$, denote $X'=f(X)$. First, it is clear that the point $X'$ lies on the line segment $SX$. Also, denote $X_1=\psi_{S,k}(X)$. From the definition of inversion and dilation, it follows that $|SX_1| \cdot |SX| =r^2$ and $|SX'| = k\cdot |SX_1| $. Therefore, $|SX'| \cdot |SX| =k\cdot r^2=\left(r\sqrt{k}\right)^2$ or $\psi_{S,r\sqrt{k}}(X)=X' =f(X)$. +Since this is true for any point $X$, we have $f=h_{S,k}\circ \psi_{S,r}=\psi_{S,r\sqrt{k}}$. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Image of a Circle or Line Under an Inversion} \label{odd9SlokaKrozPrem} + +In this section, we will determine that an inversion is not a collinearity, which means that it does not preserve the relation of collinearity of points. We will consider the images of lines and circles under inversion. Since the domain of an inversion is an Euclidean plane without the center of inversion, it makes sense to introduce the following notation: if $\Phi$ is an arbitrary figure of the Euclidean plane and $S$ is a point of this plane, then + $$\Phi^S= \Phi \setminus + \{S\}.$$ + +\bizrek \label{InverzKroznVkrozn} + Let $\psi_i$ be an inversion with respect to the circle $i(S,r)$ of the Euclidean plane. + If $p$ is a line and $k$ +is a circle of this plane, then (Figure \ref{sl.inv.9.2.1.pic}): + +(i) if $S\in p$, then $\psi_i(p^S)=p^S$, + +(ii) if $S\notin p$, then $\psi_i(p)=j^S$, where $j$ is a circle that +passes through the point $S$, + +(iii) if $S\in k$, then $\psi_i(k^S)=q$, where $q$ is a line that +does not pass through the point $S$; + +(iv) if $S\notin k$, then $\psi_i(k)=k'$, where $k'$ is a circle that +does not pass through the point $S$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.2.1.pic} +\caption{} \label{sl.inv.9.2.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} + +(\textit{i}) From the definition of inversion it follows that the image of any + point + $X \in p^S$ lies on the open ray $SX$, therefore $\psi_i(X)\in + p^S$. Because $\psi_i^{-1}=\psi_i$, any point $Y\in + p^S$ is the image of the point $\psi_i^{-1}(Y)=\psi_i(Y)\in + p^S$. Therefore $\psi_i(p^S)=p^S$. + + (\textit{ii}) Let $P$ be the orthogonal projection of the point + $S$ onto the line $p$ (Figure \ref{sl.inv.9.2.2.pic}). + Because $S\notin p$, then $P\neq S$, therefore +there exists an image of the point $P$ under the inversion $\psi_i$ -- we denote it with $P'$. +Let $X$ be any point on the line $p$ ($X\neq P$) and $X'=\psi_i(X)$. +By \ref{invPodTrik} the triangles $SPX$ and $SX'P'$ are similar, +therefore $\angle SX'P' \cong \angle SPX=90^0$. Therefore the point $X'$ lies +on the circle above the diameter $SP'$ - we denote it with $j$. Because $X'\neq +S$, it follows that $X'\in j^S$ or $\psi_i(p)\subseteq j^S$. With the reverse +procedure we prove that each point of the set $j^S$ is the image of some point +on the line $p$, therefore + $\psi_i(p)= j^S$. + + + (\textit{iii}) A direct consequence of (\textit{ii}), because $\psi_i$ is an involution, + or $\psi_i^{-1}=\psi_i$. + +(\textit{iv}) (Figure \ref{sl.inv.9.2.2.pic}) Let $P$ and $Q$ be the + intersection points of the circle $k$ with the line that goes through the point $S$, + and the center + of the circle + $k$ (for example, when the circles $i$ and $k$ are concentric, it is easy). + Without loss of generality, assume that $\mathcal{B}(S,P,Q)$ is true. + Let $P'=\psi_i(P)$, +$Q'=\psi_i(Q)$ and $X$ be any point on the circle $k$, which is +different from the points $P$ and $Q$ and $X'=\psi_i(X)$. By +the statement +\ref{invPodTrik} we have $\triangle SPX \sim \triangle SX'P'$ and $\triangle SQX +\sim \triangle SX'Q'$. Therefore, + $\angle SX'P' \cong \angle SPX$ and $\angle SX'Q'\cong \angle SQX$, which means: +$$\angle P'X'Q'=\angle SX'P'-\angle SX'Q' +=\angle SPX-\angle SQX=\angle PXQ=90^0.$$ + Then the point $X'$ lies on +the circle above the diameter $P'Q'$ - we denote it with $k'$. By reversing +the process, we can also prove that every point on the circle $k'$ +is the image of a point on the circle $k$, so $\psi_i(k)=k'$ + +We prove the statement similarly in the case when +$\mathcal{B}(P,S,Q)$ is true. + \kdokaz + +The previous statement also provides an effective way of constructing the image +of a line or a circle in different cases. We will use +the notation from this statement (\ref{InverzKroznVkrozn}). + +In case (\textit{ii}) - figure \ref{sl.inv.9.2.2a.pic} - it is enough +to determine the image of the perpendicular projection of $P$ onto the center of inversion $S$ +on the line $p$, which we map. The point $P'=\psi_i(P)$ with +the center of inversion determines the diameter of the circle $j$. But if the line +$p$ intersects the circle of inversion $i(S,r)$ for example in the points $M$ and $N$, +both points are fixed and the image of the line $p$ is the circumscribed circle +of the triangle $SMN$ (without the point $S$). The tangent of the circle of inversion +$i$ is mapped into a circle (without the point $S$), which from the inside +touches the circle $i$. + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.2.2a.pic} +\caption{} \label{sl.inv.9.2.2a.pic} +\end{figure} + +In the case of (\textit{iii}) we do the inverse of the construction from +(\textit{ii}) and we get the orthogonal projection on the line that is the image of the given circle. The other special cases (when the circle intersects or touches the circle of inversion) are the inverses of (\textit{ii}). + + Also in the case of (\textit{iv}), when the circle is mapped into a circle, the construction process is clear from the statement itself - the image $k'$ is determined by the points $P'$ and $Q'$. + We just have to be careful that in this case the center of the circle $k$ is not mapped into the center of the new circle $k'$! + Where is the image of the center of this circle located in this case? We denote with $O'$ the image of the center $O$ of the circle $k$ under the inversion $\psi_i$. Let $t_1=ST_1$ and $t_2=ST_2$ be the tangents of the circle $k$ at the points $T_1$ and $T_2$, and $T'_1=\psi_i(T_1)$ and $T'_2=\psi_i(T_2)$ (Figure \ref{sl.inv.9.2.3.pic}). It is clear that $T'_1, T'_2 \in k'$, but $t_1$ and $t_2$ are also tangents of the circle $k'$ (if, for example, $Y\in k'\cap t_1$, then also $X=\psi_i(Y)\in k\cap t_1=\{T_1\}$, i.e. $X=T'_1$). By \ref{invPodTrik} we have $\triangle ST_1O \sim \triangle SO'T'_1$ or $\angle SO'T'_1 \cong \angle ST_1O=90^0$. For the same reasons $\angle SO'T'_2 \cong \angle ST_2O=90^0$. This means that the points $T'_1$, $O'$ and $T'_2$ are collinear. So we can find the point $O'$ as the intersection of the line $T'_1T'_2$ and the segment $SO$. + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.2.3.pic} +\caption{} \label{sl.inv.9.2.3.pic} +\end{figure} + +Another question arises in case of (\textit{iv}). +Which circles are fixed under inversion? One such is of course the circle of inversion $i$ itself, since all of its points are fixed. The next theorem will reveal all other possibilities. + +\bizrek \label{InverzKroznFiks} + Let $\psi_i$ be the inversion with respect to the circle $i(S,r)$. + The only fixed circles of this inversion are the circle $i$ and the circles + that are perpendicular to this circle, that is: + $$\psi_i(k)=k \Leftrightarrow k=i \vee k\perp i.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.2.4.pic} +\caption{} \label{sl.inv.9.2.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} + + ($\Leftarrow$) If $k=i$, the statement is trivial. Let $k$ and + $i$ +be perpendicular circles (Figure \ref{sl.inv.9.2.4.pic}). With $T$ +we mark one of their intersections. The line $ST$ is tangent +to the circle $k$ (statement \ref{TangPogoj}). Let $X$ be an arbitrary point +on the circle $k$ ($X\notin i$) and $X'$ the other intersection of this +circle with the segment $SX$. Then we have: + $$p(S,k)= |SX|\cdot |SX'| = |OT|^2 = r^2,$$ +so $\psi_i(X)=X'\in k$. Similarly, every point $Y$ +on the circle $k$ is the image of some point on this circle, so +$\psi_i(k)=k$ holds. + + ($\Rightarrow$) Let $\psi_i(k)=k$. If $k\neq i$, then + there exists a point $X$ on +the circle $k$, which does not lie on the circle $i$. Let $X'=\psi_i(X)$. +It is clear that $X'\notin i$ also holds. From the definition of inversion it follows that +the points $X$ and $X'$ are on the segment with the initial point $S$, which +means that $S$ is an external point of the circle $k$ (because $\mathcal{B}(X,S,X')$ +is not true). So there exists a tangent from the point $S$ to +the circle $k$ -- with $T$ we mark the point of tangency. Then we have: + $$p(S,k)=|ST|^2 = |SX| \cdot |SX'| = r^2,$$ + so the point $T$ lies on +the circle $i$. This means that the circles $k$ and $i$ are perpendicular +(statement \ref{TangPogoj}). + \kdokaz + + We mention that a similar statement also holds for fixed lines, only that + the perpendicularity of a line with the inversion circle means that the line goes through + the center of inversion. So the only fixed lines of inversion + are those that go through the center of inversion (part (\textit{i}) of statement + \ref{InverzKroznVkrozn}). + +The previous statement also holds if instead of inversion we talk about + reflection over a line. The only fixed lines (circles) + of this reflection are the reflection axis and those lines (circles) + that are perpendicular to that axis. This will be the motivation to + equalize lines and circles in a certain way, thus also the reflection axis and + inversion. We will learn more about this in the next section. + + \bzgled Let $\gamma$ be an arc and $c$ and $p$ be lines. +Draw a triangle $ABC$, if $AB\cong c$, $\angle BCA\cong \gamma$ +and $|BA'|\cdot|BC|=p^2$ (the line $AA'$ is the altitude of the triangle $ABC$). + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.2.5.pic} +\caption{} \label{sl.inv.9.2.5.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $\triangle ABC$ be the desired triangle, +which satisfies the conditions of the task (Figure \ref{sl.inv.9.2.5.pic}). +First, from the condition $\angle BCA\cong \gamma$ it follows that the point $C$ +lies on the locus $l$, which is determined by the string $AB$ and the angular +dimension $\gamma$. Since $A'$ is the vertex of the altitude, the point $A'$ lies +on the circle $k$ above the diameter $AB$. Let $\psi_i$ be the inversion with +respect to the circle $i(B,p)$. From the condition + $|BA'|\cdot|BC|=p^2$ it follows that $\psi_i(A')=C$. Since $A'$ is also in $k$, + it follows that $C$ is in $k'$, where $k'=\psi_i(k)$ is a line (statement + \ref{InverzKroznVkrozn}). + + So if we first draw the line $AB\cong c$, we get the point + $C$ as the intersection of the line $k'=\psi_i(k)$ with the locus $l$. +\kdokaz + +\bzgled \label{MiquelKroznice}(Miquel's\footnote{\index{Miquel, A.}\textit{A. Miquel} (1816–-1851), French +mathematician, who published this statement in 1840.} statement about six +circles) + Let $k_1$, $k_2$, $k_3$ and $k_4$ be such circles that + the circles $k_1$ and $k_2$ intersect in points $A$ and +$A_0$, the circles $k_2$ and $k_3$ in points $B$ and $B_0$, +the circles $k_3$ and $k_4$ in points $C$ and $C_0$, the circles +$k_4$ and $k_1$ in points $D$ and $D_0$. If the points $A$, $B$, $C$ +and $D$ are concircular, the points $A_0$, $B_0$, $C_0$ and $D_0$ +are concircular or collinear. \index{statement!Miquel's.} + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.7a.pic} +\caption{} \label{sl.inv.9.3.7a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $k$ be the circle determined by the +points $A$, $B$, $C$ and $D$. Let $\psi_i$ be the inversion with respect +to any circle $i$ with center at point $A$ (Figure +\ref{sl.inv.9.3.7a.pic}) and + $$\psi_i:\hspace*{1mm}B,C,D,A_0,B_0,C_0,D_0 \mapsto + B',C',D',A'_0,B'_0,C'_0,D'_0.$$ + Because the circles $k$, $k_1$ and +$k_2$ go through point $A$, their images $k'$, $k'_1$ and $k'_2$ are +lines under inversion $\psi_i$ that determine the vertices of the +triangle $A_0'B'D'$. Circles $k_3$ and $k_4$, however, do not go +through point $A$, so their images $k'_3$ and $k'_4$ are circles that +intersect at points $C'\in B'D'$ and $C'_0$. Let $k'_0$ be the circle +drawn through the triangle $A'_0D'_0B'_0$. From Example +\ref{Miquelova točka} it follows that the circles $k'_3$, $k'_4$ and +$k'_0$ intersect at one point - $C'_0$, i.e. $A'_0,B'_0,C'_0,D'_0\in +k'_0$. Therefore, $A_0,B_0,C_0,D_0\in k_0=\psi_i(k'_0)$. By Theorem +\ref{InverzKroznVkrozn} $k_0$ is a circle or a line. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{The Inversive Plane} +\label{odd9InvRavn} + +The motivation for the following discussion is the fact that the +domain of definition is not the whole Euclidean plane under inversion - +the center of inversion does not have its image. This often makes our +work more difficult, as, for example, in the formulation of Theorem +\ref{InverzKroznVkrozn}, where we always have to be careful when a line +or a circle goes through the center of inversion. Also, the images of +a line or a circle are not always a line or a circle, because we have +to exclude the center of inversion in certain cases. + +Because of everything mentioned above, another solution is offered; instead of excluding the center of inversion from the definition domain, we can formally add one point to the Euclidean plane that will be the image of the center of inversion at this inversion. From the definition of inversion for the image $X'$ of the point $X$ at the inversion $\psi_{S,k}$ it holds: $|SX'|\cdot |SX|=r^2$. If $S=X$, then we formally get $|SX'|\cdot 0=r^2$, which means that $|SX'|=\infty$. So, the new point is intuitively seen as a "point at infinity", so we will formally mark it with the symbol $\infty$ and call it the \index{point!at infinity}\pojem{point at infinity}. The set that arises from adding this new point to the Euclidean plane is called the \index{inversive plane} \pojem{inversive plane}, which we will mark with $\widehat{E}^2$. So: $$\widehat{E}^2=E^2 \cup \{\infty\}.$$ But we must be careful - we must not mix the inversive plane with the \index{extended Euclidean plane} \pojem{extended Euclidean plane}, which we get if we add one (infinitely distant) line to the Euclidean plane and define that the parallel lines intersect on it. The extended Euclidean plane is a model of so-called projective geometry, which we mentioned in the introduction. + +In the case of the inversive plane, we have added only one point $\infty$ and required that for every inversion $\psi_{S,k}$ it holds that $\psi_{S,k}(S)=\infty$. Since $\psi_{S,k}$ is an involution (it holds that $\psi_{S,k}^{-1}=\psi_{S,k}$), by agreement it also holds that $\psi_{S,k}(\infty)=S$. + +In section (\textit{iii}) we saw the statement \ref{InverzKroznVkrozn} that +the image of a circle $k^S$ (a circle without point $S$) under an inversion $\psi_{S,k}$ is a line $q$, which does not go through point $S$. But if we add point $S$ to the circle $k^S$, we will get the set $q \cup \{\infty\}$ as the image. Therefore, we will naturally assign point $\infty$ to lines in the inversive plane. If we call such lines with added point $\infty$ $i$-circles, we can translate the statement \ref{InverzKroznVkrozn} and say that an $i$-circle is mapped to an $i$-circle under an inversion. But we also saw certain analogies with reflection over a line and inversion. Therefore, we will call both mappings in the inversive plane $i$-inversions. It is not hard to see that we can generalize the statement \ref{InverzKroznVkrozn} and write it in a simpler form. + + \bizrek \label{InverzInvRavKvK} + An $i$-circle is mapped to an $i$-circle under an $i$-inversion. + \eizrek + + The fact that every line in the inversive plane contains point + $\infty$ means that two parallels intersect in this point. Lines, + which intersect (in a regular point of the Euclidean plane) have two + common points in + the inversive plane. + Since parallels + have only one common point, we also say that they are tangent in + point $\infty$. The image of two parallels under an $i$-inversion will + either be parallels (if it is the axis of reflection) or + two circles, which are tangent in the center of inversion (if it is + an inversion). Two circles, which are tangent (because they have only one + common point), are mapped either to parallels or to + circles, which are tangent. The same goes for the case of a line and a circle being tangent. In all these cases in the inversive plane we can formulate them much shorter in the following way. + + \bizrek \label{InverzDotik} + Let $\psi_i$ be an arbitrary inversion. + If $i$-circles $k$ and $l$ are tangent, then + $i$-circles $\psi_i(k)$ and $\psi_i(l)$ are tangent. + \eizrek + +With the help of this expression, we will first prove the following important +expression. + + \bizrek \label{InverzKonf} + Inversion is a \index{conformal mapping}conformal mapping, which means: + + The angle under which the lines $p$ and $q$ intersect in the point +$A$, is equal to the angle under which their inverse images $p'$ +and $q'$ intersect in the corresponding point $A'$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.1.pic} +\caption{} \label{sl.inv.9.3.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $\psi_i$ be the inversion with respect to the +circle $i(S,r)$, in which the lines $p$ and $q$ and their +intersection $A$ are mapped to $p'$, $q'$ and $A'$ (Figure +\ref{sl.inv.9.3.1.pic}). + +The case when $A=S$, is trivial, because then the lines $p$ and +$q$ are stationary. + + We first assume that $A\notin i$. In this case it is clear that $A\neq + A'$. + + Let $P$ be the intersection of the rectangle of the line $p$ in the point $A$ and the simetral of the line segment $AA'$ and $k_p$ the circle with the center $P$, which goes through the point $A$. From the construction of the circle $k_p$ it is clear that also the point $A'$ lies on this circle and that the line $p$ is its +tangent in the point $A$. If in a similar way we mark with $Q$ the intersection of the rectangle of the line $q$ in the point $A$ and the simetral of the line segment $AA'$ and $k_q$ the circle with the center $Q$, which goes through the point $A$, it follows that the point $A'$ lies on the circle $k_q$ and that the line $q$ is its +tangent in the point $A$. + +Because both circles $k_p$ and $k_q$ go through the pair of points $A$, $A'$ of the inversion $\psi_i$, $k_p$ and $k_q$ are perpendicular to the circle of inversion $i$ (expression \ref{invPravKrozn}). According to expression \ref{InverzKroznFiks}, then $\psi_i(k_p)=k_p$ and +$\psi_i(k_q)=k_q$. + +The lines $p$ and $q$ intersect the circles $k_p$ and $k_q$ in that order. +By \ref{InverzDotik} the intersections $i$ of the circles $p'$ and +$\psi_i(k_p)=k_p$, or $q'$ and $\psi_i(k_q)=k_q$, are in correspondence. Therefore we have: + $$\angle p,q \cong \angle k_p,k_q \cong \angle p',q'.$$ + + If $A\in i$, we apply the previous + proof (with an inversion with respect to the concentric circle) using + \ref{invRazteg}, because a dilation + preserves the size of angles. + \kdokaz + + It is clear that the proof would be practically the same if $p$ and $q$ + were circles or a circle and a line. Therefore we can write the previous statement + in a more general form. + + \bizrek + Every $i$-inversion preserves the angle between two $i$-circles. + \eizrek + + A special case of the previous statement concerns right angles. From + this it follows that perpendicularity is an invariant of inversion and the following statement holds. + + \bizrek \label{InverzKonfPrav} + Every $i$-inversion maps two perpendicular $i$-circles to + a perpendicular $i$-circle. + \eizrek + + Now we will use the proven statements that hold + in the Euclidean plane. + The fact that the circle passing through the center + of inversion is mapped to a line by this inversion allows + us to solve various design tasks where instead of the sought circle + we will first construct its image - a line. This + also applies to other tasks where we will use inversion to translate a statement that + concerns a circle into an equivalent statement that + concerns a line. In both cases, it is desirable to have + at least one point of the circle, which we will then choose as the center of inversion. + + We should also mention that because of all this, we will often say + that the image of a line $p$ ($S\notin p$) under inversion $\psi_{S,r}$ is just the circle $j$ ($j\ni S$) instead of $j^S$. And vice versa - in the same + case we will write $\psi_{S,r}(j)=p$. + +\bzgled Given are the circle $k$ and the points $A$ and $B$. Draw + the circle (line) $x$, which is perpendicular to the circle $k$ and + passes through the points $A$ and $B$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.2.pic} +\caption{} \label{sl.inv.9.3.2.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $x$ be the circle, which is +perpendicular to the circle $k$ and + passes through the points $A$ and $B$, and $\psi_i$ +inversion with respect to any circle $i$ with center $A$ (Figure +\ref{sl.inv.9.3.2.pic}). + +First, we assume that $A\notin k$. In this case, $k'=\psi_i(k)$ is +the circle and $x'=\psi_i(x)$ is the line that goes through the +point $B'=\psi_i(B)$ (statement \ref{InverzKroznVkrozn}). By +statement \ref{InverzKonfPrav} from $x \perp k$ it follows that +$x' \perp k'$. Therefore, $x'$ is the line that goes through the +point $B'$ and is perpendicular to the circle $k'$ or goes through +its center. + +The line $x'$ can therefore be drawn as the line that goes through +the points $B'=\psi_i(B)$ and the center of the circle +$k'=\psi_i(k)$. In the end, $x=\psi_i(x')$. + +But if $A\in k$, then $k'$ is the line (statement +\ref{InverzKroznVkrozn}) and the line $x'$ is drawn as its +perpendicular through the point $B'$. + +Since we can always draw one single perpendicular $x'$, there is +exactly one solution also for the $i$-circle $x$. But $x$ is the +circle exactly when $S\notin x'$ or when the points $A$, $B$ and +the center of the circle $k$ are not collinear. In the case of +collinearity of these points, the solution $x$ is the line. + \kdokaz + + \bzgled \label{TriKroznInv} + If among three circles $k_1$, $k_2$ and $k_3$ + of a plane two touch each other from the outside, + the circle $k$, which is determined by the points of contact, + is perpendicular to each of the circles $k_1$, $k_2$ and $k_3$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.3.pic} +\caption{} \label{sl.inv.9.3.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $A$ be the point of intersection of +the circles $k_2$ and $k_3$, $B$ the point of intersection of the +circles $k_1$ and $k_3$, and $C$ the point of intersection of the +circles $k_2$ and $k_1$ (Figure \ref{sl.inv.9.3.3.pic}). Then the +circle $k$ is the circumscribed circle of the triangle $ABC$. + +Let $\psi_{A,r}$ be the inversion with center $A$ and arbitrary +radius $r$. The circles $k_2$ and $k_3$ touch at the center of the +inversion $A$, so they are mapped by this inversion into the lines +$k'_2$ and $k'_3$ (statement \ref{InverzKroznVkrozn}), which have +no common points, i.e. $k'_2\parallel k'_3$. The circle $k_1$, which +does not pass through the point $A$ and touches the circles $k_2$ +and $k_3$ at the points $C$ and $B$, is mapped by the inversion +$\psi_{A,r}$ into the circle $k'_1$ (statement +\ref{InverzKroznVkrozn}), which touches the lines $k'_2$ and $k'_3$ +(statement \ref{InverzDotik}) at the points $C'=\psi_{A,r}(C)$ and +$B'=\psi_{A,r}(B)$. Therefore, the lines $k'_2$ and $k'_3$ are +parallel tangents of the circle $k'_1$ at the points $C'$ and $B'$, +and the line segment $B'C'$ is their common perpendicular, the +distance $B'C'$ being the diameter of the circle $k'_1$. The line +$B'C'$ is the image of the circle $k$, which passes through the +center of the inversion $A$ and through the points $B$ and $C$ +(statement \ref{InverzKroznVkrozn}), i.e. $\psi_{A,r}(k)=k'=B'C'$. +Since the line $k'=B'C'$ is perpendicular to the line $k'_2$ and +$k'_3$ and to the circle $k'_1$, by statement \ref{InverzKonfPrav} +the circle $k$ is perpendicular to each of the circles $k_1$, +$k_2$ and $k_3$. + \kdokaz + + \bzgled Two of the four circles touch each other from the + outside at the points $A$, $B$, $C$ and $D$. Prove that the + points $A$, $B$, $C$ and $D$ are either concilic or collinear. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.3c.pic} +\caption{} \label{sl.inv.9.3.3c.pic} +\end{figure} + +\textbf{\textit{Proof.}} We mark with $k$ the circle (or line), +which goes through the points $B$, $C$ and $D$. Let $k_1$, $k_2$, $k_3$ +and $k_4$ be such circles, that $k_1$ and $k_2$ touch in the point +$A$, $k_1$ and $k_4$ in the point $B$, $k_3$ and $k_4$ in the point $C$ and +$k_3$ and $k_2$ in the point $D$. We mark with $\psi_i$ the inversion with respect to +any circle $i$ with the center in $A$ (Figure +\ref{sl.inv.9.3.3c.pic}). Let $B'$, $C'$ and $D'$ be the images of the points +$B$, $C$ and $D$ and $k'_1$, $k'_2$, $k'_3$ and $k'_4$ be the images +of the circles $k_1$, $k_2$, $k_3$ and $k_4$ in this inversion. From +\ref{InverzKroznVkrozn} and \ref{InverzDotik} it follows, that $k'_3$ +and $k'_4$ are the circles, which touch in the point $C'$, $k'_1$ and +$k'_2$ are the parallel lines, which are tangent to the circles $k'_4$ and +$k'_3$ in the points $B'$ and $D'$. From the parallelity of the lines $k'_1$ and +$k'_2$ it follows + $\angle k'_2,D'C' \cong \angle k'_1,B'C'$, so according to + \ref{ObodKotTang} + they are also consistent with the central angle $D'S_2C'$ and $B'S_1C'$. Because of this + is $\angle D'C'S_2 \cong \angle B'C'S_1$, which means, that the points $B'$, $C'$ and $D'$ are collinear or $k'$ is a line. According to + \ref{InverzKroznVkrozn} its image with respect to the inversion $k=\psi_i(k')$ goes through the center of the inversion, so $A,B,C,D\in k$. + \kdokaz + + \bzgled Let $k$, $l$ and $j$ be three mutually perpendicular + circles with the common tangents $AB$, $CD$ and $EF$. Prove that + the circumscribed circles of the triangles $ACE$ and $ADF$ touch in + the point $A$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.3a.pic} +\caption{} \label{sl.inv.9.3.3a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $AB$ be the common chord of the circles $k$ and +$l$, $CD$ be the common chord of the circles $l$ and $j$ and $EF$ be +the common chord of the circles $j$ and $k$ (Figure \ref{sl.inv.9.3.3a.pic}). +We denote by $x$ and $y$ the ocrtani circles of the triangles $ACE$ and +$ADF$ and by $\psi_i$ the inversion with respect to any circle $i$ with +center $A$. Let: + \begin{eqnarray*} + && \psi_i:\hspace*{1mm}B,C,D,E,F \mapsto B',C',D',E',F' \hspace*{2mm} + \textrm{ in + }\\ + && \psi_i:\hspace*{1mm}k,l,j,x,y \rightarrow k',l',j',x',y'. + \end{eqnarray*} + Because $A\in k, l, x, y$ and $A\notin j$, + by izreku \ref{InverzKroznVkrozn} + $k'=E'F'$, $l'=C'D'$, $x'=E'C'$ and + $y'=D'F'$ are lines, $j'\ni C',D',E',F'$ is a circle. From + the mutual perpendicularity of the circles $k$, $l$ and $j$ and + the fact that $B\in k\cap l$ it follows that the lines $k'$ and $l'$ + are perpendicular in the point $B'$, and the circle $j'$ is perpendicular to + both lines $k'$ and $l'$ (izrek \ref{InverzKonfPrav}). This + means that $C'D'$ and $E'F'$ are perpendicular diameters of the + circle $j'$, so the quadrilateral $E'C'F'D'$ is a square and $E'C' \parallel + D'F'$. From the parallelism of the lines $x'$ and $y'$ it follows that their + inverse images (circles $x$ and $y$) touch at the center + of inversion $A$. + \kdokaz + + \bzgled Let $k$ and $l$ be the circles of some plane with centers + $O$ and $S$. Let $t_i$ be the tangents of the circle $k$, + which intersect the circle $l$ in the points $A_i$ and $B_i$. Prove that + there exists a circle that touches all the ocrtani circles of the triangles + $SA_iB_i$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.4.pic} +\caption{} \label{sl.inv.9.3.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $\psi_l$ be the inversion with respect to the circle $l$ (Figure \ref{sl.inv.9.3.4.pic}). The lines $t_i$ are mapped by this inversion to the circles $t'_i$, which are the inscribed circles of the triangles $SA_iB_i$ (statement \ref{InverzKroznVkrozn}). Because the lines $t_i$ touch the circle $k$, all the circles $t'_i$ touch the circle $k'=\psi_l(k)$ (statement \ref{InverzDotik}). + \kdokaz + + \bzgled + Let $ST$ be the diameter of the circle $k$, $t$ the tangent of this circle at the point $T$, and $PQ$ and $PR$ its tangents at the points $Q$ and $R$. + Prove that if $L$, $Q'$ and $R'$ are the intersections of the altitudes $SP$, $SQ$ and + $SR$ with the line $t$, then the point $L$ is the midpoint of the segment + $Q'R'$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.5.pic} +\caption{} \label{sl.inv.9.3.5.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $\psi_i$ be the inversion with respect to the circle $i(S,ST)$ (Figure \ref{sl.inv.9.3.5.pic}). The point $T$ is +the orthogonal projection of the center of inversion $S$ onto the line $t$. Because +$\psi_i(T)=T$, it follows from statement \ref{InverzKroznVkrozn} that +$\psi_i(t)=k$ or $k'=\psi_i(k)=t$. Therefore, $\psi_i(Q)=Q'$ +and $\psi_i(R)=R'$. We denote $P'=\psi_i(P)$. From $L\in SP$ it follows that the points $S$, $P$, $L$ and $P'$ are collinear. The tangents $q=PQ$ and +$r=PR$ of the circle $k$ are mapped to the circles $q'$ and $r'$, which pass through the point $S$ and touch the line $k'$ at the points $Q'$ and $R'$ (statements +\ref{InverzKroznVkrozn} and \ref{InverzDotik}). Therefore, the line +$t=k'$ is the common tangent of the circles $q'$ and $r'$, which intersect at +the points $S$ and $P'$. The point $L$ lies on their power axis +$SP'$, so + $|LQ'|^2=|LR'|^2$, that is, the point $L$ is the midpoint of the segment $Q'R'$. +\kdokaz + +Let $a$, $b$ and $c_0$ be circles with diameters $PQ$, $PR$ and $RQ$, where $\mathcal{B}(P,R,Q)$. Let $c_0$, $c_1$, $c_2$, ... $c_n$, ... be a sequence of circles on the same side of the line $PQ$, which each circle in the sequence touches the previous one. Prove that the distance from the center of the circle $c_n$ to the line $PQ$ is $n$ times greater than the diameter of this circle\footnote{This problem was considered by \index{Pappus} \textit{Pappus of Alexandria} (4th century). The pattern that is formed by the semicircles $a$, $b$, $c_0$, was investigated by \index{Archimedes} \textit{Archimedes of Syracuse} (3rd century BC) - see Example \ref{arbelos}.}\\ +(an example from the book \cite{Cofman}). +\ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.6.pic} +\caption{} \label{sl.inv.9.3.6.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.inv.9.3.6.pic}). Let $i$ be a circle with its center in point $P$, which is perpendicular to the circle $c_n$ (circle $i$ goes through the points of tangency from point $P$ to circle $c_n$). With the inversion $\psi_i$ with respect to this circle, the line $l=PQ$ and the circles $a$ and $b$ are mapped into lines $l$, $a'$ and $b'$ (statement \ref{InverzKroznVkrozn}); in this case, the lines $a'$ and $b'$ are perpendicular to the line $l$ (statement \ref{InverzKonfPrav}). The circles $c_0$, $c_1$, $c_2$, ..., $c_n$, ... are mapped into congruent circles $c'_0$, $c'_1$, $c'_2$, ..., $c'_n$, ... with the same inversion, which are all tangent to the parallels $a'$, $b'$ (statement \ref{InverzDotik}) and are all congruent circles $c_n$, because $c'_n=\psi_i(c_n)=c_n$ (statement \ref{InverzKroznFiks}). Since $c_0'\perp l$, the center of the circle $c'_0$ lies on the line $l$, therefore the distance from the center of the circle $c_n=c'_n$ to this line is $n$ times larger than the diameter of this circle. + \kdokaz + + + + \bzgled + Let $t$ be a common external tangent circle + of the circles $k$ and $l$, which are externally tangent in + point $A$, and $c_0$, $c_1$, $c_2$, ... $c_n$, ... a sequence + of circles, each circle in + the sequence is tangent to the previous one, circle $c_0$ is also + tangent to the circle $t$. Prove that there exists a circle (or a line), which is + perpendicular to each circle from the given sequence. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.8.pic} +\caption{} \label{sl.inv.9.3.8.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $i$ be an arbitrary circle with center +$A$ and $\psi_i$ the inversion with respect to that circle (Figure +\ref{sl.inv.9.3.8.pic}). In this inversion, the circles $k$ and $l$ +are mapped to the lines $k'$ and $l'$ (by \ref{InverzKroznVkrozn}), +which are parallel because the circles $k$ and $l$ touch at the +center of inversion $A$. The circles of the sequence $c_0$, $c_1$, +$c_2$, ..., $c_n$, ... are mapped to the circles $c'_0$, $c'_1$, +$c'_2$, ..., $c'_n$, ..., where each two consecutive ones touch and +all the circles touch the parallels $k'$ and $l'$ (by +\ref{InverzDotik}). Therefore, all the circles of this sequence are +mutually tangent. The line $n'$, determined by their centers, is +the perpendicular of the lines $k'$ and $l'$. The image $n=\psi_i(n')$ +of $n'$ is perpendicular to the circles $c'_0$, $c'_1$, $c'_2$, ..., +$c'_n$, ... (by \ref{InverzKonf}). In the end, we can conclude that $n$ +is a circle exactly when the line $n'$ does not go through the point +$A$, i.e. when the circles $k$ and $l$ are not tangent, otherwise $n$ +is a line. \kdokaz + + + \bzgled Let $A$, $B$, $C$ and $D$ be four arbitrary coplanar + points. Prove that the angle, under which the circumscribed + circles of the triangles $ABC$ and $ABD$ intersect, is equal to + the angle, under which the circumscribed + circles of the triangles $ACD$ and $BCD$ intersect. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.9.pic} +\caption{} \label{sl.inv.9.3.9pic} +\end{figure} + +\textbf{\textit{Proof.}} We mark with $k_A$, $k_B$, $k_C$ and $k_D$ +the circumscribed circles of the triangles $BCD$, $ACD$, $ABD$ and +$ABC$ (Figure \ref{sl.inv.9.3.9pic}). By an inversion $\psi_i$ with +respect to an arbitrary circle $i$ with center $A$, our statement is +transformed to the equivalent \ref{ObodKotTang}. + \kdokaz + +\bzgled Let $A$, $B$ and $C$ be points that lie on the line + $l$, + and $P$ be a point outside of this line. Prove that the point $P$ and + the centers of the circumscribed circles of the triangles $APB$, $BPC$ and $APC$ + are four concircular points. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.10.pic} +\caption{} \label{sl.inv.9.3.10pic} +\end{figure} + +\textbf{\textit{Proof.}} We mark with $K_A$, $K_B$, $K_C$ +the centers of the circumscribed circles of the triangles $BPC$, $APC$ and $APB$ +(Figure \ref{sl.inv.9.3.10pic}). If $X=\mathcal{S}_{K_A}(P)$, +$Y=\mathcal{S}_{K_B}(P)$ and $Z=\mathcal{S}_{K_C}(P)$ or +$h_{P,2}(K_A)=X$, $h_{P,2}(K_B)=Y$ and $h_{P,2}(K_C)=Z$, from +the concircularity of the points $X$, $Y$, $Z$ and $P$ it follows that the points +$K_A$, $K_B$, $K_C$ and $P$ are concircular (from $X,Y,Z,P\in k$ it follows that +$K_A,K_B,K_C,P\in h^{-1}_{P,2}$). We thus prove that the points +$X$, $Y$, $Z$ and $P$ are concircular. + +Let $\psi_{P,r}$ be an inversion with the center $P$ and an arbitrary +radius $r$ and + $$\psi_{P,r}:\hspace*{1mm} A,B,C,X,Y,Z\mapsto A',B',C',X',Y',Z'.$$ +The inversion $\psi_{P,r}$ maps the circumscribed circle of the triangle $BPC$, +$APC$ and $APB$ into the lines $B'C'$, $A'C'$ and $A'B'$ (the statement +\ref{InverzKroznVkrozn}). Because $PX$, $PY$ and $PZ$ are the diameters of these +circles, the points $X'$, $Y'$ and $Z'$ are the orthogonal projections of the center +of inversion $P$ onto the lines $B'C'$, $A'C'$ and $A'B'$. The line $l$ is mapped by +the statement \ref{InverzKroznVkrozn} into the circumscribed circle of the triangle +$A'B'C'$, which also goes through the point $P$. By the statement \ref{SimpsPrem} +the points $X'$, $Y'$ and $Z'$ lie on the \index{line!Simson's} Simson's line $s$, +which means (the statement \ref{InverzKroznVkrozn}), that the points $X$, $Y$ and +$Z$ lie on the circle $\psi_{P,r}(s)$, which goes through the point $P$, therefore +the points $X$, $Y$, $Z$ and $P$ are concurrent. + \kdokaz + +%________________________________________________________________________________ + \poglavje{Metric Properties of Inversion} \label{odd9MetrInv} + +It is clear from the definition itself that inversion is not an isometry. In the +previous chapter we have found that inversion does not preserve the relation of +collinearity of points, which means that it is not even a transformation of +similarity. However, we are interested in how the distance between points changes, +although in general the distance $AB$ is mapped by inversion into the segment +$A'B'$. The answer will be given by the following important statement. + + \bizrek \label{invMetr} If $A'$ and $B'$ ($A',B'\neq S$) are the images of the points $A$ and + $B$ by the inversion $\psi_{S,r}$, then it holds: + $$|A'B'|=\frac{r^2\cdot |AB|}{|SA|\cdot |SB|}$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.4.1.pic} +\caption{} \label{sl.inv.9.4.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} We will consider two possible +cases. + +\textit{(i)} Let the points $S$, $A$ and $B$ be nonlinear (Figure +\ref{sl.inv.9.4.1.pic}). According to Theorem \ref{invPodTrik}, the +triangles $ASB$ and $B'SA'$ are similar and $A'B':AB=SB':SA$. Since +$B'$ is the image of point $B$ under the inversion $\psi_{S,r}$, it +also holds that $|SB|\cdot |SB'|=r^2$ or $|SB'|=\frac{r^2}{|SB|}$. +From these two relations it follows: + $$|A'B'|=\frac{|SB'|\cdot |AB|}{|SA|}= + \frac{r^2\cdot |AB|}{|SA|\cdot |SB|}.$$ + + +\textit{(ii)} Let $S$, $A$ and $B$ be collinear points. Without loss +of generality, we assume that $\mathcal{B}(S,A,B)$ holds. Then +$\mathcal{B}(S,B',A')$ (Theorem \ref{invUrejenost}) holds, so + $$|A'B'|=|SA'|-|SB'|= +\frac{r^2}{|SA|}-\frac{r^2}{|SB|}= \frac{(|SB|- |SA|)\cdot +r^2}{|SA|\cdot |SB|} + = \frac{r^2\cdot |AB|}{|SA|\cdot |SB|} ,$$ which had to be proven. \kdokaz + + From the previous theorem we see that the distance between the images of points $A'$ and $B'$ + increases if at least one of the originals $A$ and $B$ + approaches the center of inversion $S$, which is logical, because we saw that the image of this center is a point at infinity. + + In section \ref{odd7Ptolomej} we have already proven Ptolomej's \index{theorem!Ptolomej's general} theorem (\ref{izrekPtolomej}) which relates to + the chordal + quadrilaterals. Now we will generalize this statement. + +\bizrek + If $ABCD$ is an arbitrary convex quadrilateral, then: + $$|AB|\cdot |CD|+|BC|\cdot |AD|\geq |AC|\cdot |BD|.$$ + Equality holds if and only if $ABCD$ is a chordal +quadrilateral. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.4.2a.pic} +\caption{} \label{sl.inv.9.4.2a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $\psi_{A,r}$ be the inversion with the center in the point $A$ (Figure \ref{sl.inv.9.4.2a.pic}). With $k$ we denote the circumscribed circle of the triangle $ABD$. Let $B'$, $C'$ and $D'$ be the images of the points $B$, $C$ and $D$ and $k'$ be the image of the circle $k$ under the inversion $\psi_{A,r}$. By the theorem \ref{InverzKroznVkrozn} the $k'$ is a line, which contains the points $B'$ and $D'$. + From the previous theorem \ref{invMetr} it follows: + $$|B'C'|=\frac{r^2\cdot |BC|}{|AB|\cdot |AC|}, \hspace*{2mm} + |C'D'|=\frac{r^2\cdot |CD|}{|AC|\cdot |AD|} \hspace*{1mm} \textrm{ and } + \hspace*{1mm} |B'D'|=\frac{r^2\cdot |BD|}{|AB|\cdot |AD|}.$$ + For the points $B'$, $C'$ and $D'$ the triangle inequality + \ref{neenaktrik} holds: + $$B'C'+C'D'\geq B'D',$$ +where the equality holds exactly when the points $B'$, $C'$ and $D'$ +are collinear (and $\mathcal{B}(B',C',D')$) or when $C'\in k'$ (and +$\mathcal{B}(B',C',D')$). This is exactly the case when the circle $k$ +contains the point $C$ or when the quadrilateral $ABCD$ is cyclic (and +convex). + We can write the previous inequality +also in the form: + $$\frac{r^2\cdot |BC|}{|AB|\cdot |AC|}+\frac{r^2\cdot |CD|}{|AC|\cdot |AD|} + \geq\frac{r^2\cdot |BD|}{|AB|\cdot |AD|}, \hspace*{2mm} \textrm{or}$$ + $$|AB|\cdot |CD|+|BC|\cdot |AD|\geq |AC|\cdot |BD|,$$ which was to be proven. \kdokaz + + The next theorem will be a generalization of the example \ref{zgledTrikABCocrkrozP} or + \ref{zgledTrikABCocrkrozPPtol}. + +\bzgled \label{zgledABCPinv} Let $k$ be the circumscribed circle of the regular triangle $ABC$. If $P$ is an arbitrary point in the plane of this triangle, then the following equivalence holds: the point $P$ does not lie on the circle $k$, exactly when there exists a triangle with sides, which are congruent to +the distances $PA$, $PB$ and $PC$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.4.3.pic} +\caption{} \label{sl.inv.9.4.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} Let $\psi_{P,r}$ be the inversion with the center in +the point $P$ (Figure \ref{sl.inv.9.4.3.pic}). Let $A'$, $B'$ and +$C'$ be the images of the points $A$, $B$ and $C$ and $k'$ be the image of the circle $k$ under the inversion $\psi_{P,r}$. + From the formula \ref{invMetr} it follows: + $$|A'B'|=\frac{r^2\cdot |AB|}{|PA|\cdot |PB|}, \hspace*{2mm} + |B'C'|=\frac{r^2\cdot |BC|}{|PB|\cdot |PC|} \hspace*{1mm} \textrm{ and } + \hspace*{1mm} |A'C'|=\frac{r^2\cdot |AC|}{|PA|\cdot |PC|}.$$ +Because the triangle $ABC$ is right, from the previous three relations it follows: +$$A'B':B'C':A'C'=PC:PA:PB.$$ + +Now we can start with the proof of the desired equivalence: + \begin{eqnarray*} + P \notin k & \Leftrightarrow& \psi_{P,r}(k) \textrm{ represents a circle + \hspace*{2mm}(formula \ref{InverzKroznVkrozn})}\\ + & \Leftrightarrow& \textrm{the points }A', B', C' \textrm{ are non-collinear}\\ + & \Leftrightarrow& \textrm{the distances }A'B', B'C', A'C' \textrm{ are the sides + of a triangle}\\ + & \Leftrightarrow& \textrm{the distances }PC, PA, PB \textrm{ are the sides + of a triangle,} + \end{eqnarray*} + which was to be proven. \kdokaz + + Another generalization will be related to the Torricelli point (formula + \ref{izrekTorichelijev}). + + \bizrek \label{izrekToricheliFerma} + The point, for which the sum of the distances from + the vertices of a triangle is minimal, is +the \index{Torricelli, E.} \textit{E. Torricelli}'s\footnote{(1608--1647), Italian mathematician and physicist.} \index{točka!Torricellijeva}\index{točka!Fermatova}\pojem{Torricelli point} of this triangle. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.4.4.pic} +\caption{} \label{sl.inv.9.4.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} Use the same notation as in Theorem + \ref{izrekTorichelijev} (Figure \ref{sl.inv.9.4.4.pic}). We have +proven that the circles $k$, $l$ and $j$ intersect in Torricelli's +point $P$ of the triangle $ABC$. Because $BEC$ is a right triangle, +from Theorem \ref{zgledTrikABCocrkrozP} it follows that +$|PB|+|PC|=|PE|$ i.e.: + $$|PA|+|PB|+|PC|=|AE|.$$ + We will prove that this sum is minimal only for Torricelli's point + $P$. Let $P'\neq P$. Then the point $P'$ does not lie on any of the circles $k$, $l$ and $j$. Without loss of generality, let $P' \notin k$. From the previous Theorem \ref{zgledABCPinv} it follows that there exists a triangle with sides $P'E$, +$P'B$ and $P'C$. Because of this and the triangle inequality (Theorem +\ref{neenaktrik}) we have $|P'B| + |P'C| +> |P'E|$ i.e.: + $$|P'A| + |P'B| + |P'C| > |P'A| + |P'E| \geq |AE|=|PA|+|PB|+|PC|,$$ which was to be proven. \kdokaz + + + + + \bnaloga\footnote{37. IMO, India - 1996, Problem 2.} + Let $P$ be a point inside triangle $ABC$ such that + $$\angle APB -\angle ACB = \angle APC -\angle ABC.$$ + Let $D$, $E$ be the incentres of triangles $APB$, $APC$, respectively. Show +that $AP$, $BD$, $CE$ meet at a point. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.4.5.pic} +\caption{} \label{sl.inv.9.4.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} +Let $r>\max\{|AB|, |AC|\}$ be an arbitrary number and $\psi_{A,r}$ be +the inversion with the center in the point $A$ and the radius $r$ +and $B'$, $C'$ and $P'$ be the images of the points $B$, $C$ and $P$ +under the inversion $\psi_{A,r}$ (Figure \ref{sl.inv.9.4.5.pic}). + +According to the statement \ref{invPodTrik} the triangles $ABC$ and $AC'B'$ are similar, so: + $$\angle AC'B'\cong \angle ABC \textrm{ in } \angle +AB'C'\cong \angle ACB.$$ + From the similarity of $\triangle ABP \sim \triangle +AP'B'$ and $\triangle ACP \sim \triangle AP'C'$ it follows: + $$\angle AB'P'\cong \angle APB \textrm{ in } \angle +AC'P'\cong \angle APC.$$ + If we use the four relations of congruence of angles and + the initial condition from the task ($\angle APB -\angle ACB = \angle APC -\angle + ABC$), we get: + \begin{eqnarray*} + \angle C'B'P' &=& \angle AB'P'-\angle AB'C' = \angle APB - \angle + ACB = \\ &=& \angle APC -\angle ABC = \angle AC'P' -\angle AC'B' + =\\ + &=& \angle B'C'P' + \end{eqnarray*} +Now from $\angle C'B'P'\cong \angle B'C'P'$ it follows $P'B'\cong +P'C'$. If we use the relation from the statement \ref{invMetr}, we get + $\frac{|PB| \cdot r^2}{|AP|\cdot |AB|}=\frac{|PC| \cdot r^2}{|AP|\cdot + |AC|}$ or: + $$\frac{|PB| }{ |AB|}=\frac{|PC| }{ |AC|}.$$ + Let $BD \cap AP =X$ and $CE \cap AP =Y$. It is also necessary to + prove that $X=Y$. Because $D$ and $E$ are the centers + of inscribed circles of triangles $APB$ +and $APC$, the lines $BD$ and $CE$ are the (internal) angle bisectors of +$ABP$ and $ACP$. Therefore (statement \ref{HarmCetSimKota}) + $$ \frac{\overrightarrow{PX}}{\overrightarrow{XA}}=\frac{PB}{BA} + =\frac{PC}{CA}=\frac{\overrightarrow{PY}}{\overrightarrow{YA}}.$$ + From $ \frac{\overrightarrow{PX}}{\overrightarrow{XA}} + =\frac{\overrightarrow{PY}}{\overrightarrow{YA}}$ it follows (statement + \ref{izrekEnaDelitevDaljice}) $X=Y$, which means that the lines $AP$, $BD$ and $CE$ intersect in the same +point $X=Y$. + \kdokaz + +%________________________________________________________________________________ +\poglavje{Inversion and Harmonic Conjugate Points} +\label{odd9InvHarm} + +In section \ref{odd7Harm} we saw that a harmonic quadruple of points +is defined in two equivalent ways. For four collinear points +$A$, $B$, $C$ and $D$, $\mathcal{H}(A,B;C,D)$, if one of the two (equivalent) conditions is fulfilled: +\begin{itemize} + \item $\frac{\overrightarrow{AC}}{\overrightarrow{CB}}=- + \frac{\overrightarrow{AD}}{\overrightarrow{DB}}$, + \item there exists a quadrilateral $ABCD$, such that + $PQ \cap RS=A$, $PS \cap QR=B$, $PR \cap AB=C$ and $QS \cap AB=D$. +\end{itemize} + + +Now we will investigate some possibilities for equivalent definitions + of the relation of harmonic quadruple of points - first with the help of inversion. + + \bizrek \label{invHarm} Let $AB$ be the diameter of the circle $i$ and $C$ and $D$ the points of the line $AB$, which are different from $A$ and $B$. If $\psi_i$ + is the inversion with respect to the circle $i$, then it holds: + $$\mathcal{H}(A,B;C,D) \hspace*{1mm} \Leftrightarrow \hspace*{1mm} + \psi_i(C)=D.$$ + \eizrek +\textbf{\textit{Proof.}} Let $O$ be the center of the line $AB$ (Figure +\ref{sl.inv.9.6.1.pic}). Then it is: + \begin{eqnarray*} + \mathcal{H}(A,B;C,D) &\hspace*{1mm} \Leftrightarrow + \hspace*{1mm}& + \frac{\overrightarrow{AC}}{\overrightarrow{CB}}=- + \frac{\overrightarrow{AD}}{\overrightarrow{DB}}\\ + &\hspace*{1mm} \Leftrightarrow + \hspace*{1mm}& + \frac{\overrightarrow{OC}-\overrightarrow{OA}} + {\overrightarrow{OB}-\overrightarrow{OC}}=- + \frac{\overrightarrow{OD}-\overrightarrow{OA}} + {\overrightarrow{OB}-\overrightarrow{OD}}\\ + &\hspace*{1mm} \Leftrightarrow + \hspace*{1mm}& + \frac{\overrightarrow{OC}+\overrightarrow{OB}} + {\overrightarrow{OB}-\overrightarrow{OC}}=- + \frac{\overrightarrow{OD}+\overrightarrow{OB}} + {\overrightarrow{OB}-\overrightarrow{OD}}\\ + &\hspace*{1mm} \Leftrightarrow + \hspace*{1mm}& \overrightarrow{OC} \cdot \overrightarrow{OD} = + OB^2\\ + &\hspace*{1mm} \Leftrightarrow + \hspace*{1mm}& \psi_i(C)=D, + \end{eqnarray*} + which had to be proven. \kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.6.1.pic} +\caption{} \label{sl.inv.9.6.1.pic} +\end{figure} + +The next statement will give the fourth possibility of an equivalent definition + of the relation of harmonic quadruple of points. + +\bizrek \label{harmPravKrozn} + Let $A$, $B$, $C$ and $D$ be different collinear points and $k$ and $l$ + be circles over diameters $AB$ and +$CD$. Then the equivalence holds: +$$\mathcal{H}(A,B;C,D) \Leftrightarrow k\perp l.$$ + \eizrek + + \textbf{\textit{Proof.}} Let $O$ and $S$ be the centers of the circles $k$ and $l$ + (Figure \ref{sl.inv.9.6.1.pic}). From both sides of the equivalence + it follows that the circles +intersect. One of their intersection points is denoted by $T$. From the previous statement \ref{invHarm} it is: + $$\mathcal{H}(A,B;C,D) \hspace*{1mm} \Leftrightarrow \hspace*{1mm} + \psi_k(C)=D.$$ + So we only need to prove: +$$k\perp l \hspace*{1mm} \Leftrightarrow \hspace*{1mm} + \psi_k(C)=D.$$ + But now we have: + \begin{eqnarray*} + k\perp l \hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} + OT\perp TS \hspace*{2mm} + \textrm{ (statement \ref{pravokotniKroznici})}\\ + &\Leftrightarrow& \hspace*{1mm} OT \textrm{ is a tangent of the circle + }l \hspace*{2mm}\textrm{ (statement \ref{TangPogoj})}\\ +&\Leftrightarrow& \hspace*{1mm} \overrightarrow{OC}\cdot \overrightarrow{OD} += OT^2\hspace*{2mm}\textrm{ (statement \ref{izrekPotenca})}\\ +&\Leftrightarrow& \hspace*{1mm} \psi_k(C)=D, + \end{eqnarray*} + which was to be proven. \kdokaz + + From the previous statement \ref{harmPravKrozn} it directly follows (already proven before) that + from $\mathcal{H}(A,B;C,D)$ it follows $\mathcal{H}(C,D;A,B)$ or + $\mathcal{H}(B,A;C,D)$. + +We already know that for three collinear points $A$, $B$ and $C$, + where point $C$ is not the midpoint of line segment $AB$, + there is only one point $D$, so that $\mathcal{H}(A,B;C,D)$ holds. So for three +given points in a harmonic quadruple of points the fourth is uniquely determined. +One of the possible constructions of this point gives us the previous statement +\ref{invHarm} ($D=\psi_k(C)$). But only two points $A$ and $B$ are not +enough to determine the other pair of points $C$ and $D$. There are infinitely many such pairs of points $(C,D)$, for which $\mathcal{H}(A,B;C,D)$ holds (Figure +\ref{sl.inv.9.6.2.pic}). For their determination + another condition is needed. We will consider such conditions in + the next two examples. + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.6.2.pic} +\caption{} \label{sl.inv.9.6.2.pic} +\end{figure} + + \bzgled + Given are points $A$, $B$ and $S$. Construct such points $C$ and $D$, + that $S$ is the midpoint +of line segment $CD$ and $\mathcal{H}(A,B;C,D)$ holds. + \ezgled + + \textbf{\textit{Solution.}} +Let $k$ be a circle with diameter $AB$. If $l$ is a circle with +diameter $CD$, then $S$ is the center of this circle. From +\ref{harmPravKrozn} it follows that $k \perp l$ (Figure +\ref{sl.inv.9.6.3.pic}). So it is enough to construct the circle $l$, because +then the points $C$ and $D$ are the intersections of this circle with the line segment $AB$. But we can construct the circle $l$, if we first construct the tangent +to the circle $k$ from point $S$ in point $T$. + \kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.6.3.pic} +\caption{} \label{sl.inv.9.6.3.pic} +\end{figure} + + + \bzgled \label{harmDaljicad} + Given is a line segment $d$ and points $A$ and $B$. Construct such points $C$ + and $D$, that $\mathcal{H}(A,B;C,D)$ holds and $CD\cong d$. + \ezgled + \textbf{\textit{Solution.}} + +It is enough to construct the center $S$ of the line $CD$, and then continue as in +the previous example. The right triangle $OTS$ can +be constructed because both sides are known $|OT| = \frac{1}{2}\cdot +|AB|$ and $|ST| = \frac{1}{2}\cdot |CD|= \frac{1}{2}\cdot |d|$. From +it we get the hypotenuse $OS$ (Figure \ref{sl.inv.9.6.3.pic}). + \kdokaz + +In the next examples we will look at the use of the previous two +constructions. + + + +\bzgled + Draw a triangle with the data $v_a$, $l_a$ and $b-c$. + \ezgled + +\textbf{\textit{Solution.}} + We will use the labels from the big task \ref{velikaNaloga} +(Figure \ref{sl.inv.9.6.4.pic}). First we can draw the right triangle $AA'E$, because $AA'\cong v_a$ and $AE\cong l_a$. By +the example \ref{harmVelNal} is $\mathcal{H}(A',E;P,Pa)$. From the big +task it follows $PP_a=b -c$, so we can by the previous example +\ref{harmDaljicad} construct the points $P$ and $P_a$. Then +we draw the center $S$ of the inscribed circle of the triangle $ABC$, at +the end the inscribed circle, their tangents from the point $A$ and +mark $B$ and $C$. + \kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.6.4.pic} +\caption{} \label{sl.inv.9.6.4.pic} +\end{figure} + + + +\bzgled + Draw a triangle with the data $v_a$, $a$ and $r+r_a$. + \ezgled + +\textbf{\textit{Solution.}} Also in this case we will use +the labels from the big task \ref{velikaNaloga} (Figure +\ref{sl.inv.9.6.4.pic}). By the example \ref{harmVelNal} is +$\mathcal{H}(A,L;A',La)$. Because $AA'\cong v_a$ and $LL_a =r+r_a$, +we can by the example \ref{harmDaljicad} first draw the line +$AA'\cong v_a$, then the points $L$ and $L_a$. So we get +$r\cong LA'$ and $r_a\cong L_aA'$. + +From the great task it follows that $RR_a = a$. This means that we can draw a rectangular trapezoid $SRR_aS_a$ ($RR_a=a$, $SR=r$ and $S_aR_a=r_a$). Then we construct the inscribed circle $k(S,r)$ and the circumscribed circle $k_a(S_a,r_a)$ and finally their common tangents (two external and one internal), which are the sides of the triangle $ABC$. + \kdokaz + +%________________________________________________________________________________ + \poglavje{Feuerbach Points} +\label{odd9Feuerbach} + +In this section we will consider some properties of the inscribed and circumscribed circles of a triangle. + + \bzgled Let $ABC$ be a triangle with a semiperimeter $s$ and $D$ and $E$ such points on the line $BC$ that $|AD|=|AE|=s$. Prove that the circumscribed circle of the triangle $ADE$ and the circumscribed circle of the triangle $ABC$ touch the side $BC$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.5.0.pic} +\caption{} \label{sl.inv.9.5.0.pic} +\end{figure} + + \textbf{\textit{Proof.}} Let $l$ be the circumscribed circle of the triangle $AED$ and $P_a$, $Q_a$ and $R_a$ the points in which the circumscribed circle $k_a$ of the triangle $ABC$ touches the side $BC$ and the sides $AC$ and $AB$ (Figure \ref{sl.inv.9.5.0.pic}). With $i$ we denote the circle with center $A$ and radius $AE$ (or $AD$). +From the great task \ref{velikaNaloga} it follows that $|AR_a|=|AQ_a|=s=|AE|=|AD|$, which means that the points $R_a$ and $Q_a$ lie on the circle $i$. +By the theorem \ref{pravokotniKroznici} the circles $i$ and $k_a$ are perpendicular. Therefore, the circle $k_a$ is mapped to itself by the inversion $\psi_i$ (theorem \ref{InverzKroznFiks}), and the line $BC$ is mapped to the circle $l$ (theorem \ref{InverzKroznVkrozn}). +Since the line $BC$ touches the circle $k_a$ at the point $P_a$, their images (the circles $l$ and $k_a$) touch at the point $T=\psi_i(P_a)$ (theorem \ref{InverzDotik}). + \kdokaz + +\bzgled \label{InvOcrtVcrt} Let $P$, $Q$ and $R$ be points in which the +incircle of triangle + $ABC$ touches the sides of the triangle. Prove that the orthocenter of + triangle + $PQR$ and the centers of + the circumscribed and incircle of triangle $ABC$ are collinear points. + \ezgled + + \begin{figure}[!htb] +\centering +\input{sl.inv.9.5.0a.pic} +\caption{} \label{sl.inv.9.5.0a.pic} +\end{figure} + + \textbf{\textit{Proof.}} Let $l(O,R)$ and $k(S,r)$ be the circumscribed and + incircle of triangle + $ABC$ and $\psi_k$ be the inversion with respect to the circle $k$ + (Figure \ref{sl.inv.9.5.0a.pic}). + +We also denote $A'=SA \cap QR$, $B'=SB \cap PR$ and $C'=SC \cap PQ$. +From the similarity of triangles $ARA'$ and $AQA'$ (the \textit{SAS} theorem \ref{SKS}) +it follows that the point $A'$ is the center of the side $QR$ and that $AS +\perp RQ$. Similarly, it also follows that points $B'$ and $C'$ +are the centers of segments $PR$ and $PQ$ and that $BS \perp PR$ and $CS +\perp PQ$. + +From the process of constructing the image of a point in an inversion it follows: + $$\psi_k:A,B,C \mapsto A',B',C'.$$ + Therefore, the circumscribed circle $l$ with inversion $\psi_k$ is mapped to + the circumscribed circle of triangle $A'B'C'$ (the \ref{InverzKroznVkrozn} theorem). This is Euler's circle $e_1$ of triangle $PQR$, because it goes through + the centers of its sides. The center of this circle - point + $E_1$ - lies on Euler's line $e_p$ of triangle $PQR$, which is + determined by its orthocenter $V_1$ and the center + of the circumscribed circle $S$. + + We also prove that point $O$ lies on line $e_p$. Although the center + of circle $l$ (point $O$) is not mapped to the center of the circle + (point $E_1$) by inversion, point $E_1$ lies on line $SO$. This + also means that point $O$ lies on line $SE_1=e_p$, so points $V_1$, $O$ and $S$ are collinear. + \kdokaz + + The next simple statement is just an introduction to the so-called Feuerbach's + theorem. + +\bizrek Let $k$ be the inscribed circle of the triangle $ABC$ and $k_a$ the circle drawn through its side $BC$. If $A_1$ is the center of the side $BC$ and $i$ is the circle with center $A_1$ and radius $\frac{1}{2}|b-c|$, then +$$\psi_i:k,k_a\rightarrow k, k_a.$$ + \eizrek + \textbf{\textit{Proof.}} Let $P$ and $P_a$ be the points of tangency of the circles +$k$ and $k_a$ with its side $BC$ (Figure +\ref{sl.inv.9.5.1.pic}). From the great task \ref{velikaNaloga} +it follows that the points $P$ and $P_a$ lie on the circle $i$. Therefore, by +the theorem \ref{pravokotniKroznici} the circles $k$ and $k_a$ +are perpendicular to the circle $i$, so $\psi_i:k,k_a\rightarrow k, +k_a$ (theorem \ref{InverzKroznFiks}). + \kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.5.1.pic} +\caption{} \label{sl.inv.9.5.1.pic} +\end{figure} + + + \bizrek \index{krožnica!Eulerjeva} + Euler's circle of the triangle intersects + the + inscribed + circle and all three of the drawn + circles of this triangle\footnote{The points of intersection are called + \index{točka!Feuerbachova} \pojem{Feuerbach points} of this + triangle. \index{Feuerbach, K. W.} \textit{K. W. Feuerbach} + (1800--1834), German mathematician, who proved this theorem in 1822.}. + \eizrek + +\textbf{\textit{Proof.}} + We will use the notation from the great task + \ref{velikaNaloga}. Let $e$ be Euler's circle of this + triangle (Figure \ref{sl.inv.9.5.1.pic}). From the previous theorem the inversion $\psi_i$ with respect to + the circle $i$ with center $A_1$ and radius + $\frac{1}{2}|b-c|$ (or containing the points $P$ and $P_a$) maps the circles + $k$ and $k_a$ to themselves. We determine the image of Euler's circle $e$ + under the inversion $\psi_i$. + +The circle $e$ contains the center of inversion $A_1$, so it is mapped to some line $e'$ by inversion (statement \ref{InverzKroznVkrozn}). We only need to prove that the line $e'$ is tangent to the circles $k$ and $k_a$ (statement \ref{InverzDotik}). First, the line $e'$ contains the point $\psi_i(A')=E$, which is the intersection of the internal angle bisector at the vertex $A$ with the side $BC$. (statements \ref{harmVelNal} and \ref{invHarm}). But the line $e'$ also contains the points $\psi_i(B_1)=B'_1$ and $\psi_i(C_1)=C'_1$. By statement \ref{invPodTrik}, the triangles $A_1B_1C_1$ and $A_1C'_1B'_1$ are similar. It follows that the line $e'$ with the line $AB$ determines the angle $ACB$. Because the line $e'$ intersects the line $BC$ at the point $E$, which lies on the bisector of the angle $BAC$, $e'$ represents the second common tangent to the circles $k$ and $k_a$. We denote by $F'$ and $F'_a$ the points of tangency of the line $e'$ with the circles $k$ and $k_a$. Then the circle $e$ is tangent to the circles $k$ and $k_a$ at the points $F=\psi_i(F')$ and $F_a=\psi_i(F'_a)$. In the same way, we prove that $e$ is also tangent to the circles $k_b$ and $k_c$. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Stainer's Theorem} +\label{odd9Stainer} + +In this section, we will elegantly (with the help of inversion) prove Stainer's statement, which is related to the problem of the existence of a sequence of circles that are cyclically tangent to each other, and also tangent to two given, non-concentric, circles. It is clear that such a sequence does not exist in general (for any two non-concentric circles). But we will prove that if there is at least one such sequence for the two circles $k$ and $l$, then we can choose the initial element of this sequence as any circle that is tangent to the circles $k$ and $l$. + + We will first solve two +auxiliary tasks. + + + + \bzgled \label{StainerjevLema1} + Let $k_1(S_1,r_1)$ and $k_2(S_2,r_2)$ be two circles of some plane. Draw a circle with +center on the line $S_1S_2$, which is perpendicular to both circles. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.7.0.pic} +\caption{} \label{sl.inv.9.7.0.pic} +\end{figure} + + + \textbf{\textit{Solution.}} + Let $k_1(S_1,r_1)$ and $k_2(S_2,r_2)$ be any + circles. In the case that a solution exists, the center + of the sought circle lies +on the power line of the two circles (Figure +\ref{sl.inv.9.7.0.pic}). This is the case when the intersection $S$ +of the power line $l$ and the central line $S_1S_2$ is an external point of both +circles or when the circles $k_1$ and $k_2$ are non-concentric (from the inside or from the outside). If we draw a tangent to one of the circles from the point $S$ +(e.g. $k_1$), we also get the radius $r=ST_1$ of the circle +$k(S,r)$, where the point $T_1$ is the point of tangency of the tangent from $S$ on the +circle $k_1$. Let the point $T_2$ be the point of tangency of the tangent from $S$ on the +circle $k_2$. Since the point $S$ lies on the power line of the two circles, we have: + $$|ST_1|=p(S,k_1)=p(S,k_2)=|ST_2|$$ + the circle $k(S,r)$ is perpendicular to the circle $k_1(S_1,r_1)$ and + $k_2(S_2,r_2)$ by \ref{pravokotniKroznici}. + \kdokaz + + + %slika + + \bzgled \label{StainerjevLema2} + If $k$ is a circle inside the circle $l$, there exists an inversion that + maps the +circles $k$ and $l$ to two concentric circles. + \ezgled + + + \begin{figure}[!htb] +\centering +\input{sl.inv.9.7.1.pic} +\caption{} \label{sl.inv.9.7.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Let $p$ be a line, determined by the centers of the circles $k$ and $l$; + it is also perpendicular +to both circles. From the previous example \ref{StainerjevLema1} it follows that +there exists a circle $n$ with the center on the line $p$, which is +perpendicular to the circles $k$ and $l$ (Figure +\ref{sl.inv.9.7.1.pic}). + Because the center of the circle $n$ lies on +the line $p$, the circle $n$ is also perpendicular to the line $p$. With $I$ +and $J$ we denote the intersections of the circle $n$ with the line $p$. Let $i$ +be an arbitrary circle with the center $I$ and $\psi_i$ the inversion with respect to this circle. Because the inversion preserves angles, the line $p$ and +the circle $n$ (without the point $I$) with $\psi_i$ are mapped to the perpendicular line $p'=p$ +and $n'$, the circles $k$ and $l$ to the circles $k'$ and $l'$, which are +perpendicular to these two lines (the statement \ref{InverzKroznVkrozn} and +\ref{InverzKonf}). Therefore, the circles $k'$ and $l'$ are concentric. + \kdokaz + + + \bzgled + (Steiner's statement \index{statement!Steiner's} + \footnote{\index{Steiner, J.} + \textit{J. Steiner} (1769--1863), Swiss mathematician.}.) + Let $l$ be a circle inside the circle $k$ and $a_1$, $a_2$, +..., $a_n$ a sequence of circles, each of which touches the circles $k$ and $l$, +and each circle touches the adjacent circle in the sequence. + If the circles $a_n$ +and $a_1$ touch, this property is independent of the choice of the first circle +$a_1$ of this sequence. + \ezgled + + \begin{figure}[!htb] +\centering +\input{sl.inv.9.7.2.pic} +\caption{} \label{sl.inv.9.7.2.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + If we use the previous statement \ref{StainerjevLema2}, the proposition + is translated into the case when the circles are concentric (the touch of the circles is + an invariant of the inversion \ref{InverzDotik}). But in this case the new sequence of circles + is simply obtained from the first one by rotation with the center in the common center of the circles + $\psi_i(k)$ and $\psi_i(l)$ (Figure +\ref{sl.inv.9.7.2.pic}). + \kdokaz + + +%________________________________________________________________________________ +\poglavje{Problem of Apollonius} +\label{odd9ApolDotik} + +Now we will deal with so-called \index{problem!Apolonijev} + Apolonijev's + problems of the intersection + of circles, which can be elegantly +solved by using inversion. These are problems of the following form: + +\textit{ + \vspace*{2mm} + Draw a circle $k$, which satisfies three +conditions, each of which has one of the following forms: +\vspace*{2mm} +\begin{itemize} + \item contains a given point, + \item touches a given line, + \item touches a given circle. +\end{itemize}} + It is clear that all the points, lines and circles mentioned in the above conditions lie + in the same plane. We have already encountered some of these problems. For example, to construct a circle that + contains a given point and touches a given line. It is not difficult + to determine that there are ten Apollonius's problems. We usually list them in the following order: +\begin{enumerate} + \item draw a circle that contains three given points. $(A,B,C)$, +\item draw a circle that contains a given point and touches a given +line $(A,B,p)$, + \item draw a circle that contains a given point and touches + a given +circle $(A,B,k)$, + \item draw a circle that contains a given point and +touches two given lines $(A,p,q)$, + \item draw a circle that contains a given point and + touches a given line and a given +circle $(A,p,k)$, + \item draw a circle that contains a given point and +touches two given circles $(A,k_1,k_2)$, + \item draw a circle that touches three given lines + $(p,q,r)$, + \item draw a circle that +touches two given lines and a given circle $(p,q,k)$, + \item draw a circle that touches a given line and two given circles +$(p,k_1,k_2)$, + \item draw a circle that touches +three given circles $(k_1,k_2,k_3)$. +\end{enumerate} + +We immediately see that the first and seventh problem are trivial, + and also all the other problems can be +solved without using inversion. However, inversion gives us a general +method for solving them. This method is based on the fact that in a certain case, a circle is transformed into a line by inversion (Theorem +\ref{InverzKroznVkrozn}). We will illustrate it with an example of the fifth Apollonius' problem: + +\bzgled + Draw a circle that contains the given point $A$ and + touches the given line $p$ and the given +circle $k$. + \ezgled + + + \begin{figure}[!htb] +\centering +\input{sl.inv.9.8.1.pic} +\caption{} \label{sl.inv.9.8.1.pic} +\end{figure} + + + \textbf{\textit{Solution.}} + +Assume that $x$ is a circle that contains the point $A$ and touches the line $p$ and the circle $k$ (Figure +\ref{sl.inv.9.8.1.pic}). We will consider the general case, so that the point $A$ does not lie on neither the line $p$ nor the circle $k$. With +$i$ we denote the circle with the center $A$ and an arbitrary radius +$r$. Let $\psi_i$ be the inversion with respect to this circle and $p'$, +$k'$ and $x'$ be the images of the line $p$ and the circles $k$ and $x$ under this inversion. Because $A\notin p,k$ and $A\in x$, $p'$ and $k'$ are circles, $x'$ is a line (Theorem \ref{InverzKroznVkrozn}). +The line $p$ and the circle $x$ have exactly one common point, therefore this is also true for the images $p'$ and $x'$, thus the line $x'$ is tangent to the circle $p'$. Similarly, the line $x'$ is also tangent to the circle +$k'$. Therefore, the problem is reduced to constructing the line $x'$, which is the common tangent of the circles $p'$ and $k'$ (Example \ref{tang2ehkroz}). +Then $x= \psi_i^{-1}(x')= \psi_i(x')$. + +The task has zero, one, two, three or four solutions, depending on the mutual position of the circles $p'$ and $k'$ or the number of their common tangents. + +In the case when $A\in p$ or $A\in k$, the line $x'$ is a line that +touches one circle ($k'$ or $p'$) and is parallel to one +line ($p'$ or $k'$). + +If $A\in p \cap k$, the task has no solution or there are infinitely many, depending on whether the circles intersect or touch. In both cases, the line $x'$ is parallel to two lines $p'$ and $k'$, but in the first case the lines $p'$ and $k'$ intersect, in the second case they are parallel. + \kdokaz + +We immediately notice that in solving this problem, it was not important whether $p$ and $k$ were a straight line and a circle, but it was important that the images $p'$ and $k'$ were circles. But $p'$ and $k'$ are also circles in the case when, for example, $p$ and $k$ are two lines and $A\notin p,k$. So the fourth and sixth Apollonius problems are solved in the same way as the fifth problem that we just solved. + +We can also solve the second and third problem by using inversion with respect to any circle with center $A$. Both problems are translated into the construction of a tangent from point $B'$ to circle $p'$ (or $k'$). + + Problems 8, 9 and +10 are translated into problems 4, 5 and 6 in order. The idea is to first plan a circle that is concentric with the desired circle and contains the center of one of the given circles - the one with the smallest radius. + +In the following, we will solve a problem similar to Apollonius' problem of the intersection of circles. + +\bzgled + Given are the point $A$, the circles $k$ and $l$, and the angles $\alpha$ and $\beta$. + Draw + a circle that passes through point $A$, +the circles $k$ and $l$ and intersects at angles $\alpha$ and $\beta$. + \ezgled + + + \begin{figure}[!htb] +\centering +\input{sl.inv.9.8.2.pic} +\caption{} \label{sl.inv.9.8.2.pic} +\end{figure} + +\textbf{\textit{Solution.}} + +Let's assume that $x$ is a circle that contains point $A$ and with +circles $k$ and $l$ defines angles $\alpha$ and $\beta$ (Figure +\ref{sl.inv.9.8.2.pic}). Again, we will consider the general case, +when point $A$ does not lie on any of the circles $k$ and $l$. With +$i$ we mark the circle with center $A$ and any radius $r$ and with +$\psi_i$ the inversion with respect to that circle. Let $k'$, $l'$ +and $x'$ be the images of circles $k$, $l$ and $x$ under that +inversion. Because $A\notin k,l$ and $A\in x$, $k'$ and $l'$ are +circles and $x'$ is a line (Theorem \ref{InverzKroznVkrozn}). +Because inversion is a conformal mapping (Theorem +\ref{InverzKonf}), line $x'$ intersects circles $k'$ and $l'$ +under angles $\alpha$ and $\beta$. + +Line $x'$ can be drawn as the common tangent of two circles $k'_1$ +and $l'_1$ that are concentric with circles $k'$ and $l'$. When +drawing circles $k'_1$ and $l'_1$ we take into account the fact +that line $x'$ with circles $k'$ and $l'$ defines the chords with +central angles $2\alpha$ and $2\beta$. In the end, $x= +\psi_i^{-1}(x')= \psi_i(x')$. + +Also in this case, the number of solutions depends on the mutual +position of circles $k'_1$ and $l'_1$ i.e. the number of their +common tangents. + +In the case when $A\in k$ or $A\in l$, $x'$ is a line that with line +$k'$ (or $l'$) defines angle $\alpha$, with circle $p'$ (or $k'$) +angle $\beta$. + +If $A\in p \cap k$, the problem in general has no solution, +because it is a line $x'$, that with given lines $k'$ and $l'$ +defines angles $\alpha$ and $\beta$. The problem has infinitely +many solutions, when lines $k'$ and $l'$ define angles $|\beta \pm +\alpha|$. + \kdokaz + + +%________________________________________________________________________________ +\poglavje{Constructions With Compass Alone} \label{odd9LeSestilo} + +In Euclidean geometry constructions we always used a ruler and a +compass, which means that we used +elementary constructions\index{konstrukcije!z ravnilom in šestilom} (see section \ref{elementarneKonstrukcije}). + +However, by using a ruler and a compass, or the aforementioned elementary constructions, in Euclidean geometry it is not possible to derive all constructions. The most well-known are the following examples\footnote{All three problems were posed by the Ancient Greeks. Later, many famous mathematicians as well as laymen tried to solve these problems. The fact that the aforementioned constructions cannot be derived using only a ruler and a compass was only proven in the 19th century, when the French mathematician \index{Galois, E.} \textit{E. Galois} (1811--1832) developed \index{group}the theory of groups. The proof for the trisection of an angle and the doubling of a cube was first given by the French mathematician \index{Wantzel, P. L.}\textit{P. L. Wantzel} (1814--1848) in 1837. The fact that the quadrature of a circle is impossible is a consequence of the transcendence of the number $\pi$, which was proven by the German mathematician \index{Lindemann, C. L. F.}\textit{C. L. F. Lindemann} (1852–-1939) in 1882.}: + \begin{itemize} + \item the division of a given angle into three congruent parts (\pojem{trisection of an angle})\index{trisection of an angle}; + \item the construction of a square with the same area as a given circle (\pojem{quadrature of a circle})\index{quadrature of a circle}; + \item the construction of the edge of a cube that has twice the volume of a cube with a given edge $a$ (\pojem{doubling of a cube})\index{doubling of a cube}. + \end{itemize} + +\index{konstrukcije!pravilnih $n$-kotnikov} In addition, the problem of constructing regular $n$-gon with a ruler and a compass is also known, which cannot be solved for every $n\in\{3,4,\ldots\}$\footnote{The well-known German mathematician \index{Gauss, C. F.}\textit{C. F. Gauss} (1777--1855) proved in 1796 that with a ruler and a compass, a regular $n$-gon can be constructed exactly when $n=2^k\cdot p$, where $p$ is either 1 or a prime number that can be written in the form $2^{2^l}+1$, for $l\in\{0,1,2,\ldots\}$ (i.e. \pojem{Fermatova števila}\normalcolor, which are not all prime - for example, for $n=5$, they are named after the French mathematician \index{Fermat, P.}\textit{P. Fermat} (1601--1665)). A regular $n$-gon can therefore be constructed with a ruler and a compass if $n\in\{3,4,5,6,8,10,12,16,17,\ldots\}$, but if $n\in\{7,9,11,13,14,15,18,19,\ldots\}$, this construction is not possible.}. + + In projective geometry, in which there is neither parallelism nor metric, + we use only a ruler\index{konstrukcije!z ravnilom} or only those + elementary constructions + that can only be done with a ruler. It is clear that in this geometry it is not possible + to design (or even define) figures such as + square, + parallelogram, circle, ... + +The question arises, what can we construct if we use + only a compass or only those + elementary constructions + that can be done only with a compass. We consider that + a line is drawn if two of its points are drawn, but we cannot + use the direct construction of the intersection of two lines (with + a plane). Surprisingly, in this way - only with a compass - + we can carry out all + the constructions that can be done with a compass and with + a plane\footnote{The constructions only with the help of a compass were researched + by the Italian mathematician \index{Maskeroni, L.} + \textit{L. Maskeroni} (1750--1800) (after him + we call them \index{konstrukcije!Maskeronijeve}\index{konstrukcije!s šestilom} + \pojem{Maskeronijeve konstrukcije}) and a hundred +years before him the Danish mathematician \index{Mor, G.} \textit{G. Mor} +(1640--1697) in his book ‘‘Danish Euclid’’ from 1672. The theoretical +basis of these constructions was given by the Austrian mathematician \index{Adler, +A.} \textit{A. Adler}, who in 1890 proved that every +design problem that can be solved with the help of a plane and a compass, can +also be solved only with the use of a compass.}! + +We will continue with the following constructions, in which we will +use only a compass. In the desire to show that a plane +"can be replaced" with a compass, the main role will be played by inversion. In +the previous use of inversion in constructions, we most often +used the fact that inversion in a certain case transforms +a circle into a line. So we translated the problem of constructing the desired circle with +the help of inversion into the problem of constructing the desired line, +which is usually easier. Now we will try the opposite - problems related to the construction of a line (and the use +of a plane) we will translate with inversion into problems of constructing a circle (and the use +of a compass). + + \bzgled \label{MaskeroniNAB} + Given are points $A$ and $B$ and $n\in \mathbb{N}$. + Draw such a point $P_n$, that + $\overrightarrow{AP_n}=n\cdot \overrightarrow{AB}$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.9.1.pic} +\caption{} \label{sl.inv.9.9.1.pic} +\end{figure} + +\textbf{\textit{Solution.}} We will carry out the construction + inductively. For $n=1$ it is clear that $P_1=B$. First, let's draw the point + $P_2$, for which $\overrightarrow{AP_2}=2\cdot \overrightarrow{AB}$ + (Figure +\ref{sl.inv.9.9.1.pic}). Now let's draw the circles $k_1(A,AB)$ +and $k_2(B,BA)$. We mark one of their intersections with $Q_1$. +Then we draw the circle $l_1(Q_1,Q_1B)$. The intersection of +the circles $l_1$ and $k_2$, which is not the point $A$, we mark with $Q_2$. The point $P_2$ +we get as one of the intersections of the circles $l_2(Q_2,Q_2B)$ and $k_2$ (that which +is not $Q_1$). \normalcolor The relation $\overrightarrow{AP_2}=2\cdot +\overrightarrow{AB}$ follows from the fact that the triangles $ABQ_1$, +$Q_1BQ_2$ and $Q_2BP_2$ are all right. + + Assume that we have with the described procedure + drawn the points $P_k$ ($k\leq n-1$), for which + $\overrightarrow{AP_k}=k\cdot \overrightarrow{AB}$. We can draw the point $P_n$ + in the same way. + First, we draw the circle $k_n(P_{n-1},P_{n-1}P_{n-2})$. The intersection of the circles $l_{n-1}$ and $k_n$, which is not the point $P_{n-2}$, +we mark with $Q_n$. The point $P_n$ we get as one of the intersections of the circles $l_n$ and $k_n$ (that which +is not $Q_{n-1}$). The triangles +$P_{n-2}P_{n-1}Q_{n-1}$, $Q_{n-1}P_{n-1}Q_n$ and $Q_nP_{n-1}P_n$ are +all right. Therefore, +$\overrightarrow{P_{n-1}P_n}=\overrightarrow{P_{n-2}P_{n-1}}$. If +we use the induction assumption +$\overrightarrow{AP_k}=k\cdot \overrightarrow{AB}$ for $k=n-1$, +we get $\overrightarrow{AP_n}=n\cdot \overrightarrow{AB}$, which +means that $P_n$ is the desired point. + \kdokaz + + We have already mentioned the importance of inversion in our constructions. + Now we are ready to prove the process of constructing the image + of a point under inversion only with the help of a compass. + + \bzgled Given are the circle $i(S,r)$ and the point $X$. Draw the point + $X'=\psi_i(X)$. \label{MaskeroniInv} + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.9.2.pic} +\caption{} \label{sl.inv.9.9.2.pic} +\end{figure} + + \textbf{\textit{Solution.}} If $X\in i$ it is trivial that $X'=X$. + + First, let us assume that $X$ is an external point of the + inversion circle $i$ (Figure \ref{sl.inv.9.9.2.pic}). Because the center $S$ is its internal point, + the circle $k(X,XS)$ intersects the inversion circle in two points + (statement \ref{DedPoslKrozKroz}) - for example $P$ and $Q$, which we + can draw. Then the point $X'$ is the second intersection + of the circles $l_P(P,PS)$ and $l_Q(Q,QS)$ (the first one is the point $S$). + We will also prove that $X'=\psi_i(X)$. Because by construction $XS\cong + XP$ and $PX'\cong PS$, it follows that $\angle PX'S\cong \angle PSX'= + \angle PSX\cong\angle SPX$. This means that the triangles $PSX'$ + and $XPS$ are similar, therefore $\frac{|PX'|}{|XS|}=\frac{|SX'|}{|PS|}$ or + $|SX|\cdot |SX'|=|PX'|\cdot |PS|=r^2$, thus $X'=\psi_i(X)$. + + If $X$ is an internal point of the + inversion circle, then there exists a natural number $n$ such that + $|n\cdot \overrightarrow{SX}|>r$. Let $X_n$ be the point for which + $\overrightarrow{SX_n}=n\cdot \overrightarrow{SX}$ and + $X'_n=\psi_i(X_n)$. From the previous example and the first part of the proof + we can draw the points $X_n$ and $X'_n$. + For the point $X'=\psi_i(X)$ it holds: + $$|SX'|=\frac{r^2}{|SX|}=\frac{n\cdot r^2}{|SX_n|}=n\cdot |SX'_n|,$$ + which means that we can also draw the point $X'$ using + the previous example. + \kdokaz + + \bzgled Dani sta točki $A$ in $B$. Načrtaj središče daljice + $AB$. \label{MaskeroniSred} + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.9.3.pic} +\caption{} \label{sl.inv.9.9.3.pic} +\end{figure} + +\textbf{\textit{Solution.}} Let $\psi_i$ be the inversion with respect to the circle $i(A,AB)$ (Figure \ref{sl.inv.9.9.3.pic}). + First, we plan such a point + $X$, that $\overrightarrow{AX}=2\cdot\overrightarrow{AB}$ + (example \ref{MaskeroniNAB}), then $X'=\psi_i(X)$ (example \ref{MaskeroniInv}). + Because $X$ is an external point of the circle $i$, $X'$ is its internal point. + Therefore $\mathcal{B}(A,X',B)$. For the point $X'$ it also holds + $|AX'|=\frac{|AB|^2}{|AX|}=\frac{|AB|^2}{2\cdot |AB|}=\frac{1}{2}\cdot |AB|,$ + which means that $X'$ is the center of the line segment $AB$. + \kdokaz + + \bzgled \label{MaskeroniProj} Given are three non-collinear points $P$, $Q$ and $R$. Plan + the orthogonal projection of the point $P$ onto the line $QR$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.9.4.pic} +\caption{} \label{sl.inv.9.9.4.pic} +\end{figure} + + \textbf{\textit{Solution.}} First, we plan the centers of the line segments + $PQ$ and $PR$ (example \ref{MaskeroniSred}) and label them in order with $Q_1$ + and $R_1$ (Figure \ref{sl.inv.9.9.4.pic}). We plan the point $P'$ + as the second intersection of the circles $k(Q_1,Q_1P)$ and $l(R_1,R_1P)$. The point $P'$ lies on the circles with diameters $PQ$ and $PR$. Therefore + $\angle PP'Q=\angle PP'R=90^0$. Therefore the point $R$ lies on + the line $P'Q$, which means that the points $P'$, $Q$ and $R$ + are collinear. Therefore the point $P'$ is the orthogonal projection of the point $P$ + onto the line $QR$. + \kdokaz + + \bzgled \label{MaskeroniOcrt} Given are three non-collinear points $A$, $B$ and $C$. + Plan + the center of the circumscribed circle and the circumscribed circle of the triangle + $ABC$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.9.5.pic} +\caption{} \label{sl.inv.9.9.5.pic} +\end{figure} + +\textbf{\textit{Solution.}} + Let $l$ be the circle drawn through the triangle $ABC$ with center in + the point $O$ and $\psi_i$ the inversion with center $A$ and arbitrary + radius (Figure \ref{sl.inv.9.9.5.pic}). Because $A\in l$, from the equation + \ref{InverzKroznVkrozn} it follows, + that $l'=\psi_i(l)$ is a line. If $P'$ is the orthogonal projection of + the center of inversion $A$ onto the line $l'$ and $P=\psi_i(P')$, then + $AP$ is the diameter of the circle $l$ (a consequence of the construction in the proof + (\textit{ii}) of the equation + \ref{InverzKroznVkrozn}). + + The proven facts allow us to construct. First, we draw + an arbitrary circle $i$ with center in the point $A$ and + points $B'=\psi_i(B)$ and $C'=\psi_i(C)$ (example + \ref{MaskeroniInv}), then the orthogonal projection $P'$ of the point $A$ + onto the line $B'C'$ (example \ref{MaskeroniProj}) and $P=\psi_i(P')$. + The center of the drawn circle is obtained as the center of the segment $AP$ + (example \ref{MaskeroniSred}). The circle $l(O,A)$ is the circle drawn + through the triangle $ABC$. + \kdokaz + + One of the elementary constructions with only a compass represents the construction of + the intersection of two circles. With the help of inversion, we + will use it for the following elementary construction, which we consider to be + a construction with a ruler and + a compass. + + \bzgled \label{MaskeroniKrPr} Given are four points $A$, $B$, $C$ and $D$. + Draw the intersection of the lines $AB$ and $CD$. + \ezgled + +\textbf{\textit{Solution.}} Let's assume that the lines $AB$ and $CD$ + are not parallel. + Let $S$ be any point that does not lie + on either of the lines $AB$ and $CD$. It is clear that such a point + exists, but an effective construction of such a point can be obtained + if we use the construction of points $Q_1$ and $Q_2 $ from + the example \ref{MaskeroniNAB} with respect to the distance $AB$. + Let $\psi_i$ be the inversion with respect to any circle $i$ with + the center at the point $S$ (Figure \ref{sl.inv.9.9.6.pic}). + We draw the points $A'=\psi_i(A)$, $B'=\psi_i(B)$, $C'=\psi_i(C)$ and + $D'=\psi_i(D)$ (example \ref{MaskeroniInv}) and the inscribed circles + of the triangles $SA'B'$ and $SC'D'$ (example \ref{MaskeroniOcrt}) or + the images of the lines $AB$ and $CD$ under the inversion $\psi_i$. The point $L'$ is + the second intersection of two inscribed circles (the first intersection is the point $S$) + It exists in the case when the lines $AB$ and $CD$ + are not parallel. The intersection $L$ of the lines $AB$ and $CD$ is the image + of the intersection $L'$ of their images - that is $L=\psi_i(L')$ (example \ref{MaskeroniInv}). + \kdokaz + + + \bzgled \label{MaskeroniKtKr}Given is a circle $k$ and points $A$ and $B$. + Draw the intersection of the line $AB$ and the circle $k$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.9.7.pic} +\caption{} \label{sl.inv.9.9.7.pic} +\end{figure} + + \textbf{\textit{Solution.}} The construction can be carried out in the same + way as in the previous case, only that $k'$ is the inscribed circle + of the triangle determined by the points $C'=\psi_i(C)$, $D'=\psi_i(D)$ + and $E'=\psi_i(E)$, where $C$, $D$ and $E$ are any points + of the circle $k$ + (Figure \ref{sl.inv.9.9.7.pic}). +\kdokaz + +We have proven with the last two constructions, that all +elementary constructions related to a line and a compass (see section \ref{elementarneKonstrukcije}), +can be done only with a compass. As we have mentioned before, in doing so we count that +a line is drawn if two of its points are drawn (the same goes for a distance and a segment), but we cannot +use the direct construction of the intersection of two lines or a line and a circle (by using +a ruler). From the last two constructions (\ref{MaskeroniKrPr} and \ref{MaskeroniKtKr}), it follows that the latter is also possible only with a compass. + +Since every design task that can be solved with a ruler and a compass, +can be carried out with elementary constructions +related to a line and a compass, from the previous it follows, +that this task can only be solved with a compass. + + +%NALOGE +%________________________________________________________________________________ + +\naloge{Exercises} + + +\begin{enumerate} + + \item +Prove that the composite of two inversions $\psi_{S,r_1}$ and $\psi_{S,r_2}$ with respect to a concentric + circle represents a dilation. Determine the center and coefficient + of this dilation. + + \item Let $A$, $B$, $C$ and $D$ be four collinear points. + Construct such points $E$ and $F$, that $\mathcal{H}(A,B;E,F)$ +and $\mathcal{H}(C,D;E,F)$ are true. + + \item In a plane are given a point $A$, a line $p$ and a circle $k$. + Draw a circle that goes through the point $A$ and is + perpendicular to the line $p$ and the circle $k$. + + \item Solve the third, fourth, ninth and tenth Apollonius' problem. + + \item Let $A$ be a point, $p$ a line, $k$ a circle + and $\omega$ an angle in some plane. Draw a circle that goes through the point $A$, touches the line $p$ and with the circle $k$ +determines the angle $\omega$. + + \item Determine the geometric location of the points of tangency of two circles that + touch the arms of a given angle in two given points $A$ and $B$. + + \item Draw a triangle, if the following data are known: +\begin{enumerate} + \item $a$, $l_a$, $v_a$ + \item $v_a$, $t_a$, $b-c$ + \item $b+c$, $v_a$, $r_b-r_c$ + \end{enumerate} + +\item Let $c(S,r)$ and $l$ be a circle and a line in the same plane that do not have any common points. Let $c_1$, $c_2$, and $c_3$ also be circles in this plane that intersect each other (two at a time) and each of them intersects $c$ and $l$. Express the distance from point $S$ to line $l$ with $r$\footnote{Proposal for MMO 1982. (SL 12.)}. + +\item Let $ABCD$ be a regular tetrahedron. To any point $M$ on edge $CD$, we assign the point $P = f(M)$ which is the intersection of the rectangle through point $A$ on line $BM$ and the rectangle through point $B$ on line $AM$. Determine the geometric position of all points $P$, if point $M$ takes on all values on edge $CD$. + +\item Let $ABCD$ be a tetrahedron with perpendicular edges and $P$, $Q$, $R$, and $S$ be the points of tangency of sides $AB$, $BC$, $CD$, and $AD$ with the inscribed circle of this tetrahedron. Prove that $PR\perp QS$. + +\item Prove that the centers of a tetrahedron with perpendicular edges, the centers of the inscribed and circumscribed circles, and the intersection of its diagonals are collinear points (\index{izrek!Newtonov}Newton's Theorem\footnote{\index{Newton, I.}\textit{I. Newton} (1643--1727), English physicist and mathematician}). + +\item Let $p$ and $q$ be parallel tangents to circle $k$. Circle $c_1$ intersects line $p$ in point $P$ and circle $k$ in point $A$, circle $c_2$ intersects line $q$ and circles $k$ and $c_1$ in points $Q$, $B$, and $C$. Prove that the intersection of lines $PB$ and $AQ$ is the center of triangle $ABC$ of the circumscribed circle. + +\item Circles $c_1$ and $c_3$ intersect each other from the outside in point $P$. Similarly, circles $c_2$ and $c_4$ also intersect each other from the outside in the same point. Circle $c_1$ intersects circles $c_2$ and $c_4$ in points $A$ and $D$, while circle $c_3$ intersects circles $c_2$ and $c_4$ in points $B$ and $C$. Prove that the following holds\footnote{Proposal for MMO 2003. (SL 16.)}: +$$\frac{|AB|\cdot|BC|}{|AD|\cdot|DC|}=\frac{|PB|^2}{|PD|^2}.$$ + +\item Let $A$ be a point that lies on the circle $k$. With only a + compass, draw the square $ABCD$ (or its vertices), which is inscribed in the given +circle. + +\item Given are the points $A$ and $B$. With only + a compass, draw such a point $C$, that + $\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}$. + + \item With only the help + of a compass, divide the given line segment in the ratio $2:3$. +\end{enumerate} + + + + + + + + + +% DEL 10 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% REŠITVE IN NAPOTKI +%________________________________________________________________________________ + +\del{Solutions and Hints} +\footnotesize +%REŠITVE - Aksiomi evklidske geometrije ravnine +%________________________________________________________________________________ + +\poglavje{Introduction} + +There are no exercises in the first chapter. \color{viol4}$\ddot\smile$ \normalcolor %smile + +\poglavje{Axioms of Planar Euclidean Geometry} + +\begin{enumerate} + +\item \res{Let $P$, $Q$ and $R$ be the inner points of the sides of the triangle $ABC$. Prove that $P$, $Q$ and $R$ are non-collinear.} + +Assume the opposite, that points $P$, $Q$ and $R$ lie on + a line $l$. This means that the line $l$ intersects all three +sides of the triangle $ABC$ and by assumption does not go through any +of its vertices. This is in contradiction with the consequence of \ref{PaschIzrek} + Pasch's axiom \ref{AksPascheva}, therefore $P$, $Q$ and + $R$ are non-collinear points. + +\item \res{Let $P$ and $Q$ be points of sides $BC$ and $AC$ of the triangle $ABC$ and at the same time +different from its vertices. Prove that the line segments $AP$ and $BQ$ +intersect in one point.} + +From the assumption that the point $P$ lies on the side $BC$ of the triangle + $ABC$, according to the definition of the distance $BC$ it holds $\mathcal{B}(B,P,C)$. + From the axiom \ref{AksII2} it follows that it does not hold + $\mathcal{B}(B,C,P)$, which means that the point $B$ does not lie on + the line $PC$. In the triangle $APC$ the line $BQ$ intersects the side $AC$, + but does not intersect the sides $PC$, so by the consequence \ref{PaschIzrek} + of Pasch's axiom \ref{AksPascheva} the line $BQ$ intersects + the side $AP$ of this triangle. So the line $BQ$ intersects the line + $AP$ in some point $X$ or $BQ\cap [AP]=\{X\}$. Analogously it + is also $[BQ]\cap AP=\{\widehat{X}\}$. We prove that + $X=\widehat{X}$. We assume the opposite, that it holds $X\neq + \widehat{X}$. Because $[AP]\subset AP$ and $[BQ]\subset BQ$, + it holds $X,\widehat{X}\in AP\cap BQ$. By the axiom \ref{AksI1} $AP$ and $BQ$ represent + the same line or the points $A$, $B$, + $P$ and $Q$ lie on the same line $l$. But on this line, which + is determined by the points $B$ and $P$ (axiom \ref{AksI1}), according to the assumption also the point $C$ lies, which means that the points $A$, $B$ + and $C$ are collinear. But this is in contradiction with the assumption that $ABC$ is a triangle, so $X=\widehat{X}$. This means $X\in + [AP]\cap[BQ]=\{X\}$. If the line $[AP]$ and $[BQ]$ had one more + common point, that point would also be the second common point of the lines $AP$ + and $BQ$. We have shown that this is not possible, so + $[AP]\cap[BQ]=\{X\}$. + +\item \res{The points $P$, $Q$ and $R$ lie in order on the sides $BC$, $AC$ and $AB$ of the triangle + $ABC$ and are different from its vertices. Prove that the lines $AP$ +and $QR$ intersect in one point.} + +We use the previous task first for the lines $AP$ and $BQ$ in +the triangle $ABC$, and then once again for the line $AX$ +($\{X\}=[AP]\cap[BQ]$) and $QR$ in the triangle $ABQ$. + + + +\item \res{The line $p$, which lies in the plane of the quadrilateral, intersects its +diagonal $AC$ and does not pass through any vertex of this quadrilateral. +Prove that the line $p$ intersects exactly two sides of this +quadrilateral.} + +We use the consequence \ref{PaschIzrek} of Pasch's axiom + \ref{AksPascheva} twice for the line $p$ and the triangle $ABC$ and $ADC$. + +\item \label{nalAks3}\res{Prove that the midline is a convex figure.} + +Let $X$ and $Y$ be any points of the midline $pA$. It needs to be proven that the whole line segment $XY$ lies within this midline. By definition of the midline from $X,Y\in pA$ it follows that $X,A\ddot{-} p$ and $Y,A\ddot{-} p$. Because $\ddot{-} p$ is an equivalence relation, it is also true that $X,Y\ddot{-} p$. Let $Z$ be any point of the line segment $XY$. We assume that it is not the case that $Z,A\ddot{-} p$ or that $Z,A\div p$ or $Z,X\div p$ (or $Z \in p$) holds. In this case, the line segment $ZX$ intersects the line $p$ at some point $T$. The point $T$ is then the common point of the line $p$ and the line segment $XY$ (according to statement \ref{izrekAksIIDaljica}), which contradicts what has been proven about $X,Y\ddot{-} p$. Therefore, $Z,A\ddot{-} p$ holds, which means that $Z\in pA$, which in turn means that the midline $pA$ is a convex figure. + +\item \label{nalAks4} \res{Prove that the intersection of two convex figures +is a convex figure.} + +Let $\Phi_1$ and $\Phi_2$ be convex figures. We prove that +then $\Phi_1\cap\Phi_2$ is also a convex figure. Let $A, B\in +\Phi_1\cap\Phi_2$ be any points. In this case, $A, B\in +\Phi_1$ and $A,B\in\Phi_2$. Because the figures $\Phi_1$ and $\Phi_2$ +are convex, it follows that $[AB]\subseteq\Phi_1$ and $[AB]\subseteq\Phi_2$ +or $[AB]\subseteq\Phi_1\cap\Phi_2$, which means that $\Phi_1\cap\Phi_2$ is also a convex figure. + +\item \res{Prove that any triangle is a convex figure.} + +The triangle $ABC$ is the intersection of the midlines $ABC$, $ACB$ and +$BCA$, so according to the previous two tasks \ref{nalAks3} and +\ref{nalAks4}, it is a convex figure. + +\item \res{If $\mathcal{B}(A,B,C)$ and $\mathcal{B}(D,A,C)$, then +$\mathcal{B}(B,A,D)$ is also true. Prove it.} + +According to the statement \ref{izrekAksIIPoltrak}, the point $A$ on the line $AB$ determines two poltraks. +We denote with $p$ the poltrak $AC$ and with $p'$ its complementary (supplementary) poltrak. From $\mathcal{B}(A,B,C)$ it follows that $B,C\ddot{-} A$ from $\mathcal{B}(D,A,C)$ we have $D,C\div A$, therefore $B\in p$ and $D\in p'$. So the points $B$ and $D$ not only lie on the same poltrak with the starting point $A$, so there is no $B,D\ddot{-} A$, which means that $B,D\div A$ or $\mathcal{B}(B,A,D)$ applies. + + + +\item \res{Let $A$, $B$, $C$ and $D$ be such collinear points that +$\neg\mathcal{B}(B,A,C)$ and $\neg\mathcal{B}(B,A,D)$. Prove that +$\neg\mathcal{B}(C,A,D)$ applies.} + +From $\neg\mathcal{B}(B,A,C)$ and $\neg\mathcal{B}(B,A,D)$ it follows that $B,C\ddot{-} A$ or $B,D\ddot{-} A$. Because $\ddot{-} A$ is an equivalence relation, it is also transitive, so $C,D\ddot{-} A$ or $\neg\mathcal{B}(C,A,D)$ applies. + +\item \res{Let $A_1A_2\ldots,A_{2k+1}$ be an arbitrary polygon with an even +number of vertices. Prove that there is no line that intersects all +of its sides.} + +Assume the opposite, that some line $p$ intersects all sides +of this polygon. Prove that in this case the points +$A_1,A_3,\ldots,A_{2k+1}$ would be on the same side of the line $p$. + +\item \label{nalAks11}\res{If the isometry $\mathcal{I}$ maps the figure $\Phi_1$ and $\Phi_2$ +into the figure $\Phi'_1$ and $\Phi'_2$, then the intersection +$\Phi_1\cap\Phi_2$ with this isometry is mapped into the intersection +$\Phi'_1\cap\Phi'_2$. Prove it.} + +If for an arbitrary point $X$ it holds that $X\in \Phi_1\cap\Phi_2$, then $X\in \Phi_1$ and $X\in\Phi_2$. From this it follows that $\mathcal{I}(X)\in \mathcal{I}(\Phi_1)$ and $\mathcal{I}(X)\in \mathcal{I}(\Phi_2)$ or $X'\in \Phi'_1$ and $X'\in \Phi'_2$ (where $X'=\mathcal{I}(X)$). So it holds that $X'\in\Phi'_1\cap\Phi'_2$. In this way we have proven $\mathcal{I}(\Phi_1\cap\Phi_2)\subseteq \Phi'_1\cap\Phi'_2$. The other inclusion $\mathcal{I}(\Phi_1\cap\Phi_2)\supseteq \Phi'_1\cap\Phi'_2$ is proven in the same way using the isometry $\mathcal{I}^{-1}$. + +\item \res{Prove that any two +lines of a plane are parallel to each other.} + +We use axiom \ref{aksIII2} and theorem \ref{izrekIzoB}. + +\item \res{Prove that any two +lines of a plane are concurrent to each other.} + +We use axiom \ref{aksIII2} and \ref{aksIII1}. + + +\item \res{Let $k$ and $k'$ be two circles +of a plane with centers $O$ and $O'$ and radii $AB$ and $A'B'$. +Prove the equivalence: $k\cong k' \Leftrightarrow AB\cong A'B'$.} + +($\Leftarrow$) Let $AB\cong A'B'$. We mark with $P$ and $P'$ any points of the circles $k$ and $k'$. Because $OP$ and $O'P'$ are radii of these circles, it holds that $OP\cong O'P'$. By theorem \ref{izrekAB} there exists an isometry $\mathcal{I}$, which maps the points $O$ and $P$ to the points $O'$ and $P'$. Let $X$ be an arbitrary point of the circle $k$ and $X'=\mathcal{I}(X)$. From $\mathcal{I}:O,X,P\rightarrow O',X',P'$ it follows that $O'X'\cong OX$. Because $X,P\in k$, it also holds that $OX\cong OP$. From the previous relations it follows that $O'X'\cong O'P'$ or $X'\in k'$. So it holds that $\mathcal{I}(k)\subseteq k'$. Similarly $\mathcal{I}(k)\supseteq k'$ (we use $\mathcal{I}^{-1}$), so $\mathcal{I}(k)= k'$. + +($\Rightarrow$) Let $k\cong k'$. We assume the opposite - that $AB\not\cong A'B'$. Without loss of generality, let $AB>A'B'$. + From $k\cong k'$ it follows that there exists an isometry $\mathcal{I}$, which maps the circle $k$ to the circle $k'$. Let $PQ$ be any diameter of the circle $k$. By \ref{premerInS} we have $PQ=2\cdot AB$. Let $P'=\mathcal{I}(P)$ and $Q'=\mathcal{I}(Q)$. Therefore $P'Q'$ is the diameter of the circle, for which $P'Q'\cong PQ =2\cdot AB>2\cdot A'B'$. But the relation $P'Q'>2\cdot A'B'$ is not possible (\ref{premerNajdTetiva}), so $AB\cong A'B'$. + +\item \res{Let $\mathcal{I}$ +be a non-identical isometry of a plane with two fixed points $A$ and +$B$. Let $p$ be a line of this plane, which is parallel to the line +$AB$, and $A\notin p$. Prove that there are no fixed points of the +isometry $\mathcal{I}$ on the line $p$.} + +We assume the opposite - that there exists a fixed point $C$ of the isometry $\mathcal{I}$, which lies on the line $p$. The point $C$ does not lie on the line $AB$, because in the opposite case $AB$ and $p$ would be parallel, which is in contradiction with the assumption $A\notin p$. So we have three non-collinear fixed points of the isometry $\mathcal{I}$, so by \ref{IizrekABC2} $\mathcal{I}$ is identical. This is in contradiction with the assumption, so there are no fixed points of the isometry $\mathcal{I}$ on the line $p$. + +\item \res{ Let $S$ be the only fixed +point of the isometry $\mathcal{I}$ in some plane. Prove that if this +isometry maps the line $p$ to itself, then $S\in p$.} + +We assume that $S\notin p$. We mark with $N$ the orthogonal projection of the point $S$ on the line $p$ and $N'=\mathcal{I}(N)$. From $S\notin p$ it follows that $N\neq S$. Because $S$ is the only fixed point of this isometry, $N'\neq N$. By the assumption $\mathcal{I}:p\rightarrow p$ we also have $N'\in p$. From $\mathcal{I}:SN,p\rightarrow SN', p$ it follows that $\angle SN',p=\angle SN,p=90^0$. In this case, there would be two different perpendiculars from the point $S$ on the line $p$, which by \ref{enaSamaPravokotnica} is not possible. Therefore the assumption $S\notin p$ is false, which means that $S\in p$. + +\item \label{nalAks17}\res{Prove that any two lines of +a plane either intersect or are parallel. } + +By Axiom \ref{AksI1}, two different lines $p$ and $q$ of a plane have at most one common point. If they have one common point, they intersect by definition, if they don't have any common points, they are parallel by definition. + +\item \label{nalAks18}\res{If a +line in a plane intersects one of two parallel lines of the same plane, + then it also intersects the other parallel line. Prove.} + +Let $p\parallel q$ and $l\cap p =\{A\}$. If $p=q$, then it is clear that $l\cap q =\{A\}$. The other possibility is $p\cap q =\emptyset$. Then, by Playfair's Axiom, $l\cap q \neq\emptyset$. Because it can't be that $l=q$, then by the previous task \ref{nalAks17} the lines $l$ and $q$ intersect. + +\item \res{Prove that every isometry maps parallel lines to parallel lines.} + +Use task \ref{nalAks11}. + +\item \res{Let $p$, $q$ and $r$ be such lines of a plane that +$p\parallel q$ and $r\perp p$. Prove that $r\perp q$.} + +The statement is a direct consequence of task \ref{nalAks18} and Theorem \ref{KotiTransverzala}. + +\item \res{Prove that a convex $n$-gon can't have more than three +acute angles.} + +Use the fact that the sum of the exterior angles of a convex $n$-gon +is equal to $360^0$ (Theorem \ref{VsotKotVeckZuna}). + + +\end{enumerate} + + +%REŠITVE - Skladnost trikotnikov +%________________________________________________________________________________ + +\poglavje{Congruence. Triangles and Polygons} + +\begin{enumerate} + + + \item \res{Let $S$ be a point that lies in the angle $pOq$, and let $A$ and $B$ be the orthogonal projections of point $S$ onto the sides $p$ and $q$ of this +angle. Prove that $SA\cong SB$ if and only if the line +$OS$ is the angle bisector of the angle $pOq$.} + +Prove the congruence of triangles $OSA$ and +$OSB$ in both directions of equivalence. In the proof for ($\Rightarrow$) use Theorem \textit{SSA} +\ref{SSK}, in the proof for ($\Leftarrow$) use Theorem \textit{ASA} +\ref{KSK}. + +\item \res{Prove that the sum of the diagonals of a convex quadrilateral is greater than +the sum of its opposite sides.} + +We use the triangle inequality for the triangles $ASB$ and $CSD$, where +$S$ is the intersection of the diagonals $AC$ and $BD$ of the convex quadrilateral +$ABCD$. + + \item \res{Prove that in every triangle + no more than one side is shorter than the corresponding altitude.} + + We mark with $a$, $b$ and $c$ the sides and with $v_a$, $v_b$, $v_c$ + the corresponding altitudes of the triangle $ABC$. We assume the opposite, that + for example $a \frac{3}{4}(a + b + c)$ + \end{enumerate}} + + We use example \ref{neenTezisZgl}. + +\item \res{Let $p$ be a line that is parallel to the circle $k$. Prove that all points of this circle are on the same side of the line $p$.} + +We assume the opposite, that points $X,Y\in k$ are on different sides of the line $p$. In this case, the segment $XY$ intersects the line $p$ in some point $N$, which, according to \ref{tetivaNotrTocke}, is an interior point of the circle $k$. The line $p$, which contains the interior point $N$ of the circle $k$, is, according to \ref{DedPoslKrozPrem}, its secant, which contradicts the assumption that $p$ is parallel. + +\item \res{If the circle $k$ lies in some convex figure $\Phi$, then also the circle, which is determined by this circle, lies in this figure. Prove it.} + +It is enough to prove that any interior point $X$ of the circle $k$ lies in the figure $\Phi$. Let $AB$ be any segment that contains the point $X$ (the existence of such a segment follows from \ref{DedPoslKrozPrem}). Because $\Phi$ is a convex figure, from $A,B\in k\subset \Phi$ it follows that $X\in [AB]\subset \Phi$. + +\item \res{Let $p$ and $q$ be two different tangents to the circle $k$, which touch it in points $P$ and $Q$. Prove the equivalence: $p \parallel q$ if and only if $AB$ is the diameter of the circle $k$.} + +We use the fact $SP\perp p$ and $SQ\perp q$, where $S$ is the center of the circle $k$. + +\item \res{If $AB$ is a segment of the circle $k$, then the intersection of the line $AB$ and the circle, which is determined by the circle $k$, is equal to this segment. Prove it.} + +We denote by $\mathcal{K}$ the mentioned circle. It is necessary to prove that $[AB]=AB\cap \mathcal{K}$. The inclusion $[AB]\subseteq AB\cap \mathcal{K}$ follows directly from the axiom \ref{AksII1} and \ref{tetivaNotrTocke}. We prove the inclusion $AB\cap \mathcal{K}\subseteq [AB]$. Let $X\in AB\cap \mathcal{K}$. We assume that $X\notin [AB]$. In this case, it would be $\mathcal{B}(X,A,B)$ or $\mathcal{B}(A,B,X)$. It is not difficult to prove that in each of these two cases we would get $SX>SA$, which is not possible, because $X\in \mathcal{K}$. Therefore, $X\in[AB]$, i.e. $AB\cap \mathcal{K}\subseteq [AB]$. + +\item \res{Let $S'$ be the orthogonal projection of the center $S$ of the circle $k$ onto the line $p$. Prove that $S'$ is an external point of this circle exactly when the line $p$ does not intersect the circle.} + +We use the statement \ref{TangSekMimobKrit}. + +\item \res{Let $V$ be the altitude point of the triangle $ABC$, for which $CV \cong AB$. Determine the size of the angle $ACB$.} + +We mark with $A'$, $B'$ and $C'$ the points of the altitudes from the vertices $A$, $B$ and $C$ of the triangle $ABC$. First, the angles $C'CB$ and $A'AB$ are congruent (the angle with perpendicular legs - statement \ref{KotaPravokKraki}). From the congruence of the triangles $CVA'$ and $ABA'$ (statement \textit{ASA} \ref{KSK}) it follows that $CA'\cong AA'$. This means that $CAA'$ is an isosceles right triangle with hypotenuse $AC$, so according to statement \ref{enakokraki} $\angle ACB=\angle ACA'=45^0$. + +\item \res{Let $CC'$ be the altitude of the right triangle $ABC$ ($\angle ACB = 90^0$). If $O$ and $S$ are the centers of the inscribed circles of the triangles $ACC'$ and $BCC'$, the internal angle bisector of $ACB$ is perpendicular to the line $OS$. Prove it.} + +Let $I$ be the center of the inscribed circle of the triangle $ABC$. We prove that $I$ is the altitude point of the triangle $COS$. + +\item \res{Let $ABC$ be a triangle in which $\angle ABC = 15^0$ and $\angle ACB = 30^0$. Let $D$ be a point on the side $BC$ such that $\angle BAD=90^0$. Prove that $BD = 2AC$.} + +We mark with $S$ the center of the segment $BD$. The point $S$ is the center of the inscribed circle of the right triangle $BAD$ (statement \ref{TalesovIzrKroz2}). Therefore $SA\cong SB\cong SD$. Because $BSA$ is an isosceles triangle with the base $AB$, according to statement \ref{enakokraki} $\angle BAS\cong \angle ABS=15^0$. For the external angle $ASD$ of the triangle $BSA$ it then holds that $\angle ASD=\angle BAS + \angle ABS=30^0$ (statement \ref{zunanjiNotrNotr}). Therefore $\angle ASC=30^0=\angle ACS$, so $ASC$ is an isosceles triangle with the base $SC$ or $AS\cong AC$ (statement \ref{enakokraki}). From this it follows that $BD = 2SB=2AS=2AC$. + +\item \res{Prove that there exists a pentagon that can tile the plane.} + +We can choose a pentagon $ABCDE$, so that $ABCE$ is a square, $ECD$ is an isosceles right triangle with the base $CE$, and $A,D\div CE$. + +\item \res{Prove that there exists a decagon, with which it is possible to tile a plane.} + +We make the union of two appropriate regular hexagons - the cells of tiling $(6,3)$. + +\item \res{In some plane each point is painted red or black. Prove that there exists a right triangle, which has all vertices of the same color.} + +We use the tiling $(3,6)$. + +\item \res{Let $l_1,l_2,\ldots, l_n$ ($n > 3$) be arcs, which all lie on the same circle. The central angle of each arc is at most $180^0$. Prove that if each triple of arcs has at least one common point, there exists a point, which lies on each arc.} + +We denote with $k(S,r)$ the given circle. Let $I_i$ ($i\in \{1,2,\ldots,n\}$) be the circular segments, which are determined by the arcs $l_k$, and $J_i=I_i\setminus\{S\}$. Because for each $l_i$ the central angle is at most $180^0$, the sets $J_i$ are convex figures (prove it). By the assumption each triple of arcs $l_{i_1}$, $l_{i_2}$ and $l_{i_2}$ ($i_1,i_2,i_3\in\{1,2,\ldots,n\}$) has a common point $X_{i_1i_2i_3}$. This point also belongs to each of the figures $J_{i_1}$, $J_{i_2}$ and $J_{i_2}$. By Helly's theorem \ref{Helly} there exists a point $Y$, which belongs to each of the figures $J_1,J_2,\ldots, J_n$. If we denote with $X$ the intersection of the line segment $SY$ and the circle $k$, it follows that also the open line segment $(SX]$ belongs to each of these figures. Because for each $i$ it holds $l_i=J_i\cap k$, the point $X$ belongs to each of the arcs $l_1,l_2,\ldots, l_n$. + +\item +\res{Let $p$ and $q$ be rectangles that intersect at point $A$. If +$B, B'\in p$, $C, C'\in q$, $AB\cong AC'$, $AB'\cong AC$, +$\mathcal{B}(B,A,B')$ and $\mathcal{B}(C,A,C')$, then the rectangle +on the line $BC$ through point $A$ goes through the center of the +line $B'C'$. Prove it.} + +First, by the \textit{SAS} theorem \ref{SKS}, the triangles $BAC$ +and $C'AB'$ are congruent. From this it follows: $\angle +AC'B'\cong\angle ABC=\beta$ and $\angle AB'C'\cong\angle +ACB=90^0-\beta$. We mark with $S$ the center of the line $B'C'$, +with $P$ the intersection of the lines $BC$ and $AS$. By theorem +\ref{TalesovIzrKroz2}, $SA\cong SB'\cong SC'$, so $\angle +CAP=\angle C'AS\cong\angle AC'S=\angle AC'B'=\beta$ (theorem +\ref{enakokraki}). Because $\angle ACP=\angle ACB=90^0-\beta$, +from triangle $ACP$ by theorem \ref{VsotKotTrik} $\angle +APC=90^0$ or $AS\perp BC$. This means that the rectangle $AP$ of +the line $BC$ goes through the center of the line $B'C'$. + +\item +\res{Prove that the internal angle bisectors of a rectangle that is +not a square intersect at points that are the vertices of a +square.} + +We mark with $s_{\alpha}$, $s_{\beta}$, $s_{\gamma}$ and +$s_{\delta}$ the internal angle bisectors at the vertices $A$, $B$, +$C$ and $D$ of the rectangle $ABCD$ and $P=s_{\alpha}\cap +s_{\beta}$, $Q=s_{\gamma}\cap s_{\delta}$, $L=s_{\beta}\cap +s_{\gamma}$ and $K=s_{\alpha}\cap s_{\delta}$. We prove that +$PKQL$ is a square. First, from $\angle PAB=45^0$ and $\angle +PBA=45^0$ it follows that $ABP$ is an isosceles right triangle +with the base $AB$ (theorem \ref{enakokraki} and \ref{VsotKotTrik}). +Therefore $\angle APB=90^0$ and $AP\cong BP$. Analogously, all +the internal angles of the quadrilateral $PKQL$ are right angles. +It is enough to prove that $PK\cong PL$. From the congruence of +the triangles $AKD$ and $BLC$ (theorem \ref{KSK}) it follows that +$AK\cong BL$. If we connect this with the already proven $AP\cong +BP$, we get $PK\cong PL$. + +\item + \res{Prove that the internal angle bisectors of a parallelogram that is not a rhombus intersect at the vertices of a rectangle. Prove also that the diagonals of this rectangle are parallel to the sides of the parallelogram and are equal to the difference of the adjacent sides of the parallelogram.} + +As in the previous task, we denote by $s_{\alpha}$, +$s_{\beta}$, $s_{\gamma}$ and $s_{\delta}$ the internal angle +bisectors at the vertices $A$, $B$, $C$ and $D$ of the rectangle +$ABCD$ and $P=s_{\alpha}\cap s_{\beta}$, $Q=s_{\gamma}\cap +s_{\delta}$, $L=s_{\beta}\cap s_{\gamma}$ and $K=s_{\alpha}\cap +s_{\delta}$. We prove that $PKQL$ is a rectangle. Let $E$ and $F$ +be the midpoints of the sides $AD$ and $BC$ of the parallelogram +$ABCD$. In the triangle $PAB$ we have + $\angle APB=180^0-\angle PAB-\angle +PBA=180^0-\frac{1}{2}\left(\angle DAB+\angle CBA +\right)=180^0-\frac{1}{2}\cdot 180^0=90^0$ (by the + \ref{VsotKotTrik} and \ref{paralelogram}). Similarly, all the other internal angles of the quadrilateral $PKQL$ are right angles. By the + \ref{TalesovIzrKroz2} the point $E$ is the center of the inscribed circle + of the right triangle $AKD$ with hypotenuse $AD$ and we have + $EK\cong EA$. Therefore $\angle EKA\cong \angle EAK \cong\angle + KAB$ (by the \ref{enakokraki}) or, by the + \ref{KotiTransverzala}, $EK\parallel AB$. Similarly, $FL\parallel AB$. Since also $EF\parallel AB$ (by the \ref{srednjTrapez}) we have + $KL\parallel AB$. Without loss of generality, we assume that $AB>AD$. We denote by $T$ the intersection of the bisector $s_{\alpha}$ with the side $CD$. By the \ref{KotiTransverzala} + we have $\angle DTA\cong\angle TAB\cong\angle DAT$. This means that + $ADT$ is an isosceles triangle, therefore $AD\cong DT$ (by the + \ref{enakokraki}). Since $KL\parallel AB\parallel CT$ and + $s_{\alpha}\parallel s_{\gamma}$, the quadrilateral $KLCT$ + is a parallelogram. Therefore $PQ\cong KL\cong CT=CD-DT=CD-AB$. + + +\item +\res{Prove that the angle bisectors of two adjacent angles are perpendicular to each other.} + +The angle bisectors determine the angle that is equal to half of the corresponding extended angle. + +\item \res{Let $B'$ and $C'$ be the altitudes of the triangle $ABC$ from the vertices $B$ and $C$. Prove that $AB\cong AC \Leftrightarrow BB'\cong CC'$.} + +Prove the equivalence in both directions. In the direction ($\Rightarrow$) use the \textit{ASA} theorem \ref{KSK}, in the direction ($\Leftarrow$) use the \textit{SSA} theorem \ref{SSK}. + +\item \res{Prove that a triangle is right if and only if the center of the circle drawn through the triangle and its altitude point coincide. Does a similar statement hold for any two characteristic points of this triangle?} + +Use the fact that the sides of the triangle are perpendicular to the altitude in this case. In a similar way, one could prove that a similar statement holds for any two characteristic points of this triangle. + +\item \res{Prove that the obtuse triangles $ABC$ and $A'B'C'$ are congruent if and only if they have congruent altitudes $CD$ and $C'D'$, sides $AB$ and $A'B'$, and angles $ACD$ and $A'C'D'$.} + +From the congruence of the triangles $ABC$ and $A'B'C'$ it follows directly first that $AB\cong A'B'$, $AC\cong A'C'$, and $\angle BAC\cong \angle B'A'C'$, then that $\triangle ACD\cong\triangle A'C'D'$ (by the \textit{ASA} theorem \ref{KSK}) or $CD\cong C'D'$, and $\angle ACD\cong \angle A'C'D'$. + +Assume that $CD\cong C'D'$, $AB\cong A'B'$, and $\angle ACD\cong \angle A'C'D'$. From the congruence of the triangles $ACD$ and $A'C'D'$ (by the \textit{ASA} theorem \ref{KSK}) it follows that there exists an isometry $\mathcal{I}$, for which $\mathcal{I}:A,C,D \mapsto A',C',D'$ (by theorem \ref{IizrekABC}. This maps the segment $AD$ onto the segment $A'D'$. Because $B$ and $B'$ lie on the segments $AD$ and $A'D'$, and $AB\cong A'B'$, also $\mathcal{I}(B)=B'$. Therefore $\mathcal{I}:A,B,C \mapsto A',B',C'$, which means that $\triangle ABC\cong\triangle A'B'C'$. + +\item \res{If $ABCD$ is a rectangle and $AQB$ and $APD$ are right triangles with the same orientation, then the segment $PQ$ is congruent to the diagonal of this rectangle. Prove it.} + +We prove that $PAQ$ and $DAB$ are congruent triangles. + +\item \res{Let $BB'$ and $CC'$ be the altitudes of the triangle $ABC$ ($AC>AB$) and + $D$ be such a point on the line segment $AB$, that $AD\cong AC$. The point +$E$ is the intersection of the line $BB'$ with the line that goes through the point $D$ and is +parallel to the line $AC$. Prove that $BE=CC'-BB'$.} + +Let $F$ be the intersection of the line $ED$ with the line that is in the point $B'$ +parallel to the line $AB$. The quadrilateral $ADFB'$ is a parallelogram, +so $FB'\cong DA$ and $\angle DFB' \cong\angle DAB'$ (by the statement +\ref{paralelogram}). Because according to the assumption $AD\cong AC$, +it also holds that $FB'\cong AC$. This means that the right-angled +triangles $FEB'$ and $AC'C$ are similar (by the \textit{ASA} statement \ref{KSK}), so +$EB'\cong C'C$ or $BE=EB'-BB'=CC'-BB'$. + +\item \res{Let $ABCD$ be a convex quadrilateral, for which + $AB\cong BC\cong CD$ and $AC\perp BD$. Prove that $ABCD$ + is a rhombus.} + + Because $ABCD$ is a convex quadrilateral, its diagonals + intersect in some point $S$. Prove $\triangle ABS\cong\triangle CBS$ and + $\triangle CBS\cong\triangle CDS$. + +\item \res{Let $BC$ be the base of the isosceles triangle $ABC$. If $K$ and +$L$ are such points, that $\mathcal{B}(A,K,B)$, $\mathcal{B}(A,C,L)$ and $KB\cong LC$, then +the center of the line segment $KL$ lies on the base $BC$. Prove.} + +Let $O$ be the center of the line $KL$. We mark with $M$ the fourth vertex of the parallelogram $CKBM$, the common center of their diagonals $BC$ and $KM$ (statement \ref{paralelogram}) with $S$. If we use the fact that the triangles $ABC$ and $MLC$ are of the same base with the bases $BC$ and $ML$, from the statement \ref{enakokraki} we get $\angle ABC\cong\angle ACB$ and $\angle CML\cong\angle CLM$. From the congruence of the triangles $BKC$ and $CMB$ (statement \ref{paralelogram} and \textit{SSS} \ref{SSS}) it follows that $\angle ABC=\angle KBC\cong\angle MCB$. As $ACM$ is the external angle of the triangle $MLC$, then by the statement \ref{zunanjiNotrNotr} $\angle ACM =\angle CML+\angle CLM=2\angle CML$. Therefore $\angle CML=\frac{1}{2}\angle ACM=\frac{1}{2}\left(\angle ACB+\angle BCM\right)\frac{1}{2}\left(\angle ACB+\angle ABC\right)=\angle ABC$. This means that the lines $ML$ and $BC$ are parallel (statement \ref{KotiTransverzala}). The line $OS$ is the median of the triangle $KML$ with the base $ML$, therefore by the statement \ref{srednjicaTrik} $SO\parallel MN$. By Playfair's axiom \ref{Playfair} the lines $SO$ and $BC$ are the same (coincide), therefore the point $O$ lies on the line $BC$. By Pasch's axiom \ref{AksPascheva} for the triangle $ABC$ and the line $KL$ the point $O$ lies on the side $BC$. + +\item \res{Let $S$ be the center of the triangle $ABC$ of the inscribed circle. The line passing through the point $S$ and parallel to the side $BC$ of this triangle intersects the sides $AB$ and $AC$ in order. Prove that $BM+NC=NM$.} + +We prove that the triangles $BSM$ and $SCN$ are of the same base with the bases $BS$ and $SC$. + +\item \res{Let $ABCDEFG$ be a convex heptagon. Calculate the sum of the convex angles determined by the broken line $ACEGBDFA$.} + +We use the fact that the double sum of the external angles of the heptagon determined by the intersections of the diagonals of the heptagon $ABCDEFG$ is equal to $720^0$. Result: $540^0$. + +\item \label{nalSkl34} +\res{Prove that the centers of the sides and the vertex of any altitude of a triangle in which no two sides are congruent are the vertices of a parallelogram.} + +We denote with $A_1$, $B_1$ and $C_1$ the centers of sides $BC$, $CA$ and +$BA$ of the triangle $ABC$ and with $A'$ the intersection point of the altitude of this triangle with +the point $A$. From $|AB|\neq |AC|$ it follows that $A'\neq A_1$. We prove that +the quadrilateral $A'A_1B_1C_1$ is a parallelogram. The distance $B_1C_1$ +is the median of the triangle $ABC$ with the base $BC$, therefore by +the statement \ref{srednjicaTrik} $B_1C_1\parallel BC$. Because $A',A_1\in +BC$, also $B_1C_1\parallel A',A_1$. So the quadrilateral $A'A_1B_1C_1$ is a parallelogram. We prove also that it is a rhombus, or +$C_1A'\cong B_1A_1$. But this fact follows from +the statements \ref{TalesovIzrKroz2} and \ref{srednjicaTrik}, because: +$C_1A'=\frac{1}{2}AB= B_1A_1$. + +We mention also that by using this statement we can another way prove the statement about Euler's circle \ref{EulerKroznica}. + + \item \res{Let $ABC$ be a right angled triangle with the right angle at the point $C$. +The points $E$ and $F$ are the intersections of the internal angle bisectors at +the points $A$ and $B$ with the opposite sides, $K$ and $L$ are +the orthogonal projections of the points $E$ and $F$ on the hypotenuse of this +triangle. Prove that $\angle LCK=45^0$.} + +Let $\alpha$ and $\beta$ be the inner angles at the vertices $A$ and +$B$ of the triangle $ABC$. According to the theorem \ref{VsotKotTrik}, +$\alpha+\beta=90^0$. According to the theorem \textit{ASA} \ref{KSK}, +$\triangle ACE\cong\triangle AKE$ and $\triangle BCF\cong\triangle BLF$, therefore +$EC\cong EK$ and $FC\cong FL$. So the triangles $CEK$ and $CFL$ + are congruent with the bases $CK$ and $CL$, which means that $\angle ECK\cong\angle EKC$ and +$\angle FCL\cong\angle FLC$. If +we use the theorem \ref{zunanjiNotrNotr} for the triangle $CFL$ and +the theorem \ref{VsotKotTrik} for the triangle $ALF$, we get: $\angle +FCL=\frac{1}{2}\angle LFA=\frac{1}{2}\left(90^0-\alpha \right)$. +Similarly, $\angle ECK=\frac{1}{2}\left(90^0-\beta +\right)$. From this it follows that $\angle FCL+\angle +ECK=\frac{1}{2}\left(90^0-\alpha +90^0-\beta\right)=45^0$. So +$\angle LCK=90^0-(\angle FCL+\angle ECK)=90^0-45^0=45^0$. + +\item \res{Let $M$ be the center of the side $CD$ of the square $ABCD$ and $P$ such a point + of the diagonal $AC$, that $3AP=PC$. Prove that $\angle BPM$ +is a right angle.} + +Let $S$ be the center of the diagonal $AC$ and $V$ the center of the line $SB$. Prove that $V$ + is the altitude of the triangle $PBC$ (see also the example +\ref{zgledPravokotnik}). + + \item \res{Let $P$, $Q$ and $R$ be the centers of the sides $AB$, $BC$ and $CD$ + of the parallelogram $ABCD$. The lines $DP$ and $BR$ intersect the line +$AQ$ in the points $K$ and $L$. Prove that $KL= \frac{2}{5} AQ$.} + +Let $S$ be the center of the line $AD$ and $M$ the intersection of the lines $SC$ +and $BR$. Prove first that $CM\cong AK$, the line $LQ$ +is the median of the triangle $CBM$ and the line $PK$ is the median of the triangle +$LAB$. + + \item \res{Let $D$ be the center of the hypotenuse $AB$ of the right-angled +triangle $ABC$ ($AC>BC$). The points $E$ and $F$ are the intersections of +the altitudes $CA$ and $CB$ with a line through $D$ and perpendicular +to the line $CD$. The point $M$ is the center of the line $EF$. Prove that +$CM\perp AB$.} + +Let $T$ be the intersection of the lines $AB$ and $CM$, and $\angle +CAB=\alpha$ and $\angle CBA=\beta$. By assumption, +$\alpha+\beta=180^0$. Because $D$ is the center of the hypotenuse $AB$ +of the right triangle $ABC$, by Tales' theorem +\ref{TalesovIzrKroz2} $DC\cong DA$. This means that $\triangle +CDA$ is an isosceles triangle and $\angle DCE=\angle +DCA\cong\angle DAC=\alpha$. From $FE\perp CD$ and $FC\perp CE$ +it follows that $\angle CFE\cong\angle DCE=\alpha$ (by +\ref{KotaPravokKraki}). The point $M$ is the center of the hypotenuse $FE$ +of the right triangle $FCE$, so (similarly to the triangle $ABC$) $\angle FCM\cong\angle CFM=\angle CFE=\alpha$. +In the triangle $BCT$ the sum of the two internal angles $\angle +CBT+\angle BCT=\angle CBA+\angle FCM=\alpha+\beta=90^0$. By +\ref{VsotKotTrik} it follows that $\angle CTB=90^0$ or +$CM\perp AB$. + +\item \res{Let $A_1$ and $C_1$ be the centers of the sides $BC$ and $AB$ of the triangle $ABC$. + The perpendicular to the internal angle at the vertex $A$ intersects the line +$A_1C_1$ at the point $P$. Prove that $\angle APB$ is a right angle.} + +By \ref{KotiTransverzala}, $\angle C_1PA\cong\angle PAC$. Therefore $\angle C_1PA\cong\angle PAC\cong\angle PAC_1$, which +means that $PAC_1$ is an isosceles triangle with the base $AP$ or $C_1A\cong C_1P$ (by \ref{enakokraki}). Because $C_1$ is the center +of the side $AB$, $C_1B\cong C_1A\cong C_1P$. Therefore +the point $P$ lies on the circle with the diameter $AB$, so by Tales' theorem +\ref{TalesovIzrKroz2} $\angle APB=90^0$. + + \item \res{Let $P$ and $Q$ be such points of the sides $BC$ and $CD$ of the square $ABCD$, + that the line $PA$ is the perpendicular to the angle $BPQ$. Determine the size of the angle + $PAQ$.} + + Let $A'=pr_{\perp PQ}(A)$. Prove first that $\triangle + ABP\cong\triangle AA'P$ and $\triangle ADQ\cong\triangle AA'Q$. + Result: $\angle PAQ=45^0$. + +\item \res{Prove that the center of the circumscribed circle lies closest to the longest side +of the triangle.} + +Let $O$ be the center of the inscribed circle of the triangle $ABC$ and $A_1$, +$B_1$ and $C_1$ be the centers of its sides $BC$, $AC$ and $AB$. +It is enough to prove, for example, that from $AC>AB$ it follows that $OB_1AB$. By the theorem \ref{vecstrveckot} in this case $\angle ABC>\angle ACB$. Because $B_1C_1$ is the median of the triangle $ABC$, $B_1C_1\parallel BC$ (theorem +\ref{srednjicaTrik}). Therefore, by the theorem \ref{KotiTransverzala} +$\angle AC_1B_1\cong \angle ABC$ and $\angle AB_1C_1\cong \angle +ACB$. From $\angle ABC>\angle ACB$ it now follows: $\angle +B_1C_1O=90^0-\angle AC_1B_1= 90^0-\angle ABC<90^0-\angle +ACB=90^0-\angle AB_1C_1=\angle C_1B_1O$ or $\angle +B_1C_1O<\angle C_1B_1O$. By the theorem \ref{vecstrveckot} for +the triangle $B_1OC_1$ it is then $OB_1\gamma$ it follows that +$SB\gamma$ it follows first +of all that $\angle +SBP=\frac{1}{2}\beta>\frac{1}{2}\gamma=\angle SCP$. Let $B'$ be +the point on the segment $PC$, for which $PB'\cong PB$. From +the similarity of the triangles $SPB'$ and $SPB$ (the +\textit{SAS} theorem \ref{SKS}) it follows that $SB'\cong SB$ +and $\angle SB'P\cong \angle SBP>\angle SCP$. We prove that +$\mathcal{B}(P,B',C)$ or $PB'\angle SB'P\cong\angle SBP$ - the +\ref{zunanjiNotrNotrVecji} theorem for the triangle $SCB'$). +Since $\mathcal{B}(P,B',C)$, it follows that $\angle +SB'C=180^0-\angle SB'P=180^0-\frac{1}{2}\beta>90^0$. In the +right-angled triangle $SB'C$ it then follows that $SC\gamma$) we mark the central angles. Because according to the statement \ref{TangPogoj} $\angle ST_1P=\angle ST_2P=90^0$ we get (from the quadrilateral $PT_1ST_2$ - statement \ref{VsotKotVeck}): $\angle T_1PT_2=180^0-\gamma= + \frac{\beta+\gamma}{2}-\gamma= + \frac{\beta-\gamma}{2}$. + +\item \res{Let $L$ be the orthogonal projection of any point $K$ of the circle $k$ + on its tangent $t$ through the point $T\in k$ and $X$ the point that is +symmetrical to the point $L$ with respect to the line $KT$. Determine the geometric +position of the points $X$.} + +Let's mark the center of the circle $k$ with $S$ and $\alpha=\angle LTK$. We will first prove that $S$, $X$ and $K$ are collinear points. From the congruence of the triangles $TLK$ and $TXK$ it follows that $\angle XTK\cong \angle LTK=\alpha$, $\angle TXK\cong \angle TLK=90^0$ and $\angle TKX\cong \angle TKL=90^0-\alpha$. According to the formulas \ref{ObodKotTang} and \ref{SredObodKot} $\angle TSK=2\angle LTK=2\alpha$. Therefore, from the isosceles triangle $TSK$ (formula \ref{enakokraki} and \ref{VsotKotTrik}) we get: $\angle TKS=\frac{180^0-2\alpha}{2}=90^0-\alpha=\angle TKX$ or $\angle TKS\cong\angle TKX$, which means that $X\in SK$. In the end $\angle SXT\cong\angle KXT=90^0$, which according to the formula \ref{TalesovIzrKroz2} means that the point $X$ lies on the circle $l$ with the diameter $TS$. It is not difficult to prove that the converse is also true - each point $X\in l$ can be obtained from some point $K\in k$ according to the described procedure. + +\item \res{Let $BB'$ and $CC'$ be the altitudes of the triangle $ABC$ and $t$ +the tangent of the circumscribed circle of this triangle at point $A$. Prove that +$B'C'\parallel t$.} + +Let's mark any point of the tangent $t$ with $P$, so that $P$ and $C$ are on different sides of this tangent. According to the formula \ref{ObodKotTang} $\angle PAC'=\angle PAB\cong\angle ACB$. Because $BCC'B'$ is a trapezoid (formula \ref{TalesovIzrKroz2}), according to the formula \ref{TetivniPogojZunanji} $\angle AC'B'\cong\angle ACB$. Therefore $\angle AC'B'\cong\angle PAC'$, which according to the formula \ref{KotiTransverzala} means that $PA\parallel B'C'$ or $B'C'\parallel t$. + +\item \res{In the right triangle $ABC$ above the leg $AC$ with the angle as the diameter +the circle is drawn, which intersects the hypotenuse $AB$ at point $E$. The tangent of this +circle at point $E$ intersects the other leg $BC$ at point $D$. Prove that $BDE$ +is an isosceles triangle.} + +If we use the formula \ref{ObodKotTang}, we get $\angle DEC\cong\angle EAC$. Therefore: +$\angle BED=90^0-\angle DEC=90^0-\angle EAC=90^0-\angle BAC=\angle ABC=\angle EBD$. From +$\angle BED\cong\angle EBD$ according to the formula \ref{enakokraki} it follows that $BD\cong ED$, which means that $BDE$ is an isosceles triangle. + +\item \res{In a right angle with the top $A$ is inscribed a circle that touches the sides of this angle in +the points $B$ and $C$. Any tangent to this circle intersects the line $AB$ and $AC$ in succession in the points $M$ and $N$ (so that $\mathcal{B}(A,M,B)$). Prove that: +$$\frac{1}{3}\left(|AB|+|AC|\right) < |MB|+|NC| < +\frac{1}{2}\left(|AB|+|AC|\right).$$} + +We denote with $T$ the point of intersection of the aforementioned circle with the line $MN$ and $s=\frac{1}{2}(|MN|+|NA|+|AM|)$ the perimeter of the triangle $MAN$. The given circle is the inscribed circle of the triangle $MAN$, so according to the great theorem it holds: +\begin{eqnarray} \label{nalSkk8Eqn1} +|AB|=|AC|=s. +\end{eqnarray} +We prove first the inequality $|MB|+|NC| <\frac{1}{2}\left(|AB|+|AC|\right)$. +If we use the statement \ref{TangOdsek} and the triangle inequality \ref{neenaktrik} and the relation \ref{nalSkk8Eqn1}, we get: + \begin{eqnarray*} +|MB|+|NC| &=& |MT|+|NT| =|MN|=\\ + &=& \frac{1}{2}\left( |MN|+|MN| \right)\\ + &<& \frac{1}{2}\left( |MN|+|NA|+|AM| \right)=\\ + &=& s=|AB|=\\ + &=& \frac{1}{2}\left(|AB|+|AC|\right). +\end{eqnarray*} +We prove the second inequality: $|MB|+|NC| >\frac{1}{3}\left(|AB|+|AC|\right)$. +Since $MAN$ is a right triangle with hypotenuse $MN$, according to the statement \ref{vecstrveckotHipot} it follows: +\begin{eqnarray} \label{nalSkk8Eqn2} +|MN|>|MA| \hspace*{1mm} \textrm{ and } \hspace*{1mm} |MN|>|NA|. +\end{eqnarray} +If we use the statement \ref{TangOdsek} and the relations \ref{nalSkk8Eqn2} and \ref{nalSkk8Eqn1}, we get: + \begin{eqnarray*} +|MB|+|NC| &=& |MT|+|NT| =|MN|=\\ + &=& \frac{1}{3}\left( |MN|+|MN|+|MN| \right)\\ + &>& \frac{1}{3}\left( |MN|+|NA|+|AM| \right)=\\ + &=& \frac{1}{3}\cdot2s=\frac{1}{3}\cdot2\cdot|AB|=\\ + &=& \frac{1}{3}\left(|AB|+|AC|\right). +\end{eqnarray*} + +\item \res{Prove that in a right triangle the sum of the sides is equal to the sum of +the diameters of the inscribed and circumscribed circle.} + +Let $ABC$ be a right angled triangle with a right angle at point $C$. Use the same notation as in the big task \ref{velikaNaloga}. Because $\angle PCQ=\angle BCA=90^0$, the quadrilateral $PCQS$ is a square, so by the big task (\ref{velikaNaloga}) it follows: + \begin{eqnarray} \label{nalSkk9Eqn} +r=|SQ|=|CP|=s-c. +\end{eqnarray} + From this and Tales' theorem (\ref{TalesovIzrKroz2}) it follows: + $$R+r=\frac{c}{2}+s-c=b+c.$$ + +\item \res{The lines of symmetry of the internal angles of a convex quadrilateral intersect in six different points. +Prove that four of these points are the vertices of the angle bisector quadrilateral.} + +Use the criterion for the angle bisector - theorem \ref{TetivniPogoj}. + +\item \res{Let: $c$ be the length of the hypotenuse, $a$ and $b$ be the lengths +of the catheti, and $r$ be the radius of the inscribed circle of a right angled triangle. Prove that:} +\begin{enumerate} + \item \res{$2r + c \geq 2 \sqrt{ab}$} + +Use the relation \ref{nalSkk9Eqn} and the inequality between arithmetic and geometric mean. + + \item \res{$a + b + c > 8r$} + +Use the relation \ref{nalSkk9Eqn} and Pythagoras' theorem \ref{PitagorovIzrek}. +\end{enumerate} + +\item \res{Let $P$ and $Q$ be the centers of the shorter arcs $AB$ and $AC$ +of the inscribed circle of a regular triangle $ABC$. Prove that the sides $AB$ and $AC$ of this triangle divide +the line $PQ$ into three congruent segments.} + +Let $X$ and $Y$ be the intersection points of the chord $PQ$ with the sides $AB$ and $AC$ of the triangle $ABC$. Prove that $PX\cong XY\cong YQ$ holds. By the theorem \ref{TockaN}, $BQ$ and $CP$ are the altitudes of the internal angles $ABC$ and $ACB$ of the triangle $ABC$. By the theorem \ref{ObodObodKot} we have: + \begin{eqnarray*} +\angle PAX &=& \angle PAB\cong\angle PCB=\frac{1}{2}\angle ACB=30^0\\ +\angle APX &=& \angle APQ\cong\angle ABQ=\frac{1}{2}\angle ABC=30^0. +\end{eqnarray*} +Therefore: + $\angle PAX = \angle APX=30^0$, which means that $APX$ is an isosceles triangle with the base $AP$ (theorem \ref{enakokraki}) or $PX\cong XA$. By the theorem \ref{zunanjiNotrNotr} we also have $\angle AXY=\angle APX+\angle PAX=60^0$. Since $\angle XAY=\angle BAC= 60^0$, $AXY$ is a right triangle, therefore $XA\cong XY\cong YA$. Similarly, $AQY$ is an isosceles triangle with the base $AQ$ or $QY\cong YA$. If we connect the proven relations of congruence of the segments, we get: $PX\cong XA\cong XY\cong AY\cong YQ$. + +\item \res{Let $k_1$, $k_2$, $k_3$, $k_4$ be four circles, each of which externally touches one side and two vertices of an arbitrary convex +quadrilateral. Prove that the centers of these circles are the points of a concurrence.} + +First, prove that the two opposite angles of the quadrilateral have measures $\frac{\alpha+\beta}{2}$ and $\frac{\gamma+\delta}{2}$, where $\alpha$, $\beta$, $\gamma$ and $\delta$ are the internal angles of the initial quadrilateral, and then use theorems \ref{TetivniPogoj} and \ref{VsotKotVeck}. + +\item \res{The circles $k$ and $l$ externally touch in the point $A$. The points $B$ and $C$ are the points of +the common external tangent of these two circles. Prove that $\angle BAC$ is a right angle.} + +Let $L$ be the intersection of the common tangent of the circles $k$ and $l$ in the point $A$ with the distance $BC$. Because $XA$ and $XB$ are tangents of the circle $k$ in the points $A$ and $B$, $XA\cong XB$ (statement \ref{TangOdsek}). Similarly, $XA$ and $XC$ are tangents of the circle $l$ in the points $A$ and $C$, or $XA\cong XC$. Therefore, $X$ is the center of the circle with the diameter $BC$ and the point $A$ lies on this circle. By Tales' statement \ref{TalesovIzrKroz2}, $\angle BAC=90^0$. + +\item \res{Let $ABCD$ be a deltoid ($AB\cong AD$ and $CB\cong CD$). Prove:} +\begin{enumerate} + \item \res{$ABCD$ is a tangential quadrilateral.} + +Use statement \ref{TangentniPogoj}. + + \item \res{$ABCD$ is a tautological quadrilateral exactly when $AB\perp BC$.} + +Use statement \ref{TetivniPogoj}. +\end{enumerate} + + + +\item \res{The circles $k$ and $k_1$ touch each other from the outside in the point $T$, in which they intersect the lines $p$ and $q$. The line $p$ has +also intersections with the circles $P$ and $P_1$, the line $q$ has $Q$ and $Q_1$ intersections. Prove that $PQ\parallel P_1Q_1$.} + +Let $t$ be the common tangent of the circles $k$ and $k_1$, $X\in t$ any point of this tangent, for which $P,X\div QQ_1$ and $Y$ the point which is symmetrical to the point $X$ with respect to $T$. By statements \ref{ObodKotTang} and \ref{sovrsnaSkladna}, we have: + \begin{eqnarray*} + \angle QPP_1= \angle QPT \cong\angle QTX\cong\angle Q_1TY\cong\angle Q_1P_1T + =\angle Q_1P_1P + \end{eqnarray*} +Therefore, $\angle QPP_1 \cong\angle Q_1P_1P$, so by statement \ref{KotiTransverzala} $PQ\parallel P_1Q_1$. + +\item \res{Let $MN$ be the common tangent of the circles $k$ and $l$ ($M$ and $N$ are touch points), +which intersect in the points $A$ and $B$. Calculate the value of the sum $\angle MAN+\angle MBN$.} + +Without loss of generality, assume that $d(A,MN) ||AE|-|BE||= |\left(|AC|-|EC|\right)-\left(|BD|-|ED|\right)| = ||AC|-|BD||. + \end{eqnarray*} + + +\item \res{Let $S$ be the intersection of the altitudes of the sides $AD$ and $BC$ +of the trapezoid $ABCD$ with the base $AB$. Prove that the circumscribed circles of the triangles $SAB$ and +$SCD$ touch each other in point $S$.} + +Use \ref{TangOdsek} theorem. + +\item \res{The lines $PB$ and $PD$ touch the circle $k(O,r)$ in points $B$ and $D$. +The line $PO$ intersects the circle $k$ in points $A$ and $C$ ($\mathcal{B}(P,A,C)$). Prove that the +line $BA$ is the angle bisector of the angle $PBD$.} + + +Use \ref{enakokraki} and \ref{TangOdsek} theorems. + +\item \res{The quadrilateral $ABCD$ is inscribed in a circle with center $O$. The diagonals $AC$ and +$BD$ are perpendicular. Let $M$ be the perpendicular projection of the center $O$ +on the line $AD$. Prove that + $$|OM|=\frac{1}{2}|BC|.$$} + +From the congruence of the triangles $OMA$ and $OMD$ (the SSA theorem \ref{SSK}) it follows that $MA\cong MD$ or that the point $M$ is the center of the line segment $AD$. +We mark with $E$ the intersection of the diagonals $AC$ and $BD$. Let $D'$ be the point that is symmetrical to the point $D$ with respect to the center $O$. It is clear that $D'\in k$. By Tales' theorem \ref{TalesovIzrKroz2} we have $\angle DAD'=90^0$. The line segment $MO$ is the median of the triangle $DAD'$ for the side $AD'$, so $|OM|=\frac{1}{2}|AD'|$. It is enough to prove that $AD'\cong BC$ or (by the theorem \ref{SklTetSklObKot}) that $\angle ADD'\cong \angle BDC$. If we now use the theorem \ref{VsotKotTrik} and \ref{ObodObodKot}, we get: +\begin{eqnarray*} + \angle ADD' &=& 90^0-\angle AD'D =90^0- \angle ACD =\\ +&=& 90^0-\angle ECD =\angle EDC =\\ +&=& \angle BDC. + \end{eqnarray*} + + +\item \res{Daljici $AB$ in $BC$ sta sosednji stranici pravilnega devetkotnika, ki je včrtan krožnici $k$ s središčem $O$. +Točka $M$ je središče stranice $AB$, točka $N$ pa središče +polmera $OX$ krožnice $k$, ki je pravokoten na premico $BC$. Dokaži, da je +$\angle OMN=30^0$.} + +From $\angle AOB\cong\angle BOC=\frac{360^0}{9}=40^0$ it follows that: + $$\angle AOX=\angle AOB + \angle BOX=\angle AOB + \frac{1}{2}\angle BOC=60^0.$$ +Since $OA\cong OB$, the triangle $AOB$ is a right triangle (theorems \ref{enakokraki} and \ref{VsotKotTrik}). Its median $AN$ is also the altitude, so $\angle ONA=90^0$. Since $\angle OMA=90^0$, by Tales' theorem the points $N$ and $M$ lie on the circle with the radius $OA$ or that the quadrilateral $ONMA$ is a square. If we now use the theorem \ref{ObodObodKot}, we get: +\begin{eqnarray*} + \angle OMN=\angle OAN=\frac{1}{2}\angle OAX=30^0. + \end{eqnarray*} + +\item \res{The circles $k_1$ and $k_2$ intersect in points $A$ and $B$. Let $p$ be a line that goes through point $A$, intersects circle $k_1$ in point $C$ and circle $k_2$ in point $D$, and let $q$ be a line that goes through point $B$, intersects circle $k_1$ in point $E$ and circle $k_2$ in point $F$. Prove that $\angle CBD\cong\angle EAF$.} + +If we use \ref{VsotKotTrik} and \ref{ObodObodKot}, we get: +\begin{eqnarray*} + \angle CBD &=& 180^0-\angle BCD - \angle BDC =\\ + &=& 180^0-\angle BCA - \angle BDA =\\ + &=& 180^0-\angle BEA - \angle BFA =\\ + &=& 180^0-\angle FEA - \angle EFA =\\ + &=& \angle EAF =\\ + \end{eqnarray*} + + +\item \res{The circles $k_1$ and $k_2$ intersect in points $A$ and $B$. Draw a line $p$ that goes through point $A$, so that the length of the line segment $MN$, where $M$ and $N$ are the intersections of line $p$ with circles $k_1$ and $k_2$, is maximal.} + +Let $p$ be an arbitrary line that goes through point $A$ and intersects circles $k_1$ and $k_2$ in points $M$ and $N$. +Let $S_1$ and $S_2$ be the centers of circles $k_1$ and $k_2$ and let $P_1$ and $P_2$ be the centers of line segments $MA$ and $NA$. Because $MN=2\cdot P_1P_2$, the problem of the maximum length of line segment $MN$ can be translated to the maximum length of line segment $P_1P_2$. From the similarity of triangles $MS_1P_1$ and $AS_1P_1$ (\textit{SSS} \ref{SSS}) it follows that $\angle S_1P_1M\cong\angle S_1P_1A=90^0$ or $\angle P_2P_1S_1=90^0$. Similarly, $\angle P_2P_1S_1=90^0$. Therefore, $S_1S_2P_2P_1$ is a right trapezoid with height $P_1P_2$, so $P_1P_2\leq S_1S_2$. Equality is achieved when $S_1S_2P_2P_1$ is a rectangle or $P_1P_2\parallel S_1S_2$. This means that line $p$ (for which $|MN|$ is maximal) is drawn as a parallel to line $S_1S_2$ through point $A$. + +\item \res{Let $L$ be the orthogonal projection of an arbitrary point $K$ of the circle $k$ onto its tangent at the point $T\in k$, and let $X$ be the point that is symmetric to the point $L$ with respect to the line $KT$. Determine the geometric position of the point $X$.} + +We denote by $S$ the center of the circle $k$ and by $k_1$ the circle with diameter $ST$. +We prove that the geometric position of the point $X$ (which we denote by $\mathcal{G}$) is equal to the circle $k_1$. We need to prove the equivalence: +\begin{eqnarray*} + X\in \mathcal{G} \hspace*{1mm} \Leftrightarrow \hspace*{1mm} X\in k_1 + \end{eqnarray*} + +For the points $T$ and $S$, it is clear from the definition that both $T\in \mathcal{G}$ and $T\in k_1$ hold, as well as $S\in \mathcal{G}$ and $S\in k_1$. We assume further that $X\neq T$ and $X\neq S$. + +We first assume $X\in \mathcal{G}$. We denote $\angle LTK=\alpha$. +The triangles $XKT$ and $LKT$ are symmetric with respect to the line $KT$ (see subsection \ref{odd6OsnZrc}), so $\triangle XKT\cong \triangle LKT$ holds, i.e. $\angle XTK\cong \angle LTK=\alpha$ and $\angle KXT\cong \angle KLT=90^0$. We prove that also $\angle SXT=90^0$. It is enough to prove that the points $S$, $X$ and $K$ are collinear, i.e. that $\angle TKX=\angle TKS$ holds. First, we have (from $\triangle XKT\cong \triangle LKT$): + $\angle TKX\cong\angle TKL =90^0-\alpha$. If we use the statement \ref{enakokraki} of the congruent triangles $KST$ with the base $KT$ and the statements \ref{ObodKotTang} and \ref{SredObodKot}, we get: + $\angle TKS=\frac{1}{2}(180^0-\angle KST)=\frac{1}{2}(180^0-2\angle LTK)=90^0-\alpha$. +Therefore $\angle TKX=\angle TKS=90^0-\alpha$, which means that the points $S$, $X$ and $K$ are collinear, i.e. $\angle SXT=90^0$. From the latter it follows that $X\in k_1$. + +Let's now assume that $X\in k_1$. We'll mark the intersection of the line segment $SX$ with the circle $k$ with $K$ and the point $L$ that is symmetrical to the point $X$ with respect to the line $TK$. +From $X\in k_1$ follows from Tales' theorem \ref{TalesovIzrKroz2} that $\angle SXT=90^0$, so $\angle KXT=90^0$ as well. +To prove that $X\in \mathcal{G}$, it is enough to prove that $X$ lies on the tangent $t$ of the circle $k$ through the point $T$. Let $L'$ be an arbitrary point on the tangent $t$, for which $L,X\ddot{-} ST$ holds. It is enough to prove $\angle LTK\cong\angle L'TK$. +We'll mark $\angle LTK=\alpha$. +The triangles $XKT$ and $LKT$ are symmetrical with respect to the line $KT$, so $\triangle XKT\cong \triangle LKT$ or $\angle XTK\cong \angle LTK=\alpha$ and $\angle KLT\angle KXT\cong =90^0$. If we use the formula \ref{ObodKotTang} and \ref{SredObodKot}, we get: +$\angle L'TK=\frac{1}{2}\angle TSK=\frac{1}{2}(180^0-2\angle SKT)=90^0-\angle SKT=90^0-\angle XKT=\angle XTK=\alpha=\angle LTK$. +Therefore $L'\in t$ or $X\in \mathcal{G}$. + +\item \res{Prove that a string polygon with an even number of vertices, which has all the internal angles compliant, is a regular polygon.} + +We'll mark the center of the circumscribed circle of this polygon with $O$ and the centers of the sides $A_iB_{i+1}$ with $B_i$ ($i\in \{1,2,\ldots , 2n-1 \}$, $A_0=A_{2n-1}$ and $A_{2n}=A_1$). We'll use the fact that $A_iOA_{i+1}$ is an isosceles triangle with the base $A_iA_{i+1}$ and we'll first prove $\triangle OB_{i-1}A_i\cong \triangle OB_{i+1}A_{i+1}$. + +Let $EA$ be the common tangent of the circles $k$ (larger) and $l$ (smaller) and $D$ the other intersection of the line $AB$ with the circle $l$. Let $L$ be the other intersection of the segment $AK$ with the circle $l$, $\alpha=\angle DCB$ and $\beta=\angle DBC$. +By the theorem \ref{ObodKotTang} it follows: + \begin{eqnarray} \label{nalSkk36Eqn1} + \angle CAB=\angle CAD\cong \angle DCB=\alpha +\end{eqnarray} +and + \begin{eqnarray} \label{nalSkk36Eqn2} + \angle LDA\cong \angle EAL=\angle EAK\cong \angle ABK=\angle DBC=\beta. +\end{eqnarray} +Since $CDA$ is the exterior angle of the triangle $CDB$, it follows from the theorem \ref{zunanjiNotrNotr} that + $\angle CDA=\angle DCB+\angle CBD=\alpha+\beta$. From this and \ref{nalSkk36Eqn2} it follows: + \begin{eqnarray} \label{nalSkk36Eqn3} + \angle LDC= \angle CDA-\angle LDA=\alpha+\beta-\beta=\alpha. +\end{eqnarray} +From the relations \ref{nalSkk36Eqn3} and \ref{nalSkk36Eqn1} and the theorem \ref{ObodObodKot}: +\begin{eqnarray*} + \angle KAC= \angle LAC\cong \angle LDC=\alpha=\angle CAB. +\end{eqnarray*} +This means that the line $AC$ is the perpendicular bisector of the angle $BAK$. + +The task is a special case of a more general task \ref{nalSkk47}. + +\item \res{Let $BC$ be the diameter of the circle $k$. Determine the geometric position of the altitude points +of all triangles $ABC$, where $A$ is an arbitrary point on the circle $k$.} + +The desired geometric position of the points is the circle that passes through the points $B$ and $C$. Calculate the measure of the angle $BVC$, where $V$ is the altitude point of the triangle $ABC$, and use the theorem \ref{ObodKotGMT}. See also the theorem \ref{TockaV'a}. + +\item \res{Let $ABCDEF$ be a tetrahedral hexagon and $AB\cong DE$ and $BC\cong EF$. +Prove that $CD\parallel AF$.} + +Let $ABCDEF$ be a tetrahedral hexagon and $AB\cong DE$ and $BC\cong EF$. +Since $\angle BAD>90^0$ and $\angle BCD>90^0$, by the theorem \ref{obodKotGMTZunNotr} $A$ and $C$ are inner points of the circle $k$. This means that $AC90^0$. This means $\mathcal{B}(A',A,V)$ (where $A'$ is the altitude of this triangle) or $V$ is an external point of the circle $k$ and it holds $|\overrightarrow{OV}|>1$. +We can also look at the task \ref{OlimpVekt15} (section \ref{odd5DolzVekt}). + +\item \res{Calculate the angles determined by the vectors $\overrightarrow{OA}$, $\overrightarrow{OB}$ and $\overrightarrow{OC}$, if the points $A$, $B$ and $C$ lie on the circle with the center +$O$ and in addition it holds: +$$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{0}.$$} + +According to Hamilton's theorem \ref{Hamilton}, $\overrightarrow{0}=\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}=\overrightarrow{OV}$, where $V$ is the altitude point of the triangle $ABC$. So $V=O$, which means that $ABC$ is an equilateral triangle and all the angles sought are equal to $120^0$. + + \item \res{Let $A$, $B$, $C$ and $D$ be points that lie on the circle with the center +$O$, and it holds +$$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = \overrightarrow{0}.$$ +Prove that $ABCD$ is a rectangle.} + +We use Hamilton's theorem \ref{Hamilton}. + +\item \res{Let $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ be vectors of some plane, for which it holds $|\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{c}| =x$. Investigate, in which case it also holds $|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}| = x$.} + +First, let's assume that $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are nonlinear vectors. +Let $O$ be an arbitrary point and $A$, $B$ and $C$ be such points that $\overrightarrow{OA}=\overrightarrow{a}$, $\overrightarrow{OB}=\overrightarrow{b}$ and $\overrightarrow{OC}=\overrightarrow{c}$ (statement \ref{vektAvObst1TockaB}). From the nonlinearity of vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ it follows that points $A$, $B$ and $C$ are nonlinear. From the condition $|\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{c}| =x$ it follows that $|\overrightarrow{OA}| = |\overrightarrow{OB}| = |\overrightarrow{OC}| =x$, which means that point $O$ is the center of the circumscribed circle of triangle $ABC$. By Hamilton's statement \ref{Hamilton} it is: + \begin{eqnarray*} +x + &=& +|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}|=\\ + &=& +|\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}|=\\ + &=& |\overrightarrow{OV}|, +\end{eqnarray*} + where $V$ is the altitude point, which therefore lies on the circumscribed circle of triangle $ABC$. This is possible only when $ABC$ is a right triangle. + +If $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are collinear vectors, the given condition is satisfied only when $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are not all of the same orientation. + + + %Tales + %______________________________________________________ + + +\item \res{Divide the given line segment $AB$:} + + (\textit{a}) \res{into five equal line segments,}\\ + (\textit{b}) \res{in the ratio $2:5$,}\\ + (\textit{c}) \res{into three line segments, which are in the ratio $2:\frac{1}{2}:1$.} + +Use example \ref{izrekEnaDelitevDaljice}. + + + \item \res{Given is the line segment $AB$. Only by using a ruler with the ability to draw parallels (constructions in affine geometry), plot point $C$, if:} + +(\textit{a}) \res{$\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}$,} \hspace*{3mm} + (\textit{b}) \res{$\overrightarrow{AC}=\frac{3}{5}\overrightarrow{AB}$,} \hspace*{3mm} + (\textit{c}) \res{$\overrightarrow{AC}=-\frac{4}{7}\overrightarrow{AB}$.} + +Use task \ref{nalVekt4} and example \ref{izrekEnaDelitevDaljiceNan}. + +\item \res{The lines $a$, $b$ and $c$ are given. Draw the line $x$, so that it will:} + +(\textit{a}) \res{$a:b=c:x$;} \hspace*{3mm} +(\textit{b}) \res{$x=\frac{a\cdot b}{c}$,} \hspace*{3mm} +(\textit{c}) \res{$x=\frac{a^2}{c}$,}\hspace*{3mm}\\ +(\textit{č}) \res{$x=\frac{2ab}{3c}$,}\hspace*{3mm} +(\textit{č}) \res{$(x+c):(x-c)=7:2$.} + +We use the Pythagorean theorem \ref{TalesovIzrek}. + +\item \res{Let $M$ and $N$ be points on the arm $OX$, $P$ a point on the arm $OY$ as $XOY$ and $NQ\parallel MP$ and $PN\parallel QS$ ($Q\in OY$, $S\in OX$). Prove that $|ON|^2=|OM|\cdot |OS|$ (for the line $ON$ in this case we say that it is the \index{geometrijska sredina daljic}\pojem{geometric mean} \color{green1} of the lines $OM$ and $OS$).} + +By the Pythagorean theorem \ref{TalesovIzrek} it is: +$$\frac{|ON|}{|OM|}=\frac{|OQ|}{|OP|}=\frac{|OS|}{|ON|},$$ from which the desired relation follows. + +\item \res{Let $ABC$ be a triangle and $Q$, $K$, $L$, $M$, $N$ and $P$ such points of the sides $AB$, $AC$, $BC$, $BA$, $CA$ and $CB$, that $AQ\cong CP\cong AC$, $AK\cong BL\cong AB$ and $BM\cong CN\cong BC$. +Prove that $MN$, $PQ$ and $LK$ are parallel.} + +We use the formulas \ref{vektParamPremica} and \ref{vektDelitDaljice}. + +\item \res{Let $P$ be the center of gravity of the centroid $AA_1$ of the triangle $ABC$. The point $Q$ is the intersection of the line $BP$ +with the side $AC$. Calculate the ratios $AQ:QC$ and $BP:PQ$.} + +Let $R$ be the center of the line $QC$. By the formula \ref{srednjicaTrikVekt} (for the triangle $BQC$ and $AA_aR$) it is: +$$\overrightarrow{PQ}=\frac{1}{2}\overrightarrow{A_1R}=\frac{1}{4}\overrightarrow{BQ},$$ +so $BP:PQ=3:1$. From $\overrightarrow{PQ}=\frac{1}{2}\overrightarrow{A_1R}$ it follows that $PQ\parallel A_1R$, so by the Pythagorean theorem $AQ:QR=AP:PA_1=1:1$. Therefore $AQ\cong QR \cong RC$ or $AQ:QC=1:2$. + +See also the example \ref{TezisceSredisceZgled}. + +\item \res{The points $P$ and $Q$ lie on the sides $AB$ and $AC$ of the triangle $ABC$, and it holds that $\frac{|\overrightarrow{PB}|}{|\overrightarrow{AP}|} + +\frac{|\overrightarrow{QC}|}{|\overrightarrow{AQ}|}=1$. Prove that the centroid of this triangle lies on the line $PQ$.} + +Let $\widehat{T}$ be the intersection of the line $PQ$ with the centroidal line $AA_1$ of this triangle. It is enough to prove that $\widehat{T}=T$. +Since the points $A$, $\widehat{T}$ and $A_1$ are collinear, $\overrightarrow{A\widehat{T}}=\lambda\overrightarrow{AA_1}$ for some $\lambda\in R$. For $\widehat{T}=T$ it is enough to prove that $\lambda=\frac{2}{3}$. + +We denote $\alpha=\frac{|\overrightarrow{PB}|}{|\overrightarrow{AP}|}$ and $\beta=\frac{|\overrightarrow{QC}|}{|\overrightarrow{AQ}|}$. By the assumption, $\alpha+\beta=1$. From the given conditions it follows that + $\overrightarrow{AB}=\left(1+\alpha\right)\cdot\overrightarrow{AP}$ and + $\overrightarrow{AC}=\left(1+\beta\right)\cdot\overrightarrow{AQ}$. If we use the latter relation and the formula \ref{vektSredOSOAOB}, we get: + \begin{eqnarray*} + \overrightarrow{A\widehat{T}} &=& \lambda\overrightarrow{AA_1}=\\ + &=& +\frac{\lambda}{2}\cdot \left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\\ + &=& +\frac{\lambda}{2}\cdot \left(\left(1+\alpha\right)\cdot\overrightarrow{AP} ++\left(1+\beta\right)\cdot\overrightarrow{AQ}\right)=\\ + &=& +\frac{\lambda}{2}\cdot \left(1+\alpha\right)\cdot\overrightarrow{AP} ++\frac{\lambda}{2}\left(1+\beta\right)\cdot\overrightarrow{AQ}. + \end{eqnarray*} +Since the point $T$ lies on the line $PQ$ and since $\alpha+\beta=1$, by the formula \ref{vektParamPremica}: + \begin{eqnarray*} +1&=&\frac{\lambda}{2}\cdot \left(1+\alpha\right) ++\frac{\lambda}{2}\left(1+\beta\right)=\\ +&=&\frac{\lambda}{2}\cdot \left(1+\alpha+1+\beta\right)=\\ + &=&\frac{3\lambda}{2}\\ + \end{eqnarray*} +i.e. $\lambda=\frac{2}{3}$. + + + \item \res{Let $a$, $b$ and $c$ be three segments with a common starting point $S$ and $M$ +a point on the segment $a$. If the point $M$ "moves" along the segment $a$, + the ratio of the distances of this point from the lines $b$ and $c$ is constant.} + +We use the Tales' theorem \ref{TalesovIzrek}. + +\item \res{Let $D$ be a point lying on the side $BC$ of the triangle $ABC$ and + $F$ and $G$ points in which the line passing through the point $D$ and parallel to the median $AA_1$, intersects the lines $AB$ and $AC$. +Prove that the sum $|DF|+|DG|$ is constant if the point $D$ "moves" along the side +$BC$.} + +We prove that $|DF|+|DG|=|AA_1|$. See problem \ref{nalVekt29}. + + + \item + \res{Draw a triangle with the following data:} + +In both cases use the same notation as in the big problem \ref{velikaNaloga} and the statement \ref{velNalTockP'}. + + (\textit{a}) \res{$v_a$, $r$, $b-c$} + +First, we draw the rectangular triangle $SPP_a$ ($SP=r$, $\angle P=90^0$ and $PP_a=b-c$), the point $P'$ and finally use the fact that the vertex $A$ lies on the segment $P_aP'$. + + (\textit{b}) \res{$\beta$, $r$, $b-c$} + +Similarly to the previous case. + + +\end{enumerate} + + + + +%REŠITVE - Izometrije +%________________________________________________________________________________ + +\poglavje{Isometries} + +\begin{enumerate} + \item + \res{Given is a line $p$ and points $A$ and $B$, which lie on + opposite sides of the line $p$. Construct a point + $X$, which lies on +the line $p$, so that the difference $|AX|-|XB|$ is maximal.} + +Let $A'=\mathcal{S}_p(A)$. We prove that $X=A'B\cap p$ is the desired point (see also example \ref{HeronProbl}). + + \item \res{In the plane are given lines $p$, $q$ and $r$. Construct + an equilateral triangle $ABC$, so that + the vertex $B$ lies on the line $p$, $C$ on $q$, +the altitude from the vertex $A$ lies on the line $r$.} + +Since $\mathcal{S}_r(B)=C$ and $B\in p$, the vertex $C$ we get from +the condition $C\in \mathcal{S}_r(p)\cap q$. + +\item \res{Given is a quadrilateral $ABCD$ and a point $S$. Draw a parallelogram +with the center in the point $S$, so +that its vertices lie on the lines +of the given quadrilateral.} + +We use the central reflection $\mathcal{S}_S$. + +\item \res{Let $\mathcal{I}$ be an indirect isometry of a plane that +maps point $A$ to point $B$, +$B$ to $A$. Prove that $\mathcal{I}$ is the central reflection.} + +Let $s$ be the line of symmetry of the segment $AB$. The composition +$\mathcal{I}\circ\mathcal{S}_s$ is a direct isometry with two +fixed points $A$ and $B$, therefore by izreku \ref{izo2ftIdent} it +represents the identity or +$\mathcal{I}\circ\mathcal{S}_s=\mathcal{E}$. If we multiply the +equality by $\mathcal{S}_s$ from the right, we get +$\mathcal{I}=\mathcal{S}_s$. + +\item \res{Let $K$ and $L$ be points that are symmetric to the vertex +$A$ of the triangle $ABC$ with respect to +the lines of symmetry of the internal angles at the vertices $B$ and +$C$. Let $P$ be the point of +tangency of the inscribed circle of this triangle and the side $BC$. +Prove that $P$ is the center of the segment $KL$.} + +We denote by $s_\beta$ and $s_\gamma$ the mentioned lines of symmetry +of the internal angles at the vertices $B$ and $C$ and by $Q$ and $R$ +the points of tangency of the inscribed circle of the triangle $ABC$ +with the sides $AC$ and $AB$. By izreku \ref{TangOdsek} we have +$AQ\cong AR$, $BR\cong BP$ and $CP\cong CQ$. Therefore +$\mathcal{S}_{s_\beta}:A,R\mapsto K,P$ and +$\mathcal{S}_{s_\gamma}:A,Q\mapsto L,P$. From this it follows that +$KP\cong AR\cong AQ\cong LP$, which means that $P$ is the center of +the segment $KL$. + +\item \res{Let $k$ and $l$ be circles on different sides of a line $p$. +Draw an isosceles triangle $ABC$, so that its altitude $AA'$ lies on +the line $p$, the vertex $B$ on the circle $k$, and the vertex $C$ on +the circle $l$.} + +Since $\mathcal{S}_p(B)=C$ and $B\in k$, we get the point $C$ from +the condition $C\in \mathcal{S}_p(k)\cap l$. + +\item \res{Let $k$ be a circle and $a$, $b$ and $c$ lines in the same +plane. Draw a triangle $ABC$, which is +inscribed in the circle $k$, so that its sides $BC$, $AC$ and $AB$ +will be parallel in order to the lines $a$, $b$ and $c$.} + +First, we can plan the simetrals of sides $BC$, $AC$ and $AB$ as +lines that go through the center $O$ of the circle $k$ and are +perpendicular in turn to the lines $a$, $b$ and $c$. We mark these +simetrals with $p$, $q$ and $r$. Because the composite +$\mathcal{I}=\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p$ +has the fixed points $O$ and $A$, by Theorem \ref{izo1ftIndZrc} +$\mathcal{I}=\mathcal{S}_{OA}$. So we can draw the line $OA$ as the +simetral of the distance $XX'$, where $X$ is an arbitrary point. +The point $A$ we get as the intersection of the line $OA$ and the +circle $k$. Then $B=\mathcal{S}_p(A)$ and $C=\mathcal{S}_r(A)$. + +\item \res{Let $ABCDE$ be a rope pentagon, in which $BC\parallel +DE$ and $CD\parallel EA$. +Prove that the point $D$ lies on the simetral of the distance $AB$.} + +We mark with $O$ the center of the inscribed circle of the +pentagon $ABCDE$. Because $BC\parallel DE$, the lines $BC$ and $DE$ +have the same perpendicular from the center $S$, which is actually +the common simetral of sides $BC$ and $DE$. We mark it with $p$. +Similarly, sides $CD$ and $EA$ have the common simetral $q$. We +mark with $r$ the simetral of the distance $AB$. We prove that +$D\in r$. The composite +$\mathcal{S}_{OD}\circ\mathcal{S}_q +\circ\mathcal{S}_p\circ\mathcal{S}_r +\circ\mathcal{S}_q\circ\mathcal{S}_p$ is a direct isometry, which +has the fixed points $O$ and $D$, so by Theorem \ref{izo2ftIdent} it +represents the identity. Therefore: +$$\mathcal{S}_{OD}\circ\mathcal{S}_q +\circ\mathcal{S}_p\circ\mathcal{S}_r +\circ\mathcal{S}_q\circ\mathcal{S}_p=\mathcal{E}.$$ + +Premices $p$, $q$ and $r$ are from the same $\mathcal{X}_O$, so +the composite $\mathcal{S}_q \circ\mathcal{S}_p\circ\mathcal{S}_r$ +represents the basic reflection (statement \ref{izoSop}) or +$$\mathcal{S}_q +\circ\mathcal{S}_p\circ\mathcal{S}_r=\mathcal{S}_l +=\mathcal{S}^{-1}_l=\left(\mathcal{S}_q +\circ\mathcal{S}_p\circ\mathcal{S}_r\right)^{-1}=\mathcal{S}_r +\circ\mathcal{S}_p\circ\mathcal{S}_q.$$ + Therefore: + \begin{eqnarray*} +\mathcal{E}&=&\mathcal{S}_{OD}\circ\mathcal{S}_q +\circ\mathcal{S}_p\circ\mathcal{S}_r +\circ\mathcal{S}_q\circ\mathcal{S}_p =\\ +&=&\mathcal{S}_{OD}\circ\mathcal{S}_r +\circ\mathcal{S}_p\circ\mathcal{S}_q +\circ\mathcal{S}_q\circ\mathcal{S}_p = +\mathcal{S}_{OD}\circ\mathcal{S}_r. +\end{eqnarray*} + From +$\mathcal{E}=\mathcal{S}_{OD}\circ\mathcal{S}_r$ follows +$\mathcal{S}_{OD}=\mathcal{S}_r$ or $OD=r$ and $D\in r$. + + +\item \res{Premice $p$, $q$ in $r$ ležijo v isti ravnini. Dokaži +ekvivalenco $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p + =\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_r$ natanko + tedaj, ko premice $p$, $q$ in $r$ pripadajo istemu šopu.} + +Označimo +$\mathcal{I}=\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p$. + +($\Leftarrow$) Predpostavimo, da premice $p$, $q$ in $r$ +pripadajo istemu šopu. Po izreku \ref{izoSop} izometrija +$\mathcal{I}$ predstavlja osmo zrcaljenje. Iz tega sledi +$\mathcal{I}=\mathcal{I}^{-1}$ oz. +$\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p + =\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_r$ + +($\Rightarrow$) Let us assume that + $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p + =\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_r$. From this + condition it follows that $\mathcal{I}=\mathcal{I}^{-1}$ or +$\mathcal{I}^2=\mathcal{E}$. Let us assume the opposite - that +lines $p$, $q$ and $r$ are not from the same sheaf. According to +the statement \ref{izoZrcdrsprq} it +is $\mathcal{I}=\mathcal{G}_{2\overrightarrow{AB}}$ +($\overrightarrow{AB}\neq \overrightarrow{0}$). But in this +case $\mathcal{E}=\mathcal{I}^2=\mathcal{G}^2_{2\overrightarrow{AB}} +=\mathcal{T}_{4\overrightarrow{AB}}$ (statement +\ref{izoZrcdrsZrcdrs}), which is not possible, because according to the assumption +$\overrightarrow{AB}\neq \overrightarrow{0}$. Therefore lines +$p$, $q$ and $r$ belong to the same sheaf. + +\item \res{Let $O$, $P$ and $Q$ be three non-linear points. Construct +a square $ABCD$ (in the plane $OPQ$) with the center in point $O$, so +that points $P$ and $Q$ in order +lie on lines $AB$ and $BC$.} + +We use the fact that $Q,\mathcal{R}_{O,90^0}(P)\in BC$. + +\item \res{Let $\mathcal{R}_{S,\alpha}$ be a rotation and $\mathcal{S}_p$ +be a reflection in the same plane and $S\in p$. +Prove that the composition $\mathcal{R}_{S,\alpha}\circ\mathcal{S}_p$ +and $\mathcal{S}_p\circ\mathcal{R}_{S,\alpha}$ represent a reflection.} + +We use the statement \ref{rotacKom2Zrc} and \ref{izoSop}. + +\item \res{Given is a point $A$ and a circle $k$ in the same plane. Draw +a square $ABCD$, so that the vertices of the diagonal $BD$ lie on the circle $k$.} + +From $B,D\in k$ and $\mathcal{R}_{A,90^0}(B)=D$ it follows that $D\in k\cap +\mathcal{R}_{A,90^0}(k)$, which allows us to construct the vertex $D$. +Then $B=\mathcal{R}_{A,-90^0}(D)$ and +$C=\mathcal{R}_{D,90^0}(A)$. + +\item \res{Let $ABC$ be an arbitrary triangle. Prove: + $$\mathcal{R}_{C,2\measuredangle BCA}\circ + \mathcal{R}_{B,2\measuredangle ABC}\circ + \mathcal{R}_{A,2\measuredangle CAB}=\mathcal{E}.$$} + +According to the statement \ref{rotacKom2Zrc} it is: + \begin{eqnarray*} + &&\mathcal{R}_{C,2\measuredangle BCA}\circ + \mathcal{R}_{B,2\measuredangle ABC}\circ + \mathcal{R}_{A,2\measuredangle CAB}=\\ &&= + \mathcal{S}_{CA}\circ\mathcal{S}_{CB}\circ\mathcal{S}_{BC} + \circ\mathcal{S}_{BA}\circ\mathcal{S}_{AB}\circ\mathcal{S}_{AC}= + \mathcal{E}. + \end{eqnarray*} + +\item \res{Prove that the composite of the central reflection $\mathcal{S}_p$ +and the central reflection $\mathcal{S}_S$ ($S\in p$) +represents the central reflection.} + + Let $q$ be the rectangle of the line $p$ through the point $S$. According to the + statement \ref{izoSrZrcKom2Zrc} it is: + $$\mathcal{S}_S\circ\mathcal{S}_p= + \mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_p=\mathcal{S}_q.$$ + +\item \res{Let $O$, $P$ and $Q$ be three non-linear points. Construct +the square $ABCD$ (in the plane $OPQ$) with the center in the point $O$, so + that the points $P$ and $Q$ +lie in succession on the lines $AB$ and $CD$.} + +We use the fact that $Q,\mathcal{S}_O(P)\in CD$. + +\item \res{What does the composite of the translation and the central reflection represent?} + +We use the statements \ref{transl2sred} and \ref{izoKomp3SredZrc}. + +\item \res{Given are the line $p$ and the circles $k$ and $l$, which +lie in the same plane. Draw the line, which is parallel to +the line $p$, so that it determines the corresponding tangents on the circles $k$ and $l$.} + +Let $K$ and $L$ be the centers of the circles $k$ and $l$. Let $q$ be the desired line that is parallel to the line $p$, and the circles $k$ and $l$ intersect in such points $A$, $B$, $C$ and $D$, so that $AB \cong CD$. Let $\mathcal{B} (A, B, C, D)$. We denote by $K_q$ and $L_q$ the orthogonal projections of the centers $K$ and $L$ onto the line $q$, and $K_p$ and $L_p$ the orthogonal projections of the centers $K$ and $L$ onto the line $p$. From the congruence of the triangles $KAK_q$ and $KBK_q$ (SSA Theorem \ref{SSK}) it follows that $K_q$ is the center of the chord $AB$. Similarly, $L_q$ is the center of the chord $CD$. From $AB \cong CD$ it then follows that $K_qB \cong L_qD$, i.e. $\overrightarrow {K_qB} = \overrightarrow {L_qD}$. Therefore: +$$\overrightarrow {BD} = \overrightarrow {BD} + \overrightarrow {0} = \overrightarrow {BD} + \overrightarrow {K_qB} + \overrightarrow {DL_q} = \overrightarrow {K_qL_q} = \overrightarrow {K_pL_p}.$$ The vector $\overrightarrow {v} = \overrightarrow {K_pL_p}$ can be constructed. This means that the previous analysis allows the construction, because $\mathcal {T} _ {\overrightarrow {v}} (B) = D$, i.e. $D \in \mathcal {T} _ {\overrightarrow {v}} (k) \cap l$. + +\item \res {Let $c$ be a line that intersects the parallels $a$ and $b$, and $l$ a distance. Draw an isosceles triangle $ABC$ so that $A \in a$, $B \in b$, $C \in c$ and $AB \cong l$.} + +First, we construct an arbitrary isosceles triangle $A_1B_1C_1$ that satisfies the conditions $A_1 \in a$, $B_1 \in b$ and $A_1B_1 \cong l$, then we use the appropriate translation. + +\item \res {Prove that the composition of a rotation and an axis reflection of a plane represents a mirror glide exactly when the center of rotation does not lie on the axis of the axis reflection.} + +Use Theorems \ref {rotacKom2Zrc}, \ref {izoZrcdrsprq} and \ref {izoSop}. + +\item \res {Let $ABC$ be an isosceles triangle. Prove that the composition $\mathcal {S} _ {AB} \circ \mathcal {S} _ {CA} \circ \mathcal {S} _ {BC}$ represents a mirror glide. Also determine the vector and the axis of this glide.} + +Let $\mathcal{I}=\mathcal{S}_{AB} \circ\mathcal{S}_{CA} + \circ\mathcal{S}_{BC}$. By \ref{izoZrcdrsprq} is + $\mathcal{I}$ a mirror reflection. Let $A_1$, $B_1$ and $C_1$ be in order the center points of the lines $BC$, $AC$ and $AB$ of the triangle $ABC$. + Because it is a right triangle is $\mathcal{I}(A_1C_1)=A_1C_1$, which + means that the line $A_1C_1$ is of this mirror reflection. It is not + difficult to prove that for the point $A'_1=\mathcal{I}(A_1)$ (both + lie on the axis $A_1C_1$) is + $\overrightarrow{A_1A'_1}=3\overrightarrow{A_1C_1}$, so + $\mathcal{I}=\mathcal{G}_{3\overrightarrow{A_1C_1}}$. + +\item \res{Given are the points $A$ and $B$ on the same side of the line +$p$. +Draw the line $XY$, which lies on the line $p$ and is consistent +with the given line $l$, so that the sum +$|AX|+|XY|+|YB|$ is minimal.} + +Let $A'=\mathcal{G}_{\overrightarrow{MN}}(A)$ (where $M,N\in +p$ and $MN\cong l$). The point $Y$ is obtained as the intersection of the lines $p$ +and $X'Y$ (see also example \ref{HeronProbl}). + +\item \res{Let $ABC$ be an isosceles right triangle with a right angle at the vertex $A$. What does the composite +$\mathcal{G}_{\overrightarrow{AB}}\circ \mathcal{G}_{\overrightarrow{CA}}$ represent?} + +Let $p$ and $q$ be the simetrali of the sides $CA$ and $AB$ of the triangle +$ABC$. By \ref{izoZrcDrsKompSrOsn} is: + $$\mathcal{G}_{\overrightarrow{AB}}\circ + \mathcal{G}_{\overrightarrow{CA}}= + \mathcal{S}_q\circ\mathcal{S}_A\circ\mathcal{S}_A\circ\mathcal{S}_p= + \mathcal{S}_q\circ\mathcal{S}_p.$$ Because $ABC$ is an isosceles + right triangle with a right angle at the vertex $A$, the lines $p$ and $q$ are perpendicular and intersect at the center $S$ + of the hypotenuse $BC$. Therefore + $\mathcal{G}_{\overrightarrow{AB}}\circ + \mathcal{G}_{\overrightarrow{CA}}=\mathcal{S}_q + \circ\mathcal{S}_p=\mathcal{S}_S$. + +\item \res{In the same plane are given the lines +$a$, $b$ and $c$. +Draw the points $A\in a$ and $B\in b$ +so that $\mathcal{S}_c(A)=B$.} + +From $A\in a$ it follows that $\mathcal{S}_c(A)\in \mathcal{S}_c(a)$ or +$B\in \mathcal{S}_c(a)$. Because $B\in b$ as well, we get the point $B$ from +the condition $B\in \mathcal{S}_c(a)\cap b$. Then $A=\mathcal{S}_c(B)$ is also true. + +\item \res{Given are the lines $p$ and $q$ and the point $A$ in the same plane. +Draw the points $B$ and $C$ so that +the lines $p$ and $q$ will be the internal angle bisectors of the triangle $ABC$.} + +We use the fact that the line $BC$ is determined by the points +$\mathcal{S}_p(A)$ and $\mathcal{S}_q(A)$. + +\item \res{Let $p$, $q$ and $r$ be lines and $K$ and $L$ points in the same plane. Draw +the lines $s$ and $s'$, which go through the points $K$ and $L$ in order, +so that $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p(s)=s'$ is true.} + +We denote +$\mathcal{I}=\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p$. +The line $s'$ is determined by the points $L$ and $\mathcal{I}(K)$, +the line $s$ is determined by the points $K$ and $\mathcal{I}^{-1}(K)$. If +$\mathcal{I}(K)=L$, the task has infinitely many solutions - the line $s$ +is an arbitrary line, which goes through the point $K$. + + +\item \res{Let $s$ be the line which bisects one of the angles determined by the lines $p$ +and $q$. Prove that $\mathcal{S}_s\circ\mathcal{S}_p = +\mathcal{S}_q\circ\mathcal{S}_s$.} + +We can even prove more - that the equivalence (under +the assumption that the lines $p$ and $q$ intersect) holds. If we multiply the given equality by $\mathcal{S}_s$ from the right, we get the equivalent equality $\mathcal{S}_s\circ\mathcal{S}_p\circ\mathcal{S}_s = +\mathcal{S}_q$. By \ref{izoTransmutacija}, this is equivalent to the equality $\mathcal{S}_{\mathcal{S}_s(p)} = +\mathcal{S}_q$ or $\mathcal{S}_s(p) = q$, which is equivalent to the fact that $s$ bisects one of the angles determined by the lines $p$ and $q$. + +\item \label{nalIzo27} +\res{Let $S$ be the center of the triangle $ABC$ inscribed in the circle and $P$ the point, + in which this circle touches the side $BC$. Prove: $$\mathcal{S}_{SC} + \circ\mathcal{S}_{SA}\circ\mathcal{S}_{SB} =\mathcal{S}_{SP}.$$} + + The equality is a direct consequence of the statement \ref{izo1ftIndZrc}, because + is\\ + $\mathcal{S}_{SC}\circ\mathcal{S}_{SA}\circ\mathcal{S}_{SB}:S,P\mapsto + S,P$. + +\item \res{The lines $p$, $q$ and $r$ of a plane go through the center +$S$ of the circle $k$. +Draw a triangle $ABC$, which is inscribed in this circle, so that +the lines $p$, $q$ and $r$ will be the altitudes of the internal angles at +the vertices $A$, $B$ and $C$ of this triangle.} + +We use the previous task \ref{nalIzo27}. + +\item \res{The lines $p$, $q$, $r$, $s$ and $t$ of a plane intersect +in the point $O$, the point $M$ lies on the line $p$. +Draw such a pentagon that $M$ will be the center of one of its sides, +the lines $p$, $q$, $r$, $s$ and $t$ will be the altitudes of the sides.} + +The point $M$ lies on one of the altitudes. Without loss of +generality, let $M\in p$. Let $ABCDE$ be such a pentagon that +the lines $p$, $q$, $r$, $s$ and $t$ are in turn the altitudes of +its sides $AB$, $BC$, $CD$, $DE$ and $EA$. Let +$\mathcal{I}=\mathcal{S}_t\circ\mathcal{S}_s\circ\mathcal{S}_r +\circ\mathcal{S}_q\circ\mathcal{S}_p$. Because +$\mathcal{I}:O,A\mapsto O,A$, by the statement \ref{izo1ftIndZrc} +$\mathcal{I}=\mathcal{S}_{OA}$. The line $OA$ is obtained as the altitude of the segment $XX'$, where $X$ is an arbitrary point and +$X'=\mathcal{I}(X)$, then the vertex $A$ as the intersection of the line +$OA$ and the rectangle of the line $p$ through the point $M$. + +\item \res{The point $P$ lies in the plane of the triangle $ABC$. Prove that +the lines, which are symmetrical to +the lines $AP$, $BP$ and $CP$ with respect to the altitudes of + at the vertices $A$, $B$ and $C$ of this triangle, belong to the same set.} + +Let us denote by $s_{\alpha}$, $s_{\alpha}$ and $s_{\alpha}$ the internal angle bisectors at the vertices $A$, $B$ and $C$ of the triangle $ABC$ +and $a=\mathcal{S}_{s_{\alpha}}(AP)$, $b=\mathcal{S}_{s_{\beta}}(BP)$ and $c=\mathcal{S}_{s_{\gamma}}(CP)$. +We will prove that the lines $a$, $b$ and $c$ belong to the same pencil. Because $\mathcal{S}_{s_{\alpha}}:AC, p\rightarrow AB, + a$ is $\measuredangle CAP=\measuredangle a,AP$. Therefore $\mathcal{R}_{A,2\measuredangle CAP} + =\mathcal{R}_{A,2\measuredangle a,AP}$ or $\mathcal{S}_{AC}\circ \mathcal{S}_{AP} + =\mathcal{S}_a\circ \mathcal{S}_{AB}$. From this it follows that $\mathcal{S}_a=\mathcal{S}_{AC} + \circ\mathcal{S}_{AP}\circ \mathcal{S}_{AB}$. Similarly: $\mathcal{S}_b=\mathcal{S}_{BA} + \circ\mathcal{S}_{BP}\circ \mathcal{S}_{BC}$ and $\mathcal{S}_c=\mathcal{S}_{CB}\circ\mathcal{S}_{CP} + \circ \mathcal{S}_{CA}$. Therefore it is (from \ref{izoSop} and \ref{izoTransmutacija}): + \begin{eqnarray*} + \mathcal{I}&=&\mathcal{S}_a\circ\mathcal{S}_b\circ\mathcal{S}_c=\\ + &=& + \mathcal{S}_{AC}\circ\mathcal{S}_{AP}\circ \mathcal{S}_{AB}\circ + \mathcal{S}_{BA}\circ\mathcal{S}_{BP}\circ \mathcal{S}_{BC}\circ + \mathcal{S}_{CB}\circ\mathcal{S}_{CP}\circ \mathcal{S}_{CA}=\\ + &=& + \mathcal{S}_{AC}\circ\mathcal{S}_{AP}\circ \mathcal{S}_{BP}\circ\mathcal{S}_{CP}\circ \mathcal{S}_{CA}=\\ + &=& + \mathcal{S}_{AC}\circ\mathcal{S}_{AX}\circ \mathcal{S}_{CA}=\\ + &=& + \mathcal{S}_{\mathcal{S}_{AC}(AX)}. + \end{eqnarray*} +By \ref{izoSop} the lines $a$, $b$ and $c$ belong to the same pencil. + +\item \res{Calculate the angle determined by the lines $p$ and $q$, if it holds that: +$\mathcal{S}_p\circ\mathcal{S}_q\circ\mathcal{S}_p = +\mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_q$.} + +If we multiply the given equality from the left side in order by $\mathcal{S}_q$, $\mathcal{S}_p$ and $\mathcal{S}_q$, we get an equivalent equality: $\mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_p = \mathcal{E}$ or $\left(\mathcal{S}_q\circ\mathcal{S}_p\right)^3 = \mathcal{E}$. + +If $p=q$, the last equality is clearly fulfilled. + +In the case $p\parallel q$, $\mathcal{S}_q\circ\mathcal{S}_p$ is a translation, so it is always the case that $\left(\mathcal{S}_q\circ\mathcal{S}_p\right)^3 =\mathcal{T}^3_{\overrightarrow{v}}=\mathcal{T}_{3\overrightarrow{v}}\neq \mathcal{E}$. + +If the lines $p$ and $q$ intersect, the composite $\mathcal{S}_q\circ\mathcal{S}_p$ is a rotation $\mathcal{R}_{S,\omega}$ ($p\cap q=\{S\}$ and $\omega=2\measuredangle p,q$). In this case, it is therefore the case that $\left(\mathcal{S}_q\circ\mathcal{S}_p\right)^3 =\mathcal{R}^3_{S,\omega}=\mathcal{R}_{S,3\omega}= \mathcal{E}$. This means that $3\omega=360^0$ or $\omega=120^0$, so that the lines $p$ and $q$ determine an angle of $60^0$. + +\item \res{Let $\mathcal{R}_{A,\alpha}$ and $\mathcal{R}_{B,\beta}$ be rotations in the same plane. Determine all points $X$ in this plane for which it is true that $\mathcal{R}_{A,\alpha}(X)=\mathcal{R}_{B,\beta}(X)$.} + +The condition $\mathcal{R}_{A,\alpha}(X)=\mathcal{R}_{B,\beta}(X)$ is equivalent to the condition $\mathcal{R}_{A,\alpha}\circ\mathcal{R}^{-1}_{B,\beta}(X)=X$ or $\mathcal{R}_{A,\alpha}\circ\mathcal{R}_{B,-\beta}(X)=X$. It is therefore necessary to determine the fixed points of the isometry $\mathcal{I}=\mathcal{R}_{A,\alpha}\circ\mathcal{R}_{B,-\beta}$. We will use Theorem \ref{rotacKomp2rotac}. We will consider several cases. + + \textit{1}) If $A=B$ and $\alpha-\beta=k\cdot360^0$ (for some $k\in \mathbb{Z}$), then $\mathcal{I}=\mathcal{E}$, so the condition is clearly true for every point in this plane. + +\textit{2}) If $A=B$ and $\alpha-\beta\neq k\cdot360^0$ (for + every $k\in \mathbb{Z}$), then + $\mathcal{I}=\mathcal{R}_{A,\alpha-\beta}$ and the condition + is fulfilled only for the point $X=A$. + + \textit{3}) If $A\neq B$ and $\alpha-\beta=k\cdot360^0$ (for + some $k\in \mathbb{Z}$), then $\mathcal{I}$ is a translation, + so the condition does not apply to any point of this plane. + + \textit{4}) If $A\neq B$ and $\alpha-\beta\neq k\cdot360^0$ + (for every $k\in \mathbb{Z}$), then $\mathcal{R}_{C,\alpha-\beta}$ + (where $\measuredangle CAB=\frac{\alpha}{2}$ and + $\measuredangle BAC=\frac{\beta}{2}$), which means that the condition + applies only to the point $C$. + +\item \res{The lines $p$ and $q$ intersect at an angle of $60^0$ at the center $O$ of an isosceles + triangle $ABC$. Prove that + the segments determined by the sides + of the triangle $ABC$ on the lines are congruent segments.} + +We use the rotation $\mathcal{R}_{O,120^0}$. + +\item \res{Let $S$ be the center of a regular pentagon $ABCDE$. +Prove that: + $$\overrightarrow{SA} + \overrightarrow{SB} + \overrightarrow{SC} + + \overrightarrow{SD} + \overrightarrow{SE} = \overrightarrow{0}.$$} + +Let $\overrightarrow{SA} + \overrightarrow{SB} + + \overrightarrow{SC} + \overrightarrow{SD} + \overrightarrow{SE} + = \overrightarrow{SX}$. Because $ABCDE$ is a regular pentagon, + we have $\mathcal{R}_{S,72^0}:A,B,C,D,E,S,X\mapsto + B,C,D,E,A,S,X'$. Therefore, $ \overrightarrow{SB} + + \overrightarrow{SC} + \overrightarrow{SD} + \overrightarrow{SE} + +\overrightarrow{SA} = \overrightarrow{SX'}$. This means that + $\overrightarrow{SX}= \overrightarrow{SX'}$ or + $X=X'=\mathcal{R}_{S,72^0}(X)$. Because $S$ is the only fixed point + of the rotation $\mathcal{R}_{S,72^0}$, we have $X=S$. Therefore, + $\overrightarrow{SA} + \overrightarrow{SB} + + \overrightarrow{SC} + \overrightarrow{SD} + \overrightarrow{SE} + = \overrightarrow{SS}=\overrightarrow{0}$. + +\item \res{Prove that the diagonals of a regular pentagon +intersect at points that are also +the vertices of a regular pentagon.} + +We use the rotation with center at the center of a regular pentagon by +an angle of $72^0$. + +\item \res{Let $ABP$ and $BCQ$ be two triangles with the same orientation and $\mathcal{B}(A,B,C)$. The points $K$ and $L$ are the centers of the segments $AQ$ and $PC$. Prove that $BLK$ is a triangle.} + +By rotating $\mathcal{R}_{B,-60^0}$, the points $A$ and $Q$ are mapped to the points $P$ and $C$. Therefore, the segment $AQ$ and its center $K$ are mapped to the segment $PC$ and its center $L$. From $\mathcal{R}_{B,-60^0}(K)=L$ it follows that $BLK$ is a triangle. + +\item \res{There are three concentric circles and a line in the same plane. Draw a triangle so that its vertices are on these circles in order, one side is parallel to the given line.} + +First, we draw an arbitrary triangle without the condition that one of its sides is parallel to the given line. We can do this by choosing an arbitrary vertex $A$ on one of the circles and using the rotation $\mathcal{R}_{A,60^0}$. Then we use the appropriate rotation with the center at the center of the concentric circles, which maps the drawn triangle into a triangle whose one side is parallel to the given line. + +\item \res{The point $P$ is an internal point of a regular triangle $ABC$, so that +$\angle APB=113^0$ and $\angle BPC=123^0$. Calculate the size of the angles of the triangle whose sides are in correspondence with the segments +$PA$, $PB$ and $PC$.} + +First, it is clear that $\angle APC=360^0-113^0-123^0=124^0$. We mark $P'=\mathcal{R}_{A,60^0}(P)$. $APP'$ is a regular triangle, so $PA\cong P'P$. Since $C=\mathcal{R}_{A,60^0}(B)$, we have $\triangle ABP\cong\triangle ACP'$, so $PB\cong P'C$ and $\angle APB\cong\angle AP'C$. This means that the sides $P'P$, $P'C$ and $PC$ of the triangle $PCP'$ are in correspondence with the segments $PA$, $PB$ and $PC$ in order. In this case, $\angle P'PC=124^0-60^0=64^0$, $\angle PP'C=113^0-60^0=53^0$ and $\angle PP'C=180^0-64^0-53^0=63^0$. + +\item \res{Given are the nonlinear points $P$, $Q$ and $R$. +Draw a triangle $ABC$, so +that $P$, $Q$ and $R$ are the centers of the squares that are constructed +over the sides $BC$, $CA$ and $AB$ of this triangle.} + +We find a fixed point of the composite +$\mathcal{R}_{Q,90^0}\circ\mathcal{R}_{R,90^0}\circ\mathcal{R}_{P,90^0}$. + +\item \res{Let $A$ and $B$ be points and $p$ a line in the same +plane. Prove that +the composite $\mathcal{S}_B\circ\mathcal{S}_p\circ\mathcal{S}_A$ + is a central reflection exactly when $AB\perp p$.} + +We represent the central reflections $\mathcal{S}_A$ and $\mathcal{S}_B$ +as composites +$\mathcal{S}_A=\mathcal{S}_a\circ\mathcal{S}_{a_1}$ and +$\mathcal{S}_B=\mathcal{S}_{b_1}\circ\mathcal{S}_b$, where $a$ and $b$ are perpendicular to the line $p$. + +\item \res{Let $p$, $q$ and $r$ be tangents to the triangle $ABC$ of inscribed circles, +which are parallel to its sides +$BC$, $AC$ and $AB$. Prove that +the lines $p$, $q$, $r$, $BC$, $AC$ and $AB$ determine such + a hexagon, in which the pairs of opposite sides are congruent distances.} + +We use the central reflection $\mathcal{S}_S$, where $S$ +is the center of the inscribed circle of the triangle $ABC$. + +\item \res{Draw a triangle with the data: $\alpha$, $t_b$, $t_c$.} + +First, we can draw the altitude $BB_1\cong t_b$ and the center +$T$. Because the vertex $A$ lies on the arc $l$ over the altitude $BB_1$ and +the obtuse angle $\alpha$, the vertex $B$ lies on the circle +$k(T,\frac{2}{3}t_c)$, and we have $C=\mathcal{S}_{B_1}(A)$, we get the vertex $C$ from the condition $C\in +k(T,\frac{2}{3}t_c)\cap\mathcal{S}_{l}(A)$. + +\item \res{Let $ALKB$ and $ACPQ$ be squares that are drawn outside the triangle $ABC$ +over the sides $AB$ and +$AC$, and $X$ the center of the side $BC$. Prove that +$AX\perp LQ$ and +$|AX|=\frac{1}{2}|QL|$.} + +Let $Q'=\mathcal{S}_A(C)$. In this case it holds that +$R_{A,90^0}:Q,L\mapsto Q',B$. So $QL\cong Q'B$ and $QL\perp +Q'B$ (statement \ref{rotacPremPremKot}). Because $AX$ is the median of triangle $BCQ$ for the base $BQ'$, it follows that +$\overrightarrow{AX}=\frac{1}{2}\overrightarrow{BQ'}$, therefore $AX\perp LQ$ and $|AX|=\frac{1}{2}|QL|$. + +\item \res{Let $O$ be the center of the regular triangle $ABC$ and $D$ and +$E$ points on the sides $CA$ and $CB$, so + that $CD\cong CE$ holds. The point $F$ is the fourth vertex of the parallelogram + $BODF$. Prove +that the triangle $OEF$ is regular.} + +Let $\mathcal{I}=\mathcal{T}_{\overrightarrow{OB}}\circ +\mathcal{R}_{C,-60^0}$. By izrek \ref{izoKompTranslRot} it holds that +$\mathcal{I}$ is a rotation by the same angle $-60^0$ with the center in some +point $\widehat{O}$. So +$\mathcal{T}_{\overrightarrow{OB}}\circ +\mathcal{R}_{C,-60^0}=\mathcal{R}_{\widehat{O},-60^0}$. In this case +$\mathcal{R}_{\widehat{O},-60^0}(E)=\mathcal{T}_{\overrightarrow{OB}}\circ +\mathcal{R}_{C,-60^0}(E)=F$, therefore $\widehat{O}EF$ is a regular +triangle. It remains to be proven that $\widehat{O}=O$ or +that $O$ is a fixed point of the rotation +$\mathcal{R}_{\widehat{O},-60^0}$. If +$O'=\mathcal{R}_{C,-60^0}(O)$, then $O'$ is the center of the isosceles triangle $AB'C$ (where +$C'=\mathcal{S}_{AC}(B)$). So +$\overrightarrow{O'O}=\overrightarrow{OB}$ or +$\mathcal{T}_{\overrightarrow{OB}}(O')=O$. From this it follows that +$\mathcal{R}_{\widehat{O},-60^0}(O)=O$ or $\widehat{O}=O$, which +means that $OEF$ + is a regular triangle. + +\item \res{Let $L$ be the point in which the inscribed circle of triangle +$ABC$ touches its side $BC$. +Prove: $$\mathcal{R}_{C,\measuredangle ACB}\circ\mathcal{R}_{A,\measuredangle BAC} +\circ\mathcal{R}_{B,\measuredangle CBA} =\mathcal{S}_L.$$} + +Let $M$ and $N$ be the points where the inscribed circle of the +triangle $ABC$ touches its sides $AC$ and $AB$. By the statement in +\ref{rotacKomp2rotac}, $\mathcal{R}_{C,\measuredangle ACB} +\circ\mathcal{R}_{A,\measuredangle BAC}\circ +\mathcal{R}_{B,\measuredangle CBA}=\mathcal{S}_{\widehat{L}}$. +Since: +\begin{eqnarray*} + \mathcal{S}_{\widehat{L}}(L)&=& + \mathcal{R}_{C,\measuredangle +ACB} \circ\mathcal{R}_{A,\measuredangle BAC}\circ +\mathcal{R}_{B,\measuredangle CBA}(L)=\\ +&=& + \mathcal{R}_{C,\measuredangle +ACB} \circ\mathcal{R}_{A,\measuredangle BAC}(N)= + \mathcal{R}_{C,\measuredangle ACB}(M)= L, + \end{eqnarray*} + it follows that $\widehat{L}=L$ or $\mathcal{R}_{C,\measuredangle ACB} +\circ\mathcal{R}_{A,\measuredangle BAC}\circ +\mathcal{R}_{B,\measuredangle CBA}=\mathcal{S}_L$. + +\item \res{The points $P$ and $Q$ as well as $M$ and $N$ are the centers of two squares that are drawn outside of the opposite sides of any quadrilateral. Prove that $PQ\perp MN$ and $PQ\cong MN$.} + +Let $ABCD$ be a given quadrilateral, $P$ and $Q$ the centers of the squares constructed on the sides $AB$ and $CD$, and $M$ and $N$ the centers of the squares constructed on the sides $BC$ and $AD$. The composite $\mathcal{I}=\mathcal{R}_{N,90^0}\circ \mathcal{R}_{P,90^0}$ according to the theorem \ref{rotacKomp2rotac} represents a central reflection. Because in this case $\mathcal{I}(B)=D$, the center of this reflection is actually the center of the diagonal $BD$; we denote it with $S$. So $\mathcal{R}_{N,90^0}\circ \mathcal{R}_{P,90^0}=\mathcal{S}_S$. For the point $S$ according to the same theorem \ref{rotacKomp2rotac} it holds that $\angle NPS=\frac{1}{2}90^0=45^0$ and $\angle PNS=\frac{1}{2}90^0=45^0$. This means that $PNS$ is an isosceles right-angled triangle with the base $NP$ and the right angle at the vertex $S$. Analogously, $MSQ$ is an isosceles right-angled triangle with the base $MQ$ and the right angle at the vertex $S$. From these two facts it follows that $R_{S,90^0}: M,N\mapsto Q,P$, so $MN\cong QP$ and $MN\perp QP$ (theorem \ref{rotacPremPremKot}). + + +\item \res{Let $APB$ and $ACQ$ be two right-angled triangles outside the triangle $ABC$, constructed on the sides $AB$ and $AC$. The point $S$ is the center of the side $BC$ and $O$ is the center of the triangle $ACQ$. Prove that $|OP|=2|OS|$.} + +Let $\mathcal{I}=\mathcal{R}_{O,120^0}\circ \mathcal{R}_{P,60^0}$. According to the theorem \ref{rotacKomp2rotac} we have: + $$\mathcal{I}=\mathcal{R}_{O,120^0}\circ +\mathcal{R}_{P,60^0}=\mathcal{R}_{\widehat{S},180^0} +=\mathcal{S}_{\widehat{S}},$$ where $\widehat{S}$ is the vertex of the triangle $OP\widehat{S}$ and $\angle + \widehat{S}PO=\frac{1}{2}60^0=30^0$ and $\angle + PO\widehat{S}=\frac{1}{2}120^0=60^0$. Because $\mathcal{S}_{\widehat{S}}(S)=\mathcal{I}(S)=S$ or $\widehat{S}=S$. + + If we denote $O'=\mathcal{S}_S$, then $POO'$ is a right-angled triangle, so $|OP|=|OO'|=2|OS|$. + +\item \res{Prove that the reflection in an axis and the translation of a plane commute if and only if the axis of this reflection is parallel to the translation vector.} + +If we use the statement \ref{izoTransmutacija}, we get: + \begin{eqnarray*} +\mathcal{T}_{\overrightarrow{v}}\circ \mathcal{S}_p= +\mathcal{S}_p\circ\mathcal{T}_{\overrightarrow{v}} +&\Leftrightarrow& +\mathcal{T}_{\overrightarrow{v}}\circ \mathcal{T}_p\circ\mathcal{T}^{-1}_{\overrightarrow{v}}= + \mathcal{S}_p\\ +&\Leftrightarrow& +\mathcal{S}_{\mathcal{T}_{\overrightarrow{v}}(p)}= + \mathcal{S}_p +\Leftrightarrow +\mathcal{T}_{\overrightarrow{v}}(p)= + p +\Leftrightarrow +\overrightarrow{v}\parallel + p. +\end{eqnarray*} + +\item \res{In the same plane are given a line $p$, circles $k$ and $l$, and a distance $d$. +Draw a rhombus $ABCD$ with a side that is congruent to the distance $d$, the side $AB$ lies on +the line $p$, the vertices $C$ and $D$ are in turn on the circles $k$ and $l$.} + +Use a translation for the vector $\overrightarrow{v}$, which is +parallel to the line $p$ and $|\overrightarrow{v}|=|d|$. In this +case, $D\in \mathcal{T}_{\overrightarrow{v}}(k)\cap l$. + +\item \res{Let $p$ be a line, $A$ and $B$ be points that lie on the same side +of the line $p$, and $d$ be a distance in the same plane. +Draw points $X$ and $Y$ on the line $p$ so that $AX\cong BY$ and $XY\cong d$.} + +Let $\overrightarrow{v}$ be a vector that is parallel to the line +$p$ and $|\overrightarrow{v}|=|d|$. Let +$A'=\mathcal{T}_{\overrightarrow{v}}(A)$, $Y$ be the intersection +of the line $A'B$ with the line $p$, and +$X=\mathcal{T}^{-1}_{\overrightarrow{v}}(Y)$. + +The point $Y$ lies on the line $A'B$, so $A'Y\cong YB$. Because +$\mathcal{T}_{\overrightarrow{v}}:A,X\mapsto A',Y$ is a parallelogram, $AYYA'$ is a parallelogram, so $AX\cong A'Y\cong BY$. From $\mathcal{T}_{\overrightarrow{v}}(X)=Y$ +it follows that $\overrightarrow{XY}=\overrightarrow{v}$ or $|XY|=|\overrightarrow{v}|=|d|$. + +\item \res{Let $H$ be the orthocenter of the triangle $ABC$ and $R$ be the radius of the circumscribed circle of this triangle. Prove that $|AB|^2+|CH|^2=4R^2$.} + +Let $O$ be the center of the circle $k(O,R)$ inscribed in the triangle $ABC$ +and $A'=\mathcal{S}_O(A)$. The distance $AA'$ is the diameter of the circle $k$, +so $\angle ACA'=90^0$ or $A'C\perp AC$. Because +$BH\perp AC$, $A'C\parallel BH$. Similarly +$A'B\parallel CH$, which means that the quadrilateral $BA'CH$ +is a parallelogram. Therefore $CH\cong A'B$. Because $A'BA$ +is a right triangle, by the Pythagorean Theorem +\ref{PitagorovIzrek}: +$|AB|^2+|CH|^2=|AB|^2+|A'B|^2=|AA'|^2=4R^2$. + +\item \res{Let $EAB$ be a triangle that is drawn over the side $AB$ of the square +$ABCD$. Let also $M=pr_{\perp AE}(C)$ and $N=pr_{\perp BE}(D)$ and the point $P$ +be the intersection of the lines $CM$ and $DN$. Prove that $PE\perp AB$.} + +We use a translation for the vector $\overrightarrow{CB}$. + +\item \res{Draw an isosceles triangle $ABC$ so that its vertices in order +lie on three parallel lines $a$, $b$ and $c$ in the same plane, +the center of this triangle lies on the line $s$, which intersects +the lines $a$, $b$ and $c$ through the point $S_1$.} + + First, we draw any regular triangle $A_1B_1C_1$, where $A_1\in a$, $B_1\in b$ and $C_1\in c$, + then we use a translation for the vector $\overrightarrow{S_1S}$, where $S_1$ is the center of the triangle + $A_1B_1C_1$, the point $S$ is the intersection of the line $s$ with the parallel line through the point $S_1$. + +\item \res{If a pentagon has at least two axes of symmetry, it is regular. Prove it.} + +Let $\mathfrak{G}(\mathcal{V}_5)$ be the group of symmetries + of our pentagon $\mathcal{V}_5$ and $p$ and $q$ its axes of symmetry. It is clear that this group is finite, so according to the theorem \ref{GrupaLeonardo} it represents either the cyclic group $\mathfrak{C}_n$ or the dihedral group $\mathfrak{D}_n$. Because it contains the axes of symmetry, $\mathfrak{G}(\mathcal{V}_5)=\mathfrak{D}_n$. In this case it is clear that $n\leq 5$. Because $\mathcal{S}_p, \mathcal{S}_q \in \mathfrak{G}(\mathcal{V}_5)=\mathfrak{D}_n$, the axes $p$ and $q$ intersect in some point $S$, the composition $\mathcal{S}_q\circ \mathcal{S}_p$ represents the rotation $\mathcal{R}_{S, \alpha}$. Because the group $\mathfrak{G}(\mathcal{V}_5)=\mathfrak{D}_n$ contains at least two axes of symmetry, $n\geq 2$. Therefore $n\in \{2,3,4,5\}$. In this case the basic rotation of this group (for the smallest angle) is $\mathcal{R}_{S, \frac{360^0}{n}}$. Because the pentagon has five vertices, the number $n$ is a divisor of the number 5, which means $n=5$. Therefore $\mathfrak{G}(\mathcal{V}_5)=\mathfrak{D}_5$, therefore $\mathcal{V}_5$ is a regular pentagon. + +\item \res{Let $A$, $B$ and $C$ be three collinear points. What does the composition $\mathcal{G}_{\overrightarrow{BC}}\circ \mathcal{S}_A$ represent?} + +We denote with $p$ the line passing through the points $A$, $B$ and $C$. Let $b$ and $a$ be the perpendiculars to the line $p$ in the points $A$ and $B$ and $s$ the symmetry of the line segment $BC$. Then: + $$\mathcal{G}_{\overrightarrow{BC}}\circ \mathcal{S}_A= + \mathcal{S}_s\circ \mathcal{S}_b\circ \mathcal{S}_p + \circ \mathcal{S}_p\circ \mathcal{S}_a= + \mathcal{S}_s\circ \mathcal{S}_b\circ \mathcal{S}_a.$$ + Because the lines $a$, $b$ and $s$ are perpendicular to the line $p$, they are from the group of parallel lines, + therefore according to the theorem \ref{izoSop} the composition $\mathcal{S}_s + \circ \mathcal{S}_b\circ \mathcal{S}_a$ (or $\mathcal{G}_{\overrightarrow{BC}} + \circ \mathcal{S}_A$) is the axis of symmetry. + +\item \res{Let $p$, $q$ and $r$ be lines that are not in the same plane, and let $A$ be a point in the same plane. Draw a line $s$ that goes through the point $A$ such that $\mathcal{S}_r\circ \mathcal{S}_q\circ \mathcal{S}_p(s)=s'$ and $s\parallel s'$.} + +According to the theorem \ref{izoZrcdrsprq}, the composite $\mathcal{S}_r\circ \mathcal{S}_q\circ \mathcal{S}_p(s)=s'$ is a reflection - we will denote it with $\mathcal{G}_{2\overrightarrow{PQ}}$. The line $s$ is parallel or perpendicular to the axis of the reflection. This is determined by the centers of the lines $XX'$ and $YY'$, where $X$ and $Y$ are any points and $\mathcal{S}_r \circ \mathcal{S}_q\circ \mathcal{S}_p: X, Y\mapsto X', Y'$. + +%new tasks +%___________________________________ + +\item \res{Let $Z$ and $K$ be inner points of the rectangle $ABCD$. +Draw points $A_1$, $B_1$, $C_1$ and $D_1$, which in order lie on +the sides $AB$, $BC$, $CD$ and $DA$ of this rectangle, so that it holds +$\angle ZA_1A\cong\angle B_1A_1B$, $\angle A_1B_1B\cong\angle C_1B_1C$, +$\angle B_1C_1C\cong\angle D_1C_1D$ and $\angle C_1D_1D\cong\angle KD_1A$.} + + First, we draw points $Z'=S_{CB}\circ S_{AB}(Z)=S_B(Z)$ and $K'=S_{CD}\circ S_{AD}(Z)=S_D(K)$. + Then we prove and use the fact that points $Z'$, $B_1$, $C_1$ and $K'$ are collinear. + + +\item \res{The point $A$ lies on the line $a$, and the point $B$ lies on the line $b$. +Determine the rotation that transforms the line $a$ into the line $b$ and the point $A$ into the point $B$.} + + If the lines $a$ and $b$ are parallel, the desired rotation is the central reflection with the center that is the center of the line $AB$. If the lines $a$ and $b$ intersect in the point $O$, the center of the rotation is the intersection of the simetral of the line $AB$ and the simetral of the angle $aOb$, and the rotation is equal to the angle $aOb$. + +\item \res{In the center of the square, two rectangles intersect. +Prove that these rectangles intersect the sides of the square in the points that are +the vertices of a new square.} + +We use the rotation $\mathcal{R}_{S,90^0}$, where $S$ is the center of the square. + +\item \res{Given is a circle $k$ and lines $a$, $b$, $c$, $d$ and $e$, which lie +in the same plane. Draw a pentagon on the circle $k$ with sides, which +in order are parallel to the lines $a$, $b$, $c$, $d$ and $e$.} + + + First, we plan the simetrale $p$, $q$, $r$, $s$ and $t$ of the sides + of the desired pentagon $ABCDE$, which are actually perpendicular to + the lines $a$, $b$, $c$, $d$ and $e$ from the center $O$ + of the circle $k$. The composition $\mathcal{I}= \mathcal{S}_t\circ + \mathcal{S}_s\circ \mathcal{S}_r\circ \mathcal{S}_q\circ + \mathcal{S}_p$ is an indirect isometry with fixed points + $O$ and $A$, so according to izreku \ref{izo1ftIndZrc} + $\mathcal{I}=\mathcal{S}_{OA}$. We can plan the line $OA$ as + the simetral of the line $XX'$, where $X$ is an arbitrary point and + $X'=\mathcal{I}(X)$. This allows us to construct the vertex $A$, + and then the other vertices of the pentagon $ABCDE$. + +\item \res{The point $P$ lies inside the angle $aOb$. Draw a line $p$ through the point $P$, + which with the sides $a$ and $b$ determines a triangle with the smallest area.} + +We will prove that a triangle has the smallest area, + if the line $OP$ is its median. We will mark this + triangle with $OAB$ ($A\in a$ and $B\in b$). Let $X$ and + $Y$ be the intersections of any line (different from $AB$) through + the point $P$. We will prove that the area of the triangle $XOY$ is greater + than the area of the triangle $AOB$. Without loss of generality, let + $\mathcal{B}(O,X,A)$ or $\mathcal{B}(O,B,Y)$. In this case, + we will mark points $A$ and $B$ in the following way. Let + $a'=\mathcal{S}_P(a)$, $\{B\}=b\cap a'$ and + $A=\mathcal{S}_P(B)$. We will mark $X'=\mathcal{S}_P$. It is clear that $X'\in a'$. Because $\mathcal{S}_P:X,A,P\mapsto X',B,P$, the + triangles $XAP$ and $X'BP$ are congruent, therefore they have the same + area. If we mark $p_{V_1V_2\cdots V_n}$ the area + of any polygon $V_1V_2\cdots V_n$, it is: + \begin{eqnarray*} + p_{AOB}&=&p_{AXPB}+p_{XAP}=\\ + &=&p_{AXPB}+p_{X'BP}>p_{AXPB}+p_{X'BP}+ + p_{X'BY}=\\ + &=&p_{XOY}. + \end{eqnarray*} + +\item \res{The parallelogram $PQKL$ is inscribed in the parallelogram $ABCD$ (the vertices of the first one lie on the sides +of the second one). Prove that the parallelograms have a common center.} + + We will use the central reflection $\mathcal{S}_S$, where $S$ is the intersection of the diagonals of the parallelogram $ABCD$. + +\item \res{The arcs $l_1, l_2,\cdots , l_n$ lie on the circle $k$ and the sum of their +lengths is less than the radius of this circle. Prove that there exists such a diameter +$PQ$ of the circle $k$, that none of its endpoints lies on any of +the arcs $l_1, l_2,\cdots , l_n$.} + +Let $S$ be the center of the circle $k$. Let $\mathcal{S}_S:l_1, l_2,\cdots , l_n\rightarrow l'_1, l'_2,\cdots , l'_n$. + Assume the contrary, that for each diameter $PQ$ one of its endpoints lies + on one of the arcs $l_1, l_2,\cdots , l_n$. Because $\mathcal{S}_S(P)=Q$, + both endpoints lie on some arc $l_1, l_2,\cdots , l_n$, $l'_1, l'_2,\cdots , l'_n$. + This means that every point of the circle $k$ lies on one of the arcs $l_1, l_2,\cdots , + l_n$, $l'_1, l'_2,\cdots , l'_n$, which is not possible, since the total length of these arcs is less than the circumference of the circle $k$. + +\item \res{The circle $K(S,20)$ is given. The players $\mathcal{A}$ and $\mathcal{B}$ +alternately draw circles with radii $x_i$ ($1b$) are given lengths, draw such a length $x$, that it is true:}\\ +(\textit{a}) \res{$x=\sqrt{a^2+b^2}$,} \hspace*{3mm} +(\textit{b}) \res{$x=\sqrt{a^2-b^2}$,} \hspace*{3mm} +(\textit{c}) \res{$x=\sqrt{3ab}$,}\\ +(\textit{d}) \res{$x=\sqrt{a^2+bc}$,} \hspace*{3mm} +(\textit{e}) \res{$x=\sqrt{3ab-c^2}$,} \hspace*{3mm} +(\textit{f}) \res{$x=\frac{a\sqrt{ab+c^2}}{b+c}$.} + +Use the Pythagorean Theorem \ref{PitagorovIzrek} and the altitude theorem \ref{izrekVisinski}. + +%Stewart's theorem + +\item +\res{Let $a$, $b$ and $c$ be the sides of a triangle and let $a^2+b^2=5c^2$. Prove that the centroids + $t_a$ and $t_b$ are perpendicular to each other.} + +Let $T$ be the centroid of the given triangle $ABC$, and $t_a$ and $t_b$ be the lengths of the corresponding centroids. We prove that $ATB$ is a right triangle. +By \ref{StwartTezisc} of Stewart's theorem \ref{StewartIzrek}, we have: + \begin{eqnarray*} +t_a^2=\frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4}\\ +t_b^2=\frac{a^2}{2}+\frac{c^2}{2}-\frac{b^2}{4}. + \end{eqnarray*} +From this and from the assumption $a^2+b^2=5c^2$, we get: +\begin{eqnarray*} +|AT|^2+|BT|^2 + &=& +\left(\frac{2}{3}t_a\right)^2+\left(\frac{2}{3}t_b\right)^2=\\ + &=& +\frac{4}{9}\left(\frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4} ++\frac{a^2}{2}+\frac{c^2}{2}-\frac{b^2}{4}\right)=\\ + &=& +\frac{4}{9}\left(\frac{a^2+b^2}{4}+c^2\right)=\\ + &=& c^2=|AB|^2. + \end{eqnarray*} +By the inverse Pythagorean theorem \ref{PitagorovIzrekObrat}, $ATB$ is a right triangle with hypotenuse $AB$, which means that the corresponding centroids are perpendicular. + +\item +\res{Let $a$, $b$, $c$ and $d$ be the sides, $e$ and $f$ the diagonals, and $x$ the distance determined by the centers + of sides $b$ and $d$ of a quadrilateral. Prove: +$$x^2 = \frac{1}{4} \left(a^2 +c^2 -b^2 -d^2 +e^2 +f^2 \right).$$} + +We use the consequence \ref{StwartTezisc} of Stewart's theorem \ref{StewartIzrek}. + +\item +\res{Let $a$, $b$ and $c$ be the sides of a triangle $ABC$. Prove that the distance from the center $A_1$ of side $a$ to the vertex $A'$ of the altitude on +this side is equal to: +$$|A_1A'|=\frac{|b^2-c^2|}{2a}.$$} + +Without loss of generality, we assume that $b\geq c$. +We denote $x=|A_1A'|$, $v_a=|AA'|$ and $t_a=|AA_1|$. If we use the Pythagorean Theorem \ref{PitagorovIzrek} for the triangle $AA'A_1$ and $AA'B$, we get: + \begin{eqnarray*} + v_a^2 &=& t_a^2-x^2;\\ + v_a^2 &=& c^2- \left( \frac{a}{2} -x \right)^2. + \end{eqnarray*} + After subtracting the equations and solving the obtained equation for $x$, we get: + \begin{eqnarray*} +x=\frac{1}{a}\left( t_a^2-c^2+ \frac{a^2}{2} \right). + \end{eqnarray*} +Finally, we use the relation for $t_a^2$ from the Stewart Theorem \ref{StwartTezisc}: + \begin{eqnarray*} +x &=& \frac{1}{a}\left( t_a^2-c^2+ \frac{a^2}{4} \right)=\\ + &=& \frac{1}{a}\left( \frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4}-c^2+ \frac{a^2}{4} \right)=\\ + &=& \frac{b^2-c^2}{2a}. + \end{eqnarray*} + +%Pappus and Pascal + +\item +\res{Let ($A$, $B$, $C$) and ($A_1$, $B_1$, $C_1$) be two collinear points of a plane that are not on the same line. If $AB_1\parallel A_1B$ and $AC_1\parallel A_1C$, then $CB_1\parallel C_1B$. (\textit{Pappus' Theorem}\footnote{Pappus of Alexandria\index{Pappus} (3rd century BC), Greek mathematician. This is a generalization of Pappus' Theorem (see Theorem \ref{izrek Pappus}), if we choose points $X$, $Y$ and $Z$ at infinity.})} + +We use the Tales Theorem \ref{TalesovIzrek} and the converse Tales Theorem \ref{TalesovIzrekObr}. + +%Desargues' Theorem + + +\item \label{nalPodDesarg1} +\res{Let $P$, $Q$ and $R$ be such points of sides $BC$, $AC$ and $AB$ +of the triangle $ABC$, that the lines $AP$, $BQ$ and +$CR$ are from the same pencil. Prove: If $X=BC\cap QR$, $Y=AC\cap PR$ and $Z=AB\cap PQ$, then points +$X$, $Y$ and $Z$ +are collinear.} + +We use the Desargues Theorem \ref{izrekDesarguesEvkl} for the triangle $ABC$ and $PQR$. + +\item +\res{Let $AA'$, $BB'$ and $CC'$ be the altitudes of the triangle $ABC$ and $X=B'C'\cap BC$, $Y=A'C'\cap AC$ and $Z=A'B'\cap AB$. Prove that $X$, $Y$ and $Z$ are collinear points.} + +A direct consequence of the \ref{VisinskaTocka} theorem and the previous task \ref{nalPodDesarg1}. + +\item +\res{Let $A$ and $B$ be points outside the line $p$. Draw the intersection of the lines $p$ and $AB$ without directly drawing the line +$AB$.} + +Draw any collinear points $C$, $C'$ and $S$, then: $Y=p\cap AC$, $X=p\cap BC$, $A'=SA\cap C'Y$ and $B'=SB\cap C'X$. By the Desargues theorem \ref{izrekDesarguesEvkl} the lines $AB$ and $A'B'$ intersect at the point $Z$, which is collinear with the points $X$ and $Y$, therefore lies on the line $p$. This means that the intersection of the lines $p$ and $AB$ is obtained as the intersection of the lines $p$ and $A'B'$, that is, without directly drawing the line +$AB$. + +\item +\res{Let $p$ and $q$ be lines of a plane that intersect at the point $S$, which is "outside the paper", and $A$ +a point of this plane. Draw the line that goes through the points $A$ and $S$.} + +Use the \ref{izrekDesarguesOsNesk} theorem. + +\item +\res{Draw a triangle so that its vertices lie on three given parallel lines and the sides of the triangle pass through three given +points.} + +Use the Desargues theorem \ref{izrekDesarguesEvkl}. See example \ref{zgled 3.2}. + + +%Ppotenca + +\item +\res{The circle $k(S,r)$ is given.} + +(\textit{a}) \res{What values can the power of the point have with respect to the circle $k$?} + +Since according to \ref{izrekPotenca} the power of a point $P$ with respect to the circle $k$ is obtained as: $p(S,k)=|PS|^2-r^2$, the value of the power is on the interval $[-r^2,\infty)$. + + (\textit{b}) \res{What is the smallest value of this power and for which point is this minimum value achieved?} + +From the previous example it is clear that the minimum value of the power is $-r^2$ and is achieved for the center $S$ of the circle $k$. + +(\textit{c}) \res{Determine the set of all points for which the power with respect to the circle is equal to $\lambda\in \mathbb{R}$.} + +The condition $p(S,k)=|PS|^2-r^2=\lambda$ and $|PS|^2=r^2+\lambda$ are equivalent. + +In the case when $\lambda>-r^2$, it is a circle $k(S,r^2+\lambda)$. If $\lambda=-r^2$, the desired set is equal to $\{S \}$, and in the case $\lambda<-r^2$, the empty set $\emptyset$. + +\item \res{Let $k_a(S_a,r_a)$ be inscribed and $l(O,R)$ be the circumscribed circle of some triangle. Prove the equality\footnote{The statement is a generalization of Euler's formula for a circle (see Theorem \ref{EulerjevaFormula}). \index{Euler, L.} + \textit{L. Euler} + (1707--1783), Swiss mathematician.}: + $$S_aO^2=R^2+2r_aR.$$} + +The proof is similar to the proof of Theorem \ref{EulerjevaFormula}. + + +\item +\res{Draw a circle that passes through the given points $A$ and $B$ and touches the given circle $k$.} + +The desired circle is denoted by $x$. +We draw any circle $j$, which passes through the points $A$ and $B$, and the circle $k$ intersects in points $C$ and $D$. We use the fact that the intersection of the lines $AB$ and $CD$ is the power center of the circles $k$, $j$ and $x$. + +The task is one of the ten Apollonius' problems about the touch of circles (see Section \ref{odd9ApolDotik}). + +\item +\res{Prove that the centers of the lines that are determined by the common tangents of two circles are collinear points.} + +The mentioned points lie on the power line of two circles. + +\item +\res{Draw a circle that is perpendicular to two given circles, and the third given circle +intersects in points that determine the diameter of this third circle.} + +Let the given circles be denoted in order by $k(K, r_k)$, $j(J,r_j)$ and $l(L,r_l)$, and the desired circle by $x(X,r_x)$. + +Let $P\in x\cap k$, $Q\in x\cap l$ and $R, R_1\in x\cap j$. +By assumption, $x\perp k, l$. By Theorem \ref{TangPogoj}, $XP$ and $XQ$ are tangents of the circles $k$ and $l$ in points $P$ and $Q$. Since $XP\cong XQ$, the center $X$ of the circle $X$ lies on the power line $p(k,l)$ of the circles $k$ and $l$. + +We look for another geometric location of points $X$, so that we include the condition for the circle $j$. +By assumption, $RR_1$ is the diameter of the circle $j$. This means that $J$ is the center of the line $RR_1$, and from the similarity of the triangles $XJR$ and $XJR_1$ (by the \textit{SSS} theorem \ref{SSS}) it follows that $\angle XJR=90^0$. By the Pythagorean theorem \ref{PitagorovIzrek} (for $\triangle XJR$) we then have: + \begin{eqnarray*} + |XR|^2=|XJ|^2+r_j^2. + \end{eqnarray*} +From the right triangle $XQL$ we get by the same theorem: + \begin{eqnarray*} + |XQ|^2=r_l^2-|XL|^2. + \end{eqnarray*} +Since $XR\cong XQ$, from the previous two relations it follows that $|XJ|^2+r_j^2=r_l^2-|XL|^2$ or: +\begin{eqnarray*} + |XJ|^2+|XL|^2=r_l^2-r_j^2. + \end{eqnarray*} +Therefore, the point $X$ lies on some circle $g$, which we can plot (see theorem \ref{GMTmnl}). This means that we get the point $S$ as the intersection of the circle $g$ and the power line $p(k,l)$ of the circles $k$ and $l$. + + + +%Various + +\item +\res{Let $M$ and $N$ be the intersections of the sides $AB$ and $AC$ of the triangle $ABC$ with a line that goes through the center of the inscribed circle of this triangle and is parallel to its side $BC$. Express the length of the line $MN$ as a function of the lengths of the sides of the triangle $ABC$.} + +We use theorem \ref{HarmCetSimKota} and prove $\frac{AS}{SE}=\frac{b+c}{a}$. +Then we use \ref{TalesovIzrek} and prove $\frac{MN}{BC}=\frac{AS}{AE}$. Result: $|MN|=\frac{a(b+c)}{a+b+c}$. + +\item +\res{Let $AA_1$ be the altitude of the triangle $ABC$. The points $P$ and $Q$ are the intersections of the altitudes of the angles $AA_1B$ and $AA_1C$ with the sides $AB$ and $AC$. Prove that $PQ\parallel BC$.} + +By theorem \ref{HarmCetSimKota} we have: +$$\frac{AP}{PB}=\frac{AA_1}{A_1B}=\frac{AA1}{A_1C}=\frac{AQ}{QC}.$$ +By the converse of theorem \ref{TalesovIzrekObr} it follows that $PQ\parallel BC$. + +\item +\res{In the triangle $ABC$, let the sum (or difference) of the internal angles $ABC$ and $ACB$ be equal to the right angle. Prove that $|AB|^2+|AC|^2=4r^2$, where $r$ is the radius of the circumscribed circle of this triangle.} + +Let $\alpha=\angle BAC$, $\beta=\angle ABC$ and $\gamma=\angle ACB$ and $k(O,r)$ be the circumscribed circle of the triangle $ABC$. + +If $\beta+\gamma=90^0$, then $\alpha=90^0$ (by \ref{VsotKotTrik}), which means that $ABC$ is a right triangle with hypotenuse $BC$; the statement is trivial in this case as a consequence of \ref{TalesovIzrKroz2} and Pythagoras' theorem \ref{PitagorovIzrek}. + +Let $\beta-\gamma=90^0$ or $\gamma=\beta-90^0$. We mark $C'=\mathcal{S}_O(C)$. By \ref{TalesovIzrKroz2}, we first have $\angle CAC'=90^0$. Because $BCC'A$ is a trapezoid, by \ref{TetivniPogoj}: + \begin{eqnarray*} +\angle C'CA &=& 90^0-\angle AC'C=\\ + &=& 90^0-(180^0-\beta)=\beta-90^0=\\ + &=& \gamma =\angle BCA. + \end{eqnarray*} + +By \ref{ObodObodKot}, $AC'\cong AB$, so by Pythagoras' theorem \ref{PitagorovIzrek} we get: + \begin{eqnarray*} +|AB|^2+|AC|^2=|AC'|^2+|AC|^2=|C'C|^2=4r^2. + \end{eqnarray*} + +\item +\res{Let $AD$ be the altitude of the triangle $ABC$. Prove that the sum (or difference) of the internal angles $ABC$ and $ACB$ is equal to the right angle exactly when: +$$\frac{1}{|AB|^2}+\frac{1}{|AC|^2}=\frac{1}{|AD|^2}.$$} + +We consider the triangle $ADB$ and $CDA$. + +\item +\res{Express the distance between the centroid and the center of the circumscribed circle of the triangle as a function of the lengths of its sides and the radius of the circumscribed circle.} + +We use Stewart's theorem \ref{StewartIzrek} for the triangle $OAA_1$ and \ref{StwartTezisc}; $k(O,R)$ is the circumscribed circle, $T$ is the centroid and $A_1$ is the center of the side $BC$ of the triangle $ABC$. + + Result: +$|OT|=\sqrt{R^2-\frac{1}{9} \left(a^2+b^2+c^2 \right)}$. + +\item +\res{Prove that in a triangle $ABC$, the external angle bisector at vertex $A$ and the internal angle bisectors at vertices $B$ and $C$ intersect the opposite sides at three collinear points.} + +Use Menelaus' theorem \ref{izrekMenelaj} and theorem \ref{HarmCetSimKota}. + +\item +\res{Prove that in a triangle $ABC$, the center of the altitude $AD$, the center of the inscribed circle and the point where side $BC$ touches the circumscribed circle of this triangle are three collinear points.} + +Use theorem \ref{velNalTockP'}. + +\item +\res{Prove Simson's theorem \ref{SimpsPrem} by using Menelaus' theorem \ref{izrekMenelaj}.} + +Let $S$ be an arbitrary point of the circumscribed circle of triangle $ABC$ and $P$, $Q$ and $R$ the orthogonal projections of this point on sides $BC$, $AC$ and $BC$. Use: +$\triangle SRA\sim\triangle SPC$, $\triangle SPB\sim\triangle SQA$ and $\triangle SQC\sim\triangle SRB$. + + +\item +\res{A line through point $M$ of side $AB$ of triangle $ABC$ intersects side $AC$ at point $K$. Calculate the ratio in which line $MK$ divides side $BC$, if $AM:MB=1:2$ and $AK:AC=3:2$.} + +Let $P$ be the intersection of lines $MK$ and $BC$. By Menelaus' theorem +$$-1=\frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot +\frac{\overrightarrow{CK}}{\overrightarrow{KA}}\cdot +\frac{\overrightarrow{AM}}{\overrightarrow{MB}}= +\frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot +\frac{-1}{3}\cdot\frac{1}{2},$$ so $\frac{\overrightarrow{BP}}{\overrightarrow{PC}}=6:1$. + +\item +\res{Let $A_1$ be the center of side $BC$ of triangle $ABC$ and let $P$ and $Q$ be such points of sides +$AB$ and $AC$, that $BP:PA=2:5$ and $AQ:QC=6:1$. Calculate the ratio in which line $PQ$ divides the median $AA_1$.} + +Let $R$ be the intersection of lines $PQ$ and $BC$. Use Menelaus' theorem \ref{izrekMenelaj} first for triangle $ABC$ and line $PQ$, then for triangle $AA_1C$ and the same line. The result is $17:60$. + +\item +\res{Prove that in an arbitrary triangle, the lines determined by the vertices and +the points of tangency of one circumscribed circle with the opposite sides intersect at a common point.} + +We use Ceva's theorem \ref{izrekCeva} and the big task \ref{velikaNaloga}. + +\item +\res{What does the set of all points represent, from which the tangent of the two given circles represents two compatible lines?} + +Because in this case the power of these points with respect to the circle is equal, the desired set is part of its power line, which is outside both circles. + +\item +\res{Let $PP_1$ and $QQ_1$ be the external tangents of the circles $k(O,r)$ and $k_1(O_1,r_1)$ (points $P$, $P_1$, $Q$ and $Q_1$ are the corresponding touch points). Let $S$ be the intersection of these two tangents, $A$ one of the intersections of the circles $k$ and $k_1$ and $L$ and $L_1$ the intersections +of the line $SO$ with the lines $PQ$ and $P_1Q_1$. Prove that $\angle LAO\cong\angle L_1AO_1$.} + +Without loss of generality, let $\mathcal{B}(S,O,O_1)$. +Let $h_{S,\lambda}$ be the central extension with coefficient $\lambda=\frac{\overrightarrow{SO_1}}{\overrightarrow{SO}}$. Then $h_{S,\lambda}(k)=k_1$ and: + $$h_{S,\lambda}:\hspace*{1mm} O,P,Q,L \mapsto O_1,P_1,Q_1,L_1.$$ +Let $A_1=h_{S,\lambda}(A)$. From $h_{S,\lambda}(k)=k_1$ and $A\in k$ it follows + $A_1\in k_1$. By the theorem \ref{homotOhranjaKote} we have $\angle L_1A_1O_1\cong\angle LAO$, so it is enough to prove $\angle L_1A_1O_1\cong\angle L_1AO_1$. + +From the similarity of the right-angled triangles $SP_1O_1$ and $SL_1P_1$ (theorem \ref{PodTrikKKK}) and the theorem \ref{izrekPotenca} we get: + \begin{eqnarray*} +\overrightarrow{SO_1}\cdot \overrightarrow{SL_1}=|SP_1|^2=p(S.k_1)= +\overrightarrow{SA}\cdot \overrightarrow{SA_1}. + \end{eqnarray*} +This means that $O_1L_1AA_1$ is a string quadrilateral, so $\angle L_1A_1O_1\cong\angle L_1AO_1$. + + +\item +\res{Prove that the side of a regular pentagon is equal +to the larger part of the division of the radius of the circumscribed circle of this pentagon in the golden ratio.} + +Let $k(S,r)$ be the center of the inscribed circle and $AB$ ($a=|AB|$) one side of the regular pentagon. In the triangle $ASB$ measure the internal angles: + $\angle ASB=36^0$ and $\angle SAB=\angle SBA=72^0$. + +We denote by $P$ such a point that $\mathcal{B}(B,A,P)$ and $AP\cong AS$. Because $SAP$ is an isosceles triangle with the base $PS$, by \ref{enakokraki} $\angle SPA\cong\angle PSA$. From \ref{zunanjiNotrNotr} (for the triangle $SAP$) follows + $\angle APS=\angle PSA=\frac{1}{2}72^0=36^0$. Therefore $\triangle SPB\sim\triangle ASB$ (\ref{PodTrikKKK}), so $\frac{PB}{SB}=\frac{SB}{AB}$ or $\frac{a+r}{r}=\frac{r}{a}$, from which directly follows our claim. + + +\item \label{nalPod75} +\res{Let $a_5$, $a_6$ and $a_{10}$ be the sides of a regular pentagon, hexagon and decagon, which are inscribed in the same circle. Prove that: + $$a_5^2=a_6^2+a_{10}^2.$$} + +We use the similarity of triangles from the previous task \ref{nalPod75}, but we take into account the appropriate heights of these two triangles. + +\item + \res{Let $t_a$, $t_b$ and $t_c$ be the centroids and $s$ the semi-perimeter of a triangle. Prove that: + $$t_a^2+t_b^2+t_c^2\geq s^2.$$} % zvezek - dodatni MG + +If we use \ref{StwartTezisc2}: + \begin{eqnarray*} +t_a^2+t_b^2+t_c^2 &=& \frac{3}{4}\left(a^2+ b^2+c^2\right)=\\ + &\geq& \frac{3}{4}\cdot 3\cdot\left(\frac{a+ b+c}{3}\right)^2=\\ + &=& s^2. + \end{eqnarray*} + +\end{enumerate} + +Let $A_1$ be the center of side $BC$. Then, according to the statement \ref{PloscTrik}: $p_{AA_1B}=p_{AA_1C}$ and $p_{TA_1B}=p_{TA_1C}$. If we subtract the equality, we get $p_{AA_1B}-p_{TA_1B}$ and $p_{AA_1C}-p_{TA_1C}$ or, according to the statement \ref{ploscGlavniIzrek} \textit{4)}: $p_{ABT}=p_{ATC}$. Similarly, we also get $p_{TBC}=p_{ATC}$. + + +\item \res{Let $c$ be the hypotenuse, $v_c$ the corresponding altitude, and $a$ and $b$ the catheti of the right triangle. Prove that $c+v_c>a+b$.} % zvezek - dodatni MG + +According to the statement \ref{PloscTrik} for the area $p$ of this triangle, it holds that $p=\frac{1}{2}ab=\frac{1}{2}cv_c$, so $ab=cv_c$, but if we also use the Pythagorean statement \ref{PitagorovIzrek}: + \begin{eqnarray*} + (a+b)^2=a^2+2ab+b^2= +c^2+2cv_cCD$), so that it divides this trapezoid into two figures with equal area.} + +Let $v$ be the height of the trapezoid and $a$ and $c$ be the lengths of its bases $AB$ and $CD$. +If $X$ is a point in which the line $p$ intersects the base $AB$, from the given condition it follows: +$p_{AXD}=\frac{1}{2}p_{ABCD}$ or: +\begin{eqnarray*} + \frac{|AX|\cdot v}{2} =\frac{1}{2}\cdot \frac{a+c}{2}\cdot v. + \end{eqnarray*} + +Therefore, the distance $AX$ is equal to the median of the trapezoid or $|AX|=\frac{a+c}{2}$, which allows the construction of the point $X$ and the line $p$. + + +\item \res{Draw the line $p$ and $q$, which pass through the vertex $D$ of the square $ABCD$ and divide it into areas of equal size.} + +If $P$ and $Q$ are points in which the line $p$ or $q$ intersects the side $AB$ or $BC$ of the square $ABCD$, we prove that $AP:PB=2:1$ and $BQ:QC=1:2$. + + +\item \res{Let $ABCD$ be a square, $E$ be the center of its side $BC$ and $F$ be a point for which $\overrightarrow{AF}=\frac{1}{3}\cdot \overrightarrow{AB}$. The point $G$ is the fourth vertex of the rectangle $FBEG$. What part of the area of the square $ABCD$ does the triangle $BDG$ represent?} + +Let $a$ be the length of the side of the square $ABCD$. +Let $S$ be the center of this square and $H=\mathcal{T}_{\overrightarrow{AF}}(D)$. By formulas \ref{PloscTrik}, \ref{ploscKvadr} and \ref{ploscGlavniIzrek} \textit{4)} we have: +\begin{eqnarray*} + p_{BDG} &=& p_{BSG}+p_{SDG}= p_{FSG}+p_{SHG}=\\ + &=& p_{FSH}=\frac{1}{2}\cdot |FH| \cdot |GS|= \\ + &=& \frac{1}{2}\cdot a \cdot \left( \frac{a}{2}-\frac{a}{3} \right)=\\ + &=& \frac{a^2}{12} = \frac{1}{12}\cdot p_{ABCD}. + \end{eqnarray*} + + +\item \label{nalPloKoef} +\res{Let $\mathcal{V}$ and $\mathcal{V}'$ be similar polygons with a similarity coefficient of $k$. Prove that: + $$p_{\mathcal{V}'}=k^2\cdot p_{\mathcal{V}}.$$} + +First, we prove that the statement is true for triangles (we use the formula \ref{PloscTrik}), then we use the formula \ref{ploscGlavniIzrek} \textit{4)}. + +\item \label{nalPloHeronStirik} + \res{Let $a$, $b$, $c$ and $d$ be the lengths of the sides, $s$ the semiperimeter and $p$ the area of any quadrilateral. Prove that: + $$p=\sqrt{(s-a)(s-b)(s-c)(s-d)}.$$} %Lopandic - nal 924 + +Let $A$, $B$, $C$ and $D$ be the vertices of the given quadrilateral, so that $|AB|=a$, $|BC|=b$, $|CD|=c$ and $|DA|=d$. If $ABCD$ is a parallelogram, due to its rigidity it is a rectangle (see \ref{paralelogram} and \ref{TetivniPogoj}) and the statement is trivial. + +Without loss of generality, we assume that the sides $BC$ and $AD$ intersect in some point $P$ and that $a>c$. We also mark $|PC|=x$, $|PD|=y$ and $p'=p_{PCD}$. +According to the theorem \ref{TetivniPogojZunanji}, $\angle PCD\cong \angle BAD$, therefore $\triangle CDP \cong \triangle ABP$ (theorem \ref{PodTrikKKK}). The similarity coefficient of these two triangles is $k=\frac{AB}{CD}=\frac{a}{c}$. Thus, according to the statement from the previous task \ref{nalPloKoef}, it holds: + $$\frac{p_{ABP}}{p_{CDP}}=\frac{a^2}{c^2}$$ +or, according to the theorem \ref{ploscGlavniIzrek} \textit{4)} +$$\frac{p_{ABCD}+p_{CDP}}{p_{CDP}}=\frac{a^2}{c^2}.$$ +From this, we get: +\begin{eqnarray} \label{nalPloEqnHeronStirik1} + p=\frac{a^2-c^2}{c^2}\cdot p'. + \end{eqnarray} +First, let us calculate $p'=p_{PCD}$. From the mentioned similarity $\triangle CDP \cong \triangle ABP$, it also follows: +\begin{eqnarray*} + \frac{x+b}{y}=\frac{y+d}{x}=\frac{a}{c}. + \end{eqnarray*} +If we rearrange the equations, we get a system of equations for $x$ and $y$: + \begin{eqnarray*} + && ax-cy=cd\\ + && cx-ay=-bc + \end{eqnarray*} +and its solutions: +\begin{eqnarray*} + x &=& \frac{c(ad+bc)}{a^2-c^2}\\ + y &=& \frac{c(ab+cd)}{a^2-c^2} + \end{eqnarray*} +From this, we get: +\begin{eqnarray} \label{nalPloEqnHeronStirik2} + x+y &=& \frac{c(b+d)}{a+c}\\ + x-y &=& \frac{c(d-b)}{a-c} + \end{eqnarray} +Let $s'$ be the semi-perimeter of the triangle $PCD$. By using the relations \ref{nalPloEqnHeronStirik2}, we get: + \begin{eqnarray*} + s' &=& \frac{x+y+c}{2}=\frac{c}{a-c}\cdot (s-c)\\ + s'-c &=& \frac{x+y-c}{2}=\frac{c}{a-c}\cdot (s-a)\\ + s'-x &=& \frac{c+x+y}{2}=\frac{c}{a+c}\cdot (s-d)\\ + s'-y &=& \frac{c+x+y}{2}=\frac{c}{a+c}\cdot (s-b) + \end{eqnarray*} +From the previous relations according to Heron's formula for the area of a triangle (theorem \ref{PloscTrikHeron}), it holds: +\begin{eqnarray*} + p' &=& p_{PCD}=\sqrt{s'(s'-c)(s'-x)(s'-y)}=\\ + &=& \frac{c^2}{a^2-c^2}\sqrt{(s-a)(s-b)(s-c)(s-d)}. + \end{eqnarray*} +If we insert this into the relation \ref{nalPloEqnHeronStirik1}, we get: +\begin{eqnarray*} + p = \frac{a^2-c^2}{c^2}\cdot p'=\sqrt{(s-a)(s-b)(s-c)(s-d)}. + \end{eqnarray*} + +\item \res{Let $a$, $b$, $c$ and $d$ be the lengths of the sides and $p$ the area of the tangent quadrilateral. Prove that: + $$p=\sqrt{abcd}.$$} %Lopandic - nal 925 + +Let $s$ be the semiperimeter of the given quadrilateral. +Since it is a tangent quadrilateral, for its sides according to \ref{TangentniPogoj} it holds that $a+c=b+d$. Therefore: +\begin{eqnarray*} + s-a &=& \frac{b+c+d-a}{2}=c\\ + s-b &=& \frac{a+c+d-b}{2}=d\\ + s-c &=& \frac{a+b+d-c}{2}=a\\ + s-d &=& \frac{a+b+c-d}{2}=b. + \end{eqnarray*} +Since the quadrilateral is also a taut, we can use the statement from the previous task \ref{nalPloHeronStirik} and get: +$$p=\sqrt{(s-a)(s-b)(s-c)(s-d)}=\sqrt{abcd}.$$ + +% Veckotniki + +\item \res{Given is a rectangle $ABCD$ with sides $a=|AB|$ and $b=|BC|$. Calculate the area of the figure which represents the union of the rectangle $ABCD$ and its image after reflection over the line $AC$.} + +Let $B'=\mathcal{S}_{AC}(B)$ and $D'=\mathcal{S}_{AC}(D)$, $P$ be the intersection of the lines $AB'$ and $CD$ and $S$ be the intersection of the diagonal $AC$ and $BD$ of the rectangle $ABCD$. We also mark $d=|AC|=|BD|$ and $v=|SP|$. From the similarity of the triangles $AB'C$ and $ASP$ (\ref{PodTrikKKK}) we get $v:b=\frac{d}{2}:a$ or $v=\frac{bd}{2a}$. If we use the obtained relation, \ref{PloscTrik}, \ref{ploscPravok} and \ref{ploscGlavniIzrek} and Pythagoras' theorem \ref{PitagorovIzrek}, for the area $p$ of the sought figure it holds: + +\begin{eqnarray*} + p &=& 2\cdot \left(p_{AB'C}+p_{ADC}-p_{APC}\right)= +2\cdot \left(p_{ABC}+p_{ADC}-p_{APC}\right)=\\ +&=& 2\cdot \left(p_{ABCD}-p_{APC}\right)=2\cdot \left( ab-\frac{dv}{2} \right)=\\ +&=& 2\cdot \left( ab-\frac{bd^2}{4a} \right)=2ab-\frac{b(a^2+b^2)}{2a}=\\ +&=& \frac{b(3a^2-b^2)}{2a}. + \end{eqnarray*} + +\item \res{Let $ABCDEF$ be a regular hexagon, and let $P$ and $Q$ be the centers of its sides $BC$ and $FA$. What part of the area of this hexagon does the area of the triangle $PQD$ represent?} + +We use the formula \ref{srednjTrapez}. Result: $p_{PQD}=\frac{3}{8}p_{ABCDEF}$. + + +% KKrog + +\item \res{In a square, four congruent circles are drawn so that each circle touches two sides and two other circles. Prove that the sum of the areas of these circles is equal to the area of the square of the inscribed circle.} + +If we denote the length of the square's side with $a$, then $r=\frac{a}{2}$ or $r_1=\frac{a}{4}$ is the radius of the inscribed circle or each of the smaller inscribed circles. Then the aforementioned sum of areas is equal to: +$$4\cdot r_1^2\pi=4\left(\frac{a}{4} \right)^2\cdot \pi=\left(\frac{a}{2} \right)^2\cdot \pi=r^2\pi.$$ + +\item \res{Calculate the area of the circle inscribed in the triangle with sides of lengths 9, 12 and 15.} + +We use Heron's formula \ref{PloscTrikHeron} and the formula \ref{PloscTrikVcrt}. Result: the area of the triangle $p_{\triangle}=54$; the area of the circle $p=9\pi$. + +\item \res{Let $P$ be the center of the base $AB$ of the trapezoid $ABCD$, for which $|BC|=|CD|=|AD|=\frac{1}{2}\cdot |AB|=a$. Express the area of the figure determined by the base $CD$ and the shorter circular arcs $PD$ and $PC$ of the circles with centers $A$ and $B$, as a function of the base $a$.} + +Since, by assumption, $\overrightarrow{PB}=\overrightarrow{DC}$, the quadrilateral $PBDC$ is a parallelogram, so $BC\cong PD$. Since, by assumption, $AD\cong BC$ and $AD\cong AP$, we have $PD\cong AD\cong AP$ and $APD$ is an isosceles triangle. Therefore, $\angle PAD= 60^0$. Similarly, $\angle CBP=60^0$. The desired area $p_0$ is the difference between the area of the trapezoid $p$ and twice the area of the circular sector with central angle $60^0$: +$$p_0=\frac{a^2}{12}\left(9\sqrt{3}-4\pi \right).$$ + +\item \res{The chord $PQ$ ($|PQ|=d$) of the circle $k$ touches its concurring circle $k'$. Express the area of the trapezoid determined by the circles $k$ and $k'$, as a function of the chord $d$.} + +We use the Pythagorean Theorem \ref{PitagorovIzrek}. Result: $\frac{d^2\pi}{4}$. + +\item \res{Let $r$ be the radius of the inscribed circle of the polygon $\mathcal{V}$, which is divided into triangles $\triangle_1,\triangle_2,\ldots,\triangle_n$, so that no two triangles have common inner points. Let $r_1,r_2,\ldots , r_n$ be the radii of the inscribed circles of these triangles. Prove that: + $$\sum_{i=1}^n r_i\geq r.$$} + +Let $s$ be the semiperimeter and $p$ the area of the polygon $\mathcal{V}$, and let $s_i$ be the semiperimeter and $p_i$ ($i\in \{1,2,\ldots , n\}$) the area of the triangle $\triangle_i$. If we use the formula \ref{ploscTetVec} and \ref{PloscTrikVcrt} and the fact that for each $i\in \{1,2,\ldots , n\}$ we have $s\geq s_i$, we get: + $$sr=p= \sum_{i=1}^n p_i = \sum_{i=1}^n s_ir_i\leq \sum_{i=1}^n sr_i=s\cdot\sum_{i=1}^n r_i.$$ + + +\end{enumerate} + + + + +%REŠITVE - Inverzija +%________________________________________________________________________________ + +\poglavje{Inversion} + + +\begin{enumerate} + + \item \res{Prove that the composition of two inversions $\psi_{S,r_1}$ and $\psi_{S,r_2}$ with respect to a concentric + circle represents a dilation. Determine the center and the coefficient + of this dilation.} + + From the definition of inversion it follows that the composition of inversions + $\psi_{S,r_2}\circ\psi_{S,r_1}$ + is a dilation with center $S$ and coefficient $\frac{r_2^2}{r_1^2}$. + The statement is also a direct + consequence of the task \ref{invRazteg}. + + + \item \res{Let $A$, $B$, $C$ and $D$ be four collinear points. + Construct such points $E$ and $F$, that $\mathcal{H}(A,B;E,F)$ +and $\mathcal{H}(C,D;E,F)$ hold.} + + We use the formula \ref{harmPravKrozn} - first we draw the circle, which + is perpendicular to the circle with diameters $AB$ and $CD$, + +\item \res{In a plane, given are a point $A$, a line $p$ and a circle $k$. +Draw a circle that goes through point $A$ and is +perpendicular to line $p$ and circle $k$.} + +Use inversion with center $A$. + +\item \res{Solve the third, fourth, ninth and tenth Apollonius problem.} + +For the third and fourth Apollonius problem we use inversion +with center in one of the given points. For the ninth and tenth +Apollonius problem we first draw a circle that is +concentric with the sought circle and goes through the center of one of +the circles. In this way we translate the problem to the fifth or sixth +Apollonius problem. + +\item \res{Let: $A$ be a point, $p$ be a line, $k$ be a circle + and $\omega$ be an angle in some plane. Draw a circle that goes through point $A$, touches line $p$ and with circle $k$ +determines angle $\omega$.} + +Use inversion with center $A$. + + \item \res{Determine the geometric location of the points of intersection of two circles that + touch at the arms of a given angle in two given points $A$ and $B$.} + + Use inversion with center $B$. + + \item \res{Draw a triangle, if the following data are known: +\begin{enumerate} + \item $a$, $l_a$, $v_a$ + \item $v_a$, $t_a$, $b-c$ + \item $b+c$, $v_a$, $r_b-r_c$ + \end{enumerate}} + + Use the great problem (see Theorem \ref{velikaNaloga}) and + the appropriate harmonic quadruples of points. + + +\item \res{Let $c(S,r)$ and $l$ be a circle and a line in the same plane that +have no common points. Let $c_1$, $c_2$ and $c_3$ be circles in this +plane that touch each other (two at a time) and each of them +touches also $c$ and $l$. Express the distance of point $S$ from line $l$ with +$r$\footnote{Proposal for MMO 1982. (SL 12.)}.} + +First, we prove that there exists a circle $n$, which is perpendicular to + the line $l$ and the circle $c$. Let $Y$ be one of the intersections + of the circle $n$ and the rectangle on the line $l$ from the center of the + circle $c$. We use the composition $f=\psi_i\circ \mathcal{R}\circ \psi_i$, + where $i$ is an arbitrary circle with center $Y$, $\mathcal{R}$ + is a rotation with center in the center of the circle $l'=\psi_i(l)$, + which maps the tangent points of the circles $l'$ and $c'_3=\psi_i(c_3$) to the point + $Y$. If: + $f:\hspace*{1mm}l, c, c_1, c_2, c_3\mapsto + \widehat{l}, \widehat{c}, \widehat{c_1}, \widehat{c_2}, \widehat{c_3},$ + we prove that $\widehat{l}=l$, $\widehat{c}=c$, $\widehat{c_3}$ + is a line parallel to the line $l$, and $\widehat{c_1}$ and + $\widehat{c_2}$ are congruent circles, which touch each other and + also touch the parallels $l$ and $\widehat{c_3}$. + In the end, it follows that the distance from the center of the circle $c$ to the line $l$ + is equal to $7r$. + + +\item \res{Let $ABCD$ be a regular tetrahedron. To any point +$M$, lying on the edge $CD$, we assign the point $P = f(M)$, which is the intersection of the rectangle through the point $A$ on the line $BM$ and +the rectangle through the point $B$ on the line $AM$. Determine the geometric +location of all points $P$, if the point $M$ takes all values on the edge +$CD$.} + +The point $P$ is the altitude of the triangle $ABM$. If $S$ is the center of the +edge $AB$, first we prove that $\overrightarrow{SP}\cdot +\overrightarrow{SM}=\frac{a^2}{4}$, where $a$ is the edge +of the regular tetrahedron. Then we use the inversion +$\psi_{S,\frac{a}{2}}$ (in the plane $SCD$). The geometric location of the points +is then the image of the segment $CD$ under this inversion, i.e. the corresponding circular +arc with center $S$, with endpoints in the altitudes of the triangles +$ACD$ and $BCD$. + +\item \res{Let $ABCD$ be a tangent-chord quadrilateral and $P$, $Q$, +$R$ and $S$ the points of tangency of sides $AB$, $BC$, $CD$ and $AD$ +with the inscribed circle of this quadrilateral. Prove that $PR\perp +QS$.} + +We use inversion with respect to the inscribed circle. We prove +that the vertices of the quadrilateral $ABCD$ are the vertices of a +rectangle, and the sides of this rectangle are parallel to the +distances $PR$ and $QS$. + + +\item \res{Prove that the centers of a tangent-chord quadrilateral, the inscribed and +the circumscribed circle, and the intersection of its diagonals are +collinear points (\index{izrek!Newtonov}Newton's theorem\footnote{\index{Newton, +I.}\textit{I. Newton} (1643--1727), English physicist and mathematician}).} + +We use the previous exercise and prove that the center $G$ of the +rectangle from that exercise is also the center of the line determined +by the center of the inscribed circle of the quadrilateral $ABCD$ +(the center of inversion) and the intersection of the lines $PR$ and +$QS$. The point $G$ is in fact the center of the image of the +circumscribed circle under that inversion. + + +\item \res{Let $p$ and $q$ be parallel tangents to the circle $k$. +Circle $c_1$ is tangent to the line $p$ at point $P$ and to the +circle $k$ at point $A$, circle $k_2$ is tangent to the line $q$ and +to the circles $k$ and $k_1$ at points $Q$, $B$ and $C$. Prove that +the intersection of the lines $PB$ and $AQ$ is the center of the +triangle $ABC$ of the circumscribed circle.} + +We use inversion with center $B$ and first prove that $PB$ is a +common tangent to the circles $k$ and $k_2$, and then that the +intersection of the lines $PB$ and $AQ$ is the potent center of the +circles $k$, $k_1$ and $k_2$. + +\item \res{The circles $k_1$ and $k_3$ touch each other externally in the point $P$. The circles $k_2$ and $k_2$ also touch each other externally in the same point. The circle $k_1$ intersects the circles $k_2$ and $k_4$ also in the points $A$ and $D$, the circle $k_3$ intersects the circles $k_2$ and $k_4$ also in the points $B$ and $C$. Prove that it holds\footnote{A suggestion for MMO 2003. (SL 16.)}: + $$\frac{|AB|\cdot|BC|}{|AD|\cdot|DC|}=\frac{|PB|^2}{|PD|^2}.$$} + +Let $\psi_P$ be an inversion with an arbitrary radius $r$. This +transforms the quadrilateral $A'B'C'D'$ into a parallelogram (by +the statement \ref{InverzDotik}). So $A'B'\cong C'D'$ holds. If we +use the statement \ref{invMetr}, we get + $\frac{|AB|\cdot r^2}{|PA|\cdot|PB|}=\frac{|CD|\cdot + r^2}{|PC|\cdot|PD|}$ or + $\frac{|AB|}{|CD|}=\frac{|PA|\cdot|PB|}{|PC|\cdot|PD|}$. In a + similar way from $C'B'\cong A'D'$ we get + $\frac{|CB|}{|AD|}=\frac{|PC|\cdot|PB|}{|PA|\cdot|PD|}$. By + multiplying the two relations we get: + $\frac{|AB|\cdot|BC|}{|AD|\cdot|DC|}=\frac{|PB|^2}{|PD|^2}$. + + \item \res{Let $A$ be a point that lies on the circle $k$. With the help of a + ruler only, draw a square $ABCD$ (or its vertices), that is inscribed in the given +circle.} + + First we draw a regular hexagon $AB_1B_2CD_1D_2$, that + is inscribed in the given circle. + + \item \res{Given are the points $A$ and $B$. With the help of + a ruler only, draw such a point $C$, that + $\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}$.} + + We use a similar procedure as in the problem \ref{MaskeroniSred}. + First we draw the point $X$, for which + $\overrightarrow{AX}=3\cdot \overrightarrow{AB}$, then the desired + point + $X'=\psi_k(X)$, where $k$ is a circle with the center $A$ and the radius $AB$. + + \item \res{With the help of + a ruler only, divide the given segment in the ratio $2:3$.} + +For a given line $AB$ we first draw a point $X$, for which + $\overrightarrow{AX}=2\cdot \overrightarrow{AB}$ (similarly to + the previous task), then + a point $Y$, for which + $\overrightarrow{AY}=\frac{1}{5}\cdot \overrightarrow{AX}$ holds. $Y$ + is the desired point, because + $\overrightarrow{AY}=\frac{2}{5}\cdot \overrightarrow{AB}$ holds. +\end{enumerate} +\newpage + + +\normalsize + +%________________________________________________________________________________ +% LITERATURA - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +\begin{thebibliography}{1} + + + \bibitem{Berger} Berger, M. +\emph{Geometry}, Springer-Verlag, Berlin 1987. + + \bibitem{Cofman} Cofman, J. +\emph{What to solve?}, Oxford University Press, Oxford, 1990. + + \bibitem{CoxeterRevisited} Coxeter, H. S. M.; Greitzer, S. L. +\emph{Geometry Revisited}, Random House, New York, 1976. + + \bibitem{Djerasimovic} Djerasimovi\'c, B. +\emph{Zbirka zadataka iz geometrije}, Stručna knjiga, Beograd, 1987. + + \bibitem{MMO} Djuki\'c, D.; Jankovi\'c, V.; + Mati\'c, I.; Petrovi\'c, N. +\emph{The IMO Compendium}, Springer, New York, 2006. + + \bibitem{Efimov} Efimov, N. V. +\emph{Višja geometrija}, Nauka, Moskva, 1978. + + \bibitem{Evklid} Evklid, \emph{Elementi}, Naučna knjiga, Beograd, 1949. + + \bibitem{Fetisov} Fetisov, A. I. +\emph{O euklidskoj i neeuklidskim geometrijama}, Školska knjiga, Zagreb, 1981. + + \bibitem{KratkaZgodCasa} Hawking, S. W. +\emph{Kratka zgodovina časa}, DMFA, Ljubljana, 2003. + + \bibitem{Hilbert} Hilbert, D. \emph{Osnove geometrije}, Naučno delo, Beograd, 1957. + + \bibitem{ZutaKnjiga} Lopandi\'c, D. +\emph{Geometrija}, Naučna knjiga, Beograd, 1979. + + \bibitem{Lopandic} Lopandi\'c, D. +\emph{Zbirka zadataka iz osnova geometrije}, PMF, Beograd, 1971. + + \bibitem{Lucic} Luči\'c, Z. +\emph{Euklidska i hiperbolička geometrija}, Matematički +fakultet, Beograd, 1994. + +\bibitem{Martin} \emph{Martin, G.}, Transformation Geometry, Springer-Verlang, New York, 1982. + + \bibitem{Mitrovic} Mitrovi\'c, M. +\emph{Projective geometry}, DMFA, Ljubljana, 2009. + + \bibitem{MitrovicMG} Mitrovi\'c, M.; Ognjanovi\'c, S.; Veljkovi\'c, M.; Petkovi\'c, L.; Lazarevi\'c, N. +\emph{Geometry for the first grade of Mathematical Gymnasium}, Krug, Beograd, 1996. + + \bibitem{Nice} Niče, V. +\emph{Introduction to synthetic geometry}, Školska knjiga, Zagreb, +1956. + + \bibitem{Prasalov} Prasalov, V. V. \emph{Problems in geometry}, Nauka, Moskva, 1986. + + \bibitem{Prvanovic} Prvanovi\'c, M. \emph{The basics of geometry}, Gradjevinska knjiga, Beograd, 1987. + + \bibitem{Tosic} Toši\'c, R.; Petrovi\'c, V. +\emph{A collection of problems in the basics of geometry}, Gradjevinska +knjiga, Novi Sad, 1982. + + \bibitem{Stojanovic} Stojanović, V. \emph{Matematiskop III}, Nauka, Beograd, 1988. + + \bibitem{Struik} Struik, D. J. +\emph{A brief history of mathematics}, Državna založba Slovenije, +Ljubljana, 1978. + + \bibitem{Oblika} Weeks, J. R. +\emph{The shape of space}, DMFA, Ljubljana, 1998. + +%%%% Pregledano v celoti! Roman \ No newline at end of file diff --git a/examples/book_translation/data/geometry_slovenian.tex b/examples/book_translation/data/geometry_slovenian.tex new file mode 100644 index 0000000..3991d21 --- /dev/null +++ b/examples/book_translation/data/geometry_slovenian.tex @@ -0,0 +1,32857 @@ +\documentclass[11pt]{book} + +%PAPIR - US TRADE +\paperwidth 15.24cm +\paperheight 22.86cm + +%TEKST +\textwidth 11.9cm \textheight 19.4cm +\oddsidemargin=-0.5cm +\evensidemargin=-1.2cm +\topmargin=-15mm + +\headheight=13.86pt + +%\usepackage[slovene]{babel} +\usepackage[english]{babel} + +%\usepackage[cp1250]{inputenc} +\usepackage[utf8]{inputenc} + + +\usepackage[T1]{fontenc} +\usepackage{amsmath} +\usepackage{color} +\usepackage{amsfonts} +\usepackage{makeidx} +\usepackage{calc} +\usepackage{gclc} +%\usepackage[dvips]{hyperref} +\usepackage{amssymb} +\usepackage[dvips]{graphicx} +\usepackage{fancyhdr} + +%za slike +\usepackage{caption} +\DeclareCaptionFormat{empty} + +\def\contentsname{Vsebina} + +\makeindex + +\newcommand{\ch}{\mathop {\mathrm{ch}}} +\newcommand{\sh}{\mathop {\mathrm{sh}}} +\newcommand{\tgh}{\mathop {\mathrm{th}}} +\newcommand{\tg}{\mathop {\mathrm{tg}}} +\newcommand{\ctg}{\mathop {\mathrm{ctg}}} +\newcommand{\arctg}{\mathop {\mathrm{arctg}}} +\newcommand{\arctgh}{\mathop {\mathrm{arcth}}} + +\def\indexname{Indeks} + +\definecolor{green1}{rgb}{0,0.5,0} +\definecolor{viol}{rgb}{0.5,0,0.5} +\definecolor{viol1}{rgb}{0.2,0,0.9} +\definecolor{viol3}{rgb}{0.3,0,0.6} +\definecolor{viol4}{rgb}{0.6,0,0.6} +\definecolor{grey}{rgb}{0.5,0.5,0.5} + + \def\qed{$\hfill\Box$} +\newcommand{\kdokaz}{\color{red}\qed\vspace*{2mm}\normalcolor} + +\newcommand{\res}[1]{\color{green1}\textit{#1}\normalcolor} + +\newtheorem{izrek}{Theorem}[section] +\newtheorem{lema}{Lemma}[section] +\newtheorem{definicija}{Definition}[section] +\newtheorem{aksiom}{Axiom}[section] +\newtheorem{zgled}{Exercise}[section] +\newtheorem{naloga}{Problem} +\newtheorem{trditev}{Proposition}[section] +\newtheorem{postulat}{Postulate} +\newtheorem{ekv}{E} + + +%BARVA + +\newcommand{\pojem}[1]{\color{viol4}\textit{#1}\normalcolor} + +%\newcommand{\pojemFN}[1]{\textit{#1}} + +\newcommand{\blema}{\color{blue}\begin{lema}} +\newcommand{\elema}{\end{lema}\normalcolor} + +\newcommand{\bizrek}{\color{blue}\begin{izrek}} +\newcommand{\eizrek}{\end{izrek}\normalcolor} + +\newcommand{\bdefinicija}{\begin{definicija}} +\newcommand{\edefinicija}{\end{definicija}} + +\newcommand{\baksiom}{\color{viol3}\begin{aksiom}} +\newcommand{\eaksiom}{\end{aksiom}\normalcolor} + +\newcommand{\bzgled}{\color{green1}\begin{zgled}} +\newcommand{\ezgled}{\end{zgled}\normalcolor} + +\newcommand{\bnaloga}{\color{red}\begin{naloga}} +\newcommand{\enaloga}{\end{naloga}\normalcolor} + +\newcommand{\btrditev}{\color{blue}\begin{trditev}} +\newcommand{\etrditev}{\end{trditev}\normalcolor} + + +\newcommand{\del}[1]{\chapter{#1}} +\newcommand{\poglavje}[1]{\section{#1}} +%\newcommand{\naloge}[1]{\color{red}\section*{#1}\normalcolor} +\newcommand{\naloge}[1]{\section{#1}} +\newcommand{\ppoglavje}[1]{\subsection{#1}} + +\setlength\arraycolsep{2pt} + +\author{Milan Mitrovi\'c} +\title{\textsl{\Huge{\textbf{Euclidean Plane Geometry}}}} + +\date{} + +%_________________________________________________________________________________________ + +\begin{document} +\pagestyle{fancy} +\lhead[\thepage]{\textsl{\nouppercase{\rightmark}}} +\rhead[\textsl{\nouppercase{\leftmark}}]{\thepage} +\cfoot[]{} + + + \vspace*{-12mm} + +\hspace*{24mm} \textsl{\Huge{\textbf{Euclidean }}}\\ + +\hspace*{24mm} \textsl{\Huge{\textbf{Plane}}}\\ + +\hspace*{24mm} \textsl{\Huge{\textbf{Geometry}}} + + \vspace*{8mm}\hspace*{60mm}Milan Mitrovi\'c +% \normalcolor + + \vspace*{0mm} + +\hspace*{-21mm} +\input{sl.NASL2A4.pic} + +%\color{viol1} + \hspace*{48mm}Sevnica + + \hspace*{49mm} +2013 +% \normalcolor + %slikaNova0-1-1 +%\includegraphics[width=120mm]{slikaNaslov.pdf} + + \setcounter{section}{0} + \thispagestyle{empty} + \newpage + + +%\pagecolor{white} + +%\color{viol1} +\vspace*{17mm} + \hspace*{77mm} \textit{To Boris and Jasmina} +%\normalcolor + %slikaNova0-1-1 +%\includegraphics[width=120mm]{slikaNaslov.pdf} + + + \thispagestyle{empty} +\newpage + +%________________________________________________________________________ + +%PREDGOVOR +%{\hypertarget{Vsebina}\tableofcontents} + %\printindex +\thispagestyle{empty} + + +%_______________________________________________________________________ + + +\chapter*{Preface} + +\thispagestyle{empty} + + +\thispagestyle{empty} + +The present book is the result of the experience I gained as a professor in teaching geometry at the Mathematical Gymnasium in Belgrade for many years and preparing students in Slovenia for the International Mathematical Olympiad (IMO). + +Formally, the substance is presented in such a way that it does not rely on prior knowledge of geometry. In the book, we will deal only with planar Euclidean geometry - all definitions and statements refer to the plane. + +The first two chapters deal with the history and axiomatic design of geometry. +The consequences of the axioms of incidence, congruence and parallelism are discussed in detail, while in the other two groups (axioms of order and continuity) the consequences are mostly not proven. +Chapters three and four deal with the relation of the congruence of figures, the use of the triangle congruence theorems, and a circle. +In the fifth chapter, a vectors are defined. Thales's theorem of proportion is proven. +Chapter six deals with isometries and their use. Their classification has been performed. +Chapters 7 and 8 deal with similarity transformations, figure similarity relation, and area of figures. The ninth chapter presents the inversion. +At the end of each chapter (except the introductory one) are exercises. Solutions and instructions can be found in the last, tenth chapter. + + +The book contains 341 theorems, 247 examples and 418 solved problems (28 of them from the IMO). In this sense, the book in front of you is at the same time a preparing guide for the IMO. + +For some well-known theorems and problems, are given brief historical remarks. That can help high school and college students to better understand the development of geometry over the centuries. + +A lot of help in writing the book was selflessly offered to me by Prof. Roman Drstvenšek, who read the manuscript in its entirety. With his professional and linguistic comments, he made a great contribution to the final version of the book. +In that work, he was assisted by Prof. Ana Kretič Mamič. I kindly thank both of them for the effort and time they have generously devoted to this book. + +I would especially like to thank Prof. Kristjan Kocbek, who read the partial manuscript + and with his critical remarks contributed to + significant improvement of the book. + +%I sincerely thank prof. Gordana Kenda Kozinc, who read and proofread the introductory chapter, and thanks also to prof. Alenka Brilej, who helped Roman with the language test with quite a few useful tips. + +I also thank Prof. Dr Predrag Janiči\'c, who wrote the wonderful software package \textit {GCLC} for \ LaTeX {}. Almost all pictures in this book were made with this package. + +Last but not least, I would like to thank the students of the Bežigrad Grammar School, the 1st Grammar School in Celje and the Brežice Grammar School for their inspiration and support. Students from these schools attended the renewed course of geometry which I have already taken before +years in Belgrade. + +\thispagestyle{empty} + +\vspace*{12mm} Sevnica, December 2013 \hfill Milan +Mitrovi\'c +%\newpage + +%________________________________________________________________________ + + \tableofcontents + + %\thispagestyle{empty} + +%\newpage + + + +% DEL 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% O DEDUKTIVNI IN INDUKTIVNI METODI +%________________________________________________________________________________ + + \del{Introduction} \label{pogUVOD} + +%________________________________________________________________________________ +\poglavje{Deductive and Inductive Method} \label{odd1DEDUKT} + +Že v osnovni šoli spoznamo veliko geometrijskih pojmov, kot so: +trikotnik, krožnica, pravi kot itd. Kasneje se naučimo tudi +nekaj izrekov: izreki o skladnosti trikotnikov, Pitagorov in +Talesov izrek. V začetku izrekov ne dokazujemo, ampak dejstva +ugotavljamo na osnovi večih posameznih primerov. Ta način +sklepanja se imenuje induktivna metoda. Induktivna metoda (lat. +inductio -- uvajanje) je torej način sklepanja, pri katerem od +posameznih pridemo do splošnih zaključkov. Kasneje začnemo +posamezne izreke dokazovati. Skozi te dokaze se prvič srečamo s +t. i. deduktivnim načinom sklepanja oz. z dedukcijo. Dedukcija +(lat. deductio -- izvajanje) je način sklepanja, pri katerem se od +splošnih pride do posameznih zaključkov. Ideja pri tej +metodi torej je, da z dokazovanjem izpeljemo splošni zaključek in ga +potem uporabljamo v posameznih primerih. Ker pri induktivni +metodi ne moremo preveriti vseh primerov, saj je njihovo število +najpogosteje neskončno, lahko s to metodo pridemo tudi do +napačnih zaključkov. Z deduktivno metodo dobimo vedno pravilne +zaključke, če so le predpostavke, ki jih v dokazu uporabljamo, +pravilne. Na naslednjem primeru analizirajmo obe omenjeni metodi. +Poskusimo priti do ugotovitve: + \btrditev \label{TalesUvod} + The diameter of a circle subtends a right angle to any point on the circle. + \etrditev + + +\begin{figure}[!htb] +\centering +\input{sl.1.2.1.6.pic} +\caption{} \label{sl.sl.1.2.1.6.pic} +\end{figure} + + %slikaNova0-1-1 +%\includegraphics[width=50mm]{slikaNova0-1-1.pdf} + +Če bi uporabljali induktivno metodo, bi preverjali, ali ta trditev +velja v nekih posameznih primerih; npr. v primeru, ko je vrh +kota središče polkrožnice in podobno (Figure +\ref{sl.sl.1.2.1.6.pic}). Če bi le iz teh posameznih +primerov izpeljali splošno ugotovitev, seveda ne bi mogli biti +prepričani, da v katerem od primerov, ki ga nismo preverili, ta +trditev ne drži. + +Uporabimo sedaj deduktivno metodo. Naj bo $AB$ polmer krožnice s +središčem $O$ in $L$ poljubna točka te krožnice, različna od +točk $A$ in $B$ (Figure \ref{sl.sl.1.2.1.6.pic}). Dokažimo, da je +kot $ALB$ pravi kot. Ker je + $OA\cong OB\cong OL$, sledi, +da sta trikotnika $AOL$ in $BOL$ enakokraka, zato je + $\angle ALO\cong\angle LAO=\alpha$ in $\angle BLO\cong\angle LBO=\beta$. +Tedaj je $\angle ALB=\alpha+\beta$. + Vsota notranjih kotov v +trikotniku $ALB$ je enaka $180^0$, torej je +$2\alpha+2\beta=180^0$. Iz tega sledi: + $$\angle ALB=\alpha+\beta=90^0$$ + +Opazimo, da v primeru uporabe deduktivne metode oz. pri dokazovanju +trditve nismo obravnavali neke določene točke $L$ na krožnici, +ampak poljubno točko (v splošni legi). To pomeni, da trditev +velja za vsako točko krožnice (razen $A$ in $B$), če je seveda +dokaz pravilen. Toda ali je dokaz pravilen? V tem dokazu smo +uporabljali naslednji dve trditvi: + \btrditev + If two sides in a triangle are congruent, then the angles opposite the congruent sides are congruent angles. + \etrditev + \btrditev + The sum of the interior angles of a triangle is equal to $180^0$. + \etrditev +Uporabili smo tudi pojme, kot so: enakokraki trikotnik, skladnost +kotov; v sami trditvi pa tudi pojme: premer, krožnica, kot nad +premerom in pravi kot. Da smo prepričani, ali je trditev, ki +smo jo dokazovali, točna, moramo biti gotovi, da sta tudi trditvi, +ki smo ju uporabili v dokazu, točni. V našem primeru +predpostavljamo, da smo omenjeni dve trditvi že dokazali in da +smo vpeljali vse omenjene pojme. Jasno je, da se ta problem pojavi +pri vsaki trditvi -- tudi pri dveh, na kateri smo se sklicevali v +dokazu. To zahteva določeno sistematizacijo cele geometrije. +Zastavlja se vprašanje, kako začeti, če se v dokazu vsake +trditve zopet sklicujemo na prej dokazane. Ta proces bi se potem +lahko nadaljeval v neskončnost. Tako pridemo do potrebe po +začetnih trditvah -- \index{aksiomi} \pojem{aksiomih}. Isto velja +za pojme -- potrebujemo t. i. \pojem{začetne pojme}.\index{začetni +pojmi} Na ta način je vsaka geometrija (lahko jih je torej več), +ki jo obravnavamo, odvisna od izbire začetnih pojmov in aksiomov. +Ta pristop izgrajevanja neke geometrije imenujemo +\pojem{sintetični postopek}, sami geometriji pa pravimo, da je +\pojem{sintetična geometrija}\index{geometrija!sinteti\v{c}na}. + + +%________________________________________________________________________________ +\poglavje{Basic Terms and Basic Theorems} \label{odd1POJMI} + +V neki teoriji (kot je geometrija) vsako vpeljavo novega +pojma naredimo z \index{definicija} \pojem{definicijo}, s katero ta pojem +opišemo s pomočjo + nekih začetnih ali že definiranih pojmov. + Povezave med pojmi in tudi njihove ustrezne lastnosti so dane z izjavami, + ki jih imenujemo +\pojem{trditve teorije}. Kot smo že omenili, začetne trditve +imenujemo \index{aksiomi} \pojem{aksiomi}, trditve, ki so +iz njih izpeljane, pa \pojem{izreki} te teorije. Formalno je +\pojem{dokaz} \index{dokaz izreka} nekega izreka $\tau$ zaporedje +trditev, ki logično sledijo ena iz druge, od katerih je vsaka +ali aksiom ali iz aksiomov izpeljana trditev (izrek), zadnja v tem +zaporedju pa je ravno trditev $\tau$. + + Čeprav izbira aksiomov ni enolično določena, ta ne more biti poljubna. + Pri tej izbiri je +treba paziti, da aksiomi ne pripeljejo do protislovnih trditev +oziroma da ne pride do protislovja. To pomeni, da pri neki izbiri +aksiomov ne obstaja takšna trditev, da sta ta trditev in hkrati njena +negacija izreka v tej teoriji. Potrebno je imeti tudi dovolj +aksiomov, da bi za vsako trditev, ki jo lahko formuliramo v tej +teoriji, lahko ugotovili, ali velja ali ne. To pomeni, da je +bodisi trditev bodisi njena negacija izrek v tej teoriji. Za +sistem aksiomov, ki izpolnjuje prvo zahtevo, pravimo, da je +\index{sistem aksiomov!neprotisloven} \pojem{neprotisloven}, za +tistega, ki izpolnjuje drugo zahtevo, pa pravimo, da je \index{sistem +aksiomov!popoln} \pojem{popoln}. Pri izbiri aksiomov obstaja tudi +tretja zahteva -- da je sistem aksiomov \index{sistem +aksiomov!minimalen} \pojem{minimalen}, kar pomeni, da se noben od +aksiomov ne da izpeljati iz ostalih. Omenimo, da zadnja + zahteva ni tako pomembna, kot sta prvi dve. + +Dodati moramo še, da evklidske geometrije ne gradimo +neodvisno od algebre in logike. Uporabljali bomo namreč pojme, kot +so npr. množica, funkcija, relacija z lastnostmi, ki za njih +veljajo. Uporabljali bomo tudi t. i. pravila sklepanja, kot je +npr. metoda protislovja. Za matematične discipline, ki jih na ta +način uporabljamo pri gradnji geometrije, pravimo, da so +\pojem{predpostavljene teorije}. + + + +%________________________________________________________________________________ +\poglavje{A Brief Historical Overview of the Development of Geometry} +\label{odd1ZGOD} + +Z geometrijo so se ljudje začeli ukvarjati že v rani zgodovini. + V začetku je bilo to le opazovanje karakterističnih oblik, kot sta + krožnica ali kvadrat. Po risbah, ki so bile odkrite na stenah starih jam, sklepamo, + da so se ljudje že v prazgodovini zanimali za simetrijo likov. + +V nadaljnjem razvoju je človek ugotavljal razne lastnosti +geometrijskih likov. To je bilo zaradi praktičnih potreb, npr. +merjenja površine zemljišč -- tako je tudi nastala beseda +‘‘geometrija’’. V tem obdobju se je geometrija razvijala kot +induktivna znanost. To pomeni, da so do geometrijskih trditev +prihajali z izkušnjami -- s pomočjo meritev in s preverjanjem na +posameznih primerih. V tem smislu je bila geometrija razvita pri +vseh starih civilizacijah: kitajski, indijski in še posebej +egipčanski. + +V Egiptu se je geometrija razvijala predvsem kot učenje o +meritvah. Ker je reka Nil večkrat poplavila, je +bilo potrebno zemljišča zelo pogosto znova premeriti. Razen tega so +znanje geometrije uporabljali tudi v gradbeništvu. Poznali so +npr. formulo za izračunavanje prostornine piramide in prisekane +piramide, čeprav so do nje prišli empirično. Tako je bila geometrija za +Egipčane predvsem pragmatična disciplina. +Najstarejši zapisi o tem segajo približno v leto 1700 pr. n. š. + +Tudi v Mezopotamiji so imeli razvito geometrijo merjenja +ploščin. Geometrija trirazsežnega prostora ni bila +toliko obravnavana kot pri Egipčanih. + +O kitajski geometriji ni toliko podatkov kot o egipčanski, +čeprav vemo, da je bila tudi ta zelo razvita. V +najstarejših ohranjenih zapisih najdemo opis računanja +prostornin prizme, piramide, valja, stožca, prisekane piramide in +prisekanega stožca. + +Indijska geometrija je precej mlajša od prejšnjih treh. Datira +približno v peto stoletje pr. n. š. V njej že vidimo prve +poskuse dokazovanja. Kasneje se je razvijala vzporedno s +starogrško geometrijo. + +Preobrat v razvoju geometrije se je zgodil v Stari Grčiji. Tedaj +se je prvič v zgodovini pričela v geometriji uporabljati deduktivna metoda. +Prvi geometrijski dokazi so povezani s +Talesom\footnote{Starogrški filozof in matematik \textit{Tales} +\index{Tales} iz Mileta (640--546 pr. n. š.).}. Z njegovim imenom +povezujemo znani izrek o sorazmerju odsekov pri vzporednicah. +Dokazal je tudi izrek, da so koti nad premerom krožnice pravi, +čeprav je bila ta trditev brez dokaza znana že Babiloncem 1000 let +pred tem. Ta način razvoja geometrije so nadaljevali tudi drugi +starogrški filozofi, od katerih je bil Pitagora\footnote{Starogrški +filozof in matematik \textit{Pitagora} \index{Pitagora} z otoka +Samosa (ok. 580--490 pr. n. š.).} eden najpomembnejših. +Znamenit je seveda njegov \index{izrek!Pitagorov}\pojem{Pitagorov +izrek}. Toda ta izrek so kot dejstvo poznali že Egipčani 3000 let +pr. n. š. (mogoče je bil izrek znan celo pred tem), pač pa je +Pitagora podal prvi znani dokaz. Arhimed\footnote{Starogrški filozof +in matematik \textit{Arhimed} \index{Arhimed} iz Sirakuze (287--212 pr. n. š.).} +je prvi predstavil teoretičen izračun števila $\pi$, tako da +je obravnaval v krožnico včrtane in očrtane večkotnike s $96$ +stranicami. Dokazani so bili tudi izreki o skladnosti trikotnikov. +Ob hitrem napredku geometrije, ki se je odražal v velikem številu +dokazanih izrekov, se je pokazala potreba po sistematizaciji in s +tem po vpeljavi aksiomov. Potrebo po aksiomah sta prva opisala +Platon\footnote{Starogrški filozof in matematik \textit{Platon} +\index{Platon} (427--347 pr. n. š.).} in +Aristotel\footnote{Starogrški filozof in matematik +\textit{Aristotel} \index{Aristotel} iz Aten (384--322 pr. n. š.).}. + Platon je v matematiki znan tudi po tem, da je raziskoval pravilne poliedre: + tetraeder, kocko, oktaeder, dodekaeder in ikozaeder, zato jih po njem + imenujemo tudi platonska telesa. + +Enega prvih poskusov aksiomatične zasnove geometrije -- in iz tega +časa edinega ohranjenega -- je dal najznamenitejši geometer tega časa, +Platonov učenec Evklid\footnote{Starogrški filozof in matematik +\textit{Evklid} \index{Evklid} iz Aleksandrije (ok. 330--270 pr. n. š.).}, v svojem znanem delu \textit{Elementi}, ki je sestavljeno iz +13 knjig. V njem je sistematiziral vse dotedanje znanje +geometrije. Začetne trditve je razdelil na aksiome in t. i. +postulate, od katerih so slednji čisto geometrične vsebine (danes +tudi njih imenujemo aksiomi). \textit{Elementi} so postali ena od +najpomembnejših in najvplivnejših knjig v zgodovini matematike. +Geometrija, ki jo je na ta način razvil, z manjšimi nepomembnimi +spremembami, je tista, ki se v šolah uči še danes. Dokazi, kot je +npr. ta o središčnem in obodnem kotu, so se obdržali v praktično +nespremenjeni obliki. Navedimo postulate, kot jih je podal Evklid +(Figure \ref{sl.sl.1.3.1.9.pic}): +\color{viol3} +\begin{postulat} + We can draw a straight line from any point to any point. + \end{postulat} + \begin{postulat} +We can produce a finite straight line continuously in a straight line. + \end{postulat} + \begin{postulat} +We can describe a circle with any center and distance. + \end{postulat} + \begin{postulat} +All right angles are equal to one another. + \end{postulat} + \begin{postulat} +If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the straight lines, if produced indefinitely, will meet on that side on which the angles are less that two right angles. + \end{postulat} +\normalcolor + +\begin{figure}[!htb] +\centering +\input{sl.1.3.1.9.pic} +\caption{} \label{sl.sl.1.3.1.9.pic} +\end{figure} + + %slikaNova1-3-3 +%\includegraphics[width=100mm]{slikaNova1-3-3.pdf} + + +Vendar sistem aksiomov, ki jih je podal Evklid, ni bil popoln. V +nekaterih dokazih je določene dele sprejel kot očitne in jih ni +dokazoval. Seveda ne smemo biti preveč kritični, saj je bilo to delo za tiste +čase revolucionarno. \textit{Elementi} so bili stoletja +zgled in inspiracija matematikom in so začrtali nadaljnji razvoj +geometrije vse do danes. Za nadaljnji razvoj geometrije je bil posebej +pomemben zadnji aksiom, t. i. \index{aksiom!peti Evklidov} +\pojem{peti Evklidov aksiom}. Problem njegove neodvisnosti od +ostalih aksiomov je bil odprt naslednjih 2000 let! + +Zadnji v nizu velikih starogrških matematikov so bili +Apolonij\footnote{Starogrški matematik \textit{Apolonij} +\index{Apolonij} iz Perge (262--190 pr. n. š.).}, +Menelaj\footnote{Starogrški matematik \textit{Menelaj} +\index{Menelaj} iz Aleksandrije + (ok. 70--130).} in Pappus\footnote{Starogrški matematik \textit{Pappus} + \index{Pappus} + iz + Aleksandrije (ok. 290--350).}. Apolonij je v svoji knjigi \textit{Razprava + o presekih stožca} definiral elipso, parabolo in hiperbolo kot + preseke ravnine in krožnega (neskončnega) stožca. Tako je lahko določene + njihove + lastnosti obravnaval hkrati, kar je bil za tisti čas precej + sodoben pristop. Menelaj in Pappus sta dokazala določene izreke, + ki so postali aktualni šele v 19.~stoletju z razvojem projektivne geometrije. + Torej so bile ideje teh treh matematikov zelo sodobne in na nek način + lahko rečemo, da so bili na pragu odkritja prve neevklidske geometrije. + +Po dokončnem padcu Stare Grčije pod Rimsko cesarstvo se je +obdobje slavne starogrške geometrije končalo. Čeprav so Stari Rimljani +prevzeli velik del starogrške kulture in so gradili ceste, +vodovode in tako dalje, je zanimivo, da se nikoli niso preveč zanimali za +starogrško teoretično matematiko. Tako je njihov prispevek k razvoju +geometrije zelo skromen. + +Pomembno vlogo v nadaljnjem razvoju geometrije so prevzeli Arabci. +Najprej je treba povedati, da so nam vsa dela Starih Grkov +vključno z Evklidovimi \textit{Elementi} danes znana zato, ker so jih +takrat prevedli in tako ohranili ravno Arabci. Od ustanovitve +Bagdada leta 762 so v naslednjih 100-tih letih prevedli večino del +starogrške in indijske matematike. Naredili so tudi sintezo +starogrškega pretežno geometričnega in indijskega pretežno +algebričnega pristopa. Omenimo, da je sama beseda \pojem{algebra} +arabskega izvora. Poleg tega so Arabci nadaljevali razvoj +\pojem{trigonometrije}, ki so jo zasnovali že Stari Grki. A. R. al-Biruni\footnote{Arabski matematik \textit{ A. R. al-Biruni} \index{al-Biruni, A. R.} (973--1048).} je dokazal danes znani \pojem{sinusni izrek}. + +V Evropi se je razvoj geometrije začel v 12. stoletju, ko so preko +Španije in Sicilije znanja prinašali arabski in judovski matematiki. +(Evklidovi \textit{Elementi} so bili prevedeni iz arabskega jezika +približno leta 1200); toda pravi razcvet je Evropa doživela šele v +16. stoletju. V srednjeveškem obdobju se je namreč matematika zelo +počasi razvijala. V srednjem veku so se zahodnoevropski matematiki +šele učili starogrško geometrično dediščino iz arabskih prevodov, +vendar ta proces ni bil hiter. Ko se je to znanje akumuliralo in so +se družbeno-politični pogoji spremenili, se je v +razvoju geometrije začelo novo obdobje. Prve nove rezultate so dali italijanski +matematiki tistega časa, ki so veliko pozornosti posvetili +konstrukcijam s pomočjo ravnila in šestila. + +Kot smo že prej omenili, je imel peti Evklidov aksiom zelo velik vpliv na +nadaljnji razvoj geometrije. Zaradi svoje +formulacije, ki ni tako enostavna kot pri prejšnjih aksiomih, in +tudi zaradi pomena je veliko matematikov v tistem obdobju menilo, +da ga ni potrebno obravnavati kot aksiom, ampak se ga lahko z ostalimi aksiomi dokaže kot +izrek. Če preberemo druge Evklidove začetne trditve, zares drži, da je +peti aksiom bolj zapleten. + Problem neodvisnosti petega aksioma od ostalih je v naslednjih stoletjih okupiral mnoge + matematike. Vse do druge polovice 19. + stoletja problem ni bil rešen. V mnogih poskusih dokazovanja petega + aksioma iz preostalih so bile uporabljene trditve, katerih dokaz + je bil izpuščen. +Kasneje se je pokazalo, da se teh trditev niti ne da dokazati iz +ostalih aksiomov, če se iz njihovega seznama izpusti peti aksiom. +Podobno kot iz njih sledi peti aksiom, tudi te trditve +sledijo iz petega aksioma (seveda z uporabo ostalih aksiomov). +Zato jih imenujemo \pojem{ekvivalenti petega Evklidovega aksioma}. +Navedimo nekaj primerov teh ekvivalentov (Figure +\ref{sl.sl.1.3.1.9a.pic}): + +\color{blue} +\begin{ekv} +If $ ABCD $ is a quadrilateral with two right angles on the side $BC$ +and the sides $AB$ and $CD$ are congurent, then the two remain angles +of this quadrilateral are also right angles.\footnote{Ta ekvivalent je postavil +italijanski matematik \index{Saccheri, G. G.} \textit{G. G. +Saccheri} (1667--1733).} +\end{ekv} +\begin{ekv} +A line perpendicular to one arm of an acute angle intersects +his other arm. +\end{ekv} +\begin{ekv} +Every triangle can be circumscribed. +\end{ekv} +\begin{ekv} +If three angles of a quadrilateral are right angles, then the fourth angle is also a right angle.\footnote{\index{Lambert, J. H.}\textit{J. +H. Lambert} (1728--1777), francoski matematik.} +\end{ekv} +\begin{ekv} +The sum of the interior angles in every triangle is $180^0$.\footnote{\index{Legendre, A. M.} \textit{A. M. Legendre} +(1752--1833), francoski matematik.} +\end{ekv} +\begin{ekv} +For any given line $p$ and point $A$ not on $p$, in the plane containing both line $p$ and point $A$ there is just one line + through point $A$ that do not intersect line $p$\footnote{\index{Playfair, J.} \textit{J. Playfair} +(1748--1819), škotski matematik.}. + \end{ekv} +\normalcolor + + +\begin{figure}[!htb] +\centering +\input{sl.1.3.1.9a.pic} +\caption{} \label{sl.sl.1.3.1.9a.pic} +\end{figure} + +Matematikom je torej uspelo dokazati peti Evklidov aksiom s +pomočjo vsake od teh trditev, toda sčasoma se je pokazalo, da se + nobene od njih brez petega aksioma sploh ne da dokazati. Zato +so te trditve, kot smo že omenili, petemu aksiomu ekvivalentne. +Danes se najpogosteje uporablja Playfairjev ekvivalent, ki je bil kasneje +namesto petega aksioma dodan k Evklidovim aksiomom. + +Toda kako so matematiki ugotovili, da se peti Evklidov aksiom ne +more izpeljati iz ostalih aksiomov? Samo dejstvo, da ga niso +uspeli dokazati, še ni pomenilo, da to ni mogoče. Odgovor na to +vprašanje je prišel konec 19. stoletja in je, kot bomo videli, +za razvoj geometrije prinesel veliko več kot samo dejstvo o +nedokazljivosti petega aksioma. + +Naslednja prelomnica v razvoju geometrije je bilo odkritje +\index{geometrija!neevklidska}\pojem{neevklidskih geometrij} v 19. stoletju. Za začetnika tega razvoja +štejemo N.~I.~Lobačevskega\footnote{\index{Lobačevski, N. +I.}\textit{N. I. Lobačevski} (1792--1856), ruski matematik.}. Tudi on +je obravnaval problem neodvisnosti petega Evklidovega aksioma. +Izhajajoč iz njegove negacije oz. iz negacije Playfairjeve +ekvivalentne trditve je Lobačevski postavil predpostavko, da skozi +točko, ki ne leži na neki premici, obstajata vsaj dve premici, ki se s +to premico ne sekata, in sta komplanarni. V želji, da bi prišel do +protislovja (tako bi bil peti aksiom dokazan), je zgradil celo +zaporedje novih trditev. Ena med njimi je na primer ta, da je vsota notranjih kotov +trikotnika vedno manjša od iztegnjenega kota. Toda nobena teh trditev +ni bila v protislovju z ostalimi aksiomi, če seveda s seznama izključimo +peti Evklidov aksiom. Iz tega je dobil idejo, da je mogoče zgraditi +popolnoma novo geometrijo, ki je neprotislovna in temelji na vseh +Evklidovih aksiomih z izjemo petega, ki ga zamenjamo z njegovo +negacijo. Danes to geometrijo imenujemo +\index{geometrija!hiperbolična} \pojem{hiperbolična geometrija} +ali \pojem{geometrija Lobačevskega}. + +Neodvisno od Lobačevskega je do istih rezultatov prišel tudi J. +Bolyai\footnote{\index{Bolyai, J.} \textit{J. Bolyai} (1802--1860), +madžarski matematik.}. Kot se to pogosto dogaja, ideje Lobačevskega v času njegovega življenja žal +niso bile sprejete. Popolno potrditev teh idej +oziroma dokaz neprotislovja te nove geometrije je konec 19. stoletja, tj. šele po smrti Lobačevskega, predstavil A. +Poincar\'{e}\footnote{\index{Poincar\'{e}, J. H.} \textit{J. H. +Poincar\'{e}} (1854--1912), francoski matematik.}. +Poincar\'{e} je zgradil model, na +osnovi katerega je pokazal, da bi morebitno protislovje geometrije +Lobačevskega pomenilo hkrati protislovje Evklidove geometrije. +Kasneje je prišlo do odkritja tudi drugih neevklidskih geometrij. + +Čeprav je bil konec 19. in +v začetku 20. stoletja sistem aksiomov Evklidove geometrije že skoraj popolnoma zgrajen, je prvi +pravilen popolni sistem dal D. Hilbert\footnote{\index{Hilbert, D.} +\textit{D. Hilbert} (1862--1943), nemški matematik.} v svoji znani +knjigi \textit{Osnove geometrije}, objavljeni leta 1899. Zelo podoben +sistem aksiomov uporabljamo v skoraj nespremenjeni +obliki tudi danes. + +Vzporedno z raziskovanjem problema petega Evklidovega aksioma in +razvojem neevklidskih geometrij so se v obravnavi geometrije razvile +tudi druge pomembne +metode. Že okoli leta 1637 je R. +Descartes\footnote{\index{Descartes, R.} \textit{R. Descartes} +(1596--1650) francoski matematik.} v svoji knjigi +\textit{Geometrija} pokazal, da lahko vsako točko v ravnini opišemo z +ustreznim parom dveh realnih števil in podobno v prostoru kot +trojico treh realnih števil. To je povezal s pojmom koordinatne +predstavitve odvisnosti ene količine (funkcije) od druge +(spremenljivke), ki je bil znan že prej. Danes takšen +način določanja točk v prostoru po njem imenujemo \pojem{kartezični +koordinatni sistem}. Premice in ravnine potem lahko opisujemo kot +množice rešitev ustreznih linearnih enačb, kjer so neznanke +koordinate točk. + +Tako sta se pod vplivom idej F. +Vi\'{e}teja\footnote{\index{Vi\'{e}te, F.} \textit{F. Vi\'{e}te} +(1540--1603), francoski matematik.}, Descartesa in P. +Fermata\footnote{\index{Fermat, P.} \textit{P. Fermat} (1601--1665), +francoski matematik.} začeli razvijati dve zelo pomembni matematični +disciplini - najprej \index{geometrija!analitična} +\pojem{analitična geometrija}, nato še \index{linearna algebra} +\pojem{linearna algebra}, ki predstavljata povezavo med algebro in +geometrijo. Nadaljnji razvoj teh dveh disciplin je omogočal razvoj +\index{geometrija!večdimenzionalna} \pojem{večdimenzionalne +geometrije}, v kateri se lahko obravnavajo prostori, dimenzije, večje +od tri, saj v algebri ni takšnih omejitev, kot jih imamo v +geometrični percepciji prostora. Tako lahko definiramo t. i. +\pojem{politope} -- objekte večdimenzionalnega prostora, ki so analogija +dvodimenzionalnih večkotnikov in tridimenzionalnih poliedrov. + +Kasneje je prišlo tudi do odkritja drugih neevklidskih geometrij. +V 19. stoletju se je razvila še t. i. \index{geometrija!projektivna}\pojem{projektivna geometrija}, vendar +njen razvoj ni potekal aksiomatično kot pri hiperbolični +geometriji, ustrezen sistem aksiomov je bil postavljen šele +kasneje. V tej geometriji v ravnini ni premic, ki se ne sekata. + +Eden od prvih motivov za začetek razvoja + projektivne geometrije izvira iz slikarstva oziroma iz želje, da se + občutek trirazsežnega prostora prenese v ravnino. Že v zelo + zgodnjem slikarstvu srečamo zelo pomembno lastnost -- da sta + vzporednici na sliki predstavljeni kot premici, ki se sekata. + + V 15. stoletju so se italijanski umetniki zelo zanimali za + geometrijo prostora. Teorijo perspektive je prvi obravnaval F. + Brunellechi\footnote{\index{Brunellechi, F.} \textit{F. + Brunellechi} (1377--1446), italijanski arhitekt.} leta 1425. + Njegovo delo sta +nadaljevala L. B. Alberti\footnote{\index{Alberti, L. B.} + \textit{L. B. + Alberti} (1404--1472), italijanski matematik in slikar.} in A. + D\"{u}rer\footnote{\index{D\"{u}rer, A.} \textit{A. + D\"{u}rer} (1471--1528), nemški slikar.}. Albertijeva knjiga iz + leta + 1435 predstavlja prvo predstavitev središčne projekcije. + + Za začetek razvoja projektivne geometrije kot matematične + discipline smatramo obdobje, ko sta + J. Kepler\footnote{\index{Kepler, J.} \textit{J. Kepler} (1571--1630), + nemški astronom.} in G. Desargues\footnote{\index{Desargues, G.} + \textit{G. Desargues} + (1591--1661), francoski arhitekt.} + neodvisno drug od drugega + vpeljala pojem točk v neskončnosti. + Kepler je pokazal, da ima parabola dve gorišči, od katerih je eno + točka v neskončnosti. Desargues + je leta 1639 pisal: ‘‘Dve vzporednici imata skupni konec na + neizmerni oddaljenosti.’’ Leta 1636 je napisal knjigo o perspektivi, + in leta 1639 še o stožnicah. Znameniti \textit{Desarguesov izrek} + je objavil leta 1648. + + Z nadaljnjim razvojem projektivne geometrije povezujemo francoske matematike. + Genialni B. Pascal\footnote{\index{Pascal, B.} \textit{B. Pascal} (1623--1662), + francoski filozof in matematik.} je + že kot šestnajstletnik dokazal pomemben izrek o + stožnicah, ki ga danes po njem imenujemo \textit{Pascalov izrek}. + Ta izrek, ki je eden izmed osnovnih izrekov + projektivne geometrije, je bil objavljen leta 1640. + G. Monge\footnote{\index{Monge, G.} \textit{G. Monge} + (1746--1818), francoski matematik.} + je bil med prvimi matematiki, ki ga lahko smatramo za specialista; je + namreč prvi pravi geometer. \pojem{Opisno geometrijo}je razvil kot + posebno disciplino. V njegovem raziskovanju v opisni geometriji + najdemo veliko idej projektivne geometrije. + Najbolj originalen Mongeov učenec je + bil J. V. Poncelet\footnote{\index{Poncelet, J. V.} + \textit{J. V. Poncelet} (1788--1867), francoski matematik.}. + Čeprav je že Pappus\footnote{\index{Pappus} \textit{Pappus iz Aleksandrije} (3. stol.), starogrški matematik.} + odkril prve projektivne izreke, jih je Poncelet + s popolnoma projektivnim načinom sklepanja dokazal šele v 19. stoletju. + Leta 1822 je + Poncelet objavil svoj znani ‘‘Traktat o projektivnih lastnostih likov’’, + v katerem se pojavljajo vsi pomembni pojmi, karakteristični za + projektivno + geometrijo: harmonična četverica, perspektivnost, projektivnost, + involucija itd. Poncelet je vpeljal premico v neskončnosti za vse + ravnine, ki so vzporedne dani ravnini. Poncelet in J. D. + Gergonne\footnote{\index{Gergonne, J. D.} \textit{J. D. Gergonne} (1771--1859), francoski matematik.} sta + neodvisno + drug od drugega proučevala dualnost v projektivni geometriji, + C.~J.~Brianchon\footnote{\index{Brianchon, C. J.} \textit{C. J. Brianchon} + (1783--1864), francoski matematik.} pa + je dokazal izrek, ki je dualen Pascalovem izreku. + M.~Chasles\footnote{\index{Chasles, M.} \textit{M. Chasles} (1793--1880), francoski matematik.} je bil zadnji iz + velike šole + francoskih projektivnih geometrov tistega časa. + + Tipičen predstavnik t. i. čiste geometrije + (danes bi rekli sintetične geometrije) je bil + J. Steiner\footnote{\index{Steiner, J.} \textit{J. Steiner} + (1796--1863), + švicarski geometer.}. + Steiner je razvijal projektivno geometrijo zelo sistematično, + od perspektivnosti do projektivnosti in potem do stožnic. + + Sredi 19. stoletja so primat v razvoju projektivne geometrije + prevzeli nemški matematiki. Negovali so sintetični pristop + h geometriji. Vsi matematiki do tedaj so projektivno geometrijo + zasnovali na evklidski metrični geometriji -- z dodajanjem + točk v neskončnosti. Toda C.~G.~C.~Staudt + \footnote{\index{Staudt, K. G. C.} \textit{C. G. C. Staudt} (1798--1867), + nemški matematik.} + je bil prvi, ki jo je poskusil osamosvojiti ter + zasnovati samo na incidenčnih aksiomih, brez pomoči metrike. + Tako je prišlo do ukinitve razlike med točkami v neskončnosti in navadnimi + točkami + oziroma prehoda z razširjenega evklidskega na projektivni prostor. + + F. Klein\footnote{\index{Klein, F. C.} \textit{F. C. Klein} (1849--1925), nemški matematik.} je leta 1871 + projektivni geometriji postavil algebraične temelje s pomočjo + t. i. \pojem{homogenih koordinat}, ki sta jih leta 1827 neodvisno drug od drugega, + odkrila K. W. Feuerbach\footnote{\index{Feuerbach, K. W.} \textit{K. W. Feuerbach} + (1800--1834), nemški matematik.} + in A. F. M\"{o}bius\footnote{\index{M\"{o}bius, A. F.} \textit{A. F. M\"{o}bius} + (1790--1868), nemški matematik.}. + A. Cayley\footnote{\index{Cayley, A.} \textit{A. Cayley} (1821--1895), angleški matematik.} in + Klein + najdeta uporabo projektivne geometrije v drugih neevklidskih geometrijah. + Odkrila sta model hiperbolične geometrije in modele drugih geometrij + v projektivni. + + Prva, ki sta popolnoma aksiomatično zasnovala projektivno geometrijo, sta + bila G. Fano\footnote{\index{Fano, G.} \textit{G. Fano} (1871--1952), italijanski matematik.} + leta 1892 in M. Pieri\footnote{\index{Pieri, M.} \textit{M. Pieri} (1860--1913), italijanski matematik.} + leta 1899. + +Zaradi svoje relativne enostavnosti je bil razvoj klasične +(sintetične) projektivne geometrije konec 19. stoletja že skoraj +popolnoma zaključen. Njen razvoj se danes nadaljuje v okviru +drugih teorij -- posebej v algebri in algebraični geometriji kot +$n$-razsežna projektivna geometrija. + + + +G. F. B. Riemann\footnote{\index{Riemann, G. F. B.} \textit{G. F. +B. Riemann} (1828--1866), nemški matematik.} je že v svoji knjigi +\textit{O domnevah, ki ležijo v osnovi geometrije} definiral +prostor poljubne dimenzije, ki ni vedno konstantne ukrivljenosti. +Po njem ga danes imenujemo \pojem{Riemannov metrični +prostor}\index{Riemannovi prostori}. Evklidovo geometrijo potem +dobimo kot poseben primer: če je ukrivljenost konstantna in enaka +0; hiperbolično geometrijo dobimo, če izberemo, da je +ukrivljenost konstantna in negativna. Če je ukrivljenost +konstantna in pozitivna, dobimo t. i. +\index{geometrija!eliptična} \pojem{eliptično geometrijo}. +Slednja geometrija je pravzaprav projektivna geometrija, če ji +dodamo metriko. To raziskovanje je bilo hkrati začetek razvoja +nove discipline v matematiki, t. i. \index{geometrija!diferencialna} \pojem{diferencialne geometrije}. + +Če razmišljamo o neevklidskih geometrijah, se nam mogoče zdi čudno, +da se v matematiki sploh lahko obravnava več različnih teorij, kot +so npr. evklidska geometrija in hiperbolična geometrija, ki sta v +nasprotju druga z drugo. Za sodobno matematiko je največjega pomena, +da sta obe geometriji določeni s sistemoma aksiomov, ki sta (vsak zase) neprotislovna in popolna. Na vprašanje, katera od teh +dveh geometrij velja, je nesmiselno iskati odgovor v okviru +matematike. To je namreč odvisno od tega, za katere aksiome smo se +odločili. Takšno vprašanje bi bilo enako vprašanju, kateri aksiomi +veljajo. Toda aksiome po definiciji privzamemo brez dokaza. Seveda +lahko zastavimo vprašanje, kakšna je geometrija prostora v fizičnem +smislu in kako jo lahko opišemo z aksiomi. + +Za odgovor na to vprašanje je potrebna fizikalna interpretacija +osnovnih geometričnih pojmov. Na primer, premico je najbolj naravno +interpretirati kot svetlobni žarek. V tem smislu se izkaže, da +fizični prostor ni evklidski. Določen ni niti s hiperbolično +geometrijo. S pojavom Einsteinove\footnote{\index{Einstein, A.} +\textit{A. Einstein} (1879--1955), slaven nemški fizik.} teorije +relativinosti v začetku 20. stoletja se je izkazalo, da je v +prostoru vesoljskih razsežnosti bolj ugodno uporabljati neevklidsko +geometrijo s spremenljivo ukrivljenostjo (Riemannovi metrični +prostori!). Lahko rečemo, da je geometrija vesolja lokalno različna +v vsaki točki, odvisno od bližine in velikosti neke mase. +Einsteinova teorija nam tudi pove, da sta prostor in čas medsebojno +povezana in da niti čas (kar je seveda presenetljivo) ne poteka +enako v vsaki točki vesolja. V zvezi z omenjeno povezavo prostora in +časa je pomemben t. i. \pojem{štirirazsežni prostor +Minkowskyega}\footnote{Ta prostor je odkril \index{Minkowsky, H.} +\textit{H. Minkowsky} (1864--1909), nemški matematik.}. + + Že od 20-ih let 20. stoletja in razvoja teorije prapoka vemo, + da vesolje ni statično in da se širi. Od tega, kakšna + je njegova usoda, je odvisno, katera geometrija ga globalno najbolje + opisuje. Toda še vedno ne vemo dokončno, kakšna je oblika + vesolja niti kakšna je njegova usoda. Niti ne vemo, ali je vesolje + končno ali neskončno. Kot piše S. + Hawking\footnote{\index{Hawking, S.} \textit{S. Hawking} (1942), + angleški fizik. + Eden najbolj značilnih teoretičnih fizikov našega časa, + ki je največ vplival na sodobno predstavo o vesolju.} + v svoji znani popularni knjigi iz leta 1988 + \textit{Kratka zgodovina časa} (\cite{KratkaZgodCasa}), je celo vesolje + morda končno in neomejeno. Slednje se zdi paradoksalno, + čeprav si lahko to predstavljamo ako, de si namesto trirazsežnega zamišljamo + ‘‘dvorazsežno vesolje’’. + Tako bi bitja tega dvorazsežnega vesolja lahko + enkrat ugotovila, da njihovo vesolje pravzaprav ni + ravnina ampak sfera, ki je končna, toda neomejena. + Sfera je del trirazsežnega prostora. Tako si teoretično + lahko predstavljamo vesolje kot trirazsežno sfero v + štirirazsežnem prostoru. Trirazsežna sfera je ena od 3-mnogoterosti. + V tem primeru bi bila geometrija vesolja globalno eliptična. + + +Čeprav je samo vprašanje oblike in usode vesolja vprašanje +teoretične fizike in kozmologije, vidimo, kako je sodobna +geometrija (neevklidske geometrije, geometrija večrazsežnih +prostorov itd.) tesno povezana s tem problemom (\cite{Oblika}). +Pomembno je razumeti, da se geometrija (in vsaka druga matematična +disciplina) razvija in obravnava kot abstraktna disciplina, v kateri +so nam fizične interpretacije le inspiracija -- v tem smislu nam +ostajajo le aksiomi in začetni pojmi, na katerih potem izgrajujemo +matematično teorijo. + + +Na koncu omenimo še enega najpomembnejših geometrov 20. +stoletja H. S. M. Coxeterja\footnote{\index{Coxeter, H. S. M.} +\textit{H. S. M. Coxeter} (1907--2003), kanadski matematik. Eden +največjih geometrov 20. stoletja.}. Coxeter je naprej raziskoval politope +v poljubnih razsežnostih, predvsem pravilne politope. Razen tega se +je veliko ukvarjal z grupami izometrij v hiperbolični geometriji in +z večrazsežno hiperbolično geometrijo. + +Potrebno je dodati, da z vsem tem, kar smo +povedali, razvoj geometrije še zdaleč ni končan. Ravno obratno -- nasprotno običajni +predstavi -- geometrija in na splošno matematika se sedaj razvijata +še hitreje kot kadar koli prej. Tudi danes obstaja v matematiki (v geometriji posebej iz neevklidskih +geometrij) veliko +problemov, ki so še vedno nerešeni. + + + + + + + +% DEL 2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% AKSIOMI RAVNINSKE EVKLIDSKE GEOMETRIJE +%________________________________________________________________________________ + + +\del{Axioms of Planar Euclidean Geometry} \label{pogAKS} + +V nadaljevanju bomo ilustrirali aksiomatično zasnovo ravninske +evklidske geometrije. Navedli bomo začetne pojme in začetne izreke - +aksiome, potem pa izpeljali še nekaj novih pojmov in izrekov. +Omenimo, da smo izbrali ravninske aksiome, ker se bomo v tej knjigi +ukvarjali le z geometrijo evklidske ravnine. + +Naj bo $\mathcal{S}$ neprazna množica. Njene elemente imenujemo +\index{točka} \pojem{točke} in jih označujemo z $A, B, C, \ldots$ +Določene podmnožice množice $\mathcal{S}$ imenujemo \index{premica} +\pojem{premice} in jih označujemo z $a, b, c, \ldots$ Množico +$\mathcal{S}$ (množico vseh točk) imenujemo tudi \index{ravnina} +\pojem{ravnina}. Razen teh sta začetna pojma tudi dve relaciji na +množici $\mathcal{S}$. Prva je \index{relacija!$\mathcal{B}$} +\pojem{relacija $\mathcal{B}$} in se nanaša na tri točke. Dejstvo, +da so točke $A$, $B$ in $C$ v tej relaciji, bomo označevali z +$\mathcal{B}(A,B,C)$ in brali: Točka $B$ je med točkama $A$ in $C$. +Druga je \index{relacija!skladnosti parov točk} \pojem{relacija +skladnosti parov točk}; dejstvo, da so pari točk $A, B$ in $C, D$ v +tej relaciji, bomo označevali z $(A,B) \cong (C,D)$ in brali: Par +točk $(A,B)$ je skladen s parom točk $(C,D)$. + + S pomočjo omenjenih začetnih pojmov lahko +definiramo tudi naslednje izpeljane pojme: + +Če točka $A$ pripada premici $p$ ($A\in p$), oz. premica $p$ +vsebuje točko $A$ ($p\ni A$), bomo rekli, da + točka $A$\index{relacija!leži na premici} \pojem{leži na} premici $p$, oz. da premica $p$ \index{relacija!poteka skozi točko}\pojem{poteka + skozi} točko $A$. + Za tri ali +več točk pravimo, da so \index{kolinearne točke}\pojem{kolinearne}, +če ležijo na isti premici, sicer so +\index{nekolinearne točke}\pojem{nekolinearne}. Dve +različni premici se \pojem{sekata}, če njun presek (presek dveh podmnožic) +ni prazna množica. Njun presek imenujemo \index{presečišče +dveh premic} \pojem{presečišče} dveh premic. Poljubno neprazno +podmnožico $\Phi$ množice $\mathcal{S}$ ($\Phi\subset\mathcal{S}$) imenujemo \index{lik} \pojem{lik}. Pravimo, da lika $\Phi_1$ in $\Phi_2$ \index{lika!sovpadata}\pojem{sovpadata} (oz. sta \index{lika!identična}\pojem{identična}), če je $\Phi_1=\Phi_2$. + + Sedaj bomo +navedli tudi osnovne trditve - aksiome. Po svoji naravi so +razdeljeni v pet skupin: + +\begin{enumerate} + \item incidence axioms (three axioms), + \item ordering axioms (four axioms), + \item congruence axioms (four axioms), + \item continuity axiom (one axiom), + \item Playfair's axiom (one axiom). +\end{enumerate} + + + +%________________________________________________________________________________ + \poglavje{Incidence Axioms} + \label{odd2AKSINC} + + Ker premice kot začetni pojmi predstavljajo določene množice točk, + lahko za +točke in premice obravnavamo ustrezne relacije med elementi +in množicami: $\in$ in $\ni$ - relaciji imenujemo tudi +\index{relacija!incidencije}\pojem{relaciji incidence}. Aksiomi te +skupine opisujejo ravno osnovne lastnosti teh relacij (Figure +\ref{sl.aks.2.1.1.pic}): + +\vspace*{3mm} + + \baksiom \label{AksI1} For every pair of distinct points $A$ and $B$ + there is exactly one line $p$ such that $A$ and $B$ lie on $p$. + \eaksiom + + \baksiom \label{AksI2} + For every line there exist at least two distinct points such that both lie on. + \eaksiom + + \baksiom \label{AksI3} There exist three points that do not all lie on any one line. + \eaksiom + +\vspace*{3mm} + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.1.1.pic} +\caption{} \label{sl.aks.2.1.1.pic} +\end{figure} + + + + Iz prvih dveh aksiomov \ref{AksI1} in \ref{AksI2} sledi, da je vsaka premica določena s + svojima dvema različnima točkama. Zato +premico $p$, ki je določena s točkama $A$ in $B$, imenujemo tudi +premica $AB$. + + Iz prvega aksioma \ref{AksI1} sledi, da je presečišče dveh + premic, ki se sekata, ena sama točka. Če bi namreč dve premici + imeli še eno skupno točko, bi po tem aksiomu sovpadali (bili bi identični), + pri definiciji premic, ki se sekata, pa smo zahtevali, da + sta različni. + Dejstvo, da se premici $p$ in $q$ sekata v točki $A$, bomo + zapisali $p\cap q=\{A\}$ ali krajše $p\cap q=A$ (Figure \ref{sl.aks.2.1.2.pic}). + + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.1.2.pic} +\caption{} \label{sl.aks.2.1.2.pic} +\end{figure} + + + +Tretji aksiom \ref{AksI3} lahko povemo tudi takole: Obstajajo +vsaj tri točke, ki so nekolinearne. + + Tako smo izpeljali prve posledice aksiomov incidence; + zaradi enostavnosti jih nismo izrazili v obliki izrekov. To so skoraj vse posledice, ki izhajajo iz prve skupine aksiomov. + Zaradi tega je geometrija, ki temelji le na aksiomih incidence, + preveč enostavna. V njej bi lahko dokazali le obstoj treh točk in treh + premic. Torej potrebujemo nove aksiome. + + + +%________________________________________________________________________________ + \poglavje{Ordering Axioms} + \label{odd2AKSURJ} + +Aksiomi v tej skupini opisujejo osnovne karakteristike relacije +$\mathcal{B}$, ki smo jo navedli kot osnovni pojem. +\vspace*{3mm} + + + \baksiom \label{AksII1} If $\mathcal{B} (A, B, C)$, then $A$, $B$ + and $C$ are three distinct collinear points, and also + $\mathcal{B} (C, B, A)$ (Figure \ref{sl.aks.2.2.1.pic}). + \eaksiom + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.1.pic} +\caption{} \label{sl.aks.2.2.1.pic} +\end{figure} + + + \baksiom \label{AksII2} If $A$, $B$ and $C$ are three distinct + collinear points, + exactly one of the relations holds: $\mathcal{B}(A,B,C)$, + $\mathcal{B}(A,C,B)$, $\mathcal{B}(C,A,B)$ + (Figure \ref{sl.aks.2.2.2.pic}). + \eaksiom + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.2.pic} +\caption{} \label{sl.aks.2.2.2.pic} +\end{figure} + + + + \baksiom \label{AksII3} Given a pair of distinct points $A$ and $B$ there is a point $C$ on line $AB$, so that + is $\mathcal{B}(A,B,C)$ + (Figure \ref{sl.aks.2.2.3.pic}). + \eaksiom + +\vspace*{-1mm} + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.3.pic} +\caption{} \label{sl.aks.2.2.3.pic} +\end{figure} + + \baksiom \label{AksPascheva}\index{aksiom!Paschev} + (Pasch's\footnote{\index{Pasch, M.} + \textit{M. Pasch} + (1843--1930), nemški matematik, ki je vpeljal pojem urejenosti točk + v svojih ‘‘Predavanjih o novejši geometriji’’ iz leta 1882. Te + aksiome sta kasneje dopolnila italijanski matematik \index{Peano, G.} \textit{G. Peano} + (1858--1932), v ‘‘Načelih geometrije’’, nato pa še nemški matematik + \index{Hilbert, D.}\textit{D. Hilbert} (1862--1943) v svoji znani knjigi + ‘‘Osnove geometrije’’ iz leta + 1899.} axiom) + Let $A$, $B$ and $C$ be three noncollinear points and $l$ be a line that does not contain point $A$. + If there is a point $P$ on $l$ that is $\mathcal{B}(B,P,C)$ then either $l$ contains a point $Q$ that is $\mathcal{B}(A,Q,C)$ or $l$ contains a point $R$ that is $\mathcal{B}(A,R,B)$ (Figure \ref{sl.aks.2.2.4.pic}). + \eaksiom + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.4.pic} +\caption{} \label{sl.aks.2.2.4.pic} +\end{figure} + + +V prejšnjem aksiomu nismo posebej poudarili, da premica $l$ leži v ravnini $ABC$, ker gradimo ravninsko evklidsko geometrijo, kjer vse točke ležijo v isti ravnini. + + Na tem mestu ne bomo dokazali vseh posledic aksiomov urejenosti. +Formalna izpeljava vseh dejstev namreč ni tako enostavna in bi vzela veliko prostora. Večino dokazov lahko bralec poišče v \cite{Lucic}. + +Dokažimo prvo posledico aksiomov urejenosti. + + + \bizrek \label{izrekAksUrACB} + Given a pair of distinct points $A$ and $B$ there is a point $C$, so that + is $\mathcal{B}(A,C,B)$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.5.pic} +\caption{} \label{sl.aks.2.2.5.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Po aksiomu \ref{AksI1} obstaja natanko ena premica, ki poteka skozi točki $A$ in $B$ - označimo jo z $AB$. +Po aksiomu \ref{AksI3} obstajajo vsaj tri nekolinearne točke. +Obstaja torej vsaj ena točka izven premice $AB$ - označimo jo z $D$ + (Figure \ref{sl.aks.2.2.5.pic}). Naprej po +aksiomu \ref{AksII3} obstaja takšna točka $E$, da velja +$\mathcal{B}(B,D,E)$, nato pa še takšna točka $F$, da velja +$\mathcal{B}(A,E,F)$. $A$, $B$ in $E$ so nekolinearne točke, +saj bi sicer točka $D$ ležala na premici $AB$ (aksiom \ref{AksI1}). +Premica $FD$ ne poteka skozi točko $A$, ker bi bile po aksiomu \ref{AksI1} točke $F$, $D$, $A$ in $E$ +kolinearne, z njimi pa tudi točka $B$. To pa ni možno, ker bi iz tega sledilo, da točka $D$ leži na premici $AB$. +Uporabimo sedaj Paschev aksiom \ref{AksPascheva} na +točkah $A$, $B$ in $E$ ter premici $FD$. Premica $FD$ namreč seka premico $EB$ v takšni točki $D$, +da je $\mathcal{B}(B,D,E)$, zato seka bodisi premico $AE$ v takšni točki $F$, da je $\mathcal{B}(A,F,E)$ bodisi +premico $AB$ v takšni točki $C$, da je $\mathcal{B}(A,C,B)$. Ker je že $\mathcal{B}(A,E,F)$, po aksiomu \ref{AksII2} ne more biti tudi $\mathcal{B}(A,F,E)$. Torej premica $FD$ seka +premico $AB$ v taki točki $C$, za katero velja $\mathcal{B}(A,C,B)$. +\kdokaz + +Relacija $\mathcal{B}$ in aksiomi urejenosti, ki se nanašajo nanjo, nam omogočajo definiranje novih pojmov. + + + Naj bosta $A$ in $B$ poljubni različni točki. + \index{daljica!odprta}\pojem{Odprta daljica} $AB$ z oznako $(AB)$ je množica vseh točk + $X$, za + katere velja $\mathcal{B}(A,X,C)$ (Figure \ref{sl.aks.2.2.6.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.6.pic} +\caption{} \label{sl.aks.2.2.6.pic} +\end{figure} + + + Če odprti daljici $AB$ + dodamo točki $A$ in $B$, dobimo \index{daljica}\pojem{daljico} + (ali \index{daljica!zaprta} \pojem{zaprto daljico}) $AB$, + ki jo označimo tudi z $[AB]$. + Točki $A$ in $B$ + sta njeni \index{krajišče daljice}\pojem{krajišči}, ostale njene točke pa so \pojem{notranje točke} te daljice (Figure \ref{sl.aks.2.2.6.pic}). Še bolj formalno: daljica (oz. zaprta daljica) je unija odprte daljice in množice $\{A,B\}$ oz. $[AB]=(AB)\cup \{A,B\}$. + +Na podoben način definiramo tudi + \index{daljica!polodprta} \pojem{polodprto daljico}: $(AB]=(AB)\cup \{B\}$, oz. $[AB)=(AB)\cup \{A\}$ (Figure \ref{sl.aks.2.2.6a.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.6a.pic} +\caption{} \label{sl.aks.2.2.6a.pic} +\end{figure} + + Iz aksioma \ref{AksII1} neposrednono sledi, da sta daljici $AB$ in $BA$ isti. Iz istega aksioma sledi tudi, da je daljica $AB$ podmnožica premice $AB$. Zato pravimo, da daljica $AB$ \pojem{leži na premici} $AB$, + premico $AB$ + pa imenujemo \index{nosilka!daljice} \pojem{nosilka daljice} $AB$ (Figure \ref{sl.aks.2.2.6b.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.6b.pic} +\caption{} \label{sl.aks.2.2.6b.pic} +\end{figure} + +Po izreku \ref{izrekAksUrACB} ima daljica $AB$ razen svojih krajišč $A$ in $B$ vsaj še eno točko $C_1$. Na ta način lahko dobimo neskončno zaporedje točk $C_1$, $C_2$, ..., za katere velja $\mathcal{B}(A, C_n, C_{n-1})$ ($n\in \{2,3,\cdots\}$) (Figure \ref{sl.aks.2.2.6c.pic}). Na tem mestu ne bomo formalno dokazovali dejstva, da so vse točke iz zaporedja različne in vse ležijo na daljici $AB$. Iz te trditve pa sledi, da ima vsaka daljica (posledično tudi premica) neskončno mnogo točk. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.6c.pic} +\caption{} \label{sl.aks.2.2.6c.pic} +\end{figure} + +Definirajmo še relacijo $\mathcal{B}$, ki se nanaša na več kot tri kolinearne točke. Pravimo, da je $\mathcal{B}(A_1,A_2,\ldots,A_n)$ ($n\in\{4,5,\ldots\}$), če za vsak $k\in\{1,2,\ldots,n-2\}$ velja $\mathcal{B}(A_k,A_{k+1},A_{k+2})$ (Figure \ref{sl.aks.2.2.6d.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.6d.pic} +\caption{} \label{sl.aks.2.2.6d.pic} +\end{figure} + + Naj bo $S$ točka, ki leži na premici $p$. Na množici $p\setminus \{S\}$ (vseh točk premice $p$ brez točke $S$) definirajmo dve relaciji. +Pravimo, da sta točki $A$ in $B$ ($A,B\in p\setminus \{S\}$) \index{relacija!na različnih straneh točke} \pojem{na različnih straneh točke} $S$ (kar označimo z $A,B\div S$), če je $B(A,S,B)$, sicer sta točki $A$ in $B$ ($A,B\in p\setminus \{S\}$) \index{relacija!na isti strani točke} \pojem{na isti strani točke} $S$ (kar označimo z $A,B\ddot{-} S$). Torej za točki $A,B\in p\setminus \{S\}$ velja $A,B\ddot{-} S$, če ni $A,B\div S$ (Figure \ref{sl.aks.2.2.7.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.7.pic} +\caption{} \label{sl.aks.2.2.7.pic} +\end{figure} + + +Naj bosta $A$ in $B$ različni točki. Množico vseh takšnih točk $X$, za katere je $B,X\ddot{-} A$ vključno s točko $A$, imenujemo \index{poltrak}\pojem{poltrak} $AB$ z \pojem{začetno točko} ali \pojem{izhodiščem} $A$. Premica $AB$ je +\index{nosilka!poltraka} \pojem{nosilka poltraka} $AB$ (Figure \ref{sl.aks.2.2.8.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.8.pic} +\caption{} \label{sl.aks.2.2.8.pic} +\end{figure} + +Že iz same definicije sledi, da je poltrak podmnožica svoje nosilke oz. da leži na svoji nosilki. Iz relacije $B,X\ddot{-} A$ sledi, da so $B$, $X$ in $A$ kolinearne točke, zato točka $X$ leži na premici $AB$. + +Drugih pomembnih lastnosti daljice in poltraka, ki jih bomo kasneje uporabljali, na tem mestu ne bomo dokazovali. Povejmo nekaj teh lastnosti. + +If $C$ is the an interior point of the line segment $AB$, then that line segment can be expressed as the union +the line segments $AC$ and $CB$. + + \bizrek \label{izrekAksIIDaljica} + If $C$ is an interior point of the line segment $AB$, then that line segment can be expressed as the union + the line segments $AC$ and $CB$ (Figure \ref{sl.aks.2.2.9.pic}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.9.pic} +\caption{} \label{sl.aks.2.2.9.pic} +\end{figure} + + + \bizrek \label{izrekAksIIPoltrak} + Each point lying on the line determines exactly + two rays on it. The union of these rays is equal to that line (Figure \ref{sl.aks.2.2.9.pic}). + \eizrek + +Dokaz prejšnjega izreka temelji na dejstvu, da je relacija $\ddot{-} A$ ekvivalenčna relacija, ki ima dva razreda. Vsak od razredov je ustrezen odprti poltrak. + +Poltraka iz prejšnjega izreka, ki sta določena z isto začetno točko na premici, imenujemo \index{poltrak!komplementarni}\pojem{komplementarna poltraka}. + +Pojma daljica in poltrak nam omogočata definiranje novih pojmov. + +Naj bodo $A_1$, $A_2$, ... $A_n$ takšne točke v ravnini, da nobene tri v zaporedju niso kolinearne. Unijo daljic $A_1A_2$, $A_2A_3$,... $A_{n-1}A_n$ imenujemo \index{lomljenka} \pojem{lomljenka} $A_1A_2\cdots A_n$ ali \index{poligonska +črta}\pojem{poligonska črta} $A_1A_2\cdots A_n$ (Figure \ref{sl.aks.2.2.10.pic}). Točke $A_1$, $A_2$, ... $A_n$ so \index{oglišče!lomljenke} \pojem{oglišča lomljenke}, +daljice $A_1A_2$, $A_2A_3$,... $A_{n-1}A_n$ pa \index{stranica!lomljenke} \pojem{stranice lomljenke}. Stranici lomljenke s skupnim ogliščem sta \index{sosednji stranici!lomljenke} \pojem{sosednji stranici lomljenke}. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.10.pic} +\caption{} \label{sl.aks.2.2.10.pic} +\end{figure} + +Če stranice lomljenke nimajo skupnih točk, razen sosednjih stranic, ki imata skupno oglišče, takšno lomljenko imenujemo \index{lomljenka!enostavna} \pojem{enostavna lomljenka} (Figure \ref{sl.aks.2.2.10a.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.10a.pic} +\caption{} \label{sl.aks.2.2.10a.pic} +\end{figure} + +Lomljenka $A_1A_2\cdots A_nA_{n+1}$, pri kateri je $A_{n+1}=A_1$ in so $A_n$, $A_1$ in $A_2$ nekolinearne točke, se imenuje \index{lomljenka!sklenjena} \pojem{sklenjena lomljenka} $A_1A_2\cdots A_n$ (Figure \ref{sl.aks.2.2.10b.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.10b.pic} +\caption{} \label{sl.aks.2.2.10b.pic} +\end{figure} + +Posebej nas bodo zanimale \pojem{enostavne sklenjene lomljenke} (Figure \ref{sl.aks.2.2.10b.pic}). + +Naj bosta $p$ in $q$ dva poltraka s skupnim izhodiščem $O$ (Figure \ref{sl.aks.2.2.10c.pic}). Unijo teh dveh poltrakov imenujemo \index{kotna lomljenka} \pojem{kotna lomljenka} $pq$ (ali $pOq$). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.10c.pic} +\caption{} \label{sl.aks.2.2.10c.pic} +\end{figure} + + +Za nek lik $\Phi$ pravimo, da je \index{lik!konveksen}\pojem{konveksen}, če je za poljubni njegovi točki $A,B\in \Phi$ daljica $AB$ podmnožica tega lika oz. če velja naslednje (Figure \ref{sl.aks.2.2.10d.pic}): + $$(\forall A)(\forall B)\hspace*{1mm} (A,B\in \Phi \Rightarrow [AB]\subseteq \Phi).$$ +Za lik, ki ni konveksen, pravimo, da je \index{lik!nekonveksen}\pojem{nekonveksen} (Figure \ref{sl.aks.2.2.10d.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.10d.pic} +\caption{} \label{sl.aks.2.2.10d.pic} +\end{figure} + +Neposredno iz definicije sledi, da je premica konveksen lik. Kot posledico aksiomov te skupine je mogoče dokazati, da sta tudi daljica in poltrak konveksna lika. + + +Za nek lik $\Phi$ pravimo, da je +\index{lik!povezan}\pojem{povezan lik}, če za vsaki njegovi točki $A,B\in \Phi$ obstaja lomljenka $AT_1T_2\cdots T_nB$, ki je podmnožica tega lika oz. če velja naslednje (Figure \ref{sl.aks.2.2.10e.pic}): + $$(\forall A\in \Phi)(\forall B\in \Phi)(\exists T_1,T_2,\cdots , T_n)\hspace*{1mm} AT_1T_2\cdots T_nB\subseteq \Phi.$$ + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.10e.pic} +\caption{} \label{sl.aks.2.2.10e.pic} +\end{figure} + + Za lik, ki ni povezan, pravimo, da je \index{lik!nepovezan}\pojem{nepovezan}. + + Jasno je, da je vsak konveksen lik tudi povezan. Za lomljenko je dovolj vzeti kar daljico $AB$. Obratno seveda ne velja. Obstajajo liki, ki so povezani, niso pa konveksni, kar bomo ugotovili kasneje. + + + +Sedaj bomo definirali dve relaciji, ki sta analogni z relacijama $\ddot{-} S$ in $\div S$. + Naj bo $p$ premica, ki leži v ravnini $\alpha$ (ker aksiomatsko gradimo le evklidsko geometrijo ravnine, so pravzaprav vse točke, ki za nas obstajajo, v tej ravnini). Na množici $\alpha\setminus p$ (vseh točk razen točk premice $p$) definirajmo dve relaciji. +Pravimo, da sta točki $A$ in $B$ ($A,B\in \alpha\setminus p$) \index{relacija!na različnih bregovih premice} \pojem{na različnih bregovih premice} $p$ (kar označimo z $A,B\div p$), če ima daljica $AB$ s premico $p$ skupno točko, sicer sta točki $A$ in $B$ ($A,B\in \alpha\setminus p$) \index{relacija!na istem bregu premice} \pojem{na istem bregu premice} $p$ (kar označimo z $A,B\ddot{-} p$). Torej za točki $A,B\in \alpha\setminus p$ je $A,B\ddot{-} p$, če ni $A,B\div p$ (Figure \ref{sl.aks.2.2.11.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.11.pic} +\caption{} \label{sl.aks.2.2.11.pic} +\end{figure} + + +Naj bo $A$ točka, ki ne leži na premici $p$. Množico vseh takšnih točk $X$, za katere je $A,X\ddot{-} p$, imenujemo +\index{polravnina!odprta}\pojem{odprta polravnina} $pA$. Unija odprte polravnine $pA$ in premice $p$ je \index{polravnina!zaprta}\pojem{zaprta polravnina} oz. kar +\index{polravnina}\pojem{polravnina} $pA$. Premica $p$ je \index{rob!polravnine} \pojem{rob} te polravnine (Figure \ref{sl.aks.2.2.11a.pic}). Če točki $B$ in $C$ ležita na robu $p$ polravnine $pA$, bomo to ravnino imenovali tudi polravnina $BCA$. Razen tega bomo polravnine označevali tudi z grškimi črkami $\alpha$, $\beta$, $\gamma$,... + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.11a.pic} +\caption{} \label{sl.aks.2.2.11a.pic} +\end{figure} + +Podobno kot pri poltraku je mogoče dokazati (kot posledico aksiomov te skupine), da vsaka premica $p$ v ravnini določa dve polravnini $\alpha$ in $\alpha'$, ki imata premico $p$ za rob (Figure \ref{sl.aks.2.2.11a.pic}). Pravimo, da sta si v tem primeru $\alpha$ in $\alpha'$ \index{polravnina!komplementarna}\pojem{komplementarni polravnini}. +Izkaže se, da je unija dveh komplementarnih polravnin cela ravnina. Podobno kot pri poltraku, tudi dokaz omenjenih trditev temelji na dejstvu, da je relacija $\ddot{-} p$ ekvivalenčna relacija z dvema razredoma. Vsak od razredov je ustrezna odprta polravnina. + +Naj bo $pq$ oz. $pOq$ kotna lomljenka. Definirajmo novo relacijo na množici vseh točk ravnine razen točk, ki ležijo na lomljenki. Pravimo, da sta točki $A$ in $B$ na isti strani kotne lomljenke $pq$ (kar označimo z $A,B\ddot{-} pq$), če obstaja lomljenka $AT_1T_2\cdots T_nB$, ki kotne lomljenke $pq$ ne seka oz. z njo nima skupnih točk (Figure \ref{sl.aks.2.2.12.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.12.pic} +\caption{} \label{sl.aks.2.2.12.pic} +\end{figure} + + Tudi relacija $\ddot{-} pq$ je ekvivalenčna relacija, ki ima dva razreda. Unijo vsakega od teh dveh razredov s kotno lomljenko $pq$ imenujemo + \index{kot}\pojem{kot} $pq$, ki ga označimo z $\angle pq$, oz. $\angle pOq$. + Kotna lomljenka torej določa dva kota. Dilemo, za katerega od kotov gre pri oznaki $\angle pOq$, bomo kmalu odpravili. + Poltraka $p$ in $q$ sta +\index{krak!kota}\pojem{kraka kota} in točka $O$ \index{vrh kota}\pojem{vrh kota}. +Če sta $P\in p$ in $Q\in q$ točki, ki ležita na krakih kota $pOq$ in se razlikujeta od njegovega vrha $O$, bomo kot imenovali tudi kot $POQ$ in označili z $\angle POQ$ (Figure \ref{sl.aks.2.2.12a.pic}). Če vemo, za kateri kot gre, ga bomo označevali kar z njegovim vrhom: $\angle O$. Razen tega bomo kote označevali tudi z grškimi črkami $\alpha$, $\beta$, $\gamma$,... + +Vse točke kota $pOq$, ki ne ležijo na nobenem od obeh krakov $p$ in $q$, imenujemo \index{notranje točke!kota} \pojem{notranje točke kota}, množico vseh teh točk pa \index{notranjost!kota}\pojem{notranjost kota}. Jasno je, da gre za točke ustreznega razreda, ki jih določa relacija $\ddot{-} pq$. Točke drugega razreda so \index{zunanje!točke kota}\pojem{zunanje točke kota}, cel razred pa \index{zunanjost!kota}\pojem{zunanjost kota}. Točke, ki ležijo na krakih, oz. na kotni lomljenki $pOq$, ki kot $pOq$ določa, so \index{robne točke!kota}\pojem{robne točke kota}, cela lomljenka pa je \index{rob!kota}\pojem{rob kota}. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.12a.pic} +\caption{} \label{sl.aks.2.2.12a.pic} +\end{figure} + +Če sta kraka kota komplementarna poltraka, takšen kot imenujemo +\index{kot!iztegnjeni}\pojem{iztegnjeni kot} (Figure \ref{sl.aks.2.2.12b.pic}). Kot množica točk je ta kot v bistvu enak polravnini z robom, ki je nosilka obeh krakov. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.12b.pic} +\caption{} \label{sl.aks.2.2.12b.pic} +\end{figure} + +Če kotna lomljenka $pOq$ ne določa iztegnjenenega kota oz. ni enaka premici, se izkaže, da $pOq$ določa dva kota, ki predstavljata konveksen in nekonveknen lik - imenujemo ju \index{kot!konveksen}\pojem{konveksen (izbočeni) kot} in \index{kot!nekonveksen}\pojem{nekonveksen (vdrti) kot}. Formalen dokaz tega dejstva bomo na tem mestu izpustili. Če ne poudarimo drugače, bomo pod oznako $\angle pOq$ (oz. $\angle pq$ ali $\angle POQ$) vedno mislili na konveksen kot (Figure \ref{sl.aks.2.2.12b.pic}). V tem smislu je že iz definicije kota jasno, da (konveksna) kota $pOq$ in $qOp$ predstavljata isti kot. + +Kota $pOq$ in $qOr$, ki imata skupen krak $q$, ki je hkrati njun presek (kot množice točk), sta \index{kot!sosednji}\pojem{sosednja kota} (Figure \ref{sl.aks.2.2.12c.pic}). Če sta pri tem poltraka $p$ in $r$ še komplementarna (oz. določata iztegnjeni kot), pravimo, da sta $pOq$ in $qOr$ \index{kota!sokota}\pojem{sokota} (Figure \ref{sl.aks.2.2.12c.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.12c.pic} +\caption{} \label{sl.aks.2.2.12c.pic} +\end{figure} + +Kota $pOq$ in $rOs$ sta \index{kota!sovršna}\pojem{sovršna kota}, če sta $p$ in $r$ oz. $q$ in $s$ para komplementarnih (dopolnilnih) poltrakov (Figure \ref{sl.aks.2.2.12d.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.12d.pic} +\caption{} \label{sl.aks.2.2.12d.pic} +\end{figure} + + Naj bo $A_1A_2\cdots A_n$ ($n\in \{3,4,5,\cdots\}$) enostavna sklenjena lomljenka. + Podobno kot pri kotni lomljenki, lahko na množici vseh točk ravnine razen točk, ki ležijo na lomljenki $A_1A_2\cdots A_n$, definiramo naslednjo relacijo: pravimo, da sta točki $B$ in $C$ na isti strani enostavne sklenjene lomljenke $A_1A_2\cdots A_n$ (kar označimo z $B,C\ddot{-} A_1A_2\cdots A_n$), če obstaja lomljenka $BT_1T_2\cdots T_nC$, ki enostavne sklenjene lomljenke $A_1A_2\cdots A_n$ ne seka oz. z njo nima skupnih točk (Figure \ref{sl.aks.2.2.13.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.13.pic} +\caption{} \label{sl.aks.2.2.13.pic} +\end{figure} + + Tudi v tem primeru je mogoče dokazati, da gre za ekvivalenčno relacijo z dvema razredoma - pri tem za en razred obstaja premica, ki cela leži v njem, za drugi pa takšne premice ni. Unijo tistega razreda, ki ne vsebuje nobene premice (intuitivno - tistega, ki je omejen) in enostavne sklenjene lomljenke, $A_1A_2\cdots A_n$ imenujemo +\index{večkotnik}\pojem{večkotnik} $A_1A_2\cdots A_n$ ali +\index{$n$-kotnik}\pojem{$n$-kotnik} $A_1A_2\cdots A_n$ (Figure \ref{sl.aks.2.2.13a.pic}). Vse točke omenjenega razreda, ki ne vsebuje nobene premice, imenujemo \index{notranje točke!večkotnika} \pojem{notranje točke večkotnika}, cel razred pa \index{notranjost!večkotnika}\pojem{notranjost večkotnika}. Točke drugega razreda so \index{zunanje!točke večkotnika}\pojem{zunanje točke večkotnika}, cel razred pa \index{zunanjost!večkotnika}\pojem{zunanjost večkotnika}. Točke, ki ležijo na lomljenki $A_1A_2\cdots A_n$, ki določa večkotnik, so \index{robne točke!večkotnika}\pojem{robne točke večkotnika}, cela lomljenka pa je \index{rob!večkotnika}\pojem{rob večkotnika}. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.13a.pic} +\caption{} \label{sl.aks.2.2.13a.pic} +\end{figure} + + +Za notranje točke večkotnika velja izrek, ki je pravzaprav ekvivalenten njihovi definiciji. Izrek bomo podali brez dokaza. + + \bizrek + A point $N$ is an interior point of a polygon if and only + if any ray from the point $N$, that does not contain the vertices of the + polygon, intersects an odd number of sides of the polygon (Figure \ref{sl.aks.2.2.13b.pic}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.13b.pic} +\caption{} \label{sl.aks.2.2.13b.pic} +\end{figure} + +Točke $A_1$, $A_2$,..., $A_n$ (oglišča lomljenke) so \index{oglišče!večkotnika}\pojem{oglišča večkotnika}, daljice $A_1A_2$, $A_2A_3$, ... $A_{n-1}A_n$, $A_nA_1$ pa +\index{stranica!večkotnika}\pojem{stranice večkotnika}. Premice $A_1A_2$, $A_2A_3$, ... $A_{n-1}A_n$, $A_nA_1$ so \index{nosilka!stranice}\pojem{nosilke stranic} $A_1A_2$, $A_2A_3$, ... $A_{n-1}A_n$, $A_nA_1$. Stranici, ki vsebujeta skupno oglišče, sta \index{stranica!sosednja}\pojem{sosednji stranici}, sicer sta stranici \index{stranica!nesosednja}\pojem{nesosednji}. Če sta oglišči hkrati krajišči iste stranice, pravimo, da sta oglišči \index{oglišče!sosednje}\pojem{sosednji}, sicer sta oglišči \index{oglišče!nesosednje}\pojem{nesosednji}. +Iz definicije je jasno, da ima vsako oglišče natanko dve sosednji oglišči. Prav tako ima vsaka starnica natanko dve sosednji stranici. + Daljica, ki ju določata nesosednji oglišči, se imenuje +\index{diagonala!večkotnika}\pojem{diagonala večkotnika} (Figure \ref{sl.aks.2.2.13c.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.13c.pic} +\caption{} \label{sl.aks.2.2.13c.pic} +\end{figure} + +O diagonalah večkotnika govori naslednji izrek. + + \bizrek + The number of diagonals of an $n$-gon is $\frac{n(n-3)}{2}$. + \eizrek + +Dokaz tega izreka bomo podali v razdelku \ref{odd3Helly}, kjer bomo posebej obravnavali kombinatorne lastnosti množic točk v ravnini. + +Definirajmo še kote večkotnika. Naj bo $O$ poljubno oglišče večkotnika ter $P$ in $Q$ njegovi sosednji oglišči. Poltraka $OP$ in $OQ$ označimo s $p$ in $q$. V tem primeru kotna lomljenka $pOq$ določa dva kota. Tisti kot, za katerega velja, da vsak poltrak z začetno točko $O$, ki pripada temu kotu in ne vsebuje drugih oglišč večkotnika, seka rob večkotnika razen v točki $O$ še v lihem številu točk, imenujemo \index{kot!notranji večkotnika}\pojem{notranji kot večkotnika} ali krajše \index{kot!notranji}\pojem{kot večkotnika} ob oglišču $O$ (Figure \ref{sl.aks.2.2.13d.pic}). Če je notranji kot večkotnika konveksen, njegov sokot imenujemo \index{kot!zunanji večkotnika}\pojem{zunanji kot večkotnika} (Figure \ref{sl.aks.2.2.13d.pic}). + Kota večkotnika sta + \index{kot!sosednji}\pojem{sosednja kota}, če sta njuna vrha sosednji oglišči večkotnika, sicer sta kota \index{kot!nesosednji}\pojem{nesosednja}. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.13d.pic} +\caption{} \label{sl.aks.2.2.13d.pic} +\end{figure} + +Najbolj enostaven $n$-kotnik in hkrati eden najpogosteje uporabljenih likov v geometriji ravnine dobimo v primeru $n=3$ - \index{trikotnik}\pojem{trikotnik}. +V primeru trikotnika $ABC$ (označevali ga bomo z $\triangle ABC$) so torej točke $A$, $B$ in $C$ njegova \pojem{oglišča}, daljice $AB$, $BC$ in $CA$ pa njegove +\index{stranica!trikotnika}\pojem{stranice} (Figure \ref{sl.aks.2.2.14.pic}). Očitno sta vsaki dve stranici trikotnika sosednji. Prav tako sta sosednji tudi vsaki dve oglišči. Trikotnik torej nima nobene diagonale. Pravimo, da sta oglišče $A$ ($B$ in $C$) oz. kot $BAC$ ($ABC$ in $ACB$) \index{oglišče!nasprotno trikotnika}\pojem{nasprotno oglišče} oz. \index{kot!nasprotni trikotnika}\pojem{nasprotni kot} stranice $BC$ ($AC$ in $AB$) trikotnika $ABC$. In tudi stranica $BC$ ($AC$ in $AB$) je \index{stranica!nasprotna trikotnika}\pojem{nasprotna stranica} oglišča $A$ ($B$ in $C$) oz. kota $BAC$ ($ABC$ in $ACB$) trikotnika $ABC$. Kote (notranje) trikotnika $ABC$ ob ogliščih $A$, $B$ in $C$ pogosto označimo z $\alpha$, $\beta$ in $\gamma$, ustrezne zunanje pa z $\alpha_1$, $\beta_1$ in $\gamma_1$. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.14.pic} +\caption{} \label{sl.aks.2.2.14.pic} +\end{figure} + +Trikotnik $ABC$ lahko definiramo tudi kot presek polravnin $ABC$, $ACB$ in $BCA$. Na tem mestu ekvivalentnosti dveh definicij ne bomo dokazovali. + + Paschev aksiom v terminih trikotnikov lahko sedaj izrazimo v krajši obliki: + +If a line in the plane of a triangle intersects one of its sides +and does not pass through any of its vertices, then intersects exactly one more side of this triangle + + + + \bizrek \label{PaschIzrek} + If a line, not passing through any vertex of a triangle, intersects one side of the triangle + then the line intersects exactly one more side of this triangle (Figure \ref{sl.aks.2.2.14a.pic}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.14a.pic} +\caption{} \label{sl.aks.2.2.14a.pic} +\end{figure} + + V primeru $n=4$ za $n$-kotnik dobimo +\index{štirikotnik}\pojem{štirikotnik}. Ker ima vsako oglišče štirikotnika natanko eno nesosednje oglišče, bomo to oglišče imenovali tudi \index{oglišče!nasprotno štirikotnika}\pojem{nasprotno oglišče} štirikotnika. Podobno bosta nesosednji stranici \index{stranica!nasprotna štirikotnika}\pojem{nasprotni stranici} štirikotnika. Diagonalo štirikotnika torej določata nasprotni oglišči. + Iz same definicije je jasno, da ima štirikotnik dve diagonali\index{diagonala!štirikotnika} (Figure \ref{sl.aks.2.2.14b.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.14b.pic} +\caption{} \label{sl.aks.2.2.14b.pic} +\end{figure} + +Za nesosednja kota štirikotnika pravimo tudi, da sta + \index{kot!nasprotni štirikotnika}\pojem{nasprotna kota} štirikotnika. Kote (notranje) štirikotnika $ABCD$ ob ogliščih $A$, $B$, $C$ in $D$ ponavadi označimo z $\alpha$, $\beta$, $\gamma$ in $\delta$, ustrezne zunanje (tiste, ki obstajajo) pa z $\alpha_1$, $\beta_1$, $\gamma_1$ in $\delta_1$ (Figure \ref{sl.aks.2.2.14c.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.14c.pic} +\caption{} \label{sl.aks.2.2.14c.pic} +\end{figure} + + + +Kot posledico aksiomov urejenosti bomo na koncu tega razdelka vpeljali še pojma orientacije trikotnika in orientacije kota. + +Trikotnik $ABC$, pri katerem so oglišča urejena trojica $(A,B,C)$, imenujemo + \index{orientacija!trikotnika} \pojem{orientirani trikotnik}. +Pravimo, da sta orientirana trikotnika $ABC$ in $BCA'$ \pojem{iste orientacije}, če velja $A,A'\ddot{-} BC$, in nasprotne orientacije, če je $A,A'\div BC$ + (Figure \ref{sl.aks.2.2.15.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.15.pic} +\caption{} \label{sl.aks.2.2.15.pic} +\end{figure} + +Ko govorimo o orientaciji dveh trikotnikov, bomo v bodoče vedno mislili na orientirana trikotnika (besedo orientirana bomo pogosto izpustili). Trikotnika $ABC$ in $A'B'C'$ sta \pojem{iste orientacije} oz. sta \pojem{enako orientirana}, če obstaja takšno zaporedje trikotnikov: $\triangle ABC=\triangle P_1P_2P_3$, $\triangle P_2P_3P_4$, $\triangle P_3P_4P_5$, ..., $\triangle P_{n-2}P_{n-1}P_n=\triangle A'B'C'$, da je v tem zaporedju število sprememb orientacije dveh sosednjih trikotnikov sodo + (Figure \ref{sl.aks.2.2.15a.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.15a.pic} +\caption{} \label{sl.aks.2.2.15a.pic} +\end{figure} + +Mogoče je dokazati, da je relacija iste orientacije dveh trikotnikov ekvivalenčna relacija, ki ima dva razreda. Za dva trikotnika, ki nista v istem razredu, pravimo, da sta \pojem{različne orientacije} oz. \pojem{različno orientirana}. Vsak od dveh razredov določa \index{orientacija!ravnine}\pojem{orientacijo ravnine}. Imenujemo ju \pojem{pozitivna orientacija} in \pojem{negativna orientacija}. Za lažjo predstavo se dogovorimo, naj bo orientacija, ki ustreza smeri vrtenja urinega kazalca, negativna, njej nasprotna pa pozitivna orientacija. + Če ne bomo drugače poudarili, bomo vedno uporabljali pozitivno orientacijo ravnine oz. trikotnikov + (Figure \ref{sl.aks.2.2.15b.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.15b.pic} +\caption{} \label{sl.aks.2.2.15b.pic} +\end{figure} + + +Definirajmo še orientacijo kotov. Kota $ASB$ in $A'S'B'$, od katerih nobeden ni iztegnjeni kot, sta \pojem{iste orientacije}, če: +\begin{itemize} + \item sta oba konveksna ali oba nekonveksna, trikotnika $ASB$ in $A'S'B'$ pa sta iste orientacije (Figure \ref{sl.aks.2.2.16.pic}), + \item je en kot konveksen in drugi nekonveksen, tikotnika $ASB$ in $A'S'B'$ pa sta nasprotne orientacije (Figure \ref{sl.aks.2.2.16c.pic}). +\end{itemize} + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.16.pic} +\caption{} \label{sl.aks.2.2.16.pic} +\end{figure} + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.16c.pic} +\caption{} \label{sl.aks.2.2.16c.pic} +\end{figure} + + Če je $\angle ASB$ iztegnjeni kot, $\angle A'S'B'$ pa konveksen kot, sta kota $ASB$ in $A'S'B'$ iste orientacije, če obstaja takšna točka $C$ v notranjosti kota $ASB$, da sta kota $ASC$ in $A'S'B'$ iste orientacije (Figure \ref{sl.aks.2.2.16d.pic}). Podobno naredimo tudi, če je kot $A'S'B'$ nekonveksen ali če sta oba kota $ASB$ in $A'S'B'$ iztegnjena. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.16d.pic} +\caption{} \label{sl.aks.2.2.16d.pic} +\end{figure} + + +Tudi v tem primeru se izkaže, da je relacija iste orientacije kotov ekvivalenčna relacija, ki ima dva razreda. Pri tem pozitivno orientacijo kota predstavlja tisti razred, pri katerem ima trikotnik $ASB$ za konveksni kot + $ASB$ iz tega razreda negativno orientacijo + (Figure \ref{sl.aks.2.2.16a.pic}). +V tem smislu sta kota $ASB$ in $BSA$ nasprotno orientirana. + \index{orientacija!kota} \pojem{Orientirani kot} $ASB$ bomo označili z $\measuredangle ASB$. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.16a.pic} +\caption{} \label{sl.aks.2.2.16a.pic} +\end{figure} + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.2.16b.pic} +\caption{} \label{sl.aks.2.2.16b.pic} +\end{figure} + +Če je $C$ poljubna točka, ki ne leži na robu kota $ASB$, bomo definirali vsoto orientiranih kotov $\measuredangle ASC$ in $\measuredangle CSB$ + (Figure \ref{sl.aks.2.2.16b.pic}): + \begin{eqnarray} + \measuredangle ASC+\measuredangle CSB = \measuredangle ASB. + \label{orientKotVsota} + \end{eqnarray} + + +%________________________________________________________________________________ + \poglavje{Congruence Axioms} + \label{odd2AKSSKL} + + +Naslednji aksiomi so potrebni, da lahko vpeljemo pojem in lastnosti +skladnosti likov. S prejšnjimi aksiomi smo namreč lahko vpeljali in +obravnavali pojme: daljica, poltrak, kot, večkotnik, ... ne pa +še pojmov, ki so povezani s skladnostjo: krožnica, pravi kot, +skladnost trikotnikov,~... + +Intuitivna ideja skladnosti likov, ki smo jo uporabljali že v +osnovni šoli, je povezana z gibanjem, ki prvi lik preslika v +drugega. To idejo bomo sedaj uporabili, da bi formalno definirali +pojem skladnosti in njene lastnosti. + Najprej bomo začeli z osnovnim, že omenjenim, pojmom skladnosti + parov točk $(A,B)\cong (C,D)$ (Figure \ref{sl.aks.2.3.1.pic}) + ter formalno definirali pojem + ‘‘gibanja’’. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.1.pic} +\caption{} \label{sl.aks.2.3.1.pic} +\end{figure} + + S pomočjo skladnosti parov točk najprej definirajmo skladnost $n$-terice točk. + Pravimo, da +sta dve $n$-terici točk skladni (Figure \ref{sl.aks.2.3.2.pic}) oz. +$$(A_1 , A_2,\ldots ,A_n ) \cong ( A'_1 , A'_2 ,\ldots , A'_n ),$$ +če je: $(A_i,A_j)\cong (A'_i,A'_j)$ za vsako $i,j\in \{1,2,\ldots, +n\}$. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.2.pic} +\caption{} \label{sl.aks.2.3.2.pic} +\end{figure} + + +Bijektivna preslikava ravnine v ravnino +$\mathcal{I}:\mathcal{S}\rightarrow \mathcal{S}$ je +\index{izometrija}\pojem{izometrija} ali \pojem{izometrijska +transformacija}, če ohranja relacijo skladnosti parov točk +(Figure \ref{sl.aks.2.3.3.pic}) oz. če za vsaki dve točki $A$ in +$B$ velja: + $$(\mathcal{I}(A),\mathcal{I}(B))\cong (A,B).$$ + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.3.pic} +\caption{} \label{sl.aks.2.3.3.pic} +\end{figure} + + +Z naslednjimi aksiomi bomo vpeljali lastnosti na novo definirane +preslikave. + + +\vspace*{3mm} + + + \baksiom \label{aksIII1} Isometries preserve the relation + $\mathcal{B}$ (Figure \ref{sl.aks.2.3.4.pic}), which means that for every + isometry $\mathcal{I} $ holds: + $$\mathcal{I}: A, B,C\mapsto A',B',C'\hspace*{2mm}\wedge \hspace*{2mm} + \mathcal{B}(A,B,C) + \hspace*{1mm}\Rightarrow\hspace*{1mm} \mathcal{B}(A',B',C').$$ + \eaksiom + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.4.pic} +\caption{} \label{sl.aks.2.3.4.pic} +\end{figure} + + + \baksiom \label{aksIII2} If $ABC$ and $A'B'C'$ are two half-planes + (Figure \ref {asl.aks.2.3.5.pic}), then there is a single isometry + $\mathcal{I}$, which maps: + + \begin{itemize} + \item point $A$ to point $A'$, + \item half-line $AB$ in half-line $A'B'$, + \item half-plane $ABC$ to half-plane $A'B'C'$. + \end{itemize} + If $(A,B)\cong (A',B')$ holds, + then it is $\mathcal{I}(B)=B'$. + \\ If in addition + $(A,B,C)\cong (A',B',C')$ also holds, + then it is $\mathcal{I}(B)=B'$ and $\mathcal{I}(C)=C'$. + \eaksiom + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.5.pic} +\caption{} \label{asl.aks.2.3.5.pic} +\end{figure} + + \baksiom \label{aksIII3} For every two points $A$ and $B$ there exists + isometry such that holds $$\mathcal{I}: A, B\mapsto B,A.$$ + If $(S,A)\cong (S,B)$ and $S\in AB$, then for each isometry $\mathcal{I}$ with this property holds $\mathcal{I}(S)=S$ (Figure \ref{sl.aks.2.3.6.pic}). + \eaksiom + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.6.pic} +\caption{} \label{sl.aks.2.3.6.pic} +\end{figure} + + \baksiom \label{aksIII4} The set of all isometries with respect to the composition of mappings form a group, which means that: + \begin{itemize} + \item composition of two isometries $\mathcal{I}_2\circ \mathcal{I}_1$ is isometry, + \item identity map $\mathcal{E}$ is isometry, + \item if $\mathcal{I}$ is isometry, then its inverse transformation + $\mathcal{I}^{-1}$ is also isometry. + \end{itemize} + \eaksiom + +\vspace*{3mm} + + Omenimo, da je pri strukturi grupe zahtevana tudi lastnost + asociativnosti oz. $\mathcal{I}_1\circ (\mathcal{I}_2\circ \mathcal{I}_3)= + (\mathcal{I}_1\circ \mathcal{I}_2)\circ \mathcal{I}_3$ (za poljubne izometrije + $\mathcal{I}_1$, $\mathcal{I}_2$ in $\mathcal{I}_3$), ki pa je pri + operaciji kompozituma funkcij + avtomatično izpolnjena. Omenimo še, da je \pojem{identiteta} \index{identiteta} + $\mathcal{E}$ iz prejšnjega aksioma preslikava, za katero je + $\mathcal{E}(A)=A$ za vsako točko ravnine. Preslikava + $\mathcal{I}^{-1}$ je \pojem{inverzna preslikava} za izometrijo + $\mathcal{I}$, če velja $\mathcal{I}^{-1}\circ \mathcal{I} + =\mathcal{I}\circ\mathcal{I}^{-1}=\mathcal{E}$. Po prejšnjem + aksiomu sta torej identiteta in inverzna preslikava vsake izometrije tudi + izometriji. + + + +Dokažimo prve posledice aksiomov skladnosti. Najprej bomo +obravnavali naslednje lastnosti izometrij. + + + + \bizrek \label{izrekIzoB} Isometry maps a line to a line, a line segment to a line segment, a ray to a ray, + a half-plane to a half-plane, an angle to an angle and an $n$-gon to an $n$-gon. + \eizrek + +\textbf{\textit{Proof.}} + Po aksiomu \ref{aksIII1} izometrije ohranjajo relacijo + $\mathcal{B}$. Zato se vse točke daljice $AB$ pri izometriji $I$ preslikajo + v točke, ki ležijo na daljici $A'B'$, kjer je $A'=\mathcal{I}(A)$ in + $B'=\mathcal{I}(B)$. Ker je tudi inverzna preslikava $\mathcal{I}^{-1}$ + izometrija (aksiom \ref{aksIII4}), je vsaka točka daljice + $A'B'$ slika neke točke, ki leži na daljici $AB$. Torej se z + izometrijo $\mathcal{I}$ + daljica $AB$ preslika v daljico $A'B'$. + + Tudi preostale like iz izreka smo definirali s pomočjo relacije + $\mathcal{B}$, tako da je dokaz podoben kot za daljico. + \kdokaz + +Iz dokaza prejšnjega izreka sledi, da se krajišči daljice $AB$ +z izometrijo preslikata v krajišči slike $A'B'$. Na podoben +način se z izometrijo izhodišče poltraka preslika v izhodišče +poltraka, rob polravnine v rob polravnine, vrh kota v vrh kota in +oglišče večkotnika v oglišče večkotnika. + +Izometrije so definirane kot bijektivne preslikave, ki ohranjajo +skladnost parov točk. Ali velja tudi, da za skladne pare točk +obstaja izometrija, ki prvi par preslika v drugega? Odgovor podajmo +z naslednjim izrekom. + + + \bizrek \label{izrekAB} If $(A,B)\cong (A',B')$, then there is an isometry + $\mathcal{I}$, which maps the points $A$ and $B$ to the points + $A'$ and $B'$, + i.e.: + $$\mathcal{I}: A, B\mapsto A',B'.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.7.pic} +\caption{} \label{sl.aks.2.3.7.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Naj bo $C$ točka, ki ne leži na premici $AB$, in $C'$ točka, + ki ne leži na premici $A'B'$ (Figure \ref{sl.aks.2.3.7.pic}). + Po aksiomu \ref{aksIII2} obstaja ena sama izometrija $\mathcal{I}$, ki + preslika točko $A$ v točko $A'$, poltrak $AB$ v poltrak $A'B'$ + in polravnino $ABC$ v polravnino $A'B'C'$. Ker je po predpostavki + $(A,B)\cong (A',B')$ iz istega aksioma \ref{aksIII2}, sledi še + $\mathcal{I}(B)=B'$. + \kdokaz + +Podoben je dokaz naslednjega izreka, ki bo kasneje v drugačni obliki +podan kot prvi izrek o skladnosti trikotnikov. + + + + \bizrek \label{IizrekABC} Let $(A,B,C)$ and $(A',B',C')$ be + triplets of non-collinear points such that $$(A,B,C)\cong (A',B',C'),$$ + then there is a single isometry $\mathcal{I}$, that maps the points + $A$, $B$ and $C$ into the points $A'$, $B'$ and $C'$, i.e.: + $$\mathcal{I}: A, B,C\mapsto A',B',C'.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.5.pic} +\caption{} \label{sl.aks.2.3.5.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + Po aksiomu \ref{aksIII2} obstaja ena sama izometrija $\mathcal{I}$, ki + preslika točko $A$ v točko $A'$, poltrak $AB$ v poltrak $A'B'$ + in polravnino $ABC$ v polravnino $A'B'C'$ (Figure \ref{sl.aks.2.3.5.pic}). + Ker je po predpostavki + $(A,B,C)\cong (A',B',C')$ iz istega aksioma \ref{aksIII2}, sledi še + $\mathcal{I}(B)=B'$ in $\mathcal{I}(C)=C'$. + + Potrebno je dokazati, da je $\mathcal{I}$ edina takšna izometrija. + Predpostavimo, da obstaja takšna izometrija $\mathcal{\widehat{I}}$, da + velja $\mathcal{\widehat{I}}: A, B,C\mapsto A',B',C'$. Po + izreku \ref{izrekIzoB} izometrija $\mathcal{\widehat{I}}$ + tudi poltrak $AB$ preslika v poltrak $A'B'$ in polravnino $ABC$ + v polravnino $A'B'C'$. Iz aksioma \ref{aksIII2} sledi + $\mathcal{\widehat{I}}=\mathcal{I}$. + \kdokaz + +Direktna posledica je naslednji izrek. + + + \bizrek \label{IizrekABCident} Let $A$, $B$ and $C$ be three non-collinear points, then the identity map + $\mathcal{E}$ is the only isometry that maps points $A$, $B$, and $C$ to the same points + $A$, $B$ and $C$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.5a.pic} +\caption{} \label{sl.aks.2.3.5a.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.aks.2.3.5a.pic}) + +Najprej je identična preslikava $\mathcal{E}$, ki točke $A$, $B$ +in $C$ preslika v točke $A$, $B$ in $C$, izometrija po aksiomu +\ref{aksIII4}. Iz prejšnjega izreka \ref{IizrekABC} sledi, da je +takšna izometrija ena sama. + \kdokaz + +Za točko $A$ pravimo, da je \index{točka!fiksna} \pojem{fiksna +točka} (ali \index{točka!negibna} \pojem{negibna +točka}) izometrije $\mathcal{I}$, če velja $\mathcal{I}(A)=A$. +Prejšnji izrek nam pove, da so edine izometrije, ki imajo fiksne tri +nekolinearne točke, identitete. + + + Izometrije bomo podrobneje obravnavali v poglavju + \ref{pogIZO}, na tem mestu pa jih bomo uporabili predvsem za pomoč + pri vpeljavi skladnosti likov. Dva lika $\Phi$ in $\Phi'$ sta + \index{lika!skladna}\pojem{skladna} + (označimo + $\Phi\cong \Phi'$), + če obstaja izometrija $I$, ki lik $\Phi$ preslika v lik $\Phi'$. + + Direktna posledica aksioma \ref{aksIII4} je naslednji izrek. + + + \bizrek + Congruence of figures is an equivalence relation. \label{sklRelEkv} + \eizrek + +\textbf{\textit{Proof.}} + +\textit{Refleksivnost.} Za vsak lik $\Phi$ velja $\Phi \cong +\Phi$, ker je identična preslikava $\mathcal{E}$ izometrija +(aksiom \ref{aksIII4}) in $\mathcal{E}:\Phi\rightarrow\Phi$. + +\textit{Simetričnost.} Iz $\Phi \cong \Phi_1$ sledi, da obstaja +izometrija $\mathcal{I}$, ki preslika lik $\Phi$ v lik $\Phi_1$. +Inverzna preslikava $\mathcal{I}^{-1}$, ki je po aksiomu +\ref{aksIII4} izometrija, preslika lik $\Phi_1$ v lik $\Phi$, +zato velja $\Phi_1 \cong \Phi$. + +\textit{Tranzitivnost.} Iz $\Phi \cong \Phi_1$ in $\Phi_1 \cong +\Phi_2$ sledi, da obstajata takšni izometriji $\mathcal{I}$ in +$\mathcal{I}'$, da velja $\mathcal{I}:\Phi\rightarrow\Phi_1$ in +$\mathcal{I}':\Phi_1\rightarrow\Phi_2$. + Potem kompozitum $\mathcal{I}'\circ\mathcal{I}$, + ki je po aksiomu \ref{aksIII4} izometrija, preslika lik $\Phi$ +v lik $\Phi_2$, zato velja $\Phi \cong \Phi_2$. +\kdokaz + + + + Pojem skladnosti likov se nanaša tudi na daljice. Intuitivno + smo skladnost daljic povezovali s skladnostjo parov točk. + Sedaj bomo dokazali ekvivalentnost obeh relacij. + + \bizrek \label{izrek(A,B)} $AB \cong A'B' \Leftrightarrow + (A,B)\cong (A',B')$ + \eizrek + +\textbf{\textit{Proof.}} + + ($\Rightarrow$) Če je $(A,B)\cong +(A',B')$, po izreku \ref{izrekAB} obstaja izometrija +$\mathcal{I}$, ki preslika točki $A$ in $B$ v točki $A'$ in +$B'$. Iz izreka \ref{izrekIzoB} sledi, da izometrija $\mathcal{I}$ +preslika daljico $AB$ v daljico $A'B'$ oz. $AB \cong A'B'$. + +($\Leftarrow$) Če je $AB \cong A'B'$, obstaja izometrija +$\mathcal{I}$, ki preslika daljico $AB$ v daljico $A'B'$. Po +posledici izreka \ref{izrekIzoB} se krajišče daljice preslika v +krajišče daljice. To pomeni, da velja bodisi +$\mathcal{I}:A,B\mapsto A',B'$ bodisi $\mathcal{I}:A,B\mapsto +B',A'$. Iz prve relacije sledi $(A,B)\cong (A',B')$ iz druge pa +$(A,B)\cong (B',A')$. Toda tudi iz drugega primera dobimo +$(A,B)\cong (A',B')$, kar je posledica aksiomov \ref{aksIII3} in +\ref{aksIII4}. +\kdokaz + + Zaradi prejšnjega izreka bomo v nadaljevanju namesto + relacije $(A,B)\cong (A',B')$ vedno pisali $AB\cong + A'B'$. + + + \bizrek \label{ABnaPoltrakCX} + For each line segment $AB$ and each ray $CX$, there is exactly + one point $D$ on the ray + $CX$ that $AB\cong CD$ holds. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.5b.pic} +\caption{} \label{sl.aks.2.3.5b.pic} +\end{figure} + + + + \textbf{\textit{Proof.}} Naj bo $P$ točka, ki ne leži na premici $AB$, + in $Q$ točka, + ki ne leži na premici $CX$ (Figure \ref{sl.aks.2.3.5b.pic}). + Po aksiomu \ref{aksIII2} obstaja ena sama + izometrija $\mathcal{I}$, ki + preslika točko $A$ v točko $C$, poltrak $AB$ v poltrak $CX$ + in polravnino $ABP$ v polravnino $CXQ$. + Naj bo $D=\mathcal{I}(C)$, potem je $AB \cong CD$. + + Predpostavimo, + da na poltraku $CX$ obstaja še ena točka $\widehat{D}$, za + katero + velja $AB \cong C\widehat{D}$. Ker poltraka + $CX$ in $CD$ sovpadata, izometrija $\mathcal{I}$ pa preslika točko + $A$ v točko $C$, poltrak $AB$ v poltrak $CD$ + in polravnino $ABP$ v polravnino $CDQ$, + iz aksioma \ref{aksIII2} sledi $\mathcal{I}(C)=\widehat{D}$ oz. + $\widehat{D}=D$. + \kdokaz + + + \bizrek \label{izomEnaC'} Let $A$, $B$, $C$ be three non-collinear points + and $A'$, $B'$ points of the edge of a half-plane $\pi$ such that $AB \cong A'B'$. + Then there is exactly one point $C'$ in the half-plane $\pi$ such that $AC \cong A'C'$ and $BC \cong B'C'$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.11a.pic} +\caption{} \label{sl.aks.2.3.11a.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.aks.2.3.11a.pic}) + + Po aksiomu \ref{aksIII2} obstaja ena sama + izometrija $\mathcal{I}$, ki + preslika točko $A$ v točko $A'$, poltrak $AB$ v poltrak $A'B'$ + in polravnino $ABC$ v polravnino $\pi$ ter velja $\mathcal{I}(B)=B'$. + Naj bo $C'=\mathcal{I}(C)$, potem velja $AC \cong A'C'$ in + $BC \cong B'C'$. Predpostavimo, da obstaja takšna točka + $\widehat{C}'$, ki leži v polravnini $\pi$ ter velja $AC \cong A'\widehat{C}'$ in + $BC \cong B'\widehat{C}'$. Ker je še $AB \cong +A'B'$, po izreku \ref{IizrekABC} obstaja ena sama izometrija +$\mathcal{\widehat{I}}$, ki preslika točke $A$, $B$ in $C$ v +točke $A'$, $B'$ in $\widehat{C}'$. Toda le-ta preslika tudi +poltrak $AB$ v poltrak $A'B'$ in polravnino $ABC$ v polravnino +$A'B'\widehat{C}'=\pi$. Po aksiomu \ref{aksIII2} je +$\mathcal{\widehat{I}}=\mathcal{I}$ in zato tudi +$\widehat{C}'=\mathcal{\widehat{I}}(C)=\mathcal{I}(C)=C'$. + \kdokaz + + + \bizrek \label{izoABAB} If $\mathcal{I}$ is an isometry that maps a points $A$ and $B$ into the same points + $A$ and $B$ (i.e. $\mathcal{I}(A)=A$ and $\mathcal{I}(B)=B$), then it also holds for each point $X$ on the line + $AB$ (i.e. $\mathcal{I}(X)=X$). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.8.pic} +\caption{} \label{sl.aks.2.3.8.pic} +\end{figure} + + +% + +\textbf{\textit{Proof.}} Označimo z $X$ poljubno točko premice +$AB$. Brez škode za splošnost predpostavimo, da točka $X$ leži +na poltraku $AB$ (Figure \ref{sl.aks.2.3.8.pic}). Dokažimo, da +velja $\mathcal{I}(X)=X$. + +Naj bo $P$ točka, ki ne leži na premici $AB$ in +$P'=\mathcal{I}(P)$. Izometrija $\mathcal{I}$ + preslika točko $A$ v točko $A$, poltrak $AB$ v poltrak $AB$ + (oz. poltrak $AX$ v poltrak $AX$) + in polravnino $ABP$ v polravnino $ABP'$ + (oz. polravnino $AXP$ v polravnino $AXP'$). + Po aksiomu \ref{aksIII2} + iz $AX\cong AX$ sledi $\mathcal{I}(X)=X$. + \kdokaz + + Vpeljimo nove pojme, ki se nanašajo na daljice. + +Pravimo, da je daljica $EF$ \index{vsota!daljic}\pojem{vsota daljic} +$AB$ in $CD$, kar označimo $EF=AB+CD$, če obstaja takšna točka $P$ +na daljici $EF$, da velja $AB \cong EP$ in $CD \cong PF$ (Figure +\ref{sl.aks.2.3.9.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.9.pic} +\caption{} \label{sl.aks.2.3.9.pic} +\end{figure} + + +Daljica $EF$ je \index{razlika!daljic}\pojem{razlika daljic} $AB$ +in $CD$, kar označimo $EF=AB-CD$, če je $AB=EF+CD$ (Figure +\ref{sl.aks.2.3.9.pic}). + + Na podoben način lahko definiramo tudi + množenje daljice z naravnim in s pozitivnim racionalnim + številom. Za daljici $AB$ in $CD$ je $AB=n\cdot CD$ + ($n\in \mathbb{N}$), če obstajajo takšne točke + $X_1$, $X_2$,..., $X_{n-1}$, da je + $\mathcal{B}(X_1,X_2,\ldots,X_{n-1})$ in + $AX_1 \cong X_1X_2 \cong X_{n-1}B \cong CD$ (Figure +\ref{sl.aks.2.3.10.pic}). + V tem primeru je tudi $CD=\frac{1}{n}\cdot AB$. + + Na tem mestu ne bomo formalno dokazovali dejstva, da za vsako daljico $PQ$ in vsako naravno število $n$ obstaja daljica $AB$, za katero je $AB=n\cdot PQ$, ter daljica $CD$, za katero je $CD=\frac{1}{n}\cdot PQ$. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.10.pic} +\caption{} \label{sl.aks.2.3.10.pic} +\end{figure} + + + Množenje daljice s pozitivnim + racionalnim številom vpeljemo na naslednji način. Za + $q=\frac{n}{m} \in \mathbb{Q^+}$ je: +$$q\cdot AB=\frac{n}{m}\cdot AB = n\cdot\left(\frac{1}{m}\cdot AB\right)$$ + +Če za točko $P$ daljice $AB$ velja $AP=\frac{n}{m}\cdot PB$, +pravimo, da točka $P$ deli daljico $AB$ v \index{razmerje} +\pojem{razmerju} $n:m$, kar zapišemo $AP:PB=n:m$. + +Daljica $AB$ je \index{relacija!urejenosti daljic}\pojem{daljša} +od daljice $CD$, kar označimo $AB>CD$, če obstaja takšna točka +$P\neq B$ na daljici $AB$, da velja $CD \cong AP$ (Figure +\ref{sl.aks.2.3.11.pic}). V tem primeru pravimo tudi, da je +daljica $CD$ \pojem{krajša} od daljice $AB$ (oznaka $CDCD$ ali $AB \angle cd$), če +obstaja v kotu $ab$ poltrak $s=SX$, da velja $\angle as \cong +\angle cd$ (Figure \ref{sl.aks.2.3.17.pic}). V tem primeru je tudi +kot $cd$ \pojem{manjši} od kota $ab$ ($\angle cd< \angle ab$). Ni +težko dokazati, da za dva kota $ab$ in $cd$ velja natanko ena od +relacij: $\angle ab > \angle cd$, $\angle ab < \angle cd$ ali +$\angle ab \cong \angle cd$. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.17.pic} +\caption{} \label{sl.aks.2.3.17.pic} +\end{figure} + + +Kota sta \index{kota!suplementarna}\pojem{suplementarna}, če je +njuna vsota enaka iztegnjenem kotu (Figure +\ref{sl.aks.2.3.18.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.18.pic} +\caption{} \label{sl.aks.2.3.18.pic} +\end{figure} + + +Poltrak $s=SX$ je \index{bisektrisa kota}\pojem{bisektrisa kota} +$\angle pSq=\alpha$ (Figure \ref{sl.aks.2.3.19.pic}), če leži v tem kotu in +velja $\angle ps \cong \angle sq$. Nosilka te bisektrise je \index{simetrala!kota}\pojem{simetrala kota} $pSq$ (premica $s_{\alpha}$). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.19.pic} +\caption{} \label{sl.aks.2.3.19.pic} +\end{figure} + + + +Podobno kot za središče daljice velja za bisektriso kota naslednji izrek. + + \bizrek \label{izrekSimetralaKota} + An angle has exactly one bisector. + %(oz. eno samo simetralo). + \eizrek + +\textbf{\textit{Proof.}} + Naj bo $\alpha=pSq$ poljubni kot, $P$ poljubna točka, ki leži na kraku + $Sp$ ($P\neq S$) ter $Q$ točka, ki leži na kraku $Sq$ in velja + $SP\cong SQ$. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.20.pic} +\caption{} \label{sl.aks.2.3.20.pic} +\end{figure} + + + + Predpostavimo, da je kot $\alpha$ iztegnjeni kot + (Figure \ref{sl.aks.2.3.20.pic}), ki določa + polravnino $\pi$. Naj bo $A$ njena poljubna točka. Po izreku + \ref{izomEnaC'} v polravnini $\pi$ obstaja ena sama točka $B$, + da velja $(P,Q,A)\cong (Q,P,B)$. Iz izreka \ref{IizrekABC} sledi, + da obstaja ena sama izometrija $\mathcal{I}$, ki točke $P$, $Q$ in + $A$ preslika v točke $Q$, $P$ in $B$. Naj bo + $\mathcal{I}(B)=\widehat{A}$. Ker je + $(Q,P,B)\cong(P,Q,\widehat{A})$, je po izreku \ref{izomEnaC'} + $\widehat{A}=A$. Torej: + $$\mathcal{I}:P,Q,A,B\mapsto Q,P,B,A.$$ +Zato se središči $S$ in $L$ daljic $PQ$ in $AB$ preslikata vase +(aksiom \ref{aksIII4}), kar potem velja tudi za poltrak $s=SL$ in +vsako njegovo točko (izrek \ref{izoABAB}). Torej izometrija +$\mathcal{I}$ preslika kot $pSs$ v kot $sSq$, zato + je $pSs\cong sSq$ oz. poltrak $s$ je bisektrisa kota $pSq$. + + Dokažimo, da je $s$ edina bisektrisa kota $\alpha$. Naj bo + $\widehat{s}=S\widehat{L}$ + poltrak, ki leži v kotu $\alpha$ in velja $pS\widehat{s}\cong + \widehat{s}Sq$. Potem obstaja izometrija $\mathcal{\widehat{I}}$, ki + preslika kot $pS\widehat{s}$ v kot $\widehat{s}Sq$. Ta izometrija + preslika točko $S$ v točko $S$, poltrak $p$ v poltrak $q$ in + polravnino $\pi$ v polravnino $\pi$, zato je po aksiomu + \ref{aksIII2} $\mathcal{\widehat{I}}=\mathcal{I}$. Torej + $\mathcal{I}(\widehat{s})= + \mathcal{\widehat{I}}(\widehat{s})=\widehat{s}$. Če $\widehat{L} \notin + s$, izometrija $\mathcal{I}$ preslika tri nekolinearne točke + $S$, $L$ in $\widehat{L}$ vase in je $\mathcal{I}$ identična preslikava + (izrek \ref{IizrekABCident}), kar ni mogoče. Torej $\widehat{L} \in + s$ oz. $\widehat{s}=s$. + + + Če je $\alpha$ neiztegnjeni konveksni kot (Figure \ref{sl.aks.2.3.20.pic}), + so točke $S$, $P$ in $Q$ + nekolinearne, zato po izreku \ref{IizrekABC} obstaja ena sama + izometrija $\mathcal{I}$, ki točke $P$, $S$ in $Q$ preslika v + točke $Q$, $S$ in $P$. Z $L$ označimo središče daljice $PQ$. + Po aksiomu \ref{aksIII3} je $\mathcal{I}(L)=L$. Potem se tudi vse + točke poltraka $s=SL$ preslikajo vase (izrek \ref{izoABAB}). Kot +$\alpha$ je konveksni kot, kar pomeni, da točka $L$ in potem tudi +poltrak $s$ ležita v tem kotu. + Torej izometrija $\mathcal{I}$ preslika kot $pSs$ v kot $sSq$, zato + je $pSs\cong sSq$ oz. poltrak $s$ je bisektrisa kota $pSq$. + +Na podoben način kot v prejšnjem primeru dokažemo, da kot +$\alpha$ nima drugih bisektris. + +Če je $\alpha$ nekonveksni kot, bisektriso dobimo kot +komplementarni (dopolnilni) poltrak poltraka $s$. + \kdokaz + +Dokažimo dva izreka, ki se nanašata na sokote in sovršne kote.\index{kota!sokota} \index{kota!sovršna} + + + + \bizrek + The adjacent supplementary angles of two congruent angles are also congruent. \label{sokota} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.20a.pic} +\caption{} \label{sl.aks.2.3.20a.pic} +\end{figure} + + + +\textbf{\textit{Proof.}} Naj bosta $\alpha'=\angle P'OQ$ in $\alpha_1'=\angle P_1'O_1Q_1$ sokota + dveh skladnih kotov $\alpha=\angle POQ$ in $\alpha_1=\angle P_1O_1Q_1$ (Figure \ref{sl.aks.2.3.20a.pic}). Po aksiomu \ref{aksIII2} obstaja ena sama izometrija $\mathcal{I}$, ki preslika točko $O$ v točko $O_1$, poltrak $OP$ v poltrak $O_1P_1$ in polravnino $POQ$ v polravnino $P_1O_1Q_1$. Naj bo $Q_2=\mathcal{I}(Q)$. Potem je $\angle P_1O_1Q_2\cong \angle POQ$. Izometrija $\mathcal{I}$ preslika polravnino $POQ$ v polravnino $P_1O_1Q_1$, zato točka $Q_2$ (in tudi poltrak $O_1Q_2$) leži v polravnini $P_1O_1Q_1$. Ker je po predpostavki še $\angle POQ\cong\angle P_1O_1Q_1$, po izreku \ref{KotNaPoltrak} $OQ_1$ in $OQ_2$ predstavljata isti poltrak. Torej točka $Q_2$ leži na poltraku $O_1Q_1$. Naj bo $P_2'=\mathcal{I}(P')$. Ker izometrije poltrak preslikajo v poltrak (izrek \ref{izrekIzoB}), leži točka $P_2'$ na poltraku $O_1P_1'$. Iz $\mathcal{I}:P',O,Q\mapsto P_2',O_1,Q_2$ sledi, da izometrija $\mathcal{I}$ preslika kot $P'OQ$ v kot $P_2'O_1Q_2$ (izrek \ref{izrekIzoB}), zato je + $\angle P'OQ\cong \angle P_2'O_1Q_2=\angle P_1'O_1Q_1$. +\kdokaz + + + \bizrek \label{sovrsnaSkladna} + Vertical angles are congruent. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.20b.pic} +\caption{} \label{sl.aks.2.3.20b.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Naj bosta $\alpha=\angle POQ$ in $\alpha'=\angle P'OQ'$ sovršna kota, kjer so točke $P$, $O$, $P'$ (oz. $Q$, $O$, $Q'$) kolinearne (Figure \ref{sl.aks.2.3.20b.pic}). Kot $\beta=\angle QOP'$ je sokot za oba kota $\alpha$ in $\alpha'$. Ker je še $\beta\cong\beta$, je po prejšnjem izreku \ref{sokota} tudi $\alpha\cong\alpha'$. +\kdokaz + + + + \bizrek \label{sredZrcObstoj} + For each point $S$ there exists an isometry $\mathcal{I}$ such that $\mathcal{I}(S)=S$. + In addition, for each point $X\neq S$ the following holds:\\ if $\mathcal{I}(X)=X'$, then $S$ is the midpoint of the line segment $XX'$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.20c.pic} +\caption{} \label{sl.aks.2.3.20c.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Naj bo $P$ poljubna točka različna od $S$ (Figure \ref{sl.aks.2.3.20c.pic}). Po aksiomu \ref{AksII3} obstaja na premici $SP$ takšna točka $Q$, da velja $\mathcal{B}(P,S,Q)$. Označimo polravnini, ki ju določa rob $SP$ z $\alpha$ in $\alpha'$. Po aksiomu \ref{aksIII2} obstaja (ena sama) izometrija $\mathcal{I}$, ki preslika točko $S$ v točko $S$, poltrak $SP$ v poltrak $SQ$ in polravnino $\alpha$ v polravnino $\alpha'$. + +Označimo s $p$ premico $SP$. +Točka $P'=\mathcal{I}(P)$ leži na poltraku $SQ$ oz. premici $p$. Ker je torej $\mathcal{I}:S,P \mapsto S,P'$, se po aksiomu \ref{aksIII1} premica $SP$ preslika v premico $SP'$ oz. $\mathcal{I}:p\rightarrow p$. +Slika polravnine $\alpha'$ z robom $p$ je torej polravnina z istim robom (izrek \ref{izrekIzoB}). Ta polravnina ne more biti $\alpha'$, saj je izometrija $\mathcal{I}$ bijektivna preslikavain po predpostavki preslika polravnino $\alpha$ v polravnino $\alpha'$. Torej je $\mathcal{I}:\alpha'\rightarrow \alpha$. + +Sedaj je jasno, da je brez škode za splošnost dovolj, če izpeljemo dokaz le za točke, ki ležijo v polravnini $\alpha$ (brez roba oz. le poltraka $SP$). + +Naj bo $X\in \alpha\setminus p$ in $X'=\mathcal{I}(X)$. Takoj vidimo,da je $X'\in \alpha'\setminus p$. Po aksiomu \ref{AksII3} obstaja na premici $SX$ takšna točka $X_1$, da velja $\mathcal{B}(X,S,X_1)$. Ker sta $\angle PSX$ in $\angle P'SX_1$ sovršna kota, sta po izreku \ref{sovrsnaSkladna} tudi skladna. Toda iz $\mathcal{I}:S,P,X \mapsto S,P',X'$ sledi tudi + $\angle PSX \cong \angle P'SX'$. Torej velja $\angle P'SX_1\cong \angle P'SX'$ (izrek \ref{sklRelEkv})), zato sta po izreku \ref{KotNaPoltrak} poltraka $SX_1$ in $SX'$ identična. To pomeni, da točka $X'$ leži na poltraku $SX_1$ oz. velja $\mathcal{B}(X,S,X')$. Ker je zaradi $\mathcal{I}:S,X \mapsto S,X'$ še $SX\cong SX'$, je po definiciji točka $S$ središče daljice $XX'$. + + Naj bo na koncu $Y$ poljubna točka poltraka $SP$, ki se razlikuje od točke $S$, in $Y'=\mathcal{I}(Y)$. Točka $Y'$ leži na poltraku $SQ$, zato je $\mathcal{B}(Y,S,Y')$. Ker je zaradi $\mathcal{I}:S,Y \mapsto S,Y'$ še $SY\cong SY'$, je po definiciji točka $S$ središče daljice $YY'$. +\kdokaz + +V razdelku \ref{odd6SredZrc} bomo izometrijo, ki je omenjena v prejšnjem izreku \ref{sredZrcObstoj}, posebej obravnavali. + + + Definirajmo nove vrste kotov. + Konveksni kot je \index{kot!ostri}\pojem{ostri kot}, \index{kot!pravi} + \pojem{pravi kot} oz. + \index{kot!topi}\pojem{topi kot}, če je + manjši, enak oz. večji od svojega sokota (Figure \ref{sl.aks.2.3.21.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.21.pic} +\caption{} \label{sl.aks.2.3.21.pic} +\end{figure} + + + +Iz definicije sledi, da so ostri (oz. topi) koti tisti konveksni +koti, ki so manjši (oz. večji) od pravega kota. + +Iz izreka \ref{izrekSimetralaKota} sledi, da pravi kot +obstaja, saj bisektrisa iztegnjeni kot razdeli na dva skladna +sokota. + +Ni težko dokazati, da sta vsaka dva prava kota skladna ter da je +kot, ki je skladen s pravim kotom, tudi pravi kot. + +Če je vsota dveh kotov pravi kot, pravimo, da sta kota +\index{kota!komplementarna}\pojem{komplementarna} (Figure +\ref{sl.aks.2.3.22.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.22.pic} +\caption{} \label{sl.aks.2.3.22.pic} +\end{figure} + + + +Sedaj bomo vpeljali izjemno pomembno relacijo med premicama. Če +premici $p$ in $q$ vsebujeta kraka pravega kota, pravimo, da sta +$p$ in $q$ \pojem{pravokotni}, kar označimo $p \perp q$ (Figure +\ref{sl.aks.2.3.23.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.23.pic} +\caption{} \label{sl.aks.2.3.23.pic} +\end{figure} + +Iz same definicije je jasno, da je pravokotnost simetrična relacija +oz. iz $p \perp q$ sledi $q \perp p$. Če je $p \perp q$ in $p \cap q=S$, +pravimo, da je premica $p$ +\index{pravokotni!premici}\pojem{pravokotna} na premico $q$ v točki +$S$ oz. da je $p$ \index{pravokotnica}\pojem{pravokotnica} +premice $q$ v tej točki. + + + +Naslednji izrek je najpomembnejši izrek, ki karakterizira relacijo +pravokotnosti. + + + + \bizrek \label{enaSamaPravokotnica} + For each point $A$ and each line $p$, there is a unique line $n$ + going through the point $A$, which is perpendicular on the line $p$. + \eizrek + +\textbf{\textit{Proof.}} +Predpostavimo, da točka $A$ ne leži na premici $p$. Naj bosta +$B$ in $C$ poljubni točki, ki ležita na premici $p$ + (Figure \ref{sl.aks.2.3.24.pic}). S $\pi$ označimo polravnino $BCA$, +komplementarno polravnino pa s $\pi_1$. Po izreku \ref{izomEnaC'} +obstaja ena sama točka $A_1\in \pi_1$, za katero velja $(A,B,C) +\cong (A_1,B,C)$. Iz izreka \ref{IizrekABC} sledi, da obstaja ena +sama izometrija $\mathcal{I}$, ki preslika točke $A$, $B$ in $C$ +v točke $A_1$, $B$ in $C$. Premico $AA_1$ označimo z $n$. Ker +sta $A$ in $A_1$ na različnih bregovih premice $p$, premica $n$ +seka premico $p$ v neki točki $S$. Iz $\mathcal{I}:B,C \mapsto +B,C$ sledi $\mathcal{I}(S)=S$ (izrek \ref{izoABAB}). Torej +izometrija $\mathcal{I}$ preslika kot $ASB$ v kot $A_1SB$. Sledi, +da sta $\angle ASB$ in $\angle A_1SB$ skladna sokota, zato sta +tudi prava kota. Torej je $n \perp p$. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.24.pic} +\caption{} \label{sl.aks.2.3.24.pic} +\end{figure} + +Dokažimo, da je $n$ edina pravokotnica premice $p$ skozi točko +$A$. Naj bo $\widehat{n}$ premica, za katero je tudi $A\in +\widehat{n}$ in $\widehat{n} \perp p$. Naj bo točka $\widehat{S}$ +presečišče premic $\widehat{n}$ in $p$. Po predpostavki je +$\angle A\widehat{S}B$ pravi kot in je skladen s svojim sokotom +$\angle B\widehat{S}A_2$ ($A_2$ je takšna točka, da velja +$\mathcal{B}(A,\widehat{S},A_2)$), ki je tudi pravi kot. + +Iz $\mathcal{I}:B,C \mapsto B,C$ sledi +$\mathcal{I}(\widehat{S})=\widehat{S}$ (izrek \ref{izoABAB}). +Torej izometrija $\mathcal{I}$ preslika kot $A\widehat{S}B$ v kot +$A_1\widehat{S}B$. Sledi, da sta $\angle A\widehat{S}B$ in $\angle +A_1\widehat{S}B$ skladna, zato je tudi $\angle A_1\widehat{S}B$ +pravi kot. Torej sta kota $A_1\widehat{S}B$ in $A_2\widehat{S}B$ +prava kota in sta zato skladna. Iz tega sledi, da sta poltraka +$\widehat{S}A_1$ in $\widehat{S}A_2$ ista, zato je $A_1 \in +\widehat{S}A_2=\widehat{n}$ oz. $\widehat{n}=AA_1=n$. + + V primeru, ko točka $A$ leži na premici $p$, je pravokotnica $n$ + simetrala pripadajočega iztegnjenega kota (izrek \ref{izrekSimetralaKota}). +\kdokaz + + +Prejšnji izrek ima za posledico zelo pomembno dejstvo - obstoj parov disjunktnih premic v ravnini - oz. takšnih, ki nimata skupnih točk. To je vsebina naslednjih dveh izrekov. + + + \bizrek \label{absolGeom1} + Let $p$ and $q$ be a lines perpendicular on a line $PQ$ in the points $P$ and $Q$. + Then the lines $p$ and $q$ do not have a common points i.e. $p\cap q=\emptyset$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.25b.pic} +\caption{} \label{sl.aks.2.3.25b.pic} +\end{figure} + +\textbf{\textit{Proof.}} Izrek je direktna posledica prejšnjega izreka \ref{enaSamaPravokotnica}. Če bi se namreč premici $p$ in $q$ sekali v neki točki $S$, bi iz točke $S$ imeli dve pravokotnici na premico $PQ$ (Figure \ref{sl.aks.2.3.25b.pic}), kar je v protislovju z omenjenim izrekom. + \kdokaz + + + + \bizrek \label{absolGeom} + If $A$ is a point that does not lie on a line $p$, then there exists at least + one line (in the same plane) passing through the point $A$ and not intersecting the line + $p$ (Figure \ref{sl.aks.2.3.25a.pic}). + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.25a.pic} +\caption{} \label{sl.aks.2.3.25a.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Po izreku \ref{enaSamaPravokotnica} obstaja (natanko ena) pravokotnica $n$ premice $p$, ki poteka skozi točko $A$. Označimo z $A'$ presečišče premic $p$ in $n$. Iz istega izreka sledi, da obstaja še pravokotnica $q$ premice $n$ v točki $A$. Po prejšnjem izreku \ref{absolGeom1} je $q$ premica, ki poteka skozi točko $A$ in nima skupnih točk s premico $p$. + \kdokaz + + +Točka $A'$ je \index{nožišče}\pojem{nožišče} ali +\index{pravokotna projekcija}\pojem{pravokotna projekcija} točke +$A$ na premico $p$, če pravokotnica premice $p$ skozi točko $A$ +seka to premico v točki $A'$. Označili jo bomo z $A'=pr_{\perp +p}(A)$ (Figure \ref{sl.aks.2.3.25.pic}). + Iz prejšnjega izreka sledi, da za vsako točko in premico + obstaja ena sama pravokotna projekcija. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.25.pic} +\caption{} \label{sl.aks.2.3.25.pic} +\end{figure} + +Premico, ki poteka skozi središče $S$ daljice $AB$ in je pravokotna +na premico $AB$, imenujemo \index{simetrala!daljice}\pojem{simetrala daljice} $AB$ in jo označimo s $s_{AB}$ +(Figure \ref{sl.aks.2.3.26.pic}). Lastnosti simetrale daljice bomo +obravnavali v naslednjem poglavju. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.26.pic} +\caption{} \label{sl.aks.2.3.26.pic} +\end{figure} + + +Pravimo, da je točka $A$ \index{simetričnost!glede na premico}\pojem{simetrična} točki $B$ glede na premico $s$, če je $s$ simetrala daljice $AB$. Simetričnost glede na premico (kot preslikavo - t. i. osno zrcaljenje) bomo podrobneje obravnavali v razdelku \ref{odd6OsnZrc}. + +Naj bo $S$ točka in $AB$ daljica. Množico vseh točk $X$, za +katere velja $SX \cong AB$, imenujemo +\index{krožnica}\pojem{krožnica} s +\index{središče!krožnice}\pojem{središčem} $S$ in \index{polmer +krožnice}\pojem{polmerom} $AB$; označimo jo s $k(S,AB)$ (Figure +\ref{sl.aks.2.3.27.pic}) oz.: + $$k(S,AB)=\{X;\hspace*{1mm}SX \cong AB\}.$$ + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.27.pic} +\caption{} \label{sl.aks.2.3.27.pic} +\end{figure} + + Seveda je krožnica množica + točk v ravnini, ker v tej knjigi obravnavamo le ravninsko + geometrijo (vse točke in vsi liki pripadajo isti ravnini). + + Iz definicije je jasno, da za polmer lahko izberemo poljubno daljico, + ki je skladna z +daljico $AB$, torej katerokoli daljico $SP$, kjer je $P$ poljubna +točka na krožnici. Ker polmer ni vezan na +določeno daljico, ga običajno označujemo z malo črko $r$. Torej lahko krožnico zapišemo tudi takole: + $$k(S,r)=\{X;\hspace*{1mm}SX \cong r\}.$$ + Množico + +$$\{X;\hspace*{1mm}SX \leq r\}$$ +imenujemo \index{krog}\pojem{krog} s središčem $S$ in polmerom $r$ (Figure \ref{sl.aks.2.3.28.pic}) označimo ga s $\mathcal{K}(S,r)$. +Množica + $$\{X;\hspace*{1mm}SX < r\}$$ + je \index{notranjost!kroga} + \pojem{notranjost kroga} $\mathcal{K}(S, r)$, njene točke so pa + \pojem{notranje točke kroga}. + To pomeni, da je krog pravzaprav unija svoje notranjosti in pripadajoče krožnice. + +Množico + $$\{X;\hspace*{1mm}SX > r\}$$ + imenujemo \index{zunanjost!kroga}\pojem{zunanjost kroga} $\mathcal{K}(S, r)$ + in njene točke \pojem{zunanje točke kroga}. + + Iz praktičnih razlogov bomo notranjost kroga $\mathcal{K}(S, r)$ imenovali tudi \index{notranjost!krožnice}\pojem{notranjost} pripadajoče krožnice $k(S,r)$, zunanjost kroga $\mathcal{K}(S, r)$ pa \index{zunanjost!krožnice}\pojem{zunanjost} pripadajoče krožnice $k(S,r)$. Enako definiramo tudi \pojem{notranje} oz. \pojem{zunanje} točke krožnice. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.28.pic} +\caption{} \label{sl.aks.2.3.28.pic} +\end{figure} + + Če sta $P$ in $Q$ dve točki krožnice $k(S, r)$, daljico $PQ$ + imenujemo \index{tetiva krožnice} \pojem{tetiva} te + krožnice. Če +tetiva vsebuje središče krožnice, jo imenujemo +\index{premer krožnice}\pojem{premer} ali \index{diameter +krožnice}\pojem{diameter} te krožnice (Figure +\ref{sl.aks.2.3.29.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.29.pic} +\caption{} \label{sl.aks.2.3.29.pic} +\end{figure} + +Dokažimo naslednji izrek. + + + \bizrek \label{premerInS} + The centre $S$ of the circle $k(S, r)$ is at the same time the midpoint of each + diameter of that circle. + \eizrek + +\textbf{\textit{Proof.}} + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.3.30.pic} +\caption{} \label{sl.aks.2.3.30.pic} +\end{figure} + + Če je $PQ$ premer krožnice $k(S, r)$, točki $P$ in $Q$ ležita na +krožnici, kar pomeni: $SP \cong SQ \cong r$ (Figure +\ref{sl.aks.2.3.30.pic}). Ker točka $S$ leži na daljici $PQ$, +sledi, da je točka $S$ središče te daljice. + \kdokaz + + Iz prejšnjega izreka sledi, da je premer enak dvema polmeroma, ker je: +$PQ = PS + SQ = 2\cdot PS = 2\cdot r$. To pa pomeni, +da so vsi premeri neke krožnice med seboj skladni. + + +Naj bosta $P$ in $Q$ poljubni točki krožnice $k(S, r)$. Presek +krožnice $k$ z eno od polravnin (v ravnini te krožnice) z robom +$s=PQ$ imenujemo \index{krožni!lok} \pojem{krožni lok} $PQ$ (ali +krajše \pojem{lok}) s krajiščema $P$ in $Q$. + +Torej vsaka tetiva $PQ$ na neki krožnici $k$ določa dva loka. Predpostavimo, da središče $S$ ne leži na robu polravnine, ki generira krožni lok. +Če ta polravnina +vsebuje središče $S$ krožnice $k$, +gre za \pojem{veliki lok} $PQ$, sicer je to \pojem{mali +lok} $PQ$. + Če pa je + središče $S$ na samem robu $PQ$ polravnine, potem je vsak od obeh lokov $PQ$ + \index{polkrožnica}\pojem{polkrožnica} $PQ$ (Figure +\ref{sl.skk.4.2.1.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.1.pic} +\caption{} \label{sl.skk.4.2.1.pic} +\end{figure} + +Ker lok le s svojima krajiščema ni enolično določen, moramo na krožnici poznati še +vsaj eno točko, ki temu loku pripada oz. ne pripada. + +Na podoben način definiramo tudi določene dele kroga. + +Naj bosta $P$ in $Q$ poljubni točki krožnice $k(S, r)$. Presek +kroga $\mathcal{K}(S, r)$ z eno od polravnin (v ravnini te krožnice) z robom +$s=PQ$ imenujemo +\index{krožni!odsek} \pojem{krožni odsek}. + Torej vsaka tetiva $PQ$ na neki krožnici $k(S, r)$ določa na krogu $\mathcal{K}(S, r)$ dva krožna odseka. Predpostavimo, da središče $S$ ne leži na robu polravnine, ki generira krožni odsek. +Če ta polravnina +vsebuje središče $S$ krožnice $k$, +gre za \pojem{večji krožni odsek} $PQ$, sicer pa za \pojem{manjši +krožni odsek} $PQ$. + Če je + središče $S$ na samem robu $PQ$ polravnine, potem je vsak od obeh krožnih odsekov +\index{polkrog}\pojem{polkrog}. + Iz definicije je jasno, da je rob krožnega odseka unija tetive $PQ$ in ustreznega loka (Figure +\ref{sl.skk.4.2.1b.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.1b.pic} +\caption{} \label{sl.skk.4.2.1b.pic} +\end{figure} + + +Definirajmo še en pojem, ki je povezan s krogom. +Naj bosta $P$ in $Q$ poljubni točki krožnice $k(S, r)$. Presek +kroga $\mathcal{K}(S, r)$ z enim od kotov $PSQ$ imenujemo +\index{krožni!izsek} \pojem{krožni izsek}. Tudi v tem primeru imamo dva krožna izseka. Če je kot $PSQ$ iztegnjeni kot, dobimo dva polkroga, sicer pa konveksni in nekonveksni krožni izsek, odvisno od tega, ali je kot $PSQ$ konveksen ali nekonveksen (Figure +\ref{sl.skk.4.2.1c.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.1c.pic} +\caption{} \label{sl.skk.4.2.1c.pic} +\end{figure} + + + + + +%________________________________________________________________________________ + \poglavje{Continuity Axiom} \label{odd2AKSZVE} + + Že v osnovni šoli smo pri uvajanju številske premice in + koordinatnega sistema izvedeli, + da je možno +vzpostaviti povezavo, pri kateri vsaki točki neke premice pripada +določeno realno število in obratno, vsakemu realnemu številu lahko +priredimo točko, ki leži na tej premici. S tem je povezan naslednji +aksiom. + + \baksiom \label{aksDed}\index{aksiom!Dedekindov} + (Dedekind’s\footnote{\index{Dedekind, R.} + \textit{R. Dedekind} (1831--1916), + nemški matematik.} + axiom) + Suppose that all points on open line segment $AB$ are divided into the union of two nonempty disjoint sets $\mathcal{U}$ and +$\mathcal{V}$ such that no point of $\mathcal{U}$ is + between two points of $\mathcal{V}$ and vice versa: no point of $\mathcal{V}$ is + between two points of $\mathcal{U}$. Then there is a unique point $C$ on open line segment $AB$ such that + $B(A',C,B')$ for any two points $A'\in +\mathcal{U}\setminus{C}$ and $B'\in \mathcal{V} \setminus {C}$ +(Figure \ref{sl.aks.2.4.1.pic}). + \eaksiom + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.4.1.pic} +\caption{} \label{sl.aks.2.4.1.pic} +\end{figure} + +Povejmo brez dokazov dve pomembni posledici aksioma +zveznosti\footnote{Vse do 19. stoletja matematiki niso +čutili potrebe, da bi dokazali ti dve trditvi, oz. potrebe za +uvajanjem aksioma zveznosti. Celo \index{Evklid} Evklid iz +Aleksandrije (3. stol. pr. n. š.) v svojem znanem delu +‘‘Elementi’’, navaja konstrukcijo enakostraničnega trikotnika, pri kateri ne +dokazuje, da se določeni krožnici sekata.}. + + + + \bizrek \label{DedPoslKrozPrem} Let $k$ be a circle and $P$ a point inside that circle. + Then any line $p$ passing through the point $P$ and the circle $k$ has exactly two common points (Figure + \ref{sl.aks.2.4.2.pic}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.4.2.pic} +\caption{} \label{sl.aks.2.4.2.pic} +\end{figure} + + + + \bizrek \label{DedPoslKrozKroz} If $k$ and $l$ are circles such that $l$ contains at least one point inside and one point outside $k$, + then the circles has exactly two points (Figure \ref{sl.aks.2.4.3.pic}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.4.3.pic} +\caption{} \label{sl.aks.2.4.3.pic} +\end{figure} + +Dedekindov aksiom se v nekoliko drugačni obliki uporablja pri +zasnovi množice realnih števil. To nas spominja na že omenjeno +povezavo med množico točk neke premice in množico realnih števil. + +Operacijo množenja daljice $AB$ s poljubnim +pozitivnim racionalnim številom $q$ smo že definirali. Sedaj lahko razširimo pojem +množenja za poljubno pozitivno realno število $\lambda$. Definicija daljice +$\lambda\cdot AB$, ($\lambda \in \mathbb{R}^+$), ki jo tu formalno ne +bomo izpeljali do konca, je povezana z dvema množicama točk +na poltraku $CD$: + \begin{eqnarray*} +&& \{X\in CD;\hspace*{1mm}CX=q\cdot +AB,\hspace*{1mm}q\leq\lambda,\hspace*{1mm}q\in +\mathbb{Q}^+ \} \hspace*{1mm}\textrm{ in}\\ +&& \{X\in CD;\hspace*{1mm}CX=q\cdot +AB,\hspace*{1mm}q>\lambda,\hspace*{1mm}q\in \mathbb{Q}^+ \} + \end{eqnarray*} + ter +Dedekindovim aksiomom \ref{aksDed}. + + S pomočjo aksioma zveznosti \ref{aksDed} lahko vpeljemo tudi + pojme merjenja daljic in kotov. + + Pri merjenju daljic bomo vsaki daljici $AB$ priredili + pozitivno realno število $\textsl{m}(AB)$ na naslednji način. + Naj bo $\mathcal{D}$ množica vseh daljic in $\mathbb{R}^+$ množica vseh pozitivnih realnih števil. Preslikavo $\textsl{m}:\mathcal{D}\rightarrow\mathbb{R}^+$, ki izpolnjuje naslednje lastnosti: + \begin{itemize} + \item $(\exists A_0B_0\in\mathcal{D})\hspace*{1mm}\textsl{m}(A_0B_0)=1$, + \item $(\forall AB, CD\in\mathcal{D})\hspace*{1mm}(AB\cong CD \Rightarrow\textsl{m}(AB)=\textsl{m}(CD))$, + \item $(\forall AB, CD, EF\in\mathcal{D})\hspace*{1mm}(AB+CD=EF\Rightarrow \textsl{m}(AB)+\textsl{m}(CD)=\textsl{m}(EF))$, + \end{itemize} + imenujemo \index{dolžina!daljice}\pojem{dolžina daljice} ali \index{mera!daljice}\pojem{mera daljice}, trojico $\textsl{M}=(\mathcal{D},\mathbb{R}^+,\textsl{m})$ pa \index{sistem merjenja!daljic}\pojem{sistem merjenja daljic}. + + Dolžino daljice $AB$ (oz. $\textsl{m}(AB)$) bomo navadno označevali z $|AB|$. + + Intuitivno je jasno, da obstaja neskončno mnogo sistemov merjenja, ki pa so odvisni od izbire enotske daljice $A_0B_0$ - tiste, ki ima dolžino enako 1, oz. $\textsl{m}(A_0B_0)=1$. V nekem sistemu merjenja je torej dolžina poljubne daljice $AB$ predstavljena s pozitivnim realnim številom $x$, za katerega je $AB=x\cdot A_0B_0$. Torej je $\textsl{m}(AB)=x$ natanko tedaj, ko je $AB=x\cdot A_0B_0$ (Figure \ref{sl.aks.2.4.4.pic}). Sedaj je jasno, zakaj potrebujemo aksiom zveznosti - brez njega bi imeli težave z definicijo dolžine diagonale kvadrata, ki ima za stranico enotsko daljico (dolžine 1)\footnote{Stari Grki so si dolžino vedno predstavljali kot racionalno število, zato so stranico +in diagonalo kvadrata imenovali \pojem{neprimerljivi daljici}.}. + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.4.4.pic} +\caption{} \label{sl.aks.2.4.4.pic} +\end{figure} + + + \bizrek \label{meraDalj1} + Za poljuben sistem merjenja daljic velja: + + (\textit{i}) $AB|AB|$. + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.4.6.pic} +\caption{} \label{sl.aks.2.4.6.pic} +\end{figure} + + (\textit{ii}) Predpostavimo, da ni $AB\cong CD$. Brez škode za splošnost naj bo $AB\alpha$ in $\gamma'>\beta$. Relaciji pa sta direktna posledica dokazane neenakosti $\alpha+\beta=\gamma'$ iz prejšnjega izreka \ref{zunanjiNotrNotr}. + \kdokaz + + Glede na notranje kote lahko obravnavamo tri vrste trikotnikov. + Dokazali smo že, + da je v poljubnem trikotniku vsota notranjih kotov enaka $180^0$ + (izrek \ref{VsotKotTrik}). +To pomeni, da je največ eden od teh kotov topi kot ali pravi kot, +oz. sta vsaj dva ostra. Torej imamo tri možnosti (Figure +\ref{sl.aks.2.6.4a.pic}): +\begin{itemize} + \item Trikotnik je \index{trikotnik!ostrokotni} + \pojem{ostrokotni}, če ima vse notranje kote ostre. + \item Trikotnik je \index{trikotnik!pravokotni}\pojem{pravokotni}, + če ima en notranji kot pravi. Stranico, ki je nasproti +pravemu kotu pravokotnega trikotnika, imenujemo +\index{hipotenuza}\pojem{hipotenuza}, ostali dve stranici sta +\index{kateta}\pojem{kateti}. Ker je vsota notranjih kotov v vsakem + trikotniku enaka $180^0$, sta +kota pri hipotenuzi pravokotnega trikotnika komplementarna. + \item Trikotnik je \index{trikotnik!topokotni}\pojem{topokotni}, + če ima en notranji kot topi. + Ostala dva notranja kota sta potem ostra. +\end{itemize} + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.6.4a.pic} +\caption{} \label{sl.aks.2.6.4a.pic} +\end{figure} + +%slika 33.1 + + Dokažimo še izreka, ki se nanašata na poljubne večkotnike. + + + + \bizrek \label{VsotKotVeck} + The sum of the interior angles of any $n$-gon is equal to + $(n - 2) \cdot 180^0$. + \eizrek + + + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.5.5c.pic} +\caption{} \label{sl.aks.2.5.5c.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Predpostavimo najprej, da je $A_1A_2\ldots A_n$ konveksen $n $-kotnik (Figure \ref{sl.aks.2.5.5c.pic}). Njegovih $n-3$ diagonal $A_1A_3$, $A_1A_4$, ... $A_1A_{n-1}$ razdeli ta večkotnik na $n-2$ trikotnikov $\triangle_1$, $\triangle_2$, ..., $\triangle_{n-2}$. Ker se pri tem tudi vsak notranji kot $n $-kotnika $A_1A_2\ldots A_n$ razdeli na ustrezne notranje kote omenjenih trikotnikov, je vsota vseh notranjih kotov + $n $-kotnika $A_1A_2\ldots A_n$ enaka vsoti vseh kotov trikotnikov $\triangle_1$, $\triangle_2$, ..., $\triangle_{n-2}$. Po izreku \ref{VsotKotTrik} je na koncu ta vsota enaka ravno $(n - 2) \cdot 180^0$. + + Na tem mestu ne bomo dokazovali dejstva, da se tudi v primeru nekonveksnega $n $-kotnika $A_1A_2\ldots A_n$ le-ta lahko razdeli na $n-2$ trikotnikov. + \kdokaz + + + \bizrek \label{VsotKotVeckZuna} + The sum of the exterior angles of any $n$-gon is equal to + $360^0$. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.5.5b.pic} +\caption{} \label{sl.aks.2.5.5b.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Označimo z $\alpha_1$, $\alpha_2$,..., $\alpha_n$ notranje in $\alpha'_1$, $\alpha'_2$, ..., $\alpha'_n$ pripadajoče zunanje kote ob ogliščih $A_1$, $A_2$,...,$A_n$ konveksnega $n $-kotnika $A_1A_2\ldots A_n$. (Figure \ref{sl.aks.2.5.5b.pic}). Za ustrezni notranji in zunanji kot velja: + + \begin{eqnarray*} + & & \alpha_1+\alpha'_1=180^0\\ + & & \alpha_2+\alpha'_2=180^0\\ + & & \vdots\\ + & & \alpha_n+\alpha'_n=180^0 + \end{eqnarray*} + + Če seštejemo vse enakosti in upoštevamo dokazano enakost iz prejšnjega izreka \ref{VsotKotVeck} $$\alpha_1+\alpha_2+\cdots + \alpha_n=(n - 2) \cdot 180^0,$$ + dobimo $(n - 2) \cdot 180^0+\alpha'_1+\alpha'_2+\cdots + \alpha'_n=n\cdot 180^0$ oz. + $$\alpha'_1+\alpha'_2+\cdots + \alpha'_n=2 \cdot 180^0=360^0,$$ kar je bilo treba dokazati. \kdokaz + + Iz izreka \ref{VsotKotVeck} sledi, da je vsota vseh notranjih kotov poljubnega štirikotnika enaka $360^0$, iz izreka \ref{VsotKotVeckZuna} pa, da je tudi vsota vseh zunanjih kotov konveksnega štirikotnika enaka $360^0$ (Figure \ref{sl.aks.2.5.5a.pic}). + + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.5.5a.pic} +\caption{} \label{sl.aks.2.5.5a.pic} +\end{figure} + + + + Ker je trikotnik (kot presek treh polravnin) konveksen lik, je vsota vseh zunanjih kotov poljubnega trikotnika enaka $360^0$ (Figure \ref{sl.aks.2.5.5d.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.5.5d.pic} +\caption{} \label{sl.aks.2.5.5d.pic} +\end{figure} + + +Podobna trditev glede na izrek \ref{KotaVzporKraki}, ki se nanaša na kota z vzporednimi kraki, velja tudi za kota s pravokotnimi kraki. + + + + \bizrek \label{KotaPravokKraki} + Angles with perpendicular sides are either congruent or supplementary. + \index{kota!s pravokotnimi kraki} + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.aks.2.5.5.pic} +\caption{} \label{sl.aks.2.5.5.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Naj bosta $\angle aSb$ in $\angle a'S'b'$ takšna kota, da velja $a\perp a'$ in $b\perp b'$ (Figure \ref{sl.aks.2.5.5.pic}). Naj bo $A$ presečišče nosilk krakov $a$ in $a'$ ter $B$ presečišče nosilk krakov $b$ in $b'$. V štirikotniku $SAS'B$ merita notranja kota ob ogliščih $A$ in $B$ vsak $90^0$. Po izreku \ref{VsotKotVeck} je vsota vseh notranjih kotov tega štirikotnika enaka $360^0$. Zato notranja kota $BSA$ in $AS'B$ tega štirikotnika merita skupaj $180^0$, kar pomeni, da sta dva kota, ki ju določata nosilki poltrakov $a$ in $b$ oz. $a'$ in $b'$ suplementarna. Če pa enega od teh kotov zamenjamo z njegovim sokotom, sta ustrezna kota skladna. + \kdokaz + + +%________________________________________________________________________________ +\naloge{Exercises} +\begin{enumerate} + +\item Naj bodo $P$, $Q$ in $R$ notranje točke stranic trikotnika +$ABC$. Dokaži, da so $P$, $Q$ in $R$ nekolinearne. + +\item Naj bosta $P$ in $Q$ točki stranic $BC$ in $AC$ trikotnika $ABC$ in hkrati +različni od njegovih oglišč. Dokaži, da se daljici $AP$ in $BQ$ +sekata v eni točki. + +\item Točke $P$, $Q$ in $R$ ležijo po vrsti na stranicah $BC$, $AC$ in $AB$ trikotnika + $ABC$ in so različne od njegovih oglišč. Dokaži, da se daljici $AP$ +in $QR$ sekata v eni točki. + +\item Premica $p$, ki leži v ravnini štirikotnika, seka njegovo +diagonalo $AC$ in ne poteka skozi nobeno oglišče tega štirikotnika. +Dokaži, da premica $p$ seka natanko dve stranici tega +štirikotnika. + +\item Dokaži, da je polravnina konveksen lik. + +\item Dokaži, da je presek dveh konveksnih likov +konveksen lik. + +\item Dokaži, da je poljuben trikotnik konveksen lik. + +\item Če je $\mathcal{B}(A,B,C)$ in $\mathcal{B}(D,A,C)$, je tudi +$\mathcal{B}(B,A,D)$. Dokaži. + +\item Naj bodo $A$, $B$, $C$ in $D$ takšne kolinearne točke, da je +$\neg\mathcal{B}(B,A,C)$ in $\neg\mathcal{B}(B,A,D)$. Dokaži, da +velja $\neg\mathcal{B}(C,A,D)$. + +\item Naj bo $A_1A_2\ldots,A_{2k+1}$ poljubni večkotnik z lihim +številom oglišč. Dokaži, da ne obstaja premica, ki seka vse +njegove stranice. + +\item Če izometrija $\mathcal{I}$ preslika lika $\Phi_1$ in $\Phi_2$ +v lika $\Phi'_1$ in $\Phi'_2$, potem se presek +$\Phi_1\cap\Phi_2$ s to izometrijo preslika v presek +$\Phi'_1\cap\Phi'_2$. Dokaži. + +\item Dokaži, da sta poljubna poltraka neke +ravnine med seboj skladna. + +\item Dokaži, da sta poljubni premici neke +ravnine med seboj skladni. + + +\item Naj bosta $k$ in $k'$ dve krožnici +neke ravnine s središčema $O$ in $O'$ ter polmeroma $AB$ in $A'B'$. +Dokaži ekvivalenco: $k\cong k' \Leftrightarrow AB\cong A'B'$. + +\item Naj bo $\mathcal{I}$ +neidentična izometrija ravnine z dvema negibnima točkama $A$ in +$B$. Če je $p$ premica te ravnine, ki je vzporedna s premico +$AB$ in $A\notin p$. Dokaži, da na premici $p$ ni negibnih točk +izometrije $\mathcal{I}$. + +\item Naj bo $S$ edina negibna točka +izometrije $\mathcal{I}$ v neki ravnini. Dokaži, da če ta izometrija +preslika premico $p$ vase, je $S\in p$. + +\item Dokaži, da se poljubni dve premici +neke ravnine ali sekata ali sta vzporedni. + +\item Če neka +premica v ravnini seka eno od dveh vzporednic iste ravnine, + potem seka tudi drugo vzporednico. Dokaži. + +\item Dokaži, da vsaka izometrija preslika vzporednici v vzporednici. + +\item Naj bodo $p$, $q$ in $r$ takšne premice neke ravnine, tako da velja +$p\parallel q$ in $r\perp p$. Dokaži, da je $r\perp q$. + +\item Dokaži, da konveksen $n$-kotnik ne more imeti več kot treh +ostrih kotov. + +\end{enumerate} + + + +% DEL 3 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% SKLADNOST TRIKOTNIKOV. VEČKOTNIKI +%________________________________________________________________________________ + + \del{Congruence. Triangles and Polygons} \label{pogSKL} + +%________________________________________________________________________________ + \poglavje{Triangle Congruence Theorems} \label{odd3IzrSkl} + +Iz splošne definicije skladnosti likov sledi, da sta trikotnika +skladna, če obstaja izometrija, ki preslika prvi trikotnik v +drugega. Jasno je, da iz skladnosti dveh trikotnikov sledi +skladnost pripadajočih stranic in notranjih kotov. Nas pa zanima obraten problem: +Kdaj iz skladnosti nekaterih od +pripadajočih stranic in kotov sledi skladnost dveh trikotnikov? O +tem govorijo naslednji \index{izrek!o skladnosti +trikotnikov}\pojem{triangle congruence theorems}\footnote{Prvi, tretji in četrti izrek o skladnosti +trikotnikov pripisujejo \index{Pitagora} \textit{Pitagori z otoka +Samosa} (6. stol. pr. n. š.), za drugi izrek pa se +predpostavlja, da je bil znan že \index{Tales} \textit{Talesu iz +Mileta} (7.--6. stol. pr. n. š.). Vse štiri +navaja \index{Evklid} \textit{Evklid iz Aleksandrije} +(3. stol. pr. n. š.) v prvi knjigi svojih ‘‘Elementov’’.}: + + \bizrek \label{SSS} (\textit{SSS}) + Triangles are congruent if three sides of one triangle are congruent + to the corresponding sides of the other triangle, + i.e. (Figure + \ref{sl.skl.3.1.1.pic}): + \begin{eqnarray*} + \left. + \begin{array}{l} + AB \cong A'B'\\ + BC \cong B'C'\\ + AC \cong A'C' + \end{array} + \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC \cong \triangle A'B'C' + \end{eqnarray*} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.1.pic} +\caption{} \label{sl.skl.3.1.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Iz skladnosti daljic po izreku \ref{izrek(A,B)} je tudi $(A,B) + \cong (A',B')$, $(B,C) \cong (B',C')$ + in $(A,C) + \cong (A',C')$ oz $(A,B,C) \cong (A',B',C')$. Iz izreka + \ref{IizrekABC} sledi, da obstaja izometrija $\mathcal{I}$, ki + preslika točke $A$, $B$ in $C$ v točke $A'$, $B'$ in $C'$. + Le-ta preslika trikotnik $ABC$ v trikotnik $A'B'C'$, kar je + posledica izreka \ref{izrekIzoB}. Torej sta trikotnika $ABC$ in + $A'B'C'$ skladna. + \kdokaz + + + + \bizrek \label{SKS} (\textit{SAS}) + Triangles are congruent if two pairs of sides and the included angle of one triangle + are congruent to the corresponding sides and angle of the other triangle, + i.e. (Figure + \ref{sl.skl.3.1.2.pic}): + \begin{eqnarray*} + \left. + \begin{array}{l} + AB \cong A'B'\\ + AC \cong A'C'\\ + \angle BAC \cong \angle B'A'C' + \end{array} + \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC \cong \triangle A'B'C' + \end{eqnarray*} + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.2.pic} +\caption{} \label{sl.skl.3.1.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Ker sta kota $BAC$ in $B'A'C'$ skladna, + obstaja izometrija $\mathcal{I}$, ki preslika kot $BAC$ v kot $B'A'C'$. + Ta izometrija preslika vrh $A$ v vrh $A'$ ter kraka $AB$ in $AC$ v $A'B'$ in $A'C'$. + Naj bo $\mathcal{I}(B)=\widehat{B}'$ in $\mathcal{I}(C)=\widehat{C}'$. + Iz tega + sledi $AB \cong A'\widehat{B}'$ in $AC \cong A'\widehat{C}'$, + toda po izreku \ref{ABnaPoltrakCX} je $\widehat{B}'=B'$ in + $\widehat{C}'=C'$. + Torej izometrija $\mathcal{I}$ + preslika točke $A$, $B$ in $C$ v točke $A'$, $B'$ in $C'$, + oz. trikotnik $ABC$ v trikotnik $A'B'C'$, kar pomeni, da sta trikotnika $ABC$ in + $A'B'C'$ skladna. +\kdokaz + + + + \bizrek \label{KSK} (\textit{ASA}) + Triangles are congruent if two pairs of angles and the included side of one triangle + are congruent to the corresponding angles and side of the other triangle + i.e. (Figure + \ref{sl.skl.3.1.3.pic}): + \begin{eqnarray*} + \left. + \begin{array}{l} + AB \cong A'B'\\ + \angle BAC \cong \angle B'A'C'\\ + \angle ABC \cong \angle A'B'C' + \end{array} + \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC \cong \triangle A'B'C' + \end{eqnarray*} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.3.pic} +\caption{} \label{sl.skl.3.1.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Iz aksioma \ref{aksIII2} sledi, da obstaja izometrija $\mathcal{I}$, + ki preslika + točko $A$ v točko $A'$, poltrak $AB$ v poltrak $A'B'$ in + polravnino $ABC$ v polravnino $A'B'C'$. + Ker je $AB \cong A'B'$, iz istega aksioma sledi + $\mathcal{I}(B)=B'$. Naj bo $\mathcal{I}(C)=\widehat{C}'$. + Potem je $\angle BAC \cong \angle B'A'\widehat{C}'$ in + $\angle ABC \cong \angle A'B'\widehat{C}'$. + Ker je po predpostavki tudi $\angle BAC \cong \angle B'A'C'$ in + $\angle ABC \cong \angle A'B'C'$, iz izreka + \ref{KotNaPoltrak} sledi, da sta poltraka $A'\widehat{C}'$ in + $A'\widehat{C}'$ (oz. $B'C'$ in + $B'\widehat{C}'$) enaka. Zato je $\widehat{C}'=A'\widehat{C}'\cap + A'\widehat{C}'=A'C'\cap + A'C'=C'$. Torej $\mathcal{I}:A,B,C \mapsto A',B',C'$, zato sta + trikotnika $ABC$ in $A'B'C'$ skladna. + \kdokaz + +Dokaz četrtega izreka o skladnosti trikotnikov bomo izpustili. + + + + \bizrek \label{SSK} (\textit{SSA}) + Triangles are congruent if two pairs of sides and the angle opposite to the longer side of one triangle + are congruent to the corresponding sides and angle of the other triangle. + (Figure \ref{sl.skl.3.1.4.pic}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.4.pic} +\caption{} \label{sl.skl.3.1.4.pic} +\end{figure} + + +Ena od pomembnejših posledic izrekov o skladnosti trikotnikov je +naslednja trditev. + + + + \bizrek \label{enakokraki} + If two sides of a triangle are congruent, then angles opposite those sides are congruent.\\ + And vice versa:\\ + If angles opposite those sides are congruent, then two sides of a triangle are congruent. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.5.pic} +\caption{} \label{sl.skl.3.1.5.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + Naj bo $ABC$ takšen trikotnik, da je $AB \cong AC$ + (Figure \ref{sl.skl.3.1.5.pic}). Ker velja +še $AC \cong AB$ in $BC \cong CB$, iz izreka \textit{SSS} sledi, da +sta trikotnika $ABC$ in $ACB$ skladna (ta dva trikotnika imata +različno orientacijo). Zato je $\angle ABC \cong \angle CBA$. Na +enak način bi lahko dokazali tudi obratno trditev. V tem primeru bi +uporabljali izrek \textit{ASA}. + \kdokaz + + Trikotnik (kot je trikotnik $ABC$ iz prejšnjega izreka), + ki ima vsaj dve +stranici skladni, imenujemo +\index{trikotnik!enakokraki}\pojem{enakokraki trikotnik}. Vsaka od +dveh skladnih stranic je \index{krak!enakokrakega +trikotnika}\pojem{krak}, tretja stranica pa je +\index{osnovnica!enakokrakega trikotnika}\pojem{osnovnica} tega +trikotnika. Torej sta po prejšnjem izreku notranja kota ob +osnovnici enakokrakega trikotnika skladna. In obratno - če sta dva +notranja kota trikotnika skladna, je ta trikotnik enakokrak. + + + + \bzgled + Let the $E$ and $F$ be a points lies on the line containing the hypotenuse $AB$ of a perpendicular + triangle $ABC$ and let $B(E,A,B,F)$, $EA\cong AC$ and + $FB\cong BC$. What is the measure of the angle $ACB$? + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.6.pic} +\caption{} \label{sl.skl.3.1.6.pic} +\end{figure} + + + \textbf{\textit{Solution.}} (Figure \ref{sl.skl.3.1.6.pic}) + + Notranja kota trikotnika $ABC$ ob ogliščih + $A$ in $B$ označimo z $\alpha$ in + $\beta$. + Trikotnika $EAC$ in $CBF$ sta + enakokraka trikotnika z osnovnicama $CE$ in $CF$, zato je $\angle CEA \cong \angle ACE$ in + $\angle CFB \cong \angle BFC$ (izrek \ref{enakokraki}). Kot + $\alpha$ je zunanji kot trikotnika $EAC$, zato je po izreku \ref{zunanjiNotrNotr}: + $\alpha = 2\angle ECA$ oz. $\angle ECA = \frac{1}{2} \alpha$. + Podobno je tudi $\angle FCB = \frac{1}{2} \beta$. Torej: +\begin{eqnarray*} + \angle ECF&=&\angle ECA+\angle ACB+\angle BCF=\\ + &=&\frac{1}{2} + \cdot\alpha+90^0+ + \frac{1}{2} \cdot\beta=90^0+ + \frac{1}{2} \cdot\left(\alpha+\beta\right)=\\ + &=&90^0+ + \frac{1}{2}\cdot 90^0=135^0, + \end{eqnarray*} +oz. $\angle ECF=135^0$. \kdokaz + +Trikotnik, pri katerem so vse stranice enake, imenujemo +\index{trikotnik!enakostranični}\pojem{enakostranični trikotnik} +(Figure \ref{sl.skl.3.1.7.pic}), ki je poseben primer +enakokrakega trikotnika. Zato iz omenjenega izreka sledi, +da so vsi notranji koti enakostraničnega trikotnika enaki. Ker je +njihova vsota enaka $180^0$, vsak od teh kotov meri $60^0$. Velja +tudi obratno: če sta vsaj dva kota trikotnika enaka $60^0$ (in +zaradi tega tudi tretji), potem je ta trikotnik enakostraničen. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.7.pic} +\caption{} \label{sl.skl.3.1.7.pic} +\end{figure} + + + + + \bzgled + Let $ABC$ be an equilateral triangle and $P$, $Q$ and $R$ + points such that $\mathcal{B}(A,B,R)$, $\mathcal{B}(B,C,Q)$, $\mathcal{B}(C,A,P)$ and + $BR\cong CQ\cong AP$. Prove that $PQR$ is also an equilateral triangle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.8.pic} +\caption{} \label{sl.skl.3.1.8.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.1.8.pic}) + + Iz danih pogojev je najprej + $AR\cong BQ\cong CP$. Trikotnik $ABC$ je enakostranični trikotnik, + zato vsi trije notranji koti merijo $60^0$. Iz tega sledi + $\angle PAR\cong \angle RBQ\cong QCP$. Po izreku \textit{SAS} so + trikotniki $PAR$, $RBQ$ in $QCP$ skladni in je $PR\cong RQ\cong QP$. + To pomeni, da je tudi $PQR$ enakostranični trikotnik. + \kdokaz + + +V prejšnjem poglavju smo definirali simetralo daljice kot premico, +ki je pravokotnica daljice in poteka skozi njeno središče. Dokažimo +sedaj ekvivalentno definicijo simetrale daljice, ki bo zelo +pomembna v nadaljevanju. \index{simetrala!daljice} + + + \bizrek \label{simetrala} + The perpendicular bisector of a line segment $AB$ is the set of all points $X$ + that are equidistant from its endpoints, i.e. $AX \cong BX$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.9.pic} +\caption{} \label{sl.skl.3.1.9.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + Naj bo premica $s$ simetrala daljice +$AB$ v neki ravnini. Po definiciji je $s$ pravokotnica na daljico +$AB$ skozi njeno središče – točko $S$. Označimo z $\mathcal{M}$ +množico vseh točk $X$ te ravnine, za katere velja $AX \cong BX$. +Potrebno je dokazati, da je $s =\mathcal{M}$. To bomo dokazali z +dvema inkluzijama (Figure \ref{sl.skl.3.1.9.pic}). + +($s\subseteq \mathcal{M}$). Naj bo $X \in s$. Dokažimo, da potem +velja $X \in \mathcal{M}$. Iz relacij $AS \cong BS$, $XS \cong XS$ +in $\angle ASX \cong \angle BSX = 90^0$ sledi, da sta trikotnika +$ASX$ in $BSX$ skladna (izrek \textit{SAS}). Zato je $AX \cong BX$ +oz. $X \in \mathcal{M}$. + +($\mathcal{M}\subseteq s$). Naj bo sedaj $X \in \mathcal{M}$. +Dokažimo, da velja $X \in s$. Iz $X \in \mathcal{M}$ sledi $AX \cong BX$. +Sedaj iz $AS \cong BS$, $XS \cong XS$ in $AX \cong BX$ sledi, +da sta trikotnika $ASX$ in $BSX$ skladna (izrek \textit{SSS}). Zato sta kota +$ASX$ in $BSX$ skladna in kot sokota sta oba prava kota. To pomeni, +da je premica $XS$ pravokotnica daljice $AB$ v njenem središču. +Torej, premica $XS$ je simetrala $s$ oz. $X \in s$. + \kdokaz + +The next problem is an example of multiple use of the theorem of an isosceles triangle (\ref{enakokraki}). + + \bnaloga\footnote{42. IMO USA - 2001, Problem 5.} + In a triangle $ABC$, let $AP$ bisect $\angle BAC$, with $P$ on $BC$, and let $BQ$ bisect $\angle ABC$, with $Q$ on $CA$. +It is known that $\angle BAC=60^0$ and that $|AB|+|BP|=|AQ|+|QB|$. +What are the possible measures of interior angles of triangle $ABC$? + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.9.IMO1.pic} +\caption{} \label{sl.skk.4.9.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Označimo notranje kote trikotnika $ABC$ z +$\alpha=60^0$, $\beta$ in $\gamma$. Naj bosta $D$ in $E$ takšni +točki, da velja: $BD\cong BP$, $\mathcal{B}(A,B,D)$, $QE \cong QB$ +in $\mathcal{B}(A,Q,E)$ (Figure \ref{sl.skk.4.9.IMO1.pic}). Iz teh +pogojev sledi, da sta $DBP$ in $BQE$ enakokraka trikotnika z +osnovnicama $DP$ in $BE$. Iz danega pogoja $|AB|+|BP|=|AQ|+|QB|$ +sledi še $AD\cong AE$, kar pomeni, da je tudi $ADE$ enakokraki +trikotnik z osnovnico $DE$. + +Ker je $DBP$ enakokraki trikotnik, iz izrekov \ref{enakokraki} in +\ref{zunanjiNotrNotr} sledi: $\angle BDP\cong \angle BPD +=\frac{1}{2}\angle ABC=\frac{1}{2}\beta$. +Ker je tudi $BQE$ enakokraki trikotnik, je $\angle QBE\cong\angle QEB$. + Iz skladnosti trikotnikov $ADP$ in $AEP$ (izrek \textit{SAS} +\ref{SKS}) sledi $\angle ADP\cong\angle AEP$ in $PD\cong PE$. + +Če povežemo dokazane relacije, velja: + \begin{eqnarray*} +&& \angle AEP\cong \angle BDP +=\frac{1}{2}\beta\cong \angle QBP\\ +&&\textrm{ in } \angle AEB\cong\angle QBE + \end{eqnarray*} + + Predpostavimo najprej, da velja $QB>QC$ oz. + $\mathcal{B}(Q,C,E)$. V tem primeru je: + \begin{eqnarray*} + \angle PEB &=&\angle AEB-\angle AEP=\\ + &=&\angle QBE-\angle QBP=\\ + &=&\angle PBE. + \end{eqnarray*} +To pomeni, da je $PBE$ enakokraki trikotnik z osnovnico $BE$ oz. +$PE\cong PB$. Toda iz že dokazanega $PE\cong PD$ in predpostavke +$PB\cong BD$ sledi $PD\cong PB\cong BD$, zatorej je $BDP$ enakostranični +trikotnik. Iz tega sledi $\beta=2\angle BDP=2\cdot 60^0=120^0$, oz. $\alpha+\beta=60^0+120^0=180^0$, kar ni možno (izrek +\ref{VsotKotTrik}). Zato relacija $QB>QC$ ni mogoča. + +Na podoben način pripelje do protislovja tudi relacija $QB>QC$. To +pomeni, da je možno le $QB\cong QC$. V tem primeru je $C=E$ in +velja $\gamma=\angle ACB=\angle AEB\cong AEP=\frac{1}{2}\beta$. Iz +$\alpha+\beta+\gamma=180^0$, sledi +$60^0+\beta+\frac{1}{2}\beta=180^0$ oz. $\beta=80^0$. + +Dokazali smo, da iz pogojev iz naloge sledi $\beta=80^0$. Torej je +edina možna rešitev $\beta=80^0$. Potrebno je še dokazati, da +$\beta=80^0$ je rešitev, oz. da iz $\alpha=60^0$, $\beta=80^0$ +sledi $|AB|+|BP|=|AQ|+|QB|$. + Najprej iz $\angle QCB=\gamma=\frac{1}{2}\beta=40^0=\angle QBC$ sledi + (izrek \ref{enakokraki}) $QC\cong QB$ oz. + $|AQ|+|QB|=|AQ|+|QC|=|AC|$. +Če točko $D$ definiramo na isti način kot v prvem delu, spet +dobimo $\angle ADP\cong \angle BPD=\frac{1}{2}\angle ABC=40^0=\angle ACB=\angle ACP$. +To pomeni, da sta trikotnika +$ADP$ in $ACP$ skladna (izrek \textit{ASA} \ref{KSK}) oz. +$AD\cong AC$. Zato je na koncu: + $$|AB|+|BP|=|AB|+|BD|=|AD|=|AC|=|AQ|+|QB|,$$ kar je bilo treba dokazati. \kdokaz + + + +%________________________________________________________________________________ + \poglavje{Constructions in Geometry} \label{odd3NacrtNaloge} + +Ob naslednjem zgledu bomo opisali t. i. načrtovalne naloge. + + + \bzgled \label{načrt1odd3} + Two congruent line segments $AB$ and $A'B'$ in a plane are given. + Construct a point $C$ such that $\triangle ABC \cong \triangle A'B'C$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.10.pic} +\caption{} \label{sl.skl.3.1.10.pic} +\end{figure} + + \textbf{\textit{Solution.}} Predpostavimo, da je $C$ točka v ravnini daljic, za + katero je $\triangle ABC \cong \triangle A'B'C$. Potem je $AC\cong A'C$ + in $BC\cong B'C$ oz. točka $C$ leži na simetralah daljic $AA'$ + in $BB'$ (izrek \ref{simetrala}). To dejstvo nam omogoča konstrukcijo + (Figure \ref{sl.skl.3.1.10.pic}). + + Načrtajmo simetrali daljic $AA'$ + in $BB'$. Točko $C$ dobimo v njunem presečišču. + + Dokažimo, da je $C$ iskana točka oz. da izpolnjuje pogoje iz + naloge. Po predpostavki je že $AB\cong A'B'$. Ker smo točko $C$ dobili kot + presečišče simetral daljic $AA'$ + in $BB'$, je $AC\cong A'C$ in $BC\cong B'C$. Iz izreka \ref{SSS} (SSS) sledi, + da sta trikotnika $ABC$ in $A'B'C$ + skladna. + + Naloga ima rešitev (eno) natanko tedaj, ko se simetrali daljic + $AA'$ + in $BB'$ sekata, oz. ko premici $AA'$ + in $BB'$ nista vzporedni. + \kdokaz + + Prejšnji zgled je torej t. i. \index{načrtovalna naloga} + \pojem{načrtovalna naloga}, pri kateri + je za dane podatke potrebno načrtati oz. konstruirati nek novi element ali lik, + ki v zvezi z danimi podatki + izpolnjuje določene pogoje. \pojem{Načrtovanje} oz. \index{konstrukcije} + \pojem{konstrukcija} + pomeni + uporabo ravnila in šestila oz. uporabo + \index{konstrukcije!elementarne} + \pojem{elementarne konstrukcije}:\label{elementarneKonstrukcije} +\begin{itemize} + \item za dani točki $A$ in $B$ narišemo: + \begin{itemize} + \item premico $AB$, + \item daljico $AB$, + \item poltrak $AB$; +\end{itemize} + \item narišemo krožnico $k$: +\begin{itemize} + \item s središčem $S$, ki poteka skozi dano točko $A$, + \item s središčem $S$ in polmerom, ki je skladen z dano + daljico; +\end{itemize} +\item narišemo krožni lok z danima središčem in polmerom, +\item narišemo presečišče (oz. presečišči): +\begin{itemize} + \item dveh premic, + \item premice in krožnice, + \item dveh krožnic. +\end{itemize} +\end{itemize} + + Rešitev načrtovalne naloge (nariši lik $\Phi$, ki izpolnjuje pogoje + $\mathcal{A}$) je formalno sestavljena iz štirih korakov: +\begin{itemize} + \item \textit{analysis} - pri kateri predpostavimo, da je + lik $\Phi$ že načrtan in izpolnjuje pogoje $\mathcal{A}$, nato + pa iščemo nove pogoje $\mathcal{B}$, ki jih lik izpolnjuje. + Ti sledijo iz pogojev $\mathcal{A}$ in so bolj ugodni za + konstrukcijo lika $\Phi$. Dokažemo + $\mathcal{A}\Rightarrow \mathcal{B}$. + \item \textit{construction} - načrtamo lik $\Phi'$, ki izpolnjuje pogoje + $\mathcal{B}$. Natanko opišemo potek načrtovanja. + \item \textit{proof} - dokažemo, da je $\Phi' = \Phi$ oz. + $\mathcal{B}\Rightarrow \mathcal{A}$. + \item \textit{discussion} - raziščemo + število rešitev naloge, odvisno od pogojev $\mathcal{A}$. +\end{itemize} + + + \bzgled \label{konstrTrik1} + Line segments $a$, $l$ and an angle $\alpha$ are given. Construct a triangle + $ABC$, such that the side $BC$ is congruent to the line segment $a$, the sum of the sides $AB + AC$ + equal to the line segment $l$ and the interior angle $BAC$ congruent to the angle $\alpha$ ($a$, $b+c$, $\alpha$). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.10a.pic} +\caption{} \label{sl.skl.3.1.10a.pic} +\end{figure} + + + \textbf{\textit{Analysis.}} Naj bo $ABC$ trikotnik, pri katerem je $BC \cong a$, vsota $AB+AC$ enaka dani daljici $l$ in +$\angle BAC\cong \alpha$ (Figure \ref{sl.skl.3.1.10a.pic}). Naj bo $D$ takšna točka na poltraku $BA$, da je $AD\cong AC$ in točki $B$ in $D$ na različnih straneh +točke $A$. Torej velja $BD=BA+AD=AB+AC=l$ oz. $BD\cong l$. Trikotnik $ACD$ je enakokrak, zato sta kota $ADC$ in $ACD$ +po izreku \ref{enakokraki} skladna. Ker sta hkrati notranja +kota trikotnika $CAD$, sta oba enaka polovici zunanjega kota $BAC$ tega trikotnika (izrek \ref{zunanjiNotrNotr}), oz. + $\angle BDC=\angle ADC\cong \angle ACD=\frac{1}{2}\angle BAC=\frac{1}{2}\alpha$. + To dejstvo nam omogoča +konstrukcijo trikotnika $BCD$. + +\textbf{\textit{Construction.}} Načrtajmo najprej trikotnik $BCD$, kjer je +$\angle BDC=\frac{1}{2}\alpha$, + $BC\cong a$ in $BD\cong l$, +nato pa točko $A$ kot presečišče simetrale daljice $CD$ z daljico $BD$. Dokažimo da je $ABC$ +iskani trikotnik. + +\textbf{\textit{Proof.}} Najprej je $BC\cong a$, že po konstrukciji. Po konstrukciji točka $A$ leži na simetrali daljice +$CD$, zato je $AD\cong AC$ (izrek \ref{simetrala}). Torej $CAD$ je enakokraki trikotnik z osnovnico $CD$, zato je (izrek \ref{enakokraki}) tudi $\angle ADC\cong \angle ACD$. Zaradi tega je (izrek \ref{zunanjiNotrNotr}) $\angle BAC = 2 \cdot \angle BDC= 2\cdot\frac{1}{2}\alpha=\alpha$. Iz $AD\cong AC$ pa sledi + $AB + AC = AB + AD = BD \cong l$ +. + +\textbf{\textit{Discussion.}} Naloga ima rešitev (in sicer eno ali dve) natanko tedaj, +ko poltrak $DC$ seka krožnico $k(B,a)$ +in simetrala daljice $CD$ seka daljico $BD$. + \kdokaz + +V bodoče ne bomo pri vsaki načrtovalni nalogi izpeljevali vseh korakov. +V večini primerov bomo naredili le prvi korak in s tem nakazali +potek reševanja. + + + + +%________________________________________________________________________________ + \poglavje{Triangle inequality} \label{odd3NeenTrik} + +Najprej bomo dokazali dva pomembna izreka, ki sta posledici +izrekov o skladnosti trikotnikov. + + \bizrek \label{vecstrveckot} + One side of a triangle is longer than another side of a triangle if and only if + the measure of the angle opposite the longer side is greater than the angle opposite the shorter side. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.1.pic} +\caption{} \label{sl.skl.3.2.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $ABC$ trikotnik, v katerem je $AC > AB$ (Figure \ref{sl.skl.3.2.1.pic}). + Dokažimo, da je tedaj tudi $\angle ABC > \angle ACB$. Ker je $AC > AB$, obstaja +takšna točka $B'$ med točkama $A$ in $C$, za katero velja $AB \cong AB'$. +Tedaj je trikotnik $BAB’$ enakokrak in velja +$\angle ABB' \cong \angle AB'B$ (izrek \ref{enakokraki}). +Poltrak $BB'$ je znotraj kota $ABC$, zato je $\angle ABC > \angle ABB '$. +Potem je $\angle AB'B$ zunanji kot trikotnika $BCB'$. Po + izreku \ref{zunanjiNotrNotrVecji} je ta kot večji od njegovega nesosednjega +notranjega kota $B'CB$. Če uporabimo doslej dokazano, dobimo: + $$\angle ABC > +\angle ABB ' \cong \angle AB'B > \angle B'CB \cong \angle ACB.$$ + Torej velja $\angle ABC +> \angle ACB$. Na podoben način dokažemo, da velja tudi obratno. +\kdokaz + + Če z $a$, $b$ in $c$ označimo dolžine stranic $BC$, $AC$ in + $AB$ trikotnika $ABC$ ter z $\alpha$, $\beta$ in $\gamma$ + mere nasprotnih kotov ob ogliščih $A$, $B$ in $C$, lahko prejšnji izrek + zapišemo v obliki: + $$a > b \Leftrightarrow \alpha > \beta,$$ + izrek o enakokrakem trikotniku \ref{enakokraki} pa v obliki: + $$a = b \Leftrightarrow \alpha = \beta.$$ + + + To pomeni, da sta izraza $a-b$ in $\alpha-\beta$ oba pozitivna, + oba negativna ali oba enaka nič. Tako smo dokazali naslednjo + lastnost: + + \bzgled \label{vecstrveckotAlgeb} + For each triangle $ABC$ is: + $$(a-b)(\alpha-\beta)\geq 0, \hspace*{4mm} + (b-c)(\beta-\gamma)\geq 0, \hspace*{4mm} + (c-a)(\gamma-\alpha)\geq 0$$ + \ezgled + +Iz izreka \ref{vecstrveckot} direktno sledi naslednja +trditev: + + + \bizrek \label{vecstrveckotHipot} + The hypotenuse of a right-angled triangle is longer than its + two legs. + The longest side of an obtuse triangle is the one opposite to the obtuse angle. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.2.pic} +\caption{} \label{sl.skl.3.2.2.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.2.2.pic}) + + Vsota notranjih kotov trikotnika je enaka $180^0$. Zato je v +pravokotnem trikotniku največji kot ravno pravi kot. Hipotenuza +pravokotnega trikotnika je po prejšnjem izreku najdaljša stranica +tega trikotnika. Podobno dokažemo tudi v primeru topokotnega +trikotnika. + \kdokaz + + +Daljica $AA'$ je \index{višina!trikotnika}\pojem{višina} +trikotnika $ABC$, če je $AA'\perp BC$ in $A'\in BC$. Zadnja od +dveh relacij pomeni, da točka $A'$ leži na premici $BC$, ne pa +nujno na daljici $BC$. Relacija $\mathcal{B}(B,A',C)$ velja +natanko tedaj, ko sta notranja kota pri ogliščih $B$ in $C$ oba +ostra (Figure \ref{sl.skl.3.2.3.pic}). To je posledica izreka o +vsoti notranjih kotov poljubnega trikotnika (izrek +\ref{VsotKotTrik}). V primeru, da je $\angle ABC\geq 90^0$ in +$\mathcal{B}(B,A',C)$, bi bila vsota notranjih kotov v trikotniku $ABA'$ +večja od $180^0$. + Torej + višina trikotnika ni vedno v notranjosti trikotnika. + Pri pravokotnem trikotniku sta višini iz +dveh oglišč ob ostrih kotih enaki ustreznima katetama. Višine, +ki potekajo iz oglišč $A$, $B$ in $C$ ponavadi označimo z $v_a$, +$v_b$ in $v_c$. Iz prejšnjega izreka \ref{vecstrveckotHipot} +sledi, da je dolžina višina poljubnega trikotnika manjša ali enaka dolžini +nepripadajoče stranice tega trikotnika, npr.: $v_a\leq b$, $v_a\leq c$, ... + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.3.pic} +\caption{} \label{sl.skl.3.2.3.pic} +\end{figure} + + +Sedaj bomo rešili načrtovalno nalogo, v kateri kot podatek nastopa višina trikotnika. + + + \bzgled + Construct a triangle $ABC$ such that the sides $AB$, + $AC$ and the altitude from the vertex $B$ are congruent to the three given line segments $c$, $b$ and $v_b$. + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.4a.pic} +\caption{} \label{sl.skl.3.2.4a.pic} +\end{figure} + + + +\textbf{\textit{Analysis.}} Naj bo $ABC$ trikotnik, za katerega velja $AB\cong c$, + $AC\cong b$ in $AD\cong v_b$ (kjer je $BD$ višina tega trikotnika iz oglišča $B$). + V pravokotnem trikotniku $ABD$ +sta torej znani hipotenuza $AB\cong c$ in kateta +$AD\cong v_b$, kar pomeni, da ga lahko načrtamo. Tretje oglišče $C$ trikotnika $ABC$ leži na premici $AD$ (Figure \ref{sl.skl.3.2.4a.pic}). + +\textbf{\textit{Construction.}} Načrtajmo najprej pravokotni trikotnik +$ABD$ (s pogoji: $AB\cong c$, $\angle ADB=90^0$ in $BD\cong v_b$). Na premici $AD$ nato določimo takšno točko $C$, +da velja $AC\cong b$. Dokažimo, da je $ABC$ iskani trikotnik. + + +\textbf{\textit{Proof.}} Najprej je $AB\cong c$ in + $AC\cong b$ že po konstrukciji. Ker je še $\angle ADB=90^0$, je $BD$ višina trikotnika $ABC$ iz oglišča $B$ in je po konstrukciji skladna z daljico $v_b$. + + +\textbf{\textit{Discussion.}} Naloga ima rešitev natanko tedaj, ko je možna +konstrukcija trikotnika $ABD$ oz. $hb\leq c$. Pri konstrukciji točke $C$ +obstajata dve možnosti - na različnih straneh točke $A$, kar pomeni, da imamo dve rešitvi za trikotnik $ABC$. V primeru $hb\cong c$ sta rešitvi pravokotna in skladna trikotnika. + \kdokaz + + + + \bzgled + If $v_a$, $v_b$ and $v_c$ are altitudes corresponding + to the sides $a$, $b$ and $c$ of a triangle, then: + $$\frac{v_a}{b+c}+\frac{v_b}{a+c}+\frac{v_c}{a+b}<\frac{3}{2}.$$ + \ezgled + +\textbf{\textit{Proof.}} +Če seštejemo neenakosti $v_a\leq b$, +$v_a\leq c$ + dobimo $2v_a\leq b+c$ oz. $\frac{v_a}{b+c}\leq\frac{1}{2}$. + Analogno dobimo $\frac{v_b}{a+c}\leq\frac{1}{2}$ in + $\frac{v_c}{a+b}\leq\frac{1}{2}$. Ker vse enakosti ne morejo veljati hkrati, + s seštevanjem dobimo iskano neenakost. + \kdokaz + +Naslednjo lastnost trikotnika bomo imenovali \index{trikotniška +neenakost} \pojem{trikotniška neenakost}. + + + \bizrek \label{neenaktrik} + The sum of any two sides of a triangle is greater than the third side. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.4.pic} +\caption{} \label{sl.skl.3.2.4.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Naj bo $ABC$ poljuben trikotnik. +Dokažimo, da velja \\ $AB + AC > BC$. Z $D$ označimo takšno točko, +da je $\mathcal{B}(B,A,D)$ in $AD \cong AC$ (Figure +\ref{sl.skl.3.2.4.pic}). Po izreku \ref{enakokraki} +($\triangle CAD$ je enakokraki trikotnik z osnovnico $CD$) je tudi $\angle BDC=\angle ADC \cong \angle ACD$. Poltrak $CA$ je +znotraj kota $DCB$, zato je + $\angle ACD < \angle DCB$. Tedaj je tudi $\angle BDC < \angle DCB$. + Iz izreka + \ref{vecstrveckot} (glede na trikotnik $BCD$) sledi: +$$BC < BD = AB + AD = AB + AC,$$ kar je bilo treba dokazati. \kdokaz + +Iz trikotniške neenakosti dobimo kriterij za obstoj takšnega +trikotnika, da so njegove stranice skladne s tremi danimi daljicami. + + + \bzgled + Let $a$, $b$ and $c$ be three line segments. A triangle with sides $a$, + $b$ and $c$ exist if and only if: + $$b + c > a,\hspace*{2mm} + a + c > b \hspace*{1mm}\textrm{ in }\hspace*{1mm} + a + b > c.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.5.pic} +\caption{} \label{sl.skl.3.2.5.pic} +\end{figure} + +\textbf{\textit{Proof.}} Če takšen trikotnik obstaja, potem so tri +relacije direktna posledice izreka o trikotniški neenakosti +\ref{neenaktrik}. Predpostavimo torej, da veljajo vse tri relacije. +Brez škode za splošnost naj bo npr. $a$ najdaljša stranica tega +trikotnika (dovolj je, da ni krajša od neke druge stranice) ter $B$ +in $C$ poljubni točki, za kateri je $BC \cong a$ (Figure +\ref{sl.skl.3.2.5.pic}). Ker je po predpostavki $b + c > a$, +to pomeni, da se krožnici $k(B,c)$ in $k(C,b)$ sekata v neki točki +$A$ (posledica Dedekindovega aksioma - izrek +\ref{DedPoslKrozKroz}, ker vsaka od njiju vsebuje notranje točke +druge), ki ni na daljici $BC$. Trikotnik $ABC$ je potem iskani +trikotnik. \kdokaz + + Če vemo, katera od treh stranic je najdaljša, + je dovolj preveriti le eno neenakost, saj +sta drugi dve avtomatično izpolnjeni. Dokaz +prejšnjega izreka lahko uporabimo tudi za naslednji, ekvivalenten +kriterij. + + \bzgled \label{neenaktrik1} + Let $a$, $b$ and $c$ be three line segments, such that $a \geq b,c$. A triangle with sides $a$, + $b$ and $c$ exist if and only if $b + c > a$. + \ezgled + +Tako npr. lahko ugotovimo, da obstaja trikotnik s stranicami, ki +imajo dolžine 7, 5 in 3 (ker je $5+3>7$), trikotnik s stranicami, ki imajo dolžine 9, 6 in 2, +pa ne obstaja (ker ni $6+2>9$). + + Oglejmo si še nekatere posledice prejšnjih izrekov. + + + \bzgled + If $X$ is an arbitrary point of the side $BC$ of a triangle $ABC$, + then: + $$AX < AB + AC.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.6.pic} +\caption{} \label{sl.skl.3.2.6.pic} +\end{figure} + +\textbf{\textit{Proof.}} Če uporabimo trikotniško neenakost za +trikotnike $ABX$ in $AXC$ (Figure \ref{sl.skl.3.2.6.pic}), dobimo: + $$AX < AB + BX \hspace*{1mm} \textrm{ in }\hspace*{1mm} AX < AC + CX.$$ +Po seštevanju teh dveh neenakosti in uporabi trikotniške +neenakosti za trikotnik $ABC$ dobimo: + $$2AX < AB + AC + +BC < 2(AB + AC),$$ kar je bilo treba dokazati. \kdokaz + +%% !!! Dosegel magično stran - 100!!! Wow Bravo!!! + +Naslednja neenakost je posplošitev prejšnje. V tem smislu je +prejšnja trditev njena posledica in je ni bilo potrebno posebej +dokazovati. + + + \bzgled + Let $X$ be an arbitrary point of the side $BC$, different from the vertices $B$ and $C$ of a triangle $ABC$. + Then the line segment $AX$ is shorter than at least one of the two line segments $AB$ and $AC$ i.e.: + $$AX < \max\{AB, AC\}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.7.pic} +\caption{} \label{sl.skl.3.2.7.pic} +\end{figure} + +\textbf{\textit{Proof.}} Ker za točko $X$ velja $\mathcal{B}(B, X, C)$, +eden od sokotov $AXB$ in $AXC$ ni oster. Brez škode za +splošnost naj bo to kot $AXC$ (Figure \ref{sl.skl.3.2.7.pic}). Potem +je le-ta največji kot v trikotniku $AXB$, kar pomeni, da je \\ +$AX < AC$ (izrek \ref{vecstrveckot}). Podobno, če kot $AXB$ ni oster, +velja $AX < AB$. + \kdokaz + +Posebej bomo obravnavali primer daljice $AX$, če je točka $X$ iz +prejšnjih dveh izrekov središče stranice $BC$ (Figure +\ref{sl.skl.3.2.8.pic}). Takšno daljico, ki je določena z ogliščem +in središčem nasprotne stranice trikotnika, imenujemo +\index{težiščnica trikotnika} \pojem{težiščnica} trikotnika. +Težiščnice, ki ustrezajo ogliščem $A$, $B$ in $C$ trikotnika $ABC$, +običajno označujemo s $t_a$, $t_b$ in $t_c$. Zadnji +dve trditvi lahko uporabimo tudi za težiščnice. Toda za težiščnice +velja še dodatna lastnost, ki jo bomo sedaj dokazali. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.8.pic} +\caption{} \label{sl.skl.3.2.8.pic} +\end{figure} + + + + \bzgled \label{neenTezisZgl} If $a$, $b$, $c$ are the sides and $t_a$ + the corresponding median of a triangle $ABC$, then: + $$\frac{b+c-a}{2} AB$ in +$AA_1 + A_1C > AC$ oz.: +$$t_a+\frac{a}{2}>c \hspace*{2mm} \textrm{ in } \hspace*{2mm} + t_a+\frac{a}{2}>b.$$ +Če seštejemo te dve neenakosti, dobimo $\frac{b+c-a}{2} AD = 2AA_1 = 2t_a,$$ kar je bilo treba dokazati. \kdokaz + + Pokažimo še nekaj primerov uporabe trikotniške neenakosti. + + + \bzgled + Let $M$ be an arbitrary point of the bisector of the exterior angle at + the vertex $C$ of a triangle $ABC$. Then + $$MA + MB \geq CA + CB.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.10.pic} +\caption{} \label{sl.skl.3.2.10.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Naj bo $D$ takšna točka poltraka $BC$, da je $AC \cong CD$ + in $\mathcal{B}(A,C,D)$ (Figure +\ref{sl.skl.3.2.10.pic}). Trikotnika $ACM$ in $DCM$ sta skladna, +po izreku \textit{SAS} \ref{SKS} ($AC \cong DC$, $CM \cong CM$, $\angle ACM \cong +\angle DCM$). Zato je tudi $MA \cong MD$. Če sedaj uporabimo +trikotniško neenakost, dobimo: $$MA + MB = MD + MB \geq BD = DC + +CB = CA + CB.$$ + Seveda enakost velja v primeru, +kadar so točke $B$, $M$ in $D$ kolinearne oziroma $M = C$. + \kdokaz + + + + \bzgled + If the bisector of the interior angle at + the vertex $A$ of a triangle $ABC$ intersects the side $BC$ in the point $E$, then + $$AB > BE \hspace*{2mm} \textrm{ in }\hspace*{2mm} AC > CE .$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.11.pic} +\caption{} \label{sl.skl.3.2.11.pic} +\end{figure} + + \textbf{\textit{Proof.}} Ker je $\mathcal{B}(B,E,C)$, je kot + $AEB$ zunanji kot trikotnika $AEC$ (Figure +\ref{sl.skl.3.2.11.pic}). Zato je $\angle BEA > \angle EAC \cong +\angle BAE$ (izrek \ref{zunanjiNotrNotrVecji}). Nasproti večjega +kota v trikotniku $BAE$ je večja stranica, oziroma velja $AB > BE$ +(izrek \ref{vecstrveckot}). Podobno dokazujemo, da velja tudi +druga od dveh relacij. + \kdokaz + + + \bzgled \label{zgled3.2.9} + If $M$ is the interior point of a triangle $ABC$, then \\ + $BA + AC > BM + MC.$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.12.pic} +\caption{} \label{sl.skl.3.2.12.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $N$ presečišče premic $BM$ in +$CA$ (Figure \ref{sl.skl.3.2.12.pic}). Ker je $M$ notranja točka +trikotnika $ABC$, velja $\mathcal{B}(B,M,N)$ in +$\mathcal{B}(A,N,C)$. Če sedaj dvakrat uporabimo trikotniško +neenakost (izrek \ref{neenaktrik}), dobimo: + \begin{eqnarray*} +\hspace*{-4mm}BM + MC &<& BM + (MN + NC) = (BM + MN) + NC = BN + NC\\ + \hspace*{-4mm}&<& (BA + +AN) + NC = BA + (AN + NC) = BA + AC. + \end{eqnarray*} + +Definirajmo dva nova pojma. Vsoto vseh stranic nekega večkotnika +imenujemo njegov \index{obseg!večkotnika} \pojem{obseg}. Polovica +te vsote pa je \pojem{polobseg} tega večkotnika. + + + + \bzgled + If $M$ is the interior point and $s$ the semiperimeter of a triangle $ABC$, then + $$s < AM + BM + CM < 2s.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.13.pic} +\caption{} \label{sl.skl.3.2.13.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Prvo neenakost dobimo, če +trikrat uporabimo trikotniško neenakost za trikotnike $MAB$, +$MBC$ in $MCA$ in jih potem seštejemo. Drugo neenakost pa dobimo, +če trikrat uporabimo prejšnjo trditev (zgled \ref{zgled3.2.9}) +in seštejemo ustrezne neenakosti (Figure \ref{sl.skl.3.2.13.pic}). + \kdokaz + + + \bzgled + In each convex pentagon there exist three diagonals, + which are congruent to the sides of a triangle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.14.pic} +\caption{} \label{sl.skl.3.2.14.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Naj bo $AD$ najdaljša diagonala +petkotnika $ABCDE$ (naj ne bo krajša od nobene druge +diagonale). Dokažimo, da so $AD$, $AC$ in $BD$ iskane diagonale, +torej tiste, za katere obstaja trikotnik, čigar stranice so + s temi diagonalami skladne (Figure \ref{sl.skl.3.2.14.pic}). Ker je + $AD\geq AC$ in $AD\geq BD$, je dovolj dokazati (zgled \ref{neenaktrik1}), +da velja $AC + BD > AD$. Petkotnik $ABCDE$ je konveksen, +zato se njegovi diagonali $AC$ in +$BD$ sekata v neki točki $S$. Tedaj je: $$AC + BD > AS + SD > AD,$$ kar je bilo treba dokazati. \kdokaz + +Zelo pomembna je naslednja posledica izrekov o skladnosti +trikotnikov. Tudi v tem dokazu bomo potrebovali trikotniško +neenakost. + + + \bizrek \label{SkladTrikLema} + Let $ABC$ and $A'B'C'$ triangles such that $AB \cong A'B'$ + and $AC \cong A'C'$. Then $BC > B'C'$ if and only if + $\angle BAC > \angle B' A'C'$ i.e. + $$BC > B'C' \Leftrightarrow \angle BAC > \angle B' + A'C'.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.15.pic} +\caption{} \label{sl.skl.3.2.15.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.2.15.pic}) + + ($\Leftarrow$) Naj bo $\angle BAC > \angle B' A'C'$. Tedaj obstaja +znotraj kota $BAC$ takšen poltrak $l$, da je $\angle BA,l \cong \angle B'A'C'$. +S $C''$ označimo točko poltraka $l$, za katero je +$AC'' \cong A'C'$. Tedaj sta (po izreku $SKS$) trikotnika $ABC''$ in +$A'B'C'$ skladna in je $BC'' \cong B'C'$. Dovolj je dokazati, da +velja $BC > BC''$. Če $C''$ leži na stranici $BC$, je to trivialno +izpolnjeno. Predpostavimo, da točka $C''$ ne leži na stranici $BC$. +Naj bo točka $E$ presečišče simetrale kota $CAC''$ in stranice $BC$. +Po izreku \textit{SAS} sta skladna tudi trikotnika $ACE$ in $AC''E$, zato +je $CE \cong C''E$. Sedaj je: +$$BC = BE + EC = BE + EC'' \hspace{0.1mm} > BC'' = B'C'.$$ + ($\Rightarrow$) Naj bo $BC > B'C'$. Relacija $\angle BAC \cong + \angle B' A'C'$ ne velja, + ker bi bila tedaj (po izreku \textit{SAS}) trikotnika +$ABC$ in $A'B'C'$ skladna in potem tudi $BC \cong B'C'$. Če +bi pa veljalo $\angle BAC < \angle B' A'C'$, bi iz že dokazanega +sledilo $BC < B'C'$. Torej velja $\angle BAC > \angle B' A'C'$. + \kdokaz + + \bizrek \label{neenakIzlLin} + If $A_1A_2\ldots A_n$ ($n\in \mathbb{N}$, $n\geq 3$) is polygonal chain, then + $$|A_1A_2|+|A_2A_2|+\cdots +|A_{n-1}A_n|\geq |A_1A_n|.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.16.pic} +\caption{} \label{sl.skl.3.2.16.pic} +\end{figure} + +\textbf{\textit{Proof.}} Dokaz bomo izpeljali z indukcijo po $n$ +(Figure \ref{sl.skl.3.2.16.pic}). + + V primeru $n=3$ dobimo trikotniško neenakost - izrek + \ref{neenaktrik}. + + Predpostavimo, da neenakost velja za $n=k$ ($k\in \mathbb{N}$, $k> 3$) oz. + $|A_1A_2|+|A_2A_2|+\cdots +|A_{k-1}A_k|\geq |A_1A_k|.$ Dokažimo, + da potem neenakost velja tudi za $n=k+1$ oz. + $|A_1A_2|+|A_2A_2|+\cdots +|A_kA_{k+1}|\geq |A_1A_{k+1}|.$ Če + uporabimo najprej indukcijsko + predpostavko, nato pa trikotniško neenakost, dobimo: + \begin{eqnarray*} + && |A_1A_2|+|A_2A_2|+\cdots +|A_{k-1}A_k|+|A_kA_{k+1}|\geq\\ + && \geq|A_1A_k|+|A_kA_{k+1}|\geq |A_1A_{k+1}|, + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + +Dokažimo še eno neenakost, ki velja v poljubnem trikotniku. + + + \bzgled + If $a$, $b$, $c$ are the sides and $\alpha$, $\beta$, $\gamma$ + the opposite interior angles of a triangle, then + $$60^0\leq \frac{a\alpha+b\beta +c\gamma}{a+b+c} < 90^0.$$ + \ezgled + +\textbf{\textit{Proof.}} Dokazali bomo vsako od neenakosti +posebej. Pri tem bomo uporabili izrek o vsoti notranjih kotov +trikotnika (izrek \ref{VsotKotTrik}) Najprej bomo dokazali drugo +neenakost: + \begin{eqnarray*} + \frac{a\alpha+b\beta +c\gamma}{a+b+c} < 90^0 + &\Leftrightarrow& a\alpha+b\beta +c\gamma - 90^0(a+b+c)<0\\ + &\Leftrightarrow& a(\alpha-90^0)+b(\beta-90^0) +c(\gamma-90^0)<0\\ + &\Leftrightarrow& a(180^0-2\alpha)+b(180^0-2\beta) +c(180^0-2\gamma)>0\\ + &\Leftrightarrow& a(\beta+\gamma-\alpha)+b(\alpha+\gamma-\beta) + + c(\alpha+\beta-\gamma)>0\\ + &\Leftrightarrow& \alpha(b+c-a)+\beta(a+c-b) + + \gamma(a+b-c)>0 + \end{eqnarray*} + Zadnja neenakost je izpolnjena, zato ker po trikotniški neenakosti (izrek + \ref{neenaktrik}) velja $b+c-a>0$, $a+c-b>0$ in $a+b-c>0$. + Dokažimo še prvo neenakost: + \begin{eqnarray*} + && \frac{a\alpha+b\beta +c\gamma}{a+b+c} \geq 60^0\Leftrightarrow\\ + &\Leftrightarrow& a\alpha+b\beta +c\gamma - 60^0(a+b+c)\geq 0\\ + &\Leftrightarrow& a(\alpha-60^0)+b(\beta-60^0) +c(\gamma-60^0)\geq 0\\ + &\Leftrightarrow& a(3\alpha-180^0)+b(3\beta-180^0) +c(3\gamma-180^0)\geq 0\\ + &\Leftrightarrow& a(2\alpha-\beta-\gamma)+b(2\beta-\alpha-\gamma) + + c(2\gamma-\alpha-\beta)\geq 0\\ + &\Leftrightarrow& a(\alpha-\beta+\alpha-\gamma)+b(\beta-\alpha+\beta-\gamma) + + c(\gamma-\alpha+\gamma-\beta)\geq 0\\ + &\Leftrightarrow& a(\alpha-\beta)+a(\alpha-\gamma)+ + b(\beta-\alpha)+b(\beta-\gamma) + + c(\gamma-\alpha)+c(\gamma-\beta)\geq 0\\ + &\Leftrightarrow& (a-b)(\alpha-\beta)+(a-c)(\alpha-\gamma)+ + (b-c)(\beta-\gamma) \geq 0 + \end{eqnarray*} + Zadnja neenakost je posledica trditve \ref{vecstrveckotAlgeb}. + \kdokaz + + Naslednji izrek bo motivacija za definiranje razdalje točke od + premice. + + \bizrek Let $A'=pr_{\perp p}(A)$ be the foot of the perpendicular from a point $A$ on a line $p$. + If $X\in p$ and $X\neq A'$, then $AX>AA'$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.2.17.pic} +\caption{} \label{sl.skl.3.2.17.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Po definiciji je $AA'\perp p$ +(Figure \ref{sl.skl.3.2.17.pic}), kar pomeni, da je $AA'X$ + pravokotni trikotnik s hipotenuzo $AX$. Iz izreka + \ref{vecstrveckotHipot} sledi $AX>AA'$. + \kdokaz + + Če je $A'=pr_{\perp +p}(A)$, pravimo, da je dolžina daljice $AA'$ \index{razdalja!točke + od premice} \pojem{razdalja točke $A$ od premice $p$}. +Označimo jo z $d(A,p)$. Torej $d(A,p)=|AA'|$. + + + + +%________________________________________________________________________________ + \poglavje{Circle and Line} \label{odd3KrozPrem} + +V nadaljevanju se bomo ukvarjali s krožnico ter z medsebojno lego +krožnice in premice. Dokažimo najprej eno lastnost premera +krožnice, ki je enostavna posledica trikotniške neenakosti. + + + \bizrek \label{premerNajdTetiva} + The longest chord of a circle is its diameter. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.1.pic} +\caption{} \label{sl.skl.3.3.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bosta +$AB$ poljubna tetiva krožnice, ki ni premer, in $C$ točka na +premici $AS$, za katero velja $CS\cong SA$ in $\mathcal{B}(A,S,C)$ +(Figure \ref{sl.skl.3.3.1.pic}). Tedaj točka $C$ leži na +krožnici $k$ in je $AC$ njen premer. Dokazali smo že (posledica +izreka \ref{premerInS}), da so vsi premeri neke krožnice +medsebojno skladni. Zato je dovolj dokazati, da je $AC>AB$. To pa +sledi iz trikotniške neenakosti (trikotnik $ASB$). Velja: + $$AC=AS+SC=AS+SB>AB,$$ kar je bilo treba dokazati. \kdokaz + +Sledi še ena lastnost tetive krožnice, kot posledica izreka +\ref{vecstrveckotHipot}. + + + \bzgled \label{tetivaNotrTocke} + Every point that lies on a chord of a circle, except its endpoints, + is an interior point of the circle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.2.pic} +\caption{} \label{sl.skl.3.3.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $X$ notranja točka tetive $AB$ +s krajščema na krožnici $k(S,r)$ (Figure \ref{sl.skl.3.3.2.pic}). Kota $AXS$ in +$BXS$ sta sokota, kar pomeni, da nista oba ostra kota. Brez škode +za splošnost predpostavimo, da kot $BXS$ ni ostri kot. Tedaj je v +trikotniku $SXB$ stranica $SB$ najdaljša (izrek +\ref{vecstrveckotHipot}), kar pomeni, da je: + $$SX ST = r.$$ + Torej nobena od točk $T_1$ ($T_1 \neq T$), ki ležijo na +premici $PT$, ne leži na krožnici $k$. To pomeni, da je premica +$PT$ tangenta te krožnice. + \kdokaz + +Iz dokaza prejšnjega izreka ($\Leftarrow$) sledi, da so vse +točke, ki ležijo na tangenti krožnice (razen njenega +dotikališča), zunanje točke te krožnice. S pomočjo te +lastnosti bomo dokazali naslednjo trditev. + + + \bzgled \label{tangKrozEnaStr} + All points of a circle are on the one side of its tangent. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.7.pic} +\caption{} \label{sl.skl.3.3.7.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $T$ dotikališče krožnice $k(S, r)$ +in njene tangente $t$ (Figure \ref{sl.skl.3.3.7.pic}). Tangenta $t$ +deli ravnino, v kateri ležita $k$ in $t$, na dve polravnini. Tisto +polravnino, v kateri leži točka $S$, označimo z $\alpha_1$, drugo +polravnino pa z $\alpha_2$. Dokažimo, da vse točke krožnice $k$ +ležijo v polravnini $\alpha_1$. Naj bo $X$ poljubna točka polravnine +$\alpha_2$. Ker sta + točki $S$ in $X$ na različnih straneh premice $t$, sledi da jo odprta +daljica $SX$ seka v neki točki $Y$. Potem velja: + $$SX = SY + YX > SY \geq ST = r,$$ + kar pomeni, da točka $X$ ne leži na krožnici $k$ in je njena +zunanja točka. Torej nobena od točk polravnine $\alpha_2$ ne +leži na krožnici $k$, oz. so vse v polravnini +$\alpha_1$ z robom $t$. \kdokaz + +Direktna posledica izreka \ref{TangPogoj} je tudi ta, da v vsaki +točki krožnice lahko narišemo eno samo tangento. Če je $X$ +notranja točka krožnice $k(S, r)$, potem skozi to točko ne +poteka nobena tangenta, saj so vse premice skozi $X$ +sekante, kar je posledica Dedekindovega aksioma (izrek +\ref{DedPoslKrozPrem}). Kasneje (izrek \ref{tangentiKroznice}) +bomo ugotovili, da lahko skozi vsako zunanjo točko krožnice +narišemo natanko dve tangenti. Zaenkrat dokažimo naslednjo +trditev (bralec se bo spomnil, da gre za trditev, ki smo jo +obravnavali že na samem začetku v uvodnem poglavju - trditev +\ref{TalesUvod}). + + + \bizrek \label{TalesovIzrKroz} \index{izrek!Talesov za krožnico} + Thales' theorem for a circle\footnote{Starogrški + filozof in matematik \textit{Tales} + \index{Tales} iz Mileta (640--546 pr. n. š.) + ni prvi, ki je odkril to trditev. Kot empirično dejstvo so jo poznali + že stari Egipčani in Babilonci. Izrek imenujemo po Talesu, ki ga je prvi dokazal. + V dokazu je uporabljal lastnosti enakokrakih trikotnikov in dejstvo, da je + vsota notranjih kotov trikotnika enaka vsoti dveh pravih kotov. Torej je + dokaz enak temu, ki ga bomo izpeljali tukaj.}:\\ + Let $AB$ be a diameter of a circle $k$. Then for any point $X$ of this circle different from $A$ and $B$ + ($X\in k$ in $X\neq A$ in $X\neq B$) is $\angle AXB=90^0$ + \index{izrek!Talesov za krožnico} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.8.pic} +\caption{} \label{sl.skl.3.3.8.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $O$ središče krožnice $k$ (Figure +\ref{sl.skl.3.3.8.pic}). Ker $A,B,X\in k$, je + $OA\cong OB\cong OX$. Torej sta + trikotnika $AOX$ in $BOX$ enakokraka, zato je (izrek +\ref{enakokraki}): + $\angle AXO\cong\angle XAO=\alpha$ in $\angle BXO\cong\angle XBO=\beta$. +Tedaj je $\angle AXB=\alpha+\beta$. + Vsota notranjih kotov v +trikotniku $AXB$ je enaka $180^0$ (izrek \ref{VsotKotTrik}), torej +je $2\alpha+2\beta=180^0$. Iz tega sledi + $$\angle AXB=\alpha+\beta=90^0,$$ kar je bilo treba dokazati. \kdokaz + + Dokažimo tudi obratno trditev. + + + \bizrek \label{TalesovIzrKrozObrat} + If $A$, $B$ and $X$ are non-collinear points such that + $\angle AXB=90^0$, then the point $X$ lies on a circle with the diameter $AB$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.9.pic} +\caption{} \label{sl.skl.3.3.9.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Naj bo $O$ središče daljice $AB$ in $k$ +krožnica s središčem $O$ in polmerom $OA$ oz. premerom $AB$ +(Figure \ref{sl.skl.3.3.9.pic}). Iz $\angle AXB=90^0$ je po izreku +\ref{VsotKotTrik}: + \begin{eqnarray} + \angle XAB+ \angle XBA = 90^0 \label{relacija336} + \end{eqnarray} + +Dokažimo $X\in k$. Predpostavimo nasprotno, torej da točka $X$ ne +leži na krožnici $k$. V tem primeru je $OX\neq OA$. Naj bo $X_1$ +točka na poltraku $OX$, za katero je $OX_1\cong OA$ (izrek +\ref{ABnaPoltrakCX}). To pomeni, da točka $X_1$ leži na krožnici +$k$ in po Talesovem izreku \ref{TalesovIzrKroz} velja $\angle +AX_1B=90^0$. + +Po naši predpostavki $OX\neq OA$ je jasno, da $X\neq X_1$. +Obravnavali bomo dve možnosti: + +\textit{1)} Naj bo $OX_1OX$ oz. $\mathcal{B}(O,X,X_1)$. Podobno +kot v prvem primeru dobimo: + $$\angle X_1AB+ \angle X_1BA>\angle XAB+ \angle XBA = 90^0.$$ + V tem primeru je vsota kotov v trikotniku $AX_1B$ večja od +$180^0$, kar po izreku \ref{VsotKotTrik} ni mogoče. + +Iz tega sledi, da je $OX=OX_1$ oz. $X\in k$. + \kdokaz + + + Uporabimo prejšnja izreka za načrtovanje tangent. + + + \bzgled \label{tangKrozKonstr} + Let $A$ be an exterior point of a circle $k(S,r)$. + Construct all tangents of this circle passing through the point $A$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.10.pic} +\caption{} \label{sl.skl.3.3.10.pic} +\end{figure} + + + \textbf{\textit{Solution.}} Naj bo $l$ krožnica s premerom $SA$ +(Figure \ref{sl.skl.3.3.10.pic}). Ker je $S$ notranja, $A$ pa zunanja +točka dane krožnice $k$, imata krožnici $k$ in $l$ natanko +dve skupni točki $T_1$ in $T_2$ (izrek \ref{DedPoslKrozKroz}). +Po Talesovem izreku \ref{TalesovIzrKroz} je +$\angle ST_1A\cong \angle ST_2A=90^0$. Ker sta $ST_1$ in $ST_2$ +polmera krožnice $k$, sta $AT_1$ in $AT_2$ tangenti krožnice $k$ +skozi točko $A$ (izrek \ref{TangPogoj}). + + Dokažimo, da sta $AT_1$ in $AT_2$ edini tangenti krožnice $k$ + iz točke $A$. Če je $AT$ tangenta iz točke $A$, ki se krožnice $k$ + dotika v točki $T$, je po izreku \ref{TangPogoj} $\angle ATS=90^0$. + To pomeni, da točka $T$ leži na krožnici $l$ + (izrek \ref{TalesovIzrKrozObrat}) oz. $T\in k\cap l$. Torej $T$ je ena + od točk $T_1$ in $T_2$, zato sta $AT_1$ in $AT_2$ edini tangenti krožnice + $k$ iz točke $A$. + \kdokaz + + Iz prejšnje konstrukcije sledi naslednji izrek. + + + + \bizrek \label{tangentiKroznice} + If $V$ is an exterior point of a circle $k(S,r)$, then there are exactly + two tangents of the circle $k$ through the point $V$. + \eizrek + + + Dokažimo še nekaj lastnosti tangent krožnice. + + + \bzgled \label{TangOdsek} + If $VA$ and $VB$ are tangents of a circle $k(S,r)$ in a points + $A$ and $B$ of the circle, then the centre $S$ lies on the bisector + of the angle $AVB$ and $VA \cong VB$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.11.pic} +\caption{} \label{sl.skl.3.3.11.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Iz izreka \ref{TangPogoj} sledi: $VA +\perp AS$ in $VB \perp BS$ +(Figure \ref{sl.skl.3.3.11.pic}). Torej sta $ASV$ in $BSV$ pravokotna +trikotnika s skupno hipotenuzo $SV$. Ker je še $SA \cong SB = r$, +sta ta dva trikotnika skladna (izrek \textit{SSA} \ref{SSK}). Torej sta tudi kota +$AVS$ in $BVS$ skladna, kar pomeni, da je premica $VS$ +simetrala kota $AVB$. Iz skladnosti teh dveh trikotnikov sledi +tudi $VA \cong VB$. + \kdokaz + +Velja tudi obratna trditev. + + + \bzgled \label{SimKotaKraka} + If a point $S$ lies on the bisector of a convex angle, + then it is the centre of a circle touching both sides of this angle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.12.pic} +\caption{} \label{sl.skl.3.3.12.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bosta $A$ in $B$ pravokotni projekciji +točke $S$ na krakih danega kota z vrhom $V$ (Figure +\ref{sl.skl.3.3.12.pic}). Trikotnika $ASV$ in $BSV$ sta skladna +(izrek \textit{ASA} \ref{KSK}), ker imata skupno stranico $VS$ in +dva para skladnih kotov - iz $\angle AVS\cong \angle BVS$ in $\angle +SAV\cong \angle SBV=90^0$ sledi $\angle ASV\cong \angle BSV$. Zaradi +tega velja $SA\cong SB$ in je $k(S,SA)$ iskana krožnica. Kraka +danega kota sta namreč po izreku \ref{TangPogoj} tangenti krožnice. + \kdokaz + +Sedaj bomo dokazali še en kriterij o medsebojni legi premice in +krožnice v ravnini. + + + + \bizrek \label{TangSekMimobKrit} + Let $P$ be the foot of the perpendicular from the centre of a circle $k(S,r)$ + on a line $p$ (lying in the plane of the circle). Then the line $p$ is: + + (i) secant, if and only if $SP < r$; + + (ii) tangent, if and only if $SP \cong r$; + + (iii) non-intersecting line, if and only if $SP> r$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.3.13.pic} +\caption{} \label{sl.skl.3.3.13.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.3.13.pic}) + +Trditev (ii) sledi direktno iz kriterija +za tangento (izrek \ref{TangPogoj}). + +(i) V dokazu direktne smeri ekvivalence uporabljamo dejstvo, da je +hipotenuza pravokotnega trikotnika daljša od obeh katet (izrek +\ref{vecstrveckotHipot}). Če sta $A$ in $B$ presečišči sekante $p$ +in krožnice $k$, je $SA$ hipotenuza pravokotnega trikotnika $ASP$ in +velja: + $r \cong SA > SP$. + + V dokazu obratne smeri +ekvivalence uporabimo posledico Dedekindovega aksioma (izrek +\ref{DedPoslKrozPrem}). Ker je v tem primeru $P$ notranja točka te +krožnice, je vsaka premice te ravnine, ki gre skozi +točko $P$, sekanta krožnice $k$. + +(iii) Sledi iz dokazanega (i) in (ii). Če je namreč $SP > r$, +potem ni niti $SP < r$ niti $SP \cong r$. Iz ekvivalenc (i) in (ii) +sledi, da premica $p$ ni niti sekanta niti tangenta. Zato je $p$ +mimobežnica krožnice $k$. Na isti način dokazujemo tudi obratno smer +ekvivalence. + \kdokaz + +%________________________________________________________________________________ + \poglavje{Quadrilaterals} \label{odd3Stirik} + + +V razdelku \ref{odd2AKSURJ} smo pojem štirikotnika vpeljali kot +večkotnik, ki ima štiri stranice in štiri oglišča. +Definirali smo pojme sosednji in nasprotni stranici, sosednji in +nasprotni koti ter diagonalo. Štirikotniku $ABCD$ dolžine +njegovih stranic $AB$, $BC$, $CD$ in $DA$ običajno označimo z $a$, +$b$, $c$ in $d$, dolžini njegovih diagonal $AC$ in $BD$ pa z $e$ in +$f$ (Figure \ref{sl.skl.3.4.1.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.1.pic} +\caption{} \label{sl.skl.3.4.1.pic} +\end{figure} + + V istem razdelku smo kot pojme vpeljali notranje + in zunanje kote štirikotnika. Omenili smo tudi, da notranje kote ob ogliščih $A$, $B$, $C$ in $D$ štirikotnika $ABCD$ + navadno označimo + z $\alpha$, $\beta$, $\gamma$ in $\delta$, njegove zunanje kote + pa z $\alpha'$, $\beta'$, $\gamma'$ in $\delta'$. +Dokazali smo (posledica splošnega izreka \ref{VsotKotVeck}), da je +vsota vseh štirih notranjih kotov poljubnega štirikotnika +enaka $360^0$ (Figure \ref{sl.skl.3.4.1.pic}). Enaka je tudi vsota zunanjih kotov (v konveksnem štirikotniku). Torej: + \begin{eqnarray*} + \alpha+\beta+\gamma+\delta=360^0,\\ + \alpha'+\beta'+\gamma'+\delta'=360^0 + \end{eqnarray*} + +Dodajmo še, da za notranja kota pravimo, da sta +\index{kota!sosednja} \pojem{sosednja} oz. \index{kota!nasprotna} \pojem{nasprotna}, če sta pripadajoči oglišči sosednji oz. nasprotni. + + + +Sedaj bomo nekatere vrste štirikotnikov podrobneje obravnavali. + + Štirikotnik +$ABCD$ je \index{trapez}\pojem{trapez}, če je $AB\parallel CD$ (Figure \ref{sl.skl.3.4.2.pic}). +Stranici $AB$ in $CD$ sta \index{osnovnica!trapeza} +\pojem{osnovnici}, $BC$ in $AD$ pa +\index{krak!trapeza}\pojem{kraka} tega trapeza. +Daljica $PQ$ ($P\in AB$, $Q\in CD$ in $PQ\perp AB$) je \index{višina!trapeza}\pojem{višina trapeza}. Pogosto jo označimo z $v$. + + +Trapez je + \index{trapez!enakokraki}\pojem{enakokraki}, + Če je $BC \cong AD$ in ni $BC \parallel AD$, + oz. \index{trapez!pravokotni}\pojem{pravokotni}, če je vsaj +eden notranji kot pravi. + + + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.2.pic} +\caption{} \label{sl.skl.3.4.2.pic} +\end{figure} + + +Dva notranja kota ob istem kraku trapeza sta suplementarna, ker +sta kota z vzporednima krakoma (izrek \ref{KotiTransverzala}). Suplementarnost teh kotov je tudi +zadosten pogoj, da je štirikotnik trapez. Iz tega sledi, da ima pravokotni trapez vsaj dva prava notranja kota. + + +Kot posebno vrsto trapezov dobimo še eno skupino štirikotnikov. +To so \pojem{paralelogrami}. Možno jih je definirati na različne načine. +Izbrali bomo enega, za vse ostale pa bomo +dokazali, da so ekvivalentni. + +Štirikotnik $ABCD$ je \index{paralelogram} \pojem{paralelogram}, če +velja $AB +\parallel CD$ in $AD \parallel BC$ (Figure \ref{sl.skl.3.4.3.pic}). +Daljici $PQ$ ($P\in AB$, $Q\in CD$ in $PQ\perp AB$) in $MN$ ($M\in BC$, $N\in AD$ in $MN\perp BC$) sta + \index{višina!paralelograma}\pojem{višini paralelograma}. Pogosto ju označimo z $v_a$ in $v_b$. + + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.3.pic} +\caption{} \label{sl.skl.3.4.3.pic} +\end{figure} + + Paralelogram je torej štirikotnik, ki ima dva para +vzporednih stranic. V definiciji paralelograma ni uporabljen pojem +skladnosti. Paralelograme (tudi trapeze) lahko obravnavamo +tudi v t. i. \index{geometrija!afina} \pojem{afini geometriji}. +To je geometrija, ki je zasnovana na vseh aksiomah evklidske +geometrije, če iz seznama izključimo aksiome tretje skupine - +aksiome skladnosti. + +Dokažimo sedaj že omenjene ekvivalente za definicijo +paralelograma. + + + \bizrek \label{paralelogram} + Let $ABCD$ be a convex quadrilateral. + Then the following statements are equivalent: + \begin{enumerate} + \item The quadrilateral $ABCD$ is a parallelogram. + \item Any two adjacent interior angles of the quadrilateral $ABCD$ are supplementary. + \item Any two opposite interior angles of the quadrilateral $ABCD$ are congruent. + \item $AB \parallel CD$ and $AB \cong CD$\footnote{Ta ekvivalent + v nekoliko drugačni obliki navaja \index{Evklid} + \textit{Evklid iz Aleksandrije} (3. stol. pr. n. š.) v + prvi knjigi svojih ‘‘Elementov’’.}. + \item $AB \cong CD$ and $AD \cong BC$. + \item The diagonals of the quadrilateral $ABCD$ bisect each other, i.e. + line segments $AC$ and $BD$ have a common midpoint. + \end{enumerate} + \eizrek + + \textbf{\textit{Proof.}} +Potrebno je dokazati ekvivalentnost vseh izjav $(1)-(6)$. Da bi se +izognili dokazovanju vseh ekvivalenc (po dve implikaciji - npr. z +izjavo (1), kar bi skupaj dalo 10 implikacij), bomo dokaz nekoliko +poenostavili, tako da dokažemo implikacije po naslednji shemi. + +\vspace*{5mm} +\hspace*{25mm} +$\begin{array}{ccccccc} + \textit{(1)} & \Leftarrow & \textit{(2)} & \Leftarrow & \textit{(3)} & & \\ + \Downarrow & & & & \Uparrow & & \\ + \textit{(4)} & & \Rightarrow & & \textit{(5)} & \Leftrightarrow & \textit{(6)} +\end{array}$ + +\vspace*{5mm} + + Kot vidimo, je to dovolj, ker implikacija $\textit{(1)} +\Rightarrow \textit{(2)}$ sledi direktno iz: $\textit{(1)}\Rightarrow \textit{(4)}\Rightarrow +\textit{(5)}\Rightarrow \textit{(3)} \Rightarrow \textit{(2)}$. + +Označimo z $\alpha$, $\beta$, $\gamma$ in $\delta$ notranje kote pri +ogliščih $A$, $B$, $C$ in $D$ štirikotnika $ABCD$. Štirikotnik +$ABCD$ je konveksen, kar pomeni, da se njegovi diagonali sekata v +neki točki $S$. + +$\textit{(2)}\Rightarrow \textit{(1)}$. Naj bosta kota $\alpha$ in +$\beta$ suplementarna (Figure \ref{sl.skl.3.4.4.pic}). Potem sta to +kota ob transverzali $AB$ premic $AD$ in $BC$, zato je $AD\parallel +BC$ (izrek \ref{KotiTransverzala}). Podobno iz suplementarnosti +kotov $\beta$ in $\gamma$ sledi $AB\parallel CD$. Torej je štirikotnik +$ABCD$ paralelogram. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.4.pic} +\caption{} \label{sl.skl.3.4.4.pic} +\end{figure} + +$\textit{(3)}\Rightarrow\textit{(2)}$. Naj bo $\alpha =\gamma$ in $\beta =\delta$ (Figure \ref{sl.skl.3.4.4.pic}). +Ker je $\alpha + \beta +\gamma +\delta = 360°$ (vsota vseh +notranjih kotov v štirikotniku je $360°$ - izrek \ref{VsotKotVeck}), sledi +$\alpha + \beta =180°$ in $\beta +\gamma = 180°$. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.4a.pic} +\caption{} \label{sl.skl.3.4.4a.pic} +\end{figure} + +$\textit{(1)}\Rightarrow\textit{(4)}$. Naj bo štirikotnik $ABCD$ +paralelogram, oz. naj velja $AB \parallel CD$ in $AD \parallel BC$ +(Figure \ref{sl.skl.3.4.4a.pic}). Dokažimo, da je potem tudi $AB \cong CD$. +Premica $AC$ je transverzala vzporednic $AB$ in $CD$, kar +pomeni, da sta kota $CAB$ in $ACD$ izmenična kota na tej +transverzali in sta zato skladna. Podobno iz vzporednosti premic +$AD$ in $BC$ sledi, da sta tudi kota $ACB$ in $CAD$ skladna. Ker je +$AC \cong AC$, sta trikotnika $ACB$ in $CAD$ skladna (izrek +\ref{KSK} - \textit{ASA}). Zato je $AB \cong CD$. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.4b.pic} +\caption{} \label{sl.skl.3.4.4b.pic} +\end{figure} + + $\textit{(4)}\Rightarrow \textit{(5)}$. Naj bo $ABCD$ takšen štirikotnik, da velja + $AB \parallel CD$ in $AB \cong CD$ (Figure \ref{sl.skl.3.4.4b.pic}). + Dokažimo, da je $AD \cong BC$. + Premica $AC$ je transverzala vzporednic $AB$ in +$CD$, kar pomeni, da sta kota $CAB$ in $ACD$ izmenična kota ob tej +transverzali in sta zato skladna. Ker je še $AC \cong AC$, sta +trikotnika $ACB$ in $CAD$ skladna (izrek \ref{SKS} - \textit{SAS}). Zato je +$BC \cong AD$. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.4c.pic} +\caption{} \label{sl.skl.3.4.4c.pic} +\end{figure} + + $\textit{(5)}\Rightarrow \textit{(3)}$. Naj bo $ABCD$ takšen štirikotnik, da je $AB \cong CD$ in + $AD \cong BC$ (Figure \ref{sl.skl.3.4.4c.pic}). Dokažimo, da +je potem $\beta =\delta$ in $\alpha =\gamma$. Ker je še $AC \cong +AC$, sta trikotnika $ACB$ in $CAD$ skladna (izrek \ref{SSS} - \textit{SSS}). +Iz tega sledi $\angle ABC \cong \angle CDA$ oz. $\beta =\delta$. Na +podoben način dokažemo tudi $\alpha =\gamma$. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.4d.pic} +\caption{} \label{sl.skl.3.4.4d.pic} +\end{figure} + +$\textit{(5)}\Leftrightarrow \textit{(6)}$. Naj bo $ABCD$ takšen štirikotnik, da +je $AB \cong CD$ in $AD \cong BC$ (Figure \ref{sl.skl.3.4.4d.pic}). Dokažimo, da je točka $S$ +skupno središče njegovih diagonal $AC$ in $BD$. Ker je $AC \cong +AC$ je $\triangle ACB \cong \triangle CAD$ (izrek \ref{SSS} - \textit{SSS}). Zato +je $\angle ACB \cong \angle CAD$ oz. $\angle SCB \cong \angle +SAD$. Iz skladnosti sovršnih kotov $CSB$ in $ASD$ sledi, da sta +skladna tudi kota $SBC$ in $SDA$. Ker je še $AD \cong BC$, sledi +$\triangle CSB \cong \triangle ASD$ (izrek \ref{KSK} - \textit{ASA}). Zato je $SB +\cong SD$ in $SC \cong SA$ oz. točka $S$ je skupno središče +diagonal $AC$ in $BD$. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.4e.pic} +\caption{} \label{sl.skl.3.4.4e.pic} +\end{figure} + + +Obratno, naj bo $S$ skupno središče diagonal $AC$ in $BD$ (Figure \ref{sl.skl.3.4.4e.pic}). Tedaj +je $SB \cong SD$ in $SC \cong SA$. Skladna sta tudi sovršna kota +$CSB$ in $ASD$, zato je $\triangle CSB \cong \triangle ASD$ (izrek +\ref{SKS} - \textit{SAS}). Iz tega sledi $AD \cong BC$. Na podoben način +dokažemo tudi $AB \cong CD$. + \kdokaz + +Bralcu priporočamo, da dokaže prejšnji izrek z uporabo neke +podobne sheme. To bo dobra vaja za uporabo izrekov o skladnosti +trikotnikov. + +Definirajmo sedaj še neke vrste štirikotnikov, za katere se bo +izkazalo, da so posebni primeri paralelogramov (Figure +\ref{sl.skl.3.4.5.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.5.pic} +\caption{} \label{sl.skl.3.4.5.pic} +\end{figure} + +Štirikotnik, ki ima vse stranice skladne, imenujemo \index{romb} +\pojem{romb}. + +Štirikotnik, pri katerem so vsi notranji koti skladni (in torej +enaki $90^0$, ker je njihova vsota $360^0$), je \index{pravokotnik} +\pojem{pravokotnik}. + +Štirikotnik, pri katerem so vse stranice skladne in vsi notranji +koti skladni (in enaki $90^0$), imenujemo \index{kvadrat} +\pojem{kvadrat}. + + +Ni težko dokazati, da je vsak od teh štirikotnikov hkrati +paralelogram. To je direktna posledica prejšnjega izreka. Romb je +paralelogram zaradi $\textit{(5)}\Rightarrow\textit{(1)}$; pravokotnik pa zaradi +$\textit{(2)}\Rightarrow\textit{(1)}$ (ali $\textit{(3)}\Rightarrow\textit{(1)}$). Za kvadrat je +jasno, da je hkrati pravokotnik in romb, zato je tudi +paralelogram. + +Iz prejšnjega izreka \ref{paralelogram} - ekvivalent (\textit{5}) sledi, +da je paralelogram, ki ima dve sosednji stranici skladni, romb. Prav tako + po istem izreku iz ekvivalentov \textit{(2)} in \textit{(3)} sledi, + da je paralelogram, ki ima vsaj en pravi kot, pravokotnik. + +Naslednji izrek daje dodatne kriterije, kdaj je paralelogram hkrati romb, +pravokotnik oz. kvadrat. Ta izrek se nanaša na diagonale. Pri +paralelogramu se diagonali vedno razpolavljata, toda pri rombu, +pravokotniku in kvadratu bomo imeli še dodatne lastnosti. + + + + \bizrek \label{RombPravKvadr} $ $ (Figure \ref{sl.skl.3.4.5a.pic}) + + a) A parallelogram is a rhombus if and only if their diagonals are perpendicular. + + b) A parallelogram is a rectangle if and only if their diagonals are congruent. + + c) A parallelogram is a square if and only if their diagonals are perpendicular and congruent. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.5a.pic} +\caption{} \label{sl.skl.3.4.5a.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Naj bo $ABCD$ paralelogram in $S$ presečišče njegovih diagonal $AC$ in $BD$. Po +prejšnjem izreku \ref{paralelogram} je točka $S$ njuno skupno +središče. Iz istega izreka sledi tudi $AB \cong CD$ in $AD \cong +BC$. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.6.pic} +\caption{} \label{sl.skl.3.4.6.pic} +\end{figure} + +\textit{a)} (Figure \ref{sl.skl.3.4.6.pic}) + + Če je $ABCD$ romb, ima vse stranice skladne. Zato sta +trikotnika $ABS$ in + $ADS$ +skladna (izrek \ref{SSS} - \textit{SSS}). Potem sta skladna tudi kota $ASB$ +in $ASD$ in sta (kot sokota) oba prava kota. To pomeni, da sta +diagonali pravokotni. + +Če sta diagonali paralelograma $ABCD$ pravokotni, sta trikotnika +$ABS$ in $ADS$ skladna (izrek \ref{SKS} - \textit{SAS}). Zato sta stranici +$AB$ in $AD$ skladni. Na podoben način dokazujemo, da so skladne +vse stranice tega paralelograma, kar pomeni, da je paralelogram +romb. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.6a.pic} +\caption{} \label{sl.skl.3.4.6a.pic} +\end{figure} + + +\textit{b)} (Figure \ref{sl.skl.3.4.6a.pic}) + + Če je $ABCD$ pravokotnik, ima vse notranje kote skldne in prave. Tedaj +sta trikotnika $ABC$ in $DCB$ skladna (izrek \ref{SKS} - \textit{SAS}). +Zato je $AC \cong DB$. + +Če pri paralelogramu $ABCD$ velja $AC \cong DB$, sta trikotnika +$ABC$ in $DCB$ skladna (izrek \ref{SSS} - \textit{SSS}). Iz tega sledi, da +sta skladna notranja kota pri ogliščih $B$ in $D$ paralelograma +$ABCD$. Po prejšnjem izreku \ref{paralelogram} sta kota +suplementarna, kar pomeni, da sta oba prava kota. Analogno so pravi +tudi vsi koti tega paralelograma, zato je paralelogram pravokotnik. + + \textit{c)} Paralelogram je kvadrat natanko tedaj, ko je hkrati romb in pravokotnik. + Slednje je pa +izpolnjeno natanko tedaj, ko sta diagonali pravokotni in +skladni, kar sledi iz dokazanega (\textit{a.} in \textit{b.}). + \kdokaz + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.7.pic} +\caption{} \label{sl.skl.3.4.7.pic} +\end{figure} + + + +Ker sta diagonali pravokotnika skladni in se razpolavljata, +obstaja krožnica, ki vsebuje vsa oglišča tega pravokotnika (Figure \ref{sl.skl.3.4.7.pic}). To +je t. i. \index{očrtana krožnica!pravokotnika} \pojem{očrtana +krožnica pravokotnika}. Njeno središče je presečišče +njegovih diagonal. Če je namreč točka $S$ presečišče +diagonal pravokotnika $ABCD$, potem iz prejšnjega izreka +\ref{RombPravKvadr} sledi: + $$SA \cong SC \cong SB \cong SD.$$ +Polmer te krožnice je enak polovici diagonale pravokotnika. Ker +je kvadrat posebna vrsta pravokotnika, tudi za njega obstaja +očrtana krožnica. + +Dokažimo sedaj še pomembno lastnost enakokrakih trapezov. + + + \bizrek \label{trapezEnakokraki} + Interior base angles of an isosceles trapezium are congruent. + The diagonals of an isosceles trapezium are congruent line segments. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.8.pic} +\caption{} \label{sl.skl.3.4.8.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Naj bo $ABCD$ enakokraki trapez z osnovnico $AB$ (Figure \ref{sl.skl.3.4.8.pic}). Brez škode za splošnost predpostavimo, da je $AB>CD$. Po definiciji enakokrakega trapeza je $BC\cong AD$. Označimo s $C'$ in $D'$ pravokotni projekciji oglišč $C$ in $D$ na premici $AB$. Štirikotnik $D'C'CD$ je paralelogram s pravim kotom, zato je pravokotnik. Že iz dejstva, da je $D'C'CD$ paralelogram, sledi $CC'\cong DD'$ (izrek \ref{paralelogram}). Ker je še $\angle CC'B\cong \angle DD'A=90^0$, sta trikotnika $CC'B$ in $DD'A$ skladna (izrek \textit{SSA} \ref{SSK}), zato je $\beta=\angle CBC'\cong \angle DAD'=\alpha$. + + Dokažimo še, da sta diagonali $AC$ in $BD$ skladni. To sledi iz skladnosti trikotnikov $ABC$ in $BAD$ (izrek \textit{SAS} \ref{SKS}). + \kdokaz + + + \bzgled + Let $ABCD$, $AEBK$ and $CEFG$ be equally oriented squares in a plane. + Then $B$, $D$ and $F$ are collinear points and the point $B$ is a midpoint of the line segment $DF$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.9.pic} +\caption{} \label{sl.skl.3.4.9.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.4.9.pic}) + + Trikotnika $CAE$ in $FBE$ sta skladna, ker je $CE\cong FE$, $AE\cong BE$ in + $\angle AEC=90^0-\angle CEB=\angle BEF$ (izrek \ref{SKS} - \textit{SAS}). Zato je + $\angle EBF$ pravi kot in so točke $D$, $B$ in $F$ kolinearne. Iz + skladnosti teh dveh trikotnikov sledi tudi $BF\cong AC\cong BD$. + \kdokaz + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.10.pic} +\caption{} \label{sl.skl.3.4.10.pic} +\end{figure} + +Razen trapezov in paralelogramov bomo definirali še eno novo skupino +štirikotnikov. Štirikotnik $ABCD$ je +\index{deltoid}\pojem{deltoid}, če sta njegovi diagonali pravokotni in +ena od diagonal razpolavlja drugo (Figure \ref{sl.skl.3.4.10.pic}). +Naslednji izrek se nanaša na deltoid in je ekvivalent njegove +definicije. + + + \bzgled + A quadrilateral is a deltoid if and only if it has two pairs of congruent adjacent sides. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.11.pic} +\caption{} \label{sl.skl.3.4.11.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.4.11.pic}) + + Naj bo štirikotnik $ABCD$ deltoid. Tedaj sta diagonali +$AC$ in $BD$ pravokotni in ena od diagonal razpolavlja drugo. Brez +škode za splošnost naj diagonala $BD$ razpolavlja diagonalo $AC$. +Sledi, da sta pravokotna trikotnika $ABS$ in $CBS$ skladna (izrek +\ref{SKS} - \textit{SAS}). Potem je $AB \cong CB$. Iz skladnosti +trikotnikov $ADS$ in $CDS$ pa sledi $AD \cong CD$. + + Naj bo $ABCD$ štirikotnik, v katerem velja $AB \cong CB$ in $AD \cong CD$. + Trikotnika $ABD$ in $CBD$ sta +skladna (izrek \ref{SSS} - \textit{SSS}), zato sta skladna tudi kota $ADS$ +in $CDS$. Iz tega sledi, da sta skladna tudi trikotnika $ADS$ in +$CDS$ (izrek \ref{SKS} - \textit{SAS}). Zato je $S$ središče diagonale $AC$, kota $DSA$ in $DSC$ +pa sta prava kota, ker sta skladna sokota. + \kdokaz + + + + \bzgled + Let $k_1(S_1,r)$, $k_2(S_2,r)$, $k_3(S_3,r)$ be congruent circles and + $k_1\cap k_2=\{B,A_3\}$, $k_2\cap k_3=\{B,A_1\}$, $k_3\cap k_1=\{B,A_2\}$. + Prove that the lines $S_1A_1$, + $S_2A_2$ and $S_3A_3$ intersect at a single point . + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.12.pic} +\caption{} \label{sl.skk.4.2.12.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Dokazali bomo še več, da imajo daljice $O_1A_1$, $O_2A_2$ in $O_3A_3$ + isto središče, oziroma da so ustrezni štirikotniki +paralelogrami (Figure \ref{sl.skk.4.2.12.pic}). Ker so $k_1$, $k_2$ +in $k_3$ skladne krožnice, sta štirikotnika $O_1A_2O_3B$ in $O_2A_1O_3B$ + romba. Zaradi tega sta daljici $O_1A_2$ in $O_2A_1$ skladni in +vzporedni, kar pomeni, da je štirikotnik $O_1A_2A_1O_2$ paralelogram +(izrek \ref{paralelogram}). Iz istega izreka sledi, da imata njegovi +diagonali $O_1A_1$ in $O_2A_2$ skupno središče. Na podoben +način dokazujemo, da imata tudi daljici $O_2A_2$ in $O_3A_3$ skupno +središče, kar pomeni, da to velja tudi za vse tri daljice $O_1A_1$, +$O_2A_2$ in $O_3A_3$ hkrati. \kdokaz + + + \bzgled + Construct a rectangle $ABCD$ if its diagonals and the difference of its sides + are congruent with the two given line segments $d$ and $l$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.10a.pic} +\caption{} \label{sl.skl.3.4.10a.pic} +\end{figure} + +\textbf{\textit{Solution.}} Naj bo $ABCD$ pravokotnik, pri katerem je $AC\cong d$ in $AB-BC=l$ (Figure \ref{sl.skl.3.4.10a.pic}). Označimo z $E$ takšno točko stranice $AB$, da velja $EB\cong BC$. V tem primeru je $AE=AC-EB=AC-BC=l$. Ker je $EBC$ enakokraki pravokotni trikotnik, je $\angle CEB\cong\angle ECB=45^0$ (izreka \ref{enakokraki} in \ref{VsotKotTrik}) oz. $\angle AEC=135^0$. +To nam omogoča najprej konstrukcijo trikotnika $AEC$ ($AC\cong d$, $\angle AEC=135^0$ in $AE\cong l$), nato pa še pravokotnika $ABCD$. + \kdokaz + + + \bzgled + Construct a triangle with given $b$, $c$ and $t_a$ (sides $AC$, $AB$ and triangle median $AA_1$). + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.10b.pic} +\caption{} \label{sl.skl.3.4.10b.pic} +\end{figure} + +\textbf{\textit{Solution.}} Naj bo $ABC$ trikotnik, pri katerem je $AC\cong b$, $AB\cong c$ in $AA_1\cong t_a$, kjer je $A_1$ središče daljice $BC$ (Figure \ref{sl.skl.3.4.10b.pic}). Označimo z $D$ takšno točko poltraka $AA_1$, da je $DA_1\cong AA_1$ in $\mathcal{B}(A, A_1,D)$. To pomeni, da je $A_1$ skupno središče daljic $BC$ in $AD$, zato je po izreku \ref{paralelogram} štirikotnik $ABDC$ paralelogram. Po istem izreku je $CD\cong AB\cong c$. To nam omogoča najprej konstrukcijo trikotnika $ADC$ ($AC\cong b$, $CD\cong c$ in $AD=2t_a$), nato pa še točke $A$. + \kdokaz + + Sedaj bomo vpeljali krajšo obliko zapisa podatkov za načrtovanje trikotnikov. Podobno, kot smo imeli pri prejšnji nalogi zapis: $b$, $c$, $t_a$, bomo za elemente trikotnika $ABC$ običajno uporabljali oznake: + \begin{itemize} + \item $a$, $b$, $c$ - stranice, + \item $\alpha$, $\beta$, $\gamma$ - notranji koti, + \item $v_a$, $v_b$, $v_c$ - višine, + \item $t_a$, $t_b$, $t_c$ - težiščnice, + \item $l_a$, $l_b$, $l_c$ - daljice, ki so določene z ogliščem in s presečiščem simetrale notranjega kota v tem oglišču z nasprotno stranico; + \item $s$ - polobseg ($s=\frac{a+b+c}{2}$), + \item $R$ - polmer očrtane krožnice (glej razdelek \ref{odd3ZnamTock}), + \item $r$ - polmer včrtane krožnice (glej razdelek \ref{odd3ZnamTock}), + \item $r_a$, $r_b$, $r_c$ - polmeri pričrtanih krožnic (glej razdelek \ref{odd4Pricrt}). + \end{itemize} + + + \bzgled + Construct a trapezium if its sides are congruent with the four given line segments $a$, $b$, $c$ and $d$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.4.10c.pic} +\caption{} \label{sl.skl.3.4.10c.pic} +\end{figure} + +\textbf{\textit{Solution.}} Brez škode za splošnost predpostavimo najprej, da je $a\geq c$. Naj bo $ABCD$ trapez, v katerem so stranice $AB\cong a$, $BC\cong b$, $CD\cong c$ in $DA\cong d$ (Figure \ref{sl.skl.3.4.10c.pic}). V tem primeru je $AB\geq CD$, zato na stranici $AB$ obstaja takšna točka $E$, da velja $AE\cong CD$. Ker je še $AB\parallel CD$, je po izreku \ref{paralelogram} štirikotnik $AECD$ paralelogram, zato je po istem izreku tudi $CE\cong DA\cong d$. Velja tudi $EB=AB-AE=AB-CD=a-c$. To omogoča konstrukcijo trikotnika $EBC$ ($EB=a-c$, $BC\cong c$ in $CE\cong d$), nato pa še oglišč $A$ in $D$ (iz pogoja $AE\cong CD\cong c$). + \kdokaz + + + +%________________________________________________________________________________ + \poglavje{Regular Polygons}\label{odd3PravilniVeck} + + Pojem kvadrata se vklaplja v splošno definicijo nove vrste večkotnikov. + Večkotnik je +\index{pravilni!večkotniki} \pojem{pravilen}, če ima vse +stranice skladne in so vsi notranji koti skladni (Figure +\ref{sl.skl.3.5.1.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.1.pic} +\caption{} \label{sl.skl.3.5.1.pic} +\end{figure} + + Kvadrat je torej pravilni štirikotnik. Prav tako je enakostranični trikotnik + \index{trikotnik!pravilni}\pojem{pravilni trikotnik}. +To je namreč posledica dejstva, da so pri enakostraničnem +trikotniku tudi vsi koti enaki. + +Ugotovili smo že, da je vsota vseh notarnjih kotov poljubnega +$n$-kotnika enaka $(n - 2) \cdot 180^0$ (izrek \ref{VsotKotVeck}). Ker +so pri pravilnem $n$-kotniku vsi notranji koti skladni, lahko notranji kot +izračunamo tako, da vsoto vseh kotov delimo s številom $n$. Tako smo +dokazali naslednjo trditev (Figure \ref{sl.skl.3.5.2.pic}). + + + \bizrek \label{pravVeckNotrKot} + The measure of each interior angle of a regular $n$-gon is: + $$\frac{(n - 2)\cdot 180^0}{n}.$$ + \eizrek + + Tako notranji kot pravilnega trikotnika meri $60^0$, štirikotnika $90^0$, petkotnika $108^0$, šestkotnika $120^0$, ... + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.2.pic} +\caption{} \label{sl.skl.3.5.2.pic} +\end{figure} + +Dokažimo še dve pomembni lastnosti pravilnih večkotnikov. + + + \bizrek \label{sredOcrtaneKrozVeck} + For each regular polygon, there exists a circle passing through each of its vertices. + \eizrek + +\textbf{\textit{Proof.}} Naj bo $A_1A_2\ldots A_n$ pravilen +$n$-kotnik (Figure \ref{sl.skl.3.5.2.pic}). Potem ima vse +stranice skladne in vsi notranji koti so skladni ter enaki $\frac{(n +- 2)\cdot 180^0}{n}$. Naj bosta $s_1$ in $s_2$ simetrali stranic +$A_1A_2$ in $A_2A_3$ tega večkotnika ter točka $S$ njuno +presečišče. Iz izreka \ref{simetrala} sledi $SA_1 \cong SA_2$ +in $SA_2 \cong SA_3$ oz.: + $$SA_1 \cong SA_2 \cong SA_2 \cong SA_3.$$ + Sledi, da sta enakokraka trikotnika $A_1SA_2$ in $A_2SA_3$ +skladna (izrek \ref{SSS} - \textit{SSS}). Potem so skladni tudi koti +$SA_1A_2$, $SA_2A_1$, $SA_2A_3$ in $SA_3A_2$. Iz $\angle SA_2A_1 \cong +\angle SA_2A_3$ sledi, da sta oba kota enaka polovici notranjega +kota tega večkotnika oz. $\frac{\alpha}{2}=\frac{(n-2)\cdot +180^0}{2n}$. Zato je tudi: + $$\angle SA_3A_4 =\alpha - \frac{\alpha}{2}=\frac{\alpha}{2} = \angle + SA_3A_2.$$ +Torej sta trikotnika $A_2SA_3$ in $A_3SA_4$ skladna (izrek +\ref{SKS} - \textit{SAS}). Zaradi tega je $SA_3 \cong SA_4$ oz.: + $$SA_1 \cong SA_2 \cong SA_2 \cong SA_3\cong SA_4.$$ + Če ta postopek nadaljujemo, dobimo: + $$SA_1 \cong SA_2 \cong SA_2 \cdots \cong SA_n,$$ +kar pomeni, da je točka $S$ središče krožnice $k(S, SA_1)$, ki +vsebuje vsa njegova oglišča. + \kdokaz + +Krožnico iz prejšnjega izreka imenujemo \index{očrtana +krožnica!pravilnega večkotnika} \pojem{očrtana krožnica +pravilnega večkotnika}. Iz dokaza prejšnjega izreka je jasno, da +se njeno središče nahaja v presečišču simetral vseh njegovih +stranic. + + +Analogno dokazujemo tudi naslednji izrek. + + + \bizrek \label{sredVcrtaneKrozVeck} + For each regular polygon, there exists a circle touching each of its sides. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.3.pic} +\caption{} \label{sl.skl.3.5.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $A_1A_2\ldots A_n$ pravilen $n$-kotnik + (Figure \ref{sl.skl.3.5.3.pic}). +Točko $S$ definirajmo enako kot v dokazu prejšnjega izreka. +Dokazali smo, da velja: $SA_1 \cong SA_2 \cong SA_2 \cdots \cong +SA_n$. Iz tega po izreku \ref{SSS} - \textit{SSS} sledi skladnost +enakokrakih trikotnikov: + $$\triangle A_1SA_2 \cong \triangle A_2 SA_3 \cong \cdots \cong + \triangle A_{n-1}SA_n \cong \triangle A_nSA_1.$$ +Zaradi tega so skladni tudi vsi koti ob osnovnicah teh trikotnikov. +Torej so premice $SA_1$, $SA_2$,..., $SA_n$ simetrale notranjih +kotov večkotnika $A_1A_2\ldots A_n$. Naj bodo $P_1$, $P_2$,..., +$P_n$ nožišča višin iz oglišča $S$ omenjenih enakokrakih trikotnikov. +Iz skladnosti trikotnikov $\triangle A_1SP_1$, $\triangle A_2SP_1$, +$\triangle A_2SP_2$, ..., $\triangle A_1SP_n$ (izreka \ref{SSK} in +\ref{KSK}) sledi skladnost daljic $SP_1$, $SP_2$,..., $SP_n$. Po +izreku \ref{TangPogoj} se krožnica $k(S, SP_1)$ dotika vseh stranic +večkotnika $A_1A_2\ldots A_n$. + \kdokaz + +Krožnico iz prejšnjega izreka imenujemo \index{včrtana krožnica!pravilnega večkotnika} \pojem{včrtana krožnica pravilnega +večkotnika}. Iz dokaza tega izreka je jasno, da se središče včrtane +krožnice pravilnega večkotnika nahaja na presečišču simetral vseh +njegovih notranjih kotov. Iz dokaza je očitno tudi, da so točke, v +katerih se ta krožnica dotika stranic pravilnega večkotnika, hkrati +središča teh stranic. Središče očrtane in včrtane krožnice je ista +točka in jo zato imenujemo tudi \index{središče!pravilnega +večkotnika}\pojem{središče pravilnega večkotnika}. + + Videli smo tudi, da so vsi trikotniki, + določeni s središčem +pravilnega $n$-kotnika in z njegovimi stranicami, enakokraki in vsi +skladni. Polmera očrtane in včrtane krožnice $n$-kotnika sta enaka +kraku oz. višini vsakega od teh trikotnikov. Koti ob vrhu teh +trikotnikov so tudi skladni in ker jih je skupaj $n$ (enako kot +stranic $n$-kotnika), vsak od njih meri (Figure \ref{sl.skl.3.5.4.pic}): + $$\varphi = \frac{360^0}{n}.$$ + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.4.pic} +\caption{} \label{sl.skl.3.5.4.pic} +\end{figure} + + Pri pravilnem šestkotniku, oz. za $n = 6$, velja: + $$\varphi = \frac{360^0}{6}=60^0.$$ + + To pomeni, da so omenjeni trikotniki pravilni. +Torej je pravilni šestkotnik sestavljen iz šestih +pravilnih trikotnikov (Figure \ref{sl.skl.3.5.4.pic}). + +V nadaljevanju bomo obravnavali lastnosti pravilnih $n$-kotnikov. + + Naj bo najprej $n$ sodo število in $k = \frac{n}{2}+1$. +Pravimo, da je $A_k$ \pojem{nasprotno oglišče} oglišča $A_1$ +pravilnega $n$-kotnika $A_1A_2\ldots A_n$ (Figure +\ref{sl.skl.3.5.5.pic}). Analogno sta $A_2$ in $A_{k+1}$, $A_3$ in +$A_{k+2}$, ... , $A_{k-1}$ in $A_n$ nasprotni oglišči tega +$n$-kotnika. Podobno sta stranici $A_1A_2$ in $A_kA_{k+1}$, ... , +$A_{k-1} A_k$ in $A_nA_1$ \pojem{nasprotni stranici} večkotnika +$A_1A_2\ldots A_n$. Opazimo, da velja: + $$\angle A_1SA_k=\frac{n}{2}\varphi = \frac{n}{2}\cdot + \frac{360^0}{n}=180^0,$$ +kar pomeni, da diagonala $A_1A_k$ tega $n$-kotnika vsebuje njegovo +središče. Zato ta diagonala predstavlja premer očrtane krožnice. +Analogno to velja za vse diagonale, ki so določene z nasprotnimi +oglišči. Zaradi tega takšne diagonale imenujemo \index{velika diagonala +pravilnega $n$-kotnika} \pojem{velike diagonale} pravilnega +$n$-kotnika. Polmer očrtane krožnice je + enak polovici velike diagonale. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.5.pic} +\caption{} \label{sl.skl.3.5.5.pic} +\end{figure} + + Na podoben način se dokaže, da za sodo število $n$ središči + nasprotnih stranic pravilnega $n$-kotnika +$A_1A_2\ldots A_n$ določata premere včrtane krožnice tega +$n$-kotnika. Če upoštevamo prejšnje oznake, dobimo: + \begin{eqnarray*} + \angle P_1SP_k&=&\angle P_1SA_2 + \angle A_2SA_{k-1} + \angle + A_{k-1}SP_k\\ + &=& \frac{\varphi}{2}+\frac{n-2}{2}\cdot \varphi+\frac{\varphi}{2} + =\frac{n}{2}\cdot\varphi + =180^0. + \end{eqnarray*} + + + Daljice, ki so določene s parom središč nasprotnih stranic $n$-kotnika + $A_1A_2\ldots A_n$ oz. daljice $P_1P_k$, +$P_2P_{k+1}$, ... , $P_{k-1}P_n$, imenujemo \pojem{višine} +\index{višina!pravilnega $n$-kotnika} tega $n$-kotnika. Polmer +včrtane krožnice je enak polovici višine. + + Torej vsak pravilni $n$-kotnik, kjer je $n$ sodo število, vsebuje +$\frac{n}{2}$ velikih diagonal (enakih premeru očrtane krožnice) in +$\frac{n}{2}$ višin (enakih premeru včrtane krožnice). Vsaka od njih +gre skozi središče tega $n$-kotnika. + + + Naj bo sedaj $n$ liho število (Figure \ref{sl.skl.3.5.6.pic}) in + $k=\frac{n+1}{2}+1$. + Potem je: +\begin{eqnarray*} + \angle P_1SA_k=\angle P_1SA_2 + \angle A_2SA_k= + \frac{\varphi}{2}+\frac{n-1}{2}\cdot \varphi + =\frac{n}{2}\cdot\varphi + =180^0. + \end{eqnarray*} + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.6.pic} +\caption{} \label{sl.skl.3.5.6.pic} +\end{figure} + +To pomeni, da daljica $P_1A_k$ vsebuje središče $S$ pravilnega +$n$-kotnika $A_1A_2\ldots A_n$. To daljico imenujemo \index{višina!pravilnega $n$-kotnika} +\pojem{višina} tega $n$-kotnika, stranica +$A_1A_2$ in oglišče $A_k$ sta si \pojem{nasprotni}. Analogno +definiramo tudi preostalih $n$ višin in $n$ parov nasprotnih +stranic in oglišč. Na podoben način lahko dokažemo, da tudi ostale +višine tega $n$-kotnika vsebujejo njegovo središče. + +Pri pravilnem (enakostraničnem) trikotniku imamo torej tri +višine, ki se sekajo v njegovem središču (Figure +\ref{sl.skl.3.5.7.pic}). Če to velja pri +poljubnem trikotniku, bomo ugotovili kasneje + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.7.pic} +\caption{} \label{sl.skl.3.5.7.pic} +\end{figure} + + V kvadratu sta njegovi diagonali hkrati veliki + diagonali in se sekata v +njegovem središču (Figure \ref{sl.skl.3.5.7.pic}). Višini +kvadrata sta skladni z njegovo stranico, kar ni težko dokazati. + +Omenili smo že, da je pravilni šestkotnik sestavljen iz šestih +trikotnikov, ki se stikajo v njegovem središču. Pravilni +šestkotnik bomo obširneje obravnavali v nadaljevanju. Dokazali bomo +tudi nekaj lastnosti pravilnega petkotnika, sedemkotnika, +devetkotnika in dvanajstkotnika. Posebej zanimiv bo problem +načrtovanja pravilnih $n$-kotnikov za poljubno število $n$. + +Dokažimo sedaj zanimivo lastnost pravilnega devetkotnika. + + \bzgled + If $a$ is a side and $d$ and $e$ are the shortest and longest + diagonal of a regular nonagon ($9$-gon), then $e - d = a$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.5.8.pic} +\caption{} \label{sl.skl.3.5.8.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bosta $d = CE$ in $e = BF$ najkrajša in +najdaljša diagonala pravilnega devetkotnika $ABCDEFGHI$ s stranico +$a$ ter $P$ presečišče premic $BC$ in $FE$ (Figure +\ref{sl.skl.3.5.8.pic}). Notranji kot tega devetkotnika meri + $\angle CDE=\frac{9-2}{9}\cdot 180^0=140^0$, +zato je $\angle ECD = \angle CED = 20^0$. Iz tega sledi + $\angle BCE = \angle FEC = 120^0$, oz.: + $$\angle ECP = \angle CEP = 60^0.$$ + Torej je trikotnik $CPE$ pravilen. Ker je $CB = EF=a$ +in $\angle BPF \cong \angle CPE = 60^0$, je pravilen tudi +trikotnik $BPF$. Torej: + $$e = BF = BP = BC + CP = BC + CE = a + d,$$ kar je bilo treba dokazati. \kdokaz + + +%%________________________________________________________________________________ + \poglavje{Midsegment of Triangle} \label{odd3SrednTrik} + +Sedaj bomo obravnavali zelo pomembno lastnost trikotnika, ki jo +bomo pogosto uporabljali. Naj bosta $P$ in $Q$ središči stranic +$AB$ in $AC$ trikotnika $ABC$. Daljico $PQ$ imenujemo +\index{srednjica!trikotnika} \pojem{srednjica trikotnika} $ABC$, +ki je pripadajoča stranici $BC$ (Figure \ref{sl.skl.3.6.1.pic}). +Pravimo tudi, da je $PQ$ srednjica trikotnika $ABC$ za osnovnico +$BC$. Dokažimo osnovno lastnost, ki se nanaša na srednjico. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.1.pic} +\caption{} \label{sl.skl.3.6.1.pic} +\end{figure} + + + \bizrek \label{srednjicaTrik} + Let $PQ$ be the midsegment of a triangle $ABC$ corresponding to the side $BC$. Then: + $$ PQ = \frac{1}{2} BC\hspace*{2mm} + \textrm{ in } \hspace*{2mm} PQ \parallel BC.$$ + \eizrek + + + + \textbf{\textit{Proof.}} Naj bo $R$ takšna točka, da velja $PQ \cong QR$ in +$\mathcal{B}(P,Q,R)$ (Figure \ref{sl.skl.3.6.1.pic}). Daljici $AC$ +in $PR$ imata skupno središče, zato je štirikotnik $APCR$ +paralelogram (izrek \ref{paralelogram}). Zaradi tega sta daljici +$AP$ in $RC$ skladni in vzporedni. Točka $P$ je središče daljice +$AB$, zato sta tudi daljici $PB$ in $RC$ skladni in vzporedni. To +pomeni, da je tudi štirikotnik $PBCR$ paralelogram. Iz tega +sledi, da sta daljici $BC$ in $PR$ skladni in vzporedni. Končna +ugotovitev sledi iz dejstva, da je točka $Q$ središče daljice +$PR$. + \kdokaz + + + \bzgled + Let $AB$ and $A'B'$ be congruent line segments, $C$ and $D$ the midpoints of the line segments + $AA'$ and $BB'$. Suppose that $CD =\frac{1}{2} AB$. + What is a measure of the angle between the lines $AB$ and $A'B'$? + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.2.pic} +\caption{} \label{sl.skl.3.6.2.pic} +\end{figure} + + + +\textbf{\textit{Solution.}} Naj bo točka $S$ središče daljice +$A'B$ (Figure \ref{sl.skl.3.6.2.pic}). Daljici $CS$ in $DS$ sta +srednjici trikotnikov $A'AB$ in $BA'B'$, zato je: $$CS = +\frac{1}{2}AB = CD = \frac{1}{2}A'B'= DS,$$ + oziroma $SCD$ je pravilen trikotnik. Kota $\angle AB,A'B'$ in $\angle CSD$ + imata vzporedne krake. Torej je: +$\angle AB, A'B' \cong \angle CSD = 60^0$. + \kdokaz + +Naslednja posledica izreka \ref{srednjicaTrik} se nanaša na +trapez. + + + \bizrek \label{srednjTrapez} + Let $P$ and $Q$ be the midpoints of legs $BC$ and $DA$ of a trapezium $ABCD$. + Suppose that $M$ and $N$ are the midpoints of the diagonals $AC$ and $BD$ of that trapezium. + Then the points $M$ and $N$ lie on the line $PQ$, which is parallel to the bases $AB$ in $CD$ of the trapezium, + and also: + $$PQ = \frac{1}{2}( AB + CD) \hspace*{2mm} + \textrm{ in } \hspace*{2mm} MN=\frac{1}{2}( AB - CD).$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.3.pic} +\caption{} \label{sl.skl.3.6.3.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Daljice $PN$, $NQ$ in $PM$ so po vrsti srednjice + trikotnikov + $DAC$, $ACB$ in $ADB$ za +pripadajoče osnovnice $DC$, $AB$ in $AB$ (Figure +\ref{sl.skl.3.6.3.pic}). Zaradi tega so vse tri premice $PN$, $NQ$ +in $PM$ vzporedne z osnovnicama $CD$ in $AB$. Ker skozi vsako +točko (najprej $N$, nato $P$) obstaja le ena vzporednica s premico +$AB$ (Playfairjev\footnote{\index{Playfair, J.}\textit{J. +Playfair} (1748--1819), škotski matematik.} aksiom +\ref{Playfair}), so točke $P$, $N$, $M$ in $Q$ kolinearne. Velja +še (izrek \ref{srednjicaTrik}): + $PN = \frac{1}{2} CD$ in + $NQ = PM = \frac{1}{2} AB$. Iz tega sledi: + \begin{eqnarray*} + PQ&=& PN+NQ=\frac{1}{2}CD+ + \frac{1}{2}AB=\frac{1}{2}\left(AB+CD\right)\\ + NM&=& PM-PN=\frac{1}{2}AB- + \frac{1}{2}CD=\frac{1}{2}\left(AB-CD\right), + \end{eqnarray*} + kar je bilo potrebno dokazati. \kdokaz + +Daljico $PQ$ iz prejšnjega izreka imenujemo +\index{srednjica!trapeza} \pojem{srednjica trapeza}. + + Naslednje trditve se nanašajo na poljubni štirikotnik. + + + \bizrek \label{Varignon} + Let $ABCD$ be an arbitrary quadrilateral and $P$, $Q$, $K$ and $L$ + the midpoints of the sides $AB$, $CD$, $BC$ and $AD$, respectively. Then the quadrilateral $PKQL$ is + a parallelogram (so-called \index{paralelogram!Varignonov} + Varignon\footnote{\index{Varignon, P.} + \textit{P. Varignon} (1654--1722), + francoski matematik, + ki je prvi dokazal to lastnost. Vendar je bil izrek objavljen šele + po njegovi smrti leta 1731. Glede na enostavnost pa je prav + presenetljivo, da je ta trditev toliko časa ‘‘čakala’’ na svoje + odkritje.} parallelogram). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.4.pic} +\caption{} \label{sl.skl.3.6.4.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.6.4.pic}) +Daljici $PK$ in $LQ$ sta srednjici trikotnikov $ABC$ in $ADC$ za +isto osnovnico $AC$, zato sta skladni in +vzporedni. Torej je štirikotnik $PKQL$ paralelogram. + \kdokaz + +V posebnem primeru je lahko Varignonov paralelogram celo pravokotnik, +romb ali kvadrat. Kdaj je to možno? To vprašanje nam da idejo za +naslednji izrek. Vemo, da je paralelogram pravokotnik, če ima +vsaj en notranji kot pravi. Drugi (ekkvivalenten) pogoj pa je, +da ima skladni diagonali. Podobno obravnavo lahko uporabimo tudi +pri rombu in kvadratu. + + + \bzgled \label{VarignonPoslPravRomb} + Let $ABCD$ be an arbitrary quadrilateral and $P$, $K$, $Q$ and $L$ + the midpoints of the sides $AB$, $BC$, $CD$ and + $DA$, respectively. Then: + + a) $AC \perp BD \Leftrightarrow PQ \cong KL$; + + b) $AC \cong BD \Leftrightarrow PQ \perp KL$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.5.pic} +\caption{} \label{sl.skl.3.6.5.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.6.5.pic}) + Iz prejšnjega izreka \ref{Varignon} sledi, da je + štirikotnik $PKQL$ vedno paralelogram – Varignonov +paralelogram. Daljici $PL$ in $PK$ sta srednjici trikotnikov $ABD$ +in $ABC$ za osnovnici $AD$ oz. $AC$. Zaradi tega velja: + $PL= \frac{1}{2}BD$ in $PL \parallel BD$ ter $PK= \frac{1}{2}AC$ in $PK \parallel AC$. + Zatorej velja: + + a) $AC \perp BD \Leftrightarrow PL \perp PK +\Leftrightarrow PKQL \textrm{ pravokotnik} \Leftrightarrow PQ +\cong KL$; + + b) $AC \cong BD \Leftrightarrow PL \cong PK \Leftrightarrow + PKQL \textrm{ romb } \Leftrightarrow PQ \perp KL$. + \kdokaz + +Če je Varignonov paralelogram kvadrat, so izpolnjeni vsi štirje +pogoji iz prejšnjih ekvivalenc, oz. je v tem primeru: $AB \perp CD$, $AB \cong CD$, $PQ \perp KL$ in $PQ \cong KL$. + +Oglejmo si še eno uporabo lastnosti Varignonovega paralelograma. + + + \bzgled \label{VagnanPosl} + Let $ABCD$ be an arbitrary quadrilateral. If $P$, $Q$, $K$, $L$, $M$ and + $N$ are the midpoints of the line segments $AB$, $CD$, $BC$, $AD$, $AC$ and $BD$, respectively, + then the line segments $PQ$, $KL$ and + $MN$ have a common midpoint. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.6.pic} +\caption{} \label{sl.skl.3.6.6.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + (Figure \ref{sl.skl.3.6.6.pic}) Štirikotnik $PKQL$ je Varignonov +paralelogram (izrek \ref{Varignon}). Na podoben način dokazujemo, da +je tudi štirikotnik $LNKM$ paralelogram (srednjice trikotnikov $ADB$ +in $ACB$). Paralelograma $PKQL$ in $LPKQ$ imata skupno diagonalo +$LK$. Ker se diagonali poljubnega paralelograma razpolavljata (izrek +\ref{paralelogram}), imajo daljice $LK$, $PQ$ in $MN$ skupno +središče. + \kdokaz + +Točko iz prejšnjega primera, v kateri se daljice sekajo, imenujemo +\index{težišče!štirikotnika} \pojem{težišče štirikotnika}. Več +o tem bomo povedali v razdelku \ref{odd5TezVeck}. + +Naslednja trditev je lep primer kombiniranja neenakosti trikotnika +in srednjice trikotnika. Trditev je pravzaprav posplošitev izreka +o srednjici trapeza \ref{srednjTrapez}. + + + \bizrek + If $P$ and $Q$ are the midpoints of the sides $AB$ and $CD$ of + an arbitrary quadrilateral + $ABCD$, then: $$PQ \leq \frac{1}{2}\left( BC + AD\right).$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.7.pic} +\caption{} \label{sl.skl.3.6.7.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Naj bo $S$ središče diagonale $AC$ štirikotnika $ABCD$ + (Figure \ref{sl.skl.3.6.7.pic}). +Če uporabimo izrek o srednjici trikotnika (\ref{srednjicaTrik}) in +trikotniško neenakost (\ref{neenaktrik}), dobimo: + $$BC + AD = 2PS ++ 2SQ = 2(PS + SQ) \geq 2PQ.$$ + Enakost velja v primeru, kadar so +točke $P$, $S$ in $Q$ kolinearne, oz. ko je štirikotnik $ABCD$ +trapez z osnovnico $BC$. + \kdokaz + + Omenimo še, da +neenakost iz prejšnjega primera velja tudi, če točke $A$, $B$, +$C$ in $D$ niso v isti ravnini, oz. če je $ABCD$ +\index{tetraeder} \pojem{tetraeder}. + + + \bzgled \label{TezisceSredisceZgled} + Let $P$ be the midpoint of the median $AA_1$ of a triangle $ABC$ and + $Q$ + the intersection of the side $AC$ and the line $BP$. Determine the ratios + $AQ :QC$ and $BP : PQ$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.8.pic} +\caption{} \label{sl.skl.3.6.8.pic} +\end{figure} + + +\textbf{\textit{Solution.}} Naj bo $R$ središče daljice $QC$ + (Figure \ref{sl.skl.3.6.8.pic}). Daljica $A_1R$ je srednjica +trikotnika $BCQ$ za osnovnico $BQ$, zato je (izrek +\ref{srednjicaTrik}) $BQ = 2A_1R$ in $BQ\parallel A_1R$. Iz te +vzporednosti in definicije točke $P$ sledi, da je $PQ$ srednjica +trikotnika $AA_1R$ za osnovnico $A_1R$, zato je (izrek +\ref{srednjicaTrik} in Playfairjev aksiom \ref{Playfair}) točka $Q$ +središče daljice $AR$ in velja $A_1R = 2PQ$. Torej: $AQ \cong QR +\cong RC$ oz. $AQ:QC=1:2$. Na koncu je še $BQ=2A_1R=4PQ$ oz. +$BP:PQ=3:1$. + \kdokaz + + + + \bnaloga\footnote{19. IMO Yugoslavia - 1977, Problem 1.} + Equilateral triangles $ABP$, $BCL$, $CDM$, $DAN$ are constructed inside the + square $ABCD$. Prove that the midpoints of the segments $LM$, $MN$, $NP$, $PL$, $AN$, $LB$, + $BP$, $CM$, $CL$, $DN$, $DM$ in $AP$ + are the twelve vertices of a regular dodecagon ($12$-gon). + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.6.IMO1.pic} +\caption{} \label{sl.skl.3.6.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Označimo z $a$ dolžino stranice in +z $O$ središče kvadrata $ABCD$ (Figure \ref{sl.skl.3.6.IMO1.pic}). + +Dokažimo najprej, da je štirikotnik $MNPL$ tudi kvadrat z istim +središčem $O$. Ker so $ABP$, $BCL$, $CDM$ in $DAN$ vsi pravilni +trikotniki, ležita diagonali $MP$ in $LN$ štirikotnika $MNPL$ na +simetralah stranic $AB$ in $BC$ kvadrata $ABCD$. Iz tega sledi $MP \perp LN$. +Ker je še $d(M,AB)=d(P,CD)=a-v$ (kjer je $v$ dolžina +višine omenjenih pravilnih trikotnikov), velja tudi $OM\cong OP$. +Podobno je tudi $OL\cong ON$, kar pomeni, da je štirikotnik $MNPL$ +res kvadrat z istim središčem~$O$. + +Dokažimo sedaj, da je $LAM$ pravilni trikotnik. Ker je $AB\cong +AD\cong BL\cong DM=a$ in $\angle LBA\cong\angle MDA +=90^0-60^0=30^0$, sta trikotnika $LBA$ in $MDA$ skladna (izrek +\textit{SAS} \ref{SKS}). To pomeni, da je $LA\cong MA$ in $\angle +DAL= 90^0-\angle LAB=15^0$. Podobno je tudi $\angle BAM=15^0$ oz. +$\angle LAM = 90^0-2\cdot 15^0=60^0$. Torej je $LAM$ pravilni +trikotnik, zato ima stranica kvadrata $MNPL$ dolžino +$b=|LM|=|LA|$. + +Označimo s $S$ središče daljice $LM$. Središča stranic kvadrata $MNPL$ +ležijo na krožnici $k(O,\frac{b}{2})$, ki je včrtana krožnica tega +kvadrata. Dokažimo, da tudi točka $T$ - središče daljice $AN$ - +leži na tej krožnici. Daljica $OT$ je srednjica trikotnika $LAN$ +za osnovnico $LA$, zato je $OT\parallel LA$ in +$|OT|=\frac{1}{2}|LA|=\frac{b}{2}$. Torej točka $T$ in analogno +tudi vse točke v nalogi definiranega $12$-kotnika ležijo na +krožnici $k(O,\frac{b}{2})$. + +Dokažimo še, da je omenjeni $12$-kotnik pravilen. Brez škode za +splošnost zadošča dokazati, da je $\angle +SOT=\frac{360^0}{12}=30^0$. Toda iz že dokazanega dejstva +$OT\parallel LA$ sledi $\angle SOT\cong \angle +LAS=\frac{1}{2}\angle LAM=30^0$. + \kdokaz + +%________________________________________________________________________________ + \poglavje{Triangle Centers} \label{odd3ZnamTock} + + Našo raziskavo bomo nadaljevali s trikotnikom – najbolj enostavnim + večkotnikom, + ki pa je hkrati +lik, ki ima nepričakovano veliko zanimivih lastnosti. Nekatere bomo +obravnavali v tem razdelku, nekatere pa kasneje, ko se bomo ukvarjali +z drugimi pojmi, kot so izometrije in podobnost. + + Sedaj bomo obravnavali štiri \index{značilne točke + trikotnika}\pojem{značilne točke + trikotnika}\footnote{Te štiri točke omenjajo že Stari Grki, čeprav so + (še posebej to velja za težišče) bile znane verjetno že veliko časa pred +tem.} in + njihovo uporabo pri štiri- in večkotnikih. + +Začnimo najprej s prvo med značilnimi točkami, ki je povezana s + težiščnicami trikotnika. Težiščnico smo že definirali + v poglavju \ref{odd3NeenTrik}. + + + + \bizrek \label{tezisce} + The medians of a triangle intersect at one point. + That point divides the medians in the ratio $2:1$ + (from the vertex to the midpoint of the opposite side). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.1.pic} +\caption{} \label{sl.skl.3.7.1.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Naj bodo $AA_1$, $BB_1$ in $CC_1$ težiščnice trikotnika $ABC$ + (Figure \ref{sl.skl.3.7.1.pic}). +Zaradi Paschevega aksioma, \ref{AksPascheva} glede na trikotnik +$BCB_1$ in premico $AA_1$, premica $AA_1$ seka daljico $BB_1$. +Analogno premica $BB_1$ seka daljico $AA_1$, kar pomeni, da se +težiščnici $AA_1$ in $BB_1$ sekata v neki točki $T$. Naj bosta $A_2$ +in $B_2$ središči daljic $AT$ in $BT$. Daljici $A_1B_1$ in $A_2B_2$ +sta srednjici trikotnikov $ABC$ in $ABT$ z isto osnovnico $AB$. To +pomeni, da sta daljici $A_1B_1$ in $A_2B_2$ vzporedni in enaki +polovici stranice $AB$. Zaradi tega je štirikotnik $B_2A_1B_1A_2$ +paralelogram (izrek \ref{paralelogram}), kar pomeni da se njegovi +diagonali $A_1A_2$ in $B_1B_2$ razpolavljata in je točka $T$ +njuno skupno središče. Torej velja: + $A_1T\cong TA_2 \cong A_2A$ in $B_1T\cong TB_2 \cong B_2B$ + oz. $AT:TA_1=2:1$ in $BT:TB_1=2:1$ ter + $$A_1T=\frac{1}{3}A_1A \textrm{ in } B_1T = \frac{1}{3}B_1B.$$ + Na enak način + dokažemo, da se tudi težiščnici $AA_1$ in $CC_1$ sekata v neki +točki $T’$, za katero velja $AT':T'A_1=2:1$ in $CT':T'C_1=2:1$ oz.: + $$A_1T'=\frac{1}{3}A_1A \textrm{ in } C_1T' = \frac{1}{3}C_1C.$$ +To pomeni, da sta $T$ in $T'$ točki poltraka $A_1A$, za kateri velja +$A_1T\cong A_1T'=\frac{1}{3}A_1A$, zato je po izreku +\ref{ABnaPoltrakCX} $T = T'$, kar pomeni, da se težiščnice $AA_1$, +$BB_1$ in $CC_1$ sekajo v točki $T$ in velja +$AT:TA_1=BT:TB_1=CT:TC_1=2:1$. + \kdokaz + + +Točka iz prejšnjega izreka, v kateri se sekajo vse težiščnice +trikotnika, se imenuje \index{težišče!trikotnika}\pojem{težišče +trikotnika}. + +Težišče trikotnika v fizičnem smislu predstavlja točko, ki je +središče mase tega trikotnika. To bo še bolj jasno, ko bomo +v razdelku \ref{odd8PloTrik} dokazali dejstvo, da težišče + deli trikotnik na trikotnike z enako ploščino. + V naslednjem poglavju \ref{pogVEKT} (razdelek \ref{odd5TezVeck}) +bomo obravnavali težišče poljubnega večkotnika. + +V razdelku \ref{odd3PravilniVeck} smo ugotovili, da za vsak +pravilni večkotnik obstajata očrtana krožnica (ki vsebuje vsa njegova +oglišča) in včrtana krožnica (ki se dotika vseh njegovih stranic) z +istim središčem. Ta lastnost se prenaša tudi na pravilne oz. +enakostranične trikotnike. Kako pa je s poljubnim trikotnikom? +Dokazali bomo, da obstajata omenjeni krožnici za poljuben trikotnik, le +da imata v splošnem primeru različni središči. + + + \bizrek \label{SredOcrtaneKrozn} + The perpendicular bisectors of the sides of any triangle intersect at a single point, + which is the centre of a circle containing all its vertices. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.2.pic} +\caption{} \label{sl.skl.3.7.2.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Naj bodo $p$, $q$ in $r$ simetrale stranic $BC$, $AC$ in $AB$ + trikotnika $ABC$ (Figure \ref{sl.skl.3.7.2.pic}). + Simetrali $p$ +in $q$ nista vzporedni (ker bi bili v tem primeru po Playfairjevem aksiomu +\ref{Playfair} vzporedni tudi premici $BC$ in $AC$) in se +sekata v neki točki $O$. Ker le-ta leži na simetralah $p$ in $q$ +stranic $BC$ in $AC$, je $OB \cong OC$ in $OC \cong OA$. Iz tega +sledi $OA \cong OB$, kar pomeni, da točka $O$ leži tudi na +simetrali $r$ daljice $AB$. Torej se simetrale $p$, $q$ in $r$ +sekajo v eni točki. + +Ker je $OA \cong OB \cong OC$, je točka $O$ središče krožnice +$k(O,OA)$, ki vsebuje vsa oglišča trikotnika $ABC$. + \kdokaz + + Krožnico iz prejšnjega izreka, ki vsebuje vsa oglišča trikotnika, + imenujemo \index{očrtana krožnica!trikotnika} \pojem{trikotniku očrtana + krožnica}, njeno središče pa + \index{središče!očcrtane krožnice!trikotnika} + \pojem{središče trikotniku očrtane krožnice}. + + + + \bizrek \label{SredVcrtaneKrozn} + The bisectors of the interior angles of any triangle intersect at a single point, + which is the centre of a circle touching all its sides. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.3.pic} +\caption{} \label{sl.skl.3.7.3.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Naj bodo $p$, $q$ in $r$ simetrale notranjih kotov pri + ogliščih $A$, $B$ in $C$ + trikotnika $ABC$ (Figure \ref{sl.skl.3.7.3.pic}). + Dokažimo, da simetrali $p$ in $q$ nista vzporedni. V nasprotnem + bi bila po izreku \ref{KotiTransverzala} vsota polovic + notranjih kotov pri ogliščih $A$ in $B$ + enaka $180°$, kar po izreku \ref{VsotKotTrik} ni mogoče. Torej se +$p$ in $q$ sekata v neki točki $S$. Ker točka $S$ leži na +simetralah notranjih kotov pri ogliščih $A$ in $B$, je enako +oddaljena od nosilk stranic $AC$ in $AB$ oz. od nosilk stranic $BA$ +in $BC$ (izrek \ref{SimKotaKraka}). Iz tega sledi, da je točka $S$ +enako oddaljena od nosilk stranic $CA$ in $CB$, kar pomeni, da leži +na simetrali $r$ notranjega kota pri oglišču $C$. Torej se simetrale +$p$, $q$ in $r$ sekajo v točki $S$. + +S $P$, $Q$ in $R$ označimo pravokotne projekcije točke $S$ na +stranicah $BC$, $CA$ in $AB$. Ker velja $\frac{1}{2}\angle CBA<90^0$ +in $\frac{1}{2}\angle BCA<90^0$, je $\mathcal{B}(B,P,C)$. Podobno je +tudi $\mathcal{B}(C,Q,A)$ in $\mathcal{B}(A,R,B)$. Zaradi že +dokazane lastnosti točke $S$ velja $SP \cong SQ \cong SR$. Torej je +točka $S$ središče krožnice $l$, ki poteka skozi točke $P$, $Q$ in +$R$. Zaradi pravokotnosti polmerov $SP$, $SQ$ in $SR$ na ustrezne +stranice, so le-te tangente krožnice $l$. Ker je +$\mathcal{B}(B,P,C)$, $\mathcal{B}(C,Q,A)$ in $\mathcal{B}(A,R,B)$ +se krožnica $l$ dotika vseh + stranic trikotnika $ABC$. + \kdokaz + + + Krožnico iz prejšnjega izreka, ki se dotika vseh + stranic trikotnika, imenujemo \index{včrtana krožnica!trikotnika} \pojem{trikotniku včrtana + krožnica}, njeno središče pa + \index{središče!včcrtane krožnice!trikotnika} + \pojem{središče trikotniku včrtane krožnice}. + +Ostala je še ena od štirih omenjenih značilnih točk trikotnika. +Nanaša se na višine trikotnika. + + + \bizrek \label{VisinskaTocka} + The lines containing the altitudes of a triangle intersect at a single point. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.4.pic} +\caption{} \label{sl.skl.3.7.4.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Naj bodo $p$, $q$ in $r$ nosilke višin + $AA'$, $BB'$ in $CC'$ + trikotnika $ABC$ (Figure \ref{sl.skl.3.7.4.pic}). +Označimo z $a$, $b$ in $c$ premice, ki so v točkah $A$, $B$ in $C$ +pravokotne na ustrezne višine. Ker so premice $a$, $b$ in $c$ +vzporedne s stranicama trikotnika $ABC$, se vsaki dve med seboj +sekata. Označimo s $P$, $Q$ in $R$ po vrsti presečišča premic $b$ in +$c$, $a$ in $c$ ter $a$ in $b$. Štirikotnika $ABCQ$ in $RBCA$ sta + paralelograma, kar pomeni, da je $RA \cong BC \cong AQ$, oz. je točka +$A$ središče daljice $RQ$. Premica $AA’$ je zato simetrala +stranice $RQ$ trikotnika $PQR$. Analogno sta $BB’$ in $CC’$ +simetrali stranic $PR$ in $PQ$ istega trikotnika. Po izreku +\ref{SredOcrtaneKrozn} se simetrale $AA’$, $BB’$ in $CC’$ trikotnika +$PQR$ sekajo v neki točki $V$. Točka $V$ je torej presečišče +nosilk višin + $AA'$, $BB'$ in $CC'$ + trikotnika $ABC$. + \kdokaz + +Točko iz prejšnjega izreka, v kateri se sekajo nosilke višin, imenujemo +\index{višinska točka trikotnika} + \pojem{višinska točka trikotnika}. Trikotnik $A'B'C'$, ki ga + določajo nožišča višin trikotnika $ABC$ imenujemo +\index{trikotnik!pedalni} \pojem{pedalni trikotnik} trikotnika +$ABC$. + +Ugotovili smo, ima vsak trikotnik štiri značilne točke, +in sicer: težišče, središče očrtane krožnice, središče včrtane +krožnice in višinsko točko. Toda to niso edine značilne točke +trikotnika. Nekatere od njih bomo omenili kasneje. Za +neko točko v ravnini trikotnika na splošno rečemo, da je njegova +\pojem{značilna točka}, če je njena definicija simetrična glede na +oglišča tega trikotnika. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.5.pic} +\caption{} \label{sl.skl.3.7.5.pic} +\end{figure} + + + Jasno je, da se pri poljubnem trikotniku štiri značilne točke razlikujejo + (Figure \ref{sl.skl.3.7.5.pic}). + + Pri enakokrakem trikotniku imajo težiščnica, +višina, simetrala osnovnice in simetrala notranjega kota nasproti +osnovnice isto nosilko. Če je $A_1$ središče osnovnice +$BC$ enakokrakega trikotnika $ABC$, sta trikotnika $ABA_1$ in +$ACA_1$ skladna, kar pomeni, da je kot pri oglišču $A_1$ pravi kot +in sta kota $BAA_1$ in $CAA_1$ skladna. Torej je daljica +$AA_1$ hkrati težiščnica in višina, premica $AA_1$ pa hkrati +simetrala stranice $BC$ in simetrala notranjega kota pri oglišču $A$ +trikotnika $ABC$. Iz tega sledi, da vse štiri značilne točke tega +trikotnika ležijo na eni premici $AA_1$ (Figure +\ref{sl.skl.3.7.6.pic}). + +Če uporabimo že dokazano lastnost enakokrakega trikotnika za +enakostranični trikotnik, ugotovimo, da imajo pri njem vse ustrezne +težiščnice, višine, simetrale stranic in simetrale notranjih kotov +iste nosilke. To pomeni, da se pri enakostraničnem trikotniku +vse štiri značilne točke prekrivajo (Figure \ref{sl.skl.3.7.6.pic}). +To je pravzaprav že definirano (razdelek \ref{odd3PravilniVeck}) +središče tega enakostraničnega (oz. pravilnega) trikotnika. + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.6.pic} +\caption{} \label{sl.skl.3.7.6.pic} +\end{figure} + +Pokazali bomo še, kakšno lego imajo značilne točke glede na vrsto +trikotnika. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.7.pic} +\caption{} \label{sl.skl.3.7.7.pic} +\end{figure} + +Težiščnice so vedno v notranjosti trikotnika. Zato je tudi težišče +notranja točka vsakega trikotnika (Figure \ref{sl.skl.3.7.7.pic}). +Ista ugotovitev velja tudi za središče včrtane krožnice (Figure +\ref{sl.skl.3.7.7s.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.7s.pic} +\caption{} \label{sl.skl.3.7.7s.pic} +\end{figure} + +Pri ostrokotnem trikotniku nožišča pravokotnic iz njegovih oglišč +ležijo na stranicah tega trikotnika, kar pomeni (Paschev aksiom +\ref{AksPascheva}), da se njegove višine sekajo v notranjosti. Torej pri +ostrokotnem trikotniku leži višinska točka v njegovi notranjosti +(Figure \ref{sl.skl.3.7.7v.pic}). Pri pravokotnem trikotniku je višinska +točka oglišče pri pravemu kotu. To je zato, ker sta njegovi +kateti hkrati višini trikotnika. Višinska točka topokotnega +trikotnika leži v njegovi zunanjosti, ker + niso vse višine v njegovi notranjosti. Ustrezna nožišča +pripadajo namreč nosilkam stranic, ne pa samim stranicam. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.7v.pic} +\caption{} \label{sl.skl.3.7.7v.pic} +\end{figure} + +Središče očrtane krožnice je notranja oz. zunanja točka trikotnika, +odvisno od tega, ali je trikotnik ostrokotni oz. topokotni (Figure +\ref{sl.skl.3.7.7o.pic}). Formalni dokaz tega dejstva bomo +izpustili. Dokažimo le naslednji izrek, ki se nanaša na pravokotne +trikotnike. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.7o.pic} +\caption{} \label{sl.skl.3.7.7o.pic} +\end{figure} + + + \bizrek The circumcentre of a right-angled triangle is at the same time the midpoint of + its hypotenuse. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.8.pic} +\caption{} \label{sl.skl.3.7.8.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Označimo z $O$ središče hipotenuze $AB$ in +s $P$ središče katete $AC$ pravokotnega trikotnika $ABC$ (Figure +\ref{sl.skl.3.7.8.pic}). Daljica $OP$ je srednjica tega trikotnika, +ki ustreza kateti $BC$, zato je $OP\parallel BC$. Iz tega sledi $OP +\perp AC$. Torej sta trikotnika $OPC$ in $OPA$ skladna (izrek +\textit{SAS} \ref{SKS}) in je potem $OC \cong OA$. +Ker je še $OB \cong OA$, + je točka $O$ središče očrtane krožnice tega trikotnika. + \kdokaz + +Daljica $OC$ iz prejšnjega izreka je težiščnica trikotnika. To +pomeni, da je težiščnica na hipotenuzo pravokotnega +trikotnika enaka polmeru očrtane krožnice tega trikotnika, hkrati pa +tudi polovici njegove hipotenuze. + +Prejšnji izrek je povezan tudi s Talesovim izrekom za krožnico +\ref{TalesovIzrKroz} in njegovim obratnim izrekom +\ref{TalesovIzrKrozObrat}. Vse omenjene trditve bomo sedaj podali v +enem izreku. + + + \bizrek Thales’ theorem for a circle (several forms - Figure + \ref{sl.skl.3.7.9.pic}): + \index{izrek!Talesov za krožnico} + \label{TalesovIzrKroz2} + \begin{enumerate} + \item The circumcentre of a right-angled triangle is at the same time the midpoint of + its hypotenuse. + \item If $t_c$ is the median of a right-angled triangle for its hypotenuse $c$ and $R$ + the circumradius of that triangle, then + $R=t_c=\frac{c}{2}$. + \item If $AB$ is a diameter of a circle $k$, then for any point $X\in k$ ($X\neq A$ and $X\neq B$) is + $\angle AXB=90^0$. + \item If $A$, $B$ in $X$ are three non-collinear points, such that $\angle AXB=90^0$, then the point $X$ lies on a circle with the diameter $AB$. + \end{enumerate} + \index{izrek!Talesov za krožnico} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.9.pic} +\caption{} \label{sl.skl.3.7.9.pic} +\end{figure} + +Poznavanje značilnih točk trikotnika in lastnosti srednjice +trikotnika omogočata dokazovanje raznih drugih lastnosti tako +trikotnika kot tudi štirikotnika in $n$-kotnika. + + + \bzgled + Let $CD$ be the altitude at the hypotenuse $AB$ of a right-angled triangle $ABC$. + If $M$ and $N$ are the midpoints of the line segments $CD$ and $BD$, then $AM \perp CN$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.10.pic} +\caption{} \label{sl.skl.3.7.10.pic} +\end{figure} + +\textbf{\textit{Proof.}} Daljica $NM$ je srednjica trikotnika +$BCD$, zato je po izreku \ref{srednjicaTrik} $NM \parallel BC$ +(Figure \ref{sl.skl.3.7.10.pic}). Ker je kot pri oglišču $C$ pravi +kot, je tudi $NM \perp AC$. Zaradi tega je premica $NM$ nosilka +višine trikotnika $ANC$, Ker je še $CD$ višina tega trikotnika, je + $M$ njegova višinska točka. Torej je premica $AM$ nosilka +tretje višine tega trikotnika in velja $AM \perp CN$. + \kdokaz + + + \bzgled + If $P$ and $Q$ are the midpoints of the sides $BC$ and $CD$ of a parallelogram $ABCD$, + then the lines $AP$ and $AQ$ + divide the diagonal $BD$ of this parallelogram into three congruent line segments. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.11.pic} +\caption{} \label{sl.skl.3.7.11.pic} +\end{figure} + +\textbf{\textit{Proof.}} Označimo z $E$ in $F$ presečišči daljic +$AP$ in $AQ$ z diagonalo $BD$ paralelograma $ABCD$ ter s $S$ +presečišče njegovih diagonal $AC$ in $BD$ (Figure +\ref{sl.skl.3.7.11.pic}). Diagonali paralelograma se razpolavljata +(izrek \ref{paralelogram}), zato je točka $S$ skupno središče daljic +$AC$ in $BD$. To pomeni, da sta točki $E$ in $F$ težišči trikotnikov +$ACB$ in $ACD$, zato delita težiščnici $SB$ in $SD$ teh trikotnikov +v razmerju $2:1$ (izrek \ref{tezisce}). Torej: + \begin{eqnarray*} + BE &=& \frac{2}{3}BS = \frac{2}{3}DS = FD,\\ + EF&=& ES+ SF= \frac{1}{3}SB+ \frac{1}{3}SD= + \frac{1}{3}(SB +SD)=\frac{1}{3} BD, + \end{eqnarray*} + kar smo želeli dokazati. \kdokaz + + + \bzgled + Let $BAKL$ and $ACPQ$ be positively oriented squares + in the same plane. Prove that the lines $BP$ and $CL$ intersect at a point lying + on the line containing the altitude $AA'$ of the triangle $ABC$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.12.pic} +\caption{} \label{sl.skl.3.7.12.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Naj bo $AA'$ + višina trikotnika $ABC$ (Figure \ref{sl.skl.3.7.12.pic}). Označimo z $X$ + in $Y$ presečišči premice $AA'$ s +pravokotnicama na premico $CL$ skozi točko $B$ in na premico $BP$ skozi točko $C$. +Dokažimo da je $X = Y$. Trikotnika $BLC$ in $ABX$ sta po izreku \textit{ASA} \ref{KSK} skladna, ker je: $BL \cong AB$, +$\angle BLC\cong\angle ABX$ in $\angle BCL\cong\angle AXB$ (kota s pravokotnima krakoma - +izrek \ref{KotaPravokKraki}). Zaradi tega je $AX \cong BC$. Analogno +sta skladna tudi trikotnika $CPB$ in $ACY$, zato je $AY \cong BC$. +Torej velja $AX \cong AY$ oziroma $X = Y$. To pomeni, da so premice +$AA'$, $BP$ in $CL$ nosilke višin trikotnika $XBC$, zato se sekajo v +eni točki. + \kdokaz + + + \bzgled \label{zgledPravokotnik} + Let $K$ be the midpoint of the side $CD$ of a rectangle $ABCD$. + A point $L$ is the foot of the perpendicular from the vertex $B$ on the diagonal + $AC$ and $S$ is the midpoint of + the line segment $AL$. Prove that $\angle KSB$ is a right angle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.13.pic} +\caption{} \label{sl.skl.3.7.13.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Naj bo $V$ središče daljice $BL$ (Figure \ref{sl.skl.3.7.13.pic}). + Daljica $SV$ je srednjica + trikotnika $ABL$ za osnovnico $AB$, zato je $SV\parallel AB$ in + $SV =\frac{1}{2} AB$ (izrek \ref{srednjicaTrik}). Iz + prve relacije in $BC\perp AB$ sledi $SV\perp BC$ + (izrek \ref{KotiTransverzala}). To pomeni, da sta + $BL$ in $SV$ nosilki višin trikotnika $CSB$. Torej je $V$ višinska + točka tega trikotnika, zato je $CV$ nosilka njegove tretje višine + (izrek \ref{VisinskaTocka}) oz. velja $CV\perp SB$. Iz $SV\parallel AB$ in + $SV =\frac{1}{2} AB=KC$ sledi, da je štirikotnik $SVCK$ + paralelogram oz. $CV\parallel SK$. Iz tega in $CV\perp SB$ na + koncu sledi (izrek \ref{KotiTransverzala}) $SK\perp SB$, torej je +$\angle KSB$ pravi kot. + \kdokaz + + + + \bzgled + Let $AP$, $BQ$ and $CR$ be the altitude, the median + and the bisector of the angle $ACB$ ($R\in AB$) of a triangle $ABC$. Prove that if + the triangle $PQR$ is regular, then the triangle $ABC$ is also regular. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.14.pic} +\caption{} \label{sl.skl.3.7.14.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $PQR$ pravilni trikotnik oz. +$PQ\cong QR\cong RP$ (Figure \ref{sl.skl.3.7.14.pic}). Točka $Q$ je +središče hipotenuze $AC$ pravokotnega trikotnika $APC$, zato je $QA +\cong QC \cong QP$ (izrek \ref{TalesovIzrKroz2}. Ker je v tem +primeru tudi $QR\cong QC\cong QP$, iz istega izreka sledi, da je tudi +$\angle ARC$ pravi kot. Iz skladnosti trikotnikov $ACR$ in $BCR$ +(izrek \ref{KSK}) dobimo, da je točka $R$ središče stranice $AB$ in +da velja $AC\cong BC$. Ker je točka $R$ središče hipotenuze $AB$ +pravokotnega trikotnika $APB$, je $AB = 2RP = 2PQ = 2AQ = AC$. +Torej velja $AB\cong AC\cong BC$, kar pomeni, da je tudi $ABC$ +pravilni trikotnik. + \kdokaz + + +Ni težko dokazati, da je nek trikotnik enakokrak natanko tedaj, ko +sta ustrezni težiščnici skladni. Analogno velja tudi za višine. Ali +podobno velja tudi za t. i. odseke simetral kotov? + Daljici $BB'$ in $CC'$, kjer sta $BB'$ in $CC'$ simetrali notranjih kotov + trikotnika $ABC$ ter $B'\in AC$ in $C'\in AB$, imenujemo + \index{odsek simetrale kota} \pojem{odseka simetral kotov}. Odseke simetral ozačimo z $l_a$, $l_b$ in $l_c$. + Omenjena + trditev velja tudi v tem primeru, ampak +dokaz ni tako enostaven. O tem govori naslednji znani izrek. + + + + \bizrek \index{izrek!Steiner-Lemusov} + (Steiner-Lehmus\footnote{\textit{D. C. L. Lehmus} (1780--1863),\index{Lehmus, D. C. L.} francoski + matematik, ki je leta 1840 poslal to, na prvi pogled enostavno + trditev, slavnem švicarskemu geometru \index{Steiner, J.} \textit{J. Steinerju} (1796--1863), + ki je izpeljal zelo obsežen dokaz tega izreka. Nato je sledilo + več različnih rešitev tega problema in eno od njih je leta 1908 + objavil + francoski matematik \index{Poincar\'{e}, J. H.} \textit{J. H. Poincar\'{e}} (1854--1912).}) + Let $BB'$ ($B' \in AC$) and $CC'$ ($C' \in AB$) be the bisectors + of the interior angles of a triangle $ABC$. Then: + $$AB \cong AC \Leftrightarrow BB'\cong CC'.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.15.pic} +\caption{} \label{sl.skl.3.7.15.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.7.15.pic}) + +($\Rightarrow$) Iz $AB \cong AC$ sledi $\angle ABC \cong \angle +ACB$ (izrek \ref{enakokraki}) oz. $\angle B'BC \cong \angle C'CB$. +Po izreku \textit{ASA} \ref{KSK} sta trikotnika $B'BC$ in $C'CB$ +skladna, zato je $BB'\cong CC'$. + +($\Leftarrow$) Naj bo $BB'\cong CC'$. + Predpostavimo, da ni $AB\cong AC$. Brez škode za splošnost + naj bo $AB < AC$. V tem primeru je $\angle ACB < \angle ABC$ + (izrek \ref{vecstrveckot}) oz. +$\angle ACC'< \angle ABB'$. To pomeni, da v notranjosti kota $ABB'$ +obstaja poltrak $p$ z izhodiščem $B$, ki stranico $AC$ seka v +takšni točki $D$, da hkrati velja $\mathcal{B}(A,D,B')$ in $\angle +DBB'\cong \angle ACC'$. V trikotniku $BCD$ je $\angle ACB < \angle +DBC$ in zaradi tega tudi $BD < CD$ (izrek \ref{vecstrveckot}). +Torej obstaja takšna točka $E$, ki je med točkama $C$ in $D$, tako da je +$BD \cong CE$. Po izreku \textit{SAS} \ref{SKS} sta trikotnika +$BDB'$ in $CEC'$ skladna, zato sta skladna tudi kota $BDB'$ in +$CEC'$. Dokažimo, da to ni mogoče. Zaradi Paschevega aksioma +\ref{AksPascheva} (uporabljenega za trikotnik $AC'E$ in premico +$BD$) premica $BD$ seka daljico $C'E$ v neki točki $S$. V trikotniku +$SDE$ je kot $SEC$ (oz. kot $CEC'$) zunanji in po izreku +\ref{zunanjiNotrNotrVecji} ne more biti skladen nesosednjemu notranjemu +kotu $SDE$ (oz. kotu $BDB'$). To pomeni, da predpostavka $AB < AC$ +(analogno tudi $AB > AC$) ni mogoča. Torej je $AB\cong AC$. + \kdokaz + + + \bzgled + Let $A_1$ be the midpoint of the side $BC$ of a triangle $ABC$. + Calculate the measure of the angle $AA_1C$, if + $\angle BAC=45^0$ and $\angle ABC=30^0$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.1a.pic} +\caption{} \label{sl.skl.3.7.1a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $CC'$ višina trikotnika $ABC$ (Figure +\ref{sl.skl.3.7.1a.pic}). + +Iz $\angle CC'B = 90^0$ sledi najprej +$\angle C'CB=60^0$, nato pa še, da točka $C'$ leži na krožnici nad +premerom $CB$ in s središčem $A_1$ (izrek \ref{TalesovIzrKroz}), zato +je $A_1C'\cong A_1C\cong A_1B$. Torej je trikotnik $CC'A_1$ +enakokrak, oz. po izreku \ref{enakokraki} velja $\angle CC'A_1 +\cong\angle C'CB=60^0$. To pomeni, da je trikotnik $CC'A_1$ +pravilen in je $C'C\cong C'A_1$. Iz dejstva, da je $AC'C$ enakokraki +trikotnik ($\angle CAC'=\angle ACC'=45^0$), pa sledi $AC'\cong C'C$. Če to povežemo s prejšnjo relacijo, dobimo $AC'\cong C'A_1$, kar +pomeni, da je tudi trikotnik $AC'A_1$ enakokrak. Zato je (izreka +\ref{enakokraki} in \ref{zunanjiNotrNotr}): + $$\angle C'A_1A\cong\angle C'AA_1=\frac{1}{2}\angle A_1C'B= + \frac{1}{2}\angle C'BA_1=\frac{1}{2}\cdot 30^0=15^0.$$ + Na koncu je še: + $$\angle AA_1C=\angle C'A_1C-\angle C'A_1A =60^0-15^0=45^0,$$ kar je bilo potrebno izračunati. \kdokaz + + + \bzgled + Let $P$ be the midpoint of the side $BC$ of an isosceles triangle $ABC$ + and $Q$ the foot of the perpendicular from the point $P$ on the leg + $AC$ of that triangle. Let $S$ be the midpoint of the line segment $PQ$. Prove that + $AS \perp BQ$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.16.pic} +\caption{} \label{sl.skl.3.7.16.pic} +\end{figure} + +\textbf{\textit{Proof.}} Iz skladnosti trikotnikov $ABP$ in $ACP$ +(izrek \textit{SSS} \ref{SSS}) sledi skladnost sokotov $APB$ in +$APC$ oz. $AP\perp BC$ (Figure \ref{sl.skl.3.7.16.pic}). Označimo z +$R$ središče daljice $QC$. Daljica $SR$ je srednjica trikotnika +$QPC$ za osnovnico $PC$, zato je po izreku \ref{srednjicaTrik} +$SR\parallel CP$. Iz tega in $AP\perp BC$ sledi $SR\perp AP$. Torej je +$S$ višinska točka trikotnika $APR$, zato je po izreku +\ref{VisinskaTocka} tudi $AS\perp PR$. Toda daljica $PR$ je +srednjica trikotnika $BQC$ za osnovnico $BQ$, zato je $PR\parallel BQ$ +(izrek \ref{srednjicaTrik}). Iz $AS\perp PR$ in $PR\parallel BQ$ +dobimo $AS\perp BQ$. + \kdokaz + + + \bzgled \label{kotBSC} + If $S$ is the incentre and $\alpha$, $\beta$, + $\gamma$ the interior angles at the vertices + $A$, $B$, $C$ of a triangle + $ABC$, then + $$\angle BSC=90^0+\frac{1}{2}\cdot\alpha.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.1c.pic} +\caption{} \label{sl.skl.3.7.1c.pic} +\end{figure} + +\textbf{\textit{Proof.}} Po izreku \ref{SredVcrtaneKrozn} se +simetrale notranjih kotov trikotnika $ABC$ sekajo v središču včrtane +krožnice - v točki $S$ (Figure \ref{sl.skl.3.7.1c.pic}). Zato je +$\angle SBC =\frac{1}{2}\cdot \beta$ in $\angle SCB +=\frac{1}{2}\cdot \gamma$. Ker je po izreku \ref{VsotKotTrik} v vsakem +trikotniku vsota notranjih kotov enaka $180^0$, sledi: + $$\angle BSC = 180^0-\frac{1}{2}\cdot\left( \beta+ + \gamma\right)=180^0-\frac{1}{2}\cdot\left( 180^0- + \alpha\right)=90^0+\frac{1}{2}\cdot\alpha,$$ kar je bilo treba dokazati. \kdokaz + + + \bzgled + Construct a triangle with given $a$, $t_a$, $R$ (see the labels in section \ref{odd3Stirik}). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.16a.pic} +\caption{} \label{sl.skl.3.7.16a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Konstrukcijo lahko izpeljemo tako, da najprej narišemo očrtano krožnico $k(O,R)$, izberemo poljubno točko $B\in k$, načrtamo tetivo $BC\cong a$ te krožnice, središče $A_1$ tetive $BC$ in na koncu oglišče $A$ kot presečišče krožnic $k(O,R)$ in $k_1(A_1,t_a)$ (Figure \ref{sl.skl.3.7.16a.pic}). Jasno je, da ima naloga rešitve natanko tedaj, ko je $a\leq 2R$ in presečišče krožnic $k(O,R)$ in $k_1(A_1,t_a)$ ni prazna množica. Število rešitev je v tem primeru odvisno od števila presečišč krožnic $k(O,R)$ in $k_1(A_1,t_a)$. + \kdokaz + + \bzgled + Construct a right-angled triangle if the hypotenuse and the altitude to that hypotenuse + are congruent to the given line segments $c$ and $v_c$. + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.16b.pic} +\caption{} \label{sl.skl.3.7.16b.pic} +\end{figure} + +\textbf{\textit{Analysis.}} +Naj bo $ABC$ pravokotni trikotnik s pravim kotom v oglišču $C$, +pri katerem sta hipotenuza $AB$ in višina $CC'$ skladni z daljicama $c$ in $v_c$ (Figure \ref{sl.skl.3.7.16b.pic}). +Po izreku \ref{TalesovIzrKroz2} je središče $O$ hipotenuze $AB$ hkrati središče očrtane krožnice trikotnika $ABC$. +Torej oglišče $C$ leži na krožnici $k$ s premerom $AB$. Ker je še $CC'\cong v_c$, leži oglišče $C$ tudi na vzporednici premice $AB$, +ki je od nje oddaljena za $v_c$. Točka $C$ je potem presečišče te vzporednice in krožnice $k$. + + +\textbf{\textit{Construction.}} +Najprej načrtajmo daljico $AB$, ki je skladna z dano daljico $c$, nato še središče $O$ daljice $AB$ in krožnico $k(O,OA)$. Potem načrtajmo vzporednico $p$ premice $AB$ na razdalji $v_c$. Eno od presečišč premice $p$ in krožnice $k(O,OA)$ označimo s $C$. Dokažimo, da je $ABC$ iskani trikotnik. + + +\textbf{\textit{Proof.}} + Po konstrukciji točka $C$ leži na krožnici s polmerom $AB$, zato je po izreku \ref{TalesovIzrKroz2} $\angle ACB=90^0$, kar pomeni, da je $ABC$ pravokotni trikotnik s hipotenuzo $AB$. Po konstrukciji je le-ta skladna z daljico $c$. Naj bo $CC'$ višina trikotnika $ABC$. Po konstrukciji leži točka $C$ na premici $p$, ki je od premice $AB$ oddaljena $v_c$, zato je tudi $|CC'|=d(C,AB)=d(p,AB)$ oz. $CC'\cong v_c$. + + +\textbf{\textit{Discussion.}} + Število rešitev je odvisno od števila presečišč premice $p$ in krožnice $k(O,OA)$. + \kdokaz + + + + \bnaloga\footnote{2. IMO Romania - 1960, Problem 4.} + Construct triangle ABC, given $v_a$, $v_b$ (the altitudes from $A$ and $B$) and $t_a$, + the median from vertex $A$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.7.IMO1.pic} +\caption{} \label{sl.skl.3.7.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Naj bo $ABC$ takšen trikotnik, da sta +$AA' \cong v_a$ in $BB' \cong v_b$ njegovi višini ter $AA_1\cong t_a$ +njegova težiščnica (Figure \ref{sl.skl.3.7.IMO1.pic}). Označimo +z $A'_1$ pravokotno projekcijo točke $A_1$ na premici $AC$. +Daljica $A_1A'_1$ je srednjica trikotnika $BB'C$ za osnovnico +$BB'$, zato je po izreku \ref{srednjicaTrik}: + $$|A_1A'_1|=\frac{1}{2}\cdot|BB'|=\frac{1}{2}\cdot v_b + \hspace*{1mm} \textrm{ in } \hspace*{1mm} A_1A'_1 +\parallel BB'.$$ Iz tega sledi, da sta premici +$A_1A'_1$ in $AC$ pravokotni v točki $A'_1$, torej je premica $AC$ +tangenta krožnice $k(A_1,\frac{1}{2} v_b)$ \ref{TangPogoj}. +Dokazane lastnosti nam omogočajo konstrukcijo. + + Najprej lahko načrtamo pravokotni trikotnik $AA'A_1$ ($AA'\cong v_a$, + $AA_1\cong t_a$ in $\angle AA'A_1 = 90^0$), nato pa krožnico + $k(A_1,\frac{1}{2} v_b)$. Iz točke $A$ načrtamo tangenti na + krožnico $k(A_1,\frac{1}{2} v_b)$. Presečišče ene od tangent + s premico $A'A_1$ označimo s $C$. Na koncu načrtamo takšno točko + $B$, da velja + $BA_1 \cong CA_1$ in $\mathcal{B}(C,A_1,B)$. + + Dokažimo, da trikotnik $ABC$ izpolnjuje dane pogoje. Iz + konstrukcije je $AA'\cong v_a$ višina in $AA_1 \cong t_a$ + težiščnica (ker je $A_1$ središče daljice $BC$) trikotnika + $ABC$. Naj bo $BB'$ višina tega trikotnika. Dokažimo še $BB' \cong + v_b$. + Premica $AC$ je po konstrukciji tangenta krožnice $k(A_1,\frac{1}{2} + v_b)$. Njuno dotikališče označimo z $A'_1$. Po izreku + \ref{TangPogoj} sta premici +$A_1A'_1$ in $AC$ pravokotni v točki $A'_1$, zato je daljica +$A_1A'_1$ srednjica trikotnika $BB'C$ za osnovnico $BB'$ in velja +$|BB'|= 2\cdot |A_1A'_1|=2\cdot\frac{1}{2}\cdot v_b=v_b$. + + Naloga nima rešitev, kadar je $v_a>t_a$. Če je $v_a\leq + t_a$, je število rešitev odvisno od števila tangent, ki jih + lahko načrtamo iz točke $A$ na krožnico $k(A_1,\frac{1}{2} + v_b)$. Pri tem mora tangenta sekati premico $A'A_1$. + Če velja $\frac{1}{2} v_b4. \label{teselRelHyp} +\end{eqnarray} + Ta neenačba ima neskončno mnogo rešitev v množici $\mathbb{N}^2$, kar pomeni, +da imamo v hiperbolični geometriji neskončno mnogo pravilnih tlakovanj. +Dve od njih sta npr. $(3,7)$ in $(4,5)$ (Figure +\ref{sl.skl.3.9.2H.pic}\footnote{http://math.slu.edu/escher/index.php/Category:Hyperbolic-Tessellations}). +Pri slednji se pet kvadratov stika okrog enega oglišča. To je +mogoče, ker je v hiperbolični geometriji notranji kot pri kvadratu vedno +oster in ni konstanten. Izkaže se, da ima kvadrat z daljšo stranico +manjši notranji kot. Mogoče je izbrati takšno stranico kvadrata, +da je notranji kot enak $\frac{360^0}{5} =72^0$, kar ravno +ustreza tlakovanju $(4,5)$. + +\begin{figure}[!htb] +\centering +\includegraphics[width=0.413\textwidth]{whyptess1.eps}\hspace*{4mm} + \includegraphics[width=0.387\textwidth]{whyptess.eps} +\caption{} \label{sl.skl.3.9.2H.pic} +\end{figure} + +V eliptični geometriji, kjer je vsota kotov v trikotniku vedno +večja od $180^0$, omenjena relacija za trikotnik $OSB$ postane: +$\frac{360^0}{2n}+\frac{360^0}{2m}+90^0>180^0$, oz.: + \begin{eqnarray} +(n-2)(m-2)<4. \label{teselRelElipt} +\end{eqnarray} +Ta enačba ima v množici $\mathbb{N}^2$ rešitve $(3,3)$, $(4,3)$, +$(3,4$), $(5,3)$ in $(3,5)$. Ker se eliptična geometrija realizira +kot model na sferi, te rešitve pomenijo tlakovanja sfere s sfernimi +večkotniki. Stranice teh večkotnikov so loki velikih krožnic sfere. +Če v evklidskem prostoru z daljicami povežemo ustrezna oglišča teh +tlakovanj, dobimo t. i. \index{pravilni!poliedri} \pojem{pravilne +poliedre} (Figure +\ref{sl.skl.3.9.2E.pic}\footnote{http://www.upc.edu/ea-smi/personal/claudi/web3d/}): +\pojem{pravilni tetraeder}, \pojem{kocko} (oz. \pojem{pravilni +heksaeder}), \pojem{pravilni oktaeder}, \pojem{pravilni dodekaeder} +in \pojem{pravilni ikozaeder}. Npr. $(4,3)$ bi predstavljal kocko, +pri kateri se po trije kvadrati (pravilna 4-kotnika) stikajo v eni +točki. + + +\begin{figure}[!htb] +\centering + \includegraphics[bb=0 0 11cm 6cm]{wpoliedri.eps} +\caption{} \label{sl.skl.3.9.2E.pic} +\end{figure} + + +Vrnimo se nazaj k evklidski ravnini. Če dopustimo možnost, da pri +prekrivanju ravnine uporabimo več (končno mnogo) vrst pravilnih +večkotnikov oz. več vrst celic, ki so enako razporejene v vsakem oglišču, potem razen treh pravilnih +tlakovanj iz izreka \ref{pravilnaTlakovanja} obstaja še osem t. i. +\index{tlakovanja!Arhimedova} \pojem{Arhimedovih tlakovanj} +\footnote{Dokaz te trditve je izvedel nemški astronom, matematik in +fizik \index{Kepler, J.} \textit{J. Kepler} (1571--1630).} (Figure +\ref{sl.skl.3.9.2A.pic}\footnote{http://commons.wikimedia.org/wiki/File\%3AArchimedean-Lattice.png}). + + + +\begin{figure}[!htb] +\centering + \includegraphics[bb=0 0 10cm 7.7cm]{Archimedean.eps} +\caption{} \label{sl.skl.3.9.2A.pic} +\end{figure} + +Razen pravilnih in Arhimedovih tlakovanj obstajajo tudi druga +tlakovanja z večkotniki, ki niso pravilni. Najbolj enostaven primer +je tlakovanje s skladnimi paralelogrami (Figure +\ref{sl.skl.3.9.3a.pic}), ki ga dobimo, če pravilno +tlakovanje $(4,4)$ tako deformiramo, da se namesto kvadratov v enem oglišču stikajo štirje +paralelogrami. To tlakovanje je določeno z mrežo +dveh šopov vzporednic. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.9.3aa.pic} +\input{sl.skl.3.9.3aaa.pic} +\caption{} \label{sl.skl.3.9.3a.pic} +\end{figure} + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.9.3bb.pic} +\input{sl.skl.3.9.3cc.pic} +\caption{} \label{sl.skl.3.9.3bc.pic} +\end{figure} + +Če vse paralelograme razdelimo na +dva trikotnika z diagonalami, ki imajo isto smer, dobimo tlakovanje ravnine s skladnimi trikotniki +(Figure \ref{sl.skl.3.9.3a.pic}). Trikotnik (osnovna celica) je lahko +poljuben, saj lahko dva takšna (skladna) trikotnika po skupni +stranici vedno povežemo v paralelogram in tako dobimo tlakovanje s +paralelogrami. Tlakovanje s poljubnimi skladnimi trikotniki je +posplošitev pravilnega tlakovanja $(3,6)$ s pravilnimi trikotniki. + + +%\vspace*{-1mm} + +Kot posebna primera tlakovanja s paralelogrami dobimo tlakovanje s +pravokotniki in tlakovanje z rombi (Figure \ref{sl.skl.3.9.3bc.pic}). +Dokazali bomo še bolj splošno +trditev, ki morda ni tako samoumevna. Mogoče je namreč tudi +tlakovanje s poljubnimi skladnimi štirikotniki. + +%\vspace*{-1mm} + + + + \bzgled + Let $ABCD$ be an arbitrary quadrilateral. Prove that it is + possible to tessellate the plane with the cell $ABCD$ so that each vertex + is surrounded by four such quadrilaterals. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.9.4.pic} +\caption{} \label{sl.skl.3.9.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure +\ref{sl.skl.3.9.4.pic}) + Naj bo $ABCD$ poljuben štirikotnik v +ravnini z notranjimi koti $\alpha$, $\beta$, $\gamma$ in $\delta$. +Točki $O$ in $S$ naj bosta središči njegovih stranic $AB$ in $BC$. S središčnima +zrcaljenjema + $\mathcal{S}_O$ in $\mathcal{S}_S$ (za definicijo središčnega zrcaljenja + glej razdelek \ref{odd6SredZrc}) + se štirikotnik $ABCD$ preslika v štirikotnika +$BAC_1D_1$ in $A_2CBD_2$. Pri tem je $\angle ABD_1\cong\alpha$ in +$CBD_2\cong\gamma$, zato je tudi $\angle D_1BD_2\cong\delta$. Ker je +še $BD_1\cong AD$ in $BD_2\cong CD$, obstaja takšna točka $E$, da +sta štirikotnika $D_1ED_2B$ in $ABCD$ skladna. Torej se okrog točke +$B$ stikajo štirje štirikotniki, ki so vsi skladni štirikotniku +$ABCD$. Postopek ‘‘prekrivanja’’ ravnine lahko nadaljujemo, če +uporabljamo središčne simetrije glede na središča stranic +novonastalih štirikotnikov. + \kdokaz + + Torej obstajajo tlakovanja ravnine s poljubnim trikotnikom in poljubnim + štirikotnikom. Jasno je, da za poljubni petkotnik, šestkotnik,... + ta lastnost ne velja. +Za pravilni šestkotnik obstaja pravilno tlakovanje $(6,3)$. Če v +prejšnjem izreku sestavimo po dva ustrezna sosednja štirikotnika, +dobimo prekrivanje ravnine s skladnimi šestkotniki, ki niso nujno +pravilni, so pa vedno središčno simetrični. + + + %_______________________________________________________________________________ + \poglavje{Sets of Points in a Plane. Sylvester's Problem} + \label{odd3Silvester} + +V tem razdelku bomo raziskovali probleme, ki so povezani z množicami +točk v ravnini in s premicami, ki jih te točke določajo. Na začetku bomo +obravnavali nekaj posledic prvih dveh skupin aksiomov (incidence +in urejenosti). + Najprej bomo definirali nove pojme. Naj bo $\mathfrak{T}$ množica + $n$ ($n>2$) točk v ravnini. S $\mathcal{P}(\mathfrak{T})$ označimo + množico vseh premic, od katerih gre vsaka skozi vsaj dve točki iz + množice $\mathfrak{T}$ (Figure +\ref{sl.skl.3.10.1.pic}). Ker množica $\mathfrak{T}$ vsebuje vsaj +dve točki, + iz aksiomov incidence sledi, da je + množica $\mathcal{P}(\mathfrak{T})$ neprazna. Zastavlja se + vprašanje, koliko premic je v množici + $\mathcal{P}(\mathfrak{T})$. + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.1.pic} +\caption{} \label{sl.skl.3.10.1.pic} +\end{figure} + + + + \bizrek \label{stevPremic} + Let $\mathfrak{T}$ be a set of $n$ ($n>2$) points in the plane such that + no three of them are collinear. Then the number of lines of the set $\mathcal{P}(\mathfrak{T})$ is equal to + $$\frac{n(n-1)}{2}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.2.pic} +\caption{} \label{sl.skl.3.10.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} Skozi vsako od $n$ točk iz množice +$\mathfrak{T}$ poteka natanko $n -1$ premic iz množice +$\mathcal{P}(\mathfrak{T})$. Ker na ta način vsako premico štejemo +dvakrat (Figure \ref{sl.skl.3.10.2.pic}), moramo še deliti z 2. Torej je v množici +$\mathcal{P}(\mathfrak{T})$ ravno $\frac{n(n-1)}{2}$ premic. + \kdokaz + +Prejšnji izrek lahko rešimo tudi na naslednji način: skozi prvo +točko gre $n -1$ premic, skozi drugo $n - 2$ premic (ena manj, ker +premice, ki je določena s tema dvema točkama, ne štejemo dvakrat), $n +- 3$ premic skozi tretjo točko in tako naprej vse do ene premice +skozi predzadnjo točko. To je skupaj $(n -1) + (n - 2) +\cdots+1$ +premic. Seveda je to spet enako $\frac{n(n-1)}{2}$. Če vzamemo +$n-1= k$, dobimo znano formulo za vsoto prvih $k$ naravnih števil: + $$1+ 2+ \cdots + k=\frac{k(k+1)}{2}.$$ + + Iz prejšnjega izreka lahko izpeljemo formulo za število diagonal + poljubnega $n$-kotnika. + + + + \bizrek + If $D_n$ is the number of diagonals of an arbitrary $n$-gon, + then + $$D_n=\frac{n( n-3)}{2}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.3.pic} +\caption{} \label{sl.skl.3.10.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} Označimo s $\mathfrak{O}$ množico vseh +oglišč poljubnega $n$-kotnika. + Število diagonal je enako številu vseh premic +iz množice $\mathcal{P}(\mathfrak{O})$ (izreka \ref{stevPremic}) +zmanjšano za število njegovih stranic (Figure +\ref{sl.skl.3.10.3.pic}). Torej: + $$D_n=\frac{n(n- 1)}{2}-n=\frac{n^2-3n}{2}=\frac{n(n-3)}{2},$$ kar je bilo treba dokazati. \kdokaz + +Omenimo še, da bi lahko formulo za število diagonal $n$-kotnika +izpeljali tudi direktno - s podobno obravnavo kot v dokazu izreka +\ref{stevPremic}. Iz vsakega od $n$ oglišč $n$-kotnika lahko +narišemo $n - 3$ diagonal. Na ta način vsako diagonalo štejemo +dvakrat, zato moramo še deliti z 2 in dobimo prejšnjo formulo. + +Izrek \ref{stevPremic} se je nanašal na število premic, ki so +določene z množico točk v ravnini, ki so v takšni legi, da +nobene tri niso kolinearne. V naslednjem primeru bomo preverili, kaj se +zgodi, če dodamo nove pogoje. + + + + \bzgled + Let $\mathfrak{T}$ be a set of $n$ ($n > 2$) points in the plane which are + in such a position that $m$ ($m < n$) of them lies on the same line, but otherwise + no other three points are collinear. What is the number of lines + in the set $\mathcal{P}(\mathfrak{T})$? + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.4.pic} +\caption{} \label{sl.skl.3.10.4.pic} +\end{figure} + +\textbf{\textit{Solution.}} + (Figure \ref{sl.skl.3.10.4.pic}) + + Brez dodatnega pogoja bi v množici $\mathcal{P}(\mathfrak{T})$ + obstajalo $\frac{n(n -1)}{2}$ premic (izrek \ref{stevPremic}). + Dodatni pogoj v nalogi to +število zmanjša, in sicer za $1$ manj kot je število premic, ki jih določa +množica $m$ točk v splošni legi. To je zaradi tega, ker bi sicer +te premice šteli večkrat. Torej je število premic množice +$\mathcal{P}(\mathfrak{T})$ enako: +$$\frac{(n-1)}{2}-\frac{(m -1)}{2}+1.$$ + \kdokaz + +Naslednji enostaven primer bo uvod v zelo zanimiv problem odnosa +množice točk $\mathfrak{T}$ in množice vseh premic +$\mathcal{P}(\mathfrak{T})$, ki jo ta množica točk določa. + + + \bzgled + Construct nine points lying on ten lines + in such a way, that each of those ten lines contains exactly three of these nine points\footnote{\index{Newton, I.} + \textit{I. Newton} (1642--1727), znani angleški + matematik + in fizik, ki je zastavil ta problem v obliki: + ‘‘How can you plant 9 trees in a garden with 10 rows and each row containing exactly 3 trees?’’}. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.5.pic} +\caption{} \label{sl.skl.3.10.5.pic} +\end{figure} + +\textbf{\textit{Proof.}} Ena od možnosti je naslednja. Naj bo $ABCD$ +poljubni pravokotnik, $P$ in $Q$ središči stranic $AB$ in $CD$, $S$ +presečišče diagonal, $K$ presečišče premic $AP$ in $DQ$ ter $L$ +presečišče premic $BP$ in $CQ$ (Figure \ref{sl.skl.3.10.5.pic}). Ker +je $ABCD$ pravokotnik, so točke $P$, $Q$ in $S$ kolinearne. Prav tako +so kolinearne tudi + točke $K$, $L$ in $S$ (točki $K$ in $L$ sta središči pravokotnikov + $AQPD$ in $QBCP$). Torej imamo devet točk $A$, $B$, $C$, $D$, $P$, + $Q$, $S$, $K$ in $L$, od katerih po tri ležijo na vsaki od desetih premic $AB$, + $CD$, $PQ$, $KL$, $AC$, $BD$, +$PA$, $PB$, $QC$ in $QD$. + \kdokaz + + Če v prejšnjem zgledu množico devetih točk označimo s $\mathfrak{T}$, vidimo, + da množica desetih premic ni množica $\mathcal{P}(\mathfrak{T})$. + V množici $\mathcal{P}(\mathfrak{T})$ bi namreč imeli šestnajst + premic - naših deset in še dodatne premice $AD$, $BC$ $DL$, $AL$, $CK$ in $BK$. + Toda vsaka od teh šestih premic vsebuje +le dve točki. Za njih torej ni izpolnjen pogoj, da vsebujejo +natanko tri točke začetne množice. Sedaj je logično zastaviti +naslednje vprašanje: Ali je v ravnini sploh mogoče postaviti končno +množico nekolinearnih točk $\mathfrak{T}$ tako, da vsaka premica iz množice +$\mathcal{P}(\mathfrak{T})$ vsebuje natanko tri točke iz množice +$\mathfrak{T}$? Jasno je namreč, da je to možno, če zahtevamo, +da vsaka premica iz $\mathcal{P}(\mathfrak{T})$ vsebuje natanko dve +točki iz $\mathfrak{T}$. Najbolj enostaven primer za to so oglišča +trikotnika in njegove nosilke, ali pa poljubna množica točk iz +izreka \ref{stevPremic}. Že omenjeni problem za tri točke ni tako +enostaven, odgovor pa bomo poiskali v nadaljevanju. Omenimo, da je +negativen. Celo več - odgovor je negativen tudi v primeru, +ko zahtevamo, da vsaka premica iz $\mathcal{P}(\mathfrak{T})$ vsebuje +vsaj tri točke iz $\mathfrak{T}$. Najprej pa dokažimo eno lemo +(pomožno trditev). + + + + \bizrek \label{SylvesterLema} + Let $\mathfrak{T}$ be a finite set of points in the plane which do not all lie on the same line + and $\mathcal{P}=\mathcal{P}(\mathfrak{T})$. If $T_0\in \mathfrak{T}$ and $p_0\in \mathcal{P}$ ($T_0\notin p_0$) + are such that they determine the minimum distance, i.e. + $$(\forall T\in \mathfrak{T})(\forall p\in \mathcal{P}) + ( T\notin p\Rightarrow d(T,p)\geq d(T_0,p_0)),$$ + then the line $p_0$ contains exactly two points from the set $\mathfrak{T}$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.6.pic} +\caption{} \label{sl.skl.3.10.6.pic} +\end{figure} + +\textbf{\textit{Proof.}} Predpostavimo nasprotno. Naj premica $p_0$ +vsebuje vsaj tri različne točke $A$, $B$ in $C$ iz množice +$\mathfrak{T}$ (Figure \ref{sl.skl.3.10.6.pic}). Označimo s $T'_0$ pravokotno projekcijo točke $T_0$ +na premici $p_0$. Če se točka $T'_0$ razlikuje od točk $A$, $B$ in +$C$, sta vsaj dve od teh treh točk (naj bosta to $B$ in $C$) na premici $p_0$ na isti +strani točke $T'_0$. Brez škode za splošnost naj bo +$\mathcal{B}(T'_0,B,C)$. Ker $T_0,C \in \mathfrak{T}$, potem tudi premica +$q= CT_0$ ($q \neq p_0$, ker $T_0\notin p_0$) pripada množici +$\mathcal{P}$. Naj bo $B'$ pravokotna projekcija točke $B$ na +premici $q$. Ni težko dokazati, da v tem primeru velja: + $$d(B,q)=|BB'| BP > PA$, zato je $\angle APB$ največji kot v +trikotniku $APB$ in je zaradi tega večji od $60^0$. Podobno je $CB > +BP,BC$, zato je tudi $\angle BPC > 60^0$. Enako bi veljalo tudi za +kote $CPD$, $DPE$, $EPF$ in $EPA$, kar pa ni mogoče, saj je njihova +vsota vedno enaka ali celo večja od $360^0$, ne glede na to, ali je +$P$ notranja ali zunanja točka šestkotnika $ABCDEF$. Torej točka $P$ +ni povezana z več kot s petimi točkami. + \kdokaz + + + \bzgled + What is the maximum number of regions in the plane that can be + divided by $n$ lines\footnote{Ta problem je rešil švicarski geometer + \index{Steiner, J.} \textit{J. Steiner} (1796--1863).}? + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.10.9.pic} +\caption{} \label{sl.skl.3.10.9.pic} +\end{figure} + +\textbf{\textit{Solution.}} Opisali bomo postopek načrtovanja takšnih +$n$ premic (Figure \ref{sl.skl.3.10.9.pic}). Če je $n = 1$, oz. če +gre le za eno premico, je ravnina razdeljena na dve območji. Dve +premici, če nista vzporedni, razdelita ravnino na štiri območja. Če +dodamo tretjo premico $p_3$, ki z njima ni vzporedna in ne gre skozi +njuno skupno točko, seka začetni dve premici v dveh točkah. Ti +dve točki delita premico $p_3$ na tri dele, vsak od njih pa je v +enem od treh od prejšnjih štirih območij ravnine. Torej s premico +$p_3$ dobimo še tri nove dele ravnine, torej skupaj sedem. Postopek +nadaljujemo. Če $n -1$ premic deli ravnino na $k$ delov, z +dodajanjem $n$-te premice $p_n$ (ki ni vzporednica nobene od +prejšnjih $n -1$ premic in ne vsebuje nobenega njihovega presečišča), +dobimo najprej $n -1$ presečišč na tej premici, nato $n$ njenih +delov oz. $n$ novih območij dane ravnine. Torej je največje možno +število območij za $n$ premic enako: +\begin{eqnarray*} +2+2+3+\cdots+n&=&1+1+2+3+\cdots+n=\\ +&=&1+\frac{n(n+1)}{2}=\\&=&\frac{n^2+n+2}{2}. +\end{eqnarray*} +Formalni dokaz tega dejstva bi lahko izpeljali z matematično indukcijo. + \kdokaz + + + %_______________________________________________________________________________ + \poglavje{Helly's Theorem} +\label{odd3Helly} + +Naslednji pomemben izrek je posledica le prvih dveh skupin aksiomov +oz. aksiomov incidence in aksiomov urejenosti. Uporabili ga bomo +tudi pri nalogah, ki so povezane s skladnostjo. + + + + \bizrek \label{Helly} + Let $\Phi_1$, $\Phi_2$, ... , $\Phi_n$ ($n \geq 4$) be convex sets in the plane. + If every three of these sets have a common point, then all $n$ sets have a common point + \index{izrek!Hellyjev}(Helly's theorem\footnote{Avstrijski + matematik \index{Helly, E.} \textit{E. Helly} (1884--1943) + je odkril to trditev v splošnem primeru $n$-razsežnega + prostora $\mathbb{E}^n$ + leta 1913, objavil pa šele leta 1923. Alternativna dokaza sta + medtem podala avstrijski matematik \index{Radon, J. K. A.} + \textit{J. K. A. Radon} (1887-–1956) leta 1921 in + madžarski matematik \index{Kőnig, D.} + \textit{D. Kőnig} (1884–-1944) leta 1922.}). + \eizrek + +\begin{figure}[!htb] +\centering +\hspace*{10mm} +\input{sl.skl.3.11.1.pic} +\caption{} \label{sl.skl.3.11.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} Dokaz bomo izvedeli z indukcijo po $n$. + +\textit{(A)} Naj bo najprej $n = 4$ in (Figure +\ref{sl.skl.3.11.1.pic}): +\begin{itemize} + \item $P_4\in \Phi_1 \cap \Phi_2 \cap \Phi_3$, + \item $P_3\in \Phi_1 \cap \Phi_2 \cap \Phi_4$, + \item $P_2\in \Phi_1 \cap \Phi_3 \cap \Phi_4$, + \item $P_1\in \Phi_2 \cap \Phi_3 \cap \Phi_4$. +\end{itemize} +Dokažimo, da obstaja točka, ki leži v vsakem od likov $\Phi_1$, +$\Phi_2$, $\Phi_3$ in $\Phi_4$. Glede na medsebojno lego +točk $P_1$, $P_2$, $P_3$ in $P_4$ bomo od več možnih primerov +obravnavali samo dva najbolj +splošna (dokaz v ostalih primerih je +podoben). + +\textit{1)} Štirikotnik, ki ga določajo točke $P_1$, $P_2$, $P_3$ in +$P_4$, je nekonveksen. V tem primeru je ena od točk $P_1$, $P_2$, +$P_3$ in $P_4$ notranja točka trikotnika, ki ga določajo preostale +tri točke. Brez škode za splošnost naj bo $P_4$ notranja točka +trikotnika $P_1P_2P_3$. Oglišča tega trikotnika ležijo v liku +$\Phi_4$. Ker je $\Phi_4$ konveksni lik, v njemu ležijo tudi vse +stranice in notranje točke trikotnika $P_1P_2P_3$, prav tako tudi +točka $P_4$. V tem primeru je točka $P_4$ skupna točka likov +$\Phi_1$, $\Phi_2$, $\Phi_3$ in $\Phi_4$. + +\textit{2)} Štirikotnik, ki ga določajo točke $P_1$, $P_2$, $P_3$ +in $P_4$, je konveksen. Brez škode za splošnost naj bosta njegovi +diagonali $P_1P_2$ in $P_3P_4$. Ker je štirikotnik konveksen, se +njegovi diagonali sekata v neki točki $S$. Dani liki so konveksni, +zato iz $P_1, P_2\in \Phi_3,\Phi_4$ sledi, da diagonala $P_1P_2$ +vsa leži v likih $\Phi_3$ in $\Phi_4$. Analogno iz $P_3, P_4\in +\Phi_1,\Phi_2$ sledi, da diagonala $P_3P_4$ vsa leži v likih +$\Phi_1$ in $\Phi_2$. Točka $S$, ki leži na obeh diagonalah +$P_1P_2$ in $P_3P_4$, leži v vseh štirih likih $\Phi_1$, $\Phi_2$, +$\Phi_3$ in $\Phi_4$. + +S tem smo dokazali, da trditev velja za $n=4$. + +\textit{(B)} Predpostavimo sedaj, da trditev velja za $n = k$ ($k\in +\mathbb{N}$ in $k>4$). + Dokažimo, da trditev velja tudi za +$n = k +1$. Naj bodo $\Phi_1$, $\Phi_2$, $\ldots$ , $\Phi_{k-1}$, +$\Phi_k$ in $\Phi_{k+1}$ takšni liki, da ima vsaka poljubna trojica teh likov +vsaj eno skupno točko. Naj bo $\Phi'=\Phi_k\cap\Phi_{k+1}$. Dokažimo +najprej, da ima vsaka trojica likov $\Phi_1$, $\Phi_2$ ,$\ldots$, +$\Phi_{k-1}$, $\Phi'$ skupno točko. Za trojice likov izmed +$\Phi_1$, $\Phi_2$, $\ldots$, $\Phi_{k-1}$ je to izpolnjeno že po +predpostavki. Brez škode za splošnost je dovolj, če dokazažemo, da +imajo liki $\Phi_1$, $\Phi_2$ in $\Phi'=\Phi_k\cap\Phi_{k+1}$ +skupno točko. To pa velja (na osnovi dokazanega primera za $n = 4$), +ker ima vsaka trojica izmed likov $\Phi_1$, $\Phi_2$, $\Phi_k$ in +$\Phi_{k+1}$ skupno točko. Iz indukcijske predpostavke +(za $n=k$) sledi, da imajo liki $\Phi_1$, $\Phi_2$, $\ldots$, +$\Phi_{k-1}$, $\Phi'$ skupno točko, ta točka pa hkrati leži v +vsakem od likov $\Phi_1$, $\Phi_2$, $\ldots$, $\Phi_{k-1}$, +$\Phi_k$, $\Phi_{k+1}$. + \kdokaz + +V nadaljevanju bomo obravnavali nekaj posledic Hellyjevega izreka. + + + + \bzgled + Let $\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$ + ($n > 3$) be half-planes covering a plane $\alpha$. + Prove that there are three of these half-planes that also cover the plane $\alpha$. + \ezgled + + +\textbf{\textit{Proof.}} Naj bodo $\beta_1$, $\beta_2$, $\cdots$, +$\beta_n$ odprte polravnine, ki so določene s polravninami +$\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$ kot komplementarne +polravnine glede na ravnino $\alpha$ oz. +$\beta_k=\alpha\setminus\alpha_k$, $k\in\{1,2,\ldots,n\}$. +Za vsako točko $X$ ravnine $\alpha$ in vsak +$k\in\{1,2,\ldots,n\}$ velja ekvivalenca: + $$X\in \alpha_k \Leftrightarrow X\notin \beta_k.$$ +Predpostavimo nasprotno, da nobena trojica polravnin $\alpha_1$, +$\alpha_2$, $\cdots$, $\alpha_n$ ne prekriva ravnine $\alpha$. To + pomeni, da za vsako trojico izmed njih obstaja točka v ravnini $\alpha$, +ki ne leži na nobeni od njih, oziroma da za vsako trojico izmed polravnin +$\beta_1$, $\beta_2$, $\cdots$, $\beta_n$ obstaja točka ravnine +$\alpha$, ki leži na vsaki od njih. Ker so polravnine konveksni +liki, iz Hellyjevega izreka \ref{Helly} sledi, da obstaja točka +$X$, ki leži na vsaki od polravnin $\beta_1$, $\beta_2$, $\cdots$, +$\beta_n$. Ta točka torej leži v ravnini $\alpha$, ne leži pa v +nobeni od polravnin $\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$, +kar je v nasprotju z osnovno predpostavko, da polravnine $\alpha_1$, +$\alpha_2$, $\cdots$, $\alpha_n$ prekrivajo ravnino $\alpha$. + Torej obstaja vsaj ena trojica izmed +polravnin $\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$, ki +prekrivajo ravnino $\alpha$. + \kdokaz + + + + \bzgled \label{lemaJung} + If for every three of $n$ ($n > 3$) points of a plane + there is such a circle with radius $r$ containing these three points, then it exists + a circle of equal radius containing all these $n$ points. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.11.2.pic} +\caption{} \label{sl.skl.3.11.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.11.2.pic}) + + Naj bodo $A_1$, $A_2$,$\ldots$, $A_n$ točke z danimi lastnostmi. S +$\mathcal{K}_i$ ($i\in\{1,2,\ldots,n\}$) označimo kroge s središči $A_i$ in s +polmerom $r$. Naj bodo $A_p$, $A_q$ in $A_l$ poljubne točke iz +množice $\{A_1, A_2,\ldots, A_n\}$. Po predpostavki obstaja krog s +polmerom $r$, ki te tri točke vsebuje. Označimo središče tega kroga +z $O$. Iz tega sledi $|OA_p|, |OA_q|, |OA_l|\leq r$, kar pomeni, da +točka $O$ leži v vsakem od krogov $\mathcal{K}_p$, $\mathcal{K}_q$ in $\mathcal{K}_l$. Torej, +vsaki trije izmed krogov $\mathcal{K}_1$, $\mathcal{K}_2$,$\ldots$, $\mathcal{K}_n$ imajo vsaj eno +skupno točko. Ker so krogi konveksni liki (izrek \ref{KrogKonv}), po +Hellyjevem izreku obstaja točka $S$, ki leži v vsakem izmed krogov +$\mathcal{K}_1$, $\mathcal{K}_2$,$\ldots$, $\mathcal{K}_n$. Iz tega sledi, da je $\mathcal{K}(S, r)$ iskani +krog, saj velja $|SA_1|, |SA_2|,\ldots |SA_n|\leq r$. + \kdokaz + + Zanimiva posledica zadnje trditve \ref{lemaJung} bo podana v razdelku + \ref{odd7Pitagora} (izrek \ref{Jung}). + +%________________________________________________________________________________ +\naloge{Exercises} + +\begin{enumerate} + + \item Naj bo $S$ točka, ki leži v kotu $pOq$, točki $A$ in $B$ pa + pravokotni projekciji točke $S$ na krakih $p$ in $q$ tega +kota. Dokaži, da je $SA\cong SB$ natanko tedaj, ko je premica +$OS$ simetrala kota $pOq$. + +\item Dokaži, da je vsota diagonal konveksnega štirikotnika večja od +vsote dveh njegovih nasprotnih stranic. + + \item Dokaži, da je v vsakemu trikotniku + največ ena stranica krajša od pripadajoče višine. + + \item Naj bo $AA_1$ težiščnica trikotnika $ABC$. Dokaži, + da je od dveh kotov, ki jih težiščnica $AA_1$ določa s +stranicama $AB$ in $AC$, večji tisti, ki ga težiščnica določa s +krajšo stranico. + + \item Naj bosta $BB_1$ in $CC_1$ težiščnici trikotnika $ABC$ ter + $AB \frac{3}{4}(a + b + c)$. + \end{enumerate} + +\item Naj bo premica $p$ mimobežnica krožnice $k$. Dokaži, da so +vse točke te krožnice na istem bregu premice $p$. + +\item Če krožnica $k$ leži v nekem konveksnem liku $\Phi$, potem tudi +krog, ki je določen s to krožnico, leži v tem liku. Dokaži. + +\item Naj bosta $p$ in $q$ različni tangenti krožnice $k$, ki se jo dotikata v +točkah $P$ in $Q$. Dokaži ekvivalenco: $p \parallel q$ natanko +tedaj, ko je $AB$ premer krožnice $k$. + +\item Če je $AB$ tetiva krožnice $k$, potem je presek premice $AB$ in +kroga, ki ga krožnica $k$ določa, enak tej tetivi. Dokaži. + +\item Naj bo $S'$ pravokotna projekcija središča $S$ krožnice $k$ na +premici $p$. Dokaži, da je $S'$ zunanja točka te krožnice +natanko tedaj, ko premica $p$ krožnice ne seka. + +\item Naj bo $V$ višinska točka trikotnika $ABC$, pri katerem velja +$CV \cong AB$. Določi velikost kota $ACB$. + +\item Naj bo $CC'$ višina pravokotnega trikotnika $ABC$ ($\angle ACB = +90^0$). Če sta $O$ in $S$ središči včrtanih krožnic trikotnikov +$ACC'$ in $BCC'$, je simetrala notranjega kota $ACB$ +pravokotna na premici $OS$. Dokaži. + +\item Naj bo $ABC$ trikotnik, v katerem je $\angle ABC = 15^0$ in +$\angle ACB = 30^0$. Naj bo $D$ takšna točka stranice $BC$, da je + $\angle BAD=90^0$. Dokaži, da je $BD = 2AC$. + +\item Dokaži, da obstaja takšen petkotnik, da je mogoče prekriti ravnino + s takšnimi petkotniki, ki so mu skladni. + +\item Dokaži, da obstaja takšen desetkotnik, da je mogoče prekriti ravnino + s takšnimi desetkotniki, ki so mu skladni. + +\item V neki ravnini je vsaka točka pobarvana rdeče ali črno. + Dokaži, da obstaja pravilni trikotnik, ki ima vsa + oglišča iste barve. + +\item Naj bodo $l_1,l_2,\ldots, l_n$ ($n > 3$) loki, ki vsi ležijo na isti +krožnici. Središčni kot vsakega loka je kvečjemu enak $180^0$. + Dokaži, da obstaja točka, ki leži na vsakem loku, + če imajo vsaki trije loki vsaj eno skupno točko. + +%drugi del + +\item +Naj bosta $p$ in $q$ pravokotnici, ki se sekata v točki $A$. Če +je $B, B'\in p$, $C, C'\in q$, $AB\cong AC'$, $AB'\cong AC$, +$\mathcal{B}(B,A,B')$ in $\mathcal{B}(C,A,C')$, potem +pravokotnica na premico $BC$ skozi točko $A$ poteka skozi središče +daljice $B'C'$. Dokaži. + +\item +Dokaži, da se simetrale notranjih kotov pravokotnika, ki ni +kvadrat, sekajo v točkah, ki so oglišča kvadrata. + +\item + Dokaži, da se simetrale notranjih kotov paralelograma, ki ni +romb, sekajo v točkah, ki so oglišča pravokotnika. Dokaži še, da so diagonale +tega pravokotnika vzporedne s stranicami paralelograma in so +enake razliki sosednjih stranic tega paralelograma. + +\item +Dokaži, da sta simetrali dveh sokotov med seboj pravokotni. + +\item Naj bosta $B'$ in $C'$ nožišči višin iz oglišč $B$ in $C$ trikotnika +$ABC$. Dokaži ekvivalenco $AB\cong AC \Leftrightarrow BB'\cong +CC'$. + +\item Dokaži, da je trikotnik pravilen, +če središče trikotniku očrtane krožnice in njegova višinska točka sovpadata. +Ali podobna trditev velja za poljubni dve značilni +točki tega trikotnika? + +\item Dokaži, da sta ostrokotna trikotnika $ABC$ in $A'B'C'$ skladna natanko +tedaj, ko imata skladni višini $CD$ in $C'D'$, stranici $AB$ in +$A'B'$ ter kota $ACD$ in $A'C'D'$. + +\item Če je $ABCD$ pravokotnik ter $AQB$ in $APD$ pravilna + trikotnika z enako orientacijo, je daljica $PQ$ +skladna z diagonalo tega pravokotnika. Dokaži. + +\item Naj bosta $BB'$ in $CC'$ višini trikotnika $ABC$ ($AC>AB$) ter + $D$ takšna točka poltraka $AB$, da velja $AD\cong AC$. Točka +$E$ je presečišče premice $BB'$ s premico, ki poteka skozi točko $D$ in je +vzporedna s premico $AC$. Dokaži, da je $BE=CC'-BB'$. + +\item Naj bo $ABCD$ konveksni štirikotnik, pri katerem velja + $AB\cong BC\cong CD$ in $AC\perp BD$. Dokaži, da je $ABCD$ + romb. + +\item Naj bo $BC$ osnovnica enakokrakega trikotnika $ABC$. Če sta $K$ in +$L$ takšni točki, da je $\mathcal{B}(A,K,B)$, $\mathcal{B}(A,C,L)$ in $KB\cong LC$, potem +središče daljice $KL$ leži na osnovnici $BC$. Dokaži. + +\item Naj bo $S$ središče trikotniku $ABC$ včrtane krožnice. +Premica, ki poteka skozi točko $S$ in je vzporedna s stranico $BC$ +tega trikotnika, seka stranici $AB$ in $AC$ po vrsti v točkah +$M$ in $N$. Dokaži, da je $BM+NC=NM$. + +\item Naj bo $ABCDEFG$ konveksni sedemkotnik. Izračunaj vsoto +konveksnih kotov, ki jih določa lomljenka $ACEGBDFA$. + +\item Dokaži, da so središča stranic in nožišče poljubne višine trikotnika, +v katerem nobeni dve stranici nista skladni, oglišča +enakokrakega trapeza. + + \item Naj bo $ABC$ pravokotni trikotnik s pravim kotom pri oglišču $C$. +Točki $E$ in $F$ naj bosta presečišči simetral notranjih kotov pri +ogliščih $A$ in $B$ z nasprotnima katetama, $K$ in $L$ pa +pravokotni projekciji točk $E$ in $F$ na hipotenuzi tega +trikotnika. Dokaži, da je $\angle LCK=45^0$. + + +\item Naj bo $M$ središče stranice $CD$ kvadrata $ABCD$ in $P$ takšna točka + diagonale $AC$, da velja $3AP=PC$. Dokaži, da je $\angle BPM$ +pravi kot. + + \item Naj bodo $P$, $Q$ in $R$ središča stranic $AB$, $BC$ in $CD$ + paralelograma $ABCD$. Premici $DP$ in $BR$ naj sekata daljico +$AQ$ v točkah $K$ in $L$. Dokaži, da je $KL= \frac{2}{5} AQ$. + + \item Naj bo $D$ središče hipotenuze $AB$ pravokotnega +trikotnika $ABC$ ($AC>BC$). Točki $E$ in $F$ naj bosta presečišči +poltrakov $CA$ in $CB$ s premico, ki poteka skozi $D$ in je pravokotna +na premico $CD$. Točka $M$ naj bo središče daljice $EF$. Dokaži, da +je $CM\perp AB$. + +\item Naj bosta $A_1$ in $C_1$ središči stranic $BC$ in $AB$ trikotnika $ABC$. + Simetrala notranjega kota pri oglišču $A$ seka daljico +$A_1C_1$ v točki $P$. Dokaži, da je $\angle APB$ pravi kot. + + \item Naj bosta $P$ in $Q$ takšni točki stranic $BC$ in $CD$ kvadrata $ABCD$, + da je premica $PA$ simetrala kota $BPQ$. Določi velikost kota + $PAQ$. + +\item Dokaži, da središče očrtane krožnice leži najbliže najdaljši stranici +trikotnika. + + \item Dokaži, da je središče včrtane krožnice najbliže oglišču, ki je vrh +največjega notranjega kota trikotnika. + +\item Naj bo $ABCD$ konveksen štirikotnik. Določi točko $P$, tako da +bo vsota $AP+BP+CP+DP$ minimalna. + + \item Diagonali $AC$ in $BD$ enakokrakega trapeza $ABCD$ z osnovnico $AB$ +se sekata v točki $O$ in velja $\angle AOB=60^0$. Točke $P$, $Q$ +in $R$ so po vrsti središča daljic $OA$, $OD$ in $BC$. Dokaži, da +je $PQR$ pravilni trikotnik. + +\item Naj bo $P$ poljubna notranja točka trikotnika $ABC$, za katero velja + $\angle PBA\cong \angle PCA$. Točki $M$ in $L$ sta pravokotni +projekciji točke $P$ na stranicah $AB$ in $AC$, točka $N$ pa +središče stranice $BC$. Dokaži, da je $NM\cong +NL$\footnote{Predlog za MMO 1982 (SL 9.)).}. + +\item Naj bodo $P$, $Q$ in $R$ središča stranic $BC$, $AC$ in $AB$ + trikotnika $ABC$ ($ABr_1+r_2$; + + (ii) the circles $k_1(S_1,r_1)$ + and $k_2(S_2,r_2)$ are touching each other externally + if and only if $|S_1S_2|=r_1+r_2$; + + (iii) the circles $k_1(S_1,r_1)$ + and $k_2(S_2,r_2)$ are intersecting each other at two points + if and only if $r_1-r_2<|S_1S_2|\omega$. Če pa velja $\mathcal{B}(A,N,M)$, lahko s +podobnim sklepanjem ugotovimo, da je v tem primeru $\angle +AMB<\omega$, kar pomeni, da za nobeno točko $M\notin l \cup +l'\setminus\{A,B\}$ ne velja $\angle AMB \cong \omega$. + \kdokaz + +Iz dokaza prejšnjega izreka dobimo tudi naslednjo ugotovitev. + + + \bizrek \label{obodKotGMTZunNotr} + Let $AB$ be an arc of a circle $k$, $\omega$ the corresponding circumferential angle + of this arc and $M$ a point of the half-plane with the edge $AB$ not containing this arc (Figure + \ref{sl.skk.4.2.8.pic}). Then: + + (i) $\angle AMB >\omega$ if and only if the point $M$ is an interior point of the circle $k$; + + (ii) $\angle AMB \cong\omega$ if and only if the point $M$ lies on the circle $k$; + + (iii) $\angle AMB <\omega$ if and only if the point $M$ is an exterior point of the circle $k$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.8.pic} +\caption{} \label{sl.skk.4.2.8.pic} +\end{figure} + +V naslednjih primerih bomo videli uporabo izreka o obodnem in +središčnem kotu in njegovih posledic. + + + + \bzgled + Construct a triangle with given $a$, $\alpha$, $v_a$. \label{konstr_aalphava} + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.1d.pic} +\caption{} \label{sl.skk.4.3.1d.pic} +\end{figure} + + \textbf{\textit{Analysis.}} Točka $A$ leži hkrati na geometrijskem mestu +točk, iz katerih se daljica $BC$ vidi pod kotom $\alpha$ (unija dveh krožnih lokov - izrek \ref{ObodKotGMT}) in vzporednici premice $BC$, ki je od nje +oddaljena $v_a$ (Figure \ref{sl.skk.4.3.1d.pic}). Torej je oglišče $A$ presečišče te +vzporednice in omenjenega geometrijskega mesta točk. + +\textbf{\textit{Construction.}} Načrtajmo najprej daljico +$BC\cong a$, nato pa geometrijsko mesto točk $\mathcal{L}$, iz +katerih se ta daljica vidi pod kotom $\alpha$ (izrek \ref{ObodKotGMT}). +Nato načrtajmo vzporednico $p$ premice $BC$ na razdalji $v_a$. Z $A$ označimo presečišče premice $p$ +in omenjenega geometrijskega mesta točk $\mathcal{L}$. Dokažimo, da je $ABC$ iskani trikotnik. + +\textbf{\textit{Proof.}} Že po konstrukciji je jasno, da je $BC\cong a$. Po konstrukciji točka $A$ leži na +geometrijskem mestu točk, iz katerih se daljica $BC$ vidi pod kotom $\alpha$, zato je tudi $BAC\cong\alpha$. Višina +trikotnika $ABC$ iz oglišča $A$ je skladna daljici $v_a$, ker točka $A$ po konstrukciji leži na premici $p$, ki je na +razdalji $v_a$ vzporedna premici $BC$. + +\textbf{\textit{Discussion.}} Nujni pogoj je seveda $\alpha<180^0$. Število rešitev naloge je enako številu presečišč +premice $p$ in množice $\mathcal{L}$. + \kdokaz + + + + \bzgled + Let $p$, $q$ and $r$ be lines in the plane intersecting + at one point and divide this plane into six congruent angles. Suppose that $P$, $Q$ and $R$ + are the foots of the perpendiculars from an arbitrary point $X$ of this plane on the lines $p$, $q$ and + $r$, respectively. Prove that $PQR$ is a regular triangle. + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.9.pic} +\caption{} \label{sl.skk.4.2.9.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.2.9.pic}). + Naj bo $S$ +presečišče premic $p$, $q$ in $r$. Jasno je, da premice določajo +kote $60^0$. Ker je $\angle XPS \cong \angle XQS \cong \angle XRS = +90^0$, po izreku \ref{TalesovIzrKroz2} točke $S$, $X$, $P$, $Q$ in +$R$ ležijo na krožnici $k$ s premerom $SX$. + Če uporabimo izrek \ref{ObodObodKot} za ustrezne loke $PQ$ in $QR$, + je $\angle PRQ \cong \angle PSQ = +60^0$ in $\angle QPR \cong \angle QSR = 60^0$. Ker ima vse kote +enake $60^0$, je torej $PQR$ pravilni trikotnik. + \kdokaz + + + \bzgled + Let $k(O,R)$ and $l(S,r)$ ($R = 2r$) be circles touching each other internally + and $P$ an arbitrary point on the circle $l$. Which curve + is described by the point $P$ if the circle $l$ rolls without slipping around the circle $k$\footnote{Ta problem je rešil poljski astronom + \index{Copernicus, N.} \textit{N. Copernicus} (1473--1543). + V splošnem primeru, kadar ni nujno $R = 2r$, krivuljo, po + kateri se giblje točka $P$, imenujemo \index{hipocikloida} + \pojem{hipocikloida}. V primeru zunanjega kotaljenja krožnice + po drugi krožnici, krivuljo imenujemo \index{epicikloida} + \pojem{epicikloida}, v primeru + kotaljenja krožnice po premici pa je to \index{cikloida} + \pojem{cikloida}. Cikloido je prvi raziskoval nemški matematik in + filozof \index{Kuzanski, N.} \textit{N. Kuzanski} (1401--1464) + in kasneje francoski matematik in filozof \index{Mersenne, M.} + \textit{M. Mersenne} (1588--1648). Poimenoval jo je italijanski fizik, + matematik, astronom in filozof \index{Galilei, G.} + \textit{G. Galilei} (1564--1642) leta 1599.}? + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.10.pic} +\caption{} \label{sl.skk.4.2.10.pic} +\end{figure} + +\textbf{\textit{Solution.}} + Naj bo +$P_0$ lega točke $P$ v trenutku, ko leži na krožnici $k$ (Figure +\ref{sl.skk.4.2.10.pic}). Takrat, ko je točka $P$ v legi $P_i$, +se krožnica $l$, ki je v legi $l_i$, dotika krožnice $k$ v neki +točki $T_i$. Ker gre za ‘‘gibanje brez spodrsavanja’’, sta dolžini +ustreznih lokov $P_0T_i$ in $P_iT_i$ krožnic $k$ in $l$ med seboj +enaki. Polmer krožnice $k$ je dvakrat večji od polmera krožnice $l$, +zato za pripadajoče središčne kote omenjenih lokov velja $\angle +T_iS_iP_i= 2\angle T_iOP_0$. Toda $\angle T_iOP_i$ je ustrezni +obodni kot krožnice $l$ za isti lok $P_iT_i$, zato je $2\angle +T_iOP_i = \angle T_iS_iP_i$ (izrek \ref{ObodObodKot}). Iz prejšnjih +dveh relacij dobimo $\angle T_iOP_i= \angle T_iOP_0$, kar pomeni, +da je točka $P_i$ kolinearna s točkama $O$ in $P_0$, zato iskani tir +točke $P$ predstavlja premer $P_0P'_0$ krožnice $k$. + \kdokaz + + + + \bzgled + Let $C$ be the midpoint of an arc $AB$ and $D$ an arbitrary point of this + arc other than $C$. Prove that + $$AC + BC > AD + BD.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.11.pic} +\caption{} \label{sl.skk.4.2.11.pic} +\end{figure} + +\textbf{\textit{Proof.}} + (Figure +\ref{sl.skk.4.2.11.pic}). + Naj bosta $C'$ in $D'$ takšni točki poltrakov $AC$ in $AD$, + da velja: $CC'\cong CB \cong CA$, $DD'\cong DB$, +$\mathcal{B}(A,C,C')$ in $\mathcal{B}(A,D,D')$. Trikotnika $C'CB$ in +$D'DB$ sta enakokraka, zato po izrekih \ref{enakokraki} in \ref{zunanjiNotrNotr}, +sledi: + $$\angle CC'B \cong \angle CBC' = \frac{1}{2}\angle ACB +\hspace*{2mm}\textit{ in }\hspace*{2mm} + \angle DD'B \cong \angle DBD' += \frac{1}{2}\angle ADB.$$ + Ker sta kota $ACB$ in $ADB$ obodna kota nad tetivo $AB$, sta skladna + (izrek \ref{ObodObodKot}). +Skladna sta tudi kota $AC'B$ in $AD'B$, točki $C'$ in $D'$ pa +pripadata pripadajočemu loku $l'$ nad tetivo $AB$ (izrek +\ref{ObodKotGMT}). Ker je $CC'\cong CB \cong CA$, je daljica $AC'$ +premer krožnice, ki vsebuje lok $l'$, zato je $\angle AD'C'$ pravi +kot (izrek \ref{TalesovIzrKroz2}). Daljica $AC'$ je torej +hipotenuza pravokotnega trikotnika $AD'C'$ in zato po izreku +\ref{vecstrveckot} velja: +$$AC + CB = AC'> AD'= AD + DB,$$ kar je bilo treba dokazati. \kdokaz + + + + \bnaloga\footnote{35. IMO Hong Kong - 1994, Problem 2.} + $ABC$ is an isosceles triangle with $BC \cong AC$. Suppose that: + + (i) $D$ is the midpoint of $AB$ and $E$ is the point on the line $CD$ such that + $EA\perp AC$; + + (ii) $F$ is an arbitrary point on the segment $AB$ different from $A$ and $B$; + + (iii) $G$ lies on the line $CA$ and $H$ lies on the line $CB$ such that $G$, $F$, $H$ are + distinct and collinear. + + Prove that $EF$ is perpendicular to $HG$ if and only if $GF\cong FH$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.IMO1.pic} +\caption{} \label{sl.skk.4.2.IMO1.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Najprej iz skladnosti trikotnikov $CAE$ in +$CBE$ (izrek \textit{SAS} \ref{SKS}) sledi $\angle EBC\cong\angle +EAC=90^0$ in $EA\cong EB$ (Figure \ref{sl.skk.4.2.IMO1.pic}). +Dokažimo ekvivalenco $EF\perp HG \Leftrightarrow GF\cong FH$ v obe +smeri. + + ($\Rightarrow$) Predpostavimo, da sta premici $EF$ in $HG$ + pravokotni oz. $\angle EFG\cong\angle EFH=90^0$. + Naj bosta $k$ in $l$ krožnici s premeroma $EG$ in +$EH$. Ker je $\angle EAG\cong\angle EFG=90^0$ in $\angle +EBH\cong\angle EFH=90^0$, po izreku \ref{TalesovIzrKroz} velja +$A,F\in k$ in $B,F\in l$. Najprej iz $EA\cong EB$ sledi $\angle +EAB\cong\angle EBA$. Iz tega in izreka \ref{ObodObodKot} sledi: + $$\angle EGF\cong\angle EAF\cong\angle EBF\cong\angle EHF.$$ + Torej je trikotnik $EGH$ enakokrak, zato je njegova višina $EF$ + hkrati težiščnica (skladnost trikotnikov $EFG$ in $EFH$, izrek + \textit{ASA} \ref{KSK}) oz. velja $GF\cong FH$. + +($\Leftarrow$) Naj bo sedaj $GF\cong FH$, oz. je točka $F$ +središče daljice $GH$. Naj bo $k$ krožnica s premerom $EG$. +Poleg točke $A$ označimo s $\widehat{F}$ drugo presečišče te +krožnice s premico $AB$. Če se krožnica $k$ dotika premice +$AC$, sledi $G=A$ oz. $F=D$ in $H=B$, torej je v tem primeru že +izpolnjeno $GF\cong FH$. + +Predpostavimo, da je $\widehat{F}\neq F$. Naj bo $\widehat{H}$ +presečišče premic $G\widehat{F}$ in $CB$. Ker točka +$\widehat{F}$ leži na krožnici $k$ s premerom $EG$, je $\angle +G\widehat{F}E=90^0$, oz. velja $E\widehat{F}\perp G\widehat{H}$. +Torej so za točke $G$, $\widehat{F}$ in $\widehat{H}$ izpolnjene +predpostavke leve strani ekvivalence, zato iz že dokazanega +prvega dela trditve ($\Rightarrow$) sledi $G\widehat{F}\cong +\widehat{F}\widehat{H}$, oz. točka $\widehat{F}$ je središče +daljice $G\widehat{H}$. V trikotniku $GH\widehat{H}$ je +$\widehat{F}F$ srednjica, zato je $\widehat{F}F\parallel +\widehat{H}H$ in $AB\parallel BC$, kar pa ni mogoče. Tako +predpostavka $\widehat{F}\neq F$ odpade, torej velja $\widehat{F}= +F$, zato je tudi $\widehat{H}=H$. Na koncu iz $E\widehat{F}\perp +G\widehat{H}$ sledi $EF\perp GH$. + \kdokaz + + + + + +%________________________________________________________________________________ + \poglavje{More About Circumcircle and Incircle of a Triangle} + \label{odd4OcrtVcrt} + + +Najprej bomo obravnavali nekatere pomembne točke, ki ležijo na +očrtani krožnici trikotnika. + + + + \bizrek \label{TockaN} + The bisector of the side $BC$ and the bisector of the interior angle $BAC$ of a + triangle $ABC$ ($AB\neq AC$) intersect at the Circumcircle $l(O,R)$ + of that triangle. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.1.pic} +\caption{} \label{sl.skk.4.3.1.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.3.1.pic}). + + Naj bo točka $N$ eno od presečišč očrtane krožnice trikotnika $ABC$ in + simetrale stranice $BC$ +(takšna, da je $A,N÷BC$). Ker leži točka $N$ na simetrali stranice +$BC$, velja $NB \cong NC$. Torej je $BNC$ enakokraki trikotnik, +zato sta $\angle NBC$ in $\angle NCB$ skladna kota (izrek +\ref{enakokraki}). Ker leži točka $N$ tudi na očrtani krožnici +trikotnika $ABC$, po izreku \ref{ObodObodKot} velja: + \begin{eqnarray*} + \angle BAN +\cong \angle BCN \textrm{ (obodna kota za krajši lok }BN \textrm{) }\\ + \angle NAC \cong \angle NBC \textrm{ (obodna kota za krajši lok } + CN \textrm{).} + \end{eqnarray*} + Torej je $\angle BAN \cong \angle NAC$ oz. premica $AN$ je simetrala kota +$BAC$, s tem pa je trditev dokazana. + \kdokaz + + + Točka $N$ iz prejšnjega izreka je središče tistega loka $BC$ očrtane + krožnice trikotnika $ABC$, ki ne vsebuje oglišča $A$. + Dokažimo še eno pomembno lastnost točke $N$. + + + + \bizrek \label{TockaN.NBNC} + For the point $N$ from the previous theorem is $NB\cong NS\cong NC$, + where $S$ is the incentre of the triangle $ABC$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.2.pic} +\caption{} \label{sl.skk.4.3.2.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Označimo z $\alpha$ in $\beta$ notranja +kota trikotnika $ABC$ ob ogliščih $A$ in $B$ (Figure +\ref{sl.skk.4.3.2.pic}). $BNS$ je enakokraki trikotnik (izrek +\ref{enakokraki}), ker sta kota pri ogliščih $B$ in $S$ skladna. +Če uporabimo izreka \ref{zunanjiNotrNotr} in \ref{ObodObodKot}, +dobimo: + \begin{eqnarray*} + \angle BSN &=& \angle ABS + \angle BAS =\frac{1}{2}\alpha+\frac{1}{2}\beta,\\ +\angle SBN &=& \angle SBC +\angle CBN = \angle SBC + \angle CAN = +\frac{1}{2}\beta+\frac{1}{2}\alpha. + \end{eqnarray*} +Torej velja $NB \cong NS$ in podobno tudi $NC \cong NS$. +\kdokaz + + + + \bizrek \label{TockaNbetagama} + Let $AA'$ be the altitude from the vertices $A$ and $AE$ bisector of the interior + angle $BAC$ of a triangle $ABC$ ($A',E\in BC$). Suppose that $l(O,R)$ is the + circumcircle of that triangle. If $\angle CBA=\beta\geq\angle ACB=\gamma$, + then + $$\angle A'AE\cong \angle EAO=\frac{1}{2}\left( \beta-\gamma\right).$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.1b.pic} +\caption{} \label{sl.skk.4.3.1b.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Naj bo $N$ točka, definirana kot v v prejšnjih trditvah + (Figure \ref{sl.skk.4.3.1b.pic}). Premici + $AA'$ in $ON$ sta vzporedni, ker sta obe pravokotni na premico $BC$. + Ker je $OA\cong ON=R$, je $AON$ enakokraki trikotnik. Zaradi tega je + (izreka \ref{KotiTransverzala} in \ref{enakokraki}) najprej: + $$\angle A' AE \cong \angle ANO \cong \angle NAO = +\angle EAO,$$ + nato pa še: + $$\angle A'AE=\frac{1}{2}\alpha-\left(90^0-\beta \right)= + \frac{1}{2}\alpha-\left(\frac{\alpha+\beta+\gamma}{2}-\beta \right)= + \frac{1}{2}\left( \beta-\gamma\right),$$ kar je bilo treba dokazati. \kdokaz + + + + + \bzgled \label{tockaNtockePQR} + Let $P$, $Q$ and $R$ be the midpoints of those arcs $BC$, $AC$ and $AB$ of + the circumcircle of a triangle $ABC$ not containing the vertices $A$, $B$ and $C$ + of that triangle. + If $E$ and $F$ are intersections of the line $QR$ with sides $AB$ and $AC$, respectively + and $S$ the incentre of this triangle, then: + + (i) $AP \perp QR$, + + (ii) the quadrilateral $AESF$ is a rhombus. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.2a.pic} +\caption{} \label{sl.skk.4.3.2a.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.3.2a.pic}) + +(\textit{i}) Naj bo $L$ presecišče premic $AP$ in $QR$. Po izreku +\ref{TockaN} točke $P$, $Q$ in $R$ ležijo na simetralah $AS$, $BS$ +in $CS$ notranjih kotov trikotnika $ABC$ ($S$ je središče trikotniku +$ABC$ včrtane krožnice). Če z $\alpha$, $\beta$ in $\gamma$ označimo +notranje kote trikotnika $ABC$, potem zaradi skladnosti ustreznih +obodnih kotov (izrek \ref{ObodObodKot}) dobimo: + \begin{eqnarray*} +\angle RPL &=& \angle RPA = \angle RCA =\frac{1}{2}\gamma,\\ + \angle PRL &=& \angle PRQ = \angle PRC + \angle CRQ = +\angle PAC + \angle CBQ = \frac{1}{2}\alpha+ \frac{1}{2}\beta. + \end{eqnarray*} +Torej je vsota kotov v trikotniku $PRL$ (izrek \ref{VsotKotTrik}) +$180^0 =\frac{1}{2}\alpha+ \frac{1}{2}\beta +\frac{1}{2}\gamma ++\angle RLP = 90° + \angle RLP$. Zato je $\angle RLP = 90°$ oz. $AP +\perp QR$. + + (\textit{ii}) Po +izreku \ref{TockaN.NBNC} je $RA \cong RS$, zato iz skladnosti +pravokotnih trikotnikov $ALR$ in $SLR$ (izrek \textit{SSA} +\ref{SSK}) sledi, da je točka $L$ središče diagonale $AS$ +štirikotnika $AESF$. Iz skladnosti trikotnikov $AEL$ in $AFL$ (izrek +\textit{ASA} \ref{KSK}) sledi, da je točka $L$ tudi središče +diagonale $EF$, zato je $AESF$ paralelogram (izrek +\ref{paralelogram}. Ker sta diagonali $AS$ in $EF$ še pravokotni, je + štirikotnik $AESF$ romb (izrek \ref{RombPravKvadr}). + \kdokaz + +Iz prejšnje trditve dobimo direktno posledico. + + + \bzgled \label{PedalniLemasPQR} + Let $P$, $Q$ and $R$ be the midpoints of those arcs $BC$, $AC$ and $AB$ of + the circumcircle of a triangle $ABC$ not containing the vertices $A$, $B$ and $C$ + of that triangle. + Prove that the incentre of the triangle $ABC$ is at the same time + orthocentre of the triangle $PQR$ (Figure \ref{sl.skk.4.3.2b.pic}). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.2b.pic} +\caption{} \label{sl.skk.4.3.2b.pic} +\end{figure} + + + +Dokažimo nekaj posledic izreka \ref{ObodObodKot}, ki so povezane z +višinsko točko. + + + + \bizrek \label{TockaV'} + Points that are symmetric to the orthocentre of an acute triangle + with respect to its sides lie on the circumcircle of this triangle. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.3.pic} +\caption{} \label{sl.skk.4.3.3.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Naj bo $V$ višinska točka trikotnika $ABC$, +$l$ očrtana krožnica trikotnika $ABC$ in $V_a$ drugo presečišče +krožnice $l$ z nosilko višine $AA'$ (Figure \ref{sl.skk.4.3.3.pic}). +Dokažimo da je točka $V_a$ simetrična točki $V$ glede na premico +$BC$. Dovolj je dokazati, da je $VA'\cong V_aA'$. Kota $V_aBC$ in +$V_aAC$ sta skladna (obodna kota za tetivo $V_aC$ - izrek +\ref{ObodObodKot}), kota $V_aAC$ in $CBV$ sta skladna kota s +pravokotnima krakoma (izrek \ref{KotaPravokKraki}). Torej sta skladna + tudi kota $V_aBC$ in $CBV$, zato pa tudi trikotnika $V_aBA'$ in +$VBA'$ oz. velja $VA'\cong V_aA'$. Podobno velja tudi za drugi dve +višini. + \kdokaz + + \bizrek \label{TockaV'a} + Let $V$ be the orthocentre of a triangle + $ABC$. The circumcircles of the triangles $VBC$, $AVB$ and $ACV$ are congruent to + the circumcircle of the triangle $ABC$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.3a.pic} +\caption{} \label{sl.skk.4.3.3a.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Direktna posledica prejšnjega izreka +\ref{TockaV'}, kajti tri omenjene očrtane krožnice so simetrične +očrtani krožnici trikotnika $ABC$ glede na nosilke njegovih stranic +(Figure \ref{sl.skk.4.3.3a.pic}). + + + + \bizrek \label{TockaV1} + Points that are symmetric to the orthocentre of an acute triangle + with respect to the midpoints of its sides lie on the circumcircle of this triangle. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.3b.pic} +\caption{} \label{sl.skk.4.3.3b.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Naj bo $V$ višinska točka trikotnika $ABC$, +$l$ očrtana krožnica (s središčem $O$) trikotnika $ABC$ in $V_{A_1}$ točka, ki je simetrična točki $A$ glede na točko $O$ (Figure \ref{sl.skk.4.3.3.pic}). Iz same definicije točke $V_{A_1}$ je jasno, da ta leži na krožnici $l$. Dokažimo še, da je $V_{A_1}$ simetrična točki $V$ glede na točko $A_1$, ki je središče daljice $BC$. Ker je $AV_{A_1}$ premer krožnice $l$, je po izreku \ref{TalesovIzrKroz2} $\angle ACV_{A_1}=90^0$ oz. $V_{A_1}C\perp AC$. Premica $BV$ je nosilka višine trikotnika $ABC$, zato velja $BV\perp AC$. Iz zadnjih dveh relacij sledi $V_{A_1}C\parallel BV$. Analogno je tudi +$V_{A_1}B\parallel CV$. Torej je štirikotnik $V_{A_1}CVB$ paralelogram, zato imata njegovi diagonali $VV_{A_1}$ in $BC$ skupno središče. Središče daljice $BC$ je točka $A_1$, kar pomeni, da je $V_{A_1}$ simetrična točki $V$ glede na točko $A_1$. Točka $V_{A_1}$ pa po konstrukciji leži na krožnici $l$. +\kdokaz + + +Dokažimo še nekaj posledic izreka \ref{ObodObodKot}, ki so povezane z očrtano +krožnico trikotnika. + + + \bzgled \label{zgledTrikABCocrkrozP} + Let $k$ be the circumcircle of a regular triangle $ABC$. + If $P$ is an arbitrary point lying + on the shorter arc $BC$ of the circle $k$, then + $$|PA|=|PB|+|PC|.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.4.pic} +\caption{} \label{sl.skk.4.3.4.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Ker je $\angle ACP>60^0>\angle PAC$, je +po izreku \ref{vecstrveckot} $AP>PC$ (Figure +\ref{sl.skk.4.3.4.pic}). Zato na daljici $AP$ obstaja takšna točka +$Q$, da velja $PQ\cong PC$. Po izreku \ref{ObodObodKot} je $\angle +CPQ=\angle CPA\cong\angle CBA=60^0$, kar pomeni, da je $PCQ$ +enakostranični trikotnik, zato je tudi $CQ\cong CP$ in $\angle +PCQ=60^0$. Iz tega sledi $\angle ACQ=\angle ACP-60^0=\angle BCP$. +Po izreku \textit{SAS} (izrek \ref{SKS}) sta torej skladna trikotnika +$ACQ$ in $BCP$, zato je tudi $AQ\cong BP$. + +Na koncu je: $|PB|+|PC|=|AQ|+|PQ|=|AP|$. +\kdokaz + + + + \bzgled + Three circles of equal radii $r$ intersect at point $O$. + Furthermore each two of them intersect at one more point: $A$, $B$, and $C$. + Prove that the radius of the circumcircle of the triangle $ABC$ is also $r$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.5.pic} +\caption{} \label{sl.skk.4.3.5.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.3.5.pic}) + +Označimo s $k$, $l$, $j$ in $o$ očrtane krožnice trikotnikov $OBC$, +$OAC$, $OAB$ in $ABC$ ter s $P$ poljubno točko krožnice $k$ tako, da +sta točki $O$ in $P$ na različnih straneh premice $BC$. Po +predpostavki so krožnice $k$, $l$ in $j$ skladne. Kota $BAO$ in +$BCO$ sta tudi skladna, ker sta obodna kota skladnih krožnic $k$ in $j$ +nad tetivo $BO$ (izrek \ref{SklTetSklObKot2}). Analogno sta skladna +tudi kota $CAO$ in $CBO$. Zaradi tega je: + $$\angle BAC = \angle BAO + \angle CAO = \angle BCO ++ \angle CBO = 180° - \angle BOC = \angle BPC.$$ + Torej imata krožnici $k$ in $o$ skladna +obodna kota nad skupno tetivo $BC$, zato sta med seboj skladna. + \kdokaz + + + + \bzgled \label{KvadratKonstr4tocke} + Construct a square $ABCD$ such that given points $P$, $Q$, $R$ and $S$ + lyes on the sides $AB$, $BC$, $CD$ and $DA$ of this square, respectively. + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.1c.pic} +\caption{} \label{sl.skk.4.3.1c.pic} +\end{figure} + + +\textbf{\textit{Solution.}} +Ker sta $\angle PAS$ in $\angle QCR$ prava kota, oglišči $A$ in $B$ ležita na krožnicah $k$ in $l$ s premeroma +$PS$ in $QR$ (Figure \ref{sl.skk.4.3.1c.pic}). Nosilka diagonale $AC$ kvadrata $ABCD$ je hkrati simetrala notranjih kotov $BAD$ in $BCD$, zato gre skozi +središči $N$ in $M$ ustreznih polkrožnic, ki sta določeni s $k$ in $l$ (izrek \ref{TockaN}). +Konstrukcijo lahko torej izpeljemo tako, da najprej načrtamo krožnici $k$ in $l$, nato premico $NM$, oglišči $A$ in $C$ in na koncu še +oglišči $B$ in $D$. + \kdokaz + + + + + \bzgled + Construct a triangle $v_a$, $t_a$, $l_a$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.1e.pic} +\caption{} \label{sl.skk.4.3.1e.pic} +\end{figure} + + \textbf{\textit{Solution.}} + Naj bo $ABC$ trikotnik, v katerem so višina $AA'$, +težiščnica $AA_1$ in odsek simetrale $AE$ notranjega kota $BAC$ skladni z +daljicami $v_a$, $t_a$ in $l_a$. Z $O$ označimo središče trikotniku $ABC$ očrtane +krožnice $k$. Po izreku \ref{TockaN} +se premici $AE$ in $OA_1$ sekata v točki $N$, ki leži na +krožnici $k$ (Figure \ref{sl.skk.4.3.1e.pic}). + +Torej lahko najprej načrtamo +pravokotni trikotnik $AA'E$ s kateto $v_a$ in hipotenuzo $l_a$ ter točko $A_1$ iz pogoja $AA_1\cong t_a$. Nato +načrtamo točko $N$ kot presečišče premice $AE$ in pravokotnice premice $A'E$ skozi +točko $A_1$. Središče $O$ je presečišče premice $A_1N$ in +simetrale daljice $AN$ (ker je $AN$ tetiva krožnice $k$). Oglišči $B$ in $C$ sta presečišči +krožnice $k(O,OA)$ s premico $A'E$. + \kdokaz + + + + + \bzgled + Construct a triangle $R$, $r$, $a$. \label{konstr_Rra} + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.1a.pic} +\caption{} \label{sl.skk.4.3.1a.pic} +\end{figure} + + +\textbf{\textit{Solution.}} Naj bo $ABC$ takšen trikotnik, da velja +$BC\cong a$ in sta $l(O,R)$ in $k(S,r)$ njegova očrtana oz. včrtana +krožnica + (Figure \ref{sl.skk.4.3.1a.pic}). Označimo z $\alpha$, $\beta$ in $\gamma$ +njegove notranje kote pri ogliščih $A$, $B$ in $C$. Po izreku +\ref{SredObodKot} je $\alpha = \angle BAC = \frac{1}{2}\cdot\angle BOC$. +Iz zgleda \ref{kotBSC} sledi $\angle BSC=90^0+\frac{1}{2}\cdot\alpha$. +Iz dveh relacij dobimo enakost +$\angle BSC=90^0+\frac{1}{4}\cdot\angle BOC$, ki omogoča konstrukcijo. + +Najprej načrtamo enakokraki trikotnik $BOC$ ($BC\cong a$ in $OB\cong +OC\cong R$). Točko $S$ dobimo kot eno od presečišč loka s tetivo +$BC$ in obodnim kotom $90^0+\frac{1}{4}\cdot\angle BOC$ ter premico, +ki je na razdalji $r$ vzporedna s premico $BC$. Nato načrtamo +včrtano krožnico $k(S,r)$ in točko $A$ kot presečišče ostalih dveh tangent +te krožnice iz točk $B$ in $C$. + \kdokaz + + + + \bnaloga\footnote{47. IMO Slovenia - 2006, Problem 1.} + Let $ABC$ be a triangle with incentre $I$. A point $P$ in the interior of the + triangle satisfies + $$\angle PBA + \angle PCA = \angle PBC + \angle PCB.$$ + Show that $|AP| \geq |AI|$, and that equality holds if and only if $P = I$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skl.4.3.IMO1.pic} +\caption{} \label{sl.skl.4.3.IMO1.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Označimo z $\alpha$, $\beta$ in $\gamma$ +notranje kote trikotnika $ABC$ pri ogliščih $A$, $B$ in $C$ +(Figure \ref{sl.skl.4.3.IMO1.pic}). Pogoj $\angle PBA + \angle PCA += \angle PBC + \angle PCB$ lahko preoblikujemo v obliko +$\beta-\angle PBC + \gamma-\angle PCB = \angle PBC + \angle PCB$ +oziroma: + $$\angle PBC + \angle PCB=\frac{1}{2}\left( \beta+\gamma\right).$$ +Iz tega in dejstva, da je vsota notranjih kotov vsakega izmed +trikotnikov $BPC$ in $ABC$ enaka $180^0$ (izrek\ref{VsotKotTrik}), +sledi: + $$\angle BPC =180^0-\frac{1}{2}\left( \beta+\gamma\right)=90^0+ + \frac{1}{2} \alpha.$$ +Toda iz zgleda \ref{kotBSC} sledi $\angle BIC =90^0+ + \frac{1}{2}\cdot \alpha$, zato je $\angle BPC\cong \angle BIC$. + Torej točki $P$ in $I$ ležita na istem loku $\mathcal{L}$ s + tetivo $BC$ in obodnim kotom $90^0+ + \frac{1}{2} \alpha$. Naj bo točka $N$ presečišče + simetrale stranice $BC$ in simetrale + notranjega kota $BAC$ trikotnika $ABC$. Po izreku + \ref{TockaN} leži točka $N$ na očrtani krožnici + trikotnika $ABC$ in velja $NB\cong NI\cong NC$ (izrek + \ref{TockaN.NBNC}). To pomeni, da je $N$ središče loka + $\mathcal{L}$, zato je $NP\cong NI$ oz. $\angle NIP\cong\angle + NPI<90^0$. Ker so točke $A$, $I$ in $N$ kolinearne (ležijo na + simetrali notranjega kota $BAC$), je $\angle AIP =180^0-\angle + NIP>90^0$. Iz izreka \ref{vecstrveckot} (za trikotnik $API$) + sedaj sledi $|AP| \geq |AI|$ in enakost velja natanko tedaj, ko + trikotnika $API$ ni oz. kadar je $P=I$. + \kdokaz + + + \bnaloga\footnote{43. IMO United Kingdom - 2002, Problem 2.} + $BC$ is a diameter of a circle center $O$. $A$ is any point on + the circle with $\angle AOC>60^0$. $EF$ is the chord which is the perpendicular + bisector of $AO$. $D$ is the midpoint of the minor arc $AB$. The line through + $O$ parallel to $AD$ meets $AC$ at $J$. Show that $J$ is the incenter of triangle + $CEF$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.3.IMO4.pic} +\caption{} \label{sl.skk.4.3.IMO4.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Ker +točki $E$ in $F$ ležita na simetrali daljice $EF$, hkrati pa na +krožnici s središčem $O$, je $$AF\cong FO\cong AO\cong EO \cong EA.$$ To +pomeni, da je štirikotnik $EOFA$ romb, ki je sestavljen iz dveh enakostraničnih trikotnikov $AOF$ in $AEO$. + +Brez škode za splošnost predpostavimo, da +je $\angle COE>\angle COF$ (Figure \ref{sl.skk.4.3.IMO4.pic}). +Najprej iz pogoja $\angle + AOC>60^0$ sledi, da je $\angle COF=60^0- \angle + AOC>0^0$, zato sta točki $A$ in $F$ na isti strani premice + $BC$. + + Ker je $AOC$ enakokraki trikotnik z osnovnico $AC$, po izrekih + \ref{enakokraki} in \ref{zunanjiNotrNotr} velja + $\angle ACO =\frac{1}{2}\angle AOB$. Točka $D$ je središče + loka $BD$, zato je $\angle AOD\cong\angle DOB$ + oz. $\angle DOB=\frac{1}{2}\angle AOB$. To pomeni, da je + $\angle ACO\cong\angle DOB$ in sta po izreku + \ref{KotiTransverzala} premici $AC$ in $DO$ vzporedni oz. + $AJ\parallel DO$. Ker je po predpostavki $AD\parallel JO$, je + štirikotnik $ADOJ$ paralelogram, zato je $AJ\cong OD$. Torej + $$AJ\cong OD\cong OE\cong AF\cong AE.$$ + Iz $AF\cong AE$ sledi, da + je $AJ$ simetrala notranjega kota pri oglišču $C$ trikotnika $CEF$ + (izrek \ref{SklTetSklObKot}). Ker je še + $AJ\cong AF\cong AE$, + je po izreku \ref{TockaN.NBNC} točka $J$ središče + včrtane krožnice tega trikotnika. +\kdokaz + + +%________________________________________________________________________________ + \poglavje{Cyclic Quadrilateral} \label{odd4Tetivni} + +Za večkotnik pravimo, da je +\index{štirikotnik!tetiven}\index{večkotnik!tetiven}\pojem{tetiven}, +če zanj obstaja očrtana krožnica, oz. če obstaja krožnica, ki +vsebuje vsa njegova oglišča (Figure \ref{sl.skk.4.5.10.pic}). Za +oglišča v tem primeru pravimo, da so \index{konciklične +točke}\pojem{konciklične točke}. Ugotovili smo že, da je vsak +trikotnik tetiven (izrek \ref{SredOcrtaneKrozn}) in prav tako, da je +vsak pravilni večkotnik tetiven (izrek \ref{sredOcrtaneKrozVeck}). +Po drugi strani je jasno, da niso vsi večkotniki tetivni. Npr. romb +je štirikotnik, ki nima očrtane krožnice. +V tem razdelku se bomo torej ukvarjali s tetivnimi štirikotniki. + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.10.pic} +\caption{} \label{sl.skk.4.5.10.pic} +\end{figure} + +Ker je kvadrat pravilni večkotnik, je hkrati tetivni štirikotnik. Ni +težko dokazati, da je tudi pravokotnik vrsta tetivnega štirikotnika +- središče očrtane krožnice je presečišče njegovih diagonal, ki sta +skladni in se razpolavljata. Toda kako bi na splošno ugotovili ali +je nek štirikotnik tetiven? Jasno je, da se pri tetivnem +štirikotniku (na splošno tudi večkotniku) simetrale vseh njegovih +stranic sekajo v eni točki (Figure \ref{sl.skk.4.5.10.pic}). Ta pogoj +je zadosten, da je štirikotnik tetiven, ni pa +preveč operativen v konkretnih primerih. Za tetivnost štirikotnikov +obstaja namreč potreben in zadosten pogoj, ki je uporabnejši. + + + + \bizrek \label{TetivniPogoj} + A convex quadrilateral is cyclic if and only if + its opposite interior angles are supplementary. + Thus, if $\alpha$, $\beta$, $\gamma$ and $\delta$ are + the interior angles of a convex quadrilateral $ABCD$, + it is cyclic if and only if + $$\alpha+\gamma=180^0.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.11.pic} +\caption{} \label{sl.skk.4.5.11.pic} +\end{figure} + +\textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.5.11.pic}) + +($\Rightarrow$) Predpostavimo najprej, da je štirikotnik $ABCD$ +tetiven. Ker je konveksen, sta oglišči $A$ in $C$ na različnih +straneh premice $BD$. Po izreku \ref{ObodObodKotNaspr} je +$\alpha+\gamma=180^0$. + +($\Leftarrow)$ Predpostavimo sedaj, da sta nasprotna kota +štirikotnika $ABCD$ suplementarna oz. $\alpha+\gamma=180^0$. Naj bo +$k$ očrtana krožnica trikotnika $ABD$. V tem primeru se iz četrtega +oglišča $C$ tetiva $BD$ vidi pod kotom, ki je suplementaren kotu pri +oglišču $A$, kar pomeni, da tudi točka $C$ leži na krožnici $k$ +(izrek \ref{ObodKotGMT}). + \kdokaz + + A direct consequence is the following theorem. + + + + \bizrek \label{TetivniPogojZunanji} + A convex quadrilateral is cyclic if and only if + one of its interior angles is congruent to the opposite exterior angle. + Thus, if $\alpha$, $\beta$, $\gamma$ and $\delta$ are + the interior angles and + $\alpha'$, $\beta'$, $\gamma'$ in $\delta'$ the exterior angles + of a convex quadrilateral $ABCD$, + it is cyclic if and only if + $$\alpha\cong\gamma'.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.12.pic} +\caption{} \label{sl.skk.4.5.12.pic} +\end{figure} + +Uporabimo kriterij iz izreka \ref{TetivniPogoj} za paralelogram in +trapez. + + \bizrek \label{paralelogramTetivEnakokr} + A parallelogram is cyclic if and only if it is a rectangle. + \eizrek + + +\textbf{\textit{Proof.}} Naj bodo $\alpha$, $\beta$, $\gamma$ in +$\delta$ notranji koti paralelograma $ABCD$ + (Figure \ref{sl.skk.4.5.13.pic}). + +($\Leftarrow$) Če je paralelogram pravokotnik, je +$\alpha+\gamma=90^0+90^0=180^0$, kar pomeni, da je $ABCD$ tetiven +štirikotnik (izrek \ref{TetivniPogoj}). + + ($\Rightarrow$) Predpostavimo, da je $ABCD$ tetivni paralelogram. + Ker je $ABCD$ paralelogram, je po izreku \ref{paralelogram} + $\alpha\cong\gamma$. Ker je tudi tetiven, je po izreku \ref{TetivniPogoj} +$\alpha+\gamma=180^0$. Torej velja $\alpha\cong\gamma=90^0$, zato je + $ABCD$ pravokotnik. + \kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.13.pic} +\caption{} \label{sl.skk.4.5.13.pic} +\end{figure} + + + + \bizrek \label{trapezTetivEnakokr} + A trapezium is cyclic if and only if it is isosceles. + \eizrek + +\textbf{\textit{Proof.}} Naj bo $ABCD$ trapez z osnovnico $AB$ in z +notranjimi koti $\alpha$, $\beta$, $\gamma$ in $\delta$ + (Figure \ref{sl.skk.4.5.13.pic}). V poljubnem trapezu + velja $\alpha+\delta=180^0$ in $\beta+\gamma=180^0$. + + +($\Leftarrow$) Predpostavimo, da je trapez $ABCD$ enakokrak oz. $AD +\cong BC$. Po izreku \ref{trapezEnakokraki} je v tem primeru +$\alpha\cong\beta$. Torej $\alpha+\gamma=\beta+\gamma=180^0$, zato +je po izreku \ref{TetivniPogoj} $ABCD$ tetivni štirikotnik. + +($\Rightarrow$) Naj bo trapez $ABCD$ tetivni štirikotnik in $k$ +njegova očrtana krožnica. Osnovnici $AB$ in $CD$ sta vzporedni +tetivi te krožnice, zato imata skupno simetralo, ki poteka skozi +središče $S$ krožnice $k$ in je pravokotna na tetivi $AB$ in +$CD$. To pomeni, da sta kraka $AD$ in $BC$ simetrična glede na to +simetralo, zato sta med seboj skladna in je trapez $ABCD$ enakokrak. + \kdokaz + + Posebej so zanimivi tetivni štirikotniki s pravokotnima + diagonalama\footnote{\index{Brahmagupta}\textit{Brahmagupta} (598--660), indijski matematik, ki je + preučeval takšne štirikotnike.}. + Na te štirikotnike se nanaša +naslednji primer. + + + \bzgled \label{TetivniLemaBrahm} + Suppose that the diagonals of a cyclic quadrilateral $ABCD$ are perpendicular and intersect + at a point $S$. Prove that the foot of the perpendicular from the point $S$ on the line $AB$ + contains the midpoint of the line $CD$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.14.pic} +\caption{} \label{sl.skk.4.5.14.pic} +\end{figure} + +\textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.5.14.pic}) + +Označimo z $N$ in $M$ presečišči pravokotnice na premico $AB$ skozi +točko $S$ s stranicama $AB$ in $CD$ štirikotnika $ABCD$. Tedaj velja: + \begin{eqnarray*} + \angle CDB &\cong& \angle CAB \hspace*{3mm} + \textrm{(obodna kota za ustrezni lok } CB + \textrm{ - izrek \ref{ObodObodKot}})\\ + &\cong& \angle NSB \hspace*{3mm} + \textrm{ (kota s +pravokotnima krakoma - izrek \ref{KotaPravokKraki})}\\ + &\cong& \angle MSD \hspace*{3mm} + \textrm{(sovršna kota)} + \end{eqnarray*} + Ker je $\angle CDB\cong \angle MSD$, je $MD \cong MS$ (izrek \ref{enakokraki}). + Analogno je tudi $MC \cong MS$. Torej je + $MD \cong MC$, kar pomeni, da je $M$ središče stranice $CD$. + \kdokaz + +Eno lastnost tetivnih štirikotnikov s pravokotnima diagonalama +bomo obravnavali še v zgledu \ref{HamiltonPoslTetiv}. + + + + \bzgled \label{TetŠtirZgl0} + Let $k$ be the circumcircle of a cyclic quadrilateral $ABCD$ + and $N$, $M$, $L$ and $P$ the midpoints of those arcs $AB$, $B$C, $CD$ and $AD$ + of the circle $k$, not containing the third vertices of this quadrilateral. + Prove that $NL\perp PM$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.0.pic} +\caption{} \label{sl.skk.4.5.0.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $S$ presečišče premic $NL$ in $PM$ +(Figure \ref{sl.skk.4.5.0.pic}). + Če dvakrat uporabimo izreka \ref{ObodObodKot} in \ref{TockaN}, dobimo: + + \begin{eqnarray*} + \angle PNS &=& \angle PND +\angle DNL = +\angle PBD +\angle DBL =\\ + &=& \frac{1}{2} \angle ABD +\frac{1}{2}\angle CBD = \frac{1}{2}\angle + ABC. + \end{eqnarray*} + + Na enak način dokažemo tudi $\angle NPS = \frac{1}{2}\angle + ADC$. Zato je po izreku \ref{TetivniPogoj}: + $$\angle PNS +\angle NPS = \frac{1}{2} \left(\angle ABC+\angle + ADC\right)=90^0.$$ + Če izrek \ref{VsotKotTrik} uporabimo za trikotnik $PSN$, dobimo + $\angle PSN = 90^0$. + \kdokaz + + + \bzgled \label{TetivniVcrtana} + Let $ABCD$ be a cyclic quadrilateral. + Prove that incentres of the triangles $BCD$, $ACD$, $ABD$ and $ABC$ + are the vertices of a rectangle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.1.pic} +\caption{} \label{sl.skk.4.5.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} Označimo z $A_1$, $B_1$, $C_1$ in $D_1$ +središča včrtanih krožnic trikotnikov $BCD$, $ACD$, $ABD$ in $ABC$ +ter z $N$, $M$, $L$ in $P$ središča tistih lokov $AB$, $BC$, $CD$ in +$AD$ očrtane krožnice štirikotnika $ABCD$, ki ne vsebujejo ostalih +oglišč tega štirikotnika (Figure \ref{sl.skk.4.5.1.pic}). Iz zgleda +\ref{TockaN} sledi, da sta $BL$ in $DM$ simetrali kotov $CBD$ in +$BDC$, zato je točka $A_1$ presečišče premic $BL$ in $DM$. Analogno +je točka $B_1$ presečišče premic $CP$ in $AL$. Po zgledu +\ref{TockaN.NBNC} je + $LC\cong LA_1\cong LB_1\cong LD$, torej je $A_1LB_1$ enakokraki trikotnik + z osnovnico $A_1B_1$. Iz zgleda \ref{TockaN} sledi tudi, + da je $LN$ simetrala kota $ALB$ oz. $B_1LA_1$. V enakokrakem + trikotniku $A_1LB_1$ simetrala kota $B_1LA_1$ vsebuje višino + tega trikotnika iz točke $L$. To pomeni, da velja $LN\perp + A_1B_1$. Analogno sledi tudi $LN\perp C_1D_1$, $PM\perp A_1D_1$ + in + $PM\perp C_1B_1$. Iz prejšnjega zgleda \ref{TetŠtirZgl0} je $LN\perp PM$, + zato je štirikotnik $A_1B_1C_1D_1$ pravokotnik. + \kdokaz + +Omenili smo že, da je pravokotnik tetivni štirikotnik. Sedaj bomo +dokazali zanimivo lastnost pravokotnika, ki se nanaša na točke, +ki ležijo na njegovi očrtani krožnici. + + + + \bzgled + Let $P$ be an arbitrary point on the shorter arc $AB$ of the circumcircle of a rectangle $ABCD$. + Suppose that $L$ and $M$ are the foots of the perpendiculars from the point $P$ on the + diagonals $AC$ and $BD$, respectively. Prove that the length of the line segment $LM$ does not depend + on the position of the point $P$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.15.pic} +\caption{} \label{sl.skk.4.5.15.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Naj bo točka $O$ središče krožnice $k$ (Figure \ref{sl.skk.4.5.15.pic}). +Štirikotnik $PMOL$ je tetiven, ker je $\angle OLP + \angle OMP +=90^0+90^0= 180^0$ (izrek \ref{TetivniPogoj}). Označimo z $l$ +očrtano krožnico tega štirikotnika. Ker sta kota $OLP$ in $OMP$ oba +prava, je daljica $OP$ (oz. polmer krožnice $k$) premer krožnice +$l$. Potem je $LM$ tetiva krožnice $l$, ki pripada obodnemu kotu +$\angle LOM =\angle AOB$, ki je konstanten. Ne glede na izbiro +točke $P$ je daljica $LM$ tetiva krožnice s konstantnim premerom +$OA$, ki pripada konstantnem obodnem kotu $AOB$ (oziroma ustreznemu +konstantnemu središčnem kotu te krožnice). Tetivi, ki pripadata +skladnima središčnima kotoma skladnih krožnic, sta med seboj skladni, +zato dolžina daljice $LM$ ni odvisna od lege točke $P$. + \kdokaz + + Lastnosti tetivnega štirikotnika pogosto uporabljamo tudi za dokazovanje + različnih +lastnosti trikotnika. + + + \bzgled \label{PedalniVS} + The orthocentre of an acute triangle is the incentre of its \index{trikotnik!pedalni} pedal triangle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.16.pic} +\caption{} \label{sl.skk.4.5.16.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Naj bodo $AA'$, $BB'$ in $CC'$ višine trikotnika $ABC$, ki se sekajo + v višinski točki $V$ tega +trikotnika (Figure \ref{sl.skk.4.5.16.pic}). Če je $A_1$ središče +stranice $BC$, točki $B'$ in $C'$ ležita na krožnici $k(A_1,A_1B)$ +(izrek \ref{TalesovIzrKroz2}). Torej je štirikotnik $BC'B'C$ +tetiven, zato je po izreku \ref{TetivniPogojZunanji} $\angle +AC'B'\cong \angle ACB = \gamma$. Analogno dokažemo, da je +štirikotnik $AC'A'C$ tetiven, zato je tudi $\angle BC'A'\cong +\angle ACB = \gamma$. Torej sta kota $AC'B'$ in $BC'A'$ skladna. Ker +je $CC'\perp AB$, sta skladna tudi kota $CC'B'$ in $CC'A'$. To +pomeni, da je premica $C'C$ simetrala kota $A'C'B'$. Analogno sta +tudi premici $A'A$ in $B'B$ simetrali ustreznih notranjih kotov +trikotnika $A'B'C'$, zato je točka $V$ središče trikotniku $A'B'C'$ +včrtane krožnice. + \kdokaz + + Iz dokaza prejšnje trditve (\ref{PedalniVS}) lahko ugotovimo, da so koti, ki jih določajo stranice +pedalnega trikotnika $A'B'C'$ s stranicami trikotnika +$ABC$, enaki ustreznim kotom trikotnika $ABC$. To dejstvo bomo +uporabili v naslednjem primeru. + + + + \bzgled \label{PedalniLemaOcrtana} + Let $O$ be the circumcentre of a triangle $ABC$. + Prove that the lines $OA$, $OB$ and $OC$ are perpendicular to the corresponding sides of the + pedal triangle $A'B'C'$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.17.pic} +\caption{} \label{sl.skk.4.5.17.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.5.17.pic}). + +Z $L$ označimo presečišče premic $OA$ in $B'C'$. Dovolj je dokazati, +da je notranji kot pri oglišču $L$ trikotnika $C'LA$ pravi kot. +Izračunajmo druga dva kota tega trikotnika. Iz prejšnjega zgleda +\ref{PedalniVS} je kot pri oglišču $C'$ enak $\gamma$. Trikotnik +$AOB$ je enakokrak in $\angle AOB=2\gamma$ (izrek \ref{SredObodKot}). +Torej velja (izreka \ref{enakokraki} in \ref{VsotKotTrik}) + $\angle C' AL=\angle BAO =90^0-\gamma$, +zato je $\angle ALC'=90^0$. + \kdokaz + +Direktna posledica trditev \ref{PedalniVS} in \ref{PedalniLemasPQR} +je naslednja trditev. + + + + \bzgled \label{PedalniLemasLMN} + Let $P$, $Q$ and $R$ be the midpoints of those arcs $BC$, $AC$ and $AB$ + of the circumcircle of a triangle $ABC$ not containing the vertices $A$, $B$ and $C$. + Suppose that the point $S$ is the incentre of the triangle $ABC$ + and $L=SA\cap QR$, $M=SB\cap PR$ and $N=SC\cap PQ$. Then the triangles $LMN$ and $ABC$ have + the common incentre. + (Figure \ref{sl.skk.4.5.18.pic}). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.18.pic} +\caption{} \label{sl.skk.4.5.18.pic} +\end{figure} + + + + \bzgled \label{Miquelova točka} + Let $P$, $Q$ and $R$ be an arbitrary points on the sides $BC$, $AC$ and $AB$ + of the triangle $ABC$, respectively. Prove that the circumcircles of triangles + $AQR$, $PBR$ and $PQC$ intersect at in one point (so-called \index{točka!Miquelova} + \pojem{Miquel point}\color{green1}\footnote{Točko imenujemo po + francoskem matematiku \index{Miquel, A.} \textit{A. Miquelu} (1816–-1851), ki je + to trditev objavil leta 1838 kot članek v Liouvilleovem + (\index{Liouville, J.}\textit{J. Liouville} (1809–-1882), francoski + matematik) časopisu. Toda, kot je to pogosto pri matematiki, Miquel + ni bil prvi, ki je omenjeni izrek dokazal. Že deset let pred njim je + to dejstvo odkril in objavil znani švicarski matematik + \index{Steiner, J.} \textit{J. Steiner} (1769--1863).}). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.2.pic} +\caption{} \label{sl.skk.4.5.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.5.2.pic}) + +Označimo s $k_A$, $k_B$ in $k_C$ očrtane krožnice trikotnikov +$AQR$, $PBR$ in $PQC$ ter notranje kote trikotnika $ABC$ po vrsti z +$\alpha$, $\beta$ in $\gamma$. Naj bo $S$ drugo presečišče krožnic +$k_B$ in $k_C$ (dokaz je podoben tudi v primeru, če je $S = P$). +Štirikotnika $BPSR$ in $PCQS$ sta tetivna, zato je $\angle RSP = +180^0 - \beta$ in $\angle QSP = 180^0 -\gamma$ (izrek +\ref{TetivniPogoj}). Iz tega sledi $\angle RSQ = \beta +\gamma$ in +potem tudi $\angle RAQ + \angle RSQ =\alpha + \beta +\gamma = +180^0$. Tudi štirikotnik $ARSQ$ je tetiven (izrek +\ref{TetivniPogoj}) oz. ima svojo očrtano krožnico, ki je +pravzaprav krožnica $k_A$, ki je očrtana trikotniku $AQR$. +To pomeni, da se krožnice $k_A$, $k_B$ in $k_C$ sekajo v točki $S$. + \kdokaz + + V poglavju \ref{pogINV} bomo dokazali eno posplošitev prejšnje + trditve (zgled \ref{MiquelKroznice}). + + + + \bnaloga\footnote{45. IMO Greece - 2004, Problem 1.} + Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The +circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$, +respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of +the angles $\angle BAC$ and $\angle MON$ intersect at $L$. Prove that the circumcircles +of the triangles $BML$ and $CNL$ have a common point lying on the side +$BC$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.4.IMO1.pic} +\caption{} \label{sl.skk.4.4.IMO1.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Označimo z $E$ presečišče simetrale kota +$BAC$ s stranico $BC$ trikotnika $ABC$ (Figure +\ref{sl.skk.4.4.IMO1.pic}). Dokažimo, da je $E$ iskana točka oz. +da leži na očrtanih krožnicah obeh trikotnikov $BML$ + in $CNL$. + +Ker iz konstrukcije točk $M$ in $N$ sledi $OM\cong ON$, je $OMN$ +enakokraki trikotnik z osnovnico $MN$. To pomeni, da je simetrala +$OL$ kota $MON$ hkrati simetrala stranice $MN$ (sledi iz +skladnosti trikotnikov $MSO$ in $NSO$, kjer je $S$ središče +daljice $MN$). Torej točka $L$ leži na simetrali daljice $MN$ + trikotnika, zato po izreku \ref{TockaN} leži na očrtani krožnici + $k$ + trikotnika $AMN$. Pogoj $AB\neq AC$ nam pove, da se simetrali kota $BAC$ in + stranice $MN$ (oz. kota $MON$) razlikujeta, zato je njun presek + točka. + + Če uporabimo izreka \ref{TetivniPogojZunanji} in \ref{ObodObodKot}, + dobimo: + \begin{eqnarray*} + \angle BCA &\cong& AMN \cong\angle ALN,\\ + \angle ABC &\cong& ANM \cong\angle ALM. + \end{eqnarray*} +Iz teh relacij in izreka \ref{TetivniPogojZunanji} sledi, da sta + $NLEC$ in $LMBE$ tetivna +štirikotnika. Torej točka $E$ leži na očrtanih krožnicah +obeh trikotnikov $BML$ + in $CNL$. + \kdokaz + +%________________________________________________________________________________ + \poglavje{Tangential Quadrilateral} \label{odd4Tangentni} + + + +Za večkotnik pravimo, da je +\index{štirikotnik!tangenten}\index{večkotnik!tangenten}\pojem{tangenten}, +če za njega obstaja včrtana krožnica, oziroma če obstaja takšna +krožnica, da so nosilke vseh stranic večkotnika njene tangente +(Figure \ref{sl.skk.4.6.1.pic}). Ugotovili smo že, da je vsak +trikotnik tangenten (izrek \ref{SredVcrtaneKrozn}) in prav tako +je tangenten tudi vsak pravilni večkotnik (izrek +\ref{sredVcrtaneKrozVeck}). Po drugi strani pa je jasno, da vsi +večkotniki niso tangentni. Na primer pravokotnik je štirikotnik, ki nima +včrtane krožnice. V tem razdelku se bomo posebej ukvarjali s +tangentnimi štirikotniki. + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.1.pic} +\caption{} \label{sl.skk.4.6.1.pic} +\end{figure} + +Ker je kvadrat pravilni večkotnik, je tudi tangentni štirikotnik. + Kako pa bi na +splošno ugotovili, ali je nek štirikotnik tangenten? Jasno je, da se +pri tangentnem štirikotniku (na splošno tudi večkotniku) simetrale +vseh njegovih notranjih kotov sekajo v eni točki (Figure +\ref{sl.skk.4.6.1.pic}). Ta pogoj je za tangentnost večkotnika tudi zadosten. +Žal pa ta pogoj ni preveč uporaben v +konkretnih primerih. Obstaja namreč uporabnejši pogoj, ki je za tangentnost štirikotnikov potreben in +hkrati zadosten. + + + + \bizrek \label{TangentniPogoj} + A quadrilateral $ABCD$ is tangential if and only if + $$|AB| + |CD| = |BC| + |AD|.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.2.pic} +\caption{} \label{sl.skk.4.6.2.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.6.2.pic}) + + ($\Rightarrow$) Predpostavimo najprej, da je štirikotnik $ABCD$ tangenten in + $k$ njegova včrtana krožnica. + Naj bodo $P$, $Q$, $R$ in $S$ +dotikališča stranic $AB$, $BC$, $CD$ in $DA$ s krožnico $k$. Ker so +ustrezne tangentne daljice skladne (izrek \ref{TangOdsek}), velja: +$AP \cong AS$, $BP \cong BQ$, $CQ \cong CR$ in $DR \cong DS$. Zato +je: + \begin{eqnarray*} +|AB| + |CD|&=&|AP| + |PB| + |CR| + |RD| \\&=& |AS| + |SD| + |BQ| + +|QC|\\&=&|AD| + |BC|. + \end{eqnarray*} + + ($\Leftarrow$) Dokažimo obratno trditev. Predpostavimo, da sta v štirikotniku + $ABCD$ vsoti parov nasprotnih stranic enaki, oz. da velja + $|AB| + |CD| = |BC| + |AD|$. Obstaja krožnica $k$, ki se dotika +stranic $AB$, $BC$ in $DA$ tega štirikotnika (njeno središče je +presečišče simetral notranjih kotov pri ogliščih $A$ in $B$ tega +štirikotnika). Dokažimo še, da se ta krožnica dotika tudi stranice +$CD$ štirikotnika $ABCD$. Naj bo $D'$ presečišče druge tangente iz +točke $C$ krožnice $k$ in premice $AD$. Predpostavimo, da je $D'\neq +D$. Brez škode za splošnost naj bo $\mathcal{B}(A,D',D)$. Ker je +štirikotnik $ABCD'$ tangenten z včrtano krožnico $k$, po že dokazanem +delu izreka velja $|AB| + |CD'| = |AD'|+|BC|$. Ker pa je po +predpostavki še $|AB| + |CD| = |AD| + |BC|$, velja tudi +$|CD|-|CD'| = |DA| - |D'A|=|DD'|$ oz. $|CD|= |CD'| + |DD'|$. To pa +ni mogoče zaradi trikotniške neenakosti \ref{neenaktrik} (točke $C$, +$D$ in $D'$ ne morejo biti kolinearne, ker bi sicer veljalo $C\in +AD$). Na podoben način pridemo do protislovja tudi v primeru, če je +$\mathcal{B}(A,D,D')$. Torej velja $D'= D$, zato je $ABCD$ tangenten +štirikotnik. + \kdokaz + + Direktna posledica prejšnjega kriterija sta naslednja izreka. + + \bizrek \label{TangDeltoidRomb} + A rhombus, a deltoid, and a square are + tangential quadrilaterals (Figure \ref{sl.skk.4.6.3.pic}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.3.pic} +\caption{} \label{sl.skk.4.6.3.pic} +\end{figure} + + \bizrek \label{TangParalelogram} + A parallelogram is a tangential quadrilateral + if and only if it is a rhombus (Figure \ref{sl.skk.4.6.3.pic}). + \eizrek + +V naslednjih dveh primerih bomo hkrati obravnavali +tetivne in tangentne štirikotnike. + + + + \bzgled Let $k_A$, $k_B$, $k_C$ and $k_D$ circles with centres $A$, + $B$, $C$ and $D$, such that two in a row (also $k_A$ and $k_D$) + are touching each other externally. Prove that the quadrilateral defined by + the touching points of circles is cyclic, and the quadrilateral $ABCD$ is tangential. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.4.pic} +\caption{} \label{sl.skk.4.6.4.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Naj bodo $P$, $Q$, $R$ in $S$ po vrsti dotikališča +danih krožnic ter $p$ in $r$ skupni tangenti ustreznih +krožnic v točkah $P$ in $R$ (Figure \ref{sl.skk.4.6.4.pic}). + +Najprej velja: + \begin{eqnarray*} + |AD| + |BC| &=& |AP| + |PD| + |BR| + |RC| =\\ + &=& |AQ| + |SD| + |QB| + |SC| =\\ + &=& |AB| + +|CD|. + \end{eqnarray*} + Zaradi tega je $ABCD$ tangentni štirikotnik (izrek \ref{TangentniPogoj}). + +Tangenti $p$ in $r$ delita notranja kota pri ogliščih $P$ in $R$ +štirikotnika $PQRS$ na kote, od katerih je vsak enak polovici +ustreznega središčnega kota (izrek \ref{ObodKotTang}). Omenjeni +središčni koti pa so notranji koti štirikotnika $ABCD$. Označimo +jih z $\alpha$, $\beta$, $\gamma$ in $\delta$. Zaradi tega je (izrek +\ref{VsotKotVeck}): + \begin{eqnarray*} + \angle QPS+ \angle SRQ&=& \angle QP,p+\angle p,PS+ \angle SR,r+\angle r,RQ=\\ + &=& \frac{1}{2}\alpha+\frac{1}{2}\delta+\frac{1}{2}\gamma+\frac{1}{2}\beta=\\ + &=& \frac{1}{2}\left(\alpha+\delta+\gamma+\beta\right)=\\ + &=& \frac{1}{2}\cdot360^0=180^0, + \end{eqnarray*} +kar pomeni, da je $PQRS$ tetivni štirikotnik. + \kdokaz + + Jasno je, da je včrtana + krožnica štirikotnika $ABCD$ hkrati očrtana krožnica štirikotnika $PQRS$. + Ker so namreč po predpostavki + $PAQ$, $QBR$, $RCS$ in $SDP$ enakokraki trikotniki z osnovnicami + $PQ$, $QR$, $RS$ in $SP$, so simetrale notranjih kotov štirikotnika $ABCD$ + hkrati simetrale stranic štirikotnika $PQRS$ (Figure + \ref{sl.skk.4.6.4a.pic}). To je tudi drugi (enostavnejši) + način dokaza drugega dela prejšnjega zgleda - trditve, da je $PQRS$ + tetivni štirikotnik. + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.4a.pic} +\caption{} \label{sl.skk.4.6.4a.pic} +\end{figure} + + + + \bzgled \label{tetivTangLema} + Let $L$ be the intersection of the diagonals of a cyclic quadrilateral $ABCD$. + Prove that the foots of the perpendiculars from the point $L$ on the sides of + this quadrilateral are the vertices of a tangential quadrilateral. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.5.pic} +\caption{} \label{sl.skk.4.6.5.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + Naj bodo $P$, $Q$, $R$ in $S$ pravokotne projekcije iz točke $L$ na stranicah + $AB$, $BC$, $CD$ in $DA$ +štirikotnika $ABCD$ (Figure \ref{sl.skk.4.6.5.pic}). +Zaradi ustreznih pravih kotov sta $PBQL$ in $APLS$ +tetivna štirikotnika (izrek \ref{TetivniPogoj}). Po predpostavki je tetiven tudi štirikotnik $ABCD$. Če to +uporabimo, dobimo enakost ustreznih obodnih kotov (izrek \ref{ObodObodKot}). Torej: +$$\angle SPL \cong \angle SAL = \angle DAC \cong \angle DBC += \angle LBQ \cong \angle LPQ.$$ +Iz tega sledi, da je premica $PL$ simetrala notranjega kota pri +oglišču $P$ štirikotnika $PQRS$. Podobno so premice $QL$, $RL$ in $SL$ +simetrale ostalih treh notranjih kotov tega štirikotnika. Torej je $L$ središče včrtane krožnice štirikotnika $PQRS$, +zato je ta tangenten. + \kdokaz + +Dokažimo še eno zanimivo lastnost tangentnih štirikotnikov. + + + + \bzgled + Prove that the incircles of triangles $ABC$ and $ACD$ touch each + other if and only if $ABCD$ is a tangential quadrilateral. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.6.pic} +\caption{} \label{sl.skk.4.6.6.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.6.6.pic}). + + Dokažimo najprej nekatere relacije, ki veljajo pri poljubnem konveksnem štirikotniku $ABCD$. +Naj bodo $P$, $Q$ in $X$ točke, v katerih se včrtana krožnica $k$ trikotnika $ABC$ dotika njegovih stranic +$AB$, $BC$ in $CA$, ter $R$, $S$ in $Y$ točke, v katerih se včrtana krožnica $l$ trikotnika $ACD$ dotika njegovih stranice $CD$, $DA$ in $AC$. Najprej velja (izrek \ref{TangOdsek}): + \begin{eqnarray*} + |AX| &=& |AP|=\frac{1}{2}\left(|AX|+|AP|\right)=\frac{1}{2}\left(|AC|-|CX|+|AB|-|BP|\right)\\ + &=& \frac{1}{2}\left(|AC|-|CQ|+|AB|-|BQ|\right)= + \frac{1}{2}\left(|AC|+|AB|-|BC|\right), + \end{eqnarray*} + torej velja: + $$|AX|=\frac{1}{2}\left(|AC|+|AB|-|BC|\right).$$ + Na enak način dokažemo, da velja tudi: + $$|AY|=\frac{1}{2}\left(|AC|+|AD|-|DC|\right).$$ + Sedaj lahko začnemo z dokazovanjem ekvivalence. + +Krožnici $k$ in $l$ se dotikata natanko tedaj, ko je $X = Y$ oz. $|AX| = |AY|$. Zadnja enakost velja natanko tedaj, ko je: $$\frac{1}{2}\left(|AC|+|AB|-|BC|\right)=\frac{1}{2}\left(|AC|+|AD|-|DC|\right)$$ + oz. $|AB| + |DC| = |AD| + |BC|$, ta pa je izpolnjena natanko tedaj, ko je $ABCD$ tangenten štirikotnik (izrek \ref{TangentniPogoj}). + \kdokaz + +Posledica tega je naslednja trditev. + + + + \bzgled + Let $ABCD$ be a tangential quadrilateral. + Then the incircles of the triangles $ABC$ and $ACD$ touch each other + if and only if + the incircles of the triangles $ABD$ and $CBD$ touch each + other (Figure \ref{sl.skk.4.6.6a.pic}). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.6.6a.pic} +\caption{} \label{sl.skk.4.6.6a.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Trditvi, da se včrtani krožnici trikotnikov $ABC$ in $ACD$ oz. trikotnikov $ABD$ in $CBD$ dotikata, sta + ekvivalentni s trditvijo, da je štirikotnik $ABCD$ tangenten. To pomeni, da sta tudi začetni trditvi ekvivalentni. + \kdokaz + +%_______________________________________________________________________________ + \poglavje{Bicentric Quadrilateral} \label{odd4TetivniTangentni} + + Nekateri štirikotniki so hkrati tetivni in tangentni. Imenujemo jih \index{štirikotnik!tetivnotangentni}\pojem{tetivnotangentni} ali \index{štirikotnik!bicentrični}\pojem{bicentrični} štirikotniki. Kateri + štirikotniki so to? Kvadrat je vsekakor eden tak. Ali je edini? Odgovor je negativen. Štirikotnik, ki ga dobimo iz zgleda \ref{tetivTangLema}, je vedno tangenten. V določenem primeru bo veljalo, da je ta tudi tetiven. + Velja namreč naslednja trditev. + + + + \bizrek \label{tetivTangIzrek} + If $L$ is the intersection of the perpendicular diagonals of a cyclic quadrilateral + $ABCD$, then the foots of the perpendiculars from the point $L$ on the sides of + this quadrilateral are the vertices of a bicentric quadrilateral. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.7.2.pic} +\caption{} \label{sl.skk.4.7.2.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.7.2.pic}) + +Vpeljimo iste oznake kot v zgledu \ref{tetivTangLema}. Dokazali smo že, da je $PQRS$ tangenten štirikotnik. Dokažimo še, da je tetiven. Iz dokaza trditve iz omenjenega zgleda \ref{tetivTangLema} sledi: +\begin{eqnarray*} + \angle SPQ &=& \angle SPL+\angle LPQ = \angle SAL+\angle LBQ =\\ + &=& \angle DAC + \angle DBC + = 2\cdot\angle DBC +\end{eqnarray*} + oziroma $\angle SPQ= 2\cdot\angle DBC$. Analogno je tudi $\angle SRQ= 2\cdot\angle ACB$. + Ker je po predpostavki $AC\perp BD$, je $CLB$ pravokotni trikotnik, zato velja: + $$\angle SRQ+\angle SRQ=2\cdot(\angle DBC+\angle ACB)=2\cdot 90^0=180^0.$$ + Po izreku \ref{TetivniPogoj} je $PQRS$ tetivni štirikotnik. + \kdokaz + + Ni se težko prepričati, da velja tudi obratna trditev. + + + + \bizrek + Let $PQRS$ be a bicentric quadrilateral. Suppose that point $L$ is + the incentre of this quadrilateral and at the same time the intersection of the diagonals + of a cyclic quadrilateral $ABCD$. If $P$, $Q$, $R$ and $S$ the foots of the perpendiculars + from the point $L$ on the sides of quadrilateral $ABCD$, then $AC\perp BD$. + \eizrek + +V naslednji nalogi bomo videli, da za tri nekolinearne točke $A$, $B$ in $C$ obstaja ena sama točka $D$, tako da je $ABCD$ tetivnotangentni štirikotnik. + + + \bnaloga\footnote{4. + IMO Czechoslovakia - 1962, Problem 5.} + On the circle $k$ there are given three distinct points $A$, $B$, $C$. Construct (using + only straightedge and compasses) a fourth point $D$ on $k$ such that a circle + can be inscribed in the quadrilateral thus obtained. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.5.IMO1.pic} +\caption{} \label{sl.skk.4.5.IMO1.pic} +\end{figure} + + +\textbf{\textit{Solution.}} Brez škode za splošnost predpostavimo, +da je $AB\geq BC$. + +Naj bo $D$ točka, ki izpolnjuje pogoje iz naloge oz. takšna, da je +$ABCD$ tetivnotangentni štirikotnik (Figure +\ref{sl.skk.4.5.IMO1.pic}). Označimo z $a$, $b$, $c$ in $d$ po vrsti +stranice $AB$, $BC$, $CD$ in $DA$ tega štirikotnika ter z $\alpha$, +$\beta$, $\gamma$ in $\delta$ njegove notranje kote pri ogliščih +$A$, $B$, $C$ in $D$. Iz pogoja tetivnosti štirikotnika $ABCD$ +(izrek \ref{TetivniPogoj}) sledi $\delta=180^0-\beta$, iz njegove +tangentnosti (izrek \ref{TangentniPogoj}) pa $a+c=b+d$ oz. +$d-c=a-b$. Na ta način nalogo prevedemo v načrtovanje tretjega +oglišča $D$ trikotnika $ACD$, kjer so dani stranica $AC$, kot +$\angle ADC=180^0-\beta$ in razlika stranic $AD-CD=AB-BC=a-b$. Naj bo +$E$ točka na daljici $AD$, za katero velja $DE\cong DC$. Potem je +$AE=AD-DE=AD-CD=a-b$. Trikotnik $ECD$ je enakokrak, zato po izreku +\ref{enakokraki} sledi $\angle CED\cong\angle DCE$. Zato je $\angle +AEC=180^0-\angle DEC=180^0-\frac{1}{2}\beta$. To nam omogoča +konstrukcijo trikotnika $ACE$ oz. točke $E$. + +Najprej načrtajmo točko $E$ kot presečišče loka $l$ (glej konstrukcijo, +ki je opisana v izreku \ref{ObodKotGMT}) $180^0-\frac{1}{2}\angle ABC$) ter krožnice +$j(A,AB-BC)$ (če je $AB\cong AC$, privzamemo $E=A$). Točko $D$ +nato lahko načrtamo kot presečišče poltraka $AE$ in simetrale +$s_{EC}$ daljice $EC$ (v primeru $AB\cong AC$, oz. $E=A$ je $D$ +drugo presečišče simetrale $s_{EC}=s_{AC}$ s krožnico $k$). + +Dokažimo, da točka $D$ izpolnjuje pogoje iz naloge oz. da je $ABCD$ +tetivnotangentni štirikotnik. Najprej bomo obravnavali primer, ko +je $AB>BC$. + +Po konstrukciji točka $D$ leži na simetrali daljice $EC$, zato je +$DE\cong DC$ in tudi (izrek \ref{enakokraki}) $\angle DEC\cong\angle +DCE$. Točko $E$ smo načrtali tako, da leži na loku $l$ s tetivo $AC$ in +obodnim kotom $180^0-\frac{1}{2}\angle ABC$, zato je $\angle +AEC=180^0-\frac{1}{2}\angle ABC$. Ker je po konstrukciji +$\mathcal{B}(A,E,D)$, je $\angle DCE\cong\angle +DEC=\frac{1}{2}\angle ABC$. Iz enakokrakega trikotnika $EDC$ + po izreku \ref{VsotKotTrik} sledi $\angle ADC=\angle EDC=180^0-\angle + ABC$. Torej velja $\angle EDC+\angle ABC=180^0$, + zato je po izreku \ref{TetivniPogoj} $ABCD$ tetivni + štirikotnik oz. $D\in k$. + + Dokažimo še, da je $ABCD$ tangentni štirikotnik. V prvem delu + dokaza (tetivnost) smo že + videli, da velja $DE\cong DC$. Točka $E$ po konstrukciji leži + na krožnici $j(A,|AB-BC|)$, zato je $|AE|=|AB|-|BC|$. Ker velja še + $\mathcal{B}(A,E,D)$, je $|AD|-|CD|=|AD|-|DE|=|AE|=|AB|-|BC|$. Iz tega sledi + $|AD|+|BC|=|AB|+|CD|$ in je po izreku \ref{TangPogoj} štirikotnik + $ABCD$ tangenten. + + Če je $AB\cong BC$, točka $D$ že po + konstrukciji leži na krožnici $k$. Ker točki $B$ in $D$ obe ležita na simetrali + daljice $AC$, je štirikotnik $ABCD$ deltoid, zato je tudi + tangenten (izrek \ref{TangDeltoidRomb}). + + Raziščimo še število rešitev naloge. Krožnica $k(A,AB-AC)$ in + lok $l$ se vedno sekata v eni točki $E$. Ker je $ABC<180^0$, je + $\angle AEC=180^0-\frac{1}{2}\beta>90^0$. + To pomeni, da simetrala $s_{EC}$ vedno seka poltrak $AE$ v eni točki + $D$ in velja $\mathcal{B}(A,E,D)$. To pomeni, da ima naloga vedno eno samo + rešitev. + \kdokaz + + + + + %______________________________________________________________________________ + \poglavje{Simson Line} \label{odd4Simson} + +Dokažimo najprej osnovni izrek. + + + \bizrek \label{SimpsPrem} + The foots of the perpendiculars + from an arbitrary point lying on the circumcircle of a triangle + to the lines containing the sides of this triangle are three collinear points. + The line containing these points is the so-called + \index{premica!Simsonova} + \pojem{Simson\footnote{Premico imenujemo po škotskem matematiku + \index{Simson, R.} + \textit{R. Simsonu} (1687--1768), čeprav je to lastnost + prvi objavil škotski matematik \index{Wallace, W.} + \textit{W. Wallace} + (1768--1843) šele leta 1799.} line}\color{blue}. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.7.1a.pic} +\caption{} \label{sl.skk.4.7.1a.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $S$ poljubna točka očrtane krožnice + $k$ +trikotnika $ABC$ ter $P$, $Q$ in $R$ pravokotne projekcije točke $S$ +na nosilke stranic $BC$, $AC$ in $AB$ (Figure \ref{sl.skk.4.7.1a.pic}). Brez +škode za splošnost predpostavimo, da velja + $\mathcal{B}(B,P,C)$, $\mathcal{B}(A,Q,C)$ in $\mathcal{B}(A,B,R)$. + V tem primeru sta točki $Q$ in $R$ na različnih bregovih premice $BC$, +zato je dovolj dokazati $\angle BPR \cong \angle CPQ$. Zaradi +ustreznih pravih kotov in lege točke $S$ so po izreku +\ref{TetivniPogoj} štirikotniki $BRSP$, $ABSC$, $ARSQ$ in $SPQC$ +tetivni, zato je + (izrek \ref{TetivniPogoj}: + \begin{eqnarray*} + \angle BPR &=& \angle BSR = \angle RSC - \angle BSC=\\ +&=& \angle RSC - (180° - \angle BAC) =\\ +&=& \angle RSC - \angle RSQ = \angle CSQ = \angle CPQ, + \end{eqnarray*} + kar pomeni, da so točke $P$, $Q$ in $R$ kolinearne. \kdokaz + + V nadaljevanju bomo obravnavali nadaljnje zanimive lastnosti Simsonove + premice. Ker vsaka točka $X$, ki leži na očrtani krožnici nekega + trikotnika, določa Simsonovo premico, bomo to premico označili z + $x$. Na ta način vsak trikotnik določa eno preslikavo $X\mapsto x$. + + + + \bzgled \label{SimsZgled1} + Let $P$ be an arbitrary point of the circumcircle $k$ of a triangle + $ABC$. Suppose that $P_A$ is the intersection of the perpendicular line of + the line $BC$ through the point $P$ with the circle $k$. + Prove that the line $AP_A$ is parallel to the Simson line $p$ + of the triangle at the point $P$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.7.1b.pic} +\caption{} \label{sl.skk.4.7.1b.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Naj bodo $X$, $Y$ in $Z$ pravokotne projekcije točke $P$ na premice + $BC$, $AC$ in $AB$ (Figure \ref{sl.skk.4.7.1b.pic}). Po izreku \ref{SimpsPrem} je Simsonova premica + $p$ določena s točkami $X$, $Y$ in $Z$. Podobno kot v izreku + \ref{SimpsPrem} je +štirikotnik $PYXC$ tetiven, zato je $\angle YXP \cong \angle ACP$. +Po izreku \ref{ObodObodKot} sta kota $ACP$ in $AP_AP$ nad +tetivo $AP$ skladna, zaradi tega je tudi $\angle YXP \cong \angle AP_AP$ +oz. $XY\parallel AP$ (izrek \ref{KotiTransverzala}). + \kdokaz + + + + \bzgled \label{SimsZgled2} + Let $P$ and $Q$ be arbitrary points lying on the circumcircle + $k(O,r)$ of a triangle $ABC$ and $p$ and $q$ their Simson lines. Prove that + $$\angle pq = \frac{1}{2}\angle POQ.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.7.1c.pic} +\caption{} \label{sl.skk.4.7.1c.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Naj bosta $X_P$ in $X_Q$ nožišči pravokotnic iz točk $P$ in $Q$ + na premici $BC$ ter $P_A$ in $Q_A$ +presečišči teh pravokotnic s krožnico $k$ (Figure +\ref{sl.skk.4.7.1c.pic}). Simsonovi premici $p$ in $q$ sta po vrsti +vzporedni s premicami $AP_A$ in $AQ_A$ (zgled \ref{SimsZgled1}). +Torej je kot, ki ga določata premici $p$ in $q$, skladen obodnemu +kotu $Q_AAP_A$, ta pa je enak polovici središčnega kota $Q_AOP_A$ +(izrek \ref{SredObodKot}) oziroma polovici kota $QOP$ (ker je +trapez $PP_AQ_AQ$ tetiven in po izreku \ref{trapezTetivEnakokr} tudi +enakokrak oz. $PQ \cong P_AQ_A$). + \kdokaz + + + \bzgled \label{SimsZgled3} + Let $P$ be an arbitrary point of the circumcircle $k$ of a triangle + $ABC$, $p$ its Simson line and $V$ the orthocentre of this triangle. + Prove that the line $p$ bisects the line segment $PV$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.7.1d.pic} +\caption{} \label{sl.skk.4.7.1d.pic} +\end{figure} + + \textbf{\textit{Proof.}} +Naj bosta $P_A$ in $X$ točki, definirani kot v zgledu +\ref{SimsZgled1} (Figure \ref{sl.skk.4.7.1d.pic}). Naj bosta še $V'$ +in $P'$ točki, ki sta simetrični točkam $V$ in $P$ glede na premico +$BC$. Točka $V'$ leži na očrtani krožnici $k$ trikotnika $ABC$ +(izrek \ref{TockaV'}). Zaradi lastnosti simetrije +oz. osnega zrcaljenja (glej razdelek \ref{odd6OsnZrc}), izreka +\ref{KotiTransverzala}, izreka \ref{ObodObodKot} in zgleda +\ref{SimsZgled1} velja: + $$\angle VP'P\cong\angle V'PP'\cong\angle AV'P + \cong\angle AP_AP\cong\angle p,PP'.$$ + Torej je $\angle VP'P \cong \angle p,PP'$, zato sta po izreku + \ref{KotiTransverzala} premici $VP'$ in $p$ vzporedni. Ker +je točka $X$ središče daljice $PP'$, premica $p$ vsebuje srednjico +trikotnika $PVP'$ (izrek \ref{srednjicaTrik}) oz. središče njegove +stranice $PV$. + \kdokaz + +V razdelkih \ref{odd5Hamilton} in \ref{odd7SredRazteg} bomo dokazali še dve lastnosti +Simsonovih premic (zgleda \ref{HamiltonSimson} in \ref{SimsEuler}), ki sta povezani s Hamiltonovim izrekom oz. +Eulerjevo krožnico trikotnika. + + + + +%________________________________________________________________________________ + \poglavje{Torricelli Point} \label{odd4Torricelli} + +V tem razdelku bomo podali še eno znamenito točko trikotnika. + + + \bizrek \label{izrekTorichelijev} + On each side of a triangle $ABC$ the equilateral triangles $BEC$, $CFA$ and $AGB$ + are externally erected. Prove: + \begin{enumerate} + \item $AE$, $BF$ and $CG$ are congruent line segments; + \item the lines $AE$, $BF$ and $CG$ intersect at one point + (so-called \pojem{Torricelli\footnote{Problem je prvi zastavil francoski matematik \index{Fermat, P.} \textit{P. + Fermat} (1601--1665) kot izziv italijanskem matematiku in fiziku \index{Torricelli, E.} \textit{E. + Torricelliju} (1608--1647). Torricellijevo rešitev je objavil njegov učenec - italijanski matematik in fizik \textit{V. Viviani} (1622–-1703) - leta 1659. Omenjeno točko imenujemo tudi \index{točka!Fermatova}\pojem{Fermat + point}.} + point} \color{blue}of this triangle) and every two of them determine an angle with measure $60^0$. + \end{enumerate} + \index{točka!Torricellijeva} + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.7.1.pic} +\caption{} \label{sl.skk.4.7.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} + (Figure \ref{sl.skk.4.7.1.pic}) + + (\textit{i}) Trikotnika $AEC$ in $FBC$ sta skladna po izreku \textit{SAS} \ref{SKS} ($AC \cong FC$ , $CE \cong CB$ in +$\angle ACE \cong \angle FCB = \angle ACB + 60°$), zato je $AE\cong BF$. Analogno je tudi $AE\cong CG$. + +(\textit{ii}) Naj bodo $k$, $l$ in $j$ očrtane krožnice trikotnikov $BEC$, $CFA$ in $AGB$. Dokažimo najprej, da se te krožnice +sekajo v eni točki. S $T$ označimo drugo presečišče krožnic $k$ in $l$ ($T\neq C$). Štirikotnika $BECT$ in $CFAT$ sta tetivna, zato (izrek \ref{TetivniPogoj}) oba kota $BTC$ in +$ATC$ merita $120^0$. Torej tudi kot $ATB$ meri $120^0$, kar pomeni, da je štirikotnik $AGBT$ tetiven (izrek \ref{TetivniPogoj}) oz. točka $T$ tudi leži + na krožnici $j$. + +Dokažimo, da vsaka od daljic $AE$, $BF$, $CG$ poteka skozi točko $T$. Iz enakosti ustreznih obodnih +kotov (izrek \ref{ObodObodKot}) dobimo: + +\begin{eqnarray*} +\angle ATE&=&\angle ATF+\angle FTC+\angle CTE=\\ +&=&\angle ACF+\angle FAC+\angle CBE +=3\cdot 60^0=180^0. +\end{eqnarray*} + +Torej so $A$, $T$ in $E$ kolinearne točke, oz. točka $T$ leži na daljici $AE$. Analogno leži točka $T$ tudi +na daljicah $BF$ in $CG$. Jasno je, da velja tudi: +$\angle AE,BF\cong\angle ATF\cong\angle ACF=60^0$. + \kdokaz + + V razdelku \ref{odd9MetrInv} (izrek \ref{izrekToricheliFerma}) bomo dokazali še eno zanimivo lastnost Torricellijeve točke. + + +%________________________________________________________________________________ + \poglavje{Excircles of a Triangle} \label{odd4Pricrt} + + Dokazali smo že, da za poljubni trikotnik obstajata + očrtana in včrtana krožnica. Prva vsebuje vsa oglišča +trikotnika, druga pa se dotika vseh njegovih stranic. Sedaj bomo +pokazali, da obstajajo tudi krožnice, ki se dotikajo po ene +stranice in dveh nosilk stranic trikotnika. + + + \bizrek + The bisector of the interior angle at vertex $A$ and the bisectors of + the exterior angles at vertices $B$ and $C$ of a triangle $ABC$ intersect at one point, + which is the centre of the circle touching the side $BC$ and the lines containing the sides $AB$ and + $AC$. It is so-called \index{pričrtane krožnice trikotnika} \pojem{excircle of the triangle}\color{blue}. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.27.1.94_veliki_zadatak_lema.pic} +\caption{} \label{sl.27.1.94_veliki_zadatak_lema.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Izrek dokažemo podobno kot pri včrtani + krožnici trikotnika. Simetrali +zunanjih kotov pri ogliščih $B$ in $C$ nista vzporedni in se + sekata v neki točki - označimo jo s $S_a$ (Figure +\ref{sl.27.1.94_veliki_zadatak_lema.pic}). Ker točka $S_a$ leži na +teh dveh simetralah, velja $A,S_a\div BC$ in je $S_a$ enako oddaljena od premic $AB$, $BC$ in +$AC$. Torej $S_a$ pripada tudi simetrali notranjega kota pri +oglišču $A$ in je središče krožnice, ki se dotika stranice $BC$ ter +nosilk stranic $AB$ in $AC$. + \kdokaz + + Sedaj smo pripravljeni dokazati t. i. \index{velika naloga} + \pojem{‘‘veliko nalogo’’}, ki je zelo uporabna pri + načrtovanju trikotnikov. + + + + + \bizrek \label{velikaNaloga} + Let $P$, $Q$, $R$ be the touching points of the incircle $k(S,r)$ + of a triangle $ABC$ with the sides $BC=a$, $AC=b$, $AB=c$ ($b>c$) and $P_i$, $Q_i$, $R_i$ + ($i\in \{a,b,c\}$) the touching points of the excircles + $k_i(S_i,r_i)$ with lines $BC$, $AC$ in $AB$. Let $l(O,R)$ + be the circumcircle of this triangle with the semiperimeter + $s=\frac{a+b+c}{2}$, $A_1$ the midpoint of the line segments $BC$, $M$ and $N$, + intersections of the line $OA_1$ with the circle $l$ ($N,A\div BC$) and + $M’$, $N’$ the foots of the perpendiculars from these points on the line $AB$ (Figure + \ref{sl.27.1.94_veliki_zadatak.pic}). Then: + \vspace*{2mm} + + (\textit{i}) $AQ_a\cong AR_a=s$, \hspace*{0.4mm} (\textit{ii}) + $AQ\cong AR=s-a$, \hspace*{0.4mm} (\textit{iii}) $QQa\cong RRa\cong a$, + % \vspace*{1mm} + + (\textit{iv}) $PPa=b-c$,\hspace*{1mm} + (\textit{v}) $P_bP_c=b+c$, + % \vspace*{1mm} + + (\textit{vi}) + $A_1$ is the midpoint of the line segment $PP_a$ in $P_bP_c$, + %\vspace*{1mm} + + (\textit{vii}) $A_1N= \frac{r_a- r}{2}$,\hspace*{2mm} + (\textit{viii}) $A_1M= \frac{r_b + r_c }{2} $,\hspace*{1mm} + + + (\textit{ix}) $r_a +r_b +r_c =4R+r$,\footnote{To lastnost + trikotnika je leta 1790 odkril francoski matematik \index{L'Huilier, S. A. J.} + \textit{S. A. J. L'Huilier} + (1750--1840).} + %\vspace*{1mm} + + (\textit{x}) $NN’= \frac{r_a +r}{2}$, \hspace*{1mm} (\textit{xi}) + $MM’= \frac{r_b -r_c }{2 }$,\hspace*{1mm} (\textit{xii}) + $N’B\cong AM’=\frac{ b - c}{2}$, + % \vspace*{1mm} + + (\textit{xiii}) $AN’\cong BM’= \frac{b + c}{2} $, \hspace*{1mm} (\textit{xiv}) $M’N’\cong b$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.27.1.94_veliki_zadatak.pic} +\caption{} \label{sl.27.1.94_veliki_zadatak.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Preden začnemo z dokazovanjem, omenimo, da po izreku \ref{TockaN} + točka $N$ leži na simetrali notranjega kota $BAC$ trikotnika + $ABC$. + + (\textit{i}) Če uporabimo enakost tangentnih daljic + (izrek \ref{TangOdsek}), dobimo: + \begin{eqnarray*} + AQ_a&\cong &AR_a= \frac{1}{2}\left(AQ_a+AR_a\right)= + \frac{1}{2}\left(AB+BR_a+AC+CQ_a\right)\\ + &=& \frac{1}{2}\left(AB+BP_a+AC+CP_a\right)= +\frac{1}{2}\left(AB+BC+AC\right)=s. + \end{eqnarray*} + + (\textit{ii}) Na podoben način dobimo: + \begin{eqnarray*} + AQ&\cong &AR= \frac{1}{2}\left(AQ+AR\right)= + \frac{1}{2}\left(AB-BR+AC-CQ\right)\\ + &=& \frac{1}{2}\left(AB-BP+AC-CP\right)= +\frac{1}{2}\left(AB+AC-BC\right)=s-a. + \end{eqnarray*} + + (\textit{iii}) $QQ_a\cong AQ_a-AQ=a$. + + (\textit{iv}) Enakost dokažemo tako, da + najprej izračunamo:\\ $BP$ in $CP_a\cong CQ_a$. + + (\textit{v}) $P_bP_c\cong CP_c+BP_b-a=2s-a=b+c$. + + (\textit{vi}) Sledi iz $BP\cong CP_a=s-b$. + + (\textit{vii}) Točki $A_1$ in $N$ sta središči diagonal trapeza + $SPS_aP_a$ z + osnovnicama $SP\cong r$ in +$S_aP_a\cong r_a$. Enakost sledi iz izreka \ref{srednjTrapez}. + + (\textit{viii}) Premici $NA$ in $S_cS_b$ + sta pravokotni (simetrali notranjega + in zunanjega kota pri oglišču +$A$), zato točka $M$ pripada premici $S_cS_b$. Iskana enakost sledi +iz dejstva, da je daljica $A_1M$ srednjica trapeza +$S_cP_cP_bS_b$ (izrek \ref{srednjTrapez}). + + (\textit{ix}) Sledi neposredno iz $2\cdot R=NM=NA_1+A_1M$ in + (\textit{vii}) ter (\textit{viii}). + +(\textit{x}) Sledi iz izreka \ref{srednjTrapez} in dejstva, da je +$NN'$ srednjica trapeza $SRR_aS_a$. + +(\textit{xi}) Točki $M$ in $M'$ sta središči diagonal trapeza +$R_cS_cR_bS_b$ z osnovnicama $R_cS_c\cong r_c$ in $R_bR_b\cong r_b$. Enakost +sledi iz izreka \ref{srednjTrapez}. + + (\textit{xii}) Točka $N$ je središče daljice $RR_a$, zato je: + $$N'B=AN'-AB= \frac{1}{2}\left(AR_a+AR\right)-c= + \frac{1}{2}\left(s+(s-a)\right)-c= \frac{1}{2}\left(b-c\right).$$ + + (\textit{xiii}) in (\textit{xiv}) sledita + direktno iz dokazane enakosti (\textit{xii}). + \kdokaz + +%KONSTRUKCIJA (VN) + + \bzgled + Construct a triangle $ABC$, with given: + + (\textit{a}) $a$, $b-c$, $r$, \hspace*{3mm} (\textit{b}) + $b-c$, $r$, $v_b$, \hspace*{3mm} (\textit{c}) $a$, $b+c$, $r$, + \hspace*{3mm}(\textit{č}) $R$, $r$, $r_a$. + \ezgled + + + +\textbf{\textit{Solution.}} + Pri vsaki konstrukciji bomo uporabili veliko nalogo - \ref{velikaNaloga}. + Uporabili bomo + enake oznake. + +\begin{figure}[!htb] +\centering +\input{sl.27.1.94_veliki_zadatak_konstr.pic} +\caption{} \label{sl.27.1.94_veliki_zadatak_konstr.pic} +\end{figure} + +\begin{figure}[!htb] +\centering +\input{sl.27.1.94_veliki_zadatak_konstr2.pic} +\caption{} \label{sl.27.1.94_veliki_zadatak_konstr2.pic} +\end{figure} + + + +(\textit{a}) Ker je $PP_a=b-c$ in točka $A_1$ skupno središče +stranice $BC$ in + daljice $PP_a$, velja tudi $PA_1 = \frac{1}{2}( b - c)$ (Figure +\ref{sl.27.1.94_veliki_zadatak_konstr.pic}). +Najprej načrtamo stranico $BC$, nato pa njeno središče +$A_1$, točko $P$, včrtano krožnico trikotnika, tangenti iz +oglišč $B$ in $C$ ter na koncu še oglišče $A$. + + + +(\textit{b}) Podobno kot pri prejšnjem primeru. Najprej načrtamo +daljico $PA_1$, nato včrtano krožnico trikotnika $ABC$ (Figure +\ref{sl.27.1.94_veliki_zadatak_konstr.pic}). +Potrebno je uporabiti še pogoj višine iz oglišča $B$. Z $L$ +označimo pravokotno projekcijo iz točke $A_1$ na premici $AC$. +Daljica $A_1L$ je srednjica trikotnika $CBB'$, zato je $A_1L +=\frac{1}{2}BB'=\frac{1}{2} v_b$. Torej premico $AC$ lahko +konstruiramo kot skupno tangento včrtane krožnice in krožnice +$k(A_1, \frac{1}{2}v_b)$. Tako dobimo oglišče $C$, nato pa tudi +oglišči $B$ in $A$. + + +(\textit{c}) Vemo, da velja $RR_a\cong a$ in $AN'=\frac{1}{2}(b+c)$, +točka $N'$ pa je središče daljice $RR_a$ (Figure \ref{sl.27.1.94_veliki_zadatak_konstr2.pic}). Torej iz danih +podatkov + najprej konstruiramo točke $A$, $N'$, $R$ in $R_a$, nato pa tudi $S$, $N$ + in +$S_a$. Na koncu narišemo včrtano in pričrtano krožnico - +stranice trikotnika pa ležijo na njunih skupnih tangentah. + + +(\textit{č}) Ker je $A_1N =\frac{1}{2}(r_a -r)$ in $MN = 2R$, lahko +načrtamo najprej točke $N$, $A_1$ in $M$, nato pa še očrtano +krožnico trikotnika $ABC$ in stranico $BC$ (Figure \ref{sl.27.1.94_veliki_zadatak_konstr2.pic}). Konstrukcijo lahko +končamo na dva načina. V prvem primeru nalogo prevedemo v že znano +konstrukcijo trikotnika: $a$, $R$, $r$ (zgled \ref{konstr_Rra}), v +drugem pa uporabimo enakost $RR_a\cong a$. + \kdokaz + + + + +%________________________________________________________________________________ +\naloge{Exercises} + +\begin{enumerate} + +\item Dolžine stranic nekega trikotnika so $6$, $7$ in $9$. Naj bodo $k_1$, $k_2$ in $k_3$ krožnice s središči +v ogliščih tega trikotnika. Krožnice se med seboj dotikajo tako, da + se krožnica s središčem v oglišču najmanjšega kota trikotnika z ostalima krožnicama dotika od znotraj, +preostali dve krožnici pa se dotikata od zunaj. Izračunaj dolžine polmerov teh treh krožnic. + +\item Dokaži, da je kot, ki ga določata sečnici krožnice, ki se med seboj sekata v zunanjosti krožnice, enak polovici razlike središčnih kotov, prirejenih lokoma, ki ležita med krakoma tega kota. + +\item Vrh kota $\alpha$ je zunanja točka krožnice $k$. Med krakoma tega kota ležita na krožnici dva +loka, ki sta v razmerju $3:10$. Večji od teh lokov ustreza središčnemu kotu $40^0$. Določi +mero kota $\alpha$. + +\item Dokaži, da je kot, ki ga oklepata tangenti krožnice, enak polovici razlike središčnih kotov, prirejenih lokoma, ki ležita med krakoma tega kota. + +\item Naj bo $L$ pravokotna projekcija poljubne točke $K$ krožnice $k$ + na njeni tangenti $t$ skozi točko $T\in k$ in $X$ točka, ki je +simetrična točki $L$ glede na premico $KT$. Določi geometrijsko +mesto točk $X$. + +\item Naj bosta $BB'$ in $CC'$ višini trikotnika $ABC$ in $t$ +tangenta očrtane krožnice tega trikotnika v točki $A$. Dokaži, da +je $B'C'\parallel t$. + +\item V pravokotnem trikotniku $ABC$ je nad kateto $AC$ kot premerom +načrtana krožnica, ki seka hipotenuzo $AB$ v točki $E$. Tangenta te +krožnice v točki $E$ seka drugo kateto $BC$ v točki $D$. Dokaži, da je $BDE$ +enakokraki trikotnik. + +\item V pravi kot z vrhom $A$ je včrtana krožnica, ki se dotika krakov tega kota v +točkah $B$ in $C$. Poljubna tangenta te krožnice seka premici $AB$ in $AC$, po vrsti +v točkah $M$ in $N$ (tako, da je je $\mathcal{B}(A,M,B)$). Dokaži, da velja: +$$\frac{1}{3}\left(|AB|+|AC|\right) < |MB|+|NC| < +\frac{1}{2}\left(|AB|+|AC|\right).$$ + +\item Dokaži, da je pri pravokotnem trikotniku vsota katet enaka vsoti +premerov očrtane in včrtane krožnice. + +\item Naj se simetrale notranjih kotov konveksnega štirikotnika sekajo v šestih različnih točkah. +Dokaži, da so štiri od teh točk oglišča tetivnega štirikotnika. + +\item Naj bodo: $c$ dolžina hipotenuze, $a$ in $b$ dolžini +katet ter $r$ polmer včrtane krožnice pravokotnega trikotnika. Dokaži da velja: +\begin{enumerate} + \item $2r + c \geq 2 \sqrt{ab}$, \item $a + b + c > 8r$. +\end{enumerate} + +\item Naj bosta $P$ in $Q$ središči krajših lokov $AB$ in $AC$ +pravilnemu trikotniku $ABC$ očrtane krožnice. Dokaži, da stranici $AB$ in $AC$ tega trikotnika razdelita +tetivo $PQ$ na tri skladne daljice. + +\item Naj bodo $k_1$, $k_2$, $k_3$, $k_4$ štiri krožnice, od katerih se vsaka od zunaj dotika ene stranice in dveh nosilk stranic poljubnega konveksnega +štirikotnika. Dokaži, da so središča teh krožnic konciklične točke. + +\item Krožnici $k$ in $l$ se dotikata od zunaj v točki $A$. Točki $B$ in $C$ sta dotikališči +skupne zunanje tangente teh dveh krožnic. Dokaži, da je $\angle BAC$ pravi kot. + +\item Naj bo $ABCD$ deltoid ($AB\cong AD$ in $CB\cong CD$). Dokaži: +\begin{enumerate} + \item $ABCD$ je tangenten štirikotnik, + \item $ABCD$ je tetiven štirikotnik natanko tedaj, ko je $AB\perp BC$. +\end{enumerate} + +\item Krožnici $k$ in $k_1$ se dotikata od zunaj v točki $T$, v kateri se sekata premici $p$ in $q$. Premica $p$ ima +s krožnicama še presečišči $P$ in $P_1$, premica $q$ pa $Q$ in $Q_1$. Dokaži, da je $PQ\parallel P_1Q_1$. + +\item Naj bo $MN$ skupna tangenta krožnic $k$ in $l$ ($M$ in $N$ sta dotikališči), +ki se sekata v točkah $A$ in $B$. Izračunaj mero vsote $\angle MAN+\angle MBN$. + +\item Naj bo $t$ tangenta trikotniku $ABC$ očrtane krožnice v točki $A$. Premica, ki je vzporedna s tangento $t$, seka stranici $AB$ in $AC$ v točkah $D$ in $E$. +Dokaži, da so točke $B$, $C$, $D$ in $E$ konciklične. + +\item Naj bosta $D$ in $E$ poljubni točki polkrožnice, ki je načrtana nad premerom $AB$. Naj bo $AD\cap BE= \{F\}$ in $AE\cap BD= \{G\}$. +Dokaži, da je $FG\perp AB$. + +\item Naj bo $M$ točka krožnice $k(O,r)$. Določi geometrijsko mesto središč + vseh tetiv te krožnice, ki imajo eno krajišče v točki $M$. + +\item Naj bosta $M$ in $N$ točki, ki sta simetrični nožišču $A'$ višine $AA'$ trikotnika $ABC$ glede na stranici $AB$ in $AC$, ter $K$ presečišče premic $AB$ in $MN$. Dokaži, +da so točke $A$, $K$, $A'$, $C$ in $N$ konciklične. + +\item Naj bo $D$ nožišče višine iz oglišča $A$ ostrokotnega trikotnika $ABC$ in $O$ +središče očrtane krožnice tega trikotnika. Dokaži, da je $\angle CAD\cong \angle BAO$. + +\item Naj bo $ABCD$ tetivni štirikotnik, $E$ višinska točka trikotnika $ABD$ in $F$ + višinska točka trikotnika $ABC$. Dokaži, da je štirikotnik $CDEF$ +paralelogram. + +\item Krožnici s središčema $O_1$ in $O_2$ se sekata v točkah $A$ in $B$. Premica $p$, +ki poteka skozi točko $A$, seka ti dve krožnici še v točkah $M_1$ in $M_2$. Dokaži, da je +$\angle O_1M_1B\cong\angle O_2M_2B$. + +\item Polkrožnica s središčem $O$ je načrtana nad premerom $AB$. +Naj bosta $C$ in $D$ takšni točki na daljici $AB$, da velja $CO\cong OD$. Vzporedni premici skozi točki $C$ in $D$ sekata polkrožnico v točkah $E$ in $F$. +Dokaži, da sta premici $CE$ in $DF$ pravokotni na premico $EF$. + +\item Na tetivi $AB$ krožnice $k$ s središčem $O$ naj leži točka $C$, točka $D$ pa naj bo +drugo presečišče krožnice $k$ z očrtano krožnico trikotnika $ACO$. Dokaži, da je +$CD\cong CB$. + +\item Naj bo $AB$ mimobežnica krožnice $k$. Premici $AC$ in $BD$ naj bosta tangenti +krožnice $k$ v točkah $C$ in $D$. +Dokaži, da velja: + $$||AC|-|BD||< |AB| < |AC|+|BD|.$$ + +\item Naj bo $S$ presečišče nosilk krakov $AD$ in $BC$ +trapeza $ABCD$ z osnovnico $AB$. Dokaži, da se očrtani krožnici trikotnikov $SAB$ in +$SCD$ dotikata v točki $S$. + +\item Premici $PB$ in $PD$ se dotikata krožnice $k(O,r)$ v točkah $B$ in $D$. +Premica $PO$ seka krožnico $k$ v točkah $A$ in $C$ ($\mathcal{B}(P,A,C)$). Dokaži, da je +premica $BA$ simetrala kota $PBD$. + +\item Štirikotnik $ABCD$ je včrtan v krožnico s središčem $O$. Diagonali $AC$ in +$BD$ sta pravokotni. Naj bo $M$ pravokotna projekcija središča $O$ +na premici $AD$. Dokaži, da je + $$|OM|=\frac{1}{2}|BC|.$$ + +\item Daljici $AB$ in $BC$ sta sosednji stranici pravilnega devetkotnika, ki je včrtan krožnici $k$ s središčem $O$. +Točka $M$ je središče stranice $AB$, točka $N$ pa središče +polmera $OX$ krožnice $k$, ki je pravokoten na premico $BC$. Dokaži, da je +$\angle OMN=30^0$. + +\item Krožnici $k_1$ in $k_2$ se sekata v točkah $A$ in $B$. Naj bo $p$ premica, ki poteka skozi točko $A$, krožnico $k_1$ seka še v točki $C$, krožnico $k_2$ pa še v točki $D$, ter $q$ premica, ki poteka skozi točko $B$, krožnico $k_1$ seka še v točki $E$, krožnico $k_2$ pa še v točki $F$. Dokaži, da +je $\angle CBD\cong\angle EAF$. + +\item Krožnici $k_1$ in $k_2$ se sekata v točkah $A$ in $B$. Načrtaj premico $p$, ki poteka skozi točko $A$, tako da bo +dolžina daljice $MN$, kjer sta $M$ in $N$ presečišči premice $p$ +s krožnicama $k_1$ in $k_2$, maksimalna. + +\item Naj bo $L$ pravokotna projekcija poljubne točke $K$ krožnice $k$ na njeni +tangenti skozi točko $T\in k$ ter $X$ točka, ki je simetrična točki $L$ glede na premico +$KT$. Določi geometrijsko mesto točk $X$. + +\item Dokaži, da je tetivni večkotnik z lihim številom oglišč, ki ima vse notranje kote skladne, pravilni večkotnik. + +\item Dve krožnici se dotikata od znotraj v točki $A$. Daljica $AB$ je premer večje +krožnice, tetiva $BK$ večje krožnice pa se dotika manjše krožnice v točki $C$. Dokaži +da je premica $AC$ simetrala kota $BAK$. + +\item Naj bo $BC$ tetiva krožnice $k$. Določi geometrijsko mesto višinskih točk +vseh trikotnikov $ABC$, kjer je $A$ poljubna točka, ki leži na krožnici $k$. + +\item Imejmo štirikotnik s tremi topimi notranjimi koti. Dokaži, da daljša +diagonala poteka iz oglišča, ki pripada ostrem kotu. + +\item Naj bo $ABCDEF$ tetivni šestkotnik ter $AB\cong DE$ in $BC\cong EF$. +Dokaži, da je $CD\parallel AF$. + +\item Naj bo $ABCD$ konveksni štirikotnik, pri katerem je $\angle ABD=50^0$, $\angle ADB=80^0$, $\angle ACB=40^0$ in $\angle DBC=\angle BDC +30^0$. Izračunaj mero kota $\angle DBC$. + +\item Naj bo $M$ poljubna notranja točka kota z vrhom $A$, točki $P$ in $Q$ pravokotni projekciji točke $M$ na krakih tega kota, točka $K$ +pa pravokotna projekcija vrha $A$ na premici $PQ$. Dokaži, da je $\angle MAP\cong \angle QAK$. + + +\item Pri tetivnem osemkotniku $A_1A_2\ldots A_8$ velja $A_1A_2\parallel A_5A_6$, $A_2A_3\parallel A_6A_7$, +$A_3A_4\parallel A_7A_8$. Dokaži, da je $A_8A_1\cong A_4A_5$. + +\item Neka krožnica vsako stranico štirikotnika seka v dveh točkah in tako na vseh stranicah štirikotnika določa skladne tetive. +Dokaži, da je ta štirikotnik tangenten. + +\item Dolžine stranic tangentnega petkotnika $ABCDE$ so naravna števila in hkrati velja $|AB|=|CD|=1$. +Včrtana krožnica petkotnika se dotika stranice $BC$ v točki $K$. +Izračunaj dolžino daljice $BK$. + +\item Dokaži, da krožnica, ki poteka skozi sosednji oglišči $A$ in $B$ pravilnega +petkotnika $ABCDE$ in njegovo središče $O$, poteka tudi skozi presečišče +njegovih diagonal $AD$ in $BE$. + +\item Naj bo $H$ višinska točka trikotnika $ABC$, $l$ krožnica nad premerom $AH$ +ter $P$ in $Q$ presečišči te krožnice s stranicama $AB$ in $AC$. Dokaži, da se +tangenti krožnice $k$ skozi točki $P$ in $Q$ sekata na stranici $BC$. + +\item Krožnica $l$ se od znotraj dotika krožnice $k$ v točki $C$. Naj bo $M$ poljubna točka +krožnice $l$ (različna od $C$). Tangenta krožnice $l$ v točki $M$ seka krožnico $k$ v +točkah $A$ in $B$. Dokaži, da je $\angle ACM \cong \angle MCB$. + +\item Naj bo $k$ očrtana krožnica trikotnika $ABC$ in $R$ središče tistega loka $AB$ te krožnice, +ki ne vsebuje točke $C$. Daljici $RP$ in $RQ$ sta tetivi te krožnice. Prva je +vzporedna, druga pa pravokotna na simetralo notranjega kota $\angle BAC$. Dokaži: +\begin{enumerate} +\item premica $BQ$ je simetrala notranjega kota $\angle CBA$, +\item trikotnik, ki ga določajo premice $AB$, $AC$ in $PR$, je pravilni trikotnik. + \end{enumerate} + +\item Naj bo $X$ takšna notranja točka trikotnika $ABC$, da velja: + $\angle BXC =\angle BAC+60^0$, $\angle AXC =\angle ABC+60^0$ in $\angle AXB =\angle AC B+60^0$. Naj bodo +$P$, $Q$ in $R$ druga presečišča premic $AX$, $BX$ in $CX$ z očrtano krožnico trikotnika $ABC$. Dokaži, da je trikotnik $PQR$ pravilen. + + +\item Dokaži, da dotikališča včrtane krožnice s trikotnikom $ABC$ delijo njegove stranice na +odseke dolžin $s-a$, $s-b$ in $s-c$ ($a$, $b$ in $c$ so dolžine stranic, $s$ pa je polobseg +trikotnika). + + + \item Krožnice $k$, $l$ in $j$ se paroma od zunaj dotikajo v nekolinearnih točkah $A$, $B$ in $C$. Dokaži, da je očrtana krožnica trikotnika $ABC$ pravokotna na krožnice $k$, $l$ in $j$. + + + \item Naj bo $ABCD$ tetivni štirikotnik s središčem očrtane krožnice v točki $O$. Z $E$ označimo presečišče njegovih diagonal $AC$ in $BD$ ter z $F$, $M$ in $N$ središča daljic $OE$, $AD$ in $BC$. Če so $F$, $M$ in $N$ kolinearne točke, velja $AC\perp BD$ ali $AB\cong CD$. Dokaži. + + + +\item Načrtaj trikotnik $ABC$ (glej oznake v razdelku \ref{odd3Stirik}): + + (\textit{a}) $a$, $\alpha$, $r$, \hspace*{2mm} + (\textit{b}) $a$, $\alpha$, $r_a$, \hspace*{2mm} + (\textit{c}) $a$, $v_b$, $v_c$, \hspace*{2mm} + + (\textit{d}) $\alpha$, $v_a$, $s$, \hspace*{2mm} + (\textit{e}) $v_a$, $l_a$, $r$, \hspace*{2mm} +(\textit{f}) $\alpha$, $v_a$, $l_a$, \hspace*{2mm} + + (\textit{g}) $\alpha$, $\beta$, $R$, \hspace*{2mm} + (\textit{h}) $c$, $r$, $R$, \hspace*{2mm} + (\textit{i}) $a$, $v_b$, $R$, \hspace*{2mm} + + \item Načrtaj krožnico $k$ tako, da: + + \begin{enumerate} + \item se dotika dveh danih nevzporednih premic $p$ in $q$, tetiva, ki jo določata dotikališči, pa je skladna z dano daljico $t$, + \item je središče dana točka $S$, dana premica $p$ pa na njej določa tetivo, ki je skladna z dano daljico $t$, + \item poteka skozi dani točki $A$ in $B$, središče pa leži na dani krožnici $l$, + \item ima dan polmer $r$ in se dotika dveh danih krožnic $l$ in $j$, + \item se dotika premice $p$ v točki $P$ in poteka skozi dano točko $A$. + \end{enumerate} + + \item Načrtaj kvadrat $ABCD$, če je dano oglišče $B$ ter dve točki $E$ in $F$, ki ležita na nosilkah stranic $AD$ in $CD$. + + \item Dana je premica $CD$ ter točki $A$ in $B$ ($A,B\notin CD$). Načrtaj na premici $CD$ takšno točko $M$, da velja $\angle AMC\cong2\angle BMD$. + + + \item Načrtaj trikotnik $ABC$ s podatki: + + (\textit{a}) $v_a$, $t_a$, $\beta-\gamma$, \hspace*{2mm} + (\textit{b}) $v_a$, $l_a$, $R$, \hspace*{2mm} + (\textit{c}) $R$, $\beta-\gamma$, $t_a$, \hspace*{2mm} + + (\textit{d}) $R$, $\beta-\gamma$, $v_a$, \hspace*{2mm} + (\textit{e}) $R$, $\beta-\gamma$, $a$. \hspace*{2mm} + + +\item Na nosilki stranice $AB$ pravokotnika $ABCD$ načrtaj točko $E$, iz katere se stranici $AD$ in $DC$ vidita pod enakim kotom. Kdaj ima naloga rešitev? + + \item V konveksnem štirikotniku $ABCD$ velja $BC\cong CD$. Načrtaj ta štirikotnik, če so dani: stranici $AB$ in $AD$ ter notranja kota ob ogliščih $B$ in $D$. + + \item V dano krožnico $k$ včrtaj trikotnik $ABC$, če so dani: oglišče $A$, premica $p$, ki je vzporedna z višino $AA'$, in presečišče $B_2$ nosilke višine $BB'$ in te krožnice. + + \item Načrtaj pravilni trikotnik $ABC$, če je njegova stranica $BC$ skladna z daljico $a$, nosilki stranic $AB$ in $AC$ ter simetrala notranjega kota $BAC$ pa gredo po vrsti skozi dane točke $M$, $N$ in $P$. + + \item Načrtaj trikotnik $ABC$, če so dani: + \begin{enumerate} + \item oglišče $A$, središče očrtane krožnice $O$ in središče včrtane krožnice $S$, + \item središče očrtane krožnice $O$, središče včrtane krožnice $S$ in središče pričrtane krožnice $S_a$, + \item oglišče $A$, središče očrtane krožnice $O$ in višinska točka $V$, + \item oglišči $B$ in $C$ ter simetrala notranjega kota $BAC$, + \item oglišče $A$, središče očrtane krožnice $O$ in presečišče $E$ stranice $BC$ s simetralo notranjega kota $BAC$, + \item točke $M$, $P$ in $N$, v katerih nosilki višine in težiščnice iz oglišča $A$ ter simetrala notranjega kota $BAC$ sekajo očrtano krožnico trikotnika, + \item oglišče $A$, središče očrtane krožnice $O$, točka $N$, v kateri simetrala notranjega kota $BAC$ sekaj očrtano krožnico trikotnika, in daljica $a$, ki je skladna s stranico $BC$. + \end{enumerate} + + \item Načrtaj trikotnik $ABC$ s podatki: + + (\textit{a}) $a$, $b$, $\alpha=3\beta$, \hspace*{3mm} + (\textit{b}) $t_a$, $t_c$, $v_b$.\hspace*{3mm} + +\item Skozi točko $M$, ki leži v notranjosti dane krožnice $k$, načrtaj takšno tetivo, da je razlika njenih odsekov (od točke $M$) enaka dani daljici $a$. + +\item Načrtaj trikotnik $ABC$, če poznaš: + + (\textit{a}) $b-c$, $r$, $r_a$, \hspace*{1.8mm} + (\textit{b}) $a$, $r$, $r_a$, \hspace*{1.8mm} + (\textit{c}) $a$, $r_b+r_c$, $v_a$, \hspace*{1.8mm} + (\textit{d}) $b+c$, $r_b$, $r_c$, + + (\textit{e}) $R$, $r_b$, $r_c$, \hspace*{1.8mm} + (\textit{f}) $b$, $R$, $r+r_a$, \hspace*{1.8mm} +(\textit{g}) $a$, $v_a$, $r_a-r$, \hspace*{1.8mm} + (\textit{h}) $\alpha$, $r$, $b+c$. + + + \item Dani so: krožnica $k$, njen premer $AB$ in točka $M\notin k$. Samo z ravnilom načrtaj pravokotnico iz točke $M$ na premico $AB$. + + \item Dani so: kvadrat $ABCD$ in takšni točki $M$ in $N$ na stranicah $BC$ in $CD$, da velja $\angle MAN=45^0$. + Samo z ravnilom načrtaj pravokotnico iz točke $A$ na premico $MN$. + +\end{enumerate} + + + + + +% DEL 5 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% VEKTORJI +%________________________________________________________________________________ + + \del{Vectors} \label{pogVEKT} + + +%________________________________________________________________________________ +\poglavje{Vector Definition. The Sum of Vectors} \label{odd5DefVekt} + +Intuitivno je vektor usmerjena daljica, ki jo lahko vzporedno premikamo\footnote{Pojem vektorja je bil znan že Starim Grkom. Sodoben koncept vektorjev, ki je povezan z \index{linearna algebra} linearno algebro in \index{geometrija!analitična} analitično geometrijo, se je začel razvijati v 19. stoletju, najprej kot posplošitev kompleksnih števil. V tem smislu je angleški matematik \index{Hamilton, W. R.}\textit{W. R. Hamilton} (1805--1865) definiral t. i. \index{kvaternioni}\pojem{kvaternione} $q = w + ix + jy + kz$, $i^2 = j^2 = k^2 = -ijk = -1$ kot posplošitev kompleksnih števil v štirirazsežnem prostoru}. V tem smislu bi vektor $\overrightarrow{AB}$ predstavljala cela množica daljic, ki so skladne, vzporedne in isto usmerjene kot neka dana daljica $AB$. Lahko bi pisali tudi $\overrightarrow{CD}=\overrightarrow{AB}$ za vsako daljico $CD$ iz te množice daljic (Figure \ref{sl.vek.5.1.1.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.1.pic} +\caption{} \label{sl.vek.5.1.1.pic} +\end{figure} + +Na ta način dobimo idejo za formalno definicijo vektorjev. Najprej vpeljimo relacijo $\varrho$ na množici parov točk. Naj bodo $A$, $B$, $C$ in $D$ točke v isti ravnini +(Figure \ref{sl.vek.5.1.2.pic}). Pravimo, da je $(A,B)\varrho (C,D)$, če je izpolnjen eden od treh pogojev\footnote{Na ta način - v \index{geometrija!afina}afini geometriji (brez aksiomov skladnosti) - je vektorje definiral srbski matematik \index{Veljković, M.}\textit{M. Veljković} (1954--2008), profesor na Matematični gimnaziji v Beogradu.}\index{relacija!$\varrho$}: + +\begin{enumerate} + \item štirikotnik $ABDC$ je paralelogram, + \item obstajata takšni točki $P$ in $Q$, da sta štirikotnika $ABQP$ in $CDQP$ paralelograma, + \item $A=B$ in $C=D$. +\end{enumerate} + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.2.pic} +\caption{} \label{sl.vek.5.1.2.pic} +\end{figure} + +Intuitivno je jasno, da je drugi pogoj potrebno dodati zaradi primera, kadar so točke $A$, $B$, $C$ in $D$ kolinearne. Tretji pogoj se bo nanašal na t. i. vektor nič. + +Iz same definicije relacije $\varrho$ dobimo naslednji izrek. + + + + \bizrek \label{vektRelRo} + If $(A,B)\varrho (C,D)$ and $A\neq B$, then the line segments $AB$ and $CD$ + are congruent, parallel, and have the same direction. + \eizrek + +\textbf{\textit{Proof.}} Ker je $A\neq B$, ostaneta le prva dva pogoja iz definicije. V tem primeru sta relaciji $AB\parallel CD$ in $AB\cong CD$ direktni posledici definicije paralelograma in izreka \ref{paralelogram}. + +Glede usmerjenosti daljic bi uporabili definicijo: vzporedni daljici $XY$ in $UV$ sta \pojem{enako usmerjeni}, če je izpolnjen eden od pogojev: +\begin{itemize} + \item $Y,V\ddot{-} XU$; + \item obstajata takšni točki $S$ in $T$, da je $ST\parallel XY$, $Y,T\ddot{-} XS$ in $V,T\ddot{-} US$. +\end{itemize} +\kdokaz + +Naslednji izrek potrebujemo, da bomo lahko definirali vektorje. + + + \bizrek + The relation $\varrho$ on the set of pairs of points in the plane is an equivalence relation. + \eizrek + + +\textbf{\textit{Proof.}} Refleksivnost in simetričnost sta direktni posledici definicije relacije $\varrho$. Za tranzitivnost pa je potrebno preveriti vse možne kombinacije pogojev 1--3 iz definicije. +\kdokaz + +\pojem{Vektor} \index{vektor} sedaj definiramo kot razred ekvivalenčne relacije $\varrho$: + +$$\overrightarrow{AB}=[AB]_{\varrho}=\{(X,Y);\hspace*{1mm}(X,Y)\varrho(A,B)\}.$$ + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.3.pic} +\caption{} \label{sl.vek.5.1.3.pic} +\end{figure} + + +Točko $A$ bomo imenovali \index{začetna točka vektorja}\pojem{začetna točka}, točko $B$ pa \index{končna točka!vektorja}\pojem{končna točka} vektorja $\overrightarrow{AB}$. + +Jasno je, da sta v primeru $(A,B)\varrho (C,D)$ para $(A,B)$ in $(C,D)$ iz istega razreda, kar pomeni, da je tedaj $\overrightarrow{AB}=\overrightarrow{CD}$. Velja tudi obratno: relacija $\overrightarrow{AB}=\overrightarrow{CD}$ pomeni, da velja $(A,B)\varrho (C,D)$ +(Figure \ref{sl.vek.5.1.3.pic}). + +Če ni potrebno posebej poudariti, za katerega predstavnika razreda gre, bomo vektorje označevali tudi z $\overrightarrow{v}$, $\overrightarrow{u}$, $\overrightarrow{w}$, ... + +Vektor $\overrightarrow{AA}$, ki predstavlja množico parov sovpadajočih točk, bomo imenovali \index{vektor!nič}\pojem{vektor nič}. Vektor nič bomo označevali tudi z $\overrightarrow{0}$ +(Figure \ref{sl.vek.5.1.4.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.4.pic} +\caption{} \label{sl.vek.5.1.4.pic} +\end{figure} + +Za vektor $\overrightarrow{BA}$ bomo rekli, da je \index{vektor!nasproten}\pojem{nasproten vektor} vektorju $\overrightarrow{AB}$ in ga označili z $-\overrightarrow{AB}$. Jasno je, da definicija nasprotnega vektorja ni odvisna od izbire predstavnika razreda; nasprotni vektor dobimo, če zamenjamo začetno in končno točko vektorja (oz. vsakega para točk ustreznega razreda). Torej če označimo $\overrightarrow{v}=\overrightarrow{AB}$ je +$-\overrightarrow{v}=-\overrightarrow{AB}=\overrightarrow{BA}$ +(Figure \ref{sl.vek.5.1.4.pic}). + + +Množico vseh vektorjev ravnine označimo z $\mathcal{V}$. + +Direktna posledica izreka \ref{vektRelRo} je naslednja trditev. + + + \bizrek \label{vektVzpSkl} + If $\overrightarrow{AB}=\overrightarrow{CD}$, then the line segments $AB$ and $CD$ + are congruent, parallel, and have the same direction. + \eizrek + +Na osnovi prejšnjega izreka lahko sedaj korektno definiramo naslednje pojme: +\begin{itemize} + \item \index{smer vektorja}\pojem{smer} vektorja $\overrightarrow{AB}\neq\overrightarrow{0}$ je določena s poljubno premico $p\parallel AB$, + \item \index{dolžina!vektorja}\pojem{dolžina} ali \index{intenziteta vektorja}\pojem{intenziteta} vektorja $\overrightarrow{AB}$ je $|\overrightarrow{AB}|=|AB|$ ($|\overrightarrow{0}|=0$), + \item \index{usmerjenost vektorja}\pojem{usmerjenost} ali \index{orientacija!vektorja}\pojem{orientacija} vektorja $\overrightarrow{AB}\neq\overrightarrow{0}$ je določena z urejenim parom $(A,B)$, kjer je $A$ začetna točka, $B$ pa končna točka tega vektorja. +\end{itemize} + +Če imata vektorja $\overrightarrow{v}$ in $\overrightarrow{u}$ isto smer, pravimo tudi, da sta \index{vektorja!vzporedna}\pojem{vzporedna} ali \index{vektorja!kolinearna}\pojem{kolinearna} in označimo z $\overrightarrow{v}\parallel\overrightarrow{u}$ (pri tem pravimo tudi, da je $\overrightarrow{0}$ kolinearen z vsakim vektorjem), sicer sta vektorja \index{vektorja!nekolinearna}\pojem{nekolinearna}. + +Če za kolinearna vektorja $\overrightarrow{v}$ in $\overrightarrow{u}$ velja $\overrightarrow{v}=\overrightarrow{SA}$ in $\overrightarrow{u}=\overrightarrow{SB}$, pravimo, da sta vektorja $\overrightarrow{v}$ in $\overrightarrow{u}$ \pojem{enako usmerjena} (oznaka $\overrightarrow{u}\rightrightarrows \overrightarrow{v}$), če je $A,B\ddot{-} S$, oz. \pojem{nasprotno usmerjena} (oznaka $\overrightarrow{u}\rightleftarrows \overrightarrow{v}$), če je $A,B\div S$. + +Jasno je, da sta kolinearna vektorja $\overrightarrow{AB}$ in $\overrightarrow{CD}$ enako usmerjena natanko tedaj, ko sta enako usmerjeni daljici $AB$ in $CD$. + + +Iz teh definicij direktno sledi, da je $|-\overrightarrow{v}|=|\overrightarrow{v}|$ in $-\overrightarrow{v}\parallel\overrightarrow{v}$; vektorja $\overrightarrow{v}$ in $-\overrightarrow{v}$ sta nasprotno usmerjena (če je $\overrightarrow{v}=\overrightarrow{AB}$, je $-\overrightarrow{v}=\overrightarrow{BA}$). + +Zelo pomemben je naslednji izrek. + + + + \bizrek \label{vektABCObst1TockaD} + For every three points there $A$, $B$ and $C$ there exists exactly one point $D$, + such that $\overrightarrow{AB}=\overrightarrow{CD}$, i.e. + $$(\forall A, B, C)(\exists_1 D)\hspace*{1mm}\overrightarrow{AB}=\overrightarrow{CD}.$$ + \eizrek + + +\textbf{\textit{Proof.}} Obravnavali bomo tri različne možnosti (Figure \ref{sl.vek.5.1.5.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.5.pic} +\caption{} \label{sl.vek.5.1.5.pic} +\end{figure} + +\textit{1)} Če je $A=B$, je $D=C$ edina točka, za katero je $(A,B)\varrho (C,D)$ oz. $\overrightarrow{CD}=\overrightarrow{AB}=\overrightarrow{0}$. + +\textit{2)} Naj bo $A\neq B$ in $C\notin AB$. Potem je $D$ četrto oglišče paralelograma $ABDC$ (ki obstaja in je eno samo po posledici +Playfairjevega aksioma \ref{Playfair1}) ter edina točka, za katero je $(A,B)\varrho (C,D)$ oz. $\overrightarrow{CD}=\overrightarrow{AB}$. + +\textit{3)} Naj bo $A\neq B$ in $C\in AB$. Naj bo $P$ poljubna točka, ki ne leži na premici $AB$. Po dokazanem delu \textit{2)} obstaja natanko ena točka $Q$, za katero je $\overrightarrow{PQ}=\overrightarrow{AB}$, nato pa tudi natanko ena točka $D$, za katero je $\overrightarrow{CD}=\overrightarrow{PQ}=\overrightarrow{AB}$. +\kdokaz + +Prejšnji izrek lahko zapišemo tudi v drugačni obliki (Figure \ref{sl.vek.5.1.6.pic}). + + + + \bizrek \label{vektAvObst1TockaB} + For every point $X$ and every vector $\overrightarrow{v}$ there exists exactly one point $Y$, + such that $\overrightarrow{XY}=\overrightarrow{v}$, i.e. + $$(\forall X)(\forall \overrightarrow{v})(\exists_1 Y)\hspace*{1mm}\overrightarrow{XY}=\overrightarrow{v}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.6.pic} +\caption{} \label{sl.vek.5.1.6.pic} +\end{figure} + + +Na osnovi dokazanega je jasno, da lahko za vsak vektor $\overrightarrow{v}$ izberemo poljubnega predstavnika njegovega razreda s poljubno začetno točko. Na podoben način bi lahko dokazali, da je to možno tudi glede na končno točko. To intuitivno pomeni, da lahko vektor vzporedno premikamo na en sam način do njegove začetne oz. končne točke. + +Dva poljubna vektorja lahko torej postavimo tako, da imata isto začetno točko ali pa tako, da je začetna točka drugega hkrati končna točka prvega vektorja. +To dejstvo nam omogoča definicijo vsote dveh vektorjev. + +Če je $\overrightarrow{v}=\overrightarrow{AB}$ in $\overrightarrow{u}=\overrightarrow{BC}$, je \index{vsota!vektorjev}\pojem{vsota vektorjev} $\overrightarrow{v}$ in $\overrightarrow{u}$ vektor $\overrightarrow{v}+\overrightarrow{u}=\overrightarrow{AC}$ (Figure \ref{asl.vek.5.1.7.pic}). Torej velja $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$. + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.7.pic} +\caption{} \label{asl.vek.5.1.7.pic} +\end{figure} + + +Potrebno je dokazati korektnost definicije - da vsota vektorjev ni odvisna od izbire predstavnikov dveh razredov ($\overrightarrow{v}$ in $\overrightarrow{u}$) oz. od izbire točke $A$. + + \bizrek \label{vektKorektDefSest} + If $A$, $B$, $C$, $A'$, $B'$ and $C'$ are points such that $\overrightarrow{AB}=\overrightarrow{A'B'}$ and + $\overrightarrow{BC}=\overrightarrow{B'C'}$, then there is also $\overrightarrow{AC}=\overrightarrow{A'C'}$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.7a.pic} +\caption{} \label{sl.vek.5.1.7a.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.1.7.pic}). + +V formalnem dokazu bi obravnavali različne možnosti glede na relacijo $\varrho$. + \kdokaz + + Pravilo seštevanja vektorjev, pri katerem uporabljamo že omenjeno enakost: + +\begin{itemize} + \item $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$, +\end{itemize} + + imenujemo \index{pravilo!trikotniško}\pojem{trikotniško pravilo} ali \index{Chaslesova identiteta}\pojem{Chaslesova\footnote{\index{Chasles, M.} \textit{M. Chasles} (1793–-1880), francoski matematik.} identiteta}. + Pravilo se tako imenuje kljub temu, da velja tudi v primeru, kadar so $A$, $B$ in $C$ kolinearne točke (oz. $\overrightarrow{v}$ in $\overrightarrow{u}$ kolinearna vektorja); takrat namreč ne gre za trikotnik (Figure \ref{sl.vek.5.1.8.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.8.pic} +\caption{} \label{sl.vek.5.1.8.pic} +\end{figure} + + + +Izkaže se, da množica vseh vektorjev $\mathcal{V}$ glede na operacijo seštevanja +vektorjev predstavlja strukturo t. i. \index{group!commutative}\pojem{commutative group} (ali \index{group!Abelian}\pojem{Abelian\footnote{\index{Abel, N. H.}\textit{N. H. Abel} (1802--1829), norveški matematik.} grupe}). Pojem grupe (ki ni vedno komutativna) smo že omenili v razdelku \ref{odd2AKSSKL} pri izometrijah, podrobneje pa se bomo s to strukturo ukvarjali v razdelku \ref{odd6Grupe}. Lastnosti te strukture (komutativne grupe) so dane v zapisu +naslednjega izreka. + + + + \bizrek \label{vektKomGrupa} The ordered pair $(\mathcal{V},+)$ form a commutative group, which means that: + \begin{enumerate} + \item $(\forall \overrightarrow{v}\in\mathcal{V})(\forall \overrightarrow{v}\in\mathcal{V})\hspace*{1mm} + \overrightarrow{v}+\overrightarrow{v}\in\mathcal{V}$ + (\index{grupoidnost}\textit{closure}), + \item $(\forall \overrightarrow{u}\in\mathcal{V}) + (\forall \overrightarrow{v}\in\mathcal{V}) + (\forall \overrightarrow{w}\in\mathcal{V}) + \hspace*{1mm} + \left(\overrightarrow{u}+\overrightarrow{v}\right)+\overrightarrow{w} + = \overrightarrow{u}+\left(\overrightarrow{v}+\overrightarrow{w}\right) + $ + (\index{asociativnost}\textit{associativity}), + \item $(\exists \overrightarrow{e}\in\mathcal{V}) + (\forall \overrightarrow{v}\in\mathcal{V}) + \hspace*{1mm} + \overrightarrow{v}+\overrightarrow{e} + = \overrightarrow{e}+\overrightarrow{v} + $ + (\index{nevtralni element}\textit{identity element}), + \item $(\forall \overrightarrow{v}\in\mathcal{V}) + (\exists \overrightarrow{u}\in\mathcal{V}) + \hspace*{1mm} + \overrightarrow{v}+\overrightarrow{u} + = \overrightarrow{u}+\overrightarrow{v}=\overrightarrow{e} + $ + (\index{inverzni element}\textit{inverse element}), + \item $(\forall \overrightarrow{v}\in\mathcal{V}) + (\forall \overrightarrow{u}\in\mathcal{V}) + \hspace*{1mm} + \overrightarrow{v}+\overrightarrow{u} + = \overrightarrow{u}+\overrightarrow{v} + $ + (\index{komutativnost}\textit{commutativity}). + \end{enumerate} + \eizrek + +\textbf{\textit{Proof.}} + + \textit{1)} Vsota dveh poljubnih vektorjev je vektor, kar sledi iz same definicije vsote dveh vektorjev. + + \textit{2)} Naj bodo $\overrightarrow{u}$, $\overrightarrow{v}$ in $\overrightarrow{w}$ poljubni vektorji ter $A$, $B$, $C$ in $D$ takšne točke, da velja $\overrightarrow{u}=\overrightarrow{AB}$, $\overrightarrow{v}=\overrightarrow{BC}$ in $\overrightarrow{w}=\overrightarrow{CD}$ (izrek \ref{vektAvObst1TockaB}) (Figure \ref{sl.vek.5.1.9a.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.9a.pic} +\caption{} \label{sl.vek.5.1.9a.pic} +\end{figure} + + + Po definiciji seštevanja vektorjev (in korektnosti te definicije - izrek \ref{vektKorektDefSest}) je: + \begin{eqnarray*} + & & \left(\overrightarrow{u}+\overrightarrow{v}\right)+\overrightarrow{w}= + \left(\overrightarrow{AB}+\overrightarrow{BC}\right)+\overrightarrow{CD}= + \overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AD}\\ + & & \overrightarrow{u}+\left(\overrightarrow{v}+\overrightarrow{w}\right) =\overrightarrow{AB}+\left(\overrightarrow{BC}+\overrightarrow{CD}\right)= + \overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AD} + \end{eqnarray*} + Iz tega pa sledi asociativnost seštevanja vektorjev. + + \textit{3)} V primeru seštevanja vektorjev je nevtralni element kar vektor nič oz. $\overrightarrow{e}=\overrightarrow{0}$, ker za vsak vektor $\overrightarrow{v}=\overrightarrow{AB}$ velja: + + \begin{eqnarray*} + & & \overrightarrow{v}+\overrightarrow{0}= + \overrightarrow{AB}+\overrightarrow{BB}= + \overrightarrow{AB}=\overrightarrow{v}\\ + & & \overrightarrow{0}+\overrightarrow{v}= + \overrightarrow{AA}+\overrightarrow{AB}= + \overrightarrow{AB}=\overrightarrow{v}. + \end{eqnarray*} + + + \textit{4)} V primeru seštevanja vektorjev je za poljubni vektor $\overrightarrow{v}=\overrightarrow{AB}$ inverzni elament nasproten vektor $\overrightarrow{u}=-\overrightarrow{v}=\overrightarrow{BA}$: + + \begin{eqnarray*} + & & \overrightarrow{v}+\left(-\overrightarrow{v}\right)= + \overrightarrow{AB}+\overrightarrow{BA}= + \overrightarrow{AA}=\overrightarrow{0}\\ + & & -\overrightarrow{v}+\overrightarrow{v}= + \overrightarrow{BA}+\overrightarrow{AB}= + \overrightarrow{BB}=\overrightarrow{0}. + \end{eqnarray*} + + \textit{5)} Naj bosta $v$ in $u$ poljubna vektorja ter $A$, $B$ in $C$ takšne točke, da velja + $\overrightarrow{v}=\overrightarrow{AB}$ in $\overrightarrow{u}=\overrightarrow{BC}$ (izrek \ref{vektAvObst1TockaB}). Obravnavali bomo dva primera (Figure \ref{sl.vek.5.1.9b.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.9b.pic} +\caption{} \label{sl.vek.5.1.9b.pic} +\end{figure} + + \textit{a)} Predpostavimo, da so točke $A$, $B$ in $C$ nekolinearne oz. $\overrightarrow{v}$ in $\overrightarrow{u}$ nekolinearna vektorja. Označimo z $D$ četrto oglišče paralelograma $ABCD$. Po definiciji relacije $\varrho$ je $(A,B)\varrho (D,C)$ in $(B,C)\varrho (A,D)$, iz definicije vektorjev pa nato sledi $\overrightarrow{AB}=\overrightarrow{DC}$ in $\overrightarrow{BC}=\overrightarrow{AD}$. Torej: + $$\overrightarrow{v}+\overrightarrow{u}= + \overrightarrow{AB}+\overrightarrow{BC}= + \overrightarrow{AC}= + \overrightarrow{AD}+\overrightarrow{DC}= + \overrightarrow{BC}+\overrightarrow{AB} + = \overrightarrow{u}+\overrightarrow{v}.$$ + + \textit{a)} Naj bodo $A$, $B$ in $C$ kolinearne točke oz. $\overrightarrow{v}$ in $\overrightarrow{u}$ kolinearna vektorja. Izrazimo vektorja $\overrightarrow{v}$ in $\overrightarrow{u}$ kot vsoti $\overrightarrow{v}=\overrightarrow{v}_1+\overrightarrow{v}_2$ in $\overrightarrow{u}=\overrightarrow{u}_1+\overrightarrow{u}_2$, tako da nobena dva od vektorjev $\overrightarrow{v}_1$, $\overrightarrow{v}_2$, $\overrightarrow{u}_1$ in $\overrightarrow{u}_2$ nista kolinearna. Če sedaj uporabimo dokazano v primeru \textit{a)}, dobimo: + $$\overrightarrow{v}+\overrightarrow{u}= + \overrightarrow{v}_1+\overrightarrow{v}_2+ + \overrightarrow{u}_1+\overrightarrow{u}_2= + \overrightarrow{u}_1+\overrightarrow{u}_2+ + \overrightarrow{v}_1+\overrightarrow{v}_2= + \overrightarrow{u}+\overrightarrow{v},$$ kar je bilo treba dokazati. \kdokaz + +Naslednja trditev je posledica prejšnjega izreka. + + \bzgled \label{vektABCD_ACBD} + For arbitrary points $A$, $B$, $C$ and $D$ is + $$\overrightarrow{AB}=\overrightarrow{CD}\Rightarrow \overrightarrow{AC}=\overrightarrow{BD}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.10.pic} +\caption{} \label{sl.vek.5.1.10.pic} +\end{figure} + +\textbf{\textit{Proof.}} Po definiciji seštevanja vektorjev in izreka \ref{vektKomGrupa} (komutativnost) je (Figure \ref{sl.vek.5.1.10.pic}): +$\overrightarrow{AC}= +\overrightarrow{AB}+ \overrightarrow{BC}= +\overrightarrow{CD}+ \overrightarrow{BC}= +\overrightarrow{BC}+\overrightarrow{CD}= +\overrightarrow{BD}.$ + \kdokaz + + Še ena posledica komutativnosti seštevanja vektorjev (izrek \ref{vektKomGrupa}) je naslednje pravilo seštevanja za nekolinearna vektorja. + + \begin{itemize} + \item \textit{Za vsake tri nekolinearne točke $A$, $B$ in $C$ velja $\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AD}$ natanko tedaj, ko je $ABDC$ paralelogram.} + \end{itemize} + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.11.pic} +\caption{} \label{sl.vek.5.1.11.pic} +\end{figure} + + + To pravilo imenujemo \index{pravilo!paralelogramsko}\pojem{paralelogramsko pravilo}\footnote{Paralelogramsko pravilo je bilo verjetno znano že Starim Grkom. Predpostavlja se, da ga je omenjal starogrški matematik in filozof \index{Aristotel} \textit{Aristotel} (384--322 pr. n. š.)} (Figure \ref{sl.vek.5.1.11.pic}). + + Posledica asociativnosti iz izreka \ref{vektKomGrupa} je naslednje pravilo za seštevanje vektorjev: + \begin{itemize} + \item $\overrightarrow{A_1A_2}+\overrightarrow{A_2A_3}+\cdots +\overrightarrow{A_{n-1}A_n}=\overrightarrow{A_1A_n}$, + \end{itemize} + +ki ga imenujemo \index{pravilo!poligonsko}\pojem{poligonsko pravilo} (Figure \ref{asl.vek.5.1.12.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.12.pic} +\caption{} \label{asl.vek.5.1.12.pic} +\end{figure} + +Direktna posledica tega pravila je naslednja trditev (Figure \ref{sl.vek.5.1.13.pic}). + + \bzgled + For arbitrary points $A_1$, $A_2$,..., $A_n$ in a plane is + $$\overrightarrow{A_1A_2}+\overrightarrow{A_2A_3}+\cdots +\overrightarrow{A_{n-1}A_n}+\overrightarrow{A_nA_1}=\overrightarrow{0}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.13.pic} +\caption{} \label{sl.vek.5.1.13.pic} +\end{figure} + +Z naslednjim izrekom bomo podali ekvivalentno definicijo središča daljice. + + \bzgled \label{vektSredDalj} + A point $S$ is the midpoint of a line segment $AB$ if and only if $\overrightarrow{SA}=-\overrightarrow{SB}$ i.e. $\overrightarrow{SA}+\overrightarrow{SB}=\overrightarrow{0}$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.14.pic} +\caption{} \label{sl.vek.5.1.14.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.1.14.pic}) + + ($\Leftarrow$) Predpostavimo, da velja $\overrightarrow{SA}=-\overrightarrow{SB}$ oz. $\overrightarrow{AS}=\overrightarrow{SB}$. To pomeni, da sta $AS$ in $SB$ vzporedni, enako usmerjeni in skladni daljici (izrek \ref{vektVzpSkl}) oz. $SA\cong BS$ in $\mathcal{B}(A,S,B)$. Torej je $S$ središče daljice $AB$. + +($\Rightarrow$) Naj bo sedaj $S$ središče daljice $AB$. Dovolj je dokazati, da velja $\overrightarrow{SA}=\overrightarrow{BS}$ oz. $\overrightarrow{BS}=\overrightarrow{SA}$. + Predpostavimo, da velja $\overrightarrow{BS}=\overrightarrow{SA'}$. + Toda v tem primeru je $SA'\cong BS$ in $\mathcal{B}(A',S,B)$. Po izreku \ref{ABnaPoltrakCX} je $A=A'$, zato je $\overrightarrow{BS}=\overrightarrow{SA'}=\overrightarrow{SA}$. +\kdokaz + +Definirajmo še operacijo odštevanja dveh vektorjev. \pojem{Razlika vektorjev} \index{razlika!vektorjev} $\overrightarrow{v}$ in $\overrightarrow{u}$ je vsota vektorjev $\overrightarrow{v}$ in $-\overrightarrow{u}$ oz. +$$\overrightarrow{v}-\overrightarrow{u}=\overrightarrow{v}+(-\overrightarrow{u}).$$ + + \bzgled \label{vektOdsev} + For arbitrary three points $O$, $A$ and $B$ is + $$\overrightarrow{OB}-\overrightarrow{OA}=\overrightarrow{AB}.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.15.pic} +\caption{} \label{sl.vek.5.1.15.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.1.15.pic}) + + Iz same definicije odštevanja in seštevanja vektorjev ter komutativnosti seštevanja + (izrek \ref{vektKomGrupa})sledi: + $$\overrightarrow{OB}-\overrightarrow{OA}= + \overrightarrow{OB}+(-\overrightarrow{OA})= + \overrightarrow{OB}+\overrightarrow{AO}= +\overrightarrow{AO}+\overrightarrow{OB}= + \overrightarrow{AB},$$ kar je bilo treba dokazati. \kdokaz + + + + \bzgled \label{vektOPi} + Suppose that a point $O$ lies on a line $p$ and let $P_1, P_2, \cdots, P_n$ be points + lying in the same half-plane $\alpha$ with the edge $p$. + If: + $$\overrightarrow{OS_n}=\overrightarrow{OP_1}+\overrightarrow{OP_2}+ + \cdots+\overrightarrow{OP_n},$$ + then the point $S_n$ also lies in the half-plane $\alpha$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.1a.pic} +\caption{} \label{sl.vek.5.2.1a.pic} +\end{figure} + + \textbf{\textit{Solution.}} Dokaz bomo izpeljali z indukcijo + glede na $n$ (Figure \ref{sl.vek.5.2.1a.pic}). + + (\textit{i}) Za $n=1$ je jasno $\overrightarrow{OS_1}=\overrightarrow{OP_1}$, + in $P_1\in\alpha\Rightarrow S_1=P_1\in\alpha$. + +(\textit{ii}) Predpostavimo, da trditev velja za $n=k$. Dokažimo, da potem + velja tudi za $n=k+1$. Naj bodo $P_1, P_2, \cdots, P_k, + P_{k+1}\in\alpha$. Dokažimo, da je potem tudi $S_{k+1}\in + \alpha$. Velja: + $$\overrightarrow{OS_{k+1}}=\overrightarrow{OP_1}+\overrightarrow{OP_2}+ + \cdots+\overrightarrow{OP_k}+\overrightarrow{OP_{k+1}}= + \overrightarrow{OS_k}+\overrightarrow{OP_{k+1}}.$$ +Točka $S_k$ po indukcijski predpostavki leži v polravnini +$\alpha$. Ker $S_k,P_{k+1}\in \alpha$ in $O\in p$, je kot $\angle +S_kOP_{k+1}$ konveksen in cel leži v polravnini $\alpha$. To +pomeni, da tudi točka $S_{n+1}$ (kot četrto oglišče +paralelograma $P_{k+1}OP_kS_{k+1}$) leži v notranjosti kota $\angle +S_kOP_{k+1}$ oz. v polravnini $\alpha$. + + \kdokaz + + + +%________________________________________________________________________________ +\poglavje{Linear Combination of Vectors} \label{odd5LinKombVekt} + +S pomočjo pojma seštevanja vektorjev lahko definiramo pojem množenja vektorja s poljubnim naravnim številom: + $$1\cdot \overrightarrow{v} =\overrightarrow{v},\hspace*{2mm} (n+1)\cdot \overrightarrow{v}=n\cdot\overrightarrow{v}+\overrightarrow{v},\hspace*{2mm} (n\in \mathbb{N}).$$ + Definicijo lahko razširimo tudi na cela števila: $0\cdot \overrightarrow{v}=\overrightarrow{0}$ in $-n\cdot \overrightarrow{v}=n\cdot(-\overrightarrow{v})$. + Jasno je, da sta v tem primeru vektorja $\overrightarrow{v}$ in $l\cdot \overrightarrow{v}$ ($l\in \mathbb{Z}$) vedno kolinearna in velja + $|l\cdot \overrightarrow{v}|=|l|\cdot |\overrightarrow{v}|$. + Na ta način dobimo idejo za definicijo množenja vektorja s poljubnim realnim številom $\lambda\in \mathbb{R}$. + + Najprej za $\lambda=0$ definirajmo $0\cdot \overrightarrow{v}=\overrightarrow{0}$. Če + je $\lambda\neq 0$ in $\overrightarrow{v}=\overrightarrow{AB}$, je $\lambda\cdot\overrightarrow{v}=\overrightarrow{AC}$, kjer je $C$ takšna točka na premici $AB$, da velja (Figure \ref{sl.vek.5.2.1.pic}): + + \begin{itemize} + \item $|AC|=|\lambda| \cdot|AB|$, + \item $C,B\ddot{-} A$, če je $\lambda>0$, + \item $C,B\div A$, če je $\lambda<0$. + \end{itemize} + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.1.pic} +\caption{} \label{sl.vek.5.2.1.pic} +\end{figure} + + + + Jasno je, da je točka $C$ za vsak $\lambda\neq 0$ enolično določena, vektor $\lambda\cdot \overrightarrow{v}$ pa ni odvisen od izbire točke $A$. To pomeni, da je definicija množenja vektorja z realnim številom korektna. + + Iz same definicije sledi, da za vsak vektor $\overrightarrow{v}$ in vsako realno število $\lambda\in \mathbb{R}$ velja $|\lambda \cdot \overrightarrow{v}|=|\lambda| \cdot |\overrightarrow{v}|$. + +Množenje vektorja $\overrightarrow{v}$ z realnim številom $\lambda$ bomo imenovali tudi \index{množenje vektorja s skalarjem}\pojem{množenje vektorja $\overrightarrow{v}$ s skalarjem $\lambda$}. + +Podobno kot pri množenju algebrskih izrazov bomo oznako $\cdot$ množenja vektorja z realnim številom običajno izpuščali, oz. bomo namesto $\lambda\cdot\overrightarrow{v}$ raje pisali kar $\lambda\overrightarrow{v}$. + + Naslednja trditev nam da potreben in zadosten pogoj za kolinearnost dveh vektorjev. + + + + \bizrek \label{vektKriterijKolin} + Vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ ($\overrightarrow{v},\overrightarrow{u}\neq \overrightarrow{0}$) are collinear if and only if there is such $\lambda\in \mathbb{R}$, that is $\overrightarrow{u}=\lambda\cdot\overrightarrow{v}$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.2.pic} +\caption{} \label{sl.vek.5.2.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.2.pic}) + + Naj bosta $\overrightarrow{v}$ in $\overrightarrow{u}$ poljubna vektorja ($\overrightarrow{v},\overrightarrow{u}\neq \overrightarrow{0}$), $A$ poljubna točka ter $B$ in $C$ takšni točki, da velja $\overrightarrow{AB}=\overrightarrow{v}$ in $\overrightarrow{AC}=\overrightarrow{u}$ (izrek \ref{vektAvObst1TockaB}). + + + ($\Leftarrow$) Predpostavimo najprej, da velja $\overrightarrow{u}=\lambda\cdot\overrightarrow{v}$ za nek $\lambda\in \mathbb{R}$. Po definiciji točka $C$ leži na premici $AB$, kar pomeni, da sta vektorja $\overrightarrow{AB}$ in $\overrightarrow{AC}$ oz. $\overrightarrow{v}$ in $\overrightarrow{u}$ kolinearna. + + ($\Rightarrow$) Predpostavimo sedaj, da sta vektorja $\overrightarrow{v}$ in $\overrightarrow{u}$ oz. $\overrightarrow{AB}$ in $\overrightarrow{AC}$ kolinearna. + V tem primeru je: + \begin{eqnarray*} + \overrightarrow{u}&=&\overrightarrow{AC}=\frac{|AC|}{|AB|}\cdot \overrightarrow{AB}= + \frac{|\overrightarrow{u}|}{|\overrightarrow{v}|}\cdot \overrightarrow{v}, + \hspace*{2mm} \textrm{ če je } C,B\ddot{-} A;\\ + \overrightarrow{u}&=&\overrightarrow{AC}=-\frac{|AC|}{|AB|}\cdot \overrightarrow{AB}= + -\frac{|\overrightarrow{u}|}{|\overrightarrow{v}|}\cdot \overrightarrow{v}, + \hspace*{2mm} \textrm{ če je } C,B\div A, + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + +Če kombiniramo operaciji seštevanja vektorjev in množenja vektorja s skalarjem, dobimo t. i. linearno kombinacijo vektorjev. Še natančneje - za poljubno $n$-terico vektorjev $(\overrightarrow{a_1},\overrightarrow{a_2},\ldots,\overrightarrow{a_n})$ in poljubno $n$-terico realnih števil $(\alpha_1,\alpha_2,\ldots,\alpha_n)\in \mathbb{R}^n$ je vektor + $$\overrightarrow{v}=\alpha_1\cdot \overrightarrow{a_1}+ + \alpha_2\cdot \overrightarrow{a_2} + +\cdots +\alpha_n\cdot \overrightarrow{a_n}$$ + \index{linearna kombinacija vektorjev}\pojem{linearna kombinacija vektorjev} $\overrightarrow{a_1}$, $\overrightarrow{a_2}$, $\ldots$, $\overrightarrow{a_n}$. + + + + \bizrek \label{vektLinKombNicLema} + Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two non-collinear non-zero vectors + and $(\alpha,\beta)\in \mathbb{R}^2$ two real numbers. + If $\alpha\cdot\overrightarrow{a}= + \beta\cdot\overrightarrow{b}$, then $\alpha=\beta=0$. + \eizrek + + \textbf{\textit{Proof.}} Predpostavimo nasprotno - brez škode za splošnost, da velja $\alpha\neq0$. Potem je $\overrightarrow{a}= + \frac{\beta}{\alpha}\cdot\overrightarrow{b}$, kar pomeni (izrek \ref{vektKriterijKolin}), da je vektor $\overrightarrow{b}$ kolinearen z vektorjem $\overrightarrow{a}$. To je v nasprotju s predpostavko, zato je $\alpha=\beta=0$. +\kdokaz + +Direktna posledica prejšnjega izreka je naslednja trditev. + + + + \bizrek \label{vektLinKombNic} + If for the linear combination of two non-collinear non-zero vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ holds + $$\alpha\cdot\overrightarrow{a}+ + \beta\cdot\overrightarrow{b}=\overrightarrow{0},$$ + then $\alpha=\beta=0$. + \eizrek + + \textbf{\textit{Proof.}} Če relacijo $\alpha\cdot\overrightarrow{a}+ + \beta\cdot\overrightarrow{b}=\overrightarrow{0}$ zapišemo v obliki $\alpha\cdot\overrightarrow{a}= + -\beta\cdot\overrightarrow{b}$, vidimo, da je trditev direktna posledica prejšnjega izreka. +\kdokaz + + Zelo pomemben je naslednji izrek, ki se nanaša na predstavljanje poljubnega vektorja kot linearne kombinacije dveh nekolinearnih neničelnih vektorjev. + + + \bizrek \label{vektLinKomb1Razcep} + Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two non-collinear non-zero vectors in the same plane. + Each vector $\overrightarrow{v}$ in this plane can be express in a single way as a linear combination + of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, i.e. there is exactly + one pair of real numbers $(\alpha,\beta)\in \mathbb{R}^2$, such that + $$\overrightarrow{v}=\alpha\cdot\overrightarrow{a}+ + \beta\cdot\overrightarrow{b}.$$ + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.3.pic} +\caption{} \label{sl.vek.5.2.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.3.pic}) + + Naj bo $O$ poljubna točka ter $A$, $B$ in $V$ takšne točke, da velja $\overrightarrow{OA}=\overrightarrow{a}$, $\overrightarrow{OB}=\overrightarrow{b}$ in $\overrightarrow{OV}=\overrightarrow{v}$ (izrek \ref{vektAvObst1TockaB}). Označimo z $A'$ presečišče premice $OA$ in vzporednice premice $OB$ skozi točko $V$ ter z $B'$ presečišče premice $OB$ in vzporednice premice $OA$ skozi točko $V$. Torej je štirikotnik $OA'VB'$ paralelogram. Po izreku \ref{vektKriterijKolin} obstajata takšni $\alpha,\beta\in \mathbb{R}$, da velja + $\overrightarrow{OA'}= + \alpha\cdot\overrightarrow{OA}= + \alpha\cdot\overrightarrow{a}$ in + $\overrightarrow{OB'}= + \beta\cdot\overrightarrow{OB}= + \beta\cdot\overrightarrow{b}$. Po paralelogramskem pravilu je: + + $$\overrightarrow{v}= + \overrightarrow{OV}=\overrightarrow{OA'}+\overrightarrow{OB'}= + \alpha\cdot\overrightarrow{a}+\beta\cdot\overrightarrow{b}.$$ + +Dokažimo, da je $(\alpha,\beta)\in \mathbb{R}^2$ edini tak par. Predpostavimo, da za +$(\alpha',\beta')\in \mathbb{R}^2$ velja $\overrightarrow{v}= + \alpha'\cdot\overrightarrow{a}+\beta'\cdot\overrightarrow{b}$. Iz + $\overrightarrow{v}= + \alpha'\cdot\overrightarrow{a}+\beta'\cdot\overrightarrow{b}= + \alpha\cdot\overrightarrow{a}+\beta\cdot\overrightarrow{b}$ sledi + $(\alpha'-\alpha)\cdot\overrightarrow{a}+(\beta'-\beta)\cdot\overrightarrow{b}= +\overrightarrow{0}$. Po prejšnjem izreku \ref{vektLinKombNic} je + $\alpha'-\alpha=\beta'-\beta=0$ oz. $\alpha'=\alpha$ in $\beta'=\beta$. +\kdokaz + + + + \bizrek \label{vektVektorskiProstor} + The set $\mathcal{V}$ of all vectors in the plane form so-called + \index{vector space} vector space \footnote{ Prvi, ki je leta 1888 na sodoben način definiral pojem vektorskega prostora, + je bil italijanski matematik \index{Peano, G.}\textit{G. Peano} (1858–-1932).} over the field $\mathbb{R}$, which means that: + \begin{enumerate} + \item the ordered pair $(\mathcal{V},+)$ form a commutative group, + \item $(\forall \alpha \in \mathbb{R})(\forall \overrightarrow{v}\in \mathcal{V})\hspace*{1mm} \alpha\cdot \overrightarrow{v} \in \mathcal{V}$, + \item $(\forall \alpha,\beta \in \mathbb{R})(\forall \overrightarrow{v}\in \mathcal{V})\hspace*{1mm} \alpha\cdot (\beta\cdot\overrightarrow{v})=(\alpha\beta)\cdot \overrightarrow{v}$, + \item $(\forall \overrightarrow{v}\in \mathcal{V})\hspace*{1mm} 1\cdot \overrightarrow{v}=\overrightarrow{v}$, + \item $(\forall \alpha,\beta \in \mathbb{R})(\forall \overrightarrow{v}\in \mathcal{V})\hspace*{1mm} (\alpha+ \beta)\cdot\overrightarrow{v}=\alpha\cdot \overrightarrow{v}+\beta\cdot \overrightarrow{v}$, + \item $(\forall \alpha \in \mathbb{R})(\forall \overrightarrow{v},\overrightarrow{u}\in \mathcal{V})\hspace*{1mm} \alpha\cdot(\overrightarrow{v}+\overrightarrow{u})= + \alpha\cdot \overrightarrow{v}+\alpha\cdot \overrightarrow{u}$. + \end{enumerate} + \eizrek + + + + \textbf{\textit{Proof.}} Trditev $\textit{1}$ je v bistvu že dokazan izrek \ref{vektKomGrupa}. Trditve $\textit{2}-\textit{5}$ so direktna posledica definicije množenja vektorja s skalarjem. Dokaz trditve $\textit{6}$ pa ni tako enostavna - v dokazu bi morali uporabiti tudi Dedekindov aksiom zveznosti \ref{aksDed}. + \kdokaz + + +Za strukturo vektorskega prostora na množici vseh vektorjev ravnine $\mathcal{V}$ nad obsegom $\mathbb{R}$ bomo uporabljali oznako $\mathcal{\overrightarrow{V}}^2$. + Če bi obravnavali vektorje v evklidskem prostoru, bi prav tako dobili vektorski prostor. Toda vektorski prostor lahko obravnavamo tudi bolj abstraktno kot poljubni urejeni par $(\mathcal{\overrightarrow{V}},\mathcal{F})$, ker $(\mathcal{\overrightarrow{V}},+)$ predstavlja komutativno grupo, $(\mathcal{F},+,\cdot)$ obseg in so izpolnjeni vsi pogoji $\textit{2}-\textit{6}$ iz prejšnjega izreka. + +Ker lahko vsak vektor v ravnini izrazimo kot linearno kombinacijo dveh nekolinearnih neničelnih vektorjev $\overrightarrow{a}$ in $\overrightarrow{b}$ te ravnine, pravimo, da sta takšna dva vektorja $\overrightarrow{a}$ in $\overrightarrow{b}$ t. i. \index{baza vektorskega prostora}\pojem{baza} vektorskega prostora $\mathcal{\overrightarrow{V}}^2$. Če za nek vektor $\overrightarrow{v}\in \mathcal{V}$ velja $\overrightarrow{v}=\alpha\cdot\overrightarrow{a}+ + \beta\cdot\overrightarrow{b}$, pravimo, da par $(\alpha,\beta)$ predstavlja \index{koordinate vektorja}\pojem{koordinati} vektorja $\overrightarrow{v}$ v bazi $(\overrightarrow{a},\overrightarrow{b})$. Iz izreka \ref{vektLinKomb1Razcep} sledi, da so v poljubni bazi koordinate vsakega vektorja enolično določene. + + + +V vektorskem prostoru $\mathcal{\overrightarrow{V}}^2$ je torej neskončno mnogo baz, število vektorjev v poljubni bazi je vedno 2, zato pravimo, da ima vektorski prostor $\mathcal{\overrightarrow{V}}^2$ \pojem{dimenzijo} 2 oz. je ravnina \pojem{dvorazsežna}. + +Jasno je, da je dimenzija vektorskega prostora $\mathcal{\overrightarrow{V}}^3$, ki ga določajo vektorji v evklidskem prostoru, enaka 3. Evklidski prostor je torej \pojem{trirazsežen}, vsak vektor v neki bazi je določen s trojico realnih števil, ki predstavljajo koordinate tega vektorja. + +V poljubnem vektorskem prostoru na podoben način določamo bazo in dimenzijo. Ker v splošnem primeru dimenzija prostora ni omejena na 3, nam to omogoča raziskovanje \pojem{$n$-razsežnih} evklidskih prostorov (za poljubno $n\in \mathbb{N}$). + +Zelo uporabni so naslednji zgledi. + + \bzgled \label{vektSredOSOAOB} + Let $O$, $A$ and $B$ be arbitrary points and $S$ the midpoint of the line segment $AB$. Then + $$\overrightarrow{OS}=\frac{1}{2}\cdot\left(\overrightarrow{OA}+ + \overrightarrow{OB}\right).$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.4.pic} +\caption{} \label{sl.vek.5.2.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.4.pic}) + + Po izreku \ref{vektSredDalj} je $\overrightarrow{SA}=-\overrightarrow{SB}$ oz. $\overrightarrow{AS}=-\overrightarrow{BS}$. Po trikotniškem pravilu za seštevanje vektorjev je + $\overrightarrow{OS}=\overrightarrow{OA}+\overrightarrow{AS}$ in + $\overrightarrow{OS}=\overrightarrow{OB}+\overrightarrow{BS}$. Če seštejemo enakosti in upoštevamo $\overrightarrow{AS}=-\overrightarrow{BS}$, dobimo: $2\cdot\overrightarrow{OS}=\overrightarrow{OA}+\overrightarrow{AS} + +\overrightarrow{OB}+\overrightarrow{BS}=\overrightarrow{OA} + +\overrightarrow{OB}$ + oz. +$\overrightarrow{OS}=\frac{1}{2}\cdot\left(\overrightarrow{OA}+ + \overrightarrow{OB}\right)$. +\kdokaz + + \bzgled \label{vektDelitDaljice} + Let $O$, $A$ and $B$ be arbitrary points and $P$ a point in the line segment $AB$ such that $|AP|:|PB|=n:m$. Then + $$\overrightarrow{OP}=\frac{1}{n+m}\left(m\cdot\overrightarrow{OA}+ + n\cdot\overrightarrow{OB}\right).$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.5.pic} +\caption{} \label{sl.vek.5.2.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.5.pic}) + + Najprej iz $|AP|:|PB|=n:m$ sledi $|AP|=\frac{n}{n+m}\cdot|AB|$ in $|BP|=\frac{m}{n+m}\cdot|AB|$ oz. $\overrightarrow{AP}=\frac{n}{n+m}\cdot\overrightarrow{AB}$ in $\overrightarrow{BP}=\frac{m}{n+m}\cdot\overrightarrow{BA}$ (ker je $\mathcal{B}(A,P,B)$). +Nato je: +\begin{eqnarray*} + \overrightarrow{OP}&=&\overrightarrow{OA}+\overrightarrow{AP}=\overrightarrow{OA} + +\frac{n}{n+m}\cdot\overrightarrow{AB}\hspace*{3mm} \textrm{ in}\\ + \overrightarrow{OP}&=&\overrightarrow{OB}+\overrightarrow{BP}=\overrightarrow{OB} + +\frac{m}{n+m}\cdot\overrightarrow{BA}. +\end{eqnarray*} + Če prvo enakost množimo z $m$, drugo z $n$ in nato seštejemo, dobimo: +\begin{eqnarray*} +(n+m)\cdot\overrightarrow{OP}&=&m\cdot\overrightarrow{OA} + +n\cdot\overrightarrow{OB}+\frac{nm}{n+m}\cdot\left(\overrightarrow{AB}+ + \overrightarrow{BA}\right)=\\ + &=&m\cdot\overrightarrow{OA} + +n\cdot\overrightarrow{OB}. +\end{eqnarray*} + Če dobljeno enakost delimo z $n+m$, dobimo iskano relacijo. +\kdokaz + + + \bzgled \label{vektParamDaljica} + Let $O$, $A$ and $B$ be arbitrary points. A point $X$ lies on the line segment $AB$ if and only if for some scalar $0\leq\lambda\leq1$ is + $$\overrightarrow{OX}=(1-\lambda)\cdot\overrightarrow{OA}+ + \lambda\cdot\overrightarrow{OB}.$$ + \ezgled + + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.6.pic}) + + Predpostavimo najprej, da točka $X$ leži na daljici $AB$. Potem za nek $0\leq\lambda\leq1$ velja $\overrightarrow{AX}=\lambda\cdot \overrightarrow{AB}$. + + \begin{eqnarray*} + \overrightarrow{OX}&=&\overrightarrow{OA}+\overrightarrow{AX}=\\ + &=& \overrightarrow{OA}+\lambda\cdot \overrightarrow{AB}=\\ + &=& \overrightarrow{OA}+\lambda\cdot \left(\overrightarrow{OB}-\overrightarrow{OA}\right)=\\ + &=& (1-\lambda)\cdot\overrightarrow{OA}+ + \lambda\cdot\overrightarrow{OB}. +\end{eqnarray*} + +Predpostavimo sedaj, da za nek $0\leq\lambda\leq1$ velja + $\overrightarrow{OX}=(1-\lambda)\cdot\overrightarrow{OA}+ + \lambda\cdot\overrightarrow{OB}$. Potem je: + + \begin{eqnarray*} + \overrightarrow{AX}&=&\overrightarrow{AO}+\overrightarrow{OX}=\\ + &=&\overrightarrow{AO}+(1-\lambda)\cdot\overrightarrow{OA}+ + \lambda\cdot\overrightarrow{OB}=\\ + &=&-\lambda\cdot\overrightarrow{OA}+ + \lambda\cdot\overrightarrow{OB}=\\ + &=&\lambda\cdot\left(\overrightarrow{AO}+ + \overrightarrow{OB}\right)=\\ + &=&\lambda\cdot\overrightarrow{AB}. +\end{eqnarray*} +Torej za nek $0\leq\lambda\leq1$ velja $\overrightarrow{AX}=\lambda\cdot \overrightarrow{AB}$, + kar pomeni, da točka $X$ leži na daljici $AB$. +\kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.6.pic} +\caption{} \label{sl.vek.5.2.6.pic} +\end{figure} + +A direct consequence is the following theorem. + + $O$, $A$ in $B$ . A point $X$ lies on the line segment $AB$ if and only if for some + + \bzgled + Let $O$, $A$ and $B$ be arbitrary points. A point $X$ lies on the line segment $AB$ if and only if for some scalars $\alpha, \beta\in [0,1]$ ($\alpha+\beta=1$) is + $$\overrightarrow{OX}=\alpha\cdot\overrightarrow{OA}+ + \beta\cdot\overrightarrow{OB}.$$ + \ezgled + +V naslednjem izreku bomo podali \index{vektorska enačba premice} \pojem{enačbo premice v vektorski obliki}. + + \bzgled \label{vektParamPremica} + Let $O$, $A$ in $B$ be arbitrary points. A point $X$ lies on the line $AB$ if and only if for some scalar $\lambda\in \mathbb{R}$ is + $$\overrightarrow{OX}=(1-\lambda)\cdot\overrightarrow{OA}+ + \lambda\cdot\overrightarrow{OB}.$$ + \ezgled + + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.6.pic}) + +Dokaz je enak kot pri prejšnjem izreku, le da uporabimo dejstvo, da točka $X$ leži na premici $AB$ natanko tedaj, ko sta vektorja $\overrightarrow{AX}$ in $\overrightarrow{AB}$ kolinearna, oz. za nek +$\lambda\in \mathbb{R}$ velja $\overrightarrow{AX}=\lambda\cdot \overrightarrow{AB}$ (izrek \ref{vektKriterijKolin}). +\kdokaz + +\index{količnik kolinearnih vektorjev}\index{razmerje!kolinearnih vektorjev} +Če za kolinearna vektorja $\overrightarrow{v}$ in $\overrightarrow{u}$ velja $\overrightarrow{v}=\lambda \overrightarrow{u}$ ($\lambda\in\mathbb{R}$) in $\overrightarrow{u}\neq \overrightarrow{0}$, potem lahko definiramo \pojem{količnik kolinearnih vektorjev} ali \pojem{razmerje kolinearnih vektorjev} (Figure \ref{sl.vek.5.2.7.pic}): +$$\overrightarrow{v}:\overrightarrow{u} +=\frac{\overrightarrow{v}}{\overrightarrow{u}}=\lambda.$$ + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.7.pic} +\caption{} \label{sl.vek.5.2.7.pic} +\end{figure} + +Iz definicije množenja vektorja z realnim številom dobimo naslednjo trditev. + + \bizrek \label{vektKolicnDolz} + Let $\overrightarrow{v}$ and $\overrightarrow{u}$ be collinear vectors and $\overrightarrow{u}\neq \overrightarrow{0}$. Then + $$\frac{\overrightarrow{v}}{\overrightarrow{u}}= + \left\{ \begin{array}{lll} + \frac{|\overrightarrow{v}|}{|\overrightarrow{u}|}, & + \textrm{ if } \overrightarrow{u}\rightrightarrows \overrightarrow{v}\\ + -\frac{|\overrightarrow{v}|}{|\overrightarrow{u}|}, & + \textrm{ if } \overrightarrow{u}\rightleftarrows\overrightarrow{v}\\ + 0, & \textrm{ if } \overrightarrow{v}=\overrightarrow{0} + \end{array}\right.$$ + \eizrek + + +Tudi naslednji izrek se nanaša na novo definirani pojem. + + \bizrek + \label{izrekEnaDelitevDaljiceVekt} + For any line segment $AB$ and any $\lambda\in\mathbb{R}\setminus\{-1\}$ there exists + exactly one such point $P$ on the line $AB$, that is + $$\frac{\overrightarrow{AP}}{\overrightarrow{PB}}=\lambda.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.8.pic} +\caption{} \label{sl.vek.5.2.8.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.8.pic}) + + Naj bo $P$ takšna točka, da velja: $$\overrightarrow{AP}=\frac{\lambda}{1+\lambda}\cdot\overrightarrow{AB}.$$ +Ker je $\lambda\neq-1$, takšna točka vedno obstaja. Ker je še: +$$\overrightarrow{PB}=\overrightarrow{PA}+\overrightarrow{AB}= +-\frac{\lambda}{1+\lambda}\cdot\overrightarrow{AB}+\overrightarrow{AB}= +\frac{1}{1+\lambda}\overrightarrow{AB},$$ +velja tudi $\frac{\overrightarrow{AP}}{\overrightarrow{PB}}=\lambda$. + +Če za neko drugo točko $P'$ velja $\frac{\overrightarrow{AP'}}{\overrightarrow{P'B}}=\lambda$, iz +$\overrightarrow{AP'}=\lambda\cdot\overrightarrow{P'B}$ in +$\overrightarrow{P'B}=\overrightarrow{P'A}+\overrightarrow{AB}$ z enostavnim računanjem dobimo $\overrightarrow{AP'}=\frac{\lambda}{1+\lambda}\cdot\overrightarrow{AB}$. Torej + $\overrightarrow{AP'}=\overrightarrow{AP}$ oz. + $\overrightarrow{P'P}=\overrightarrow{P'A}+\overrightarrow{AP}= + -\overrightarrow{AP}+\overrightarrow{AP}=\overrightarrow{0}$. Zato je $P'=P$, kar pomeni, da obstaja ena sama točka $P$, za katero velja $\frac{\overrightarrow{AP}}{\overrightarrow{PB}}=\lambda$. + \kdokaz + + Pravimo, da točka $P$ iz prejšnjega izreka deli daljico $AB$ v \pojem{razmerju} $\lambda$. + + + + \bzgled + Let $ABCD$ be a trapezium with the base $AB$. Calculate the ratio in which the line + segment $PD$ divides the diagonal $AC$, if $|AB| = 3\cdot |CD|$ and $P$ is the midpoint of the line segment $AB$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.2.9.pic} +\caption{} \label{sl.vek.5.2.9.pic} +\end{figure} + + \textbf{\textit{Solution.}} Presečišče daljice $DP$ in diagonale $AC$ označimo s $S$ (Figure \ref{sl.vek.5.2.9.pic}) +in naj bo + $\overrightarrow{u}=\overrightarrow{AP}$ in $\overrightarrow{v}=\overrightarrow{AD}$. +Vektorja $\overrightarrow{AS}$ in $\overrightarrow{AC}$ sta kolinearna, zato je po izreku \ref{vektKriterijKolin} $\overrightarrow{AS}=\lambda\overrightarrow{AC}$, za nek $\lambda\in \mathbb{R}$. Prav tako sta kolinearna tudi vektorja $\overrightarrow{PS}$ in $\overrightarrow{PD}$, oz. za nek $\mu\in \mathbb{R}$ velja $\overrightarrow{PS}=\mu\overrightarrow{PD}$. Vektorja $\overrightarrow{AC}$ in $\overrightarrow{AS}$ zapišemo kot linearno kombinacijo vektorjev $\overrightarrow{u}$ in $\overrightarrow{v}$: + \begin{eqnarray*} +\hspace*{-1.8mm} \overrightarrow{AC}&=&\overrightarrow{AD}+\overrightarrow{DC} + =\overrightarrow{v}+\frac{2}{3}\overrightarrow{u} + =\frac{2}{3}\overrightarrow{u}+\overrightarrow{v},\\ + \hspace*{-1.8mm}\overrightarrow{AS}&=&\overrightarrow{AP}+\overrightarrow{PS} + =\overrightarrow{u}+\mu\overrightarrow{PD} + =\overrightarrow{u}+\mu(-\overrightarrow{u}+\overrightarrow{v}) + =(1-\mu)\overrightarrow{u}+\mu\overrightarrow{v}. + \end{eqnarray*} +Ker je še $\overrightarrow{AS}=\lambda\overrightarrow{AC}$, dobimo: +\begin{eqnarray*} + \overrightarrow{AS}&=& + \frac{2}{3}\lambda\overrightarrow{u}+\lambda\overrightarrow{v};\\ + \overrightarrow{AS}&=& + (1-\mu)\overrightarrow{u}+\mu\overrightarrow{v}. + \end{eqnarray*} + Ker sta $\overrightarrow{u}$ in $\overrightarrow{v}$ kolinearna vektorja, je po izreku \ref{vektLinKomb1Razcep} $\frac{2}{3}\lambda=1-\mu$ in $\lambda=\mu$. Če rešimo ta enostaven sistem enačb, dobimo $\lambda=\mu=\frac{3}{5}$. Torej je $\overrightarrow{AS}=\frac{3}{5}\overrightarrow{AC}$, oz. $AS:SC=3:2$. + \kdokaz + +%________________________________________________________________________________ +\poglavje{Vector Length} \label{odd5DolzVekt} + +Dolžino vektorja smo definirali že v razdelku \ref{odd5DefVekt}. Iz definicije množenja vektorja s skalarjem v prejšnjem razdelku \ref{odd5LinKombVekt} smo videli, da za vsak vektor $\overrightarrow{v}$ in vsako realno število $\lambda$ velja $|\lambda \cdot \overrightarrow{v}|=|\lambda| \cdot |\overrightarrow{v}|$. V tem razdelku bomo obravnavali še nekatere lastnosti dolžine vektorja. + +Najprej iz definicije seštevanja vektorjev in trikotniške neenakosti dobimo naslednjo trditev. + + \bizrek \label{neenakTrikVekt} + For any two vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ is + $$|\overrightarrow{v}+\overrightarrow{u}|\leq |\overrightarrow{v}|+|\overrightarrow{u}|.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.7.pic} +\caption{} \label{sl.vek.5.1.7.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.1.7.pic}) + + Naj bo $A$ poljubna točka ter $B$ in $C$ takšni točki, da velja $\overrightarrow{AB}=\overrightarrow{v}$ in $\overrightarrow{BC}=\overrightarrow{u}$ (izrek \ref{vektAvObst1TockaB}). Iz definicij seštevanja vektorjev in dolžine vektorja ter trikotniške neenakosti (izrek \ref{neenaktrik}) dobimo: + \begin{eqnarray*} + |\overrightarrow{v}+\overrightarrow{u}|&=&|\overrightarrow{AB}+\overrightarrow{BC}|=\\ + &=&|\overrightarrow{AC}|=|AC|\leq + |AB|+|AC|= + |\overrightarrow{AB}|+|\overrightarrow{AC}|=\\ + &=&|\overrightarrow{v}|+|\overrightarrow{u}|, + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + +Na podoben način kot količnik lahko definiramo tudi produkt kolinearnih vektorjev. +Če sta $\overrightarrow{v}$ in $\overrightarrow{u}$ dva kolinearna vektorja, potem definiramo operacijo \pojem{množenja dveh kolinearnih vektorjev}\footnote{Gre za poseben primer t. i. \textit{skalarnega produkta}, ki se v linearni algebri definira za poljubna dva vektorja. V evklidskem prostoru je $\overrightarrow{v}\cdot\overrightarrow{u}= +|\overrightarrow{v}|\cdot|\overrightarrow{u}|\cdot\cos +\angle \overrightarrow{v},\overrightarrow{u}$.}: + \begin{eqnarray*} + \overrightarrow{v}\cdot \overrightarrow{u}= +\left\{ + \begin{array}{ll} + |\overrightarrow{v}|\cdot|\overrightarrow{u}|, & +\overrightarrow{v},\overrightarrow{u}\rightrightarrows ; \\ + -|\overrightarrow{v}|\cdot|\overrightarrow{u}|, & \overrightarrow{v},\overrightarrow{u}\rightleftarrows. + \end{array} +\right. +\end{eqnarray*} + Rezultat te operacije, ki je realno število, imenujemo \index{produkt kolinearnih vektorjev} \pojem{produkt kolinearnih vektorjev}. +Že iz definicije sledi: + \begin{eqnarray} \label{eqnMnozVektDolzina} + \overrightarrow{v}\cdot \overrightarrow{v}=|\overrightarrow{v}|^2 + \end{eqnarray} + Jasno je tudi, da za tri kolinearne točke $A$, $B$ in $L$ velja ekvivalenca (Figure \ref{sl.vek.5.3.1.pic}): + + \begin{eqnarray} \label{eqnMnozVektRelacijaB} + \overrightarrow{LA}\cdot \overrightarrow{LB}<0 +\hspace*{1mm}\Leftrightarrow\hspace*{1mm} \mathcal{B}(A,L,B) +\hspace*{3mm} \textrm{(}A,B,L \textrm{ so kolinearne)}. + \end{eqnarray} + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.3.1.pic} +\caption{} \label{sl.vek.5.3.1.pic} +\end{figure} + + + + + \bnaloga\footnote{15. IMO USSR - 1973, Problem 1.} + Point $O$ lies on line $g$; + $\overrightarrow{OP_1},\overrightarrow{OP_2}, + \cdots,\overrightarrow{OP_n}$ + are unit vectors such that points $P_1, P_2, \cdots, P_n$ + all lie in a plane containing $g$ and on one side of $g$. Prove that + if $n$ is odd, + $$|\overrightarrow{OP_1}+\overrightarrow{OP_2}+ + \cdots+\overrightarrow{OP_n}|\geq 1.$$ \label{OlimpVekt15} + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.3.IMO1.pic} +\caption{} \label{sl.vek.5.3.IMO1.pic} +\end{figure} + + \textbf{\textit{Solution.}} Naj bo $k$ enotska krožnica s središčem + v točki $O$, ki seka premico $g$ v točkah $A$ in $B$. Vse + točke $P_1, P_2, \cdots, P_n$ ležijo na pripadajoči + polkrožnici, ki jo na krožnici $k$ določata točki $A$ in $B$. + Brez škode za splošnost lahko + predpostavimo, da so vektorji označeni tako, da velja: + $\angle BOP_1\leq \angle BOP_2\leq\cdots\leq \angle BOP_n$ + (Figure \ref{sl.vek.5.3.IMO1.pic}). + Naj bo $s$ simetrala kota $\angle P_1OP_n$ in $g'$ + pravokotnica premice $s$ v točki $O$. Ker je $\angle P_1OP_n$ konveksni + kot, cel leži na istem bregu premice $g'$. Potem so tudi vse + točke $P_i$ ($i\in \{1,2,\cdots,n\}$ na istem bregu premice $g'$, + ker ležijo v tem kotu. Tako lahko našo predpostavko iz naloge, da so + vse točke $P_1, P_2, \cdots, P_n$ na istem bregu neke premice + $g$ oz. $g'$, + zaostrimo z zahtevo (ki sledi iz te predpostavke), + da sta prvi in zadnji vektor zaporedja + simetrična glede na pravokotnico $s$ premice $g'$ skozi točko $O$. To + dejstvo bomo uporabljali v dokazu. + + Po predpostavki je $n$ liho število. Naj bo $n=2k-1$, $k\in \mathbb{N}$. + Dokaz bomo izpeljali z indukcijo po $k$. + + (\textit{i}) Če je $k=1$ oz. $n=1$, je že $|\overrightarrow{OP_1}|=1$ + oz. velja tudi $|\overrightarrow{OP_1}|\geq1$ in je trditev + izpolnjena. + + (\textit{ii}) Predpostavimo, da trditev velja za $k=l$ oz. za vsako zaporedje + $n=2l-1$ vektorjev. Dokažimo, da potem trditev velja tudi za $k=l+1$ oz. v + primeru zaporedja $n=2l+1$ vektorjev. Naj bodo + $\overrightarrow{OP_1},\overrightarrow{OP_2}, + \cdots,\overrightarrow{OP_{2l+1}}$ + enotski vektorji, kjer so $P_1, P_2, \cdots, P_{2l+1}$ + točke, ki + ležijo v isti polravnini z robom $g'$ (v polravnini $\alpha$) + in sta vektorja $\overrightarrow{OP_1}$ + in $\overrightarrow{OP_{2l+1}}$ simetrična glede + na premico $s$. Iz tega sledi, da vektor + $\overrightarrow{OU}=\overrightarrow{OP_1}+\overrightarrow{OP_{2l+1}}$ + leži na premici $s$ oz. $U\in s$. Naj bo + $\overrightarrow{OV}=\overrightarrow{OP_2}+\overrightarrow{OP_3}+ + \cdots+\overrightarrow{OP_{2l}}$. Po indukcijski predpostavki + je $|\overrightarrow{OV}|=|\overrightarrow{OP_2}+\overrightarrow{OP_3}+ + \cdots+\overrightarrow{OP_{2l}}|\geq 1$. Torej: + \begin{eqnarray*} + |\overrightarrow{OP_1}+\overbrace{\overrightarrow{OP_2}+ + \cdots+\overrightarrow{OP_{2l}}}+\overrightarrow{OP_{2l+1}}|= + |\overrightarrow{OU}+\overrightarrow{OV}|. + \end{eqnarray*} + Naj bo + $\overrightarrow{OW}=\overrightarrow{OU}+\overrightarrow{OV}$. Štirikotnik $VOUW$ je paralelogram (ali pa so točke $O$, $U$, $V$ in $W$ + kolinearne). + Po zgledu \ref{vektOPi} sta točki $U$ in $V$ v polravnini $\alpha$. Točka + $U$ leži na premici $s$, ki je simetrala iztegnjenega kota, ki + ga določata premica $g'$ in vrh $O$. Zato je $0\leq\angle + UOV\leq 90^0$. Iz paralelograma $VOUW$ je potem $\angle OVW > + 90^0$. Iz trikotniške neenakosti \ref{neenaktrik} (za trikotnik + $OVW$) sledi $|\overrightarrow{OW}|>|\overrightarrow{OV}|$. + Neenakost velja tudi v primeru, kadar so točke $O$, $U$, $V$ in $W$ + kolinearne. Sedaj imamo: + \begin{eqnarray*} + |\overrightarrow{OP_1}+\overrightarrow{OP_2}+ + \cdots+\overrightarrow{OP_{2l}}+\overrightarrow{OP_{2l+1}}|= + |\overrightarrow{OU}+\overrightarrow{OV}|= + |\overrightarrow{OW}|>|\overrightarrow{OV}|\geq 1, + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + +%________________________________________________________________________________ + \poglavje{Further Use of Vectors} \label{odd5UporabVekt} + + Najprej si oglejmo trditev, ki je + direktna posledica definicije vektorjev (razdelek \ref{odd5DefVekt}). + + \bizrek \label{vektParalelogram} + A quadrilateral $ABCD$ is a parallelogram if and only if + $\overrightarrow{AB}=\overrightarrow{DC}$. + \eizrek + + + Omenimo tudi, da lahko v terminih vektorjev že +dokazano lastnost srednjice trikotnika - izrek \ref{srednjicaTrik} - +sedaj izrazimo v naslednji obliki. + + \bizrek + \label{srednjicaTrikVekt} + Let $PQ$ be the midsegment of a triangle $ABC$, corresponding + to the side + $BC$. Then + $$ \overrightarrow{PQ} = \frac{1}{2} \overrightarrow{AB}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.4.1c.pic} +\caption{} \label{sl.vek.5.4.1c.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.4.1c.pic}) + Trditev je direktna posledica izreka \ref{srednjicaTrik} + \kdokaz + +Videli smo, da lahko vektorja seštevamo po paralelogramskem pravilu +(kadar imata skupni začetek) ali po trikotniškem pravilu (kadar je +začetek drugega na koncu prvega). Sedaj bomo vpeljali še eno +pravilo za seštevanje dveh vektorjev, ki sta v poljubni legi. Gre za t. i. \index{pravilo!splošno za seštevanje vektorjev}\pojem{splošno pravilo za seštevanje vektorjev}. + + \bizrek \label{vektSestSplosno} + If $S_1$ and $S_2$ + are the midpoints of line segments $A_1B_1$ and $A_2B_2$, then + $$\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}= + 2\overrightarrow{S_1S_2}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.4.1b.pic} +\caption{} \label{sl.vek.5.4.1b.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.4.1b.pic}) + +Ker sta $S_1$ in $S_2$ + središči daljic $A_1B_1$ in $A_2B_2$, je + $\overrightarrow{S_1A_1}=-\overrightarrow{S_1B_1}$ in $\overrightarrow{S_2A_2}=-\overrightarrow{S_2B_2}$ oz. $\overrightarrow{S_1A_1}+\overrightarrow{S_1B_1}=\overrightarrow{0}$ in $\overrightarrow{S_2A_2}+\overrightarrow{S_2B_2}=\overrightarrow{0}$. + + Če vektor $\overrightarrow{S_1S_2}$ razstavimo na dva načina po poligonskem pravilu za seštevanje vektorjev, dobimo najprej: + \begin{eqnarray*} + \overrightarrow{S_1S_2}&=&\overrightarrow{S_1A_1}+\overrightarrow{A_1A_2}+ + \overrightarrow{A_2S_2}\\ + \overrightarrow{S_1S_2}&=&\overrightarrow{S_1B_1}+\overrightarrow{B_1B_2}+ + \overrightarrow{B_2S_2}, + \end{eqnarray*} + nato pa s seštevanjem teh še: + \begin{eqnarray*} + 2\cdot\overrightarrow{S_1S_2}&=&\overrightarrow{S_1A_1}+\overrightarrow{A_1A_2}+ + \overrightarrow{A_2S_2}+\\ + &+&\overrightarrow{S_1B_1}+\overrightarrow{B_1B_2}+ + \overrightarrow{B_2S_2}=\\ + &=& \overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}, + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + +Pogosto bomo relacijo iz prejšnjega izreka uporabljali tudi v +obliki: + $$\overrightarrow{S_1S_2}= + \frac{1}{2}(\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}),$$ + kar je posplošitev lastnosti srednjice trapeza in trikotnika. + + + + Na tem mestu bomo še enkrat ponovili vsa tri pravila za seštevanje dveh vektorjev. V prvem primeru torej predstavnika vektorjev izberemo tako, da je začetek drugega na koncu prvega, v drugem primeru imata vektorja skupno začetno točko, v tretjem sta predstavnika v splošni legi (Figure \ref{sl.vek.5.4.1a.pic}): + + + \begin{itemize} + \item \textit{za vsake tri točke $A$, $B$ in $C$ velja $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ (trikotniško pravilo),} + \item \textit{za vsake tri nekolinearne točke $A$, $B$ in $C$ velja $\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AD}$ natanko tedaj, ko je $ABDC$ paralelogram (paralelogramsko pravilo),} + \item \textit{če sta $S_1$ in $S_2$ + središči daljic $A_1B_1$ in $A_2B_2$, je + $\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}= + 2\overrightarrow{S_1S_2}$ (splošno pravilo).} + \end{itemize} + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.4.1a.pic} +\caption{} \label{sl.vek.5.4.1a.pic} +\end{figure} + + + Razen teh treh imamo še poligonsko pravilo, ki se nanaša na seštevanje več vektorjev (Figure \ref{sl.vek.5.1.12.pic}): + \begin{itemize} + \item \textit{$\overrightarrow{A_1A_2}+\overrightarrow{A_2A_3}+\cdots +\overrightarrow{A_{n-1}A_n}=\overrightarrow{A_1A_n}$ (poligonsko pravilo).} + \end{itemize} + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.1.12.pic} +\caption{} \label{sl.vek.5.1.12.pic} +\end{figure} + + + \bzgled \label{vektPetkoinikZgled} + Points $M$, $N$, $P$ and $Q$ are the midpoints of the sides $AB$, $BC$, $CD$ and $DE$ of a pentagon $ABCDE$. Prove + that the line segment $XY$, determined by the midpoints of the line segments $MP$ and $NQ$, is parallel to the line $AE$ + and calculate $\frac{\overrightarrow{XY}}{\overrightarrow{AE}}$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.4.2.pic} +\caption{} \label{sl.vek.5.4.2.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.4.2.pic}) + Če uporabimo izreka \ref{vektSestSplosno} in \ref{srednjicaTrikVekt} dobimo: + \begin{eqnarray*} + \overrightarrow{XY}&=&\frac{1}{2}\left(\overrightarrow{MQ}+\overrightarrow{PN} \right)=\\ + &=&\frac{1}{2}\left(\frac{1}{2}\left(\overrightarrow{AE}+\overrightarrow{BD}\right) + +\frac{1}{2}\overrightarrow{DB} \right)=\\ + &=&\frac{1}{4}\overrightarrow{AE} + \end{eqnarray*} + Torej sta vektorja $\overrightarrow{XY}$ in $\overrightarrow{AE}$ kolinearna in $\frac{\overrightarrow{XY}}{\overrightarrow{AE}}=\frac{1}{4}$. + \kdokaz + + + \bzgled + Let $O$ be an arbitrary point in the plane of a triangle $ABC$ and + $D$ and $E$ points of the sides $AB$ and $BC$ such that + $$\frac{\overrightarrow{AD}}{\overrightarrow{DB}}= + \frac{\overrightarrow{BE}}{\overrightarrow{EC}}=\frac{m}{n}.$$ + Let $F$ be the intersection of the line segments $AE$ and $CD$. Express $\overrightarrow{OF}$ as + a function of $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$, $m$ and $n$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.4.3.pic} +\caption{} \label{sl.vek.5.4.3.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.4.3.pic}) + +Dovolj je izraziti vektor $\overrightarrow{BF}$ kot linearno kombinacijo vektorjev $\overrightarrow{BA}$ in $\overrightarrow{BC}$, kajti: +\begin{eqnarray*} + \overrightarrow{OF}&=&\overrightarrow{OB}+\overrightarrow{BF}\\ + \overrightarrow{BA}&=&\overrightarrow{OA}-\overrightarrow{OB}\\ + \overrightarrow{BC}&=&\overrightarrow{OC}-\overrightarrow{OB}. + \end{eqnarray*} +Po \ref{vektParamPremica} je: +\begin{eqnarray*} + \overrightarrow{BF}&=&\lambda\overrightarrow{BD}+(1-\lambda)\overrightarrow{BC}= + \lambda\frac{n}{n+m}\overrightarrow{BA}+(1-\lambda)\overrightarrow{BC};\\ + \overrightarrow{BF}&=&\mu\overrightarrow{BA}+(1-\mu)\overrightarrow{BE}= + \mu\overrightarrow{BA}+(1-\mu)\frac{m}{n+m}\overrightarrow{BC} + \end{eqnarray*} +za neki števili $\lambda,\mu\in \mathbb{R}$. Ker sta vektorja $\overrightarrow{BA}$ in $\overrightarrow{BC}$ nekolinearna, po izreku \ref{vektLinKomb1Razcep} dobimo sistem: +\begin{eqnarray*} + & & \lambda\frac{n}{n+m}=\mu\\ + & & 1-\lambda=(1-\mu)\frac{m}{n+m}, + \end{eqnarray*} + ki ga rešimo po $\lambda$ in $\mu$ v funkciji $m$ in $n$. Dovolj je, če izračunamo le $\lambda$ in uvrstimo v $\overrightarrow{BF}= + \lambda\frac{n}{n+m}\overrightarrow{BA}+(1-\lambda)\overrightarrow{BC}$. + \kdokaz + + + \bzgled + Let $A_1$, $A_2$, ..., $A_n$ be points on a line $p$ and $B_1$, $B_2$, ..., $B_n$ points on a line $q$ ($n\geq 3$), such that + $$\overrightarrow{A_1A_2}: + \overrightarrow{A_2A_3}:\cdots: + \overrightarrow{A_{n-1}A_n}= + \overrightarrow{B_1B_2}: + \overrightarrow{B_2B_3}:\cdots: + \overrightarrow{B_{n-1}B_n}.$$ + Prove that the midpoints of line segments + $A_1B_1$, $A_2B_2$, ..., $A_nB_n$ lie on one line. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.4.4.pic} +\caption{} \label{sl.vek.5.4.4.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.4.4.pic}) + + Dovolj je dokazati, da so tri poljubna središča po vrsti kolinearne točke. +Brez škode za splošnost dokažimo le, da so točke $S_1$, $S_2$, $S_3$ kolinearne. Naj bo: +$$\frac{\overrightarrow{A_1A_2}}{\overrightarrow{A_2A_3}}= +\frac{\overrightarrow{B_1B_2}}{\overrightarrow{B_2B_3}}=\lambda.$$ + Po izreku \ref{vektSestSplosno} je: + \begin{eqnarray*} + \overrightarrow{S_1S_2}&=&\frac{1}{2}\left(\overrightarrow{A_1A_2}+ + \overrightarrow{B_1B_2}\right)=\\ + &=&\frac{1}{2}\left(\lambda\overrightarrow{A_2A_3}+ + \lambda\overrightarrow{B_2B_3}\right)=\\ + &=&\frac{\lambda}{2}\overrightarrow{S_2S_3}, + \end{eqnarray*} + kar pomeni, da sta $\overrightarrow{S_1S_2}$ in $\overrightarrow{S_2S_3}$ kolinearna vektorja (izrek \ref{vektKriterijKolin}), zato so kolinearne tudi točke $S_1$, $S_2$ in $S_3$. + \kdokaz + + + +%________________________________________________________________________________ + \poglavje{Centroid of a Polygon With Respect to Its Vertices} \label{odd5TezVeck} + +Sedaj bomo posplošili pojem težišča trikotnika na poljubne večkotnike. V razdelku \ref{odd3ZnamTock} smo definirali pojem težišča trikotnika. Naslednja trditev se nanaša na eno dodatno lastnost tega pojma, ki je povezana s pojmom vektorja. + + + \bizrek \label{tezTrikVekt} + If $T$ is the centroid of a triangle $ABC$, then + $$\overrightarrow{TA}+\overrightarrow{TB}+ + \overrightarrow{TC}=\overrightarrow{0}.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.1.pic} +\caption{} \label{sl.vek.5.5.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} Označimo z $A_1$ središče stranice $BC$ (Figure \ref{sl.vek.5.5.1.pic}). Po izreku \ref{vektSredOSOAOB} je $\overrightarrow{TA_1}= + \frac{1}{2}\left(\overrightarrow{TB}+\overrightarrow{TC}\right)$, po izreku \ref{tezisce} pa za težišče $T$ trikotnika $ABC$ velja $|AT|:|TA_1|=2:1$ oz. $\overrightarrow{TA}=-2\cdot\overrightarrow{TA_1}$. Torej: +$$\overrightarrow{TA}+\overrightarrow{TB}+ + \overrightarrow{TC}=\overrightarrow{TA}+2\cdot\overrightarrow{TA_1}= + \overrightarrow{TA}-\overrightarrow{TA}= + \overrightarrow{0},$$ kar je bilo treba dokazati. \kdokaz + + +Velja tudi obratna trditev: + + + + \bizrek \label{tezTrikVektObr} + Let $A$, $B$ and $C$ be three non-collinear points. If $X$ is a point such that + $$\overrightarrow{XA}+\overrightarrow{XB}+ + \overrightarrow{XC}=\overrightarrow{0},$$ + then + $X$ is the centroid of the triangle $ABC$. + \eizrek + +\textbf{\textit{Proof.}} Naj bo $T$ težišče trikotnika. Po +prejšnjem izreku \ref{tezTrikVekt} je +$\overrightarrow{TA}+\overrightarrow{TB}+ +\overrightarrow{TC}=\overrightarrow{0}$. Po predpostavki +je tudi $\overrightarrow{XA}+\overrightarrow{XB}+ +\overrightarrow{XC}=\overrightarrow{0}$. Če odštejemo +dve enakosti, dobimo +$3 \cdot \overrightarrow{TX}=\overrightarrow{0}$ oz. $X=T$. + \kdokaz + +Dokazana lastnost težišča trikotnika nam da idejo za definicijo težišča poljubnega večkotnika (Figure \ref{sl.vek.5.5.2.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.2.pic} +\caption{} \label{sl.vek.5.5.2.pic} +\end{figure} + +Točka $T$ je \index{težišče!večkotnika}\pojem{težišče večkotnika $A_1A_2\ldots A_n$ glede na njegova oglišča}, če velja: +$$\overrightarrow{TA_1}+\overrightarrow{TA_2}+\cdots +\overrightarrow{TA_n}=\overrightarrow{0}.$$ +Prejšnjo relacijo lahko zapišemo tudi v obliki: +$$\sum_{k=1}^n\overrightarrow{TA_k}=\overrightarrow{0}.$$ + +Omenimo še, da težišče poljubnega lika $\Phi$ v ravnini definiramo kot točko $T$, za katero velja: +$$\sum_{X\in \Phi}\overrightarrow{TX}=\overrightarrow{0}.$$ + +U primeru našega večkotnika $A_1A_2\ldots A_n$ smo pravzaprav dobili težišče lika, ki predstavlja unijo vseh oglišč tega večkotnika $\{A_1,A_2,\ldots, A_n\}$. Zato smo poudarili, da gre za težišče večkotnika glede na njegova oglišča. Razen tega bi lahko govorili o \pojem{težišču večkotnika glede na vse njegove točke}, kar bi bolj ustrezalo splošni definiciji težišča poljubnega lika. + +Če obravnavamo težišče lika v fizikalnem smislu kot - središče mase - prva varianta težišča predstavlja središče mase, pri čemer je vsa masa v ogliščih in ima vsako oglišče enako maso. V drugem primeru pa gre za središče mase večkotnika, kjer je masa homogeno razporejena po njegovi celotni notranjosti\footnote{\index{Arhimed} + \textit{Arhimed + iz Sirakuze} (3. st. pr. n. š.) starogrški matematik + je prvi ustvaril koncept težišča, +ki ga je uporabljal v mnogih svojih zapisih o mehaniki, ampak +lahko le ugibamo, kaj točno je imel v mislih, ko je +obravnaval težišče, ker nobeden od njegovih ohranjenih zapisov ne vsebuje +jasne opredelitve pojma. Težišče kot središče mase v fizikalnem smislu je igralo pomembno vlogo v Newtonovi (\index{Newton, +I.}\textit{I. Newton} (1643-1727), angleški fizik in matematik) mehaniki, kjer večja telesa pogosto obravnavamo kot točke z določeno maso.}. + +Če govorimo o splošnem večkotniku, sta omenjeni težišči enaki le pri poljubnem trikotniku, že pri poljubnem štirikotniku se težišči razlikujeta. +V nadaljevanju bomo obravnavali le težišče večkotnika glede na njegova oglišča, zato ga bomo imenovali kar \pojem{težišče večkotnika}. + +Najprej bomo obravnavali težišče štirikotnika. + + \bizrek + The centroid of a parallelogram $ABCD$ is its circumcentre $S$, i.e. + the intersection of its diagonals. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.3.pic} +\caption{} \label{sl.vek.5.5.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.3.pic}) + + Po izreku \ref{paralelogram} je točka $S$ skupno središče njegovih diagonal $AC$ in $BD$, zato iz izreka \ref{vektSredDalj} sledi $\overrightarrow{SA}+\overrightarrow{SC}=\overrightarrow{0}$ in $\overrightarrow{SB}+\overrightarrow{SD}=\overrightarrow{0}$. + Torej velja: + $$\overrightarrow{SA}+\overrightarrow{SB}+ + \overrightarrow{SC}+\overrightarrow{SD}=\overrightarrow{0},$$ +kar pomeni, da je $S$ težišče paralelograma $ABCD$. +\kdokaz + + + \bizrek \label{vektVarignon} + The centroid of a quadrilateral $ABCD$ is the centroid of its Varignon + parallelogram (see theorem \ref{Varignon}). + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.1a.pic} +\caption{} \label{sl.vek.5.5.1a.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.1a.pic}) + +Naj bodo $P$, $K$, $Q$ in $L$ središča stranic $AB$, $BC$, $CD$ in $DA$ oz. $PKQL$ Varignonov paralelogram štirikotnika $ABCD$ (štirikotnik $PKQL$ je paralelogram po izreku \ref{Varignon}). Po prejšnjem izreku \ref{vektVarignon} je presečišče diagonal $PQ$ in $LK$ (označimo ga s $T$) hkrati težišče paralelograma $PKQL$. Potem je (izreka \ref{vektSredOSOAOB}) in \ref{paralelogram}): + \begin{eqnarray*} + \overrightarrow{TA}+\overrightarrow{TB}+ + \overrightarrow{TC}+\overrightarrow{TD}&=& + 2\cdot\overrightarrow{TP}+ + 2\cdot\overrightarrow{TQ}=\\ + &=& + 2\cdot\left(\overrightarrow{TP}+ +\overrightarrow{TQ}\right)=\\ + &=&\overrightarrow{0}, + \end{eqnarray*} +kar pomeni, da je $T$ težišče štirikotnika $ABCD$. +\kdokaz + + + + \bizrek \label{vektTezVeckXT} + If $X$ is an arbitrary point and $T$ the centroid of a polygon $A_1A_2\ldots A_n$, then + $$\overrightarrow{XT}=\frac{1}{n}\left(\overrightarrow{XA_1} + +\overrightarrow{XA_2}+\cdots +\overrightarrow{XA_n}\right).$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.4.pic} +\caption{} \label{sl.vek.5.5.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.4.pic}) + +Najprej je $\overrightarrow{XT}=\overrightarrow{XA_k}+\overrightarrow{A_kT}$ (za vsak $k\in\{1,2,\ldots,n\}$). Če seštejemo vseh $n$ relacij, dobimo: + +\begin{eqnarray*} + n\cdot\overrightarrow{XT} + &=&\sum_{k=1}^{n}\left(\overrightarrow{XA_k}+\overrightarrow{A_kT}\right)=\\ + &=&\sum_{k=1}^{n}\overrightarrow{XA_k}+\sum_{k=1}^{n}\overrightarrow{A_kT}=\\ + &=& + \sum_{k=1}^{n}\overrightarrow{XA_k}-\sum_{k=1}^{n}\overrightarrow{TA_k}=\\ + &=& + \sum_{k=1}^{n}\overrightarrow{XA_k}-\overrightarrow{0}=\\ + &=&\sum_{k=1}^{n}\overrightarrow{XA_k}. + \end{eqnarray*} + +Zato je: +$$\overrightarrow{XT}=\frac{1}{n}\cdot\sum_{k=1}^{n}\overrightarrow{XA_k},$$ kar je bilo treba dokazati. \kdokaz + + +Kot posledica prejšnjega izreka je posebej koristna relacija, ki velja za težišče poljubnega trikotnika. + + + + \bizrek \label{vektTezTrikXT} + If $X$ is an arbitrary point and $T$ the centroid of a triangle $ABC$, then + $$\overrightarrow{XT}=\frac{1}{3}\left(\overrightarrow{XA} + +\overrightarrow{XB}+\overrightarrow{XC}\right).$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.5.pic} +\caption{} \label{sl.vek.5.5.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.5.pic}) + +Direktna posledica prejšnjega izreka za $n=3$ +\kdokaz + + + +% Postopek za n-kotnik (n-1)-kotnik ... + + Zaenkrat smo imeli efektiven postopek za določanje težišča trikotnikov in štirikotnikov. Za poljubni štirikotnik pravzaprav še ne vemo, ali težišče sploh obstaja. To vprašanje bomo obravnavali v nadaljevanju. + + Idejo bomo iskali v naslednjih dejstvih. Če daljico $AB$ obravnavamo kot degenerirani $2$-kotnik, je njegovo težišče središče daljica $AB$; označimo ga s $T_2$. Za težišče $T_3$ trikotnika $ABC$ potem velja + $\overrightarrow{CT_3}=\frac{2}{3}\cdot \overrightarrow{CT_2}$ (izrek \ref{tezisce}). To idejo bomo posplošili v naslednjem izreku. + + + \bizrek \label{vektTezVeck} + Every polygon $A_1A_2\ldots A_n$ has exactly one centroid. If + $T_{n-1}$ is the centroid of the polygon $A_1A_2\ldots A_{n-1}$ and $T_n$ a point such that $$\overrightarrow{A_nT_n}=\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}},$$ + then $T_n$ is the centroid of the polygon $A_1A_2\ldots A_n$. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.2a.pic} +\caption{} \label{sl.vek.5.5.2a.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.2a.pic}) + + Najprej iz relacije $\overrightarrow{A_nT_n}=\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}}$ sledi + $\overrightarrow{T_nA_n}=-\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}}$ in $\overrightarrow{T_nT_{n-1}}=\frac{1}{n}\cdot\overrightarrow{A_nT_{n-1}}$. + Ker je $T_{n-1}$ težišče večkotnika $A_1A_2\ldots A_{n-1}$, po izreku \ref{vektTezVeckXT} velja $\overrightarrow{T_nA_1}+\overrightarrow{T_nA_2}+\cdots +\overrightarrow{T_nA_{n-1}}=\left(n-1\right)\cdot \overrightarrow{T_nT_{n-1}}$. + + Torej: + \begin{eqnarray*} + & & \overrightarrow{T_nA_1}+\overrightarrow{T_nA_2}+\cdots + +\overrightarrow{T_nA_{n-1}}+\overrightarrow{T_nA_n}=\\ + &=&\left(n-1\right)\cdot \overrightarrow{T_nT_{n-1}} +\overrightarrow{T_nA_n}=\\ + &=&\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}} -\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}}=\\ + &=&\overrightarrow{0}, + \end{eqnarray*} + + kar pomeni, da je $T_n$ težišče večkotnika $A_1A_2\ldots A_n$. + + Predpostavimo, da ima večkotnik $A_1A_2\ldots A_n$ še eno težišče $T'$. Toda v tem primeru je (izrek \ref{vektTezVeckXT}): + \begin{eqnarray*} + \overrightarrow{0}= + \overrightarrow{T'A_1}+\overrightarrow{T'A_2}+\cdots + \overrightarrow{T'A_n}= + n\cdot\overrightarrow{T'T_n}. + \end{eqnarray*} + Torej $\overrightarrow{T'T_n}=\overrightarrow{0}$ oz. $T'=T_n$, kar pomeni, da ima večkotnik $A_1A_2\ldots A_n$ eno samo težišče. + \kdokaz + + Prejšnji izrek \ref{vektTezVeck} nam omogoča efektivno konstrukcijo težišča večkotnika $A_1A_2\ldots A_n$, tako da po vrsti najprej konstruiramo težišča večkotnikov $A_1A_2$, $A_1A_2A_3$,$A_1A_2A_3A_4$, ... in na koncu $A_1A_2\ldots A_n$ (Figure \ref{sl.vek.5.5.2a.pic}): + \begin{itemize} + \item točka $T_2$ je središče daljice $A_1A_2$, + \item $T_3$ je takšna točka, da velja: $\overrightarrow{A_3T_3}=\frac{2}{3}\cdot \overrightarrow{A_3T_2}$, + \item $T_4$ je takšna točka, da velja: $\overrightarrow{A_4T_4}=\frac{3}{4}\cdot \overrightarrow{A_4T_3}$,\\ + $\vdots$ + \item $T_n$ je takšna točka, da velja $\overrightarrow{A_nT_n}=\frac{n-1}{n}\cdot \overrightarrow{A_nT_{n-1}}$. + \end{itemize} + + + Še lažji postopek za določanje težišča večkotnika dobimo, če uporabimo relacijo + $$\overrightarrow{XT}=\frac{1}{n}\left(\overrightarrow{XA_1} + +\overrightarrow{XA_2}+\cdots +\overrightarrow{XA_n}\right)$$ + iz izreka \ref{vektTezVeckXT}. Torej za poljubno točko $X$ enostavno načrtamo vektor $\overrightarrow{XT}$ in dobimo točko $T$. + + Poudarimo še, da v obeh primerih konstrukcije težišča večkotnika potrebujemo postopek načrtovanja točke, ki dano daljico deli v določenem razmerju. Ta postopek bomo obravnavali v razdelku \ref{odd5TalesVekt} (zgleda \ref{izrekEnaDelitevDaljice} in \ref{izrekEnaDelitevDaljiceNan}). + + + + V dokazu trditve \ref{vektTezVeck} nismo uporabljali +dejstva, da so točke $A_1$, $A_2$, ..., $A_n$ v isti ravnini. +Trditev velja tudi v primeru, če je $ABCD$ ($n=4$) t. i. \pojem{tetraeder} +\index{tetraeder}. Omenjeno točko tedaj imenujemo +\index{težišče!tetraedra} \pojem{težišče tetraedra}. Mogoče je +namreč dokazati analogni izrek za tetraeder; daljice, ki so +določene z oglišči tetraedra in težišči nasprotnih ploskev, +potekajo skozi težišče tega tetraedra, ki jih deli v razmerju $3 :1$. + + Tudi v splošnem primeru, če je $n\in \mathbb{N}$, lahko govorimo o t. i. \index{simpleks}\pojem{simpleksu} (posplošitev: točka, daljica, trikotnik, tetraeder, ...), ki leži v $(n-1)$-razsežnem evklidskem prostoru. + + \bzgled + Let $A$, $B$ and $C$ be the centroids of a triangles $OMN$, + $ONP$ and $OPM$, then the centroid $T$ of the triangle $MNP$, + the centroid $T_1$ of the triangle $ABC$ and the point $O$ are + three collinear points. Furthermore, it is $OT_1:T_1T=2:1$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.6.pic} +\caption{} \label{sl.vek.5.5.6.pic} +\end{figure} + + \textbf{\textit{Proof.}} Označimo z $P_1$, $M_1$ in $N_1$ središča stranic $MN$, $NP$ in $PM$ trikotnika $PMN$ (Figure \ref{sl.vek.5.5.6.pic}). + Če uporabimo izreke \ref{vektTezTrikXT}, \ref{tezisce} in \ref{vektSredOSOAOB}, dobimo: + + \begin{eqnarray*} + \overrightarrow{OT_1}&=&\frac{1}{3}\left( + \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC} \right)=\\ + &=&\frac{1}{3}\left( + \frac{2}{3}\overrightarrow{OP_1}+\frac{2}{3}\overrightarrow{OM_1}+ + \frac{2}{3}\overrightarrow{ON_1} \right)=\\ + &=&\frac{1}{3}\left( + \frac{1}{3}\left(\overrightarrow{OM}+\overrightarrow{ON}\right) + +\frac{1}{3}\left(\overrightarrow{ON}+\overrightarrow{OP}\right)+ + \frac{1}{3}\left(\overrightarrow{OP}+\overrightarrow{OM}\right) \right)=\\ + &=&\frac{2}{9}\left( + \overrightarrow{OM}+\overrightarrow{ON}+ + \overrightarrow{OP} \right)=\\ + &=&\frac{2}{3}\overrightarrow{OT} + \end{eqnarray*} + +Iz $\overrightarrow{OT_1}=\frac{2}{3}\overrightarrow{OT}$ pa sledi $\overrightarrow{OT_1}=2\overrightarrow{T_1T}$, kar pomeni, da sta vektorja $\overrightarrow{OT_1}$ in $\overrightarrow{T_1T}$ kolinearna (tudi točke $O$, $T_1$ in $T$) in velja $OT_1:T_1T=2:1$. +\kdokaz + + \bzgled + The centroid of a regular $n$-gon $A_1A_2...A_n$ is its centre (i.e. incentre and circumcentre). + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.5.7.pic} +\caption{} \label{sl.vek.5.5.7.pic} +\end{figure} + + \textbf{\textit{Proof.}} Označimo s $S$ središče pravilnega $n$-kotnika $A_1A_2...A_n$ (Figure \ref{sl.vek.5.5.7.pic}). +Dovolj je dokazati, da velja: +$$\overrightarrow{SA_1}+\overrightarrow{SA_2}+ +\cdots+\overrightarrow{SA_n}=\overrightarrow{0}.$$ +Čeprav je v primeru, ko je $n$ sodo število, trditev trivialna, bomo dokaz izpeljali za splošno vrednost $n$ (sodo in liho). +Predpostavimo, da je $\overrightarrow{SA_1}+\overrightarrow{SA_2}+\cdots+\overrightarrow{SA_n}=\overrightarrow{SX}$, kjer je $X\neq S$. +Toda če zavrtimo večkotnik okoli središča $S$ za kot $\theta=\frac{360}{n}$ (glej razdelek o rotaciji \ref{odd6Rotac}), se vsota vektorjev na levi strani enakosti ne spremeni, rezultat na desni pa postane vektor $\overrightarrow{SX'}$, kjer je $X'$ točka, ki jo dobimo iz $X$ z isto rotacijo. Ker mora ostati tudi desna stran enakosti nespremenjena, dobimo $\overrightarrow{SX'}=\overrightarrow{SX}$ oz. $X'=X$. To je možno edino v primeru, ko je $X=S$ oz. $\overrightarrow{SX}=\overrightarrow{0}$. +\kdokaz + + + + + + +%________________________________________________________________________________ + \poglavje{Hamilton's Theorem} \label{odd5Hamilton} + +Sedaj bomo nadaljevali z lastnostmi, ki se nanašajo na značilne točke trikotnika (razdelek \ref{odd3ZnamTock}). + + \bizrek \label{HamiltonLema} + If $O$, $V$ and $A_1$ are the circumcentre, the orthocentre and the midpoint + of the side $BC$ of a triangle $ABC$, respectively, then + $$\overrightarrow{AV}=2\cdot \overrightarrow{OA_1}.$$ + \eizrek + + + \textbf{\textit{Proof.}} +Označimo še z $B_1$ središče stranice $AC$ (Figure \ref{sl.vek.5.6.2.pic}). + +Vektorja $\overrightarrow{OA_1}$ in $\overrightarrow{AV}$ sta kolinearna, zato je $\overrightarrow{OA_1}=\alpha \cdot \overrightarrow{AV}$ za nek $\alpha \in \mathbb{R}$ (izrek \ref{vektKriterijKolin}). Prav tako je iz istih razlogov $\overrightarrow{OB_1}=\beta \cdot \overrightarrow{BV}$ (oz. $\overrightarrow{B_1O}=\beta \cdot \overrightarrow{VB}$) za nek $\beta \in \mathbb{R}$. Po izreku \ref{srednjicaTrikVekt} (srednjica trikotnika) je: + $$\overrightarrow{B_1A_1}=\frac{1}{2}\overrightarrow{AB}= + \frac{1}{2}\left(\overrightarrow{AV}+\overrightarrow{VB}\right)= + \frac{1}{2}\overrightarrow{AV}+\frac{1}{2}\overrightarrow{VB}.$$ + Hkrati velja tudi: + $$\overrightarrow{B_1A_1}=\overrightarrow{B_1O}+\overrightarrow{OA_1}= +\beta\overrightarrow{VB}+\alpha\overrightarrow{AV}= + \alpha\overrightarrow{AV}+\beta\overrightarrow{VB}.$$ + +Ker sta vektorja $\overrightarrow{AV}$ in $\overrightarrow{VB}$ nekolinearna, iz izreka \ref{vektLinKomb1Razcep} sledi $\alpha=\frac{1}{2}$ in $\beta=\frac{1}{2}$. Zato je $\overrightarrow{OA_1}=\alpha \overrightarrow{AV}=\frac{1}{2}\overrightarrow{AV}$ oz. $\overrightarrow{AV}=2\cdot \overrightarrow{OA_1}$. +\kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.6.2.pic} +\caption{} \label{sl.vek.5.6.2.pic} +\end{figure} + + Zelo uporaben je naslednji izrek. + + \bizrek \label{Hamilton}\index{izrek!Hamiltonov} + (Hamilton's\footnote{\index{Hamilton, W. R.}\textit{W. R. Hamilton} (1805--1865), angleški matematik.} theorem) + If $O$ and $V$ are circumcentre and orthocentre of a triangle $ABC$, respectively then + $$\overrightarrow{OA}+\overrightarrow{OB} + +\overrightarrow{OC}=\overrightarrow{OV}.$$ + + \eizrek + + + + \textbf{\textit{Proof.}} Označimo z $A_1$ središče stranice $BC$ (Figure \ref{sl.vek.5.6.2.pic}). Če uporabimo izreka \ref{vektSredOSOAOB} in \ref{HamiltonLema}, dobimo: + +$$\overrightarrow{OA}+\overrightarrow{OB} + +\overrightarrow{OC} + =\overrightarrow{OA}+2\cdot \overrightarrow{OA_1}= + \overrightarrow{OA}+\overrightarrow{AV}= + \overrightarrow{OV},$$ kar je bilo treba dokazati. \kdokaz + + Nadaljevali bomo s posledicami prejšnjih dveh izrekov. + + + \bzgled \label{HamiltonPoslTetiv} + A quadrilateral $ABCD$ is inscribed in a circle with a centre $O$. + The diagonals $AC$ and $BD$ are perpendicular. + If $M$ is the foot of the perpendicular from the centre $O$ on the line $CD$, then + $$|OM|=\frac{1}{2}|AB|.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.6.3a.pic} +\caption{} \label{sl.vek.5.6.3a.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $V$ višinska točka trikotnika $BCD$ + (Figure \ref{sl.vek.5.6.3a.pic}). + Ker je $AC\perp BD$, točka $V$ leži na diagonali $AC$. Po izreku + \ref{HamiltonLema} je $\overrightarrow{OM}=\frac{1}{2} \overrightarrow{BV}$, + zato je tudi $|OM|=\frac{1}{2}|BV|$. Ker je še (izreka \ref{ObodObodKot} in + \ref{KotaPravokKraki}): + $$\angle BAV=\angle BAC\cong\angle BDC\cong\angle AVB,$$ +sledi $BV\cong AB$ (izrek \ref{enakokraki}) oz. +$|OM|=\frac{1}{2}|AB|$. + \kdokaz + + + \bzgled Let $V$ be the orthocentre and $O$ the circumcentre of a triangle $ABC$ and $AV\cong AO$. + Prove that $\angle BAC=60^0$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.6.1a.pic} +\caption{} \label{sl.vek.5.6.1a.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Naj bo $A_1$ središče stranice $BC$ trikotnika $ABC$ (Figure \ref{sl.vek.5.6.1a.pic}). Po izreku \ref{HamiltonLema} je +$|AV|=2\cdot|OA_1|$. Ker je še $OA\cong OC$, sledi, da v pravokotnemu trikotniku $OA_1C$ velja $|OC|=2\cdot|OA_1|$. Z $O'$ označimo točko, ki je simetrična točki $O$ glede na točko $A_1$. Iz $\triangle OA_1C\cong \triangle O'A_1C$ (izrek \textit{SAS} \ref{SKS}) sledi $OC\cong O'C\cong OO'$, kar pomeni, da je $\triangle OO'C$ enakostranični trikotnik oz. $\angle A_1OC=60^0$. Iz izreka \ref{SredObodKot} in skladnosti trikotnikov $BOA_1$ in $COA_1$ (izrek \textit{SSS} \ref{SSS}) na koncu sledi +$\angle BAC=\frac{1}{2}\angle BOC=\angle A_1OC=60^0$. + \kdokaz + + \bzgled \label{TetivniVisinska} + Let $ABCD$ be a cyclic quadrilateral and: + $V_A$ the orthocentre of the triangle $BCD$, + $V_B$ the orthocentre of the triangle $ACD$, + $V_C$ the orthocentre of the triangle $ABD$ and + $V_D$ the orthocentre of the triangle $ABC$. + Prove that:\\ + a) the line segments $AV_A$, + $BV_B$, $CV_C$ and $DV_D$ + has a common midpoint,\\ + b) the quadrilateral $V_AV_BV_CV_D$ is congruent + to the quadrilateral $ABCD$. + \ezgled + + +\begin{figure}[htp] +\centering +\input{sl.vek.5.6.3b.pic} +\caption{} \label{sl.vek.5.6.3b.pic} +\end{figure} + + +\begin{figure}[htp] +\centering +\input{sl.vek.5.6.3.pic} +\caption{} \label{sl.vek.5.6.3.pic} +\end{figure} + + + \textbf{\textit{Solution.}} Naj bo $O$ središče očrtane krožnice tetivnega štirikotnika $ABCD$ (Figure \ref{sl.vek.5.6.3.pic}). Jasno je, da je točka $O$ hkrati središče skupne očrtane krožnice trikotnikov $BCD$, $ACD$, $ABD$ in $ABC$. + Po Hamiltonovem izreku \ref{Hamilton} sledi: + +\begin{eqnarray*} + \overrightarrow{OV_A}&=&\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}\\ +\overrightarrow{OV_B}&=&\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OD} + \end{eqnarray*} + Nato dobimo: +\begin{eqnarray*} + \overrightarrow{V_BV_A}&=&\overrightarrow{V_BO}+\overrightarrow{OV_A}=\\ +&=&\overrightarrow{OV_A}-\overrightarrow{OV_B}=\\ + &=&\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD} +-(\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OD})=\\ +&=&\overrightarrow{OB}-\overrightarrow{OA}=\\ +&=&\overrightarrow{AB}. + \end{eqnarray*} + + +Torej velja + $\overrightarrow{V_BV_A}=\overrightarrow{AB}$. Po izreku \ref{vektParalelogram} je štirikotnik $ABV_AV_B$ paralelogram, po izreku \ref{paralelogram} pa imata njegovi diagonali $AV_A$ in $BV_B$ skupno središče - označimo ga s $S$ + (Figure \ref{sl.vek.5.6.3b.pic}). Na podoben način imata tudi vsak od parov daljic $AV_A$ in $CV_C$ oz. $AV_A$ in $DV_D$ skupno središče. Ker gre za središče daljice $AV_A$, sledi, da imajo vse štiri daljice $AV_A$, + $BV_B$, $CV_C$ in $DV_D$ + skupno središče - točko $S$. + + Na podoben način kot $\overrightarrow{V_BV_A}=\overrightarrow{AB}$ sledi tudi $\overrightarrow{V_CV_B}=\overrightarrow{BC}$, $\overrightarrow{V_DV_C}=\overrightarrow{CD}$ in $\overrightarrow{V_DV_A}=\overrightarrow{DA}$. + To pomeni, da imata štirikotnika $V_AV_BV_CV_D$ in $ABCD$ (izreka \ref{vektVzpSkl} in \ref{KotaVzporKraki}) vse istoležne stranice in notranje kote skladne. Torej velja $V_AV_BV_CV_D\cong ABCD$. Za formalni dokaz tega lahko uporabimo izometrijo $\mathcal{I}:A,B,C\mapsto V_A,V_B,V_C$ in dokažemo $\mathcal{I}(D)=V_D$. + \kdokaz + + + \bzgled \label{HamiltonSimson}\index{premica!Simsonova} + Let $ABCD$ be a cyclic quadrilateral and: $a$ is the Simson + line with respect to the triangle $BCD$ and the point $A$, $b$ is the Simson + line with respect to the triangle $ACD$ and the point $B$, $c$ is the Simson + line with respect to the triangle $ABD$ and the point $C$ ter $d$ S is the Simson + line with respect to the triangle $ABC$ and the point $D$. + Prove that the lines $a$, $b$, $c$ and + $d$ intersect at a single point. + \ezgled + + +\begin{figure}[htp] +\centering +\input{sl.vek.5.6.4.pic} +\caption{} \label{sl.vek.5.6.4.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.6.4.pic}). + + Trditev je direktna posledica izrekov + \ref{SimsZgled3} in \ref{TetivniVisinska} - premice $a$, $b$, $c$ in + $d$ se sekajo v točki $S$ (iz izreka \ref{TetivniVisinska}). + \kdokaz + + + +%________________________________________________________________________________ + \poglavje{Euler Line} \label{odd5EulPrem} + +Sedaj bomo dokazali pomembno lastnost, ki se nanaša na tri značilne točke trikotnika. + + \bizrek \label{EulerjevaPremica} + The circumcentre $O$, the centroid $T$ and the orthocentre $V$ + of an arbitrary triangle lies on the same line. Besides that it is + $$|OT|:|TV|=1:2.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.7.1.pic} +\caption{} \label{sl.vek.5.7.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.7.1.pic}) + + Če uporabimo izrek \ref{vektTezTrikXT} in Hamiltonov izrek \ref{Hamilton}, dobimo: + $$\overrightarrow{OT}=\frac{1}{3}\left(\overrightarrow{OA} + +\overrightarrow{OB}+\overrightarrow{OC}\right)= + \frac{1}{3}\overrightarrow{OV}.$$ + +Vektorja $\overrightarrow{OT}$ in $\overrightarrow{OV}$ sta torej kolinearna in velja $\overrightarrow{OT}:\overrightarrow{OV}=1:3$. To pomeni, da so točke $O$, $T$ in $V$ kolinearne in velja $|OT|:|TV|=1:2$. + \kdokaz + + Premico iz prejšnjega izreka, na kateri ležijo tri značilne točke, imenujemo \index{premica!Eulerjeva} \pojem{Eulerjeva\footnote{Lastnost, ki jo je leta 1765 dokazal švicarski matematik \index{Euler, L.}\textit{L. Euler} (1707--1783).} premica}. + + V naslednjem izreku bomo videli povezavo med Eulerjevo premico in Eulerjevo krožnico, ki smo jo obravnavali v razdelku \ref{odd3EulKroz}. + + + + + \bizrek \label{EulerKrozPrem1}\index{krožnica!Eulerjeva} + The centre of Euler of an arbitrary triangle lies on + The Euler line of this triangle. + Furthermore, it is the midpoint of the line segment determined by + the orthocentre and the circumcentre of this triangle. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.7.2.pic} +\caption{} \label{sl.vek.5.7.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Naj bodo $AA'$, $BB'$ in $CC'$ višine ter $A_1$, $B_1$ in $C_1$ središča stranic $BC$, $AC$ in $AB$ trikotnika $ABC$. Označimo z $O$ središče očrtane krožnice, z $V$ višinsko točko tega trikotnika ter z $V_A$, $V_B$ in $V_C$ središča daljic $VA$, $VB$ in $VC$ (Figure \ref{sl.vek.5.7.2.pic}). + + Po izreku \ref{EulerKroznica} ležijo točke $A'$, $B'$, $C'$, $A_1$, $B_1$, $C_1$, $V_A$, $V_B$ in $V_C$ na eni krožnici - t. i. Eulerjevi krožnici. Označimo središče te krožnice z $E$. Ker je $\angle V_AA'A_1\cong \angle AA'C=90^0$, je po izreku \ref{TalesovIzrKroz2} daljica $V_AA_1$ premer te krožnice oz. je točka $E$ središče daljice $V_AA_1$. + + Po izreku \ref{HamiltonLema} velja: + $$\overrightarrow{OA_1}=\frac{1}{2}\cdot \overrightarrow{AV}=\overrightarrow{V_AV}.$$ + Torej $\overrightarrow{OA_1}=\overrightarrow{V_AV}$, kar pomeni, da je štirikotnik $A_1OV_AV$ paralelogram (izrek \ref{vektParalelogram}). Njegovi diagonali $VO$ in $V_AA_1$ se razpolavljata (\ref{paralelogram}), zato je točka $E$ središče daljice $OV$ in leži na Eulerjevi premici trikotnika $ABC$ (izrek \ref{EulerjevaPremica}). + \kdokaz + +V razdelku \ref{odd7SredRazteg} (izrek \ref{EulerKroznicaHomot}) bomo videli še nadaljevanje prejšnje trditve, ki se nanaša na Eulerjevo krožnico. + + +%________________________________________________________________________________ + \poglavje{Thales' Theorem - Basic Proportionality Theorem} \label{odd5TalesVekt} + + Že v razdelku \ref{odd5LinKombVekt} smo ugotovili, da lahko pri dveh kolinearnih vektorjih govorimo o njunem razmerju ali količniku. + Za kolinearne vektorje $\overrightarrow{v}$ in $\overrightarrow{u}$ ($\overrightarrow{u}\neq \overrightarrow{0}$) smo definirali njuno razmerje oz. količnik + $$\overrightarrow{v}:\overrightarrow{u} +=\frac{\overrightarrow{v}}{\overrightarrow{u}}=\lambda,$$ + če za nek $\lambda\in\mathbb{R}$ velja $\overrightarrow{v}=\lambda \overrightarrow{u}$. + +Na podoben način kot pri številih lahko definiramo tudi sorazmerje dveh parov kolinearnih vektorjev. Če sta $\overrightarrow{a}$ in $\overrightarrow{b}$ ($\overrightarrow{b}\neq \overrightarrow{0}$) par kolinearnih vektorjev oz. $\overrightarrow{c}$ in $\overrightarrow{d}$ ($\overrightarrow{d}\neq \overrightarrow{0}$) par kolinearnih vektorjev, pravimo, da sta para vektorjev \index{sorazmerje kolinearnih vektorjev}\pojem{sorazmerna}, če velja: + + $$\frac{\overrightarrow{a}}{\overrightarrow{b}} + =\frac{\overrightarrow{c}}{\overrightarrow{d}}.$$ + + Naslednji zelo pomemben izrek se nanaša na definirani pojem sorazmerja. + + \bizrek + \label{TalesovIzrek}(Thales'\footnote{Starogrški filozof in matematik \textit{Tales} + \index{Tales} iz Mileta (640--546 pr. n. š.) je obravnaval sorazmerje ustreznih daljic, + ki jih dobimo, če dve premici presekamo z dvema vzporednicama, pri tem pa ni omenjal + vektorske oblike.} theorem - Basic Proportionality Theorem)\\ + Let $a$, $b$, $p$ and $p'$ be lines in the same plane, and $O=a\cap b$, $A=a\cap p$, $A'=a\cap p'$, $B=b\cap p$ and $B'=b\cap p'$.\\ + If $p\parallel p'$, then + $$\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= + \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}= + \frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}.$$ + \index{izrek!Talesov o sorazmerju} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.1.pic} +\caption{} \label{sl.vek.5.8.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.1.pic}) + +Ker je po predpostavki $p\parallel p'$, sta vektorja $\overrightarrow{A'B'}$ in $\overrightarrow{AB}$ kolinearna. Po izreku \ref{vektKriterijKolin} je $\overrightarrow{A'B'}=\lambda\overrightarrow{AB}$ za nek $\lambda\in \mathbb{R}$. Na podoben način iz kolinearnosti vektorjev $\overrightarrow{OA'}$ in $\overrightarrow{OA}$ oz. $\overrightarrow{OB'}$ in $\overrightarrow{OB}$ sledi + $\overrightarrow{OA'}=\alpha\overrightarrow{OA}$ za nek $\alpha\in \mathbb{R}$ oz. + $\overrightarrow{OB'}=\beta\overrightarrow{OB}$ za nek $\beta\in \mathbb{R}$. + Iz tega dobimo: + $$\frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}=\lambda,\hspace*{2mm} + \frac{\overrightarrow{OA'}}{\overrightarrow{OA}}=\alpha,\hspace*{2mm} + \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}=\beta.$$ + Dovolj je še dokazati, da velja $\alpha=\beta=\lambda$. + Če uporabimo pravilo za odštevanje vektorjev \ref{vektOdsev} in izrek \ref{vektVektorskiProstor} (točka $\textit{6}$), dobimo: + \begin{eqnarray*} + \overrightarrow{A'B'}&=&\overrightarrow{OB'}-\overrightarrow{OA'} + =\beta\overrightarrow{OB}-\alpha \overrightarrow{OA};\\ + \overrightarrow{A'B'}&=&\lambda\overrightarrow{AB}= + \lambda\left(\overrightarrow{OB}-\overrightarrow{OA}\right) + =\lambda\overrightarrow{OB}-\lambda\overrightarrow{OA}. + \end{eqnarray*} + Ker sta $\overrightarrow{OA}$ in $\overrightarrow{OB}$ nekolinearna vektorja je po izreku \ref{vektLinKomb1Razcep} + $\alpha=\beta=\lambda$ oz. $\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= + \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}= + \frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}$. +\kdokaz + + + Direktna posledica je Talesov izrek v obliki sorazmerja daljic, ki ni v vektorski obliki (Figure \ref{sl.vek.5.8.2.pic}). + + + \bizrek \label{TalesovIzrekDolzine} + Let $a$, $b$, $p$ and $p'$ be lines in the same plane, and + $O=a\cap b$, $A=a\cap p$, $A'=a\cap p'$, $B=b\cap p$ and $B'=b\cap p'$.\\ + If $p\parallel p'$, then + $$\frac{OA'}{OA}= + \frac{OB'}{OB}= + \frac{A'B'}{AB}$$ + and also + $$\frac{OA'}{OB'}= + \frac{OA}{OB}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.2.pic} +\caption{} \label{sl.vek.5.8.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} Trditev je direktna posledica izrekov \ref{TalesovIzrek} in \ref{vektKolicnDolz}. + \kdokaz + + Dokazali bomo, da velja tudi obratno, oz. da iz ustreznega sorazmerja sledi vzporednost pripadajočih premic. + + + \bizrek \label{TalesovIzrekObr}(Converse Thales' proportionality theorem)\\ + Let $a$, $b$, $p$ and $p'$ be lines in the same plane, and $O=a\cap b$, $A=a\cap p$, $A'=a\cap p'$, $B=b\cap p$ and $B'=b\cap p'$.\\ If + $$\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= + \frac{\overrightarrow{OB'}}{\overrightarrow{OB}},$$ + then $p\parallel p'$ and also + $$\frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}= + \frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= + \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}.$$ + \index{izrek!Talesov obratni o sorazmerju} + \eizrek + + + \textbf{\textit{Proof.}} Označimo $\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= + \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}=\lambda$. V tem primeru je najprej $\overrightarrow{OA'}=\lambda\overrightarrow{OA}$ in $\overrightarrow{OB'}=\lambda\overrightarrow{OB}$. Zato je (izreka \ref{vektOdsev} in \ref{vektVektorskiProstor}): + $$\overrightarrow{A'B'}=\overrightarrow{OB'}-\overrightarrow{OA'} + =\lambda\overrightarrow{OB}-\lambda\overrightarrow{OA} + =\lambda\left(\overrightarrow{OB}-\overrightarrow{OA}\right) + =\lambda\overrightarrow{AB}.$$ + Ker je torej $\overrightarrow{A'B'}=\lambda\overrightarrow{AB}$, sta po izreku \ref{vektKriterijKolin} vektorja $\overrightarrow{A'B'}$ in $\overrightarrow{AB}$ kolinearna. To pomeni, da velja $AB\parallel A'B'$ oz. $p\parallel p'$. + + Na koncu iz izreka \ref{TalesovIzrek} sledi relacija + $\frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}= + \frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= + \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}$. + \kdokaz + +Navedimo še nekaj posledic Talesovega izreka \ref{TalesovIzrek}. + + \bizrek \label{TalesPosl1} + If parallel lines $p_1$, $p_2$, $p_3$ intersect a line $a$ at points $A_1$, $A_2$, + $A_3$ and a line $b$ at points $B_1$, $B_2$, + $B_3$, then + $$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3} + \hspace*{1mm} + \textrm{ and } \hspace*{1mm} + \frac{A_1A_2}{A_2A_3}=\frac{B_1B_2}{B_2B_3}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.4.pic} +\caption{} \label{sl.vek.5.8.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.4.pic}) + + Brez škode za splošnost naj bo $\mathcal{B}(A_1,A_2,A_3)$ in $\mathcal{B}(B_1,B_2,B_3)$. + +Označimo s $c$ vzporednico premice $b$ skozi točko $A_1$ ter s $C_2$ in $C_3$ presečišči premice $c$ s premicama $p_2$ in $p_3$. Štirikotnika $B_1B_3C_3A_1$ in $B_2B_3C_3C_2$ sta paralelograma, zato je $\overrightarrow{B_1B_3}=\overrightarrow{A_1C_3}$ in $\overrightarrow{B_2B_3}=\overrightarrow{C_2C_3}$. + +Če uporabimo še izrek \ref{TalesovIzrekDolzine}, dobimo: + \begin{eqnarray*} + \frac{|A_1A_2|}{|A_2A_3|}&=& + \frac{|A_1A_3|-|A_2A_3|}{|A_2A_3|}= +\frac{|A_1A_3|}{|A_2A_3|}-1=\\ + &=&\frac{|A_1C_3|}{|C_2C_3|}-1= + \frac{|B_1B_3|}{|B_2B_3|}-1=\\ + &=&\frac{|B_1B_3|-|B_2B_3|}{|B_2A_3|}= +\frac{|B_1B_2|}{|B_2B_3|}, + \end{eqnarray*} +oz. $\frac{A_1A_2}{A_2A_3}=\frac{B_1B_2}{B_2B_3}$ in +$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3}$. + \kdokaz + + + \bizrek \label{TalesPosl2} + If parallel lines $p_1$, $p_2$,..., $p_n$ intersect a line $a$ at points $A_1$, $A_2$,..., + $A_n$ and a line $b$ at points $B_1$, $B_2$,..., + $B_n$, then + $$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3}=\cdots= + \frac{A_{n-1}A_n}{B_{n-1}B_n}\hspace*{1mm} + \textrm{ and } \hspace*{1mm}$$ + $$A_1A_2:A_2A_3:\cdots :A_{n-1}A_n= B_1B_2:B_2B_3:\cdots :B_{n-1}B_n.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.5.pic} +\caption{} \label{sl.vek.5.8.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.5.pic}) + +Trditev je direktna posledica izreka \ref{TalesPosl1}. + \kdokaz + + + \bizrek \label{TalesPosl3} + Let $p_1$, $p_2$ and $p_3$ be lines that intersect at a point $O$. If $a$ and $b$ are + parallel lines that intersect the line $p_1$ at points $A_1$ and $B_1$, + the line $p_2$ at points $A_2$ and $B_2$, and + the line $p_3$ at points $A_3$ and $B_3$, then + $$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3} + \hspace*{1mm} + \textrm{ and } \hspace*{1mm} + \frac{A_1A_2}{A_2A_3}=\frac{B_1B_2}{B_2B_3}.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.6.pic} +\caption{} \label{sl.vek.5.8.6.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.6.pic}) + + Po izreku \ref{TalesovIzrekDolzine} dobimo: + $$\frac{A_1A_2}{B_1B_2}= + \frac{OA_2}{OB_2}=\frac{A_2A_3}{B_2B_3},$$ kar je bilo treba dokazati. \kdokaz + + + + \bizrek \label{TalesPosl4} + Let $p_1$, $p_2$,..., $p_n$ be lines that intersect at a point $O$. If $a$ and $b$ are + parallel lines that intersect the line $p_1$ at points $A_1$ and $B_1$, + the line $p_2$ at points $A_2$ and $B_2$,..., + the line $p_n$ at points $A_n$ and $B_n$, then + $$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3}=\cdots= + \frac{A_{n-1}A_n}{B_{n-1}B_n}\hspace*{1mm} + \textrm{ and } \hspace*{1mm}$$ + $$A_1A_2:A_2A_3:\cdots :A_{n-1}A_n= B_1B_2:B_2B_3:\cdots :B_{n-1}B_n.$$ + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.8.pic} +\caption{} \label{sl.vek.5.8.8.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.8.pic}) + +Trditev je direktna posledica izreka \ref{TalesPosl3}. + \kdokaz + + Zelo znani in koristni sta naslednji načrtovalni nalogi. + + + + \bzgled + \label{izrekEnaDelitevDaljiceNan} + \index{delitev daljice!na enake dele} + Construct points that divide a line segment $AB$ + into $n$ congruent line segments. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.9.pic} +\caption{} \label{sl.vek.5.8.9.pic} +\end{figure} + + +\textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.8.9.pic}) + +Naj bo $X$ poljubna točka, ki ne leži na premici $AB$ in $Q_1$, $Q_2$, ..., $Q_n$ takšne točke na poltraku $AX$, da velja + $\overrightarrow{AQ_1}=\overrightarrow{Q_1Q_2}=\cdots =\overrightarrow{Q_{n-1}Q_n}$. + S $P_1$, $P_2$, ..., $P_{n-1}$ označimo presečišče premice $AB$ z vzporednicama premice $BQ_n$ skozi točke $Q_1$, $Q_2$, ..., $Q_{n-1}$. + + Dokažimo, da so $P_1$, $P_2$, ..., $P_{n-1}$ iskane točke. Po izreku \ref{TalesPosl2} velja: + $$\frac{AP_1}{AQ_1}=\frac{P_1P_2}{Q_1Q_2}=\dots=\frac{P_{n-1}B}{Q_{n-1}Q_n}.$$ + Ker je po predpostavki $|\overrightarrow{AQ_1}|=|\overrightarrow{Q_1Q_2}|=\cdots =|\overrightarrow{Q_{n-1}Q_n}|$, je tudi +$|AP_1|=|P_1P_2|=\cdots =|P_{n-1}B|$. + \kdokaz + + + \bzgled + \label{izrekEnaDelitevDaljice} + \index{delitev daljice!v razmerju} + Divide a given line segment $AB$ in the ratio $n:m$ ($n,m\in \mathbb{N}$), + i.e. determine such point $T$ on the line $AB$ that + $$\frac{\overrightarrow{AT}}{\overrightarrow{TB}}=\frac{n}{m}.$$ + Prove that there is the only one solution for such point $T$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.10.pic} +\caption{} \label{sl.vek.5.8.10.pic} +\end{figure} + + +\textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.8.10.pic}) + + + Po izreku \ref{izrekEnaDelitevDaljiceVekt} obstaja ena sama točka $T$, za katero velja $\frac{\overrightarrow{AT}}{\overrightarrow{TB}}=\frac{n}{m}$. + + Sedaj bomo opisali postopek konstrukcije točke $T$. Naj bo $\overrightarrow{v}$ poljubni vektor, ki ni kolinearen z vektorjem $\overrightarrow{AB}$, ter $P$ in $Q$ takšni točki, da velja $\overrightarrow{AP}=n\overrightarrow{v}$ in $\overrightarrow{BQ}=-m\overrightarrow{v}$. S $T$ označimo presečišče premic $AB$ in $PQ$ (presečišče obstaja, ker je $P,Q\div AB$). Potem velja: + \begin{eqnarray*} + \frac{\overrightarrow{AT}}{\overrightarrow{TB}}= + -\frac{\overrightarrow{TA}}{\overrightarrow{TB}}= + -\frac{\overrightarrow{AP}}{\overrightarrow{BQ}}= + -\frac{n\overrightarrow{v}}{-m\overrightarrow{v}}=\frac{n}{m}, + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + V razdelku \ref{odd7Harm} bomo naprej raziskovali vprašanje delitve daljice v danem razmerju. + + + + \bzgled \label{vektTrapezZgled} + A line parallel to bases of a trapezium intersects its legs and diagonals in four points + and determine three line segments. Prove that two of them are congruent. \\ + After that, construct a line parallel to the bases of that trapezium such that all + three mentioned line segments are congruent. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.11.pic} +\caption{} \label{sl.vek.5.8.11.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Naj bo $l$ vzporednica nosilke osnovnice +$AB$ trapeza $ABCD$, ki seka +stranici $AD$ in $BC$ ter diagonali $AC$ in $BD$ +po vrsti v točkah $M$, $Q$, $N$ in $P$ (Figure \ref{sl.vek.5.8.11.pic}). Po Talesovem izreku \ref{TalesovIzrekDolzine} velja: +$MN:DC=MA:DA$ oz. +$PQ:DC=BQ:BC$. Ker je po +posledici \ref{TalesPosl1} Talesovega izreka $MA:AD=BQ:BC$, je tudi +$MN:DC=PQ:DC$ oz. $MN\cong PQ$. + +Če je $E$ središče osnovnice $AB$ in $N_0$ presečišče premic $DE$ in $AC$, iskana +premica $l_0$ poteka skozi točko $N_0$ in je vzporedna s premico $AB$. Če so $M_0$, $P_0$ in $Q_0$ presečišča premice $l_0$ s premicami $AD$, +$BD$ in $CB$, potem po posledici \ref{TalesPosl3} Talesovega +izreka velja $M_0N_0\cong N_0P_0$. +\kdokaz + + + \bzgled If $r$ is the inradius and $r_a$, $r_b$ and $r_c$ exradii + of an arbitrary triangle, then $$\frac{1}{r_a} + +\frac{1}{r_b} +\frac{1}{r_c}= \frac{1}{r}.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.12.pic} +\caption{} \label{sl.vek.5.8.12.pic} +\end{figure} + +\textbf{\textit{Proof.}} Če uporabimo oznake iz velike naloge +\ref{velikaNaloga} (Figure \ref{sl.vek.5.8.12.pic}), iz Talesovega izreka \ref{TalesovIzrek} sledi: +$$ \frac{r}{r_a} = \frac{SQ}{S_aQ_a} = \frac{AQ}{AQ_a} = \frac{s-a}{s}.$$ + Analogno je tudi: + $$ \frac{r}{r_b} = \frac{s-b}{s}\hspace*{2mm} \textrm{ in }\hspace*{2mm} + \frac{r}{r_c} = \frac{s-c}{s}.$$ + Po seštevanju treh enakosti dobimo iskano relacijo. + \kdokaz + + + + \bizrek \label{velNalTockP'} + Suppose that the incircle and the excircle of a triangle $ABC$ touch its side $BC$ in points $P$ and $P_a$. + If $PP'$ is a diameter of the incircle $k(S,r)$ ($P'\in k$), then $Pa$, $P'$ and $A$ are collinear + points. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.vek.5.8.13.pic} +\caption{} \label{sl.vek.5.8.13.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Uporabimo oznake iz velike naloge +\ref{velikaNaloga} (Figure \ref{sl.vek.5.8.13.pic}). Ker je daljica $PP'$ + premer včrtane krožnice, je točka $S$ njeno središče. + + Naj bo $\widehat{P'}$ presečišče poltraka $PS$ s premico $AP_a$. + Dokažimo, da + je $\widehat{P'}=P'$, oziroma da +velja $S\widehat{P'}=r$. Iz Talesovega izreka sledi: + $$\frac{S\widehat{P'}}{S_aP_a} = \frac{AS}{AS_a} =\frac{SQ}{S_aQ_a}.$$ + Ker je $S_a P_a = S_aQ_a = r_a$, + je tudi $S\widehat{P'}=SQ=r$. + \kdokaz + +Na podoben način kot \ref{velNalTockP'} se dokažeta tudi +naslednja izreka. + + + + + \bizrek \label{velNalTockP'1} + Suppose that the incircle and the excircle of a triangle $ABC$ touch its side $BC$ in points $P$ and $P_a$. + If $P_aP_a'$ is a diameter of the excircle $k_a(S_a,r_a)$ ($P_a'\in k_a$), then $P_a'$, $P$ and $A$ are collinear + points + (Figure \ref{sl.vek.5.8.14.pic}). + \eizrek + + + + +\begin{figure}[htp] +\centering +\input{sl.vek.5.8.14.pic} +\caption{} \label{sl.vek.5.8.14.pic} +\end{figure} + +%\vspace*{10mm} + + + + \bizrek \label{velNalTockP'2} + Suppose that the excircles $k_b(S_b,r_b)$ and $k_c(S_c,r_c)$ of a triangle $ABC$ + touch the line $BC$ in points $P_b$ and $P_c$. + If $P_bP_b'$ and $P_cP_c'$ are diameters of the excircles + $k_b$ ($P_b'\in k_b$) and $k_c$ ($P_c'\in k_c$), then $P_c'$, $P_b$ and $A$ are collinear + points and also $P_b'$, $P_c$ and $A$ are collinear + points (Figure \ref{sl.vek.5.8.15.pic}). + \eizrek + + +\begin{figure}[htp] +\centering +\input{sl.vek.5.8.15.pic} +\caption{} \label{sl.vek.5.8.15.pic} +\end{figure} + + + + + + +Zadnje tri trditve lahko uporabimo pri konstrukcijah trikotnikov, +kar bomo ilustrirali v naslednji nalogi. + + \bzgled + Construct a triangle $ABC$, with given $r$, $b-c$, $t_a$. + \ezgled + + +\begin{figure}[htp] +\centering +\input{sl.vek.5.8.16.pic} +\caption{} \label{sl.vek.5.8.16.pic} +\end{figure} + + \textbf{\textit{Solution.}} + Naj bo $ABC$ iskani trikotnik oziroma trikotnik, pri katerem je +polmer včrtane krožnice $r$, ustrezna težiščnica $AA_1$ skladna s +$t_a$ in razlika stranic $AC$ in $AB$ enaka $b-c$. Uporabimo +oznake iz velike naloge \ref{velikaNaloga} in izreka +\ref{velNalTockP'} (Figure \ref{sl.vek.5.8.16.pic}). Vemo, da velja $PP_a=b-c$, točka $A_1$ pa je +skupno središče stranice $BC$ in daljice $PP_a$. Iz izreka +\ref{velNalTockP'} sledi, da so točke $P_a$, $P'$ in $A$ +kolinearne. + +Torej najprej konstruiramo pravokotni trikotnik $P'PP_a$, nato pa + krožnico $k$ s premerom $PP'$ oz. včrtano krožnico + trikotnika $ABC$ in točko $A_1$ kot +središče daljice $PP_a$. Oglišče $A$ je presečišče poltraka +$P_aP'$ s krožnico $k_1(A_1,t_a)$, oglišči $B$ in $C$ pa sta +presečišči premice $BC$ s tangentama krožnice $k$ iz točke +$A$. + \kdokaz + + \bzgled + Let $A_1$ be the midpoint of the line segment $BC$, $k(S,r)$ + the incircle and $AA'$ the altitude of a triangle $ABC$. + Suppose that $L$ is the intersection of the lines $A_1S$ and $AA'$. + Prove that $AL\cong + r$. + \ezgled + + +\begin{figure}[htp] +\centering +\input{sl.vek.5.8.3.pic} +\caption{} \label{sl.vek.5.8.3.pic} +\end{figure} + + +\textbf{\textit{Solution.}} Označimo s $P$, $Q$ in $R$ +dotikališča krožnice $k$ s + stranicami $BC$, $AC$ in $AB$. + Naj bo $k(S_a,r_a)$ pričrtana krožnica, ki se dotika stranice + $BC$ v točki $P_a$, nosilk $AB$ in $AC$ pa v točkah $R_a$ in + $Q_a$. Označimo še s $P'$ presečišče premic $AP_a$ in $PS$ + (Figure \ref{sl.vek.5.8.3.pic}). + Iz trditve \ref{velNalTockP'} in velike naloge \ref{velikaNaloga} + sledi: +\begin{itemize} + \item točka $P'$ leži na krožnici $k$, + \item točka $A_1$ je središče daljice $PP_a$. +\end{itemize} +Dokažimo, da velja $AL\cong + r$. Daljica $SA_1$ je srednjica trikotnika +$PP'P_a$, zato je $SA_1\parallel P'P_a$ oz. $LS\parallel AP'$. +Ker je še $AL\parallel P'S$, je štirikotnik $AP'SL$ paralelogram +in velja $AL\cong P'S\cong r$. + \kdokaz + + + +%________________________________________________________________________________ +\naloge{Exercises} + +\begin{enumerate} + + %Vsota in razlika vektorjev + %_____________________________________ + + \item Načrtaj poljubne vektorje $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ in $\overrightarrow{d}$ tako, da je njihova vsota enaka: + + (\textit{a}) enem od teh štirih vektorjev,\\ + (\textit{b}) razliki dveh od teh štirih vektorjev. + + \item Naj bo $ABCDE$ petkotnik v neki ravnini. Dokaži, da v tej ravnini obstaja +petkotnik s stranicami, ki določajo enake vektorje, kot jih določajo + diagonale petkotnika $ABCDE$. + + \item Naj bodo $A$, $B$, $C$ in $D$ poljubne točke v ravnini. Ali splošno velja: + + (\textit{a}) $\overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AD}+\overrightarrow{BC}$?\\ + (\textit{b}) $\overrightarrow{AB}=\overrightarrow{DC}\hspace*{1mm}\Rightarrow \hspace*{1mm} \overrightarrow{AC}+\overrightarrow{BD}=2\overrightarrow{BC}$? + + \item Dana je daljica $AB$. Samo z ravnilom z možnostjo risanja vzporednic (konstrukcije v afini geometriji) načrtaj točko $C$, tako da bo: + + (\textit{a}) $\overrightarrow{AC}=-\overrightarrow{AB}$, \hspace*{3mm} + (\textit{b}) $\overrightarrow{AC}=5\overrightarrow{AB}$, \hspace*{3mm} + (\textit{c}) $\overrightarrow{AC}=-3\overrightarrow{AB}$. + + \item Naj bo $ABCD$ štirikotnik in $O$ poljubna točka v ravnini tega štirikotnika. Izrazi vektorje stranic in +diagonal tega štirikotnika z vektorji $\overrightarrow{a}=\overrightarrow{OA}$, $\overrightarrow{b}=\overrightarrow{OB}$, $\overrightarrow{c}=\overrightarrow{OC}$ in $\overrightarrow{d}=\overrightarrow{OD}$. + + \item Naj bo $ABCD$ štirikotnik in $O$ poljubna točka v ravnini tega štirikotnika. Ali velja ekvivalenca, da je štirikotnik $ABCD$ paralelogram natanko tedaj, ko je $\overrightarrow{OA}+\overrightarrow{OC}= + \overrightarrow{OB}+\overrightarrow{OD}$? + + \item Naj bo $ABCD$ paralelogram, $S$ +presečišče njegovih diagonal in $M$ poljubna točka v ravnini tega paralelograma. Dokaži, da velja: + $$\overrightarrow{MS} = \frac{1}{4}\cdot + \left( \overrightarrow{MA}+\overrightarrow{MB} + +\overrightarrow{MC} + \overrightarrow{MD} \right).$$ + + \item Naj bodo $ABB_1A_2$, +$BCC_1B_2$ in $CAA_1C_2$ paralelogrami, ki so načrtani nad stranicami trikotnika $ABC$. Dokaži, da velja: +$$\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}+ +\overrightarrow{C_1C_2}=\overrightarrow{0}.$$ + + \item Pravokotni premici $p$ in $q$, ki se sekata v točki $M$, sekata krožnico $k$ s središčem $O$ v +točkah $A$, $B$, $C$ in $D$. Dokaži, da velja: +$$\overrightarrow{OA}+ \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = 2\overrightarrow{OM}.$$ + +\item Naj bodo $A$, $B$, $C$ in $D$ poljubne točke v neki ravnini. Ali lahko vseh šest daljic, ki jih določajo te točke, orientiramo tako, da je vsota ustreznih šestih vektorjev enaka vektorju nič? + +\item Naj bodo $A_1$, $B_1$ in $C_1$ središča stranic $BC$, $AC$ in $AB$ trikotnika $ABC$ ter $M$ poljubna točka. Dokaži: + + (\textit{a}) $\overrightarrow{AA_1}+\overrightarrow{BB_1}+ +\overrightarrow{CC_1}=\overrightarrow{0}$,\\ + (\textit{b}) $\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}= + \overrightarrow{MA_1}+\overrightarrow{MB_1}+\overrightarrow{MC_1}$,\\ + (\textit{c}) Obstaja tak trikotnik $PQR$, da za njegova oglišča velja:\\ \hspace*{7mm} $\overrightarrow{PQ}=\overrightarrow{CC_1}$, $\overrightarrow{PR}=\overrightarrow{BB_1}$ in $\overrightarrow{RQ}=\overrightarrow{AA_1}$. + + \item Naj bodo $M$, $N$, $P$, $Q$, $R$ in $S$ po vrsti središča stranic poljubnega šestkotnika. +Dokaži, da velja: +$$\overrightarrow{MN}+\overrightarrow{PQ}+ +\overrightarrow{RS}=\overrightarrow{0}.$$ + + + %Linearna kombinacija vektorjev + %_____________________________________ + + \item Naj bo $ABCDEF$ konveksni šestkotnik, pri katerem je $AB\parallel DE$, točki $K$ in $L$ pa sta središči daljic, ki jih +določajo središča preostalih parov nasprotnih stranic. Dokaži, da je $K=L$ natanko tedaj, ko je $AB\cong DE$. + + \item Naj bosta $P$ in $Q$ takšni točki stranic $BC$ in $CD$ paralelograma $ABCD$, da je $BP:PC=2:3$ in +$CQ:QD=2:5$. Točka $X$ je presečišče daljic $AP$ in $BQ$. Izračunaj razmerji, v katerih točka $X$ deli +daljici $AP$ in $BQ$. + +\item Naj bodo $A$, $B$, $C$ in $D$ poljubne točke neke ravnine. Točka $E$ je središče daljice $AB$, $F$ in +$G$ takšni točki, da velja $\overrightarrow{EF} = \overrightarrow{BC}$ in $\overrightarrow{EG} = \overrightarrow{AD}$, ter $S$ središče daljice $CD$. Dokaži, da so $G$, $S$ in $F$ +kolinearne točke. + +\item Naj bosta $K$ in $L$ takšni točki stranice $AD$ in diagonale $AC$ paralelograma $ABCD$, da velja $\frac{\overrightarrow{AK}}{\overrightarrow{KD}}=\frac{1}{3}$ in + $\frac{\overrightarrow{AL}}{\overrightarrow{LC}}=\frac{1}{4}$. Dokaži, da so $K$, $L$ in $B$ kolinearne točke. + +\item Naj bosta $X_n$ in $Y_n$ ($n\in \mathbb{N}$) takšni točki stranic $AB$ in $AC$ trikotnika $ABC$, da velja $\overrightarrow{AX_n}=\frac{1}{n+1}\cdot \overrightarrow{AB}$ in + $\overrightarrow{AY_n}=\frac{1}{n}\cdot \overrightarrow{AC}$. Dokaži, da obstaja točka, ki leži na vseh premicah +$X_nY_n$ ($n\in \mathbb{N}$). + + \item Naj bodo $M$, $N$, $P$ in $Q$ središča stranic $AB$, $BC$, $CD$ in $DA$ štirikotnika + $ABCD$. Ali velja ekvivalenca, da je štirikotnik $ABCD$ paralelogram natanko tedaj, ko je: + + (\textit{a}) $2\overrightarrow{MP}=\overrightarrow{BC}+\overrightarrow{AD}$ in + $2\overrightarrow{NQ}=\overrightarrow{BA}+\overrightarrow{CD}$?\\ + (\textit{b}) $2\overrightarrow{MP}+2\overrightarrow{NQ}= + \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}$? + + \item Naj bodo $E$, $F$ in $G$ središča stranic $AB$, $BC$ in $CD$ paralelograma $ABCD $, premici $BG$ in $DE$ pa sekata premico $AF$ v točkah $N$ in $M$. Izrazi $\overrightarrow{AF}$, $\overrightarrow{AM}$ in $\overrightarrow{AN}$ kot linearno kombinacijo vektorjev $\overrightarrow{AB}$ in $\overrightarrow{AD}$. Dokaži, da točki $M$ in $N$ delita daljico $AF$ v razmerju $2:2:1$. + + \item Točke $K$, $L$, $M$ in $N$ ležijo na stranicah $AB$, $BC$, $CD$ in $DA$ +štirikotnika $ABCD$. Če je štirikotnik $KLMN$ paralelogram in velja +$$\frac{\overrightarrow{AK}}{\overrightarrow{KB}}= +\frac{\overrightarrow{BL}}{\overrightarrow{LC}} +=\frac{\overrightarrow{CM}}{\overrightarrow{MD}}= +\frac{\overrightarrow{DN}}{\overrightarrow{NA}}=\lambda$$ za nek $\lambda\neq\pm 1$, + je tudi štirikotnik $ABCD$ +paralelogram. Dokaži. + +\item Naj bo $M$ središče stranice $DE$ pravilnega šestkotnika $ABCDEF$. Točka +$N$ je središče daljice $AM$, točka $P$ pa središče stranice $BC$. Izrazi $\overrightarrow{NP}$ kot linearno kombinacijo vektorjev +$\overrightarrow{AB}$ in $\overrightarrow{AF}$. + + + %Dolžina vektorja + %_____________________________________ + + \item Dokaži, da za poljubne točke $A$, $B$ in $C$ velja: + + (\textit{a}) $|\overrightarrow{AC}|\leq|\overrightarrow{AB}|+|\overrightarrow{BC}|$ \hspace*{6mm} + (\textit{b}) $|\overrightarrow{AC}|\geq|\overrightarrow{AB}|-|\overrightarrow{BC}|$\\ + Pod katerimi pogoji velja enakost? + + \item Naj bosta $M$ in $N$ takšni točki, ki ležita na daljicah $AD$ oz. $BC$, tako da velja $\frac{\overrightarrow{AM}}{\overrightarrow{MD}}\cdot \frac{\overrightarrow{CN}}{\overrightarrow{NB}}=1$. Dokaži, da je: + $$|MN|\leq\max\{|AB|, |CD|\}.$$ + + %Težišče + %_____________________________________ + + \item Točki $T$ in $T'$ sta težišči $n$-kotnikov $A_1A_2...A_n$ in $A'_1A'_2...A'_n$. Izračunaj: +$$\overrightarrow{A_1A'_1}+\overrightarrow{A_2A'_2}+\cdots+\overrightarrow{A_nA'_n}.$$ + + \item Dokaži, da imata štirikotnika $ABCD$ in $A'B'C'D'$ skupno težišče natanko tedaj, ko je: +$$\overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}+ +\overrightarrow{DD'}=\overrightarrow{0}.$$ + + \item Naj bodo $P$, $Q$, $R$ in $S$ težišča trikotnikov $ABD$, $BCA$, $CDB$ in $DAC$. Dokaži, da imata +štirikotnika $PQRS$ in $ABCD$ skupno težišče. + + \item Naj bo $A_1A_2A_3A_4A_5A_6$ poljubni šestkotnik in $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ in $B_6$ po vrsti težišča trikotnikov $A_1A_2A_3$, $A_2A_3A_4$, $A_3A_4A_5$, $A_4A_5A_6$, $A_5A_6A_1$ in $A_6A_1A_2$. +Dokaži, da ta težišča določajo šestkotnik s tremi pari vzporednih stranic. + + \item Naj bodo $A$, $B$, $C$ in $D$ štiri različne točke. Točke $T_A$, $T_B$, $T_C$ in $T_D$ + so težišča trikotnikov $BCD$, $ACD$, $ABD$ in $ABC$. Dokaži, da se daljice $AT_A$, $BT_B$, $CT_C$ in $DT_D$ sekajo v eni točki $T$. V katerem razmerju točka $T$ deli te daljice? + + \item Naj bo $CC_1$ težiščnica trikotnika $ABC$ in $P$ poljubna točka na stranici +$AB$ tega trikotnika. Vzporednica $l$ premice $CC_1$ skozi točko $P$ seka premici $AC$ in $BC$ v točkah $M$ in $N$. Dokaži, da velja: +$$\overrightarrow{PM} + \overrightarrow{PN}= \overrightarrow{AC} + \overrightarrow{BC}.$$ + + + %Hamilton in Euler + %______________________________________________________ + + + \item Naj bodo $A$, $B$, $C$ točke neke ravnine, ki ležijo na isti strani premice $p$, in $O$ točka na +premici $p$, za katero velja $|\overrightarrow{OA}| = |\overrightarrow{OB}| = |\overrightarrow{OC}| =1$. Dokaži, da potem velja tudi: $$|\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}| \geq 1.$$ + +\item Izračunaj kote, ki jih določajo vektorji $\overrightarrow{OA}$, $\overrightarrow{OB}$ in $\overrightarrow{OC}$, če točke $A$, $B$ in $C$ ležijo na krožnici s središčem +$O$ in dodatno velja še: +$$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{0}.$$ + + \item Naj bodo $A$, $B$, $C$ in $D$ točke, ki ležijo na krožnici s središčem +$O$, in velja +$$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = \overrightarrow{0}.$$ +Dokaži, da je $ABCD$ pravokotnik. + +\item Naj bodo $\overrightarrow{a}$, $\overrightarrow{b}$ in $\overrightarrow{c}$ vektorji neke ravnine, za katere velja $|\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{c}| =x$. Razišči, v katerem primeru velja tudi $|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}| = x$. + + + %Tales + %______________________________________________________ + + + \item Dano daljico $AB$ razdeli: + + (\textit{a}) na pet enakih daljic,\\ + (\textit{b}) v razmerju $2:5$,\\ + (\textit{c})na tri daljice, ki so v razmerju $2:\frac{1}{2}:1$. + + + \item Dana je daljica $AB$. Samo z uporabo ravnila z možnostjo risanja vzporednic (konstrukcije v afini geometriji) načrtaj točko $C$, če je: + +(\textit{a}) $\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}$ \hspace*{3mm} + (\textit{b}) $\overrightarrow{AC}=\frac{3}{5}\overrightarrow{AB}$ \hspace*{3mm} + (\textit{c}) $\overrightarrow{AC}=-\frac{4}{7}\overrightarrow{AB}$ + + \item Dane so daljice $a$, $b$ in $c$. Načrtaj daljico $x$, tako da bo: + + (\textit{a}) $a:b=c:x$ \hspace*{3mm} + (\textit{b}) $x=\frac{a\cdot b}{c}$ \hspace*{3mm} + (\textit{c}) $x=\frac{a^2}{c}$\hspace*{3mm}\\ + (\textit{č}) $x=\frac{2ab}{3c}$\hspace*{3mm} + (\textit{č}) $(x+c):(x-c)=7:2$ + + \item Naj bosta $M$ in $N$ točki na kraku $OX$, $P$ točka na kraku $OY$ kota +$XOY$ ter $NQ\parallel MP$ in $PN\parallel QS$ ($Q\in OY$, $S\in OX$). Dokaži, da je +$|ON|^2=|OM|\cdot |OS|$ (za daljico $ON$ v tem primeru pravimo, da je \index{geometrijska sredina daljic}\pojem{geometrijska sredina} daljic $OM$ +in $OS$). + + +\item Naj bo $ABC$ trikotnik in $Q$, $K$, $L$, $M$, $N$ in $P$ takšne točke poltrakov $AB$, $AC$, $BC$, +$BA$, $CA$ in $CB$, da velja $AQ\cong CP\cong AC$, $AK\cong BL\cong AB$ in $BM\cong CN\cong BC$. +Dokaži, da so $MN$, $PQ$ in $LK$ tri vzporednice. + +\item Naj bo $P$ središče težiščnice $AA_1$ trikotnika $ABC$. Točka $Q$ je presečišče premice $BP$ +s stranico $AC$. Izračunaj razmerji $AQ:QC$ in $BP:PQ$. + +\item Točki $P$ in $Q$ ležita na stranicah $AB$ in $AC$ trikotnika $ABC$, pri tem pa velja $\frac{|\overrightarrow{PB}|}{|\overrightarrow{AP}|} + +\frac{|\overrightarrow{QC}|}{|\overrightarrow{AQ}|}=1$. Dokaži, da težišče tega trikotnika leži na daljici $PQ$. + + \item Naj bodo $a$, $b$ in $c$ trije poltraki s skupnim izhodiščem $S$ ter $M$ +točka na poltraku $a$. Če se točka $M$ “giblje” po poltraku $a$, + je razmerje razdalj te točke od premic $b$ in $c$ konstantno. Dokaži. + + \item Naj bo $D$ točka, ki leži na stranici $BC$ trikotnika $ABC$ ter + $F$ in $G$ točki, v katerih premica, ki poteka skozi točko $D$ in je vzporedna s težiščnico $AA_1$, seka premici $AB$ in $AC$. +Dokaži, da je vsota $|DF|+|DG|$ konstantna, če se točka $D$ “giblje” po stranici +$BC$. + + + \item + Načrtaj trikotnik s podatki: + + (\textit{a}) $v_a$, $r$, $b-c$ \hspace*{3mm} + (\textit{b}) $\beta$, $r$, $b-c$ + + +\end{enumerate} + + + + + + +% DEL 6 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% IZOMETRIJE +%________________________________________________________________________________ + + \del{Isometries} \label{pogIZO} + + +%________________________________________________________________________________ +\poglavje{Isometries. Identity Map} \label{odd6Ident} + +Izometrije oz. izometrijske transformacije ravnine +$\mathcal{I}:\mathbb{E}^2\rightarrow \mathbb{E}^2$ smo formalno že +definirali v razdelku \ref{odd2AKSSKL} kot transformacije ravnine, ki +ohranjajo relacijo skladnosti parov točk. Kasneje smo jih uporabili +za definiranje relacije skladnosti likov. Intuitivno jih zaznamujemo +kot gibanja ravnine. Nekatere so nam znane že od prej (v tej +knjigi jih še nismo formalno vpeljali) – rotacija in +translacija (Figure \ref{sl.izo.6.1.1.pic}). Tudi zrcaljenje čez +premico je izometrija. Toda ta se + razlikuje od omenjenih dveh izometrij, ker ni pravo gibanje (v +ravnini). Da bi jo videli kot gibanje, je potrebno +preiti v prostor, kjer lahko ravnino zavrtimo za $180^0$. S tem +spremenimo orientacijo ravnine. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.1.1.pic} +\caption{} \label{sl.izo.6.1.1.pic} +\end{figure} + +Za takšne izometrije, ki spremenijo orientacijo ravnine, pravimo, +da so \index{izometrija!indirektna} \pojem{indirektne}. +Zrcaljenje čez premico je torej indirektna izometrija. Za tiste +izometrije, ki ohranjajo orientacijo ravnine, pravimo, da so +\index{izometrija!direktna}\pojem{direktne}. Rotacija in +translacija sta primera direktnih izometrij. Na tem mestu ne bomo +dokazovali dejstva, da je vsaka izometrija ravnine bodisi direktna +bodisi indirektna, oz. da če neka izometrija ohranja orientacijo +enega lika, ohranja orientacijo tudi vseh drugih likov. + + Jasno je, da kompozitum dveh direknih ali dveh indirektnih + izometrij predstavlja direktno izometrijo. Prav tako je + kompozitum ene direktne in ene idirektne izometrije indirektna + izometrija. + +Intuitivno lahko pri direktnih izometrijah lik in njegovo sliko z +gibanjem v ravnini pripeljemo do prekrivanja. Za indirektne +izometrije to ni mogoče narediti s prostim gibanjem v ravnini - +potrebno je uporabiti gibanje v trirazsežnemu prostoru. + +Razen pogoja direktnosti in indirektnosti izometrij je njihova zelo +pomembna lastnost tudi število fiksnih točk. Ponovimo, da za točko +pravimo, da je fiksna točka pri neki izometriji, če se s to +izometrijo preslika sama vase (razdelek \ref{odd2AKSSKL}). Intuitivno ima +rotacija natanko eno fiksno točko – svoje središče. Translacija +nima fiksnih točk. Zrcaljenje čez premico jih ima neskončno mnogo, vendar +so vse fiksne točke na osi tega zrcaljenja. + Ali je možno, da ima izometrija tri fiksne nekolinearne točke? Intuitivno je + jasno dejstvo, ki smo ga tudi formalno dokazali (izrek \ref{IizrekABCident}), da + so za takšno izometrijo tudi vse ostale točke v ravnini fiksne. + Takšno izometrijo smo + imenovali \index{izometrija!identična} identična izometrija ali + identiteta in jo označili z $\mathcal{E}$. Očitno je tudi identiteta + direktna izometrija, saj ohranja vse like. + Zaradi pomembnosti + bomo že dokazani izrek \ref{IizrekABCident} še enkrat zapisali v nekoliko spremenjeni + obliki. + + + + + + \bizrek \label{IizrekABC2} + The identity map $\mathcal{E}$ is the only + isometry of a plane having three non-collinear fixed points. + \eizrek + + Z naslednjim izrekom bomo podali še en kriterij za identitete. + + + + \bizrek \label{izo2ftIdent} + A direct isometry of a plane that has at least + two fixed points is the identity map. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.1.2.pic} +\caption{} \label{sl.izo.6.1.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Predpostavimo, da ima direktna izometrija +ravnine $\mathcal{I}$ vsaj dve fiksni točki $A$ in $B$ oz. +$A'=\mathcal{I}(A)=A$ in $B'=\mathcal{I}(B)=B$ (Figure +\ref{sl.izo.6.1.2.pic}). Naj bo $X$ poljubna točka, ki ne leži na +premici $AB$. Naj bo $X'=\mathcal{I}(X)$. Ker je $\mathcal{I}$ +direktna izometrija, sta trikotnika $AXB$ in $AX'B$ enako +orientirana, zato točka $X'$ leži v polravnini z robom $AB$ in +s točko $X$. Iz $\mathcal{I}:A,B,X\mapsto A,B,X'$ sledi + $XA\cong X'A$, $XB\cong X'B$ in $AB\cong AB$. Ker je +tudi $XA\cong XA$ in $XB\cong XB$, je po izreku + \ref{izomEnaC'} $X=X'$. Torej je tudi $X$ fiksna točka. Ker so $A$, $B$ in $X$ tri + nekolinearne točke izometrije $\mathcal{I}$ je $\mathcal{I}=\mathcal{E}$ + (izrek \ref{IizrekABC2}). + \kdokaz + +V nadaljevanju bomo formalno vpeljali in obravnavali različne +vrste izometrij. + +%________________________________________________________________________________ + \poglavje{Reflections} \label{odd6OsnZrc} + +Čeprav nam je osno zrcaljenje intuitivno že znana preslikava, bomo +najprej podali formalno definicijo. + +Naj bo $s$ premica neke ravnine. Transformacija te ravnine, pri +kateri je vsaka točka premice $s$ fiksna in pri kateri se vsaka +točka $X$, ki ne leži na premici $s$, slika v takšno točko $X'$, + da je $s$ simetrala daljice $XX'$, imenujemo + \index{zrcaljenje!čez premico} \pojem{osno zrcaljenje} + ali \index{zrcaljenje!osno}\pojem{zrcaljenje čez premico} + $s$ in jo označimo s $\mathcal{S}_s$ (Figure \ref{sl.izo.6.2.1.pic}). + Premica $s$ pa je + \index{os!zrcaljenja} \pojem{os zrcaljenja}. + + Ker je iz oznake $\mathcal{S}_s$ + že jasno, da gre za zrcaljenje čez premico $s$, ga bomo imenovali + tudi krajše: zrcaljenje $\mathcal{S}_s$. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.1.pic} +\caption{} \label{sl.izo.6.2.1.pic} +\end{figure} + + Če se lik $\phi$ z zrcaljenjem $\mathcal{S}_s$ preslika v lik + $\phi'$ oz. $\mathcal{S}_s: \phi \rightarrow \phi'$, bomo rekli, + da sta lika $\phi$ in $\phi'$ + \pojem{simetrična} glede na os $s$ (Figure + \ref{sl.izo.6.2.2.pic}). Os $s$ pa je \pojem{os simetrije} likov $\phi$ in + $\phi'$. + + Če je še $\phi=\phi'$ oz. $\mathcal{S}_s: \phi \rightarrow + \phi$, pravimo, da je lik $\phi$ \index{lik!osnosimetričen} + \pojem{osno simetričen} oz. \pojem{osno someren}. Premica $s$ je + \index{os!simetrije lika}\pojem{os simetrije} ali \index{somernica} + \pojem{somernica} tega lika (Figure \ref{sl.izo.6.2.2.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.2.pic} +\caption{} \label{sl.izo.6.2.2.pic} +\end{figure} + + Iz definicije je že jasno, da vse fiksne točke zrcaljenja $\mathcal{S}_s$ + ležijo na osi $s$ tega zrcaljenja. + + Čeprav je intuitivno jasna, je potrebno dokazati naslednjo lastnost + definirane preslikave. + + + + \bizrek \label{izozrIndIzo} + A reflection is an opposite isometry. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.3.pic} +\caption{} \label{sl.izo.6.2.3.pic} +\end{figure} + + +\textbf{\textit{Proof.}} + Naj bo $\mathcal{S}_s$ zrcaljenje čez premico $s$ (Figure +\ref{sl.izo.6.2.3.pic}). Iz definicije je jasno, da predstavlja bijektivno preslikavo. Potrebno je še + dokazati, da za poljubni točki + $X$ in $Y$, ki ju $\mathcal{S}_s$ preslika v točki $X'$ in $Y'$, velja + $XY\cong X'Y'$. Obravnavali bomo več primerov. + + Če je $X=X'$ in $Y=Y'$, je ta relacija + avtomatično izpolnjena (izrek \ref{sklRelEkv}). + + Naj bo $X\neq + X'$. Označimo z $X_s$ središče daljice $XX'$. Ker je $s$ simetrala + daljice $XX'$, je po definiciji $X_s\in s$ in $XX'\perp s$. + Če je $Y=Y'$, sta trikotnika $XX_sY$ in $X'X_sY$ (oz. $X'X_sY'$) + skladna (izrek + \textit{SAS} \ref{SKS}), zato je tudi $XY\cong X'Y'$. + + Ostane nam še primer $X\neq X'$ in $Y\neq Y'$. Podobno za + središče + $Y_s$ daljice $YY'$ velja $Y_s\in s$ in $YY'\perp s$. + Trikotnika $XX_sY_s$ in $X'X_sY_s$ sta + skladna (izrek + \textit{SAS}, zato obstaja takšna izometrija $\mathcal{I}$, da + velja $\mathcal{I}: X, X_s,Y_s\mapsto X', X_s,Y_s$. Le-ta + preslika polravnino $X_sY_sX$ v polravnino $X_sY_sX'$ + (aksiom \ref{aksIII2}). Ker izometrija $\mathcal{I}$ pravi kot + preslika v pravi kot, se z njo poltrak $Y_sY$ preslika $Y_sY'$, + točka $Y$ pa v točko $Y'$. Iz + $\mathcal{I}: X, Y\mapsto X', Y'$ na koncu sledi $XY\cong X'Y'$. + + Naj bo $A,B\in s$ in $C\notin s$. Potem je $\mathcal{S}_s(A)=A$, $\mathcal{S}_s(B)=A$ in $\mathcal{S}_s(C)=C'\neq C$. To pomeni, da se trikotnik $ABC$ z osnim zrcaljenjem $\mathcal{S}_s$ preslika v trikotnik $ABC'$. Ker je $C,C' \div s$ oz. $C,C' \div AB$, sta omenjena trikotnika različno orientirana, zato je osno zrcaljenje $\mathcal{S}_s$ indirektna izometrija. + \kdokaz + +Jasno je, da v dokazu prejšnjega izreka velja tudi +$\mathcal{I}=\mathcal{S}_s$. To lahko ugotovimo šele na koncu, ko +dokažemo, da je $\mathcal{S}_s$ izometrija in uporabimo izrek +\ref{IizrekABC}. + +Dokažimo še nekaj enostavnih lastnosti zrcaljenja čez premico. + + + \bizrek \label{izoZrcPrInvol} + For an arbitrary line $p$ is $$\mathcal{S}_p^2=\mathcal{E}\hspace*{1mm}\textrm{ i.e. } + \hspace*{1mm}\mathcal{S}_p^{-1}=\mathcal{S}_p.$$ + \eizrek + + +\textbf{\textit{Proof.}} Dovolj je dokazati, da velja +$\mathcal{S}_p^2(X)=X$ za vsako točko $X$ ravnine. Naj bo +$\mathcal{S}_p(X)=X'$. Če je $X\in p$ oz. $X=X'$, je relacija +$\mathcal{S}_p^2(X)=X$ avtomatično izpolnjena. Če $X\notin p$, je +po definiciji premica $p$ simetrala daljice $XX'$ (tudi $X'X$), +zato je $\mathcal{S}_p(X')=X$ oz. $\mathcal{S}_p^2(X)=X$. + \kdokaz + + Vsako izometrijo (tudi vsako preslikavo) $f:\mathbb{E}^2\rightarrow \mathbb{E}^2$, + za katero velja $f^2=\mathcal{E}$, imenujemo + \pojem{involucija}\index{involucija}. + Osno zrcaljenje je torej involucija. + + + + \bizrek \label{zrcFiksKroz} + Let $l$ be an arbitrary circle + with the centre $S$ and $p$ a line in the plane. + If $S\in p$, then $\mathcal{S}_p(l)=l$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.4.pic} +\caption{} \label{sl.izo.6.2.4.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Naj bo $X\in l$ poljubna točka krožnice +$l$ in $\mathcal{S}_p(X)=X'$ (Figure \ref{sl.izo.6.2.4.pic}). Ker +$\mathcal{S}_p:S,X\mapsto S, X'$, je $SX\cong SX'$ oz. $X'\in l$. +Torej je $\mathcal{S}_p(l)\subseteq l$. Podobno je poljubna točka +$Y$ krožnice $l$ slika točke +$Y'=\mathcal{S}_p^{-1}(Y)=\mathcal{S}_p(Y)$, ki leži na tej +krožnici. Zato je tudi $\mathcal{S}_p(l)\supseteq l$. Torej velja +$\mathcal{S}_p(l)=l$. + \kdokaz + + \bzgled \label{izoSimVekt} + If + $\mathcal{S}_s:A, B\mapsto A', B'$, the vector + $\overrightarrow{v}=\overrightarrow{AB}+\overrightarrow{A'B'}$ + is parallel to the line $s$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.5.pic} +\caption{} \label{sl.izo.6.2.5.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Naj bosta $A_s$ in $B_s$ središči daljic +$AA'$ in $BB'$ (Figure \ref{sl.izo.6.2.5.pic}). Ker je $p$ +simetrala teh daljic, je $A_s, B_s\in p$. Zato je vektor + \begin{eqnarray*} + \overrightarrow{v}&=&\overrightarrow{AB}+\overrightarrow{A'B'}=\\ + &=&(\overrightarrow{AA_s}+\overrightarrow{A_sB_s}+\overrightarrow{B_sB})+ + (\overrightarrow{A'A_s}+\overrightarrow{A_sB_s}+\overrightarrow{B_sB'})=\\ + &=& 2\overrightarrow{A_sB_s} + \end{eqnarray*} +vzporeden s premico $s$. + \kdokaz + + Z lastnostmi osnega zrcaljenja bomo nadaljevali v naslednjem razdelku, + sedaj pa si oglejmo uporabo te izometrije. + + + + \bzgled \label{HeronProbl} + (Heron's\footnote{Ta problem je + zastavil \index{Heron}\textit{Heron iz Aleksandrije} (20--100). V + svojem delu ‘‘Catoprica’’ je formuliral zakon, ki pravi, da žarek, ki gre iz + točke $A$ in se odbije od premice $p$ skozi točko $B$, prehaja + najkrajšo možno pot.} problem) \index{problem!Heronov} + Two points $A$ and $B$ are given on the same side of a line $p$. + Find a point $X$ on the line $p$ such that the sum $|AX|+|XB|$ is minimal. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.6.pic} +\caption{} \label{sl.izo.6.2.6.pic} +\end{figure} + + +\textbf{\textit{Solution.}} + Naj bo $A'$ slika točke $A$ pri +zrcaljenju $\mathcal{S}_p$ (Figure \ref{sl.izo.6.2.6.pic}). Z $X$ +označimo presečišče premice $p$ in $A'B$ (točki $A'$ in $B$ sta na +različnih bregovih premice $p$). Dokažimo da je $X$ iskana točka. + +Naj bo $Y\neq X$ poljubna točka premice $p$. Ker je zrcaljenje +$\mathcal{S}_p$ izometrija, ki daljici $AX$ in $AY$ slika v +daljici $A'X$ in $A'Y$, velja $AX\cong A'X$ in $AY\cong A'Y$. Če +uporabimo še trikotniško neenakost - izrek \ref{neenaktrik} (za +trikotnik $A'YB$), dobimo: $$|AX| + |XB| = |A'X| + |XB| = |A'B| < +|A'Y| + |YB| = |AY| + |YB|,$$ kar je bilo treba dokazati. \kdokaz + + + + \bzgled + Let $k$ and $l$ be circles on the same side of a line $p$ in the same plane. + Construct a point $S$ on the line $p$ such that + the tangents from this point to the circles $k$ and $l$ + determine the congruent angles with the line $p$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.7.pic} +\caption{} \label{sl.izo.6.2.7.pic} +\end{figure} + + +\textbf{\textit{Solution.}} Naj bosta $q$ in $r$ po vrsti tangenti +krožnic $k$ in $l$, ki se sekata na premici $p$ v točki $S$ in z +njo določata skladna kota (Figure \ref{sl.izo.6.2.7.pic}). Premica +$p$ je simetrala kota, ki ga tangenti $q$ in $r$ določata, zato je +$\mathcal{S}_p(r)=q$. Premica $q$ je torej tangenta tudi krožnice +$l'=\mathcal{S}_p(l)$. To pomeni, da lahko premico $q$ načrtamo +kot skupno tangento krožnic $k$ in $l'$ (glej zgled +\ref{tang2ehkroz}). Nato pa je še $S=q\cap p$ in $r=\mathcal{S}_p(q)$. + \kdokaz + + + \bzgled \label{FagnanLema} + Let $AP$, $BQ$ and $CR$ be the altitudes of a triangle $ABC$. + If $P'=\mathcal{S}_{AB}(P)$ and + $P''=\mathcal{S}_{AC}(P)$, prove that $P'$, $R$, $Q$ and $P''$ + are collinear points. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.9a.pic} +\caption{} \label{sl.izo.6.2.9a.pic} +\end{figure} + + +\textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.2.9a.pic}) Najprej +iz $\angle BRC=90^0$ in $\angle BQC=90^0$ po Talesovem izreku +\ref{TalesovIzrKroz} sledi, da točki $R$ in $Q$ ležita na krožnici +s premerom $BC$. Torej je $BRQC$ tetivni štirikotnik in po +izreku \ref{TetivniPogojZunanji} velja $\angle ARQ\cong \angle BCQ = +\gamma$. Analogno je (iz tetivnosti štirikotnika $CARP$) tudi +$\angle BRP\cong\gamma$, zato je $\angle ARQ\cong \angle BRP$. Ker +velja $\mathcal{S}_{AB}:P, R, B\mapsto P', R, B$, je $\angle BRP\cong +\angle BRP'$. Iz prejšnjih relacij sledi $\angle ARQ\cong \angle +BRP'$. Trikotnik $ABC$ je ostrokotni, kar pomeni, da sta točki $P$ +in $Q$ na istem bregu premice $AB$. Iz tega sledi, da sta točki +$Q$ in $P'$ na različnih straneh premice $AB$. Iz dokazane +relacije $\angle ARQ\cong \angle BRP'$ sedaj sledi, da so $P'$, +$R$ in $Q$ kolinearne točke. Analogno so tudi točke $P''$, $R$ in +$Q$ kolinearne, kar pomeni, da vse štiri točke $P'$, $R$, $Q$ in +$P''$ ležijo na isti premici. + \kdokaz + + + + \bizrek \label{FagnanLema1} + Let $A$ be a point not lying on lines $p$ and $q$ in the same plane. + Construct points $B\in p$ and $C\in q$ such that + the perimeter of the triangle $ABC$ is minimal. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.8.pic} +\caption{} \label{sl.izo.6.2.8.pic} +\end{figure} + + +\textbf{\textit{Solution.}} + Naj bo $A'=\mathcal{S}_p(A)$, $A''=\mathcal{S}_q(A)$ ter $B$ in $C$ + presecišči premice $A'A''$ s premicama $p$ in $q$ + (Figure \ref{sl.izo.6.2.8.pic}). + +Dokažimo, da je $ABC$ iskani trikotnik. Če sta $B_1$ in $C_1$ +poljubni točki (kjer je $B_1\neq B$ ali $C_1\neq C$), ki ležita na +premicah $p$ in $q$, je obseg trikotnika $AB_1C_1$ enak dolžini +lomljenke $A'B_1C_1A''$ ($AB_1\cong A'B_1$ in $AC_1\cong +A'' C_1$), obseg trikotnika $ABC$ pa dolžini daljice $A'A''$ +($AB\cong A'B$ in $AC\cong A'' C$). Iz izreka \ref{neenakIzlLin} +sledi, da je prva dolžina večja od druge, zato je obseg trikotnika +$ABC$ manjši od obsega trikotnika $AB_1C_1$. + \kdokaz + + + \bzgled + For a given acute triangle $ABC$ determine the inscribed triangle of minimal perimeter + \index{problem!Fagnano}(Fagnano's problem\footnote{\textit{G. F. Fagnano} + \index{Fagnano, G. F.} (1717--1797), italijanski matematik, + je postavil ta problem leta 1775 in ga rešil + z metodami + diferencialnega računa. Bolj + elementarno rešitev tega problema je kasneje podal madžarski + matematik \index{Fejer, L.} \textit{L. Fejer} (1880--1959).}). + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.9.pic} +\caption{} \label{sl.izo.6.2.9.pic} +\end{figure} + + +\textbf{\textit{Solution.}} Označimo za $AP$, $BQ$ in $CR$ višine +trikotnika $ABC$ (Figure \ref{sl.izo.6.2.9.pic}). + Če je $X$ +poljubna točka stranice $BC$, tedaj minimalen obseg trikotnika +$XYZ$, kjer oglišči $Y$ in $Z$ ležita na stranicah $AC$ in $AB$, +dobimo na način, ki je opisen v prejšnjem izreku +\ref{FagnanLema1}. Točki $Y$ in $Z$ sta torej presečišči stranic +$AC$ in $AB$ s premico $X'X''$ ($X'=\mathcal{S}_{AB}(X)$, +$X''=\mathcal{S}_{AC}(X)$). Pri tem je obseg trikotnika $XYZ$ enak +dolžini daljice $X'X''$. + +Torej nam ostaja, da ugotovimo, za katero točko $X$ stranice +$BC$ je obseg trikotnika $XYZ$ najmanjši oz. daljica $X'X''$ +najkrajša. Iz lastnosti zrcaljenja sledi $AX'\cong AX\cong AX''$, +$\angle X'AB\cong \angle BAX$ in $\angle X''AC\cong \angle CAX$ +oz. $\angle X'AX''=2\angle BAC$. Kot $X'AX''$ enakokrakega +trikotnika $X'AX''$ je torej konstanten (glede na točko $X$), zato +je njegova osnovnica $X'X''$ najkrajša v primeru, kadar je +najkrajši njen krak $AX'$ (izrek \ref{SkladTrikLema}) oz. daljica +$AX\cong AX'\cong AX''$. To pa je izpolnjeno, ko je $X$ nožišče +višine trikotnika $ABC$ iz oglišča $A$ oz. $X=P$. Iz zgleda +\ref{FagnanLema} sledi $Y=Q$ in $Z=R$. Torej je včrtani trikotnik z +najmanjšim obsegom pravzaprav pedalni trikotnik trikotnika +$ABC$. + \kdokaz + + + + \bzgled + Let $P$ and $Q$ be interior points of an angle $aOb$. + Construct a point $X$ on the side $a$ of this angle such that + the rays $XP$ and $XQ$ intersect the side $b$ at points $Y$ and $Z$ such that $XY\cong XZ$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.10.pic} +\caption{} \label{sl.izo.6.2.10.pic} +\end{figure} + + +\textbf{\textit{Solution.}} + Naj bodo $X$, $Y$ in $Z$ iskane točke (Figure \ref{sl.izo.6.2.10.pic}). + Naj bosta še: $S$ središče daljice $YZ$ in $P'=\mathcal{S}_a(P)$. + Mero kota +$aOb$ označimo z $\omega$. Trikotnik $XYZ$ je enakokrak, zato je +(zaradi skladnosti trikotnikov $XSY$ in $XSZ$ - izrek +\textit{SSS} \ref{SSS}) + težiščnica $XS$ hkrati višina in simetrala notranjega kota $XYZ$ + tega +trikotnika. Torej velja $\angle OSX=90^0$ in $\angle YXS\cong +\angle ZXS$. Iz $\mathcal{S}_a:P, X, O\mapsto P', X, O$ pa sledi +$\angle PXO\cong \angle P'XO$. Sedaj lahko računamo mero kota +$P'XQ$: + \begin{eqnarray*} +\angle P' XQ &=& \angle P' XP + \angle PXQ = 2\angle OXP + 2\angle +PXS\\ &=& 2\angle OXS = 2(90° - \omega) =180° - 2\omega. + \end{eqnarray*} +Ker sta nam točki $P'$ in $Q$ znani, točko $X$ dobimo kot +presečišče kraka $a$ in loka s tetivo $P'Q$ z obodnim kotom $180° +- 2\omega$ (izrek \ref{ObodKotGMT}). + \kdokaz + + + + \bzgled + Let $ABCD$ be a rectangle such that $|AB|=3|BC|$. Suppose that $E$ and $F$ are + points on the side $AB$ such that + $AE\cong EF\cong FB$. + Prove $$\angle AED+\angle AFD+\angle + ABD=90^0.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.11.pic} +\caption{} \label{sl.izo.6.2.11.pic} +\end{figure} + + +\textbf{\textit{Solution.}} Naj bo $F'=\mathcal{S}_{CD}(F)$ in +$B'=\mathcal{S}_{CD}(B)$ + (Figure \ref{sl.izo.6.2.11.pic}). + Tedaj je $DF'\cong DF$ in $\angle DF'F\cong \angle DFF'$. Iz skladnosti +trikotnikov $DAF$ in $F'B'B$ (izrek \textit{SAS} \ref{SKS}) sledi +$DF\cong F'B$ in $\angle DFA\cong \angle B'BF'$. Velja tudi +$\angle FF'B\cong \angle B'BF'$ (izrek \ref{KotiTransverzala}). +Torej velja: + $$\angle DF'B=\angle DF'F+\angle FF'B=\angle DFF'+\angle DFA=\angle AFF'=90^0.$$ + Ker je $DF'\cong F'B$, je +$DF'B$ enakokraki pravokotni trikotnik z osnovnico $BD$, zato je +(izrek \ref{enakokraki}) $\angle DBF'=\angle BDF'=45^0$. Ker je +tudi $DAE$ enakokraki pravokotni trikotnik z osnovnico $DE$, je +$\angle AED=45^0$ oz. $\angle AED=\angle DBF'$. Zatorej je: +$$\angle AED+\angle AFD+\angle ABD=\angle DBF'+\angle F'BB'+ +\angle ABD=\angle ABC=90^0,$$ kar je bilo treba dokazati. \kdokaz + + Ob naslednjem zgledu bomo opisali standardni postopek uporabe + izometrij pri načrtovalnih nalogah. Predpostavimo, da je potrebno konstruirati + točki $X\in k$ in $Y\in l$, kjer sta $k$ in $l$ dani krožnici + (lahko tudi premici ali krožnica in premica), vemo pa, da za neko + izometrijo $\mathcal{I}$ velja $\mathcal{I}(X)=Y$. V tem primeru + iz $X\in k$ sledi $Y=\mathcal{I}(X)\in \mathcal{I}(k)$. Ker je + tudi $Y\in l$, lahko točko $Y$ konstruiramo iz pogoja $Y\in + l\cap \mathcal{I}(k)$. + + + + \bzgled + Two circles $k$ and $l$ on different sides of a line $p$ are given. + Construct a square $ABCD$ such that $A\in k$, $C\in l$ and $B,D\in p$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.12.pic} +\caption{} \label{sl.izo.6.2.12.pic} +\end{figure} + + +\textbf{\textit{Solution.}} + (Figure \ref{sl.izo.6.2.12.pic}) + + Naj bo $ABCD$ iskani kvadrat. Potem je $\mathcal{S}_p(C)=A$. +Točka $C$ leži na krožnici $l$, zato njena slika -- točka $A$ -- pri +zrcaljenju $\mathcal{S}_p$ leži na sliki krožnice $l$ -- krožnici +$l'$, ki jo lahko načrtamo. Toda točka $A$ leži tudi na krožnici +$k$, zatorej je $A\in l'\cap k$. Na ta način dobimo najprej +oglišče $A$, nato pa $C=\mathcal{S}_p(A)$. Središče kvadrata $S$ +dobimo kot središče diagonale $AC$. Na koncu sta oglišči $B$ in +$D$ presečišči premice $p$ in krožnice s središčem $S$ in polmerom +$SA$. + +Število rešitev je odvisno od števila presečišč krožnic $l'$ in +$k$. \kdokaz + + + + \bnaloga\footnote{38. IMO Argentina - 1997, Problem 2.} + The angle at $A$ is the smallest angle of triangle $ABC$. The points $B$ and $C$ + divide the circumcircle of the triangle into two arcs. Let $U$ be an interior point + of the arc between $B$ and $C$ which does not contain $A$. The perpendicular + bisectors of $AB$ and $AC$ meet the line $AU$ at $V$ and $W$, respectively. The lines + $BV$ and $CW$ meet at $T$. Show that $$|AU| =|TB| + |TC|.$$ + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.IMO1.pic} +\caption{} \label{sl.izo.6.2.IMO1.pic} +\end{figure} + + \textbf{\textit{Solution.}} Označimo s $q$ in $p$ simetrali + stranic $AB$ in $AC$ trikotnika $ABC$ + (Figure \ref{sl.izo.6.2.IMO1.pic}). Njuno presečišče -- točka + $O$ -- je središče očrtane krožnice $l$ tega trikotnika. Po + predpostavki je $V=AU\cap q$ in $W=AU\cap p$. Naj bo $D$ drugo + presečišče poltraka $CW$ s krožnico $l$. S $\mathcal{S}_p$ + označimo zrcaljenje čez + premico $p$. $\mathcal{S}_p$ preslika točke $A$, $C$, $W$ in $O$ + po vrsti v točke $C$, $A$, $W$ in $O$ ter poltrak $AW$ + v poltrak $CW$. Ker $O\in p$, je po izreku \ref{zrcFiksKroz} + $\mathcal{S}_p(l) =l$. Iz tega sledi: + $$\mathcal{S}_p(U) =\mathcal{S}_p(AW\cap l)= + \mathcal{S}_p(AW)\cap \mathcal{S}_p(l)= + CW\cap l=D.$$ + Torej $\mathcal{S}_p:\hspace*{1mm} A,U\mapsto C,D$, zato je + $AU\cong CD$ in $AD\cong CU$. Iz skladnosti tetiv $AD$ + in $CU$ sledi skladnost pripadajočih obodnih kotov + (izrek \ref{SklTetSklObKot}) oz. velja + $\angle ABD\cong\angle UAC$. Nad tetivo $BC$ pa sta skladna + tudi kota $BDC$ in $BAC$ (izrek\ref{ObodObodKot}). + + Ker točka $V$ leži na simetrali $q$ daljice $AB$, je + $\mathcal{S}_q:\hspace*{1mm} A,B,V\mapsto B,A,V$. To pomeni, da je + $\angle BAV\cong\angle ABV$. + + Iz dokazane skladnosti ustreznih kotov sledi: + \begin{eqnarray*} \angle BDT&=&\angle BDC\cong\angle BAC=\\ + &=&\angle BAU+ + \angle UAC=\\ + &=&\angle ABV+\angle ABD=\\ + &=&\angle DBT. + \end{eqnarray*} + Torej je trikotnik $DTB$ enakokrak in po izreku \ref{enakokraki} + velja $TD\cong TB$. Na koncu dobimo: + $$|AU|= |CD|=|CT|+|TD|=|CT|+|TB|,$$ kar je bilo treba dokazati. \kdokaz + + + \bnaloga \footnote{50. IMO Germany - 2009, Problem 4.} + Let $ABC$ be a triangle with $|AB|=|AC|$. The angle bisectors of $\angle CAB$ and $\angle ABC$ + meet the sides $BC$ and $CA$ at $D$ and $E$, respectively. Let $K$ be the incentre of triangle $ADC$. + Suppose that $\angle BEK=45^0$. Find all possible values of $\angle CAB$. + \enaloga + + + + \textbf{\textit{Solution.}} Označimo s $S$ središče včrtane krožnice + trikotnika $ABC$. + To pomeni, da je $S$ presečišče simetral $AD$ in $BE$ + kotov $\alpha=\angle CAB$ in $\beta=\angle ABC$. Točka $S$ leži tudi na + simetrali +$CF$ ($F\in AB$) kota $\gamma=\angle ACB$. Iz $|AB|=|AC|$ sledi +$\beta\cong\gamma$, zato je tudi $\angle SBC\cong\angle SCB$. + +Po predpostavki je točka $K$ središče včrtane krožnice + trikotnika $ADC$, zato točka $K$ leži na simetrali $CS$ kota + $\angle ACD=\angle ACB$. Poltrak $DK$ je simetrala kota + $ADC$. Iz skladnosti trikotnikov $ADB$ in $ADC$ (izrek \textit{SAS} \ref{SKS}) + je $\angle + ADC\cong\angle ADB=90^0$. Iz tega sledi $\angle SDK=45^0$. + +Naj bo $E'=\mathcal{S}_{SC}(E)$. Ker je $CS$ simetrala kota $ACB$, +se krak $CA$ z zrcaljenjem $\mathcal{S}_{SC}$ preslika v poltrak +$CB$. Zato je $E'\in CB$. Ker $S,K,C\in SC$, je +$\mathcal{S}_{SC}:\hspace*{1mm}S,K,C\mapsto S,K,C$. + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.IMO2a.pic} +\caption{} \label{sl.izo.6.2.IMO2a.pic} +\end{figure} + +Obravnavali bomo dva primera. Če je $E'=D$ (Figure +\ref{sl.izo.6.2.IMO2a.pic}), je $\angle SEC\cong \angle SDC=90^0$. +Trikotnika $ABE$ in $CBE$ sta v tem primeru skladna (izrek \textit{ASA} +\ref{KSK}), zato je $AB\cong BC$. Trikotnik $ABC$ je +enakostraničen oz. velja $\angle BAC=60^0$. + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.2.IMO2.pic} +\caption{} \label{sl.izo.6.2.IMO2.pic} +\end{figure} + +Naj bo $E'\neq D$ (Figure \ref{sl.izo.6.2.IMO2.pic}). Iz +$\mathcal{S}_{SC}:\hspace*{1mm}S,K\mapsto S,K$ sledi $\angle SE'K +\cong \angle SEK=45^0$. Ker je torej $\angle SDK \cong \angle +SE'K=45^0$, so po izreku \ref{ObodKotGMT} točke $S$, $K$, $D$ in +$E'$ konciklične. Iz izreka \ref{TalesovIzrKroz} potem sledi +$\angle SKE'\cong \angle SDE'=90^0$. Ker zrcaljenje +$\mathcal{S}_{SC}$ preslika kot $SKE$ v kot $SKE'$, je $\angle +SKE\cong \angle SKE'=90^0$ (točke $E$, $K$ in $E'$ so zato +kolinearne). Iz tega in izreka \ref{VsotKotTrik} za trikotnik +$SKE$ sledi $\angle KSE=45^0$. Le-ta je zunanji kot trikotnika +$SBC$, zato je po izreku \ref{zunanjiNotrNotr} $\angle SBC +\angle +SCB=45^0$ oz. $\beta+\gamma=90^0$. Iz slednje relacije pa sledi +$\angle BAC=90^0$. + + Možni velikosti kota + $\angle CAB$ sta torej $60^0$ in $90^0$. +\kdokaz + + +%________________________________________________________________________________ + \poglavje{More on Reflections} \label{odd6Sopi} + +V začetni raziskavi o izometrijah smo ugotovili, da vsaka direktna +izometrija, ki ima vsaj dve fiksni točki, predstavlja identiteto. +Intuitivno je jasno, da indirektna izometrija z vsaj dvema fiksnima +točkama predstavlja osno zrcaljenje. Dokazali bomo še močnejšo +trditev. + + + + \bizrek \label{izo1ftIndZrc} + An opposite isometry of the plane with at least one fixed point is a reflection. + The axis of this reflection passes through this point. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.1.pic} +\caption{} \label{sl.izo.6.3.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} + (Figure \ref{sl.izo.6.3.1.pic}) + +Naj bo $A$ fiksna točka indirektne izometrije $\mathcal{I}$ neke +ravnine. Ker je indirektna, izometrija $\mathcal{I}$ ni +identiteta, zato obstaja takšna točka $B$ te ravnine, da velja +$\mathcal{I}(B)=B'\neq B$. Naj bo $p$ simetrala daljice $BB'$. +Torej je $\mathcal{I}$ izometrija, ki točki $A$ in $B$ preslika v +točki $A$ in $B'$, zato je $AB\cong AB'$. To pomeni, da točka $A$ +leži na simetrali $p$ daljice $BB'$. Dokažimo, da izometrija +$\mathcal{I}$ predstavlja zrcaljenje čez premico $p$ oz. +$\mathcal{I} = \mathcal{S}_p$. Kompozitum $\mathcal{S}_p \circ +\mathcal{I}$ (dveh indirektnih izometrij) je direktna izometrija, +z dvema fiksnima točkama $A$ in $B$, zato na osnovi že omenjenega +izreka \ref{izo2ftIdent} predstavlja identiteto. Torej velja +$\mathcal{S}_p \circ \mathcal{I}=\mathcal{E}$. Iz tega sledi $ +\mathcal{I}=\mathcal{S}_p^{-1} \circ\mathcal{E}$ oz. $\mathcal{I} += \mathcal{S}_p$. + \kdokaz + + Zelo koristen je naslednji izrek. + + + \bizrek \label{izoTransmutacija} + Let $\mathcal{I}$ be an isometry and $p$ an arbitrary line in the plane. + Suppose that this isometry maps the line $p$ to a line $p'$. If $\mathcal{S}_p$ is + a reflection with axis $p$, then + $$\mathcal{I}\circ \mathcal{S}_p\circ \mathcal{I}^{-1} = \mathcal{S}_{p'}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.2.pic} +\caption{} \label{sl.izo.6.3.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Naj bo $Y$ poljubna točka premice $p'$ in $X = \mathcal{I}^{-1}(Y)$ + (Figure \ref{sl.izo.6.3.2.pic}). + + Ker je $\mathcal{I}(p)=p'$ oz. $\mathcal{I}^{-1}(p')=p$, + točka $X$ leži na premici $p$ in je + $\mathcal{S}_p(X)=X$. + Zatorej velja: + $$\mathcal{I}\circ\mathcal{S}_p\circ\mathcal{I}^{-1}(Y)= + \mathcal{I}\circ\mathcal{S}_p(X)= + \mathcal{I}(X)=Y.$$ + Torej je kompozitum $\mathcal{I}\circ\mathcal{S}_p\circ\mathcal{I}^{-1}$ + indirektna izometrija s fiksno točko $Y$, zato po prejšnjemu izreku + \ref{izo1ftIndZrc} predstavlja osno zrcaljenje. Toda $Y$ je poljubna + točka premice $p'$, kar pomeni, da je os tega zrcaljenja ravno premica + $p'$ oz. $\mathcal{I}\circ \mathcal{S}_p\circ \mathcal{I}^{-1} + = \mathcal{S}_{p'}$. + \kdokaz + +Transformacijo +$\mathcal{I}\circ\mathcal{S}_p\circ\mathcal{I}^{-1}$ iz prejšnjega +izreka imenujemo \index{transmutacija!zrcaljenja čez premico} +\pojem{transmutacija} zrcaljenja $\mathcal{S}_p$ z izometrijo +$\mathcal{I}$. + + Dokažimo enostavno posledico prejšnjega +izreka. + + + + \bzgled \label{izoZrcKomut} + Two reflections in the plane commute if and only if + their axis are perpendicular or coincident, i.e. + $$\mathcal{S}_p \circ \mathcal{S}_q + = \mathcal{S}_q \circ \mathcal{S}_p \Leftrightarrow + (p\perp q \vee p=q).$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.3.pic} +\caption{} \label{sl.izo.6.3.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} + (Figure \ref{sl.izo.6.3.3.pic}) + + Najprej je $\mathcal{S}_p \circ \mathcal{S}_q + = \mathcal{S}_q \circ \mathcal{S}_p + \Leftrightarrow \mathcal{S}_p= \mathcal{S}_q \circ + \mathcal{S}_p \circ \mathcal{S}_q^{-1}$. Če uporabimo prejšnji izrek + \ref{izoTransmutacija} za $\mathcal{I} = \mathcal{S}_q$, +dobimo, da je zadnja enakost ekvivalentna z $\mathcal{S}_p = +\mathcal{S}_{p'}$, kjer je $p' = \mathcal{S}_q(p)$. To pa velja +natanko tedaj, ko je $p=p'=\mathcal{S}_q(p)$ oz. natanko tedaj, +ko je $p\perp q$ ali $p=q$. + \kdokaz + +Kompozitum dveh osnih zrcaljenj $\mathcal{I}=\mathcal{S}_p \circ +\mathcal{S}_q$ bo zelo pomemben v nadaljnjem raziskovanju +izometrij. Takoj ugotovimo, da kompozitum dveh +indirektnih izometrij predstavlja direktno izometrijo. +Če osi teh zrcaljenj sovpadata, smo že +ugotovili, da gre za identiteto (izrek \ref{izoZrcPrInvol}). +Zanimivo je vprašanje, kaj $\mathcal{I}$ predstavlja v splošnem +primeru, kadar sta osi $p$ in $q$ teh zrcaljenj različni in +komplanarni; še posebej v primeru, ko se sekata ali sta vzporedni. +Odgovor na to vprašanje bomo izvedeli v naslednjih razdelkih. V +naslednjem izreku pa bomo izvedeli nekaj o fiksnih točkah kompozituma +dveh osnih zrcaljenj. + + \bizrek \label{izoKomppqX} + Let $p$ and $q$ be two distinct coplanar lines. + Then + $$\mathcal{S}_q\circ\mathcal{S}_p(X)=X \Leftrightarrow p\cap q=\{X\}.$$ + \eizrek + +\textbf{\textit{Proof.}} + +($\Leftarrow$) Trivialno, kajti iz $X\in p$ in $X\in q$ sledi +$\mathcal{S}_p(X)= \mathcal{S}_q(X)=X$ oz. +$\mathcal{S}_q\circ\mathcal{S}_p(X)=X$. + +($\Rightarrow$) Naj bo $\mathcal{S}_q\circ\mathcal{S}_p(X)=X$ in +$\mathcal{S}_p(X)=X'$. Iz tega sledi $\mathcal{S}_q(X')=X$. +Predpostavimo, da je $X\neq X'$. V tem primeru sta po definiciji +osnega zrcaljenja $p$ in $q$ simetrali daljice $XX'$. To pa ni +možno, saj ima daljica eno samo simetralo, $p$ in $q$ pa sta po +predpostavki različni. Zatorej je $X=X'$. Iz tega sledi +$\mathcal{S}_p(X)=\mathcal{S}_q(X)=X$, kar pomeni, da velja $X\in +p$ in $X\in q$ oz. $X\in p\cap q$. Ker sta $p$ in $q$ različni +premici, je po aksiomu \ref{AksI1} $p\cap q=\{X\}$. + \kdokaz + + Če se premici $p$ in $q$ sekata, iz prejšnjega izreka sledi, da ima kompozitum + $\mathcal{S}_q\circ\mathcal{S}_p$ + eno samo fiksno točko, ki je njuno presečišče. Če sta $p$ in +$q$ vzporednici, kompozitum $\mathcal{S}_q\circ\mathcal{S}_p$ nima +fiksnih točk. + +Za nadaljevanje raziskave izometrij je zelo značilen naslednji +pojem. + +Množico vseh premic neke ravnine, ki potekajo skozi eno točko $S$ te +ravnine, + imenujemo \index{šop!konkurentnih premic} + \pojem{šop konkurentnih premic} (ali \index{šop!eliptični}\pojem{eliptični šop}) s središčem $S$ in jo označimo z + $\mathcal{X}_S$. + +Množico vseh premic neke ravnine, ki so vzporedne z neko premico +$s$ te ravnine, imenujemo \index{šop!vzporednic} \pojem{šop +vzporednic} (ali \index{šop!parabolični}\pojem{parabolični šop} ali \index{snop premic}\pojem{snop premic}) in jo označimo z + $\mathcal{X}_s$. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.4.pic} +\caption{} \label{sl.izo.6.3.4.pic} +\end{figure} + + + + V obeh primerih bomo govorili o \pojem{šopu premic} + (Figure \ref{sl.izo.6.3.4.pic}). Če ne bomo posebej poudarili, ali gre za šop + konkurentnih premic oz. šop vzporednic, bomo ta šop označevali + le z $\mathcal{X}$. + +V zvezi z vpeljanimi pojmi govori naslednji izrek. + + + \bizrek \label{izoSop} + Let $p$, $q$ and $r$ be lines in the plane. + The product $\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p$ + is a reflection if and only if + the lines $p$, $q$ and $r$ belong to the same family of lines. + The axis of the new reflection also belongs to this family. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.5.pic} +\caption{} \label{sl.izo.6.3.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} + (Figure \ref{sl.izo.6.3.5.pic}) + + ($\Leftarrow$) Predpostavimo najprej, da premice $p$, $q$ in $r$ + pripadajo istemu šopu $\mathcal{X}$. Obravnavali bomo dva + primera. + +\textit{1)} Naj bo $\mathcal{X}$ šop konkurentnih premic s +središčem $S$ oz. $p, q, r\in \mathcal{X}_S$. V tem primeru +je kompozitum $\mathcal{I}=\mathcal{S}_r \circ \mathcal{S}_q \circ +\mathcal{S}_p$ indirektna izometrija s fiksno točko $S$, zato po +izreku \ref{izo1ftIndZrc} predstavlja neko zrcaljenje +$\mathcal{S}_t$ z osjo, ki poteka skozi točko $S$ oz. $t\in +\mathcal{X}_S$. + + \textit{2)} Naj bo $\mathcal{X}$ šop vzporednic in $n$ poljubna skupna + pravokotnica premic $p$, $q$ in $r$. V tem primeru je kompozitum + $\mathcal{I}=\mathcal{S}_r \circ \mathcal{S}_q \circ +\mathcal{S}_p$ indirektna izometrija in velja $\mathcal{I}(n)=n$. +Dokažimo, da na premici $n$ obstaja + neka fiksna točka te izometrije. Naj bo $A$ poljubna točka + premice + $n$ in $\mathcal{I}(A)=A'$. Če $A$ ni fiksna točka izometrije + $\mathcal{I}$, z $O$ + označimo središče daljice $AA'$ in z $O'=\mathcal{I}(O)$. + $\mathcal{I}$ je indirektna izometrija, zato sta + $AO$ in $A'O'$ skladni in različno orientirani daljici na premici $n$. + Iz tega in iz dejstva, da je $O$ središče daljice $AA'$, sledi + $\overrightarrow{A'O}=\overrightarrow{OA}=\overrightarrow{A'O'}$. + Torej $O=O'$ oz. $O$ je fiksna + točka izometrije $\mathcal{I}$, zato predstavlja neko osno zrcaljenje + $\mathcal{S}_t$ (izrek \ref{izo1ftIndZrc}). Pri tem je + $\mathcal{S}_t(n)=\mathcal{I}(n)=n$. To pomeni, da je $t=n$ ali + $t\perp n$. Prva možnost odpade, ker iz $\mathcal{S}_r + \circ \mathcal{S}_q \circ +\mathcal{S}_p=\mathcal{S}_n$ sledi $\mathcal{S}_q \circ +\mathcal{S}_p=\mathcal{S}_r\circ \mathcal{S}_n$. Zadnja relacija +ni možna, ker sta premici $p$ in $q$ vzporedni, premici $r$ in $n$ +sta pravokotni in se sekata (izrek \ref{izoKomppqX}). Zatorej je +$t\perp n$, kar pomeni, da je tudi $t$ premica iz šopa vzporednic +$\mathcal{X}$. + + + ($\Rightarrow$) Predpostavimo sedaj, da je + $\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p=\mathcal{S}_t$ + za neko premico $t$. V tem primeru velja + $\mathcal{S}_q \circ \mathcal{S}_p=\mathcal{S}_r + \circ\mathcal{S}_t$. Spet bomo obravnavali dva primera. + + \textit{1)} Predpostavimo, da se $p$ in $q$ sekata v neki točki $S$. + V tem primeru je je $\mathcal{S}_r + \circ\mathcal{S}_t(S) = \mathcal{S}_q \circ \mathcal{S}_p(S)=S$. Po + izreku \ref{izoKomppqX} se tudi premici $r$ in $t$ sekata v točki $S$, zato je + $p, q, r, t\in \mathcal{X}_S$. + + \textit{2)} Predpostavimo, da sta premici $p$ in $q$ vzporedni. + Iz že omenjenega izreka \ref{izoKomppqX} sledi, da + kompozitum $\mathcal{S}_q \circ \mathcal{S}_p$ nima fiksnih točk. + Tedaj niti kompozitum $\mathcal{S}_r \circ \mathcal{S}_t$ + nima fiksnih točk, zato je tudi $r\parallel t$. Naj bo $n$ + skupna pravokotnica premic $p$ in $q$. Tedaj je $\mathcal{S}_r + \circ\mathcal{S}_t(n) = \mathcal{S}_q \circ \mathcal{S}_p(n)=n$. + Iz tega sledi $\mathcal{S}_t(n)=\mathcal{S}_r(n)=n'$. + Če je $n\neq n'$, sta obe vzporednici $p$ in $q$ osi + simetrij premic $n$ in $n'$, kar ni možno. Zato je $n=n'$ in + $\mathcal{S}_t(n)=\mathcal{S}_r(n)=n$. + Torej velja $r,t\perp n$, kar pomeni, da premice $p$, $q$, $r$ in + $t$ + pripadajo šopu vzporednic, + ki so vse pravokotne s premico $n$. + \kdokaz + +Posledica dokazane trditve je naslednji izrek. + + + + \bizrek \label{izoSop2n+1} + The product of an odd number of reflections with the axes + from the same family of lines is also a reflection. + \eizrek + + \textbf{\textit{Proof.}} Dokaz bomo izpeljali z indukcijo po + številu $n\geq 1$, kjer je $m=2n+1$ (liho) število premic. + + Za $n=1$ oz. $m=3$, je trditev direktna posledica prejšnjega izreka + \ref{izoSop}. + + Predpostavimo, da za vsak $k\in \mathbb{N}$ ($k\geq 1$) + in vsako $(2 k+1)$-terico + premic + $p_1,p_2,\ldots ,p_{2k+1}$, ki so vse iz istega šopa, kompozitum + $\mathcal{S}_{p_{2k+1}}\circ + \cdots \circ\mathcal{S}_{p_2}\circ\mathcal{S}_{p_1}$ predstavlja + neko osno zrcaljenje $\mathcal{S}_p$ (indukcijska predpostavka). + + Potrebno je dokazati, da potem tudi za $k+1$ + in vsako $(2(k+1)+1)$-terico + premic + $p_1,p_2,\ldots ,p_{2(k+1)+1}$, ki so vse iz istega šopa, kompozitum + $\mathcal{I}=\mathcal{S}_{p_{2(k+1)+1}}\circ + \cdots \circ\mathcal{S}_{p_2}\circ\mathcal{S}_{p_1}$ predstavlja + neko osno zrcaljenje $\mathcal{S}_{p'}$. Toda: + \begin{eqnarray*} + \mathcal{I}&=&\mathcal{S}_{p_{2(k+1)+1}}\circ + \cdots \circ\mathcal{S}_{p_2}\circ\mathcal{S}_{p_1}=\\ + &=&\mathcal{S}_{p_{2k+3}}\circ\mathcal{S}_{p_{2k+2}}\circ + \mathcal{S}_{p_{2k+1}}\circ + \cdots \circ\mathcal{S}_{p_2}\circ\mathcal{S}_{p_1}=\\ + &=& \mathcal{S}_{p_{2k+3}}\circ\mathcal{S}_{p_{2k+2}}\circ + \mathcal{S}_p= \hspace*{14mm} \textrm{ (po indukcijski predpostavki)}\\ + &=& \mathcal{S}_{p'} \hspace*{48mm} \textrm{ (po izreku \ref{izoSop}),} + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + + Dokažimo še nekaj direktnih posledic izreka \ref{izoSop}. + + \bzgled \label{iropqrrqp} + If $p$, $q$ and $r$ are lines from the same family of lines $\mathcal{X}$, then + $$\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p = + \mathcal{S}_p \circ \mathcal{S}_q \circ \mathcal{S}_r.$$ + \ezgled + +\textbf{\textit{Proof.}} Ker so $p$, $q$ in $r$ premice istega +šopa $\mathcal{X}$, po prejšnjem izreku \ref{izoSop} sledi +$\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p +=\mathcal{S}_t$ za neko os $t\in\mathcal{X}$. Zaradi tega je: + $$(\mathcal{S}_r \circ +\mathcal{S}_q \circ \mathcal{S}_p)^2 +=\mathcal{S}_t^2=\mathcal{E},\hspace*{3mm}\textrm{oz.}$$ + $$\mathcal{S}_r \circ +\mathcal{S}_q \circ \mathcal{S}_p \circ\mathcal{S}_r \circ +\mathcal{S}_q \circ \mathcal{S}_p=\mathcal{E}.$$ + Če zadnjo +relacijo množimo po vrsti z leve s $\mathcal{S}_r$, +$\mathcal{S}_q$ in $\mathcal{S}_p$, dobimo iskano enakost. + \kdokaz + + \bzgled \label{izoSopabc} + If $a$, $b$ and $c$ are lines such that $\mathcal{S}_a(b)=c$, + then those lines are from the same family of lines. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.6a.pic} +\caption{} \label{sl.izo.6.3.6a.pic} +\end{figure} + + \textbf{\textit{Proof.}} + (Figure \ref{sl.izo.6.3.6a.pic}) + +Iz izreka o transmutaciji \ref{izoTransmutacija} sledi +$\mathcal{S}_a\circ\mathcal{S}_b\circ\mathcal{S}_a= +\mathcal{S}_{\mathcal{S}_a(b)}= \mathcal{S}_c$. Po množenju +dobljene relacije +$\mathcal{S}_a\circ\mathcal{S}_b\circ\mathcal{S}_a= \mathcal{S}_c$ +najprej z desne s $\mathcal{S}_a$, nato pa še z leve s +$\mathcal{S}_c$ dobimo +$\mathcal{S}_c\circ\mathcal{S}_a\circ\mathcal{S}_b=\mathcal{S}_a$. +Po izreku \ref{izoSop} premice $a$, $b$ in $c$ pripadajo istemu +šopu. + \kdokaz + + Jasno je, da je premica $a$ iz prejšnjega izreka os simetrije + premic $b$ in $c$. + + Z naslednjim izrekom bomo določili premico $t$ iz relacije + $\mathcal{S}_r \circ +\mathcal{S}_q \circ \mathcal{S}_p =\mathcal{S}_t$ za + premice $p$, $q$ in $r$ iz istega šopa. + + + \bizrek \label{izoSoppqrt} + If $p$, $q$, $r$ and $t$ are lines such that + $\mathcal{S}_r \circ + \mathcal{S}_q \circ \mathcal{S}_p =\mathcal{S}_t$, then the axis of symmetry of the lines + $p$ and $r$ is also + the axis of symmetry of the lines $q$ and $t$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.6.pic} +\caption{} \label{sl.izo.6.3.6.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Iz $\mathcal{S}_r \circ +\mathcal{S}_q \circ \mathcal{S}_p =\mathcal{S}_t$ sledi, da +premice $p$, $q$, $r$ in $t$ + pripadajo istemu šopu $\mathcal{X}$ (izrek \ref{izoSop}). + Naj bo $s$ os simetrije premic $p$ in $r$ (Figure \ref{sl.izo.6.3.6.pic}). + To pomeni, da velja $\mathcal{S}_s(p)=r$, + zato je po izreku o transmutaciji \ref{izoTransmutacija} tudi + $\mathcal{S}_s\circ\mathcal{S}_p\circ\mathcal{S}_s + =\mathcal{S}_{\mathcal{S}_s(p)} + =\mathcal{S}_r$. Iz tega naprej sledi: + $$\mathcal{S}_t = \mathcal{S}_r \circ\mathcal{S}_q \circ \mathcal{S}_p + = (\mathcal{S}_s\circ\mathcal{S}_p\circ\mathcal{S}_s)\circ + \mathcal{S}_q\circ\mathcal{S}_p = +\mathcal{S}_s\circ(\mathcal{S}_p\circ \mathcal{S}_s\circ +\mathcal{S}_q)\circ \mathcal{S}_p.$$ + Ker je $\mathcal{S}_s(p)=r$, iz prejšnjega zgleda \ref{izoSopabc} sledi, + da premice $s$, $p$ in $r$ +pripadajo istemu šopu, zato je $s\in \mathcal{X}$. Iz izreka +\ref{iropqrrqp} sledi $\mathcal{S}_p\circ \mathcal{S}_s\circ +\mathcal{S}_q = \mathcal{S}_q\circ \mathcal{S}_s\circ +\mathcal{S}_p$. Če uporabimo še izrek o transmutaciji +\ref{izoTransmutacija}, dobimo: +$$\mathcal{S}_t = +\mathcal{S}_s\circ\mathcal{S}_p\circ \mathcal{S}_s\circ +\mathcal{S}_q\circ \mathcal{S}_p= +\mathcal{S}_s\circ\mathcal{S}_q\circ \mathcal{S}_s\circ +\mathcal{S}_p\circ \mathcal{S}_p= +\mathcal{S}_s\circ\mathcal{S}_q\circ \mathcal{S}_s= +\mathcal{S}_{\mathcal{S}_s(q)}.$$ + Iz $\mathcal{S}_t =\mathcal{S}_{\mathcal{S}_s(q)}$ sledi + $\mathcal{S}_s(q)=t$, kar pomeni, da je premica $s$ os simetrije +premic $q$ in $t$. + \kdokaz + + Ob naslednji načrtovalni nalogi bomo ilustrirali dve metodi + reševanja. + + + \bizrek + Let $p$, $q$ and $r$ be lines in the plane. + Construct a triangle $ABC$ such that its vertices $B$ and $C$ + lie on the line $p$ and the lines $q$ and $r$ are perpendicular + bisectors of the sides $AB$ and $AC$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.7.pic} +\caption{} \label{sl.izo.6.3.7.pic} +\end{figure} + + \textbf{\textit{Solution.}} (prvi način) + + Naj bosta $q$ in $r$ simetrali stranic $AB$ in $AC$ + (Figure \ref{sl.izo.6.3.7.pic}\textit{a}). Potem je + $\mathcal{S}_q(B)=A$ in $\mathcal{S}_r(C)=A$. +Točki $B$ in $C$ ležita na premici $p$, zato točka $A$ leži na +slikah te premice glede na zrcaljenji $\mathcal{S}_q$ in +$\mathcal{S}_r$. Torej lahko oglišče $A$ konstruiramo kot +presečišče premic $p'_q=\mathcal{S}_q(p)$ in +$p'_r=\mathcal{S}_r(p)$. + + \textbf{\textit{Solution.}} (drugi način) + + + Presečišče premic $q$ in $r$ je središče očrtane krožnice tega trikotnika + (izrek \ref{SredOcrtaneKrozn}); +označimo jo z $O$ (Figure \ref{sl.izo.6.3.7.pic}\textit{b}). Torej +lahko konstruiramo tudi tretjo simetralo $s$ stranice $BC$, ki je +pravokotnica premice $p$ v točki $O$. Kompozitum $\mathcal{S}_q +\circ \mathcal{S}_r \circ \mathcal{S}_s$ po izreku \ref{izoSop} +predstavlja neko osno zrcaljenje $\mathcal{S}_l$ (ker $q, r, s\in +\mathcal{X}_O$). Fiksni točki tega kompozituma sta $O$ in $B$, +zato je $\mathcal{S}_q \circ \mathcal{S}_r \circ \mathcal{S}_s = +\mathcal{S}_{OB}$. Premico $OB=l$ lahko torej konstruiramo kot +simetralo daljice $XX'$, kjer je $X$ poljubna točka in +$X'=\mathcal{S}_q \circ \mathcal{S}_r \circ \mathcal{S}_s(X)$. V +primeru, če je $X=X'$, je $l=OX$. Na koncu dobimo točko $B$ kot +presečišče premic $p$ in $l$. + \kdokaz + + + + \bzgled + Let $ABCDE$ be a pentagon with a right angle + at the vertex $A$. + The perpendicular bisectors of the sides $AE$, $BC$, + and $CD$ (lines $p$, $q$, and $r$) intersect + at a point $O$ and the perpendicular bisectors of the sides $AB$ and $DE$ + (lines $x$ and $y$) intersect at a point $S$ + ($S\neq O$). Prove that one of the vertices of a triangle $KLM$, + such that $p$, $q$ + and $r$ are the perpendicular bisectors of its sides, + belongs to the line $OS$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.3.8.pic} +\caption{} \label{sl.izo.6.3.8.pic} +\end{figure} + + \textbf{\textit{Solution.}} + (Figure \ref{sl.izo.6.3.8.pic}) + +Naj bo $\mathcal{I} = \mathcal{S}_y \circ\mathcal{S}_r +\circ\mathcal{S}_q \circ\mathcal{S}_x \circ\mathcal{S}_p$. +$\mathcal{I}$ je indirektna izometrija, ki ima fiksno točko $E$, +zato po izreku \ref{izo1ftIndZrc} predstavlja neko zrcaljenje +$S_l$. Ker je $\angle EAB$ pravi kot, sta simetrali $p$ in $x$ +pravokotni, zato zrcaljenji $\mathcal{S}_p$ in $\mathcal{S}_x$ +komutirata (zgled \ref{izoZrcKomut}). Torej velja: + $$\mathcal{I} =\mathcal{S}_l= + \mathcal{S}_y \circ\mathcal{S}_r +\circ\mathcal{S}_q \circ\mathcal{S}_x \circ\mathcal{S}_p= + \mathcal{S}_y \circ (\mathcal{S}_r +\circ\mathcal{S}_q \circ\mathcal{S}_p) \circ\mathcal{S}_x.$$ + Premice $p$, $q$ in $r$ pripadajo istemu šopu +$\mathcal{X}_O$, zato kompozitum $\mathcal{S}_r \circ\mathcal{S}_q +\circ\mathcal{S}_p$ predstavlja neko zrcaljenje $\mathcal{S}_t$, +kjer velja tudi $t\in\mathcal{X}_O$ oz. os $t$ poteka skozi točko +$O$ (izrek \ref{izoSop}). + Torej velja $\mathcal{S}_l= + \mathcal{S}_y +\circ\mathcal{S}_t \circ\mathcal{S}_x$, zato premice $x$, $t$ in +$y$ pripadajo istem šopu, oz. premica $t$ poteka skozi točko $S=x\cap +y$. Ker premica $t$ poteka skozi različni točki $O$ in $S$, je +$t=OS$. Zato je $\mathcal{S}_{OS}=\mathcal{S}_t=\mathcal{S}_r +\circ\mathcal{S}_q \circ\mathcal{S}_p$. Naj bodo $p$, $q$ in $r$ +simetrale stranic $KL$, $LM$ in $MK$ trikotnika $KLM$. V tem +primeru je $\mathcal{S}_{OS}(K)=\mathcal{S}_t(K)=\mathcal{S}_r +\circ\mathcal{S}_q \circ\mathcal{S}_p(K)=K$, torej točka $K$ +leži na premici $OS$. + \kdokaz + + + +%________________________________________________________________________________ + \poglavje{Rotations} \label{odd6Rotac} + + + Do sedaj smo spoznali izometrije, ki imajo več kot eno fiksno točko. + Dokazali smo, da obstajata dve takšni vrsti izometrij, in sicer + identiteta kot direktna izometrija z vsaj dvema fiksnima + točkama (izrek \ref{izo2ftIdent}) in zrcaljenje kot indirektna izometrija z vsaj eno + fiksno točko (izrek \ref{izo1ftIndZrc}). V tem razdelku bomo + spoznali vrsto direktnih izometrij, ki imajo natanko eno + fiksno točko. + +Naj bo $S$ poljubna točka in $\omega\neq 0$ orientirani kot +evklidske ravnine. Transformacija te ravnine, v kateri je točka +$S$ fiksna, vsako drugo točko $X\neq S$ te ravnine pa preslikava v +takšno točko $X'$, da velja $\measuredangle XSX'\cong \omega$ in +$SX'\cong SX$, imenujemo rotacija s središčem $S$ za kot $\omega$; +označimo jo z $\mathcal{R}_{S,\omega}$ (Figure +\ref{sl.izo.6.4.1.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.1.pic} +\caption{} \label{sl.izo.6.4.1.pic} +\end{figure} + +Iz same definicije direktno sledijo naslednji izreki. + + + + \bizrek \label{RotacFiksT} + The only fixed point of a rotation $\mathcal{R}_{S,\omega}$ + is its centre $S$ i.e. + $$\mathcal{R}_{S,\omega}(X)=X + \hspace*{1mm} \Leftrightarrow \hspace*{1mm} X=S.$$ + \eizrek + + + \bizrek \label{rotacEnaki} + Rotations $\mathcal{R}_{S,\alpha}$ and $\mathcal{R}_{B,\beta}$ + are equal if and only if $A=B$ and $\alpha=\beta$ i.e. + $$\mathcal{R}_{S,\alpha}=\mathcal{R}_{B,\beta} + \hspace*{1mm} \Leftrightarrow \hspace*{1mm} A=B \hspace*{1mm} + \wedge\hspace*{1mm} \alpha=\beta.$$ + \eizrek + + + + \bizrek + The inverse transformation of a rotation is a rotation with the same centre + and congruent angle with opposite orientation, i.e. + $$\mathcal{R}_{S,\omega}^{-1}=\mathcal{R}_{S,-\omega}.$$ + \eizrek + + + + Čeprav je intuitivno jasno, je potrebno dokazati, da je rotacija + izometrija. + + \bizrek + Rotations are isometries of the plane. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.2.pic} +\caption{} \label{sl.izo.6.4.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $\mathcal{R}_{S,\omega}$ poljubna + rotacija. Iz definicije je jasno, da predstavlja bijektivno preslikavo. Potrebno je še dokazati, da za poljubni točki $X$ in $Y$ + ter njuni sliki $X'$ in $Y'$ ($\mathcal{R}_{S,\omega}:X,Y\mapsto X', Y'$) velja + $X'Y'\cong XY$ (Figure \ref{sl.izo.6.4.2.pic}). + +Če je ena od točk $X$ ali $Y$ enaka središču rotacije $S$ (npr. +$X=S$), je relacija $X'Y'\cong XY$ avtomatično izpolnjena, saj v +tem primeru po definiciji rotacije velja $SY'\cong SY$. + +Predpostavimo, da nobena od točk $X$ in $Y$ ni središče rotacije +$S$. Najprej je po definiciji rotacije $SX'\cong SX$ in $SY'\cong +SY$. Nato je še (iz relacije \ref{orientKotVsota}): + \begin{eqnarray*} + \measuredangle Y'SX'&=& \measuredangle Y'SX+\measuredangle XSX'= + \measuredangle Y'SX + \omega=\\ + &=& \measuredangle Y'SX + \measuredangle YSY'= + \measuredangle YSY'+ \measuredangle Y'SX=\\ &=&\measuredangle YSX + \end{eqnarray*} + To pomeni, da sta trikotnika $X'SY'$ in $XSY$ skladna (izrek \textit{SAS} \ref{SKS}), + zato je + $X'Y'\cong XY$. + \kdokaz + + Kot vsaka izometrija, tudi rotacija preslika premico v premico. + V naslednjem izreku bomo videli, kakšen je odnos premice in + njene slike pri rotaciji. + + + + \bizrek \label{rotacPremPremKot} + A line and its rotated image determine an angle congruent + to the angle of this rotation. + If the angle of the rotation measures $180^0$, + the line and its image are parallel. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.3.pic} +\caption{} \label{sl.izo.6.4.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $p'$ slika premice $p$ pri rotaciji + $\mathcal{R}_{S,\omega}$ (Figure \ref{sl.izo.6.4.3.pic}). + + Če je $S\in p$, je dokaz trivialen. Predpostavimo, da $S\notin p$. + Označimo s $P$ pravokotno projekcijo središča $S$ na + premici $p$ in njeno sliko $P'=\mathcal{R}_{S,\omega}(P)$. Ker $S\notin + p$, + je $P\neq S$ oz. $P'\neq P$. Iz + $P\in p$ sledi $P'\in p'$. Ker izometrija ohranja kote, iz + $SP\perp p$ sledi $SP'\perp p'$. + + Če je $\omega=180^0$, so točke $P$, $S$ in $P'$ kolinearne in + imata premici $p$ in $p'$ skupno pravokotnico + $PP'$, kar pomeni $p\parallel p'$. + + V primeru $\omega\neq 180^0$ se premici sekata (v nasprotnem + bi iz $p\parallel p'$ sledilo $SP\parallel SP'$, kar pa ni možno). + Naj bo njuno presečišče točka $V$. Kot, ki ga določata premici + $p$ in $p'$, je enak kotu, ki ga določata premici $SP$ in + $SP'$ (kota s pravokotnimi kraki - izrek \ref{KotaPravokKraki}), + ki pa je ravno kot rotacije $\omega$. + \kdokaz + + + V dokazu prejšnjega izreka je opisan tudi postopek načrtovanja slike + $p'$ + premice $p$ pri rotaciji $\mathcal{R}_{S,\omega}$. + Najprej narišemo pravokotno projekcijo $P$ središča $S$ na + premici $p$, nato pa njeno sliko $P=\mathcal{R}_{S,\omega}(P')$. + Premico $p'$ dobimo kot pravokotnico premice $SP'$ v točki + $P'$. + Drugi način bi seveda bil, da rotiramo dve poljubni točki + premice $p$. + + V naslednjem pomembnem izreku bomo videli, kako lahko izrazimo + rotacijo s pomočjo osnih zrcaljenj. + + + + + \bizrek \label{rotacKom2Zrc} + Any rotation + $\mathcal{R}_{S,\omega}$ can be expressed as the product of two reflections + $\mathcal{S}_p$ and $\mathcal{S}_q$ where + $p$ and $q$ are arbitrary lines, such that $S=p\cap q$ + and $\measuredangle pq=\frac{1}{2}\omega$.\\ The reverse is also true - + the product of two reflections $\mathcal{S}_p$ and $\mathcal{S}_q$ ($S=p\cap q$) + is a rotation with the centre $S$ and the angle + $\omega=2\cdot\measuredangle pq$, i.e.\\ + $$\mathcal{R}_{S,\omega}=\mathcal{S}_q\circ\mathcal{S}_p + \hspace*{1mm} \Leftrightarrow \hspace*{1mm} S=p\cap q \hspace*{1mm} + \wedge\hspace*{1mm} \measuredangle pq=\frac{1}{2}\omega.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.4.pic} +\caption{} \label{sl.izo.6.4.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} + + ($\Leftarrow$) Predpostavimo, da sta $p$ + in $q$ premici, ki se sekata v točki $S$. Naj bo $\omega + =2\cdot\measuredangle pq$ + (Figure \ref{sl.izo.6.4.4.pic}). Dokažimo, da velja + $\mathcal{R}_{S,\omega}=\mathcal{S}_q\circ\mathcal{S}_p$, + oz. $\mathcal{R}_{S,\omega}(X)=\mathcal{S}_q\circ\mathcal{S}_p(X)$ za vsako +točko $X$ ravnine. Obravnavali bomo dva primera. + + \textit{1)} Če je $X=S$, je $\mathcal{S}_p(X)=\mathcal{S}_q(X)=X$ (ker +se premici $p$ in $q$ sekata v točki $S$). Torej je +$\mathcal{R}_{S,\omega}(S)=S=\mathcal{S}_q\circ\mathcal{S}_p(S)$. + + \textit{2)} Naj bo $X\neq S$ in $\mathcal{S}_q\circ\mathcal{S}_p(X)=X'$. + Dokažimo, da je tudi $\mathcal{R}_{S,\omega}(X)=X'$. Naj bo +$\mathcal{S}_p(X)=X_1$. Potem je $\mathcal{S}_q(X_1)=X'$. Zaradi +tega je najprej $SX\cong SX_1\cong SX'$, nato pa tudi: + \begin{eqnarray*} + \measuredangle XSX'&=& \measuredangle XSX_1+ \measuredangle + X_1SX'=\\ + &=& 2\cdot\measuredangle p,SX_1+2\cdot\measuredangle SX_1,q=\\ + &=& 2\cdot(\measuredangle p,SX_1+ \measuredangle SX_1,q) =\\ + &=& 2\cdot\measuredangle pq=\omega. + \end{eqnarray*} + Po definiciji rotacije je $\mathcal{R}_{S,\omega}(X)=X'$. + + ($\Rightarrow$) Naj bo + $\mathcal{R}_{S,\omega}=\mathcal{S}_q\circ\mathcal{S}_p$. Potem + je + $\mathcal{S}_q\circ\mathcal{S}_p(S)=\mathcal{R}_{S,\omega}(S)=S$. + Po izreku \ref{izoKomppqX} je $S=p\cap q$. Potrebno je dokazati še + $\measuredangle pq=\frac{1}{2}\omega$ oz. $\omega + =2\cdot\measuredangle pq$. Označimo $\widehat{\omega} + =2\cdot\measuredangle pq$. Po prvem delu dokaza ($\Leftarrow$) + je + $\mathcal{S}_q\circ\mathcal{S}_p=\mathcal{R}_{S,\widehat{\omega}}$. + Torej velja $\mathcal{R}_{S,\omega}=\mathcal{S}_q\circ\mathcal{S}_p= + \mathcal{R}_{S,\widehat{\omega}}$. Iz izreka \ref{rotacEnaki} + sledi $\omega =\widehat{\omega} + =2\cdot\measuredangle pq$. + \kdokaz + + + + \bizrek \label{RotacDirekt} + A rotation is a direct isometry. + \eizrek + + \textbf{\textit{Proof.}} Direktna posledica prejšnjega izreka + \ref{rotacKom2Zrc}, ker je kompozitum dveh osnih zrcaljenj, ki + sta indirektni izometriji. + \kdokaz + +Podobno, kot pri osnem zrcaljenju, tudi za rotacijo velja izrek o +transmutaciji\index{transmutacija!rotacije}. + + + \bizrek \label{izoTransmRotac} + For an arbitrary rotation $\mathcal{R}_{O,\alpha}$ + and an arbitrary isometry $\mathcal{I}$ is + $$\mathcal{I}\circ + \mathcal{R}_{O,\alpha}\circ\mathcal{I}^{-1}= + \mathcal{R}_{\mathcal{I}(O),\alpha'},$$ + where: $\alpha'=\alpha$, if $\mathcal{I}$ a direct isometry, or + $\alpha'=-\alpha$, if $\mathcal{I}$ is an opposite isometry. + \eizrek + + \textbf{\textit{Proof.}} Po izreku \ref{rotacKom2Zrc} lahko rotacijo + $\mathcal{R}_{O,\alpha}$ + zapišemo kot kompozitum dveh osnih +zrcaljenj, in sicer +$\mathcal{R}_{O,\omega}=\mathcal{S}_q\circ\mathcal{S}_p$, kjer je + $O=p\cap q$ in $\measuredangle pq=\frac{1}{2}\alpha$. +Če uporabimo izrek o transmutaciji za osna zrcaljenja, dobimo: + \begin{eqnarray*} + \mathcal{I}\circ + \mathcal{R}_{O,\alpha}\circ\mathcal{I}^{-1}&=& + \mathcal{I}\circ\mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{I}=\\ + &=& + \mathcal{I}\circ\mathcal{S}_q\circ\mathcal{I}^{-1} + \circ\mathcal{I}\circ\mathcal{S}_p\circ\mathcal{I}^{-1}=\\ + &=&\mathcal{S}_{q'}\circ\mathcal{S}_{p'}=\\ + &=&\mathcal{R}_{O',\alpha'}, + \end{eqnarray*} +kjer sta $p'$ in $q'$ sliki premic $p$ in $q$ pri izometriji +$\mathcal{I}$, $O'=p'\cap q'$ in $\alpha'=2\cdot\measuredangle +p'q'$. Ker je $O=p\cap q$, velja $\mathcal{I}(O)= +\mathcal{I}(p\cap q)=\mathcal{I}(p)\cap \mathcal{I}(q)=p'\cap +q'=O'$. Ker izometrije ohranjajo relacijo skladnosti kotov, je + $\alpha'=2\cdot\measuredangle p'q'=2\cdot\measuredangle pq=\alpha$, če je + $\mathcal{I}$ direktna, oz. + $\alpha'=2\cdot\measuredangle p'q'=-2\cdot\measuredangle pq=-\alpha$, če je + $\mathcal{I}$ indirektna. + \kdokaz + + V naslednjem izreku bomo dokazali pomembno dejstvo, da za skladni daljici, ki nista vzporedni, obstaja rotacija, ki preslika prvo daljico v drugo. + + + + \bizrek + Let $AB$ and $A'B'$ be congruent line segments that are not parallel. + There is exactly one rotation that maps the line segment $AB$ to the line segment $A'B'$, + such that the point $A$ maps to the point $A'$ and the point $B$ to the point $B'$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.skl.3.1.10Rotac.pic} +\caption{} \label{sl.skl.3.1.10Rotac.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $S$ presečišče simetral daljic $AA'$ in $BB'$ ter $\omega=\measuredangle AB, A'B'$. Dokazali bomo, da je $\mathcal{R}_{S,\omega}$ iskana rotacija. + Ker točka $S$ leži na simetralah daljic $AA'$ in $BB'$, je $SA\cong SA'$ in $SB\cong SB'$ (Figure \ref{sl.skl.3.1.10Rotac.pic}). Potem iz $AB\cong A'B'$ po izreku \textit{SSS} \ref{SSS} sledi $\triangle SAB\cong \triangle SA'B'$ (glej tudi zgled \ref{načrt1odd3}). Torej velja $\measuredangle ASB\cong \measuredangle A'SB'$, oziroma: + $$\measuredangle ASA'=\measuredangle ASB+\measuredangle BSA'= + \measuredangle A'SB'+\measuredangle BSA'= \measuredangle BSB'.$$ + Po definiciji rotacije je $\mathcal{R}_{S,\measuredangle ASA'}:A,B\mapsto A',B'$. Po izreku \ref{rotacPremPremKot} je $\measuredangle ASA'=\measuredangle AB, A'B'=\omega$, zato je $\mathcal{R}_{S,\omega}:A,B\mapsto A',B'$. + Ker rotacija kot izometrija ohranja relacijo $\mathcal{B}$, se daljica $AB$ s to rotacijo preslika v daljico $A'B'$. + +Naj bo $\mathcal{R}_{\widehat{S},\widehat{\omega}}$ še ena rotacija, za katero velja $\mathcal{R}_{\widehat{S},\widehat{\omega}}:A,B\mapsto A',B'$. Potem je +$\mathcal{R}^{-1}_{S,\omega} \circ \mathcal{R}_{\widehat{S},\widehat{\omega}}$ direktna izometrija z dvema fiksnima točkama $A$ in $B$, zato po izreku \ref{izo2ftIdent} predstavlja identiteto $\mathcal{E}$ oz. velja +$\mathcal{R}_{\widehat{S},\widehat{\omega}} = \mathcal{R}_{S,\omega}$. + \kdokaz + + + V naslednjih primerih bomo uporabljali dejstvo, da je pozitivno orientiran + trikotnik + $ABC$ enakostraničen natanko tedaj, ko je + $\mathcal{R}_{A,60^0}(B)=C$. + + + + \bzgled + Let $P$ be an interior point of an equilateral triangle + $ABC$.\\ + a) Prove that + $|PA|+|PB|\geq |PC|.$\\ + b) Suppose that $\angle BPA=\mu$, + $\angle CPA= \nu$ and $\angle BPC= \xi$. Calculate the interior angles of a triangle, + with sides that are congruent to the line segments $PA$, $PB$ in $PC$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.5.pic} +\caption{} \label{sl.izo.6.4.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} Če je $P'=\mathcal{R}_{A,60^0}(P)$, + je trikotnik $APP'$ pravilen (Figure \ref{sl.izo.6.4.5.pic}). + Ker je še $\mathcal{R}_{A,60^0}(B)=C$, se + daljica $BP$ pri tej rotaciji preslika v +skladno daljico $CP'$. Torej ima trikotnik $PP'C$ stranice, ki so +skladne z daljicami $PA$, $PB$ in $PC$. Po trikotniški neenakosti +- izrek \ref{neenaktrik} - je $|PA|+|PB|=|PP'|+|P'C|\geq PC.$ + +Izračunajmo še kote trikotnika $PP'C$. Rotacija +$\mathcal{R}_{A,60^0}$ preslika trikotnik $ABP$ v trikotnik +$ACP'$, zato sta trikotnika skladna in je $\angle BPA=\angle CP'A$. Iz +tega in iz izreka \ref{VsotKotTrik} sledi: + \begin{eqnarray*} + \angle CP'P &=& \angle CP'A-\angle PP'A=\angle + BPA-60^0=\mu-60^0,\\ + \angle CPP' &=& \angle CPA-60^0=\nu-60^0,\\ + \angle PCP' &=& 180^0 + -(\mu-60^0)-(\nu-60^0)=300^0-(\mu+\nu)=\\ + &=& 300^0-(360^0-\xi)=\xi-60^0, + \end{eqnarray*} + kar je bilo treba izračunati. \kdokaz + + + + \bzgled + Let $ABCD$ be a rhombus, such that the interior angle at the vertex $A$ + is equal to $60^0$. + If a line $l$ intersects the sides $AB$ and $BC$ of this rhombus at points $P$ and + $Q$ such that $|BP|+|BQ|=|AB|$, then $PQD$ is a regular triangle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.6.pic} +\caption{} \label{sl.izo.6.4.6.pic} +\end{figure} + + \textbf{\textit{Proof.}} Iz $AB\cong AD$ in $\angle DAB=60^0$ + sledi, da je trikotnik $ABD$ pravilen (Figure \ref{sl.izo.6.4.6.pic}). + Podobno je pravilen tudi + trikotnik $BCD$. To pomeni, da $\mathcal{R}_{D,60^0}:A,B\mapsto + B,C$. Ker je $|BP|+|BQ|=|AB|$ in $|BP|+|AP|=|AB|$, je $AP\cong BQ$. + Toda rotacija $\mathcal{R}_{D,60^0}$ preslika poltrak $AB$ +v poltrak $BC$, zato zaradi pogoja $AP\cong BQ$ preslika tudi +točko $P$ v točko $Q$. Torej je $\mathcal{R}_{D,60^0}(P)=Q$, kar +pomeni, da je trikotnik $DPQ$ pravilen. + \kdokaz + +Ob naslednjem zgledu bomo ilustrirali uporabo rotacije pri +načrtovalnih nalogah. + + + + \bzgled + Let $A$ be an interior point of an angle $pOq$. Construct points + $B$ and $C$ on the sides $p$ and $q$ + such that $ABC$ is a regular triangle. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.7.pic} +\caption{} \label{sl.izo.6.4.7.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.4.7.pic}). +Naj bo $ABC$ takšen pravilni trikotnik, da njegovi oglišči $B$ in +$C$ ležita na krakih $p$ in $q$. Potem je +$\mathcal{R}_{A,60^0}(B)=C$. Točka $B$ leži na poltraku $p$, zato +njena slika - točka $C$ - leži na poltraku +$p'=\mathcal{R}_{A,60^0}(p)$. Oglišče $C$ lahko dobimo kot +presečišče poltrakov $p'$ in $q$, nato pa oglišče $B$ kot +$B=\mathcal{R}_{A,-60^0}(C)$. Če se poltraka $p'$ in $q$ ne +sekata, ni rešitve. Če pa ležita na isti premici, ima naloga + neskončno mnogo rešitev. + \kdokaz + +Če bi bil v prejšnji nalogi pogoj, da je trikotnik $ABC$ +enakokraki in pravokoten s pravim kotom pri oglišču $A$, bi +uporabili rotacijo s središčem $A$ za kot $45^0$. Podobno bi bilo, +če bi bilo potrebno načrtati kvadrat $PQRS$ s središčem v točki +$A$ pri pogoju $P\in p$, $Q\in q$ (Figure \ref{sl.izo.6.4.8.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.8.pic} +\caption{} \label{sl.izo.6.4.8.pic} +\end{figure} + + + + \bnaloga\footnote{1. IMO Romania - 1959, Problem 5.} + An arbitrary point $M$ is selected in the interior of the segment $AB$. The + squares $AMCD$ and $MBEF$ are constructed on the same side of $AB$, with + the segments $AM$ and $MB$ as their respective bases. The circles circumscribed + about these squares, with centres $P$ and $Q$, intersect at $M$ and also + at another point $N$. Let $N'$ denote the point of intersection of the straight + lines $AF$ and $BC$. + + (a) Prove that the points $N$ and $N'$ coincide. + + (b) Prove that the straight lines $MN$ pass through a fixed point $S$ independent + of the choice of $M$. + + (c) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between + $A$ and $B$ + \enaloga + + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.IMO1.pic} +\caption{} \label{sl.izo.6.4.IMO1.pic} +\end{figure} + + \textbf{\textit{Solution.}} Označimo s $k$ in $l$ + očrtani krožnici kvadratov + $AMCD$ in $MBEF$ + (Figure \ref{sl.izo.6.4.IMO1.pic}). + + (\textit{i}) Rotacijo $\mathcal{R}_{M,-90^0}$ preslika točki + $A$ in $F$ po vrsti v točki $C$ in $B$ oz. premico $AF$ v + premico $CB$. Po izreku \ref{rotacPremPremKot} + ti dve premici določata kot rotacije, + zato je $AF\perp CB$. Torej $\angle AN'C\cong\angle BN'F =90^0$, kar pomeni + (izrek + \ref{TalesovIzrKroz}), da točka $N'$ leži na obeh krožnicah + $k$ in $l$ + s premeroma $AC$ in $BF$. Zaradi tega je + $N'=N$. + + (\textit{ii}) Naj bo $j$ krožnica nad premerom $AB$ in $S$ središče + tistega loka (polkrožnice) $AB$, ki je na nasprotnem bregu premice + $AB$ glede na kvadrata $AMCD$ in $MBEF$. Točka $S$ ni odvisna od + izbire točke $M$. Dokažimo, da gredo vse premice $MN$ skozi točko $S$. + Ker je $\angle ANB=90^0$, po izreku \ref{TalesovIzrKroz} tudi točka $N$ leži + na krožnici $j$. Iz izreka \ref{ObodObodKot} + sledi $\angle ANM =\angle ADM=45^0$ in $\angle MNB =\angle + MEB=45^0$, zato je $NM$ simetrala kota $ANB$. Po izreku + \ref{TockaN} simetrala $NM$ poteka skozi točko $S$. + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.4.IMO1a.pic} +\caption{} \label{sl.izo.6.4.IMO1a.pic} +\end{figure} + + (\textit{iii}) Naj bo $O$ središče daljice $PQ$. + Označimo s $P'$, $Q'$ in $O'$ pravokotne projekcije točk + $P$, $Q$ in $O$ na premici $AB$ (Figure + \ref{sl.izo.6.4.IMO1a.pic}). Daljica $OO'$ je srednjica pravokotnega trapeza + $P'Q'QP$, zato je po izreku \ref{srednjTrapez}: + $$|OO'|=\frac{1}{2}\left(|PP'|+|QQ'| \right)= + \frac{1}{2}\left(\frac{1}{2}|AM|+\frac{1}{2}|MB| \right)=\frac{1}{4}|AB|.$$ + Torej je razdalja točke $O$ od premice $AB$ konstantna oziroma + neodvisna od izbire točke $M$. + + Naj bo $L$ središče kvadrata $ABGH$, ki je na istem bregu + premice $AB$ kot kvadrata $AMCD$ in $MBEF$. Z $O_A$ in $O_B$ + označimo središča daljic $LA$ in $LB$. Ker je $O_AO_B$ + srednjica trikotnika $ALB$ z osnovnico $AB$, je $O_AO_B \parallel + AB$. + Iz $|AO_A|=\frac{1}{2}|AL|=\frac{1}{4}|AG|$ pa sledi, da je + razdalja vzporednic $O_AO_B$ in $AB$ enaka $\frac{1}{4}|AB|$. Iz + že dokazanega dejstva $d(O,AB)=\frac{1}{4}|AB|$ sledi, da točka + $O$ leži na premici $O_AO_B$. Toda, ko se točka $M$ giblje po + notranjosti daljice $AB$, se točki $P$ in $Q$ gibljeta po + notranjosti daljic $AL$ oz. $BL$. To pomeni, da je točka $O$ v + notranjosti trikotnika $ALB$ oz. $O$ leži na odprti daljici + $O_AO_B$. + + Dokažimo še, da je poljubna točka $O$ daljice $O_AO_B$ + središče neke daljice $PQ$ za določeno izbiro kvadratov + $AMCD$ in $MBEF$ oz. točke $M$. Točko $M$ v tem primeru + dobimo iz pogoja $|AO'|=\frac{1}{2}|AM|+\frac{1}{4}|AB|$ oz. + $|AM|=2|AO'|-\frac{1}{2}|AB|$. Takšna točka $M$ vedno + obstaja, če je $\frac{1}{4}|AB|<|AO'|<\frac{3}{4}|AB|$, oz. + kadar točka $O$ leži na odprti daljici $O_AO_B$. + + Iskano geometrijsko mesto točk $M$ je torej daljica $O_AO_B$. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Half-Turn} \label{odd6SredZrc} + +Rotacijo za kot $180^0$ bomo obravnavali kot +posebno vrsto izometrij. + +Rotacijo $\mathcal{R}_{S,\omega}$ ravnine za kot $\omega=180^0$ +imenujmo \index{zrcaljenje!središčno}\pojem{središčno zrcaljenje} +ali \index{zrcaljenje!čez točko}\pojem{zrcaljenje čez točko} s +\index{središče!središčnega zrcaljenja}\pojem{središčem} $S$ in ga +označimo $\mathcal{S}_S$ (Figure \ref{sl.izo.6.5.1.pic}). Torej +$\mathcal{S}_S=\mathcal{R}_{S,180^0}$. + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.1.pic} +\caption{} \label{sl.izo.6.5.1.pic} +\end{figure} + +Iz definicije je jasno, da za poljubno točko $X\neq S$ velja +$\mathcal{S}_S(X)=X'$ natanko tedaj, ko je $S$ središče daljice +$XX'$. + +Direktno iz definicije sledi tudi naslednja trditev. + + \bizrek \label{izoSredZrcInv} + A half-turn is an involution, i.e. + $\mathcal{S}^2_S=\mathcal{E}$. \eizrek + + Ker je središčno zrcaljenje vrsta rotacije, ima tudi vse + lastnosti rotacije. Edina fiksna točka + središčnega zrcaljenja je torej središče tega zrcaljenja. + + Enako kot rotacijo lahko tudi središčno zrcaljenje + $\mathcal{S}_S$ + predstavimo kot kompozitum dveh osnih zrcaljenj. Kot, ki ga osi + določata, je enak polovici kota rotacije (izrek \ref{rotacKom2Zrc}) - + v našem primeru je + to $\frac{180^0}{2}=90^0$. To pomeni, da sta osi dveh + zrcaljenj v primeru središčnega zrcaljenja pravokotni. V tem primeru po + izreku \ref{izoZrcKomut} osni zrcaljenji komutirata + (Figure \ref{sl.izo.6.5.2.pic}). + Torej velja naslednji izrek. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.2.pic} +\caption{} \label{sl.izo.6.5.2.pic} +\end{figure} + + + + \bizrek \label{izoSrZrcKom2Zrc} + Any half-turn around a point $S$ + can be expressed as the product of two reflections + $\mathcal{S}_p$ and $\mathcal{S}_q$ where + $p$ and $q$ are arbitrary perpendicular lines intersecting at the point $S$.\\ + The reverse is also true - + the product of two reflections + $\mathcal{S}_p$ and $\mathcal{S}_q$ ($S=p\cap q$ and + $p\perp q$) + is the half-turn around the point $S$, i.e.\\ + $$\mathcal{S}_S=\mathcal{S}_q\circ\mathcal{S}_p= + \mathcal{S}_p\circ\mathcal{S}_q + \hspace*{1mm} \Leftrightarrow \hspace*{1mm} S=p\cap q \hspace*{1mm} + \wedge\hspace*{1mm} p\perp q.$$ + \eizrek + + + + \bizrek \label{izoKomp3SredZrc} + The product of three half-turns is a half-turn. + If the centres of these half-turns are three non-collinear points, + then the centre of the new half-turn is the fourth vertex of a parallelogram. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.3.pic} +\caption{} \label{sl.izo.6.5.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} + Naj bodo $\mathcal{S}_A$, $\mathcal{S}_B$ in $\mathcal{S}_C$ tri središčne +simetrije s središči $A$, $B$ in $C$ (Figure +\ref{sl.izo.6.5.3.pic}). Označimo s $p$ premico $AB$, z $a$, $b$ +in $c$ po vrsti pravokotnice premice $p$ skozi točke $A$, $B$ in +$C$ ter s $c'$ pravokotnico premice $c$ skozi točko $C$. Potem je +po izreku \ref{izoSrZrcKom2Zrc}: + $$\mathcal{S}_C \circ \mathcal{S}_B \circ \mathcal{S}_A= + \mathcal{S}_{c'} \circ \mathcal{S}_c \circ \mathcal{S}_b + \circ \mathcal{S}_p \circ \mathcal{S}_p \circ \mathcal{S}_a + = \mathcal{S}_{c'} \circ \mathcal{S}_c \circ \mathcal{S}_b \circ \mathcal{S}_a.$$ + Osi $a$, $b$ in $c$ pripadajo istem šopu vzporednic $\mathcal{X}_a$, +ker so vse pravokotne na premico $p$. Kompozitum +$\mathcal{S}_c \circ \mathcal{S}_b \circ \mathcal{S}_a$ torej +predstavlja osno zrcaljenje $S_d$, kjer os $d$ pripada istem šopu +$\mathcal{X}_a$ (izrek \ref{izoSop}). Potem sta tudi premici $c'$ +in $d$ pravokotni, zato je: + $$\mathcal{S}_C \circ \mathcal{S}_B \circ \mathcal{S}_A= + \mathcal{S}_{c'} \circ \mathcal{S}_d=\mathcal{S}_D,$$ +kjer je $D=d\cap c'$. Po izreku \ref{izoSoppqrt} imata para premic $a$, +$c$ in $b$, $d$ skupno somernico. To pomeni, da če +so $A$, $B$ in $C$ tri nekolinearne točke, je štirikotnik $ABCD$ +paralelogram. + \kdokaz + +Posledica dokazane trditve je naslednji izrek. + + + + + \bizrek \label{izoKomp2n+1SredZrc} + The product of an odd number of half-turns is a half-turn. + \eizrek + + \textbf{\textit{Proof.}} Dokaz bomo izpeljali z indukcijo po + številu $n\geq 1$, kjer je $m=2n+1$ (liho) število točk. + + Za $n=1$, oz. $m=3$, je trditev direktna posledica prejšnjega izreka + \ref{izoKomp3SredZrc}. + + Predpostavimo, da za vsak $k\in \mathbb{N}$ ($k\geq 1$) + in vsako $(2k+1)$-terico + komplanarnih točk + $A_1,A_2,\ldots ,A_{2k+1}$ kompozitum + $\mathcal{S}_{A_{2k+1}}\circ + \cdots \circ\mathcal{S}_{A_2}\circ\mathcal{S}_{A_1}$ predstavlja + neko središčno zrcaljenje $\mathcal{S}_A$ (indukcijska predpostavka). + + Potrebno je dokazati, da potem tudi za $k+1$ + in vsako $(2(k+1)+1)$-terico + komplanarnih točk + $A_1,A_2,\ldots ,A_{(2(k+1)+1)}$ kompozitum + $\mathcal{I}=\mathcal{S}_{A_{2(k+1)+1}}\circ + \cdots \circ\mathcal{S}_{A_2}\circ\mathcal{S}_{A_1}$ predstavlja + neko središčno zrcaljenje $\mathcal{S}_{A'}$. Torej: + + \begin{eqnarray*} + \mathcal{I}&=&\mathcal{S}_{A_{2(k+1)+1}}\circ + \cdots \circ\mathcal{S}_{A_2}\circ\mathcal{S}_{A_1}=\\ + &=&\mathcal{S}_{A_{2k+3}}\circ\mathcal{S}_{A_{2k+2}}\circ + \mathcal{S}_{A_{2k+1}}\circ + \cdots \circ\mathcal{S}_{A_2}\circ\mathcal{S}_{A_1}=\\ + &=& \mathcal{S}_{A_{2k+3}}\circ\mathcal{S}_{A_{2k+2}}\circ + \mathcal{S}_A= \hspace*{14mm} \textrm{ (po indukcijski predpostavki)}\\ + &=& \mathcal{S}_{A'} \hspace*{48mm} \textrm{ (po izreku \ref{izoKomp3SredZrc}),} + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + +Direktna posledica izreka \ref{izoTransmRotac}, ki se nanaša na rotacijo, je naslednja trditev ozirma izrek o transmutaciji središčnega zrcaljenja\index{transmutacija!središčnega zrcaljenja}. + + + \bizrek \label{izoTransmSredZrc} + For an arbitrary half-turn + $\mathcal{S}_{O}$ + and an arbitrary isometry $\mathcal{I}$ is + $$\mathcal{I}\circ + \mathcal{S}_{O}\circ\mathcal{I}^{-1}= + \mathcal{R}_{\mathcal{I}(O)}.$$ + \eizrek + + + Podobno kot osno zrcaljenje tudi središčno + zrcaljenje pogosto uporabljamo pri načrtovalnih + nalogah. + + \bzgled + Let $S$ be an interior point of an angle $pOq$. Construct a square + $ABCD$ with the centre $S$ such that the vertices + $A$ and $C$ lie on the sides $p$ and $q$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.4.pic} +\caption{} \label{sl.izo.6.5.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $ABCD$ kvadrat, ki izpolnjuje dane pogoje - + $S$ je njegovo središče, $A\in p$ in $C\in q$ (Figure \ref{sl.izo.6.5.4.pic}). + Ker je točka $S$ središče diagonale + $AC$, je $\mathcal{S}_S(A)=C$. Iz $A\in p$ sledi $C\in + \mathcal{S}_S(p)$. + Točko $C$ torej dobimo iz pogoja $C\in q\cap\mathcal{S}_S(p)$. + Na koncu je še $A=\mathcal{S}_S(C)$, $D=\mathcal{R}_{S,90^0}(C)$ in + $B=\mathcal{R}_{S,-90^0}(C)$. + \kdokaz + + \bzgled + Let $A$ be one of the two intersections of circles $k$ and $l$. Construct + a common secant of these two circles passing through the point $A$ and determine + a two congruent chord with these circles. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.5.pic} +\caption{} \label{sl.izo.6.5.5.pic} +\end{figure} + + \textbf{\textit{Solution.}} + Označimo z $B$ drugo presečišče krožnic $k$ in $l$ + (Figure \ref{sl.izo.6.5.5.pic}). Potem je ena rešitev naloge + premica $AB$, saj je daljica $AB$ + skupna tetiva obeh krožnic. + + Naj bo $s\ni A$ skupna sekanta krožnic $k$ in $l$, ki ju seka še + v točkah $K$ in $L$, tako da velja $AK\cong AL$. V primeru $K=L$ + je $K=L=B$, kar je že omenjena rešitev. Če je $K\neq L$, iz + pogoja $A, K, L\in s$ sledi, da je $A$ središče daljice $KL$. + Torej je $\mathcal{S}_A(K)=L$, kar pomeni, da točko $L$ dobimo + iz pogoja $L\in l\cap \mathcal{S}_A(k)$. Na koncu je + $K=\mathcal{S}_A(L)$. + \kdokaz + + + \bzgled + Construct a quadrilateral $ABCD$ with the sides that are congruent to + the four given line segments $a$, $b$, $c$, and $d$ and + the line segment defined by the midpoints of + the two opposite sides $AD$ and $BC$ that is congruent to the given line segment $l$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.6.pic} +\caption{} \label{sl.izo.6.5.6.pic} +\end{figure} + + \textbf{\textit{Solution.}} +Naj bosta $P$ in $Q$ središči stranic $AD\cong d$ in $BC\cong b$ +štirikotnika $ABCD$, tako da velja še $PQ\cong l$, $AB\cong a$ in +$CD\cong c$ (Figure \ref{sl.izo.6.5.6.pic}). Naj bo $S$ +središče diagonale $AC$. Trikotnik $PQS$ lahko konstruiramo, saj +je $PQ\cong l$, $QS=\frac{1}{2}a$ in $PS =\frac{1}{2}c$ (izrek +\ref{srednjicaTrik}). Točki $C$ in $A$ ležita po vrsti na +krožnicah $k_d=k(P,\frac{1}{2}d)$ in $k_b=k(Q,\frac{1}{2}b)$. Ker +je $C=\mathcal{S}_S(A)$, točka $C$ leži tudi na sliki $k'_d$ +krožnice $k_d$ pri središčnem zrcaljenju $\mathcal{S}_S$. +Točka $C$ je torej eno od presečišč krožnic $k'_d$ in $k_b$. Nato je še +$A=\mathcal{S}_S(C)$, $D=\mathcal{S}_P(A)$ in +$B=\mathcal{S}_Q(C)$. + \kdokaz + + + \bzgled + Let $P$, $Q$, $R$, $S$ and $T$ be points in the plane. Construct + a pentagon $ABCDE$ such that the points $P$, $Q$, $R$, $S$ and $T$ are + the midpoints of its sides $AB$, $BC$, $CD$, $DE$ and $EA$, respectively. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.7.pic} +\caption{} \label{sl.izo.6.5.7.pic} +\end{figure} + + \textbf{\textit{Solution.}} +Naj bo $ABCDE$ iskani petkotnik + (Figure \ref{sl.izo.6.5.7.pic}) in: + $$\mathcal{I} = \mathcal{S}_T \circ \mathcal{S}_S + \circ \mathcal{S}_R \circ \mathcal{S}_Q \circ \mathcal{S}_P.$$ + Iz izreka izreka \ref{izoKomp2n+1SredZrc} sledi, da je izometrija + $\mathcal{I}$ neko središčno +zrcaljenje $\mathcal{S}_O$. Ker je +$\mathcal{S}_O(A)=\mathcal{I}(A)=A$, je $O=A$ oz. $\mathcal{I}=\mathcal{S}_A$. Točko +$A$ torej lahko konstruiramo kot središče daljice $XX'$, kjer je +$X$ poljubna točka dane ravnine in $X'= \mathcal{I}(X)$. + \kdokaz + + + \bzgled + Let $A_iB_i$ ($i\in \{1,2,3\}$) + be parallel chords of a circle $k$. Suppose that $S_1$, $S_2$ and $S_3$ + are the midpoints of the line segments $A_2A_3$, $A_1A_3$ and $A_1A_2$, + respectively. Prove that + $C_i=\mathcal{S}_{S_i}(B_i)$ are three collinear points. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.5.4a.pic} +\caption{} \label{sl.izo.6.5.4a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $S$ središče krožnice $k$ +(Figure \ref{sl.izo.6.5.4a.pic}). Ker je $SA_i\cong SB_i$ ($i\in +\{1,2,3\}$), potekajo vse simetrale $s_i$ tetiv $A_iB_i$ ($i\in +\{1,2,3\}$) skozi točko $S$. Iz $s_i\perp A_iB_i$ in dejstva, da +so tetive medseboj vzporedne, sledi, da so vse simetrale $s_i$ +med seboj enake - označimo jih s $s$. Torej je $s$ skupna simetrala +tetiv $A_iB_i$, zato velja $\mathcal{S}_s(A_i)=B_i$ ($i\in +\{1,2,3\}$). + + Ker so $S_1$, $S_2$ in $S_3$ + središča daljic $C_1B_1$, $C_2B_2$ in $C_3B_3$ ter stranic $A_2A_3$, $A_1A_3$ in $A_1A_2$ + trikotnika $A_1A_2A_3$, je po izreku \ref{vektSestSplosno} +$\overrightarrow{S_1S_2}= +\frac{1}{2}(\overrightarrow{C_1C_2}+\overrightarrow{B_1B_2})$ oz. +$\overrightarrow{C_1C_2}= +2\overrightarrow{S_1S_2}-\overrightarrow{B_1B_2} +=\overrightarrow{A_2A_1}-\overrightarrow{B_1B_2}=\overrightarrow{A_2A_1} ++\overrightarrow{B_2B_1}$. + Ker se točki $A_2$ in $A_1$ s premico $s$ +prezrcalita v točki $B_2$ in $B_1$, je vektor +$\overrightarrow{C_1C_2}=\overrightarrow{A_2A_1} ++\overrightarrow{B_2B_1}$ kolinearen s premico $s$ (zgled +\ref{izoSimVekt}). Analogno je tudi vektor $\overrightarrow{C_2C_3}$ +kolinearen s premico $s$, zatorej točke $C_1$, $C_2$ in $C_3$ +ležijo na isti premici, ki je vzporedna s premico $s$. + \kdokaz + + + +%________________________________________________________________________________ + \poglavje{Translations} \label{odd6Transl} + + Do sedaj smo obravnavali izometrije, ki imajo vsaj eno fiksno točko. Od direktnih smo imeli identiteto in rotacijo, od indirektnih pa osno zrcaljenje. Sedaj bomo vpeljali novo vrsto izometrij, ki nimajo fiksnih točk. + + Naj bo $\overrightarrow{v}$ + poljubni vektor evklidske ravnine ($\overrightarrow{v}\neq \overrightarrow{0}$). Transformacija te ravnine, pri kateri se +točka $X$ preslika v takšno točko $X'$, da je $\overrightarrow{XX'}=\overrightarrow{v}$, imenujemo \index{translacija} \pojem{translacija} ali \index{vzporedni premik} \pojem{vzporedni premik} za +vektor $\overrightarrow{v}$ in jo označimo s $\mathcal{T}_{\overrightarrow{v}}$ (Figure \ref{sl.izo.6.6.1.pic}). Vektor $\overrightarrow{v}$ je\index{vektor!translacije} \pojem{vektor translacije}. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.1.pic} +\caption{} \label{sl.izo.6.6.1.pic} +\end{figure} + +Dokažimo prve osnovne lastnosti translacije. + + + + \bizrek \label{translEnaki} + Two translations are equal if and only + the vectors of these translations are equal, i.e. + $$\mathcal{T}_{\overrightarrow{v}}= + \mathcal{T}_{\overrightarrow{u}}\Leftrightarrow \overrightarrow{v}=\overrightarrow{u}.$$ + \eizrek + + \textbf{\textit{Proof.}} Del ($\Leftarrow$) je trivialen. Dokažimo del ($\Rightarrow$). Za poljubno točko $X$ je torej + $\mathcal{T}_{\overrightarrow{v}}(X)= + \mathcal{T}_{\overrightarrow{u}}(X)=X'$. V tem primeru je + $\overrightarrow{v}=\overrightarrow{u}=\overrightarrow{XX'}$. + \kdokaz + + + \bizrek + A translation has no fixed points. + \eizrek + +\textbf{\textit{Proof.}} Če je $X$ fiksna točka translacije $\mathcal{T}_{\overrightarrow{v}}$ oz. $\mathcal{T}_{\overrightarrow{v}}(X)=X$, je $\overrightarrow{v}=\overrightarrow{XX}=\overrightarrow{0}$, kar po definiciji translacije ni možno. +\kdokaz + + + \bizrek + The inverse transformation of a translation is a translation + with the opposite vector, i.e. + $\mathcal{T}^{-1}_{\overrightarrow{v}}= + \mathcal{T}_{-\overrightarrow{v}}$ + \eizrek + +\textbf{\textit{Proof.}} Naj bo $Y$ poljubna točka in $X$ takšna točka, da velja $\overrightarrow{XY}=\overrightarrow{v}$. Iz tega sledi $\mathcal{T}_{\overrightarrow{v}}(X)=Y$. Iz $\overrightarrow{YX}=-\overrightarrow{v}$ pa sledi $\mathcal{T}_{-\overrightarrow{v}}(Y)=X$. Ker je še $\mathcal{T}^{-1}_{\overrightarrow{v}}(Y)=X$, sledi $\mathcal{T}^{-1}_{\overrightarrow{v}}(Y)= + \mathcal{T}_{-\overrightarrow{v}}(Y)$. Ker to velja za poljubno točko $Y$, je $\mathcal{T}^{-1}_{\overrightarrow{v}}= + \mathcal{T}_{-\overrightarrow{v}}$. + \kdokaz + + \bizrek + Translations are isometries of the plane. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.2.pic} +\caption{} \label{sl.izo.6.6.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $\mathcal{T}_{\overrightarrow{v}}$ poljubna translacija. Iz definicije je jasno, da predstavlja bijektivno preslikavo. Potrebno je še dokazati, da za poljubni točki $X$ in $Y$ + ter njuni sliki $\mathcal{T}_{\overrightarrow{v}}:X,Y\mapsto X', Y'$ velja + $X'Y'\cong XY$ (Figure \ref{sl.izo.6.6.2.pic}). + Iz $\overrightarrow{XX'}=\overrightarrow{v}$ in $\overrightarrow{YY'}=\overrightarrow{v}$ sledi $\overrightarrow{XX'}=\overrightarrow{YY'}$. Po izreku \ref{vektABCD_ACBD} je potem tudi $\overrightarrow{XY}=\overrightarrow{X'Y'}$ oz. $X'Y'\cong XY$. + \kdokaz + +Sedaj bomo pokazali, da lahko vsako translacijo, podobno kot rotacijo, predstavimo kot +kompozitum dveh osnih zrcaljenj. + + + + \bizrek \label{translKom2Zrc} + Any translation $\mathcal{T}_{\overrightarrow{v}}$ + can be expressed as the product of two reflections + $\mathcal{S}_p$ and $\mathcal{S}_q$ where + $p$ and $q$ are arbitrary parallel lines with a common + perpendicular line ($P\in p$ and $Q\in q$) + such that $\overrightarrow{v} + =2\overrightarrow{PQ}$, i.e. + $$\mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_q\circ\mathcal{S}_p + \hspace*{1mm} \Leftrightarrow \hspace*{1mm} p\parallel q \hspace*{1mm} + \wedge\hspace*{1mm} \overrightarrow{v} + =2\overrightarrow{PQ}\hspace*{1mm} (PQ\perp p,\hspace*{1mm} + P\in p,\hspace*{1mm} Q\in q).$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.3.pic} +\caption{} \label{sl.izo.6.6.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.6.3.pic}) + + ($\Leftarrow$) Predpostavimo, da velja $p\parallel q$, $\overrightarrow{v}=2\overrightarrow{PQ}$, $PQ\perp p$, + $P\in p$ in $Q\in q$. + Potrebno je dokazati, da velja + $\mathcal{T}_{\overrightarrow{v}}(X)= + \mathcal{S}_q\circ\mathcal{S}_p(X)$ za poljubno točko $X$ te ravnine. + Naj bo $\mathcal{S}_q\circ\mathcal{S}_p(X)=X'$. Dokažimo, da je tudi $\mathcal{T}_{\overrightarrow{v}}(X)=X'$ oz. $\overrightarrow{XX'}= \overrightarrow{v}$. + Označimo +$\mathcal{S}_p(X)=X_1$. V tem primeru je jasno $\mathcal{S}_q(X_1)=X'$. Zaradi tega je najprej premica $XX'$ skupna pravokotnica +vzporednic $p$ in $q$, kar pomeni, da je vektor $\overrightarrow{XX'}$ vzporeden z vektorjem $\overrightarrow{v}$. Ne glede na lego točke $X$ je v +vsakem primeru dolžina daljice $XX'$ dvakrat večja od razdalje med +vzporednicama $p$ in $q$, ta pa je enaka dolžini daljice $PQ$. Torej je $\overrightarrow{XX'} =2\overrightarrow{PQ}= \overrightarrow{v}$. S tem smo dokazali, da je +$\mathcal{T}_{\overrightarrow{v}}(X)=X'$. + +($\Rightarrow$) Naj bo sedaj $\mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_q\circ\mathcal{S}_p$. +Premici $p$ in $q$ sta v tem primeru vzporedni, ker bi bila sicer skupna točka teh dveh premic fiksna tačka kompozituma $\mathcal{S}_q\circ\mathcal{S}_p$, translacija $\mathcal{T}_{\overrightarrow{v}}$ pa nima nobene fiksne točke. +Iz prvega dela dokaza ($\Leftarrow$) je kompozitum $\mathcal{S}_q\circ\mathcal{S}_p$ enak translaciji $\mathcal{T}_{\overrightarrow{v_1}}$ za vektor $\overrightarrow{v_1}$, kjer velja $\overrightarrow{v_1}=2\overrightarrow{PQ}$, $PQ\perp p$, + $P\in p$ in $Q\in q$. Toda iz +$\mathcal{T}_{\overrightarrow{v}}= +\mathcal{S}_q\circ\mathcal{S}_p=\mathcal{T}_{\overrightarrow{v_1}}$ + sledi $\overrightarrow{v}=\overrightarrow{v_1}$ (izrek \ref{translEnaki}) oz. $\overrightarrow{v}=2\overrightarrow{PQ}$. + \kdokaz + +Iz prejšnjega izreka sledi, da je translacija kot kompozitum +dveh indirektnih izometrij direktna izometrija. + + + \bizrek + A translation is a direct isometry. + \eizrek + +Dokazali bomo, da lahko translacijo predstavimo tudi kot +kompozicijo dveh središčnih zrcaljenj. + + \bizrek \label{transl2sred} + The product of two half-turns is a translation, such that + $$\mathcal{S}_Q\circ\mathcal{S}_P= + \mathcal{T}_{2\overrightarrow{PQ}}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.4.pic} +\caption{} \label{sl.izo.6.6.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $X$ poljubna točka, $\mathcal{S}_Q\circ\mathcal{S}_P(X)=X'$ in $\mathcal{S}_P(X)=X_1$ (Figure \ref{sl.izo.6.6.4.pic}). Potem je tudi $\mathcal{S}_Q(X_1)=X'$. Po definiciji središčnih zrcaljenj sta točki $P$ in $Q$ središči daljic $XX_1$ in $X_1X'$, kar pomeni, da je daljica $PQ$ srednjica trikotnika $XX_1X'$ za osnovnico $XX'$. Po izreku \ref{srednjicaTrikVekt} je + $\overrightarrow{XX'}=2\overrightarrow{PQ}$, zato je $\mathcal{T}_{2\overrightarrow{PQ}}(X)=X'= + \mathcal{S}_Q\circ\mathcal{S}_P(X)$. Ker to velja za vsako točko $X$, je $\mathcal{S}_Q\circ\mathcal{S}_P= + \mathcal{T}_{2\overrightarrow{PQ}}$. + \kdokaz + + + + \bizrek \label{translKomp} + The product of two translations is a translation for the vector + that is the sum of the vectors of these two translations, i.e. + $$\mathcal{T}_{\overrightarrow{u}}\circ + \mathcal{T}_{\overrightarrow{v}}= + \mathcal{T}_{\overrightarrow{v}+\overrightarrow{u}}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.5.pic} +\caption{} \label{sl.izo.6.6.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.6.5.pic}). + Naj bo $P$ poljubna točka, $Q$ takšna točka, da velja + $\overrightarrow{PQ} = \frac{1}{2}\overrightarrow{v}$, +in $R$ točka, za katero je +$\overrightarrow{QR} = \frac{1}{2}\overrightarrow{u}$. + Ker velja $\overrightarrow{v} + \overrightarrow{u} + = 2\overrightarrow{PQ} + 2\overrightarrow{QR} = 2\overrightarrow{PR}$, + po prejšnjem izreku \ref{transl2sred} sledi: + $$\mathcal{T}_{\overrightarrow{u}}\circ + \mathcal{T}_{\overrightarrow{v}}= + \mathcal{T}_{2\overrightarrow{QR}}\circ + \mathcal{T}_{2\overrightarrow{PQ}}= + \mathcal{S}_R\circ\mathcal{S}_Q\circ + \mathcal{S}_Q\circ\mathcal{S}_P= + \mathcal{S}_R\circ\mathcal{S}_P= + \mathcal{T}_{2\overrightarrow{PR}}= + \mathcal{T}_{\overrightarrow{v}+\overrightarrow{u}},$$ kar je bilo treba dokazati. \kdokaz + + Posledica dokazanega je naslednji izrek. + + \bizrek + The product of translations is commutative, i.e. + $$\mathcal{T}_{\overrightarrow{u}}\circ + \mathcal{T}_{\overrightarrow{v}}= + \mathcal{T}_{\overrightarrow{v}}\circ + \mathcal{T}_{\overrightarrow{u}}$$ + \eizrek + + \textbf{\textit{Proof.}} Po prejšnjem izreku \ref{translKomp} je: + $$\mathcal{T}_{\overrightarrow{u}}\circ + \mathcal{T}_{\overrightarrow{v}}= + \mathcal{T}_{\overrightarrow{v}+\overrightarrow{u}}= + \mathcal{T}_{\overrightarrow{u}+\overrightarrow{v}}= + \mathcal{T}_{\overrightarrow{v}}\circ + \mathcal{T}_{\overrightarrow{u}},$$ kar je bilo treba dokazati. \kdokaz + + +Podobno kot pri osnem zrcaljenju, rotaciji in središčnem zrcaljenju tudi za translacijo velja izrek o +transmutaciji\index{transmutacija!translacije}. + + + \bizrek \label{izoTransmTrans} + For an arbitrary translation + $\mathcal{T}_{2\overrightarrow{PQ}}$ + and an arbitrary isometry $\mathcal{I}$ is + $$\mathcal{I}\circ + \mathcal{T}_{2\overrightarrow{PQ}}\circ\mathcal{I}^{-1}= + \mathcal{T}_{2\overrightarrow{\mathcal{I}(P)\mathcal{I}(Q)}}.$$ + \eizrek + + \textbf{\textit{Proof.}} Po izreku \ref{transl2sred} lahko translacijo + $\mathcal{T}_{2\overrightarrow{PQ}}$ + zapišemo kot kompozitum dveh središčnih +zrcaljenj, in sicer +$\mathcal{T}_{2\overrightarrow{PQ}}=\mathcal{S}_Q\circ\mathcal{S}_P$. +Če uporabimo izrek o transmutaciji za središčne simetrije \ref{izoTransmSredZrc}, dobimo: + \begin{eqnarray*} + \mathcal{I}\circ + \mathcal{T}_{2\overrightarrow{PQ}}\circ\mathcal{I}^{-1}&=& + \mathcal{I}\circ\mathcal{S}_Q\circ\mathcal{S}_P\circ\mathcal{I}=\\ + &=& + \mathcal{I}\circ\mathcal{S}_Q\circ\mathcal{I}^{-1} + \circ\mathcal{I}\circ\mathcal{S}_P\circ\mathcal{I}^{-1}=\\ + &=&\mathcal{S}_{\mathcal{I}(Q)}\circ\mathcal{S}_{\mathcal{I}(P)}=\\ + &=&\mathcal{T}_{2\overrightarrow{\mathcal{I}(P)\mathcal{I}(Q)}}, + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + + + \bizrek \label{izoKompTranslRot} + The product of a translation and a rotation (also a rotation and a translation) + is a rotation with the same angle. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.5a.pic} +\caption{} \label{sl.izo.6.6.5a.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $\mathcal{I}=\mathcal{R}_{S,\omega}\circ + \mathcal{T}_{\overrightarrow{v}}$ kompozitum translacije + $\mathcal{T}_{\overrightarrow{v}}$ in rotacije + $\mathcal{R}_{S,\omega}$. Naj bo $b$ premica, ki poteka skozi točko + $S$ in je pravokotna z vektorjem $\overrightarrow{v}$ + (Figure \ref{sl.izo.6.6.5a.pic}). Po izreku + \ref{translKom2Zrc} za določeno premico $a$, $a\parallel b$, velja + $\mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_b\circ + \mathcal{S}_a$. Naj bo $c$ premica skozi točko $S$, za katero + je $\measuredangle b,c=\frac{1}{2}\omega$. Po izreku + \ref{rotacKom2Zrc} je $\mathcal{R}_{S,\omega}=\mathcal{S}_c\circ + \mathcal{S}_b$. Torej: + $$\mathcal{I}=\mathcal{R}_{S,\omega}\circ + \mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_c\circ + \mathcal{S}_b\circ\mathcal{S}_b\circ + \mathcal{S}_a=\mathcal{S}_c\circ + \mathcal{S}_a.$$ + Po Playfairovem aksiomu se premici $a$ in $c$ sekata v neki točki + $S_1$, po izreku \ref{KotiTransverzala} pa je + $\measuredangle a,c=\measuredangle b,c=\frac{1}{2}\omega$. + Če še enkrat uporabimo izrek \ref{rotacKom2Zrc}, dobimo: +$$\mathcal{R}_{S,\omega}\circ + \mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_c\circ + \mathcal{S}_a=\mathcal{R}_{S_1,\omega}.$$ + Na podoben način se dokaže, da je tudi kompozitum + rotacije in + translacije $\mathcal{T}_{\overrightarrow{v}} + \circ\mathcal{R}_{S,\omega}$ rotacija za isti kot. + \kdokaz + +V nadaljevanju bomo obravnavali uporabo translacije. Najprej bomo +videli, kako lahko tudi translacijo uporabimo pri načrtovalnih +nalogah. + + + \bzgled + Let $A$ and $B$ be interior points of an angle $pOq$. + Construct points $C$ and $D$ on the sides $p$ and $q$ + such that the quadrilateral $ABCD$ is a parallelogram. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.6.pic} +\caption{} \label{sl.izo.6.6.6.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.6.6.pic}) + Ker je $ABCD$ paralelogram, je $\overrightarrow{AB}= \overrightarrow{DC} = \overrightarrow{v}$. + Torej je $\mathcal{T}_{\overrightarrow{v}}(D)= C$. Iz danih +pogojev točka $D$ leži na poltraku $p$, zato njena slika $C$ pri +translaciji $\mathcal{T}_{\overrightarrow{v}}$ leži na sliki $p'$ +poltraka $p$ pri tej translaciji (sliko $p'$ lahko dobimo, če +narišemo sliko $O'$ vrha $O$, nato pa vzporedni poltrak poltraka $p$ +iz točke $O'$). Ker leži točka $C$ tudi na poltraku $q$, jo lahko +narišemo iz pogoja $C\in p'\cap q$. Število rešitev naloge je +odvisna od tega, ali imata poltraka $p'$ in $q$ kaj skupnih točk. Ker +poltraka $p$ in $q$ nista vzporedna (gre za kraka kota), poltraka +$p'$ in $q$ nimata več kot eno skupno točko. Torej ima naloga +eno ali nobene rešitve. + \kdokaz + + + + \bzgled + Let $AB$ be a chord of a circle $k$, $P$ and $Q$ points of this circle lying on the + same side of the line $AB$ and $d$ a line segment in the plane. + Construct a point $L$ on the circle + $k$ such that $XY\cong d$, where $X$ and $Y$ + are intersections of the lines $LP$ and $LQ$ with the chord $AB$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.7.pic} +\caption{} \label{sl.izo.6.6.7.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.6.7.pic}) +Čeprav točki $X$ in $Y$ nista znani, je znan vektor +$\overrightarrow{v}=\overrightarrow{XY}$, ki ima enako dolžino +kot daljica $d$ ter je vzporeden in enako usmerjen kot vektor +$\overrightarrow{AB}$. Tudi kot $\omega=\angle PLQ$ je znani kot, +ker je obodni kot za tetivo $PQ$ (izrek \ref{ObodKotGMT}). Naj bo +$P'=\mathcal{T}_{\overrightarrow{v}}(P)$. Ker je +$\overrightarrow{PP'}=\overrightarrow{v}=\overrightarrow{XY}$, je +štirikotnik $PP'YX$ paralelogram, zato sta kota $P'YQ$ in $PLQ$ +skladna (izrek \ref{KotaVzporKraki}). Točko $Y$ lahko torej +konstruiramo kot presečišče tetive $AB$ z ustreznim lokom za tetivo +$P'Q$ in obodnimm kotom $\omega$. Nato je +$X=\mathcal{T}^{-1}_{\overrightarrow{v}}(Y)$. + \kdokaz + + + + \bzgled \footnote{Predlog za MMO 1996. (SL 10.)} + Let $H$ be the orthocentre of a triangle $ABC$ and $P$ the point lying on + the circumcircle of this triangle different from its vertices. + $E$ is the foot of the altitude from the vertex $B$ of the triangle $ABC$. + Suppose that the quadrilaterals $PAQB$ and $PARC$ are parallelograms. The lines + $AQ$ and $HR$ intersect at a point $X$. Prove that $EX\parallel AP$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.6.8.pic} +\caption{} \label{sl.izo.6.6.8.pic} +\end{figure} + + \textbf{\textit{Solution.}} Označimo z $O$ središče očrtane + krožnice trikotnika $ABC$ in $\mathcal{T}_{\overrightarrow{PA}}$ + translacijo za vektor $\overrightarrow{PA}$ (Figure \ref{sl.izo.6.6.8.pic}). + + Ker sta $PAQB$ in $PARC$ paralelograma, je $\overrightarrow{BQ}=\overrightarrow{PA}=\overrightarrow{CR}$. + To pomeni, da translacija $\mathcal{T}_{\overrightarrow{PA}}$ preslika trikotnik $BPC$ + v trikotnik $QAR$, višinsko točko $H_1$ trikotnika $BPC$ pa v višinsko točko $H'_1$ + trikotnika $QAR$. Dokažimo, da je $H'_1=H$ oz. $H=\mathcal{T}_{\overrightarrow{PA}}(H_1)$ + oz. $\overrightarrow{H_1H}=\overrightarrow{PA}$. Če uporabimo Hamiltonov izrek \ref{Hamilton} + za trikotnika $ABC$ in $PBC$, ki imata skupno središče očrtane krožnice, dobimo + $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC} + =\overrightarrow{OH}$ in $\overrightarrow{OP}+\overrightarrow{OB}+\overrightarrow{OC} + =\overrightarrow{OH_1}$. Iz tega sledi najprej $\overrightarrow{OH}-\overrightarrow{OA}= + \overrightarrow{OH_1}-\overrightarrow{OP}$, nato pa $\overrightarrow{OH}-\overrightarrow{OH_1}= + \overrightarrow{OA}-\overrightarrow{OP}$ oz. $\overrightarrow{H_1H}=\overrightarrow{PA}$. + + Torej je $H$ višinska točka trikotnika $ARQ$, kar pomeni, da je $RH\perp AQ$ oz. $\angle AXH=90^0$. + Ker je še $\angle AEH=90^0$, točki $X$ in $E$ ležita na krožnici + s premerom $AH$ (izrek \ref{TalesovIzrKrozObrat}), zato je $AXEH$ tetivni štirikotnik. Torej: + \begin{eqnarray*} + \angle AXE&=&180^0-\angle AHE \hspace*{4mm} \textrm{(izrek \ref{TetivniPogoj})}\\ + &=&180^0-\angle ACB \hspace*{4mm} \textrm{(izrek \ref{KotaPravokKraki})}\\ + &=&180^0-\angle APB \hspace*{4mm} \textrm{(izrek \ref{ObodObodKot})}\\ + &=&\angle QAP \hspace*{4mm} \textrm{(po izreku \ref{paralelogram}, ker je }PAQB\textrm{ paralelogram)} + \end{eqnarray*} + + Po izreku \ref{KotiTransverzala} je $EX\parallel AP$. + \kdokaz + +%________________________________________________________________________________ + \poglavje{Composition of Two Rotations} \label{odd6KompRotac} + +V prejšnjih razdelkih smo že obravnavali kompozitume nekaterih preslikav. +Ugotovili smo, da kompozitum dveh osnih zrcaljenj predstavlja direktno +izometrijo, in sicer identiteto ali rotacijo ali translacijo, odvisno od +tega, ali sta premici enaki ali se sekata ali pa sta vzporedni. V prejšnjem +razdelku smo dokazali, da kompozitum dveh središčnih zrcaljenj in prav +tako kompozitum dveh translacij predstavlja translacijo. + +V tem razdelku bomo raziskovali kompozitum dveh rotacij. Začnimo z naslednjim izrekom. + + + \bizrek \label{rotacKomp2rotac} + Let $R_{A,\alpha}$ and $R_{B,\beta}$ be rotations of the plane + and $\mathcal{I}=R_{B,\beta}\circ R_{A,\alpha}$. + Then: + \begin{enumerate} + \item if $\alpha+\beta\notin\{k\cdot 360^0;k\in \mathbb{Z}\}$ and $A=B$, + the product $\mathcal{I}$ is a rotation for the angle $\alpha+\beta$ with the centre $A$, + \item if $\alpha+\beta\notin\{k\cdot 360^0;k\in \mathbb{Z}\}$ and $A\neq B$, + the product $\mathcal{I}$ is a rotation for the angle $\alpha+\beta$ + with the centre $C$,\\ where $\measuredangle CAB=\frac{1}{2}\alpha$ + and $\measuredangle ABC=\frac{1}{2}\beta$, + \item if $\alpha+\beta\in\{k\cdot 360^0;k\in \mathbb{Z}\}$ and $A=B$, + the product $\mathcal{I}$ is the identity map, + \item if $\alpha+\beta\in\{k\cdot 360^0;k\in \mathbb{Z}\}$ and $A\neq B$, + the product $\mathcal{I}$ is a translation. + \end{enumerate} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.1.pic} +\caption{} \label{sl.izo.6.7.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.7.1.pic}) + +Predpostavimo, da sta središči $A$ in $B$ različni (primer +$A=B$ je trivialen). + +Naj bosta $p$ in $q$ takšni premici te ravnine, da je +$\measuredangle p,AB=\frac{1}{2}\alpha$ in $\measuredangle +AB,q=\frac{1}{2}\beta$. Če vsako od rotacij razstavimo s pomočjo +osnih zrcaljenj (izrek \ref{rotacKom2Zrc}), dobimo: +$$\mathcal{I}=R_{B,\beta}\circ R_{A,\alpha}= +\mathcal{S}_q\circ\mathcal{S}_{AB}\circ\mathcal{S}_{AB}\circ +\mathcal{S}_p=\mathcal{S}_q\circ +\mathcal{S}_p.$$ + +Če je $\alpha+\beta\in\{0^0,360^0,-360^0\}$, +je $\frac{1}{2}\left(\alpha+\beta\right)\in\{0^0,180^0,-180^0\}$, +kar pomeni, da sta premici $p$ in $q$ vzporedni (izrek \ref{KotiTransverzala}). + V tem primeru je kompozitum $\mathcal{I}=\mathcal{S}_q\circ\mathcal{S}_p$ + translacija (izrek \ref{translKom2Zrc}). + +Če je $\alpha+\beta\notin\{0^0,360^0,-360^0\}$, +je $\frac{1}{2}\left(\alpha+\beta\right)\notin\{0^0,180^0,-180^0\}$ +ter se premici $p$ in $q$ sekata v neki točki $C$ (izrek \ref{KotiTransverzala}). +Kot $\measuredangle pCq$ je zunanji kot trikotnika $ABC$, zato je enak vsoti +kotov $\measuredangle p,AB$ in $\measuredangle AB,q$ (izrek \ref{zunanjiNotrNotr}). +Torej $\measuredangle pCq=\frac{1}{2}\left(\alpha+\beta\right)$. +Kompozitum +$\mathcal{I}=\mathcal{S}_q\circ\mathcal{S}_p$ v tem primeru predstavlja +rotacijo (izrek \ref{rotacKom2Zrc}) $\mathcal{R}_{C,2\measuredangle pCq}=\mathcal{R}_{C,\alpha+\beta}$. + \kdokaz + + V nadaljevanju bomo videli uporabo prejšnjega izreka v različnih nalogah. + + + \bzgled + Let $P$, $Q$ and $R$ be non-collinear points. + Construct a triangle $ABC$ such that $ARB$, $BPC$ and $CQA$ are regular triangles + with the same orientation. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.2.pic} +\caption{} \label{sl.izo.6.7.2.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.7.2.pic}) + + Če so $A$, $B$ in $C$ točke, ki izpolnjujejo dane +pogoje, velja: +$$\mathcal{R}_{Q,60^0} \circ \mathcal{R}_{R,60^0} \circ \mathcal{R}_{P,60^0}(C)=C.$$ +Toda po prejšnjem izreku \ref{rotacKomp2rotac} je +kompozitum $\mathcal{R}_{Q,60^0} \circ \mathcal{R}_{R,60^0} \circ \mathcal{R}_{P,60^0}$ +središčna simetrija (ker je $60^0+60^0+60^0=180^0$) s fiksno +točko $C$, zato je: +$$\mathcal{R}_{Q,60^0} \circ \mathcal{R}_{R,60^0} \circ \mathcal{R}_{P,60^0}=\mathcal{S}_C.$$ +Oglišče $C$ lahko torej konstruiramo kot središče daljice $XX'$, +kjer je $X'$ slika poljubne točke $X$ pri kompozitumu +$\mathcal{R}_{Q,60^0} \circ \mathcal{R}_{R,60^0} \circ +\mathcal{R}_{P,60^0}$. Nato sta še $B=\mathcal{R}_{P,60^0}(C)$ in +$A=\mathcal{R}_{R,60^0}(B)$. + \kdokaz + + + + \bzgled \label{rotKompZlato} + Let $BALK$ and $ACPQ$ be squares with the same orientation and + $Z$ the midpoint of the line segment $PK$. + Prove that $BZC$ is an isosceles right triangle with the hypotenuse $BC$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.3.pic} +\caption{} \label{sl.izo.6.7.3.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.7.3.pic}) + + Naj bo $\mathcal{I}=\mathcal{R}_{B,90^0}\circ \mathcal{R}_{C,90^0}$. Ker je +$90°+90°= 180^0$, po izreku \ref{rotacKomp2rotac} +izometrija $\mathcal{I}$ predstavlja središčno zrcaljenje $\mathcal{S}_Y$, +kjer velja $\measuredangle YCB=\frac{1}{2}\cdot 90^0=45^0$ in $\measuredangle CBY=\frac{1}{2}\cdot 90^0=45^0$. +Torej je točka $Y$ oglišče enakokrakega pravokotnega +trikotnika $BYC$ s hipotenuzo $BC$. Ker pa velja še +$\mathcal{S}_Y(P)=\mathcal{I}(P)=K$, je $Y=Z$. + \kdokaz + + + + + \bzgled \label{rotZgl1} + On each side of a quadrilateral $ABCD$ squares + $BALK$, $CBMN$, $DCSR$ and $ADPQ$ + are externally erected. + Let $E$, $F$, $G$ and $H$ be the midpoints of the line segments + $PK$, $MR$, $LN$ and $SQ$. + Prove that the quadrilaterals $BFDE$ and $AGCH$ are also squares. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.4.pic} +\caption{} \label{sl.izo.6.7.4.pic} +\end{figure} + + \textbf{\textit{Solution.}} Po prejšnji trditvi \ref{rotKompZlato} + sta trikotnika $DEB$ in $BFD$ enakokraka in pravokotna s skupno + hipotenuzo $BD$ (Figure \ref{sl.izo.6.7.4.pic}). Iz tega sledi, da je štirikotnik + $BFDE$ kvadrat. Na enak način dobimo, da je tudi štirikotnik $AGCH$ kvadrat. + \kdokaz + + + + \bzgled \label{rotZgl2} + Let $A_1$, $B_1$ and $C_1$ be the centres of squares + externally erected on the sides $BC$, $AC$ and $AB$ of an arbitrary + triangle $ABC$ + and $P$ the midpoint of the side $AC$. Prove that: + \begin{enumerate} + \item $C_1PA_1$ is an isosceles right triangle, + \item $C_1B_1$ and $A_1A$ are perpendicular and congruent line segments, + \item the line segments $AA_1$, $BB_1$ and $CC_1$ intersect at + a single point\footnote{To je drugi od treh planimetričnih + izrekov, ki jih je objavil italijanski matematik + \index{Bellavitis, G.} \textit{G. + Bellavitis} (1803--1880) na kongresu v Milanu 1844.}. + \end{enumerate} + \ezgled + + +\textbf{\textit{Solution.}} + +\textit{1)} Naj bo $\mathcal{I}= \mathcal{R}_{C_1,90^0} +\mathcal{R}_{A_1,90^0}$ (Figure \ref{sl.izo.6.7.5.pic}). Ker je +$90^0+90^0= 180^0$, izometrija $\mathcal{I}$ predstavlja središčno +zrcaljenje $\mathcal{S}_Y$, kjer je točka $Y$ oglišče enakokrakega +pravokotnega trikotnika $C_1YA_1$ s hipotenuzo $C_1A_1$ (izrek +\ref{rotacKomp2rotac}). Ker je še +$\mathcal{S}_Y(C)=\mathcal{I}(C)=A$, sledi $Y=P$. + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.5.pic} +\caption{} \label{sl.izo.6.7.5.pic} +\end{figure} + + + +\textit{2)} Z rotacijo $\mathcal{R}_{P,90^0}$ se daljica $B_1C_1$ +preslika v daljico $AA_1$ (Figure \ref{sl.izo.6.7.5a.pic}). Zato sta +daljici skladni in pravokotni (izrek \ref{rotacPremPremKot}). + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.5a.pic} +\caption{} \label{sl.izo.6.7.5a.pic} +\end{figure} + +\textit{3)} Iz dokazanega dela \textit{(2)} sledi, da so premice $AA_1$, $BB_1$, $CC_1$ +nosilke višin trikotnika $A_1B_1C_1$ (Figure \ref{sl.izo.6.7.5a.pic}), zato se +sekajo v njegovi višinski točki (izrek \ref{VisinskaTocka}). + \kdokaz + + + + \bzgled \label{rotZgl3} + Let $ABCD$ and $AB_1C_1D_1$ be squares with the same orientation, + $P$ and $Q$ the midpoints of the line segments $BD_1$ and $B_1D$. + Suppose that $O$ and $S$ are the centres of these squares. + Prove that the quadrilateral $POQS$ is also a square. + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.6.pic} +\caption{} \label{sl.izo.6.7.6.pic} +\end{figure} + +\textbf{\textit{Solution.}} + Trditev je očitna, če dvakrat uporabimo del \textit{(1)} + iz prejšnjega zgleda \ref{rotZgl2} (Figure \ref{sl.izo.6.7.6.pic}). + \kdokaz + + + + Če povežemo dejstva iz zgledov \ref{rotZgl1} in \ref{rotZgl3}, + dobimo naslednjo trditev (Figure \ref{sl.izo.6.7.4a.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.4a.pic} +\caption{} \label{sl.izo.6.7.4a.pic} +\end{figure} + + + + \bzgled \label{rotZgl4} + Let $BALK$, $CBMN$, $DCSR$ and $ADPQ$ be the + squares externally erected on the four sides of an arbitrary quadrilateral $ABCD$. + Then all quadrilaterals defined by the following vertices are also squares: + \begin{enumerate} + \item the point $B$, the midpoint of the line segment $MR$, the point $D$ and the midpoint of the line segment $PK$, + \item the point $A$, the midpoint of the line segment $LN$, the point $C$ and the midpoint of the line segment $SQ$, + \item the midpoints of the line segments $QL$, $LB$, $BD$ and $DQ$, + \item the midpoints of the line segments $KM$, $MC$, $CA$ and $AK$, + \item the midpoints of the line segments $NS$, $SD$, $DB$ and $BN$, + \item the midpoints of the line segments $RP$, $PA$, $AC$ and $CR$. + \end{enumerate} + + \ezgled + + V naslednjem primeru bomo uporabljali kompozitum rotacij v situaciji, ko je vsota kotov teh rotacij enaka $360^0$. + + + \bzgled \label{rotZgl5} + Let $ABC$ and $A'B'C'$ be isosceles triangles of the same orientation with the bases $BC$ and + $B'C'$ and $\angle BAC\cong\angle B'A'C'=\alpha$. + Suppose that $A_0$, $B_0$ and $C_0$ are the midpoints of the line segments $AA'$, $BB'$ and $CC'$. + Prove that $A_0B_0C_0$ is also an isosceles triangle + with the base $B_0C_0$ and $\angle B_0A_0C_0\cong\alpha$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.7.pic} +\caption{} \label{sl.izo.6.7.7.pic} +\end{figure} + + \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.7.7.pic}). + +Naj bo $\mathcal{I}= \mathcal{S}_{C_0} +\circ \mathcal{R}_{A',\alpha} \circ \mathcal{S}_{B_0} \circ +\mathcal{R}_{A,-\alpha}$. Kompozitum $\mathcal{I}$ je direktna +izometrija. Ker je vsota ustreznih +kotov rotacij enak $360^0$ in $\mathcal{I}(C)=C$, mora biti +$\mathcal{I}=\mathcal{E}$ (izrek \ref{rotacKomp2rotac}). Če uporabimo izrek o +transmutaciji rotacije \ref{izoTransmRotac}, dobimo + $\mathcal{S}_{A_0} \circ\mathcal{R}_{A,\alpha} \circ \mathcal{S}_{A_0} =\mathcal{R}_{A',\alpha}$. + Iz dokazanega in iz izreka \ref{transl2sred} sledi: + \begin{eqnarray*} + \mathcal{E}&=&\mathcal{I}= \mathcal{S}_{C_0} + \circ \mathcal{R}_{A',\alpha} \circ \mathcal{S}_{B_0} + \circ \mathcal{R}_{A,-\alpha}=\\ + &=& \mathcal{S}_{C_0} \circ (\mathcal{S}_{A_0} \circ\mathcal{R}_{A,\alpha} + \circ \mathcal{S}_{A_0}) \circ \mathcal{S}_{B_0} \circ \mathcal{R}_{A,-\alpha}=\\ + &=& (\mathcal{S}_{C_0} \circ \mathcal{S}_{A_0}) \circ\mathcal{R}_{A,\alpha} + \circ (\mathcal{S}_{A_0} \circ \mathcal{S}_{B_0}) \circ \mathcal{R}_{A,-\alpha}=\\ + &=& \mathcal{T}_{2\overrightarrow{A_0C_0}} \circ\mathcal{R}_{A,\alpha} + \circ \mathcal{T}_{2\overrightarrow{B_0A_0}} \circ \mathcal{R}_{A,-\alpha} + \end{eqnarray*} + Torej velja $\mathcal{E}=\mathcal{T}_{2\overrightarrow{A_0C_0}} + \circ\mathcal{R}_{A,\alpha} \circ \mathcal{T}_{2\overrightarrow{B_0A_0}} + \circ \mathcal{R}_{A,-\alpha}$, oz. $\mathcal{T}_{2\overrightarrow{C_0A_0}}= + \mathcal{R}_{A,\alpha} \circ \mathcal{T}_{2\overrightarrow{B_0A_0}} \circ + \mathcal{R}_{A,-\alpha}$. Če uporabimo izrek o transmutaciji translacije + \ref{izoTransmTrans}, dobimo: + $$\mathcal{T}_{2\overrightarrow{C_0A_0}}= \mathcal{R}_{A,\alpha} + \circ \mathcal{T}_{2\overrightarrow{B_0A_0}} \circ + \mathcal{R}_{A,-\alpha}=\mathcal{T}_{2\overrightarrow{B'_0A'_0}},$$ + kjer je $\mathcal{R}_{A,\alpha}:A_0, B_0\mapsto A'_0, B'_0$. Torej + $\overrightarrow{C_0A_0}=\overrightarrow{B'_0A_0'}$ oz. + $\overrightarrow{A_0C_0}=\overrightarrow{A'_0B'_0}$. + Vektor $\overrightarrow{A_0B_0}$ se z rotacijo $\mathcal{R}_{A,\alpha}$ + preslika v vektor $\overrightarrow{A'_0B'_0}=\overrightarrow{A_0C_0}$, + zato je $|A_0B_0|=|A_0C_0|$ in $\measuredangle B_0A_0C_0=\measuredangle\overrightarrow{A_0B_0}, + \overrightarrow{A_0C_0}=\alpha$. + \kdokaz + + + \bzgled \label{RotacZglVeck} + Let $A_1A_2...A_n$ and $B_1B_2...B_n$ be regular + $n$-gons with the same orientation. Suppose that + $S_1$, $S_2$, ..., $S_n$ are the midpoints of the line segments + $A_1B_1$, $A_2B_2$, ..., $A_nB_n$. Prove that $S_1S_2...S_n$ is also a regular $n$-gon. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.7.8.pic} +\caption{} \label{sl.izo.6.7.8.pic} +\end{figure} + + \textbf{\textit{Solution.}} +Ker sta $n$-kotnika $A_1A_2...A_n$ in $B_1B_2...B_n$ +pravilna in enako orientirana, sta $A_1A_2A_3$ in $B_1B_2B_3$ +enakokraka in enako orientirana +trikotnika z osnovnicama $A_1A_3$ in $B_1B_3$ (Figure \ref{sl.izo.6.7.8.pic}). +Pri tem je še $\angle A_1A_2A_3\cong \angle B_1B_2B_3=\frac{(n - 2)\cdot 180^0}{n}$ (izrek \ref{pravVeckNotrKot}). + +Iz prejšnjega zgleda \ref{rotZgl5} je +tudi $S_1S_2S_3$ enakokraki trikotnik in velja +$\angle S_1S_2S_3=\frac{(n - 2)\cdot 180^0}{n}$. +Torej sta stranici $S_1S_2$ in $S_2S_3$ skladni, notranji kot pri +oglišču $S_2$ pa skladen s kotom pravilnega $n$-kotnika. +Analogno dokažemo, da so vse +stranice večkotnika $S_1S_2...S_n$ skladne in vsi njegovi +notranji koti skladni, kar pomeni, da je ta večkotnik pravilen. + \kdokaz + +Eno zanimivo posledico izreka o kompozitumu rotacij pa bomo +obravnavali še v razdelku \ref{odd7Napoleon}. + + +%________________________________________________________________________________ + \poglavje{Glide Reflections} \label{odd6ZrcDrs} + +Do sedaj smo spoznali nekaj vrst izometrij - tri direktne izometrije +(identiteta, rotacija in translacija) ter eno indirektno izometrijo (osno zrcaljenje). +Sedaj bomo definirali še eno indirektno izometrijo, ki nima fiksnih točk. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.8.1.pic} +\caption{} \label{sl.izo.6.8.1.pic} +\end{figure} + +Naj bo $\mathcal{T}_{\overrightarrow{v}}$ translacija + za vektor $\overrightarrow{v} = 2\overrightarrow{PQ}$ +in $\mathcal{S}_{PQ}$ zrcaljenje čez premico $PQ$. Kompozitum +$\mathcal{S}_{PQ}\circ \mathcal{T}_{\overrightarrow{v}}$ se imenuje +\index{zrcalni zdrs}\pojem{zrcalni zdrs} z osjo $PQ$ za vektor $\overrightarrow{v} = 2\overrightarrow{PQ}$ + (Figure \ref{sl.izo.6.8.1.pic}). Označimo ga z +$\mathcal{G}_{2\overrightarrow{PQ}}$. + + + Že iz definicije je jasno, da je zrcalni zdrs določen s svojo osjo in vektorjem. + + Dokažimo osnovne lastnosti zrcalnega zdrsa. + + + + \bizrek \label{IzoZrcDrs1} + Any glide reflection can be expressed as the product of three reflections.\\ + The reflection and translation in the product, as the presentation of glide reflection, commute, i.e. + $$\mathcal{G}_{2\overrightarrow{PQ}}= + \mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}= + \mathcal{T}_{2\overrightarrow{PQ}}\circ \mathcal{S}_{PQ}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.8.2.pic} +\caption{} \label{sl.izo.6.8.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $\mathcal{G}_{2\overrightarrow{PQ}}= + \mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}$ z osjo $PQ$ + in vektorjem $2\overrightarrow{PQ}$. Naj bosta $p$ in $q$ pravokotnici + premice $PQ$ v točkah $P$ in $Q$ (Figure \ref{sl.izo.6.8.2.pic}). + Ker je $p,q\perp PQ$ in $p\parallel q$, lahko translacijo + $\mathcal{T}_{2\overrightarrow{PQ}}$ predstavimo kot + kompozitum $\mathcal{S}_q\circ\mathcal{S}_p$. Pri tem zrcaljenje $\mathcal{S}_{PQ}$ + komutira z zrcaljenji $\mathcal{S}_p$ in $\mathcal{S}_q$ (izrek \ref{izoZrcKomut}). Torej: + $$\mathcal{G}_{2\overrightarrow{PQ}}= + \mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}= + \mathcal{S}_{PQ}\circ\mathcal{S}_q\circ\mathcal{S}_p= + \mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_{PQ}= + \mathcal{T}_{2\overrightarrow{PQ}}\circ\mathcal{S}_{PQ},$$ kar je bilo treba dokazati. \kdokaz + + + \bizrek + \label{izoZrcdrsZrcdrs} + Product of a glide reflection with itself is a translation (Figure \ref{sl.izo.6.8.3.pic}). + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.8.3.pic} +\caption{} \label{sl.izo.6.8.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} Uporabimo prejšnji izrek \ref{IzoZrcDrs1} + in izrek \ref{translKomp}: + $$\mathcal{G}^2_{2\overrightarrow{PQ}}= + \mathcal{T}_{2\overrightarrow{PQ}}\circ \mathcal{S}_{PQ}\circ + \mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}= + \mathcal{T}^2_{2\overrightarrow{PQ}}= + \mathcal{T}_{4\overrightarrow{PQ}},$$ kar je bilo treba dokazati. \kdokaz + + Iz dokaza izreka \ref{IzoZrcDrs1} sledi, da lahko vsak zrcalni zdrs predstavimo + kot kompozitum treh osnih +zrcaljenj, kjer je os enega zrcaljenja pravokotna na osi drugih dveh. +Zrcalni zdrs je torej indirektna +izometrija kot kompozitum treh +osnih zrcaljenj - indirektnih izometrij. Isto dejstvo sledi tudi iz +same definicije zdrsa, ker je ta kompozitum ene direktne in ene indirektne izometrije. + + \bizrek + A glide reflection has no fixed points. + \eizrek + +\textbf{\textit{Proof.}} Po izreku \ref{IzoZrcDrs1} lahko zrcalni zdrs + predstavimo kot kompozitum treh osnih zrcaljenj: +$$\mathcal{G}_{2\overrightarrow{PQ}}= + \mathcal{S}_{PQ}\circ\mathcal{S}_q\circ\mathcal{S}_p,$$ + kjer sta $p$ in $q$ pravokotnici premice $PQ$ v točkah $P$ in $Q$. + Če bi zrcalni zdrs $\mathcal{G}_{2\overrightarrow{PQ}}$ oz. + kompozitum $\mathcal{S}_{PQ}\circ\mathcal{S}_q\circ\mathcal{S}_p$ + imel fiksno točko, bi $\mathcal{S}_{PQ}\circ\mathcal{S}_q\circ\mathcal{S}_p$ + kot indirektna izometrija predstavljal osno zrcaljenje (izrek \ref{izo1ftIndZrc}). + Po izreku \ref{izoSop} bi v tem primeru premice $p$, $q$ in $PQ$ pripadale istemu + šopu, kar pa ni mogoče. Zrcalni zdrs $\mathcal{G}_{2\overrightarrow{PQ}}$ torej nima fiksnih točk. + \kdokaz + +Uporabili smo že dejstvo iz izreka \ref{IzoZrcDrs1}, +da lahko zrcalni zdrs vedno predstavimo kot kompozitum treh osnih +zrcaljenj, kjer osi teh zrcaljenj niso v istem šopu. Ali velja obratno +- da je kompozitum treh osnih zrcaljenj, katerih osi niso iz istega šopa, vedno zrcalni zdrs? +S tem v zvezi bo naslednji izrek, ki je zelo pomemben tudi za klasifikacijo + izometrij, ki jo bomo izpeljali v naslednjem razdelku. + + + \bizrek \label{izoZrcdrsprq} + If lines $p$, $q$ and $r$ in the plane + are not from the same family of lines, then product + $\mathcal{S}_p \circ + \mathcal{S}_q\circ \mathcal{S}_r$ is a glide reflection. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.8.4.pic} +\caption{} \label{sl.izo.6.8.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.8.4.pic}) + Premica $q$ seka bodisi premico $p$ bodisi premico +$r$, ker bi sicer vse tri premice pripadale enemu šopu vzporednih +premic. Brez škode za splošnost naj bo $q\cap r=\{A\}$. Premice $p$, +$q$ in $r$ niso iz istega šopa, zato $A\notin p$. Naj bo $s$ +premica, ki poteka skozi točko $A$ in je pravokotna na premico $p$ v +točki $B$. Premice $s$, $q$ in $r$ pripadajo istemu šopu +$\mathcal{X}_A$, zato je po izreku \ref{izoSop} $\mathcal{S}_r \circ +\mathcal{S}_q \circ \mathcal{S}_ s = \mathcal{S}_t$, kjer je tudi +$t\in \mathcal{X}_A$. Iz tega sledi $\mathcal{S}_r \circ +\mathcal{S}_q = \mathcal{S}_t \circ \mathcal{S}_s$ oziroma (če +uporabimo še izrek \ref{izoSrZrcKom2Zrc}): +$$\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p = \mathcal{S}_t +\circ \mathcal{S}_s \circ \mathcal{S}_p = \mathcal{S}_t \circ \mathcal{S}_B.$$ +Označimo s $C$ pravokotno projekcijo točke $B$ na premici $t$ +(pri tem je $B\neq C$, ker bi se v nasprotnem primeru premici $s$ +in $t$, posledično pa tudi $r$ in $q$ prekrivali) in z $b$ premico, ki je v točki $B$ pravokotna na premico $BC$. +Potem je po izrekih \ref{izoSrZrcKom2Zrc} in \ref{translKom2Zrc}: +$$\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p = \mathcal{S}_t \circ \mathcal{S}_B= +\mathcal{S}_t \circ \mathcal{S}_b \circ \mathcal{S}_{BC}= +\mathcal{T}_{2\overrightarrow{BC}} \circ \mathcal{S}_{BC}= +\mathcal{G}_{2\overrightarrow{BC}},$$ kar je bilo treba dokazati. \kdokaz + + \bzgled \label{izoZrcDrsKompSrOsn} + Any glide reflection can be expressed as the product of a reflection and a half-turn, specifically + $$\mathcal{G}_{2\overrightarrow{PQ}}=\mathcal{S}_q\circ \mathcal{S}_P,$$ + where $q$ perpendicular to the line $PQ$ in the point $Q$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.8.5.pic} +\caption{} \label{sl.izo.6.8.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.8.5.pic}) + + Po definiciji je + $\mathcal{G}_{2\overrightarrow{PQ}}=\mathcal{S}_{PQ} + \circ \mathcal{T}_{2\overrightarrow{PQ}}$. Če + uporabimo izreka \ref{transl2sred} in \ref{izoSrZrcKom2Zrc}, je: + $$\mathcal{G}_{2\overrightarrow{PQ}}=\mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}= + \mathcal{S}_{PQ}\circ\mathcal{S}_Q\circ\mathcal{S}_P= + \mathcal{S}_{PQ}\circ\mathcal{S}_{PQ}\circ + \mathcal{S}_q\circ\mathcal{S}_P= + \mathcal{S}_q\circ\mathcal{S}_P,$$ kar je bilo treba dokazati. \kdokaz + + +%________________________________________________________________________________ + \poglavje{Classification of Plane Isometries. Chasles' Theorem} \label{odd6KlasifIzo} + +V prejšnjih razdelkih smo raziskovali določene vrste izometrij. +Postavi se vprašanje, ali so to edine izometrije ali +pa obstaja še kakšna druga vrsta izometrij, ki jo še nismo +obravnavali. V tem razdelku bomo dokazali, da je odgovor na to +vprašanje negativen in naredili končno klasifikacijo vseh +izometrij ravnine. + +Najprej bomo obravnavali direktne izometrije. Doslej smo omenili +identiteto, rotacijo in translacijo. Dokazali bomo, da so to edine +vrste direktnih izometrij. Spomnimo se najprej, da lahko vsako od +teh izometrij predstavimo kot kompozitum dveh zrcaljenj čez +premico. Odvisno od tega, ali se osi prekrivata, sekata ali sta +vzporedni, smo dobili identiteto, rotacijo in translacijo. To +so hkrati vse možnosti medsebojne lege dveh premic v +ravnini. Ali se lahko morda vse direktne izometrije predstavijo +kot kompozitum dveh zrcaljenj čez premico? Potemtakem bi bile tri +omenjene direktne izometrije res edine. To idejo bomo +uporabili v naslednjem izreku. + + + \bizrek \label{Chaslesov+} + Any direct isometry can be expressed as the product of two reflections. + The only direct isometries are + identity map, rotations and + translations. + \eizrek + + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.10.1.pic}) + + Naj bo $\mathcal{I} : \mathbb{E}^2\rightarrow \mathbb{E}^2$ + direktna izometrija ravnine. Dokaz bomo izpeljali po številu +fiksnih točk. + +\textit{1)} Če ima direktna izometrija $\mathcal{I}$ vsaj dve fiksni točki, +je po izreku \ref{izo2ftIdent} identiteta. Lahko jo +predstavimo kot kompozitum $\mathcal{I} = \mathcal{S}_p \circ +\mathcal{S}_p$ (izrek \ref{izoZrcPrInvol}) za poljubno premico +$p$. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.10.1.pic} +\caption{} \label{sl.izo.6.10.1.pic} +\end{figure} + + \textit{2)} Predpostavimo, da ima izometrija +$\mathcal{I}$ natanko eno fiksno točko $S$. Naj bo $p$ +poljubna premica, ki poteka skozi + točko $S$. Kompozitum $\mathcal{S}_p\circ \mathcal{I}$ + je indirektna izometrija s fiksno +točko $S$, zato po izreku \ref{izo1ftIndZrc} predstavlja osno +zrcaljenje - denimo $\mathcal{S}_q$, kjer os $q$ poteka skozi točko +$S$. Torej velja $\mathcal{S}_p\circ \mathcal{I} = \mathcal{S}_q$ +oz. $\mathcal{I} +=\mathcal{S}_p^{-1}\circ\mathcal{S}_q=\mathcal{S}_p\circ\mathcal{S}_q$ +(izrek \ref{izoZrcPrInvol}). Premici $p$ in $q$ se sekata v točki +$S$ (iz $p=q$ sledi $\mathcal{I}=\mathcal{S}_p \circ +\mathcal{S}_p=\mathcal{E}$), zato po izreku \ref{rotacKom2Zrc} +$\mathcal{I}$ predstavlja rotacijo s središčem v točki $S$. + +\textit{3)} Naj bo $\mathcal{I}$ izometrija brez fiksnih točk. +Potem je za poljubno točko $A$ te ravnine $\mathcal{I}(A) = A'\neq +A$. Naj bo $p$ simetrala daljice $AA'$ in $S_p$ zrcaljenje čez +premico $p$. V tem primeru je $\mathcal{S}_p\circ \mathcal{I}(A)= +\mathcal{S}_p(A')=A$. Kompozitum $\mathcal{S} \circ +\mathcal{I}_p$ je torej indirektna izometrija s fiksno točko $A$, zato +po izreku \ref{izo1ftIndZrc} predstavlja zrcaljenje čez neko +premico $q$ - $\mathcal{S}_q$, kjer os $q$ poteka skozi točko $A$. +Enako kot v prejšnjem primeru je +$\mathcal{I}=\mathcal{S}_p\circ\mathcal{S}_q$. Premici $p$ in $q$ +se niti ne sekata niti nista enaki, ker bi sicer iz že dokazanega +$\mathcal{I}$ predstavljala identiteto ali rotacijo in bi imela +vsaj eno fiksno točko. Torej ostane le še možnost, da sta premici +$p$ in $q$ vzporedni. V tem primeru je $\mathcal{I}$ translacija +(izrek \ref{translKom2Zrc}). + \kdokaz + +Ostanejo nam še indirektne izometrije. Edini doslej omenjeni vrsti +sta osno zrcaljenje in zrcalni zdrs. +Ali sta tudi sicer edini? Odgovor bomo podali v naslednjem + izreku. + + + + \bizrek \label{Chaslesov-} + Any opposite isometry is either a reflection + either it can be represented as the product of three reflections. + The only opposite isometries are + reflections and + glide reflections. + \eizrek + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.10.2.pic}) + +Naj bo $\mathcal{I} : \mathbb{E}^2\rightarrow \mathbb{E}^2$ +indirektna izometrija ravnine. Dokaz bomo spet izpeljali glede +na število fiksnih točk. + +\textit{1)} Če ima izometrija $\mathcal{I}$ vsaj eno fiksno +točko, je $\mathcal{I}$ po izreku \ref{izo1ftIndZrc} zrcaljenje čez +premico. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.10.2.pic} +\caption{} \label{sl.izo.6.10.2.pic} +\end{figure} + + \textit{2)} Predpostavimo, da je $\mathcal{I}$ izometrija brez fiksnih točk. + V tem primeru za poljubno točko $A$ te ravnine velja +$\mathcal{I}(A) = A'\neq A$. Naj bo $p$ simetrala daljice $AA'$ in +$\mathcal{S}_p$ zrcaljenje čez premico $p$. Potem je +$\mathcal{S}_p\circ \mathcal{I}(A)= \mathcal{S}_p(A')=A$. Zato je +kompozitum $\mathcal{S}_p \circ \mathcal{I}$ direktna izometrija s +fiksno točko $A$ pa po prejšnjem izreku predstavlja rotacijo ali +identiteto (translacija nima fiksnih točk). Drugi primer odpade, +saj bi sicer bilo $\mathcal{I} = \mathcal{S}_p$ in bi +izometrija $\mathcal{I}$ imela fiksne točke, kar pa nasprotuje +začetni predpostavki. Torej je kompozitum $\mathcal{S} \circ +\mathcal{I}_p$ rotacija s središčem $A$, ki jo lahko predstavimo +kot kompozitum dveh zrcaljenj čez premici $q$ in $r$, ki se sekata +v točki $A$. Zato je $\mathcal{S}_p\circ \mathcal{I}=\mathcal{S}_q\circ +\mathcal{S}_r$ oz. $\mathcal{I}=\mathcal{S}_p\circ \mathcal{S}_q\circ +\mathcal{S}_r$. Ker je $A'\neq A$ in $p$ simetrala daljice $AA'$, +točka $A$ ne leži na premici $p$. To pomeni, da premice $p$, $q$ +in $r$ niso iz istega šopa. Po izreku \ref{izoZrcdrsprq} je zato +$\mathcal{I}=\mathcal{S}_p\circ \mathcal{S}_q\circ \mathcal{S}_r$ +zrcalni zdrs. + \kdokaz + + Zaradi prejšnjih dveh izrekov lahko rečemo, + da so omenjene izometrije edine vrste izometrij. +Lahko rečemo tudi, da lahko vsako izometrijo ravnine predstavimo +kot kompozitum osnih simetrij, pri čemer lahko osi izberemo tako, da + v kompozitumu niso več kot tri. Iz dokaza omenjenih izrekov je +jasno, da lahko vsako izometrijo določimo le na osnovi števila +fiksnih točk in tega, ali je izometrija direktna oz. indirektna. +Ta dejstva bomo formulirali v naslednjih dveh izrekih. + + + + \bizrek \label{IzoKompZrc} + Any isometry of the plane can be expressed as the product of + one, two or three reflections. + \eizrek + + + + \bizrek \label{Chaslesov} \index{izrek!Chaslesov} + (Chasles’\footnote{\index{Chasles, M.} + \textit{M. Chasles} (1793--1880), + francoski geometer, ki je to klasifikacijo izpeljal leta 1831.}) + The only isometries of the plane + $\mathcal{I} : E^2 \rightarrow E^2$ are: + identity map, reflections, rotations, + translations and glide reflections. Specifically:\\ + \hspace*{3mm}(i) if $\mathcal{I}$ is + a direct isometry and has at least two fixed points, then $\mathcal{I}$ + is the identity map,\\ + \hspace*{3mm}(ii) if $\mathcal{I}$ is a direct isometry and has exactly one fixed point, + then $\mathcal{I}$ is a rotation (or specially a half-turn),\\ + \hspace*{3mm}(iii) if $\mathcal{I}$ is + a direct isometry and has no fixed points, + then $\mathcal{I}$ is a translation,\\ + \hspace*{3mm}(iv) if $\mathcal{I}$ is + an opposite isometry and has at least one fixed point, + then $\mathcal{I}$ is a reflection,\\ + \hspace*{3mm}(v) + if $\mathcal{I}$ is + an opposite isometry and has no fixed points, + then $\mathcal{I}$ is a glide reflection. + \eizrek + + Vse, kar smo v tem razdelku povedali o klasifikaciji izometrij, je +ilustrirano v naslednji tabeli (Figure \ref{IzoKlas.eps}): + +\vspace*{-2mm} + +%\begin{figure}[!htb] +%\centering +%\input{sl.izo.6.10.3.pic} +%\caption{} \label{sl.izo.6.10.3.pic} +%\end{figure} + +\begin{figure}[!htb] +\centering + \includegraphics[width=0.85\textwidth]{IzoKlas.eps} +\caption{} \label{IzoKlas.eps} +\end{figure} + +%________________________________________________________________________________ + \poglavje{Hjelmslev's Theorem} \label{odd6Hjelmslev} + +The following theorem, which refers to opposite isometries, is very useful. + + + \bizrek \label{Chasles-Hjelmsleva} + (Hjelmslev's\footnote{ + \index{Hjelmslev, J. T.} + \textit{J. T. Hjelmslev} (1873--1950), danski + matematik.}) \index{theorem!Hjelmslev's} + The midpoints of all line segments defined by corresponding pairs + of points of an arbitrary indirect isometry lie on the same line. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.11.1.pic} +\caption{} \label{sl.izo.6.11.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.11.1.pic}) + +Naj bo $\mathcal{I}$ indirektna izometrija in $X'=\mathcal{I}(X)$ za +poljubno točko $X$ ter $X_s$ središče daljice $XX'$. Iz Chaslesovega +izreka \ref{Chaslesov}sledi, da sta edini vrsti indirektnih izometrij ravnine +osno zrcaljenje in zrcalni zdrs. Dokažimo, da je v obeh primerih +iskana premica ravno os zrcaljenja čez premico oz. zrcalnega zdrsa. V +prvem primeru, kjer je $\mathcal{I}=\mathcal{S}_s$, je trivialno, da je +$X_s\in s$. Naj bo $\mathcal{I}=\mathcal{G}_{2\overrightarrow{PQ}}= +\mathcal{S}_s\circ \mathcal{T}_{2\overrightarrow{PQ}}$, kjer je +$s=PQ$. Če označimo z $X_1=\mathcal{T}_{2\overrightarrow{PQ}}(X)$ in +$X_2$ središče daljice $X_1X'$, je daljica $X_sX_2$ srednjica +trikotnika $XX_1X'$ za osnovnico $XX_1$. Torej je (izrek +\ref{srednjicaTrik}) $X_sX_2\parallel XX_1\parallel s$ oz. $X_s\in +s$ (Playfairjev aksiom \ref{Playfair}). + \kdokaz + +Dokazani izrek lahko uporabimo, če imamo indirektno podobna lika +ali pa če najdemo indirektno izometrijo, ki preslika eno množico +točk v drugo. To bomo ilustrirali z naslednjima primeroma. + + \bzgled + Let $ABC$ and $A'B'C'$ be congruent triangles with the opposite orientation. + Prove that the midpoints of the line segments + $AA$', $BB'$ and $CC'$ lie on the same line. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.11.2.pic} +\caption{} \label{sl.izo.6.11.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.11.2.pic}) + Ker sta $ABC$ in $A'B'C'$ skladna različno orientirana + trikotnika, obstaja indirektna izometrija $\mathcal{I}$, ki trikotnik $ABC$ +preslika v trikotnik $A'B'C'$. V tem primeru so $A$ in $A'$, $B$ in +$B'$ ter $C$ in $C'$ pari te izometrije, zato po izreku + \ref{Chasles-Hjelmsleva} središča daljic $AA'$, +$BB'$ in $CC'$ ležijo na eni premici. + \kdokaz + + \bzgled + Let $A$ be a point and $p$ a line in the plane. Suppose that points $X_i$ + lie on the line $p$ and $AX_iY_i$ are regular triangles with the same orientation. + Prove that the midpoints of the line segments + $X_iY_i$ lie on the same line. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.11.3.pic} +\caption{} \label{sl.izo.6.11.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.11.3.pic}) +Izometrija $\mathcal{R}_{A,60^0}\circ\mathcal{S}_p$ je indirektna izometrija, ki preslika točke $X_i$ v točke $Y_i$, + zato po izreku \ref{Chasles-Hjelmsleva} središča + daljic $X_iY_i$ ležijo +na isti premici. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Isometry Groups. Symmetries of Figures} \label{odd6Grupe} + +V razdelku \ref{odd2AKSSKL} smo ugotovili, da množica + $\mathfrak{I}$ vseh izometrij neke ravnine skupaj z operacijo kompozituma +preslikav predstavlja t. i. +strukturo \index{grupa}\pojem{grupe}\footnote{Teorijo grup je +odkril genialni mlad francoski matematik \index{Galois, E.} +\textit{E. Galois} (1811--1832).}. + To pomeni, da so izpolnjene naslednje lastnosti: +\begin{enumerate} + \item $(\forall f\in \mathfrak{I})(\forall g\in \mathfrak{I}) + \hspace*{1mm}f\circ g\in \mathfrak{I}$, + \item $(\forall f\in \mathfrak{I})(\forall g\in \mathfrak{I}) + (\forall h\in \mathfrak{I}) + \hspace*{1mm}(f\circ g)\circ h=f\circ (g\circ h)$, + \item $(\exists e\in \mathfrak{I})(\forall f\in \mathfrak{I}) + \hspace*{1mm}f\circ e=e\circ f=f$, + \item $(\forall f\in \mathfrak{I})(\exists g\in \mathfrak{I}) + \hspace*{1mm}f\circ g=g\circ f=e$. +\end{enumerate} + + Lastnost (2) velja splošno za +kompozitum preslikav. Lastnosti (1), (3) in (4) smo vpeljali z +aksiomom \ref{aksIII4}. Lastnost (1) pomeni, da je kompozitum dveh +izometrij spet izometrija, (3) in (4) pa se nanašata na identiteto +in inverzno izometrijo. Omenjeno grupo, ki jo določa množica +$\mathfrak{I}$ vseh izometrij neke ravnine glede na operacijo +kompozituma preslikav, imenujemo \index{grupa!izometrij} +\pojem{grupa vseh izometrij ravnine}. Tudi njo bomo označili z $\mathfrak{I}$. + +Obstajajo tudi druge grupe izometrij, ki jih dobimo, če +vzamemo ustrezno podmnožico vseh izometrij ravnine. Lastnost (1) +nam pove, da ta množica ne more biti poljubna. Npr. množica vseh +rotacij ravnine ni grupa, saj kompozitum dveh rotacij ni vedno +rotacija (lahko je tudi translacija). + +Iz lastnosti translacij sledi, da množica vseh translacij +skupaj z identiteto $\mathcal{E}$ predstavlja grupo - t. i. +\index{grupa!translacij}\pojem{grupo translacij} z oznako +$\mathfrak{T}$. Za njo pravimo, da je \index{podgrupa} +\pojem{podgrupa} grupe $\mathfrak{I}$ vseh izometrij ravnine (to +dejstvo označimo s $\mathfrak{T}<\mathfrak{I}$). Pravzaprav je vsaka +grupa izometrij ravnine podgrupa grupe $\mathfrak{I}$ vseh izometrij te ravnine. +Vendar grupa translacij ni edina podgrupa grupe +$\mathfrak{I}$. Vse direktne izometrije ravnine namreč tvorijo eno +takšno podgrupo; označimo jo z $\mathfrak{I}^+$. To pomeni, da je +$\mathfrak{I^+}<\mathfrak{I}$. Ker so translacije direktne +izometrije, je tudi $\mathfrak{T}<\mathfrak{I^+}$. Jasno je, da +množica vseh indirektnih izometrij ne določa grupe, saj je kompozitum +dveh indirektnih izometriji direktna izometrija (tudi identiteta +ni indirektna izometrija). + +Obstajajo tudi končne podgrupe grupe $\mathfrak{I}$ (ki imajo +končno število izometrij). Primer takšne podgrupe je t. i. +Kleinova\footnote{\index{Klein, F. C.} \textit{F. C. Klein} +(1849--1925), nemški matematik, ki je na svojem predavanju leta 1872 na +univerzi v Erlangenu predstavil idejo, po kateri je neka +geometrija določena z grupo ustreznih transformacij na množici +(točk) in z invariantami te grupe na tej množici.} grupa +$\mathfrak{K}$ ($\mathfrak{K}<\mathfrak{I}$) (ali \textit{Kleinov četverec}), ki je določena z +množico izometrij $\{\mathcal{E}, \mathcal{S}_p, \mathcal{S}_q, +\mathcal{S}_O\}$, kjer sta $p$ in $q$ pravokotnici, ki se sekata v +točki $O$. To grupo lahko predstavimo tudi s tabelo: + +\vspace*{5mm} + +\hspace*{22mm}\begin{tabular}{|c||c|c|c|c|} \hline + % after \\ : \hline or \cline{col1-col2} \cline{col3-col4} ... + $\circ$ & $\mathcal{E}$ & $\mathcal{S}_p$ & $\mathcal{S}_q$& $\mathcal{S}_O$ + \\ \hline \hline + $\mathcal{E}$ & $\mathcal{E}$& $\mathcal{S}_p$& $\mathcal{S}_q$& $\mathcal{S}_O$ + \\ \hline + $\mathcal{S}_p$ & $\mathcal{S}_p$ & $\mathcal{E}$ & $\mathcal{S}_O$& $\mathcal{S}_q$ + \\ \hline + $\mathcal{S}_q$ & $\mathcal{S}_q$ & $\mathcal{S}_O$ & $\mathcal{E}$& $\mathcal{S}_p$ + \\ \hline + $\mathcal{S}_O$ & $\mathcal{S}_O$ & $\mathcal{S}_q$ &$\mathcal{S}_p$ & $\mathcal{E}$ + \\ \hline +\end{tabular} + + +\vspace*{5mm} + +Ker Kleinova grupa $\mathfrak{K}$ vsebuje osno zrcaljenje, ki je +indirektna izometrija, $\mathfrak{K}$ ni podgrupa grupe +$\mathfrak{I}^+$. + + Grupo izometrij, ki je sestavljena le iz + identitete $\{\mathcal{E}\}$, bomo imenovali \index{grupa!trivialna} + \pojem{trivialna grupa} z oznako $\mathfrak{E}$. Ta grupa je očitno podgrupa + vsake grupe + izometrij; npr. $\mathfrak{E}<\mathfrak{T}<\mathfrak{I^+}<\mathfrak{I}$ ali + $\mathfrak{E}<\mathfrak{K}<\mathfrak{I}$. + +Če vzamemo le identiteto $\{\mathcal{E}$ in +eno osno zrcaljenje $\mathcal{S}_p\}$ dobimo še eno končno grupo izometrij. +Isto strukturo dobimo tudi, če namesto zrcaljenja čez premico vzamemo +zrcaljenje čez točko, torej $\{\mathcal{E}$ in $\mathcal{S}_O\}$. Čeprav sta +množici različni, je struktura grupe ista, kar ponazorimo s tabelama. +Pravimo, da sta v tem primeru grupi +\index{grupa!izomorfna}\pojem{izomorfni}. + +\vspace*{5mm} + +\hspace*{12mm}\begin{tabular}{|c||c|c|} \hline + % after \\ : \hline or \cline{col1-col2} \cline{col3-col4} ... + $\circ$ & $\mathcal{E}$ & $\mathcal{S}_p$ + \\ \hline \hline + $\mathcal{E}$ & $\mathcal{E}$& $\mathcal{S}_p$ + \\ \hline + $\mathcal{S}_p$ & $\mathcal{S}_p$ & $\mathcal{E}$ + \\ \hline +\end{tabular} +\hspace*{22mm} +\begin{tabular}{|c||c|c|} \hline + % after \\ : \hline or \cline{col1-col2} \cline{col3-col4} ... + $\circ$ & $\mathcal{E}$ & $\mathcal{S}_O$ + \\ \hline \hline + $\mathcal{E}$ & $\mathcal{E}$& $\mathcal{S}_O$ + \\ \hline + $\mathcal{S}_O$ & $\mathcal{S}_O$ & $\mathcal{E}$ + \\ \hline +\end{tabular} + +\vspace*{5mm} + +Dokazali smo že, da lahko vsako izometrijo ravnine predstavimo +kot kompozitum končnega števila zrcaljenj čez premico (vedno +lahko izberemo največ tri zrcaljenja) - izrek +\ref{IzoKompZrc}. Pravimo, da zrcaljenja čez premico +\pojem{generirajo} grupo $\mathfrak{I}$ vseh izometrij neke +ravnine oz. da so \pojem{generatorji} grupe $\mathcal{I}$. + Spomnimo +se, da kompozitum sodega števila zrcaljenj čez premico predstavlja direktno izometrijo, +pri indirektni izometriji pa je število zrcaljenj čez premico liho. + + Obstaja še ena pomembna vrsta grup izometrij, ki jo bomo + predstavili v naslednjem izreku. + + + + + \bizrek + Let $\phi$ be a figure in the plane. The set of all isometries + of that plane that map the figure $\phi$ to itself forms a group. + \eizrek + +\textbf{\textit{Proof.}} + Naj bo torej $\mathfrak{G}$ množica vseh izometrij + ravnine, ki lik $\phi$ preslikajo vase. Dokažimo, + da ta množica določa grupo. Za poljubni izometriji + $f,g\in \mathfrak{G}$ velja $f(\phi)=\phi$ in +$g(\phi)=\phi$. Toda v tem primeru velja tudi $g \circ f (\phi) = +\phi$ in $f^{-1}(\phi) = \phi$. Torej sta izpolnjena pogoja +(1) in (4). Glede pogoja (2) smo že +povedali, da je za operacijo kompozituma preslikav vedno izpolnjen. +Tudi pogoj (3) je v našem primeru izpolnjen, saj +velja $\mathcal{E}(\phi)=\phi$. + \kdokaz + + + Grupo iz prejšnjega izreka - množico + vseh izometrij + te ravnine, ki lik $\phi$ preslikajo vase, + imenujemo \index{grupa!simetrij}\pojem{grupa simetrij} lika + $\phi$ z oznako $\mathfrak{G}(\phi)$. Jasno je, da za vsak lik $\phi$ + velja $\mathfrak{G}(\phi)<\mathfrak{I}$. + Omenimo, da je pri dokazovanju, da podmnožica neke grupe + (v našem primeru grupe $\mathfrak{I}$) predstavlja grupo oz. + njeno podgrupo, dovolj preveriti le pogoja (\textit{i}) in + (\textit{iv}). Pogoj (\textit{ii}) je namreč vedno izpolnjen, + pogoj (\textit{iii}) pa sledi direktno iz pogojev (\textit{i}) in + (\textit{iv}). + + Obstaja še ena grupa, ki je podgrupa grupe simetrij $\mathfrak{G}(\phi)$. + To grupo tvori + množica vseh direktnih izometrij iz $\mathfrak{G}(\phi)$. Imenujemo jo + \index{grupa!rotacij}\pojem{grupa rotacij} lika + $\phi$ in jo označimo $\mathfrak{G}^+(\phi)$. Jasno je, da je za + vsak lik njegova grupa rotacij podgrupa njegove grupe + simetrij oz. $\mathfrak{G}^+(\phi)<\mathfrak{G}(\phi)$. + + V naslednjem zgledu bomo določili grupe simetrij različnih + likov. + + + + \bzgled \label{grupeSimPrimeri} + Determine the symmetry group and the rotation group of: + (i) a square, (ii) a rectangle, (iii) a trapezium, + (iv) a line, (v) a ray, (vi) a circle. + \ezgled + + +\textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.12.1.pic}) + +(\textit{i}) Kvadrat s središčem $O$ ima štiri somernice, ki se +sekajo v točki $O$, in štiri rotacije (vključno z identiteto), +zato je: + + $\mathfrak{G}(\phi)=\{\mathcal{S}_p, \mathcal{S}_q, +\mathcal{S}_r, \mathcal{S}_s, \mathcal{E}, \mathcal{R}_{O,90^0} , +\mathcal{R}_{O,180^0} , \mathcal{R}_{O,270^0} \}$ in + +$\mathfrak{G}^+(\phi)=\{\mathcal{E}, \mathcal{R}_{O,90^0} , +\mathcal{R}_{O,180^0} , \mathcal{R}_{O,270^0} \}$. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.1.pic} +\caption{} \label{sl.izo.6.12.1.pic} +\end{figure} + +(\textit{ii}) Pravokotnik s središčem $O$ ima le dve somernici, +zato je $\mathfrak{G}(\phi)=\{\mathcal{S}_p, \mathcal{S}_q, +\mathcal{E}, \mathcal{S}_O \}$ in +$\mathfrak{G}^+(\phi)=\{\mathcal{E}, \mathcal{S}_O \}$. Zanimivo +je, da je prva grupa pravokotnika $\mathfrak{G}(\phi)$ +pravzaprav Kleinova grupa $\mathfrak{K}$. + + (\textit{iii}) Če je $\phi$ poljubni trapez, je identiteta + edina, ki ga preslikava vase. To pomeni, da je + $\mathfrak{G}(\phi)=\mathfrak{G}^+(\phi) + =\{\mathcal{E}\}$ grupa, ki smo jo že poimenovali trivialna grupa. + +(\textit{iv}) Grupi simetrij in rotacij premice $p$ sta neskončni +grupi in sicer: + \begin{eqnarray*} + \mathfrak{G}(p)&=&\{\mathcal{S}_l;l\perp p\}\cup + \{\mathcal{S}_p\}\cup + \{\mathcal{S}_O;O\in p\}\cup + \{ \mathcal{T}_{\overrightarrow{v}};\overrightarrow{v} \parallel p\}\cup + \{\mathcal{E}\}\\ + \mathfrak{G}^+(p)&=& + \{\mathcal{S}_O;O\in p\}\cup + \{ \mathcal{T}_{\overrightarrow{v}};\overrightarrow{v} \parallel p\}\cup + \{\mathcal{E}\} + \end{eqnarray*} + +(\textit{v}) Za poltrak $h=OA$ je $\mathfrak{G}(h)=\{\mathcal{E}, +\mathcal{S}_{h}\}$ in $\mathfrak{G}^+(h)=\{\mathcal{E}\}$. + +(\textit{vi}) Pri krožnici $k(O,r)$ so v grupi simetrij vsa +zrcaljenja z osmi skozi točko $O$ in vse rotacije s središčem +$O$ (vključno z identiteto). Grupa rotacij pa je sestavljena samo +iz omenjenih rotacij. Torej: + \begin{eqnarray*} + \mathfrak{G}(k)&=&\{\mathcal{S}_l;l\ni O\}\cup + \{\mathcal{R}_{O,\alpha};\alpha \textrm{ poljuben kot}\}\cup + \{\mathcal{E}\}\\ + \mathfrak{G}^+(k)&=&\{\mathcal{R}_{O,\alpha};\alpha \textrm{ poljuben kot}\}\cup + \{\mathcal{E}\}, + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + +Videli smo, da ima pravokotnik natanko dve somernici in je tudi +središčno simetričen. Sedaj bomo dokazali splošno trditev. + + + + \bzgled + If a figure in the plane has exactly two axes of symmetry, + then it also has a centre of symmetry. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.2.pic} +\caption{} \label{sl.izo.6.12.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bosta $p$ in $q$ edini somernici lika +$\phi$. Dokažimo, da je $p\perp q$ (Figure \ref{sl.izo.6.12.2.pic}). +Ker je $\mathcal{S}_p(\phi)=\mathcal{S}_q(\phi)=\phi$, je po +izreku \ref{izoTransmutacija} o transmutaciji tudi +$$\mathcal{S}_{\mathcal{S}_p(q)}(\phi)= +\mathcal{S}_p\circ\mathcal{S}_q\circ\mathcal{S}_p^{-1}(\phi)=\phi.$$ +Premici $p$ in $q$ sta edini somernici lika $\phi$, zato je bodisi +$\mathcal{S}_p(q)=p$ bodisi $\mathcal{S}_p(q)=q$. V obeh primerih +sledi $p=q$ oz. $p\perp q$. Toda prva možnost odpade, saj sta $p$ +in $q$ po predpostavki različni. Torej sta premici $p$ in $q$ +pravokotni in se sekata v neki točki $O$, zato je: +$$\mathcal{S}_O(\phi)= +\mathcal{S}_q\circ\mathcal{S}_p(\phi)=\mathcal{S}_q(\mathcal{S}_p(\phi))= +\phi,$$ kar pomeni, da je lik središčno simetričen. + \kdokaz + +V terminih grup lahko prejšnjo trditev zapišemo še v naslednji +obliki. + + + + \bzgled If the symmetry group of some figure contains exactly two reflections + then it also contains a half-turn. + \ezgled + +Najprej ugotovimo, da imata lahko različna lika isto grupo +simetrij. Npr. grupa simetrij pravokotnika in daljice je Kleinova +grupa $\mathfrak{K}$. Tudi poltrak, enakokraki trikotnik in enakokraki trapez +imajo enako grupo simetrij $\{\mathcal{E}, \mathcal{S}_{p}\}$. + +Tudi prve tri in peta grupa simetrij iz primera +\ref{grupeSimPrimeri} so končne grupe, četrta in šesta pa sta +neskončni. Iz tega primera vidimo tudi, da neskončnost grupe +simetrije nekega lika ni v povezavi z omejenostjo tega lika. +Grupa simetrij omejenega lika je lahko končna, lahko pa +tudi neskončna. Enako velja za neomejene like. Zaradi nadaljnje +uporabe bomo pojem omejenosti natančneje definirali. + +Lik $\phi$ je \index{lik!omejen} \pojem{omejen}, če obstaja takšna +daljica $AB$, da za poljubni točki $X$ in $Y$ tega lika velja +$XY|XX_n|=|n\overrightarrow{v}|$. +To naj bi veljalo za vsak $n\in \mathbb{N}$, kar seveda ne +drži. Zadnja relacija nas torej pripelje do protislovja, kar +pomeni, da v grupi $\mathfrak{G}(\phi)$ ni translacij. + \kdokaz + + Iz prejšnjega izreka sledi, da grupa simetrij + omejenega lika tudi zrcalnega zdrsa ne vsebuje, saj +je njegov kvadrat translacija (zgled \ref{izoZrcdrsZrcdrs}). Kako +pa je z zrcaljenji in rotacijami? + + + + \bizrek \label{GrupaSomer1} If a bounded figure has more axes of symmetry, + then they all intersect at one point. + \eizrek + +\textbf{\textit{Proof.}} Naj bosta $p$ in $q$ somernici omejenega +lika $\phi$. Premici $p$ in $q$ nista vzporedni, saj bi bil sicer +kompozitum ustreznih zrcaljenj $\mathcal{S}_p \circ \mathcal{S}_q$ +translacija. Torej se $p$ in $q$ sekata v neki točki $S$. Če je +$r$ tretja poljubna somernica, vsebuje točko $S$, ker bi bil +v nasprotnem primeru (izrek \ref{izoZrcdrsprq}) kompozitum $\mathcal{S}_p \circ +\mathcal{S}_q\circ \mathcal{S}_r$ zrcalni zdrs. + \kdokaz + + + + \bizrek \label{GrupaSomerRot} If a bounded figure has at least one axe of symmetry + and at least one centre of rotation, then + this centre lies on the axe of symmetry. + \eizrek + + \textbf{\textit{Proof.}} + Predpostavimo nasprotno - naj bo $p$ somernica omejenega lika $\phi$ in + $S$ + središče rotacije +$\mathcal{R}_{S,\alpha}$, ki ta lik slika vase, vendar +$S\notin p$. Rotacijo $\mathcal{R}_{S,\alpha}$ lahko +predstavimo kot kompozitum dveh zrcaljenj čez premico z osema, ki +gresta skozi točko $S$. Torej velja $\mathcal{S}_p \circ +\mathcal{R}_{S,\alpha} = \mathcal{S}_p \circ \mathcal{S}_q\circ +\mathcal{S}_r$. Premice $p$, $q$ in $r$ niso v istem šopu, zato +(izrek \ref{izoZrcdrsprq}) kompozitum $\mathcal{S}_p +\circ \mathcal{R}_{S,\alpha}$ predstavlja zrcalni zdrs, ki pa ne more +biti v grupi $\mathfrak{G}(\phi)$, kar pomeni, da središče $S$ leži +na somernici $p$. + \kdokaz + + + + \bizrek \label{GrupaRot} All rotations of a bounded figure + have the same centre. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.4.pic} +\caption{} \label{sl.izo.6.12.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.12.4.pic}). + Predpostavimo nasprotno - naj bosta $\mathcal{R}_{O,\alpha}$ in + $\mathcal{R}_{S,\beta}$ +rotaciji, ki omejen lik $\phi$ preslikata vase in velja $O\neq +S$. Tedaj je tudi $O'=\mathcal{R}_{S,\beta}(O)\neq S$. Po izreku +\ref{izoTransmRotac} je +$\mathcal{R}_{O',\alpha}=\mathcal{R}_{S,\beta}\circ +\mathcal{R}_{O,\alpha}\circ\mathcal{R}_{S,\beta}^{-1}$, zato je + $\mathcal{R}_{O',\alpha}\in \mathfrak{G}(\phi)$ in nato tudi + $\mathcal{I}=\mathcal{R}_{O',\alpha}^{-1}\circ + \mathcal{R}_{O,\alpha}\in \mathfrak{G}(\phi)$. + Naj bo $p$ premica, ki s premico $OO'$ v točki +$O$ določa kot $\frac{1}{2}\alpha$, in $q$ njena vzporednica skozi +točko $O'$. Po izreku \ref{rotacKom2Zrc} je $\mathcal{I} = +\mathcal{S}_q \circ \mathcal{S}_{OO'} \circ \mathcal{S}_{OO'} +\circ \mathcal{S}_p = \mathcal{S}_q \circ \mathcal{S}_p$, zato +je $\mathcal{I}$ translacija, kar pa ni mogoče. To pomeni, da +$O\neq S$ ne velja, torej imajo vse rotacije omejenega lika + isto središče. + \kdokaz + + Iz prejšnjih izrekov dobimo naslednjo trditev. + + + + + \bizrek \label{GrupaOmejenLik} If symmetry group of a bounded figure $\phi$ is not trivial, + then there is a point $S$ such that the only possible isometries in this group are: + \\ + (i) the identity map,\\ + (ii) rotations with the centre $S$,\\ + (iii) reflections with axes, + which all contains the point $S$. + \eizrek + + Na ta način smo ugotovili, kakšne so vse možne grupe omejenih likov. + Te grupe niso nujno +končne. Kakšne so potem končne grupe? Pa raziščimo +nekaj primerov grup simetrij. + + + \bzgled \label{GrupaDiederska} Determine the symmetry group and the rotation group + of a regular $n$-gon. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.5.pic} +\caption{} \label{sl.izo.6.12.5.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.12.5.pic}) + Ne glede na to, ali je $n$ sodo ali liho število, ima pravilni $n$-kotnik + natanko $n$ +somernic (glej razdelek \ref{odd3PravilniVeck}). Osnovni kot +rotacije, ki $n$-kotnik preslika vase je $\theta = \frac{360^0}{n}$. +Če torej z $\mathcal{D}_n$ in $\mathcal{C}_n$ označimo grupo +simetrij in grupo rotacij + pravilnega $n$-kotnika, z $O$ pa njegovo središče, dobimo: + \begin{eqnarray*} + \mathfrak{D}_n&=&\{ \mathcal{S}_{p_1}, \mathcal{S}_{p_2},\ldots, + \mathcal{S}_{p_n}, + \mathcal{E}, \mathcal{R}_{O,\theta}, \mathcal{R}_{O,2\theta},\ldots + \mathcal{R}_{O,(n-1)\theta}\}\\ + \mathfrak{C}_n&=&\{ \mathcal{E}, \mathcal{R}_{O,\theta}, + \mathcal{R}_{O,2\theta},\ldots \mathcal{R}_{O,(n-1)\theta}\}, + \end{eqnarray*} + kar je bilo treba določiti. \kdokaz + + Grupo simetrij $\mathfrak{D}_n$ pravilnega $n$-kotnika iz + prejšnjega zgleda + imenujemo + \index{grupa!diedrska} \pojem{diedrska grupa}, grupo rotacij + pravilnega $n$-kotnika $\mathfrak{C}_n$ pa + \index{grupa!ciklična} \pojem{ciklična grupa}. Jasno je, da + velja $\mathfrak{C}_n<\mathfrak{D}_n$. + + Grupe $\mathfrak{D}_n$ in $\mathfrak{C}_n$ ($n\geq 3$) + lahko posplošimo tudi za primere $n < 3$, čeprav v tem primeru ne govorimo več + o $n$-kotniku. $\mathfrak{D}_2$ je tako +pravzaprav Kleinova grupa $\mathfrak{K}$ (grupa simetrije daljice +oziroma nekakšnega ‘‘$2$-kotnika’’). Grupo $\mathfrak{C}_2$ +sestavljata identiteta in ena središčna simetrija. To grupo si lahko +predstavimo kot grupo rotacij daljice. Grupa $\mathfrak{D}_1$ +vsebuje identiteto in osno zrcaljenje, je pa izomorfna grupi +$\mathfrak{C}_2$ (primer, ki smo ga že omenili na začetku tega +razdelka). Grupa $\mathfrak{C}_1$ je trivialna grupa $\mathfrak{E}$. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.5a.pic} +\caption{} \label{sl.izo.6.12.5a.pic} +\end{figure} + +Omenimo, da grupe $\mathfrak{D}_n$ in $\mathfrak{C}_n$ ($n\in +\mathbb{N}$) niso grupe simetrij le za pravilne $n$-kotnike (Figure +\ref{sl.izo.6.12.5a.pic}). Lahko so grupe simetrij tudi neomejenih +likov. Npr. Kleinova grupa $\mathfrak{D}_2$ je tudi grupa simetrij +hiperbole z enačbo $x\cdot y= 1$ v kartezičnem pravokotnem sistemu +$O_{xy}$. Somernici sta tedaj določeni z enačbami $y = x$ in $y = +-x$, središče simetrije pa je izhodišče $O$. $\mathfrak{D}_2$ je +tudi grupa simetrij elipse, ki je dana z enačbo +$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. + +Jasno je, da so vse grupe $\mathfrak{D}_n$ in $\mathfrak{C}_n$ +($n\in \mathbb{N}$) končne grupe izometrij. Ugotovili bomo, da so +to edine končne grupe izometrij! + + + + \bizrek \label{GrupaLeonardo} \index{izrek!Leonarda da Vincija} + (Leonardo da Vinci\footnote{Slavni italijanski slikar, arhitekt in + izumitelj \index{Leonardo da Vinci} + \textit{Leonardo da Vinci} (1452--1519) je proučeval vse možne + simetrije središčne stavbe in možne razporeditve kapel okoli + nje, ki ohranjajo osnovno simetrijo.}) The only finite groups of isometries are $\mathfrak{D}_n$ and $\mathfrak{C}_n$. + \eizrek + + + +\textbf{\textit{Proof.}} Naj bo $\mathfrak{G}$ končna grupa +izometrij. Če bi vsebovala translacijo +$\mathcal{T}_{\overrightarrow{v}}$, bi imela neskončno +podgrupo $\{ \mathcal{E}, \mathcal{T}_{\overrightarrow{v}}, +\mathcal{T}_{\overrightarrow{v}}^2, +\mathcal{T}_{\overrightarrow{v}}^3,\ldots \}$, kar ni mogoče. Torej +je $\mathfrak{G}$ grupa brez translacij. Na podoben način kot prej +(izrek \ref{GrupaOmejenLik}) lahko dokažemo, da so edine možne +izometrije v tej grupi (razen identične preslikave) rotacije z +istim središčem $S$ in zrcaljenja čez premico z osmi, ki gredo +skozi točko $S$. Če grupa nima nobene rotacije, je možno kvečjemu +eno osno zrcaljenje (že dve bi generirali rotacijo). V tem primeru +sta edini možni grupi $\mathfrak{D}_1$ in $\mathfrak{C}_1$. + +Brez škode za splošnost predpostavimo, da so vsi koti rotacij +pozitivni. Ker je grupa $\mathfrak{G}$ končna, obstaja rotacija +$\mathcal{R}_{O,\theta}$ z najmanjšim kotom $\theta$. Iz istega +razloga obstaja naravno število $n$, za katero je +$\mathcal{R}_{O,\theta}^n = \mathcal{E}$ (sicer bi končna +grupa $\mathfrak{G}$ imela neskončno podgrupo $\{ \mathcal{E}, +\mathcal{R}_{O,\theta}, \mathcal{R}_{O,\theta}^2, +\mathcal{R}_{O,\theta}^3,\ldots \}$). + Torej je $\theta = +\frac{360^0}{n}$. + + Naj bo $\mathcal{R}_{O,\delta}$ poljubna rotacija iz + $\mathfrak{G}$. +Dokažimo, da je kot $\delta$ večkratnik kota $\theta$ oz. +$\delta=k\cdot \theta$ za nek $k\in \mathbb{N}$. Predpostavimo +nasprotno, da tak $k$ ne obstaja. Ker je $\delta>\theta$, za +nek $l\in \mathbb{N}$ velja $l\theta<\delta<(l+1)\theta$. Če +preoblikujemo, dobimo $0<\delta-l\theta<\theta$. Toda +$\mathcal{R}_{O,\delta-l\theta}=\mathcal{R}_{O,\delta}\circ +\left(\mathcal{R}_{O,\theta}^{-1}\right)^l\in \mathfrak{G}$, kar je +protislovno s tem, kako smo definirali kot $\theta$. Torej so +vse rotacije $\mathfrak{G}$ grupe oblike +$\mathcal{R}_{O,\theta}^k$, $k\in \{ 1,2,\ldots, n\}$. + +Če v grupi ni osnih zrcaljenj, je torej $\mathfrak{G}$ ciklična +grupa oz. $\mathfrak{G}=\mathfrak{C}_n$ za nek $n\in \mathbb{N}$. Če so v grupi +$\mathfrak{G}$ še zrcaljenja čez premico, so njihovi kompozitumi +(vsakih dveh zrcaljenj) rotacije, zato jim vsaki par osi določa kot, ki je +polovica nekega kota rotacij. V tem primeru je +$\mathfrak{G}$ diederska grupa oz. $\mathfrak{G}=\mathfrak{D}_n$ za nek $n\in \mathbb{N}$. + \kdokaz + + Še enkrat poudarimo, da končna grupa simetrij in grupa simetrij + omejenega lika nista isti pojem. + Grupa simetrij omejenih likov namreč ni nujno končna + (npr. grupa simetrij krožnice). Prav tako imajo lahko neomejeni liki +končno grupo simetrij (npr. poltrak). Seveda obstajajo omejeni +liki s končno grupo simetrij (npr. pravilni večkotnik) in +neomejeni liki z neskončno grupo simetrij (npr. premica). Ta +dejstva bomo ponazorili z naslednjim diagramom (Figure +\ref{sl.izo.6.12.6.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.6.pic} +\caption{} \label{sl.izo.6.12.6.pic} +\end{figure} + + + Omenili bomo še eno vrsto neskončnih grup izometrij, ki ohranjajo + določena tlakovanja ravnine (glej razdelek + \ref{odd3Tlakovanja}). Torej ne gre le za grupe simetrij + teh tlakovanj, ampak tudi za vse podgrupe teh grup. + + Če npr. izberemo + tlakovanja + ravnine s poljubnim paralelogramom $ABCD$ (Figure + \ref{sl.izo.6.12.7.pic}), je + neskončna grupa simetrij tega tlakovanja generirana z + zrcaljenji glede na oglišče $A$ ter središči daljic $AB$ in $AD$. V tej + grupi so potem tudi zrcaljenja glede na oglišča, središča stranic in + presečišča + diagonal vseh + paralelogramov tega tlakovanja ter vse translacije, ki so kompozitumi + po dveh od + teh zrcaljenj. + Toda ena njena podgrupa, ki tudi ohranja isto tlakovanje, je grupa + vseh omenjenih translacij. Generirana je z + dvema + translacijama $\mathcal{T}_{\overrightarrow{AB}}$ + in $\mathcal{T}_{\overrightarrow{AD}}$. + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.7aa.pic} +\input{sl.izo.6.12.7bb.pic} +\caption{} \label{sl.izo.6.12.7.pic} +\end{figure} + + Grupa simetrij v + primeru tlakovanja s pravilnim trikotnikom $ABC$ (Figure + \ref{sl.izo.6.12.7.pic}) + oz. tlakovanja $(3,6)$ vsebuje + rotacije v ogliščih mreže za kote $k\cdot 60^0$, + zrcaljenja čez premice, ki so + določene s stranicami in višinami teh trikotnikov, ter translacije, + ki jih generirata translaciji $\mathcal{T}_{\overrightarrow{AB}}$ + in $\mathcal{T}_{\overrightarrow{AC}}$. Ta grupa ima več + različnih podgrup, ki vse ohranjajo tlakovanje $(3,6)$. + + Vse takšne grupe, ki ohranjajo določena tlakovanja, imenujemo + \index{grupa!diskretna} \pojem{diskretne grupe izometrij}. + Formalno jih definiramo na + naslednji način: Grupa izometrij $\mathfrak{G}$ je diskretna grupa izometrij, + če obstaja tak + $\varepsilon>0$, da so dolžine vektorjev vseh translacij in + mere kotov vseh rotacij iz te grupe večji od $\varepsilon$. + + + + Razlikujemo dve vrsti diskretnih grup izometrij: \index{grupa!frizna} + \pojem{frizne + grupe}\footnote{Termin \textit{friz}, ki + so ga uporabljali že Stari Grki, + je pomenil ponavljajoč se vzorec bordur.} in \index{grupa!tapetna}\pojem{tapetne grupe}. Pri friznih + grupah je podgrupa + vseh translacij generirana z eno samo translacijo, pri tapetnih + grupah pa z dvema translacijama, ki ju določata nekolinearna + vektorja. Dokazano je, da obstaja natanko 7 friznih grup in 17 + tapetnih grup\footnote{Vseh 17 grup oz. ustreznih vrst ornamentov je + bilo znanih že Egipčanom, + pogosto pa so jih uporabljali tudi muslimanski umetniki. V dvorcu Alhambra (Španija) so + Mavri v 14. stoletju + naslikali vseh 17 vrst ornamentov. Formalni dokaz, da obstaja + natanko + 17 tapetnih grup, je prvi podal ruski matematik, + kristalograf in minerolog \index{Fedorov, E.} + \textit{E. Fedorov} + (1853–-1919) leta 1891, + nato je leta 1924 to delo dopolnil in nadaljeval madžarski matematik + \index{Pólya, G.}\textit{G. Pólya} (1887–-1985), ki je bolj znan po svoji + znameniti knjigi ‘‘Kako rešujemo matematične probleme’’.}. + Frizne grupe določajo 7 različnih vrst \index{bordura} + \pojem{bordur} - + trak s ponavljajočimi se vzorci (Figure \ref{sl.izo.6.12.7}a\footnote{http://mathworld.wolfram.com/WallpaperGroups.html}), tapetne + pa 17 različnih vrst \pojem{ornamentov} - prekrivanje ravnine + z enakimi vzorci (Figure \ref{sl.izo.6.12.7}b\footnote{http://www.quadibloc.com/math/tilint.htm}). + +\begin{figure}[!h] +\centering + \includegraphics[bb=0 0 7cm 7cm]{bands.eps}\\ +\vspace*{15mm} +\includegraphics[width=1\textwidth]{wall17_phpbtUJbf.eps} +\caption{} \label{sl.izo.6.12.7} +\end{figure} + +%v bmp +%\begin{figure}[!h] +%\centering + %\includegraphics[bb=0 0 7cm 7cm]{bands.bmp}\\ +%\vspace*{8mm} +%\includegraphics[bb=0 0 12cm 9.85cm]{wall17_phpbtUJbf.bmp} +%\caption{} \label{sl.izo.6.12.7} +%\end{figure} + + + + \bnaloga\footnote{40. IMO Romania - 1999, Problem 1.} + Determine all finite sets $\mathbf {S}$ of at least three points in the plane which satisfy the + following condition:\\ + for any two distinct points $A$ and $B$ in $\mathbf {S}$, the perpendicular bisector + of the line segment $AB$ is an axis of symmetry for $\mathbf {S}$. + \enaloga + + +\textbf{\textit{Solution.}} + Naj bo $\mathbf {S}=\{A_1,A_2,\ldots +A_n \}$ in $\mathcal{M}=\{s_{A_iA_j};\hspace*{1mm}A_i,A_j\in +\mathcal{S}\}$ množica vseh simetral. Ker je $\mathbf {S}$ +omejena množica (oz. lik), po izreku \ref{GrupaSomer1} vse +somernice te množice oz. simetrale iz množice $\mathcal{M}$ potekajo skozi +eno točko - označimo jo z $O$. Jasno je, da je pri tem $O$ +presečišče poljubnih dveh simetral iz $\mathcal{M}$, zato je tudi +$O=s_{A_1A_2}\cap s_{A_2A_3}$. + +Naj bo $k$ krožnica s središčem $O$, ki poteka skozi točko $A_1$. +Dokažimo najprej, da vse točke množice $\mathbf {S}$ ležijo na tej +krožnici. Naj bo $A_i\in \mathbf {S}$ poljubna točka. Iz +dokazanega $O\in s_{A_1A_i}$ in $A_1 \in k$ sledi +$A_i=\mathcal{S}_{s_{A_1A_i}}(A_1)\in +\mathcal{S}_{s_{A_1A_i}}(k)=k$ (Figure \ref{sl.izo.6.12.IMO1.pic}). + + + +\begin{figure}[!htb] +\centering +\input{sl.izo.6.12.IMO1.pic} +\caption{} \label{sl.izo.6.12.IMO1.pic} +\end{figure} + +Brez škode za splošnost lahko predpostavimo, da so točke $\mathbf +{S}$ urejene po vrsti na krožnici $k$, tako da velja $\angle +A_1OA_2<\angle A_1OA_3<\cdots \angle A_1OA_n$ (drugače lahko +naredimo novo označevanje teh točk). To pomeni, da $A_1A_2\ldots +A_n$ predstavlja večkotnik, ki je včrtan krožnici $k$. Dokažimo +še, da je ta večkotnik pravilen. Ker je ta večkotnik koncikličen, je dovolj +dokazati še, da ima vse stranice skladne. Ker je $s_{A_1A_3}$ +somernica množice $\mathbf {S}$ (zato tudi večkotnika $A_1A_2\ldots +A_n$), je $A_1A_2\cong A_2A_3$. Podobno je tudi $s_{A_iA_{i+2}}$ +($i\in \{2,3,\ldots n-1\}$) somernica večkotnika $A_1A_2\ldots A_n$ +in velja $A_iA_{i+1}\cong A_{i+1}A_{i+2}$, kar pomeni, da je +$A_1A_2\ldots A_n$ pravilni večkotnik. + +Množica $\mathbf {S}$ torej predstavlja oglišča pravilnega +večkotnika. + \kdokaz + +%\vspace*{31mm} + +\newpage + +\naloge{Exercises} + +\begin{enumerate} + + \item Dana je premica $p$ ter točki $A$ in $B$, ki ležita na + nasprotnih straneh premice $p$. Konstruiraj točko + $X$, ki leži na +premici $p$, tako da bo razlika $|AX|-|XB|$ maksimalna. + + \item V ravnini so dane premice $p$, $q$ in $r$. Konstruiraj + enakostranični trikotnik $ABC$, tako da + oglišče $B$ leži na premici $p$, $C$ na $q$, +višina iz oglišča $A$ pa na premici $r$. + +\item Dan je štirikotnik $ABCD$ in točka $S$. Načrtaj paralelogram +s središčem v točki $S$, tako +da njegova oglišča ležijo na nosilkah +stranic danega štirikotnika. + +\item Naj bo $\mathcal{I}$ indirektna izometrija ravnine, ki + preslika točko $A$ v točko $B$, +$B$ pa v $A$. Dokaži, da je $\mathcal{I}$ osno zrcaljenje. + +\item Naj bosta $K$ in $L$ točki, ki sta simetrični z ogliščem +$A$ trikotnika $ABC$ glede na +simetrali notranjih kotov ob ogliščih $B$ in $C$. Točka $P$ naj bo +dotikališče včrtane krožnice tega trikotnika in stranice $BC$. +Dokaži, da je $P$ središče daljice $KL$. + +\item Naj bosta $k$ in $l$ krožnici na različnih bregovih premice +$p$. Načrtaj enakostranični trikotnik $ABC$, tako da njegova višina +$AA'$ leži na premici $p$, oglišče $B$ na krožnici $k$, oglišče +$C$ pa na krožnici $l$. + +\item Naj bo $k$ krožnica ter $a$, $b$ in $c$ premice v isti ravnini. +Načrtaj trikotnik $ABC$, ki je +včrtan krožnici $k$, tako da bodo njegove stranice $BC$, $AC$ in + $AB$ vzporedne po vrsti s premicami $a$, $b$ in $c$. + +\item Naj bo $ABCDE$ tetivni petkotnik, v katerem je $BC\parallel +DE$ in $CD\parallel EA$. +Dokaži, da oglišče $D$ leži na simetrali daljice $AB$. + +\item Premice $p$, $q$ in $r$ ležijo v isti ravnini. Dokaži +ekvivalenco $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p + =\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_r$ natanko + tedaj, ko premice $p$, $q$ in $r$ pripadajo istemu šopu. + +\item Naj bodo $O$, $P$ in $Q$ tri nekolinearne točke. Konstruiraj +kvadrat $ABCD$ (v ravnini $OPQ$) s središčem v točki $O$, tako +da točki $P$ in $Q$ po vrsti +ležita na premicah $AB$ in $BC$. + +\item Naj bosta $\mathcal{R}_{S,\alpha}$ rotacija in $\mathcal{S}_p$ +osno zrcaljenje v isti ravnini in $S\in p$. +Dokaži, da kompozituma $\mathcal{R}_{S,\alpha}\circ\mathcal{S}_p$ +in $\mathcal{S}_p\circ\mathcal{R}_{S,\alpha}$ predstavljata osno +zrcaljenje. + +\item Dani sta točka $A$ in krožnica $k$ v isti ravnini. Načrtaj +kvadrat $ABCD$, tako da krajišči diagonale $BD$ ležita na krožnici $k$. + +\item Naj bo $ABC$ poljubni trikotnik. Dokaži: + $$\mathcal{R}_{C,2\measuredangle BCA}\circ + \mathcal{R}_{B,2\measuredangle ABC}\circ + \mathcal{R}_{A,2\measuredangle CAB}=\mathcal{E}.$$ + +\item Dokaži, da kompozitum osnega zrcaljenja $\mathcal{S}_p$ +in središčnega zrcaljenja $\mathcal{S}_S$ ($S\in p$) +predstavlja osno zrcaljenje. + +\item Naj bodo $O$, $P$ in $Q$ tri nekolinearne točke. Konstruiraj +kvadrat $ABCD$ (v ravnini $OPQ$) s središčem v točki $O$, tako + da točki $P$ in $Q$ +ležita po vrsti na premicah $AB$ in $CD$. + +\item Kaj predstavlja kompozitum translacije in središčnega zrcaljenja? + +\item Dani so premica $p$ ter krožnici $k$ in $l$, ki +ležijo v isti ravnini. Načrtaj premico, ki je vzporedna s +premico $p$, tako da na krožnicah $k$ in $l$ določa skladni tetivi. + +\item Naj bo $c$ premica, ki seka vzporednici $a$ in $b$, ter $l$ + daljica. Načrtaj enakostranični trikotnik $ABC$, tako da velja + $A\in a$, $B\in b$, $C\in c$ in $AB\cong l$. + +\item Dokaži, da kompozitum rotacije in osnega zrcaljenja +neke ravnine +predstavlja zrcalni zdrs natanko tedaj, ko središče +rotacije ne leži na osi osnega zrcaljenja. + +\item Naj bo $ABC$ enakostranični trikotnik. Dokaži, +da kompozitum $\mathcal{S}_{AB} + \circ\mathcal{S}_{CA} + \circ\mathcal{S}_{BC}$ +predstavlja zrcalni zdrs. Določi še vektor in os tega zdrsa. + +\item Dani sta točki $A$ in $B$ na istem bregu premice +$p$. +Načrtaj daljico $XY$, ki leži na premici $p$ in je skladna +z dano daljico $l$, tako da bo vsota +$|AX|+|XY|+|YB|$ minimalna. + +\item Naj bo $ABC$ enakokraki pravokotni trikotnik s pravim +kotom pri oglišču $A$. Kaj predstavlja kompozitum: +$\mathcal{G}_{\overrightarrow{AB}}\circ \mathcal{G}_{\overrightarrow{CA}}$? + +\item V isti ravnini so dane premice $a$, $b$ in $c$. +Načrtaj točki $A\in a$ in $B\in b$ +tako, da bo $\mathcal{S}_c(A)=B$. + +\item Dani sta premici $p$ in $q$ ter točka $A$ v isti ravnini. +Načrtaj točki $B$ in $C$ tako, +da bosta premici $p$ in $q$ simetrali notranjih kotov pri +ogliščih $B$ in $C$ trikotnika $ABC$. + +\item Naj bodo $p$, $q$ in $r$ premice ter $K$ in $L$ točki v +isti ravnini. Načrtaj +premici $s$ in $s'$, ki gresta po vrsti skozi točki $K$ in $L$, +tako da velja $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p(s)=s'$. + +\item Naj bo $s$ simetrala enega od kotov, ki jih določata premici $p$ +in $q$. Dokaži, da je $\mathcal{S}_s\circ\mathcal{S}_p = +\mathcal{S}_q\circ\mathcal{S}_s$. + +\item Naj bo $S$ središče trikotniku $ABC$ včrtane krožnice in $P$ točka, + v kateri se ta krožnica dotika stranice $BC$. Dokaži: $$\mathcal{S}_{SC} + \circ\mathcal{S}_{SA}\circ\mathcal{S}_{SB} =\mathcal{S}_{SP}.$$ + +\item Premice $p$, $q$ in $r$ neke ravnine potekajo skozi središče +$S$ krožnice $k$. +Načrtaj trikotnik $ABC$, ki je očrtan tej krožnici, tako da +bodo premice $p$, $q$ in $r$ simetrale notranjih kotov pri +ogliščih $A$, $B$ in $C$ tega trikotnika. + +\item Premice $p$, $q$, $r$, $s$ in $t$ neke ravnine se sekajo +v točki $O$, točka $M$ pa leži na premici $p$. +Načrtaj tak petkotnik, da je $M$ središče ene njegove stranice, +premice $p$, $q$, $r$, $s$ in $t$ pa simetrale stranic. + +\item Točka $P$ leži v ravnini trikotnika $ABC$. Dokaži, da +premice, ki so simetrične s +premicami $AP$, $BP$ in $CP$ glede na simetrale notranjih kotov + ob ogliščih $A$, $B$ in $C$ tega trikotnika, pripadajo istemu šopu. + +\item Izračunaj kot, ki ga določata premici $p$ in $q$, če velja: +$\mathcal{S}_p\circ\mathcal{S}_q\circ\mathcal{S}_p = +\mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_q$. + +\item Naj bosta $\mathcal{R}_{A,\alpha}$ in $\mathcal{R}_{B,\beta}$ +rotaciji v isti ravnini. Določi vse točke $X$ v tej ravnini, za +katere velja +$\mathcal{R}_{A,\alpha}(X)=\mathcal{R}_{B,\beta}(X)$. + +\item Premici $p$ in $q$ se pod kotom $60^0$ sekata v središču $O$ enakostraničnega + trikotnika $ABC$. Dokaži, da + sta odseka, ki jih na premicah določata stranici + trikotnika $ABC$, skladni daljici. + +\item Točka $S$ naj bo središče pravilnega petkotnika $ABCDE$. +Dokaži, da velja: + $$\overrightarrow{SA} + \overrightarrow{SB} + \overrightarrow{SC} + + \overrightarrow{SD} + \overrightarrow{SE} = \overrightarrow{0}.$$ + +\item Dokaži, da se diagonale pravilnega petkotnika +sekajo v točkah, ki so tudi +oglišča pravilnega petkotnika. + +\item Naj bosta $ABP$ in $BCQ$ pravilna trikotnika z +isto orientacijo in $\mathcal{B}(A,B,C)$. Točki $K$ in $L$ sta +središči daljic $AQ$ in $PC$. Dokaži, da je tudi $BLK$ pravilni +trikotnik. + +\item Dane so tri koncentrične krožnice in premica v isti ravnini. +Načrtaj pravilni trikotnik tako, da njegova oglišča +po vrsti ležijo na teh krožnicah, ena stranica pa bo vzporedna +dani premici. + +\item Točka $P$ je notranja točka pravilnega trikotnika +$ABC$, tako da velja +$\angle APB=113^0$ in $\angle BPC=123^0$. Izračunaj velikosti +kotov trikotnika, čigar stranice so skladne z daljicami +$PA$, $PB$ in $PC$. + +\item Dane so nekolinearne točke $P$, $Q$ in $R$. +Načrtaj trikotnik $ABC$, +da bodo $P$, $Q$ in $R$ središča kvadratov, ki so konstruirani +nad stranicami $BC$, $CA$ in $AB$ tega trikotnika. + +\item Naj bosta $A$ in $B$ točki ter $p$ premica v isti +ravnini. Dokaži, da je +kompozitum $\mathcal{S}_B\circ\mathcal{S}_p\circ\mathcal{S}_A$ + osno zrcaljenje natanko tedaj, ko je $AB\perp p$. + +\item Naj bodo $p$, $q$ in $r$ tangente trikotniku $ABC$ včrtane krožnice, +ki so vzporedne z njegovimi stranicami +$BC$, $AC$ in $AB$. Dokaži, +da premice $p$, $q$, $r$, $BC$, $AC$ in $AB$ določajo tak + šestkotnik, v katerem so pari nasprotnih stranic skladne daljice. + +\item Načrtaj trikotnik s podatki: $\alpha$, $t_b$, $t_c$. + +\item Naj bosta $ALKB$ in $ACPQ$ kvadrata, ki sta zunaj trikotnika $ABC$ načrtana +nad stranicama $AB$ in +$AC$ ter $X$ središče stranice $BC$. Dokaži, da +je $AX\perp LQ$ in +$|AX|=\frac{1}{2}|QL|$. + +\item Naj bo $O$ središče pravilnega trikotnika $ABC$ ter $D$ in +$E$ točki stranic $CA$ in $CB$, tako + da velja $CD\cong CE$. Točka $F$ je četrto oglišče paralelograma + $BODF$. Dokaži, +da je trikotnik $OEF$ pravilen. + +\item Naj bo $L$ točka, v kateri se trikotniku +$ABC$ včrtana krožnica dotika njegove stranice $BC$. +Dokaži: $$\mathcal{R}_{C,\measuredangle ACB}\circ\mathcal{R}_{A,\measuredangle BAC} +\circ\mathcal{R}_{B,\measuredangle CBA} =\mathcal{S}_L.$$ + +\item Točke $P$ in $Q$ ter $M$ in $N$ so središča po dveh kvadratov, +ki so zunaj načrtani nad +nasprotnimi stranicami poljubnega štirikotnika. Dokaži, da je +$PQ\perp MN$ in $PQ\cong MN$. + +\item Naj bosta $APB$ in $ACQ$ pravilna trikotnika, ki sta zunaj trikotnika $ABC$ +načrtana nad +stranicama $AB$ in $AC$. Točka $S$ je središče +stranice $BC$ in $O$ središče trikotnika $ACQ$. Dokaži, da je +$|OP|=2|OS|$. + +\item Dokaži, da osno zrcaljenje in translacija neke ravnine +komutirata natanko tedaj, ko je os tega zrcaljenja vzporedna z +vektorjem translacije. + +\item V isti ravnini so dani premica $p$, krožnici $k$ in $l$ ter daljica $d$. +Načrtaj romb $ABCD$ s stranico, ki je skladna daljici $d$, stranica $AB$ leži na +premici $p$, oglišči $C$ in $D$ pa po vrsti ležita na krožnicah $k$ in $l$. + +\item Naj bo $p$ premica, $A$ in $B$ pa točki, ki ležita na istem bregu +premice $p$, ter $d$ daljica v isti ravnini. +Načrtaj točki $X$ in $Y$ na premici $p$ tako, da bo $AX\cong BY$ in $XY\cong d$. + +\item Naj bo $H$ višinska točka trikotnika $ABC$ in $R$ polmer očrtane krožnice tega +trikotnika. Dokaži, da je $|AB|^2+|CH|^2=4R^2$. + +\item Naj bo $EAB$ trikotnik, ki je načrtan nad stranico $AB$ kvadrata +$ABCD$. Naj bo tudi $M=pr_{\perp AE}(C)$ in $N=pr_{\perp BE}(D)$ ter točka $P$ +presečišče premic $CM$ in $DN$. Dokaži, da je $PE\perp AB$. + +\item Načrtaj enakostranični trikotnik $ABC$ tako, da njegova oglišča po vrsti +ležijo na treh vzporednicah $a$, $b$ in $c$ v isti ravnini, +središče tega trikotnika pa leži na premici $s$, ki seka +premice $a$, $b$ in $c$. + +\item Če ima petkotnik vsaj dve osi simetrije, je pravilen. Dokaži. + +\item Naj bodo $A$, $B$ in $C$ tri kolinearne točke. Kaj predstavlja +kompozitum $\mathcal{G}_{\overrightarrow{BC}}\circ \mathcal{S}_A$? + +\item Naj bodo $p$, $q$ in $r$ premice, ki niso iz istega šopa, in $A$ točka v isti +ravnini. Načrtaj premico $s$, ki poteka skozi točko $A$, tako da velja +$\mathcal{S}_r\circ \mathcal{S}_q\circ \mathcal{S}_p(s)=s'$ in $s\parallel s'$. + +%nove naloge +%___________________________________ + +\item Naj bosta $Z$ in $K$ notranji točki pravokotnika $ABCD$. +Načrtaj točke $A_1$, $B_1$, $C_1$ in $D_1$, ki po vrsti ležijo na +stranicah $AB$, $BC$, $CD$ in $DA$ tega pravokotnika, tako da velja +$\angle ZA_1A\cong\angle B_1A_1B$, $\angle A_1B_1B\cong\angle C_1B_1C$, +$\angle B_1C_1C\cong\angle D_1C_1D$ in $\angle C_1D_1D\cong\angle KD_1A$. + +\item Točka $A$ leži na premici $a$, točka $B$ pa na premici $b$. +Določi rotacijo, ki preslika premico $a$ v premico $b$ in točko $A$ v točko $B$. + +\item V središču kvadrata se sekata dve pravokotnici. +Dokaži, da ti pravokotnici sekata stranice kvadrata v točkah, ki so +oglišča novega kvadrata. + +\item Dana je krožnica $k$ in premice $a$, $b$, $c$, $d$ in $e$, ki ležijo +v isti ravnini. Krožnici $k$ včrtaj petkotnik s stranicami, ki +so po vrsti vzporedne s premicami $a$, $b$, $c$, $d$ in $e$. + +\item Točka $P$ leži v notranjosti kota $aOb$. Načrtaj premico $p$ skozi točko $P$, + ki s krakoma $a$ in $b$ določa trikotnik z najmanjšo ploščino. + +\item Paralelogram $PQKL$ naj bo včrtan v paralelogram $ABCD$ (oglišča prvega ležijo na stranicah +drugega). Dokaži, da imata paralelograma skupno središče. + +\item Loki $l_1, l_2,\cdots , l_n$ ležijo na krožnici $k$ in je vsota njihovih +dolžin manjša od polobsega te krožnice. Dokaži, da obstaja tak premer +$PQ$ krožnice $k$, da nobeno od njegovih krajišč ne leži na nobenem od +lokov $l_1, l_2,\cdots , l_n$. + +\item Dana je krožnica $k(S,20)$. Igralca $\mathcal{A}$ in $\mathcal{B}$ +izmenično rišeta krožnici s polmeri $x_i$ ($1AB$, +oz. $b>c$. Ker je $\angle AMB =\omega$, po izreku +\ref{ObodKotGMT} točka $M$ leži na loku $l$ s tetivo $AB$ in obodnim kotom +$\omega$. Razteg $h_{B,2}$ preslika točke $B$ in $M$ po vrsti v +točke $B$ in $C$, lok $l$ pa v lok $l'$. Iz $M\in l$ sledi $C\in +l'$. Ker je še $AC=b$, sledi da točka $C$ leži tudi na +krožnici $k(A,b)$. Torej $C\in l'\cap k(A,b)$. Dokazana dejstva +omogočajo konstrukcijo. + + Načrtajmo najprej daljico $AB=c$ ter lok $l$ s to tetivo in obodnim kotom +$\omega$, nato pa lok $l'=h_{B,2}(l)$ ter na koncu točko $C$ kot +eno od presečišč loka $l'$ s krožnico $k(A,b)$. + +Dokažimo, da konstruirani trikotnik $ABC$ izpolnjuje pogoje iz +naloge. Po konstrukciji je takoj $AB=c$. Iz $C\in k(A,b)$ sledi +$AC=b$. Naj bo $M=h^{-1}_{B,2}(C)$. Ker je $C\in l'=h_{B,2}(l)$, je +$M \in l$. Ker je po konstrukciji $l$ lok s to tetivo in obodnim +kotom $\omega$, je $\angle AMB=\omega$. Potrebno je še dokazati, +da je točka $M$ središče daljice $BC$, kar sledi direktno iz +relacije $M=h^{-1}_{B,2}(C)$. + +Poiščimo pogoje za število rešitev naloge. Omenili smo že, da +mora zaradi pogoja $\omega<90^0$ biti $b>c$. V primeru $b\leq c$ + rešitve ni. Število rešitev naloge je naprej odvisno od + števila presečišč loka $l'$ s krožnico $k(A,b)$. + \kdokaz + + + + + +%________________________________________________________________________________ + \poglavje{Classification of Similarity Transformations} \label{odd7PrezentTransPod} + +V razdelku \ref{odd6KlasifIzo} smo klasificirali izometrije, tu bomo pa na podoben način naredili klasifikacijo transformacij podobnosti. Tudi pri tej klasifikaciji +bo pomembno število fiksnih točk in dejstvo, ali je transformacija podobnosti direktna ali indirektna. + +V prejšnjih dveh razdelkih smo ugotovili, da vse izometrije in središčni raztegi predstavljajo transformacije podobnosti. Prav tako je njun kompozitum transformacija podobnosti (izrek \ref{TransPodGrupa}). Sedaj bomo dokazali, da velja tudi obratno. + + + + \bizrek \label{TransPodKompHomIzo} + Each similarity transformations $f$ with coefficient $k$ can be expressed as the product + of one isometry and one homothety with an arbitrary centre: + $$f=h_{S,k}\circ\mathcal{I}_1=\mathcal{I}_2\circ h_{S,k}.$$ + \eizrek + +\textbf{\textit{Proof.}} + +Naj bo $f$ poljubna transformacija podobnosti s koeficientom $k$. +Če s $h_{S,\frac{1}{k}}$ označimo središčni razteg s poljubnim središčem $S$ in +koeficientom $\frac{1}{k}$, potem kompozitum $h_{S,\frac{1}{k}}\circ f$ predstavlja +transformacijo podobnosti s koeficientom $k\cdot \frac{1}{k}=1$ (izrek \ref{TransPodGrupa}) oz. izometrijo. Torej $h_{S,\frac{1}{k}}\circ f=\mathcal{I}_1$, kjer je $\mathcal{I}_1$ neka izometrija. Po izreku \ref{homotGrupa} je $f=h_{S,\frac{1}{k}}^{-1}\circ\mathcal{I}_1=h_{S,k}\circ\mathcal{I}_1$. +Prav tako je $f\circ h_{S,\frac{1}{k}}=\mathcal{I}_2$, kjer je $\mathcal{I}_2$ neka izometrija, zato je $f=\mathcal{I}_2\circ h_{S,k}$. + \kdokaz + + + \bizrek \label{TransPodOhranjaKote} + Similarity transformations preserve the measure of angles, i.e. there map an angle to the congruent angle. + \eizrek + + +\textbf{\textit{Proof.}} + Trditev je direktna posledica izrekov \ref{TransPodKompHomIzo} in \ref{homotOhranjaKote}. +\kdokaz + + \bizrek \label{homotTransm} + For each isometry $\mathcal{I}$ and each homothety $h_{S,k}$ + it is: + $$\mathcal{I}\circ h_{S,k}\circ \mathcal{I}^{-1}=h_{\mathcal{I}(S),k}$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.2.pic} +\caption{} \label{sl.pod.7.1p.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure +\ref{sl.pod.7.1p.2.pic}) + +Naj bo $\mathcal{I}(S)=S_1$. Označimo z $f=\mathcal{I}\circ h_{S,k}\circ \mathcal{I}^{-1}$. Potrebno je dokazati, da je $f=h_{S_1,k}$ oz. $f(X_1)=h_{S_1,k}(X_1)$ za poljubno točko $X_1$ te ravnine. + + +Označimo še $X=\mathcal{I}^{-1}(X_1)$, $X'=h_{S,k}(X)$ in +$X'_1=\mathcal{I}(X_1)$. Potem je: +\begin{eqnarray*} +f(X_1)&=& \mathcal{I}\circ h_{S,k}\circ \mathcal{I}^{-1}(X_1)\\ + &=& \mathcal{I}\circ h_{S,k}(X)\\ + &=& \mathcal{I}(X')\\ + &=& X'_1 +\end{eqnarray*} + + Torej $f(X_1)=X'_1$. Iz $X'=h_{S,k}(X)$ sledi $\overrightarrow{SX'}=k\cdot\overrightarrow{SX}$. Ker je $\mathcal{I}$ izometrija in $\mathcal{I}:\hspace*{1mm}S,X,X'\mapsto S_1,X_1,X'_1$, je tudi +$\overrightarrow{S_1X'_1}=k\cdot\overrightarrow{S_1X_1}$ oz. $h_{S_1,k}(X_1)=X'_1$. +To pomeni, da za poljubno točko $X_1$ velja $f(X_1)=X'_1=h_{S_1,k}(X_1)$, zato je +$f=h_{S_1,k}$. +\kdokaz + + + + \bizrek \label{homotIzomKom} + An Isometry $\mathcal{I}$ and a homothety $h_{S,k}$ commute + if and only if + the centre of this homothety is a fixed point of the isometry $\mathcal{I}$, i.e.: + $$\mathcal{I}\circ h_{S,k}=h_{S,k}\circ\mathcal{I}\hspace*{1mm} + \Leftrightarrow\hspace*{1mm}\mathcal{I}(S)=S.$$ + \eizrek + + +\textbf{\textit{Proof.}} + Po prejšnjem izreku (\ref{homotTransm}) je: + \begin{eqnarray*} +\mathcal{I}\circ h_{S,k}=h_{S,k}\circ\mathcal{I} +\hspace*{1mm} + &\Leftrightarrow& \hspace*{1mm} \mathcal{I}\circ h_{S,k}\circ\mathcal{I}^{-1}=h_{S,k}\\ +\hspace*{1mm} + &\Leftrightarrow& \hspace*{1mm} h_{\mathcal{I}(S),k}=h_{S,k}\\ +\hspace*{1mm} + &\Leftrightarrow& \hspace*{1mm} \mathcal{I}(S)=S, +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + +Če za izometrijo iz izreka \ref{TransPodKompHomIzo} izberemo rotacijo z istim središčem kot središčni razteg, dobimo zelo koristno vrsto transformacij podobnosti. + +Kompozitum rotacije in središčnega raztega z istim središčem imenujemo \index{rotacijski razteg} \pojem{rotacijski razteg} (Figure +\ref{sl.pod.7.1p.1.pic}): +$$\rho_{S,k,\omega}=h_{S,k}\circ \mathcal{R}_{S,\omega}$$ +s \index{središče!rotacijskega raztega}\pojem{središčem} $S$, \index{koeficient!rotacijskega raztega}\pojem{koeficientom} $k$ in \index{kot!rotacijskega raztega}\pojem{kotom} $\omega$. + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.1.pic} +\caption{} \label{sl.pod.7.1p.1.pic} +\end{figure} + +Po izreku \ref{homotIzomKom} je: +$$\rho_{S,k,\omega}=h_{S,k}\circ \mathcal{R}_{S,\omega}= +\mathcal{R}_{S,\omega}\circ h_{S,k}.$$ + + +Ker sta središčni razteg in rotacija direktni transformaciji (izreka \ref{homotDirekt} in \ref{RotacDirekt}), je tudi rotacijski razteg direktna transformacija podobnosti. + +Jasno je, da tudi središčni razteg lahko obravnavamo kot vrsto rotacijskega raztega, če privzamemo, da je identiteta rotacija za kot $0^0$: +$$h_{S,k}=\rho_{S,k,0^0}.$$ + +Prav tako lahko tudi rotacijo vidimo kot vrsto rotacijskega raztega: +$$\mathcal{R}_{S,\omega}=\rho_{S,1,\omega}.$$ + + + \bizrek \label{rotacRaztKot} + An arbitrary line and its image under a stretch rotation + determine an oriented angle which is congruent to the angle of this stretch rotation: + $$\rho_{S,k,\omega}(p)=p'\hspace*{1mm} \Rightarrow + \hspace*{1mm} \angle p,p'=\omega.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.3.pic} +\caption{} \label{sl.pod.7.1p.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Naj bo $p'=\rho_{S,k,\omega}(p)$ slika premice $p$ pri rotacijskem raztegu $\rho_{S,k,\omega}=h_{S,k}\circ \mathcal{R}_{S,\omega}$ (Figure \ref{sl.pod.7.1p.3.pic}) ter $p_1=\mathcal{R}_{S,\omega}(p)$. Potem je $h_{S,k}(p_1)=p'$. Po izreku \ref{rotacPremPremKot} je $\measuredangle p,p_1=\omega$. Po izreku \ref{RaztPremica} je $p_1\parallel p'$. Torej $\measuredangle p,p'=\angle p,p_1=\omega$ (izrek \ref{KotiTransverzala1}). +\kdokaz + + + + \bizrek \label{rotacRaztKompSredZrc} + The product of a half-turn and a stretch rotation + with the same centre is a stretch rotation. Furthermore: + $$\rho_{S,k,\omega}\circ \mathcal{S}_S= + \mathcal{S}_S\circ\rho_{S,k,\omega}=\rho_{S,-k,\omega}.$$ + \eizrek + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.3a.pic}) + +Kot smo že omenili v razdelku \ref{odd7SredRazteg}, je $\mathcal{S}_S=h_{S,-1}$. +Če uporabimo izrek \ref{homotGrupa}, dobimo: +\begin{eqnarray*} +\rho_{S,k,\omega}\circ \mathcal{S}_S= +\mathcal{R}_{S,\omega}\circ h_{S,k}\circ h_{S,-1}= +\mathcal{R}_{S,\omega}\circ h_{S,-k}=\rho_{S,-k,\omega}. +\end{eqnarray*} +Na enak način dobimo tudi: +\begin{eqnarray*} +\mathcal{S}_S\circ\rho_{S,k,\omega}= +h_{S,-1}\circ h_{S,k}\circ \mathcal{R}_{S,\omega}= +h_{S,-k}\circ\mathcal{R}_{S,\omega}=\rho_{S,-k,\omega}, +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.3a.pic} +\caption{} \label{sl.pod.7.1p.3a.pic} +\end{figure} + + A direct consequence is the following theorem. + + \bizrek \label{rotacRaztNegKoefk} + For each stretch rotation is: + $$\rho_{S,-k,\omega}=\rho_{S,k,180^0+\omega}.$$ + \eizrek + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.3a.pic}) + +Po prejšnjem izreku \ref{rotacRaztKompSredZrc} in izreku \ref{rotacKomp2rotac} je: +\begin{eqnarray*} +\rho_{S,-k,\omega}= +\mathcal{S}_S\circ\rho_{S,k,\omega}= +\mathcal{S}_S\circ\mathcal{R}_{S,\omega}\circ h_{S,k} +=\mathcal{R}_{S,180^0+\omega}\circ h_{S,k}= +\rho_{S,k,180^0+\omega}, +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + +Če za izometrijo iz izreka \ref{TransPodKompHomIzo} izberemo zrcaljenje čez premico, ki poteka skozi središče središčnega raztega, dobimo še eno vrsto transformacij podobnosti. + +Kompozitum osnega zrcaljenja $s$ in središčnega raztega $h_{S,k}$ s središčem $S\in s$ imenujemo \index{osni razteg} \pojem{osni razteg} (Figure +\ref{sl.pod.7.1p.1a.pic}): +$$\sigma_{S,k,s}=h_{S,k}\circ \mathcal{S}_s;\hspace*{2mm} (S\in s)$$ +s \index{središče!osnega raztega}\pojem{središčem} $S$, \index{koeficient!rotacijskega raztega}\pojem{koeficientom} $k$ in \index{os!osnega raztega}\pojem{osjo} $s$. + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.1a.pic} +\caption{} \label{sl.pod.7.1p.1a.pic} +\end{figure} + +Po izreku \ref{homotIzomKom} je: +$$\sigma_{S,k,s}=h_{S,k}\circ \mathcal{S}_s= +\mathcal{S}_s\circ h_{S,k}.$$ + +Ker je središčni razteg direktna in rotacija indirektna transformacija (izreka \ref{homotDirekt} in \ref{izozrIndIzo}), je osni razteg indirektna transformacija podobnosti. + + + + + \bizrek \label{transPod1FixTocLema} + Let $f$ be a similarity transformation that is not an isometry. + If $f$ maps each line to its parallel line, then $f$ is a homothety. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.1bb.pic} +\caption{} \label{sl.pod.7.1p.1bb.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.1bb.pic}) + +Označimo s $k$ koeficient podobnosti transformacije $f$. Ker po predpostavki $f$ ni izometrija, je $k\neq 1$. + +Po predpostavki $f$ preslika vsako premico v vzporedno premico. Dokažimo najprej, da obstajata vsaj dve premici, ki se sekata in se ne preslikata vase. +Naj bodo $X$, $Y$ in $Z$ poljubne nekolinearne točke ravnine ter $p$, $q$ in $r$ premice, določene s točkami $X$, $Y$ in $Z$: $p=XY$, $q=YZ$ in $r=XZ$. Označimo s $p'=f(p)$, $q'=f(q)$ in $r'=f(r)$. Po predpostavki je $p\parallel p'$, $q\parallel q'$ in $r\parallel r'$. Dokažimo, da se vsaj ena od treh premic ne preslika vase. Predpostavimo nasprotno, da je $p=p'$, $q=q'$ in $r=r'$. Toda v tem primeru je $f(X)=f(p\cap q)=f(p)\cap f(q)=p\cap q=X$ in podobno npr. $f(Y)=Y$. Daljica $XY$ bi se v tem primeru preslikala vase, kar ni možno, saj je $k\neq 1$. Brez škode za splošnost naj bo $p\neq p'$. Na isti način z uporabo nekega trikotnika, pri katerem nobena nosilka stranic ni vzporedna s premico $p$, lahko dokažemo, da obstaja še ena premica, ki seka premico $p$ in se s preslikavo $f$ ne preslika vase. + +Torej obstajata premici $b$ in $c$, ki se sekata v točki $A$ ter za $b'=f(b)$ in $c'=f(c)$ velja $b\parallel b'$, $c\parallel c'$, $b\neq b'$ in $c\neq c'$. Naj bosta $B\in b$ ter $C\in c$ poljubni točki, ki sta različni od točke $A$. +Označimo $A'=b\cap c$, $B'=f(B)$ in $C'=f(C)$. Najprej je $f(A)=f(b\cap c)=f(b)\cap f(c)=b'\cap c'=A'$, $B'\in b'$ in $C'\in c'$. +Ker je $b\parallel b'$, je tudi $AB\parallel A'B'$. Daljica $AB$ se s transformacijo $f$ slika v daljico $A'B'$, zato je $A'B'=k\cdot AB$. +Premici $AA'$ in $BB'$ nista vzporedni. V nasprotnem bi bil štirikotnik $ABB'A'$ paralelogram oz. $AB\cong A'B'$ (izrek \ref{paralelogram}), kar ni možno, ker je $k\neq 1$. Označimo s $S$ presečišče premic $AA'$ in $BB'$. Ker je $b\parallel b'$, je po Talesovem izreku: +$$\frac{SA'}{SA}=\frac{SB'}{SB}=\frac{A'B'}{AB}=k.$$ + To pomeni, da središčni razteg $h_{S,k_1}$ s središčem $S$ in koeficientom $k_1=k$ (ali $k_1=-k$) preslika točki $A$ in $B$ v točki $A'$ in $B'$. + Naj bo $\widehat{C'}=h_{S,k_1}(C)$. Ker je po predpostavki $C'=f(C)$ oz. $A'C'=k\cdot AC$, je po izreku \ref{RaztPremica} $\widehat{C'}=C'$ oz. $h_{S,k_1}(C)=C'$. + + Preslikava +$$g=f^{-1}\circ h_{S,k_1}$$ + je po izreku \ref{TransPodGrupa} transformacija podobnosti s koeficientom podobnosti $\frac{1}{k}\cdot |k_1|=\frac{1}{k}\cdot k=1$, zato predstavlja izometrijo. +Toda $g(A)=f^{-1}\circ h_{S,k_1}(A)=f^{-1}(A')=A$, oz. $A$ je fiksna točka izometrije $g$. Na podoben način dokažemo, da sta tudi $B$ in $C$ fiksni točki izometrije $g$, kar pomeni, da je $g=\mathcal{E}$ identiteta (izrek \ref{IizrekABC2}). Torej $f^{-1}\circ h_{S,k_1}=g=\mathcal{E}$ oz. $f=h_{S,k_1}$. +\kdokaz + + Sedaj smo pripravljeni na naslednji pomemben izrek. + + \bizrek \label{transPod1FixToc} + Each similarity transformation other than isometry has exactly one fixed point. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.1c.pic} +\caption{} \label{sl.pod.7.1p.1c.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.1c.pic}) + +Naj bo $f$ transformacija podobnosti s koeficientom $k$. Po predpostavki je $k\neq 1$. +Podobno kot v dokazu prejšnjega izreka \ref{transPod1FixTocLema} se hitro prepričamo, da $f$ ne more imeti dveh fiksnih točk, ker bi bil $k=1$. + +Dokažimo, da ima $f$ fiksno točko. +Po prejšnjem izreku (\ref{transPod1FixTocLema}) lahko predpostavimo, da obstaja vsaj ena premica $p$, ki se ne preslika v svojo vzporednico. Res, če predpostavimo, da se vsaka premica ravnine slika v svojo vzporednico, je po omenjenem izreku \ref{transPod1FixTocLema} preslikava $f$ središčni razteg, ki ima fiksno točko (njegovo središče). +Naj bo torej $p'=f(p)$ in $ p'\nparallel p$ ($\neg p'\parallel p$). Označimo z $A$ presečišče premic $p$ in $p'$. Naj bo $A'=f(A)$. Če je $A'=A$, je $A$ fiksna točka transformacije $f$ in je dokaz končan. Naj bo torej $A'\neq A$. Iz $A\in p$ sledi $A'\in p'$. Označimo z $q$ vzporednico premice $p$ skozi točko $A'$. Ker je $A'\neq A$, je $q\neq p$. Naj bo $q'=f(q)$. Ker je $p\parallel q$ in $f:\hspace*{1mm}p,q\rightarrow p',q'$, je tudi $p'\parallel q'$ (če bi se $p'$ in $q'$ sekali v neki točki $T$, bi se tudi premici $p$ in $q$ sekali v točki $f^{-1}(T)$). Na enak način iz $q\neq p$ sledi $q'\neq p'$. To pomeni, da premice $p$, $q$, $p'$ in $q'$ določajo paralelogram $AA'BC$, kjer je + $B=q\cap q'$ in $C=p\cap q'$. Naj bo $B'=f(B)$. Iz $B\in q$ sledi $B'\in q'$. Če je $B'=B$, je $B$ fiksna točka in je dokaz končan. Tako lahko naprej predpostavimo $B'\neq B$. Iz $p'\parallel q'$ sledi $AA'\parallel BB'$. V primeru $AB\parallel A'B'$ bi bil štirikotnik $AA'B'B$ paralelogram oz. $AB\cong A'B'$, kar ni možno, ker je $f:\hspace*{1mm}A,B\rightarrow A',B'$ in $k\neq 1$. Torej se premici $AB$ in $A'B'$ sekata v neki točki $S$. Označimo $S'=f(S)$. Dokažimo, da je $S$ fiksna točka oz. $S'=S$. + Po Talesovem izreku je: +\begin{eqnarray} \label{eqnTransfPod1Ft1} + \frac{AS}{SB}=\frac{A'S}{SB'} +\end{eqnarray} + Ker $f:\hspace*{1mm}A,B,S\rightarrow A',B',S'$, je tudi (izrek \ref{TransPodOhranjajoRazm}): +\begin{eqnarray} \label{eqnTransfPod1Ft2} + \frac{AS}{SB}=\frac{A'S'}{S'B'} +\end{eqnarray} +Prav tako iz $S\in AB$ sledi $S'\in A'B'$. +Če povežemo zadnji dve enakosti \ref{eqnTransfPod1Ft1} in + \ref{eqnTransfPod1Ft2}, je: +\begin{eqnarray} \label{eqnTransfPod1Ft3} + \frac{A'S}{SB'}=\frac{A'S'}{S'B'}, +\end{eqnarray} +kjer sta $S$ in $S'$ točki premice $A'B'$. Toda transformacije podobnosti ohranjajo relacijo $\mathcal{B}$ (izrek \ref{TransPodB}), oziroma: +\begin{eqnarray*} + \mathcal{B}(A,S,B)\hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} \mathcal{B}(A',S',B');\\ + \mathcal{B}(S,A,B)\hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} \mathcal{B}(S',A',B');\\ +\mathcal{B}(A,B,S)\hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} \mathcal{B}(A',B',S'). +\end{eqnarray*} + Iz tega in iz relacije \ref{eqnTransfPod1Ft3} sledi: +\begin{eqnarray*} + \frac{\overrightarrow{A'S}}{\overrightarrow{SB'}}= +\frac{\overrightarrow{A'S'}}{\overrightarrow{S'B'}}, +\end{eqnarray*} + zato je po izreku \ref{izrekEnaDelitevDaljiceVekt} $S'=S$ oz. $S$ je fiksna točka transformacije podobnosti $f$. +\kdokaz + +Sedaj lahko naredimo napovedano klasifikacijo transformacij podobnosti. + + + + \bizrek \label{transPodKlasif} + The only similarity transformations of the plane are: + \begin{itemize} + \item isometries, + \item homotheties, + \item stretch rotations, + \item stretch reflections. + \end{itemize} + \eizrek + +\textbf{\textit{Proof.}} +Naj bo $f$ poljubna transformacija podobnosti s koeficientom $k$. + +Če je $k=1$, je $f$ izometrija. + +Predpostavimo, da je $k\neq 1$ oz. $f$ ni izometrija. Po izreku \ref{transPod1FixToc} ima $f$ natanko eno fiksno točko - označimo jo s $S$. Torej $f(S)=S$. Po izreku \ref{TransPodKompHomIzo} lahko transformacijo podobnosti $f$ predstavimo kod kompozitum središčnega raztega s poljubnim središčem (izberimo za središče ravno točko $S$) in koeficientom $k$ ter ene izometrije $\mathcal{I}$: +$$f=\mathcal{I}\circ h_{S,k}.$$ + Ker je $S$ fiksna točka transformacije podobnosti $f$ in središčnega raztega $h_{S,k}$, +je: +$$S=f(S)=\mathcal{I}\circ h_{S,k}(S)=\mathcal{I}(S).$$ +Torej $\mathcal{I}(S)=S$ oz. $\mathcal{I}$ je izometrija s fiksno točko $S$. Po izreku \ref{Chaslesov} je $\mathcal{I}$ lahko: identiteta, rotacija s središčem $S$ ali zrcaljenje čez premico, ki poteka skozi točko $S$: +\begin{eqnarray*} +\mathcal{I}=\left\{ + \begin{array}{l} + \mathcal{E}, \\ + \mathcal{R}_{S,\omega}, \\ + \mathcal{S}_s (S\in s) + \end{array} + \right. +\end{eqnarray*} +Zato je: +\begin{eqnarray*} +f=\mathcal{I}\circ h_{S,k}=\left\{ + \begin{array}{l} + h_{S,k}, \\ + \rho_{S,k,\omega}, \\ + \sigma_{S,k,s} + \end{array} + \right. +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + A direct consequence is the following theorem. + + + \bizrek \label{transPodKlasifDirInd} + The only direct similarity transformations of the plane other than isometries are: + \begin{itemize} + \item homotheties, + \item stretch rotations. + \end{itemize} + The only opposite similarity transformations of the plane other than isometries are: + \begin{itemize} + \item stretch reflections. + \end{itemize} + \eizrek + + +Zelo koristna je lastnost kompozituma dveh rotacijskih raztegov. + + + + \bizrek \label{RotRazKomoz} + The product of two stretch rotations $\rho_{S_1,k_1,\omega_1}$ and $\rho_{S_2,k_2,\omega_2}$ + is a direct isometry, a homothety or a stretch rotation. Furthermore (Figure \ref{sl.pod.7.1p.5a.pic}): + \begin{eqnarray*} + \rho_{S_2,k_2,\omega_2}\circ \rho_{S_1,k_1,\omega_1}= + \left\{ + \begin{array}{ll} + \mathcal{R}_{S,\omega}, & k_1k_2=1, \hspace*{2mm} + \omega=\omega_1+\omega_2\neq n\cdot 180^0; \\ + \mathcal{T}_{\overrightarrow{v}}, & k_1k_2=1, \hspace*{2mm} + \omega=\omega_1+\omega_2=n\cdot 180^0; \\ + h_{S,k}, & k=k_1k_2\neq 1, \hspace*{2mm} + \omega=\omega_1+\omega_2=n\cdot 180^0; \\ + \rho_{S,k,\omega}, & k=k_1k_2\neq 1, \hspace*{2mm} + \omega=\omega_1+\omega_2\neq n\cdot 180^0. + \end{array} + \right. + \end{eqnarray*} + for $n\in \mathbb{Z}$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.5a.pic} +\caption{} \label{sl.pod.7.1p.5a.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.5.pic}) + +Brez škode za splošnost predpostavimo, da velja $k_1>0$ in $k_2>0$. Če je npr. $k_1<0$, lahko po izreku \ref{rotacRaztNegKoefk} zapišemo + $\rho_{S,k_1,\omega}=\rho_{S,-k_1,180^0+\omega}$, kjer je $-k_1>0$. + +Označimo $f=\rho_{S_2,k_2,\omega_2}\circ \rho_{S_1,k_1,\omega_1}$. Po izrekih \ref{RaztTransPod} in \ref{TransPodGrupa} je $f$ transformacija podobnosti +s koeficientom $k=k_1\cdot k_2$. Le-ta je direktna transformacija kot kompozitum dveh direktnih transformacij. Po izreku \ref{transPodKlasifDirInd} je $f$ lahko izometrija, središčni razteg ali rotacijski razteg. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1p.5.pic} +\caption{} \label{sl.pod.7.1p.5.pic} +\end{figure} + + +Naj bo $p$ poljubna premica in $f(p)=p'$. Izračunajmo mero orientiranega kota $\measuredangle p,p'$. Naj bo $\rho_{S_1,k_1,\omega_1}(p)=p_1$ in posledično $\rho_{S_2,k_2,\omega_2}(p_1)=p'$. Po izreku \ref{rotacRaztKot} je $\angle p,p_1=\omega_1$ in $\angle p_1,p'=\omega_2$. +Če je $\omega=\omega_1+\omega_2\neq n\cdot 180^0$ za vsak $n\in \mathbb{Z}$, je +$\angle p,p'=\omega_1+\omega_2$ (izrek \ref{zunanjiNotrNotr}). +Če je $\omega=\omega_1+\omega_2=n\cdot 180^0$ za nek $n\in \mathbb{Z}$, je $p\parallel p'$ (izrek \ref{KotiTransverzala}). + + +Če je $k_1\cdot k_2=1$, je kompozitum $f$ transformacija podobnosti s koeficientom $k=1$ in predstavlja direktno izometrijo $f=\mathcal{I}\in \mathfrak{I}^+$. Le-ta je lahko translacija ali rotacija (v posebnem primeru identiteta) (izrek \ref{Chaslesov+}). + Če je $\omega=\omega_1+\omega_2\neq n\cdot 180^0$ oz. +$\angle p,p'=\omega_1+\omega_2$, gre za rotacijo za kot $\omega=\omega_1+\omega_2$ (izrek \ref{rotacPremPremKot}). + Če je $\omega=\omega_1+\omega_2=n\cdot 180^0$ oz. $p\parallel p'$, je $\mathcal{I}$ translacija (ali identiteta). + +Predpostavimo sedaj, da velja $k_1\cdot k_2\neq 1$. Kot smo že omenili, je v tem primeru $f$ lahko središčni razteg ali rotacijski razteg s koeficientom $k=k_1\cdot k_2$. + Če je $\omega=\omega_1+\omega_2\neq n\cdot 180^0$ oz. +$\angle p,p'=\omega_1+\omega_2$, gre za rotacijski razteg za kot $\omega=\omega_1+\omega_2$ (izrek \ref{rotacRaztKot}). + Če je $\omega=\omega_1+\omega_2=n\cdot 180^0$ oz. $p\parallel p'$, je $f$ po izreku \ref{RaztPremica} središčni razteg (ali središčno zrcaljenje, če je $k_1\cdot k_2=-1$). +\kdokaz + +A direct consequence is the following theorem. + + + \bizrek + + The product of two homotheties + is a direct isometry or a homothety. Furthermore: + \begin{eqnarray*} + h_{S_2,k_2}\circ h_{S_1,k_1}= + \left\{ + \begin{array}{ll} + \mathcal{T}_{\overrightarrow{v}}, & k_1k_2=1; \\ + h_{S,k}, & k=k_1k_2\neq \pm 1. + \end{array} + \right. + \end{eqnarray*} + \eizrek + +\textbf{\textit{Proof.}} +Trditev sledi direktno iz prejšnjega izreka, če zapišemo +$h_{S_1,k_1}=\rho_{S_1,k_1,0^0}$ in $h_{S_2,k_2}=\rho_{S_2,k_2,0^0}$. +\kdokaz + + + + \bnaloga\footnote{17. IMO Bulgaria - 1975, Problem 3.} + On the sides of an arbitrary triangle $ABC$, triangles $ABR$, $BCP$, $CAQ$ are + constructed externally with $\angle CBP\cong\angle CAQ=45^0$, $\angle BCP\cong\angle ACQ=35^0$, + $\angle ABR\cong\angle BAR=15^0$. Prove that $\angle QRP=90^0$ and $QR\cong RP$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.1.IMO2.pic} +\caption{} \label{sl.pod.7.1.IMO2.pic} +\end{figure} + +\textbf{\textit{Solution.}} Naj bo $S$ tretje oglišče +enakostraničnega trikotnika $BAS$, ki je načrtan nad stranico $BA$ +trikotnika $ABC$ (Figure \ref{sl.pod.7.1.IMO2.pic}). Ker je $\angle +RBS\cong\angle SAR=45^0$ in $\angle BSR\cong\angle ASR=30^0$, so +si trikotniki $BPC$, $AQC$, $BRS$ in $ARS$ med seboj podobni. + +Naj bosta $f_1$ in $f_2$ rotacijska raztega: + \begin{eqnarray*} + &&f_1=\rho_{B,k,45^0}=h_{B,k}\circ \mathcal{R}_{B,45^0}\\ + &&f_2=\rho_{A,\frac{1}{k},45^0}=h_{A,\frac{1}{k}}\circ + \mathcal{R}_{A,45^0}, + \end{eqnarray*} +kjer je: +$$k=\frac{|BC|}{|BP|}=\frac{|AC|}{|AQ|}=\frac{|BS|}{|BR|} +=\frac{|AS|}{|AR|}.$$ + Naj bo $f=f_2\circ f_1$. Po izreku \ref{RotRazKomoz} je: + $$f=f_2\circ f_1=\rho_{A,\frac{1}{k},45^0}\circ + \rho_{B,k,45^0}=\rho_{T,1,90^0}=\mathcal{R}_{T,90^0}.$$ + Naprej velja: + \begin{eqnarray*} + &&\mathcal{R}_{T,90^0}(P)=f(P)=f_2\circ f_1(P)=f_2(C)=Q\\ + &&\mathcal{R}_{T,90^0}(R)=f(R)=f_2\circ f_1(R)=f_2(S)=R. + \end{eqnarray*} + Iz prejšnje (druge) relacije vidimo, da je $\mathcal{R}_{T,90^0}(R)=R$, kar pomeni, da je + $R=T$ (izrek \ref{RotacFiksT}) oz. + $\mathcal{R}_{T,90^0}=\mathcal{R}_{R,90^0}$. + Iz tega in iz prve relacije sedaj sledi: + $$\mathcal{R}_{R,90^0}(P)=Q,$$ + kar pomeni, da velja $\angle QRP=90^0$ + in $QR\cong RP$. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Similar Figures. Similarity of Triangles} \label{odd7PodobTrik} + + +Kot smo že napovedali v uvodu tega poglavja, nam transformacije podobnosti omogočajo definicijo pojma podobnosti likov. + +Pravimo, da je lik $\Phi$ \index{lika!podobna}\pojem{podoben} liku $\Phi'$ iz iste ravnine (oznaka $\Phi\sim \Phi'$), + če obstaja transformacija podobnosti $f$ te ravnine, ki lik $\Phi$ preslika v lik $\Phi'$ oz. $f(\Phi)=\Phi'$ (Figure \ref{sl.pod.7.2.1.pic}). + Koeficient podobnosti transformacije $f$ je hkrati \index{koeficient!podobnosti likov}\pojem{koeficient podobnosti likov} $\Phi$ in $\Phi'$. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.1.pic} +\caption{} \label{sl.pod.7.2.1.pic} +\end{figure} + +Jasno je, da za $k=1$ dobimo skladnost likov kot poseben primer podobnosti. Skladna lika sta si torej tudi podobna, obratno pa ne velja, torej: +$$\Phi\cong \Phi' \hspace*{1mm}\Rightarrow \hspace*{1mm}\Phi\sim \Phi'.$$ + +Dokažimo najpomembnejšo lastnost relacije podobnosti likov. + + \bizrek + The similarity of figures is an equivalence relation. + \eizrek + + \textbf{\textit{Proof.}} Potrebno (in dovolj) je dokazati, da je relacija podobnosti likov refleksivna, simetrična in tranzitivna. + + + (\textit{R}) Za vsak lik $\Phi$ velja $\Phi\sim \Phi$, ker je identiteta $\mathcal{E}$, ki preslika lik $\Phi$ vase, transformacija podobnosti s koeficientom $k=1$ (Figure \ref{sl.pod.7.2.1r.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.1r.pic} +\caption{} \label{sl.pod.7.2.1r.pic} +\end{figure} + + (\textit{S}) Predpostavimo, da za lika $\Phi_1$ in $\Phi_2$ velja $\Phi_1\sim \Phi_2$ (Figure \ref{sl.pod.7.2.1s.pic}). Po definiciji obstaja transformacija podobnosti $f$, ki lik $\Phi_1$ preslika v lik $\Phi_2$, oz. $f:\hspace*{1mm}\Phi_1\rightarrow \Phi_2$. + Po izreku \ref{TransPodGrupa} je inverzna preslikava $f^{-1}$, za katero je $f^{-1}:\hspace*{1mm}\Phi_2\rightarrow \Phi_1$, prav tako transformacija podobnosti. Torej velja $\Phi_2\sim \Phi_1$. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.1s.pic} +\caption{} \label{sl.pod.7.2.1s.pic} +\end{figure} + + (\textit{T}) Predpostavimo, da za like $\Phi_1$, $\Phi_2$ in $\Phi_3$ velja $\Phi_1\sim \Phi_2$ in $\Phi_2\sim \Phi_3$ (Figure \ref{sl.pod.7.2.1t.pic}). Dokažimo, da je potem tudi $\Phi_1\sim \Phi_3$. Po definiciji obstajata takšni transformaciji podobnosti $f$ in $g$, da velja $f:\hspace*{1mm}\Phi_1\rightarrow \Phi_2$ in $g:\hspace*{1mm}\Phi_2\rightarrow \Phi_3$. Toda po izreku \ref{TransPodGrupa} je kompozitum preslikav $g\circ f:\hspace*{1mm}\Phi_1\rightarrow \Phi_3$ + tudi transformacija podobnosti, zato velja $\Phi_1\sim \Phi_3$. +\kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.1t.pic} +\caption{} \label{sl.pod.7.2.1t.pic} +\end{figure} + + + +Ker je iz prejšnjega izreka relacija podobnosti likov simetrična, bomo v primeru $\Phi\sim \Phi'$ rekli, da sta si lika $\Phi$ in $\Phi'$ podobna. + +V nadaljevanju bomo posebej obravnavali podobnost trikotnikov. Predpostavimo, da sta si trikotnika $ABC$ in $A'B'C'$ podobna oz. $\triangle ABC\sim\triangle A'B'C'$. To pomeni, da obstaja transformacija podobnosti, ki preslika trikotnik $ABC$ v trikotnik $A'B'C'$. V primeru trikotnikov (tudi večkotnikov) bomo dodatno zahtevali, da se oglišča s transformacijo podobnosti preslikajo po vrsti, torej $f:\hspace*{1mm}A,B,C\mapsto A',B',C'$. +Ker transformacija podobnosti preslika daljice v daljice in kote v kote (izrek \ref{TransPodKol}), se s transformacijo $f$ stranice $AB$, $BC$ in $CA$ preslikajo v stranice $A'B'$, $B'C'$ in $C'A'$, notranji koti $BAC$, $ABC$ in $ACB$ trikotnika $ABC$ pa v notranje kote $B'A'C'$, $A'B'C'$ in $A'C'B'$ trikotnika $A'B'C'$. Za pare elementov v tej preslikavi bomo rekli, da so \pojem{ustrezni} ali \pojem{istoležni}. + +Po izreku \ref{TransPodOhranjajoRazm} transformacije podobnosti ohranjajo razmerje daljic, kar pomeni, da so ustrezne stranice sorazmerne,torej\footnote{Z uporabo podobnosti enakokrakih pravokotnih trikotnikov s pomočjo ustreznega sorazmerja je \index{Tales}\textit{Tales iz Mileta} (7.--6. st. pr. n. š.) izračunal višino Keopsove piramide.} (Figure \ref{sl.pod.7.2.2.pic}): + \begin{eqnarray} \label{eqnPodTrik1} + \triangle ABC\sim\triangle A'B'C'\hspace*{1mm} \Rightarrow \hspace*{1mm} \frac{A'B'}{AB}=\frac{A'C'}{AC}=\frac{B'C'}{BC}=k, + \end{eqnarray} +kjer je $k$ koeficient podobnosti. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.2.pic} +\caption{} \label{sl.pod.7.2.2.pic} +\end{figure} + +Po izreku \ref{TransPodOhranjaKote} transformacije podobnosti ohranjajo kote, torej (Figure \ref{sl.pod.7.2.2.pic}): + \begin{eqnarray} \label{eqnPodTrik2} + \triangle ABC\sim\triangle A'B'C'\hspace*{1mm} \Rightarrow \hspace*{1mm} + \left\{ + \begin{array}{l} + \angle B'A'C'\cong\angle BAC; \\ + \angle A'B'C'\cong\angle ABC; \\ + \angle A'C'B'\cong\angle ACB. + \end{array} + \right. + \end{eqnarray} + + + + Iz dejstva, da transformacija podobnosti ohranja kote (izrek \ref{TransPodOhranjaKote}) in razmerje daljic (izrek \ref{TransPodOhranjajoRazm}), sledi tudi, da se višine, težiščnice,... enega trikotnika preslikajo v višine, težiščnice,... drugega trikotnika, pri tem pa se ohranja razmerje ustreznih elementov. + +Pri podobnosti trikotnikov imamo enak problem kot pri skladnosti - zelo pogosto ni enostavno dokazati podobnosti trikotnikov direktno po definiciji. Tako, analogno kot pri skladnosti, dobimo tudi \index{izrek!o podobnosti trikotnikov}\pojem{izreke o podobnosti trikotnikov}. +Ugotovili smo že, da iz podobnosti dveh trikotnikov $ABC$ in $A'B'C'$ dobimo sorazmerje ustreznih stranic \ref{eqnPodTrik1} oz. skladnost ustreznih kotov \ref{eqnPodTrik2}. Naslednji izreki govorijo o tem, kateri pogoji zadoščajo, da sta trikotnika podobna. + + + + + \bizrek \label{PodTrikSKS} + Triangles $ABC$ and $A'B'C'$ are similar + if two pairs of the sides of the triangles are proportional and the pair of included angles between the sides is congruent, i.e.: + \begin{eqnarray*} + \left. + \begin{array}{l} + \angle B'A'C'\cong \angle BAC\\ + \frac{A'B'}{AB}=\frac{A'C'}{AC} + \end{array} + \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC\sim\triangle A'B'C' + \end{eqnarray*} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.3.pic} +\caption{} \label{sl.pod.7.2.3.pic} +\end{figure} + + + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.3.pic}) +Naj bo: + $$k=\frac{A'B'}{AB}=\frac{A'C'}{AC}.$$ +Naj bosta še $B''$ i $C''$ takšni točki poltrakov +$A'B'$ in $A'C'$, da velja $A'B''\cong AB$ in +$A'C''\cong AC$. Trikotnika + $A'B''C''$ in $ABC$ sta skladna (izrek \textit{SAS} \ref{SKS}), zato obstaja izometrija $\mathcal{I}$, ki trikotnik $ABC$ preslika v trikotnik +$A'B''C''$. Ker je + $$\frac{A'B'}{A'B''}=\frac{A'B'}{AB}=k\hspace*{1mm}\textrm{ in }\hspace*{1mm}\frac{A'C'}{A'C''}=\frac{A'C'}{AC}=k,$$ oz. (ker točki $B''$ in $C''$ ležita na poltrakih +$A'B'$ i $A'C'$) $\overrightarrow{A'B'}=k\cdot \overrightarrow{A'B''}$ in $\overrightarrow{A'C'}=k\cdot \overrightarrow{A'C''}$, središčni razteg $h_{A',k}$ preslika točke $A'$, $B''$ in $C''$ v točke $A'$, $B'$ in $C'$ oz. trikotnik $A'B''C''$ v trikotnik $A'B'C'$. Torej kompozitum $f=h_{A',k}\circ \mathcal{I}$, ki je transformacija podobnosti, preslika trikotnik $ABC$ v trikotnik $A'B'C'$, kar pomeni, da velja $\triangle ABC\sim\triangle A'B'C'$. + \kdokaz + + + + \bizrek \label{PodTrikKKK} + Triangles $ABC$ and $A'B'C'$ are similar + if two pairs of the angles of the triangles are congruent, i.e.: + \begin{eqnarray*} + \left. + \begin{array}{l} + \angle B'A'C'\cong\angle BAC\\ + \angle A'B'C'\cong\angle ABC + \end{array} + \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC\sim\triangle A'B'C' + \end{eqnarray*} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.4.pic} +\caption{} \label{sl.pod.7.2.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.4.pic}) +Označimo: + $$k=\frac{A'B'}{AB}.$$ +Naj bosta $B''$ i $C''$ takšni točki poltrakov +$A'B'$ in $A'C'$, da velja $A'B''\cong AB$ in +$A'C''\cong AC$. Potem je tudi $\frac{A'B'}{A'B''}=k$. Trikotnika + $A'B''C''$ in $ABC$ sta skladna (izrek \textit{SAS} \ref{SKS}), zato obstaja izometrija $\mathcal{I}$, ki preslika trikotnik $ABC$ v trikotnik +$A'B''C''$. Iz te skladnosti sledi še $\angle A'B''C''\cong\angle ABC$. Ker je po predpostavki $\angle A'B'C'\cong\angle ABC$, je tudi $\angle A'B''C''\cong\angle A'B'C'$. Po izreku \ref{KotiTransverzala} je $B''C''\parallel B'C'$. Po Talesovem \ref{TalesovIzrek} izreku je +$$\frac{A'C'}{A'C''}=\frac{A'B'}{A'B''}=k.$$ Ker sta $B''$ in $C''$ točki poltrakov +$A'B'$ in $A'C'$, je tudi $\overrightarrow{A'B'}=k\cdot \overrightarrow{A'B''}$ in $\overrightarrow{A'C'}=k\cdot \overrightarrow{A'C''}$. To pomeni, da središčni razteg $h_{A',k}$ preslika točke $A'$, $B''$ in $C''$ v točke $A'$, $B'$ in $C'$ oz. trikotnik $A'B''C''$ v trikotnik $A'B'C'$. Kompozitum $f=h_{A',k}\circ \mathcal{I}$, ki je transformacija podobnosti, torej preslika trikotnik $ABC$ v trikotnik $A'B'C'$, zato je $\triangle ABC\sim\triangle A'B'C'$. + \kdokaz + + + + + \bizrek \label{PodTrikSSS} + Triangles $ABC$ and $A'B'C'$ are similar + if three pairs of the sides of the triangles are proportional, i.e. + \begin{eqnarray*} + \frac{A'B'}{AB}=\frac{A'C'}{AC}=\frac{B'C'}{BC} + \hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC\sim\triangle A'B'C' + \end{eqnarray*} + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.5.pic} +\caption{} \label{sl.pod.7.2.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.5.pic}) +Označimo: + $$k=\frac{A'B'}{AB}=\frac{A'C'}{AC}=\frac{B'C'}{BC}.$$ +Naj bosta $B''$ in $C''$ takšni točki poltrakov +$A'B'$ in $A'C'$, da velja $A'B''\cong AB$ in +$A'C''\cong AC$. Potem je tudi $$\frac{A'B'}{A'B''}=\frac{A'C'}{A'C''}=k.$$ +Po obratnem Talesovem izreku \ref{TalesovIzrekObr} je $B''C''\parallel B'C'$ in $$\frac{B'C'}{B''C''}=\frac{A'B'}{A'B''}.$$ +Torej $$\frac{B'C'}{B''C''}=\frac{A'B'}{A'B''}=\frac{A'C'}{A'C''}=k=\frac{B'C'}{BC},$$ +kar pomeni, da velja $B''C''\cong BC$. Iz tega sledi, da sta +trikotnika + $A'B''C''$ in $ABC$ skladna (izrek \textit{SSS} \ref{SSS}), zato obstaja izometrija $\mathcal{I}$, ki preslika trikotnik $ABC$ v trikotnik +$A'B''C''$. Prav tako (podobno kot v dokazu prejšnjih dveh izrekov) središčni razteg $h_{A',k}$ preslika točke $A'$, $B''$ in $C''$ v točke $A'$, $B'$ in $C'$ oz. trikotnik $A'B''C''$ v trikotnik $A'B'C'$. Na koncu kompozitum $f=h_{A',k}\circ \mathcal{I}$, ki je transformacija podobnosti, preslika trikotnik $ABC$ v trikotnik $A'B'C'$, zato je $\triangle ABC\sim\triangle A'B'C'$. + \kdokaz + +Naslednji izrek bomo podali brez dokaza (Figure \ref{sl.pod.7.2.6.pic}). + + + + + \bizrek \label{PodTrikSSK} + Triangles $ABC$ and $A'B'C'$ are similar + if two pairs of the sides of the triangles are proportional and the pair of the angles opposite to the longer sides is congruent. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.6.pic} +\caption{} \label{sl.pod.7.2.6.pic} +\end{figure} + +Nadaljevali bomo z uporabo izrekov o podobnosti trikotnikov. + + + + + + \bzgled + Let $k$ be the circumcircle of a triangle $ABC$ and $B'$ and $C'$ + the foots of the perpendiculars from the vertices $B$ and $C$ on the tangent of the circle $k$ in the vertex $A$. + Prove that the altitude $AD$ of this triangle is the geometric mean of the line segments $BB'$ and $CC'$, i.e.: + $$|AD|=\sqrt{|BB'|\cdot |CC'|}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.7.pic} +\caption{} \label{sl.pod.7.2.7.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.7.pic}) + +Kot $BAB'$ je skladen obodnemu kotu s +tetivo $AB$ oz. $\angle BAB'\cong\angle ACD$ (izrek \ref{ObodKotTang}). Ker je še +$\angle AB'B\cong\angle CDA=90^0$, je +$\triangle AB'B\sim\triangle CDA$ (izrek\ref{PodTrikKKK}). Iz tega sledi: +\begin{eqnarray} \label{eqnPodTrikZgl1a} +AD:BB'=AC:BA. +\end{eqnarray} + +Na enak način dokažemo tudi $\triangle AC'C\sim\triangle BDA$ oz. +\begin{eqnarray} \label{eqnPodTrikZgl1b} + CC':AD=AC:BA. +\end{eqnarray} + Iz relacij \ref{eqnPodTrikZgl1a} in \ref{eqnPodTrikZgl1b} +sledi $AD:BB'=CC':AD$ oz. $|AD|=\sqrt{|BB'|\cdot |CC'|}$. + \kdokaz + + + \bzgled \label{izrekSinusni} + Let $v_a$ be the length of the altitude $AD$, $k(O,R)$ the circumcircle of a triangle $ABC$ + and $b=|AC|$ and $c=|AB|$. Prove that\footnote{Trditev iz tega zgleda v trigonometriji predstavlja t. + i. \index{izrek!Sinusni}\textit{sinusni izrek}, ki ga je dokazal arabski + matematik \index{al-Biruni, A. R.}\textit{A. R. al-Biruni} (973--1048). Če + namreč v dano enakost vstavimo $\frac{v_a}{c}=\sin \beta$, dobimo + $\frac{b}{\sin\beta}=2R$ in analogno + $\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2R$.}: + $$bc=2R\cdot v_a.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.8a.pic} +\caption{} \label{sl.pod.7.2.8a.pic} +\end{figure} + + \textbf{\textit{Proof.}} Označimo še z $A_0=\mathcal{S}_O(A)$ (Figure \ref{sl.pod.7.2.8a.pic}). Daljica $AA_0$ je premer krožnice $k$, zato je po izreku \ref{TalesovIzrKroz2} $\angle ACA_0=90^0=\angle ADB$. Obodna kota $CBA$ in $CA_0A$ nad tetivo $AC$ krožnice $k$ sta skladna (izrek \ref{ObodObodKot}), oz. $\angle DBA=\angle CBA\cong\angle CA_0A$. Po izreku \ref{PodTrikKKK} je $\triangle ABD\sim\triangle AA_0C$, zato je: +$$\frac{AB}{AA_0}=\frac{AD}{AC},$$ +oz. +$bc=2R\cdot v_a$. +\kdokaz + + + + \bzgled + Suppose that the bisector of the interior angle $BA$C of a triangle $ABC$ intersects + its side $BC$ at the point $E$ and the circumcircle of this triangle at the point $N$ + ($N\neq A$). Let's denote $b=|AC|$, $c=|AB|$ and $l_a=|AE|$. Prove that: + $$|AN|=\frac{bc}{l_a}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.8.pic} +\caption{} \label{sl.pod.7.2.8.pic} +\end{figure} + + \textbf{\textit{Proof.}} Označimo še z $AD$ višino ($v_a=|AD|$), s $k(O,R)$ očrtano krožnico trikotnika $ABC$ in $M=\mathcal{S}_O(N)$ (Figure \ref{sl.pod.7.2.8.pic}). Po izreku \ref{TockaN} leži točka $N$ na simetrali stranice $BC$, kar pomeni, da je $NM$ premer krožnice $k$ in $MN\perp BC$. Po izreku \ref{TalesovIzrKroz2} je $\angle NAM=90^0=\angle EDA$. Iz $AD,MN\perp BC$ sledi $AD\parallel MN$, zato je po izreku \ref{KotiTransverzala} $\angle DAE\cong\angle MNA$. Trikotnika $ADE$ in $NAM$ sta si torej podobna (izrek \ref{PodTrikKKK}), zato je: +$$\frac{AD}{AN}=\frac{AE}{NM}.$$ +Če uporabimo prejšnji zgled \ref{izrekSinusni}, dobimo: + $$|AN|=\frac{2R\cdot v_a}{l_a}=\frac{bc}{l_a},$$ kar je bilo treba dokazati. \kdokaz + + + +%________________________________________________________________________________ + \poglavje{The Theorems of Ceva and Menelaus} \label{odd7MenelCeva} + + + Na tem mestu bomo obravnavali t. i. \index{izrek!izreka dvojčka}\pojem{izreka dvojčka}. + + + + + \bizrek \label{izrekCeva}\index{izrek!Cevov} + (Ceva\footnote{\index{Ceva, G.} \textit{G. Ceva} (1648--1734), italijanski matematik, ki + je ta izrek dokazal leta 1678.}) + Let $P$, $Q$ and $R$ be points lying on the lines containing the sides $BC$, $CA$ and $AB$ + of a triangle $ABC$. Then the lines $AP$, $BQ$ and $CR$ belong + to the same family of lines if and only if: + \begin{eqnarray}\label{formulaCeva} + \frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=1. + \end{eqnarray} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.5.1.pic} +\caption{} \label{sl.pod.7.5.1.pic} +\end{figure} + +%slikaNova1-3-1 +%\includegraphics[width=100mm]{slikaNova1-3-1.pdf} + + \textbf{\textit{Proof.}} +($\Rightarrow$) Predpostavimo, da premice $AP$, $BQ$ in $CR$ +pripadajo istemu eliptičnemu šopu, torej da se sekajo v neki +točki $S$ (bralcu bomo pustili, da dokaže izrek v primeru, če +so premice vzporedne). Premico, ki vsebuje točko $A$ in je +vzporedna s premico $BC$, označimo s $p$, njena presečišča s +premicama $BQ$ in $CR$ označimo s $K$ in $L$ (Figure +\ref{sl.pod.7.5.1.pic}). Sedaj iz Talesovega izreka \ref{TalesovIzrek} sledi + $\frac{\overrightarrow{BP}}{\overrightarrow{KA}}= + \frac{\overrightarrow{SP}}{\overrightarrow{SA}}= + \frac{\overrightarrow{PC}}{\overrightarrow{AL}}.$ +Iz tega dobimo: + \begin{eqnarray*} + \frac{\overrightarrow{BP}}{\overrightarrow{PC}}= + \frac{\overrightarrow{KA}}{\overrightarrow{AL}}. + \end{eqnarray*} + Analogno je tudi: + \begin{eqnarray*} + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}= + -\frac{\overrightarrow{QC}}{\overrightarrow{QA}}= + -\frac{\overrightarrow{CB}}{\overrightarrow{AK}}= + \frac{\overrightarrow{CB}}{\overrightarrow{KA}},\\ + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= + -\frac{\overrightarrow{RA}}{\overrightarrow{RB}}= + -\frac{\overrightarrow{AL}}{\overrightarrow{BC}}= + \frac{\overrightarrow{AL}}{\overrightarrow{CB}}, + \end{eqnarray*} + kar pomeni, da velja: + $$\frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= + \frac{\overrightarrow{KA}}{\overrightarrow{AL}}\cdot + \frac{\overrightarrow{CB}}{\overrightarrow{KA}}\cdot + \frac{\overrightarrow{AL}}{\overrightarrow{CB}}=1.$$ + ($\Leftarrow$) Predpostavimo sedaj, da velja relacija + (\ref{formulaCeva}) in da je + $S$ presečišče premic $BQ$ in $CR$ (primer, ko sta $BQ$ in $CR$ vzporedni, + spet prepuščamo bralcu). Če sedaj s $P'$ označimo presečišče premice $AS$ + s premico $BC$, iz prvega dela dokaza sledi (Figure \ref{sl.pod.7.5.1.pic}): + $$\frac{\overrightarrow{BP'}}{\overrightarrow{P'C}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=1.$$ + Po predpostavki (\ref{formulaCeva}) je potem + $\frac{\overrightarrow{BP}}{\overrightarrow{PC}}= + \frac{\overrightarrow{BP'}}{\overrightarrow{P'C}}$ oz. $P=P'$ (izrek \ref{izrekEnaDelitevDaljiceVekt}). + Torej se premice $AP$, $BQ$, $CR$ sekajo v točki $S$. + \kdokaz + + Opazimo, da izrek velja v primeru, ko se premice $AP$, $BQ$ in $CR$ + sekajo v eni točki (eliptični šop), kot tudi v primeru, ko so te premice + vzporedne (parabolični šop). Zdi se, da bi izrek laže dokazali, + če bi vpeljali + točke v neskončnosti in s tem pojem (eliptičnega) šopa posplošili + tudi na primer, ko gre za vzporedne premice. To so ideje, ki so pripeljale do razvoja t. i. \index{geometrija!projektivna}\pojem{projektivne geometrije}. + + + + + \bizrek \index{izrek!Menelajev}\label{izrekMenelaj} + (Menelaj\footnote{\index{Menelaj} \textit{Starogrški matematik Menelaj iz Aleksandrije} (1. st.) + je ta izrek dokazal v svojem delu \textit{Sferika}, in sicer + v primeru sferičnih trikotnikov. Trditev v primeru trikotnikov v ravnini + omenja kot že znano. Ker prejšni dokumenti o tem niso ohranjeni, izrek + po njem imenujemo Menelajev izrek. Zaradi svoje podobnosti se Menelajev + in Cevov izrek imenujeta tudi \textit{izreka dvojčka}. Obdobje 1500 let, ki + loči ti + dve odkritji, priča o tem, kako se je v tem + obdobju geometrija počasi razvijala.}) + Let $P$, $Q$ and $R$ be points lying on the lines containing the sides $BC$, $CA$ and $AB$ + of a triangle $ABC$. Then the points $P$, $Q$ and $R$ are collinear if and only if: + \begin{eqnarray}\label{formulaMenelaj} + \frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=-1. + \end{eqnarray} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.5.2.pic} +\caption{} \label{sl.pod.7.5.2.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + ($\Rightarrow$) Naj bodo $P$, $Q$ in $R$ točke neke premice $l$. + Z $A'$, $B'$ in $C'$ označimo nožišča pravokotnic iz oglišč + $A$, $B$ in $C$ trikotnika na premico $l$ (Figure \ref{sl.pod.7.5.2.pic}). + Z uporabo Talesovega izreka (\ref{TalesovIzrek}) dobimo: + \begin{eqnarray*} + \frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= + -\frac{\overrightarrow{PB}}{\overrightarrow{PC}}\cdot + \frac{\overrightarrow{QC}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{RA}}{\overrightarrow{RB}}= + -\frac{\overrightarrow{BB'}}{\overrightarrow{CC'}}\cdot + \frac{\overrightarrow{CC'}}{\overrightarrow{AA'}}\cdot + \frac{\overrightarrow{AA'}}{\overrightarrow{BB'}}= + -1. + \end{eqnarray*} + ($\Leftarrow$) Naj bo sedaj izpolnjena relacija (\ref{formulaMenelaj}). S $P'$ + označimo presečišče premic $QR$ in $BC$. Če bi bili premici $QR$ in $BC$ + vzporedni, bi iz tega po Talesovem izreku sledilo + $\frac{\overrightarrow{CQ}}{\overrightarrow{QA}}= + \frac{\overrightarrow{RB}}{\overrightarrow{AR}}$ + in nato iz (\ref{formulaMenelaj}) še + $\frac{\overrightarrow{BP}}{\overrightarrow{PC}}=-1$ oz. + $\overrightarrow{BP}=\overrightarrow{CP}$, kar + ni mogoče. Naj bo torej $P'=QR\cap BC$. Tedaj so točke + $P'$, $Q$, in $R$ kolinearne. Po dokazanem v prvem delu + izreka + imamo: + \begin{eqnarray*} + \frac{\overrightarrow{BP'}}{\overrightarrow{P'C}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=-1. + \end{eqnarray*} + Iz tega in predpostavljene relacije (\ref{formulaMenelaj}) sledi + $\frac{\overrightarrow{BP}}{\overrightarrow{PC}}= + \frac{\overrightarrow{BP'}}{\overrightarrow{P'C}}$ oz. $P=P'$ (izrek \ref{izrekEnaDelitevDaljiceVekt}), + kar pomeni, da so točke $P$, $Q$ in $R$ kolinearne. + \kdokaz + + +Nadaljujmo z uporabo dveh dokazanih izrekov. + + + + \bizrek + The lines, joining the vertices of a triangle to the tangent points of the incircle, intersect at one point (so-called \index{točka!Gergonova}\pojem{Gergonne\footnote{To trditev je dokazal \index{Gergonne, J. D.}\textit{J. D. Gergonne} (1771--1859), francoski matematik. V 19. stoletju je bila posvečena posebna + pozornost metričnim lastnostim trikotnika, tako so bile odkrite tudi druge karakteristične točke trikotnika (glej naslednja zgleda).} point} \color{blue} of this triangle). + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.5.3.pic} +\caption{} \label{sl.pod.7.5.3.pic} +\end{figure} + + + \textbf{\textit{Proof.}} +Naj bodo $P$, $Q$ in $R$ točke, v katerih se včrtana krožnica trikotnika $ABC$ (glej izrek \ref{SredVcrtaneKrozn}) dotika njegovih +stranic $BC$, $CA$ in $AB$ (Figure \ref{sl.pod.7.5.3.pic}). Dokažimo, da se premice $AP$, $BQ$ in $CR$ sekajo v eni točki. Po izreku \ref{TangOdsek} so skladne + ustrezne tangentne daljice: $BP\cong BR$, $CP\cong CQ$ in +$AQ\cong AR$. Ker je $\mathcal{B}(B,P,C)$, $\mathcal{B}(C,Q,A)$ in $\mathcal{B}(A,R,B)$, je $\frac{\overrightarrow{BP}}{\overrightarrow{PC}} +\cdot \frac{\overrightarrow{CQ}}{\overrightarrow{QA}} +\cdot \frac{\overrightarrow{AR}}{\overrightarrow{RB}} >0$. Torej: +$$\frac{\overrightarrow{BP}}{\overrightarrow{PC}} +\cdot \frac{\overrightarrow{CQ}}{\overrightarrow{QA}} +\cdot \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= +\frac{|BP|}{|PC|}\cdot\frac{|CQ|}{|QA|}\cdot\frac{|AR|}{|RB|}=1.$$ + Po Cevovem izreku \ref{izrekCeva} premice $AP$, $BQ$ in $CR$ pripadajo enem šopu. Ker se po posledici Pashovega aksioma \ref{PaschIzrek} premici $AP$ in $BQ$ sekata, gre za eliptični šop, kar pomeni, da se premice $AP$, $BQ$ in $CR$ sekajo v eni točki. + \kdokaz + + + + \bzgled + Prove that the lines, joining the vertices of a triangle to the tangent points of the excircles to the opposite sides, intersect at one point (so-called \index{točka!Nagelova}\pojem{Nagel\footnote{\index{Nagel, C. H.}\textit{C. H. Nagel} (1803--1882), nemški matematik, ki je objavil to trditev leta 1836. Vsaka od premic v tej trditvi deli obseg trikotnika na dva enaka dela, zato to trditev imenujemo tudi \index{izrek!o polobsegu trikotnika}\pojem{izrek o polobsegu trikotnika}.} + point} \color{green1} of this triangle). + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.5.6.pic} +\caption{} \label{sl.pod.7.5.6.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + Uporabimo oznake iz velike naloge (\ref{velikaNaloga}). Dokažimo, da se premice $AP_a$, $BQ_b$ in $CR_c$ sekajo v eni točki (Figure \ref{sl.pod.7.5.6.pic}). Če uporabimo dejstvo $\frac{\overrightarrow{BP_a}}{\overrightarrow{P_aC}} +\cdot \frac{\overrightarrow{CQ_b}}{\overrightarrow{Q_bA}} +\cdot \frac{\overrightarrow{AR_c}}{\overrightarrow{R_cB}} >0$ in relacije iz velike naloge (\ref{velikaNaloga}), dobimo: +\begin{eqnarray*} +\frac{\overrightarrow{BP_a}}{\overrightarrow{P_aC}} +\cdot \frac{\overrightarrow{CQ_b}}{\overrightarrow{Q_bA}} +\cdot \frac{\overrightarrow{AR_c}}{\overrightarrow{R_cB}}&=& +\frac{|BP_a|}{|P_aC|}\cdot\frac{|CQ_b|}{|Q_bA|}\cdot\frac{|AR_c|}{|R_cB|}=\\&=& +\frac{s-c}{s-b}\cdot\frac{s-a}{s-c}\cdot\frac{s-b}{s-a}= +1 +\end{eqnarray*} +Po Cevovem izreku \ref{izrekCeva} premice $AP_a$, $BQ_b$ in $CR_c$ pripadajo enemu šopu. Ker se po posledici Pashovega aksioma \ref{PaschIzrek} premici $AP_a$ in $BQ_b$ sekata, gre za eliptični šop, kar pomeni, da se premice $AP_a$, $BQ_b$ in $CR_c$ sekajo v eni točki. + \kdokaz + + + + \bzgled + Prove that the lines, joining the vertices of a triangle to the + points dividing opposite sides in the ratio of squares of adjacent sides, + intersect at one point (so-called \index{točka!Lemoinova} + \pojem{Lemoine\footnote{\index{Lemoine, E. M. H.}\textit{E. M. H. Lemoine} (1751--1816), francoski matematik.} point} \color{green1} of this triangle). + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.5.5.pic} +\caption{} \label{sl.pod.7.5.5.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Naj bodo $X$, $Y$ in $Z$ takšne točke stranic $BC$, $AC$ in $AB$ trikotnika $ABC$, da velja: $\frac{|BX|}{|XC|}=\frac{|BA|^2}{|AC|^2}$, $\frac{|CY|}{|YA|}=\frac{|CB|^2}{|BA|^2}$ in $\frac{|AZ|}{|ZB|}=\frac{|AC|^2}{|CB|^2}$ + (Figure \ref{sl.pod.7.5.5.pic}). + Ker je $\frac{\overrightarrow{BX}}{\overrightarrow{XC}} +\cdot \frac{\overrightarrow{CY}}{\overrightarrow{YA}} +\cdot \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}} >0$, je: +\begin{eqnarray*} +\frac{\overrightarrow{BX}}{\overrightarrow{XC}} +\cdot \frac{\overrightarrow{CY}}{\overrightarrow{YA}} +\cdot \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}}&=& +\frac{|BX|}{|XC|}\cdot\frac{|CY|}{|YA|}\cdot\frac{|AZ|}{|ZB|}=\\ +&=& +\frac{|BA|^2}{|AC|^2}\cdot\frac{|CB|^2}{|BA|^2}\cdot\frac{|AC|^2}{|CB|^2}= +1. +\end{eqnarray*} + Po Cevovem izreku \ref{izrekCeva} premice $AX$, $BY$ in $CZ$ pripadajo enem šopu. Ker se po posledici Pashovega aksioma \ref{PaschIzrek} premici $AX$ in $BY$ sekata, gre za eliptični šop, kar pomeni, da se premice $AX$, $BY$ in $CZ$ sekajo v eni točki. + \kdokaz + + + + \bzgled + Let $Z$ and $Y$ be points of the sides $AB$ and $AC$ of a triangle + $ABC$ such that $AZ:ZB=1:3$ and $AY:YC=1:2$, and $X$ the point, + in which the line $YZ$ intersects the line containing the side $BC$ of this triangle. Calculate + $\overrightarrow{BX}:\overrightarrow{XC}$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.5.4.pic} +\caption{} \label{sl.pod.7.5.4.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + (Figure \ref{sl.pod.7.5.4.pic}) + +Ker so točke $X$, $Y$ in $Z$ kolinearne, po Menelajevem izreku \ref{izrekMenelaj} sledi: +$$\frac{\overrightarrow{BX}}{\overrightarrow{XC}} +\cdot\frac{\overrightarrow{CY}}{\overrightarrow{YA}} +\cdot\frac{\overrightarrow{AZ}}{\overrightarrow{ZB}}=-1.$$ +Če uporabimo dane pogoje, dobimo: +$$\frac{\overrightarrow{BX}}{\overrightarrow{XC}} +\cdot\frac{2}{1} +\cdot\frac{1}{3}=-1$$ +oziroma $\overrightarrow{BX}:\overrightarrow{XC}=-3:2$. +\kdokaz + + + + \bzgled + Prove that the tangents of the circumcircle of an arbitrary scalene triangle $ABC$ at its + vertices $A$, $B$ and $C$ intersect the lines containing the opposite sides of this triangle + at three collinear points\footnote{Izrek je poseben primer Pascalovega izreka \ref{izrekPascalEvkl} (razdelek \ref{odd7PappusPascal}). \index{Pascal, B.} \textit{B. Pascal} (1623--1662), francoski matematik in filozof.}. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.2a.pic} +\caption{} \label{sl.pod.7.12.2a.pic} +\end{figure} + +\textbf{\textit{Solution.}} + Naj bodo $X$, $Y$ in $Z$ presečišča + tangent očrtane krožnice $k$ trikotnika $ABC$ v ogliščih $A$, $B$ in $C$ + z nosilkami stranic $BC$, $AC$ in $AB$ tega + trikotnika (Figure \ref{sl.pod.7.12.2a.pic}). Dokažimo, da so $X$, $Y$ in $Z$ kolinearne točke. Po izreku \ref{ObodKotTang} je $\angle XAB\cong\angle ACB$. Ker imata trikotnika $AXB$ in $CXA$ še skupni notranji kot ob oglišču $X$, je $\triangle AXB\sim \triangle CXA$ (izrek \ref{PodTrikKKK}), zato je: + $$\frac{|AX|}{|CX|}=\frac{|AB|}{|AC|}=\frac{|BX|}{|AX|}.$$ +Iz $\frac{|AX|}{|CX|}=\frac{|BX|}{|AX|}$ sledi $|XB|\cdot |XC|=|XA|^2$. Če slednjo relacijo delimo z $|XC|^2$, dobimo $\frac{|BX|}{|CX|}=\frac{|AX|^2}{|CX|^2}$. Ker je še $\frac{|AX|}{|CX|}=\frac{|AB|}{|AC|}$, na koncu dobimo: + \begin{eqnarray*} +&& \frac{|BX|}{|XC|}=\frac{|BA|^2}{|AC|^2} + \end{eqnarray*} + in podobno + \begin{eqnarray*} +&& \frac{|CY|}{|YA|}=\frac{|CB|^2}{|BA|^2},\\ +&& \frac{|AZ|}{|ZB|}=\frac{|AC|^2}{|CB|^2}. + \end{eqnarray*} + Ker točke $X$, $Y$ in $Z$ ne ležijo na stranicah $BC$, $AC$ in $AB$, je $\frac{\overrightarrow{BX}}{\overrightarrow{XC}} +\cdot \frac{\overrightarrow{CY}}{\overrightarrow{YA}} +\cdot \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}} <0$, zato je: +\begin{eqnarray*} +\frac{\overrightarrow{BX}}{\overrightarrow{XC}} +\cdot \frac{\overrightarrow{CY}}{\overrightarrow{YA}} +\cdot \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}}&=& +-\frac{|BX|}{|XC|}\cdot\frac{|CY|}{|YA|}\cdot\frac{|AZ|}{|ZB|}=\\ +&=& +-\frac{|BA|^2}{|AC|^2}\cdot\frac{|CB|^2}{|BA|^2}\cdot\frac{|AC|^2}{|CB|^2}= +-1. + \end{eqnarray*} + Po Menelajevem izreku \ref{izrekMenelaj} so točke $X$, $Y$ in $Z$ kolinearne. + \kdokaz + + Naslednja lastnost štirikotnikov je nadaljevanje + zgledov + \ref{TetivniVcrtana} in \ref{TetivniVisinska}. Gre namreč za sledeči problem. + Dan je tetivni štirikotnik $ABCD$. Kaj predstavlja štirikotnik, ki ima za oglišča določeno značilno točko trikotnikov $BCD$, $ACD$, $ABD$ in $ABC$? Odgovor je trivialen, če gre za središče očrtane krožnice, saj je $ABCD$ tetivni štirikotnik, zato so v tem primeru vsa središča očrtanih krožnic ena točka in iskanega štirikotnika ni. Primera središč včrtanih krožnice in višinskih točk sta obravnavana v omenjenih zgledih \ref{TetivniVcrtana} in \ref{TetivniVisinska}. Naslednji zgled pa nam da odgovor za primer težišč in sicer v primeru poljubnih štirikotnikov. + + + + + \bzgled \label{TetivniTezisce} + Let $ABCD$ be an arbitrary quadrilateral and + $T_A$ the centroid of the triangle $BCD$, + $T_B$ the centroid of the triangle $ACD$, + $T_C$ the centroid of the triangle $ABD$ and + $T_D$ the centroid of the triangle $ABC$. + Prove that:\\ + a) a) the lines $AT_A$, $BT_B$, $CT_C$ and $DT_D$ intersect at one point which is the common centroid + of the quadrilaterals $ABCD$ and $T_AT_BT_CT_D$,\\ + b) the quadrilateral $T_AT_BT_CT_D$ is similar to the quadrilateral $ABCD$ with the coefficient + of similarity equal to $\frac{1}{3}$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.2.9.pic} +\caption{} \label{sl.pod.7.2.9.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.9.pic}) + +\textit{a)} +Označimo s $P$, $K$, $Q$, $L$, $M$ in $N$ središča daljic $AB$, $BC$, $CD$, $DA$, $AC$ in $BD$. +Ker je točka $T_A$ težišče trikotnika $BCD$, je presečišče njegovih težiščnic $BQ$, $CN$ in $DK$, pri tem še velja $QT_A:T_AB=1:2$ (izrek \ref{tezisce}) oz. $\overrightarrow{QT_A}=\frac{1}{3}\overrightarrow{QB}$. Analogno je iz trikotnika $ACD$ $\overrightarrow{QT_B}=\frac{1}{3}\overrightarrow{QA}$. Iz slednjih dveh relacij in izreka \ref{vektVektorskiProstor} dobimo: $$\overrightarrow{T_BT_A}=\overrightarrow{T_BQ}+\overrightarrow{QT_A} +=\frac{1}{3}\overrightarrow{AQ}+\frac{1}{3}\overrightarrow{QB}= +\frac{1}{3}\left(\overrightarrow{AQ}+\overrightarrow{QB} \right)=\frac{1}{3}\overrightarrow{AB}.$$ +Torej: +\begin{eqnarray} \label{eqnCevaTez1} +\overrightarrow{T_BT_A}=\frac{1}{3}\overrightarrow{AB}, +\end{eqnarray} +zato je tudi $T_BT_A\parallel AB$. Označimo s $T$ presečišče daljic $AT_A$ in $BT_B$. Po Talesovem izreku in \ref{eqnCevaTez1} je: +\begin{eqnarray} \label{eqnCevaTez2} +\frac{\overrightarrow{TT_A}}{\overrightarrow{TA}} +=\frac{\overrightarrow{TT_B}}{\overrightarrow{TB}}= +\frac{\overrightarrow{T_AT_B}}{\overrightarrow{AB}}=-\frac{1}{3}. +\end{eqnarray} +Daljici $AT_A$ in $BT_B$ se torej sekata v točki $T$, ki ju deli v razmerju $2:1$. Analogno se tudi daljici $AT_A$ in $CT_C$ oz. daljici $AT_A$ in $DT_D$ sekata v točki, ki ju deli v razmerju $2:1$, to pa je (zaradi daljice $AT_A$) ravno točka $T$. To pomeni, da se premice $AT_A$, $BT_B$, $CT_C$ in $DT_D$ sekajo v točki $T$. + +Dokažimo še, da je točka $T$ skupno težišče štirikotnikov $ABCD$ in + $T_AT_BT_CT_D$. +Ker je +$$\frac{\overrightarrow{QT_B}}{\overrightarrow{T_BA}}\cdot +\frac{\overrightarrow{AP}}{\overrightarrow{PB}}\cdot +\frac{\overrightarrow{BT_A}}{\overrightarrow{T_AQ}}= +\frac{1}{2}\cdot\frac{1}{1}\cdot\frac{2}{1}=1, +$$ +se po Cevovem izreku \ref{izrekCeva} za trikotnik $QAB$ premice $AT_A$, $BT_B$ in $PQ$ sekajo v eni točki - točki $T$. točka $T$ torej leži na premici $PQ$. Analogno točka $T$ leži na premici $KL$, kar pomeni, da je točka $T$ presečišče diagonal $PQ$ in $KL$ štirikotnika $PKQL$. Ker gre za Varignonov paralelogram štirikotnika $ABCD$, je točka $T$ po izreku \ref{vektVarignon} težišče štirikotnika $ABCD$. Ker je iz \ref{eqnCevaTez2} še: +$$\overrightarrow{TT_A}+\overrightarrow{TT_B}+\overrightarrow{TT_C}+\overrightarrow{TT_D}= +-\frac{1}{3}\cdot\left( +\overrightarrow{TA}+\overrightarrow{TB}+\overrightarrow{TC}+\overrightarrow{TD} +\right)=-\frac{1}{3}\cdot \overrightarrow{0}=\overrightarrow{0},$$ +je točka $T$ težišče tudi štirikotnika $T_AT_BT_CT_D$. + +\textit{b)} Iz \ref{eqnCevaTez2} sledi $\overrightarrow{TT_A}=-\frac{1}{3}\overrightarrow{TA}$, +$\overrightarrow{TT_B}=-\frac{1}{3}\overrightarrow{TB}$, +$\overrightarrow{TT_C}=-\frac{1}{3}\overrightarrow{TC}$ in +$\overrightarrow{TT_D}=-\frac{1}{3}\overrightarrow{TD}$ oz. +$h_{T,-\frac{1}{3}}:\hspace*{1mm}A,B,C,D\mapsto T_A,T_B,T_C,T_D$, kar pomeni, da je štirikotnik $T_AT_BT_CT_D$ podoben štirikotniku + $ABCD$ s koeficientom podobnosti $\frac{1}{3}$. +\kdokaz + +%________________________________________________________________________________ + \poglavje{Harmonic Conjugate Points. Apollonius Circle} + \label{odd7Harm} + + +Ker je zelo pomemben za nadaljevanje, bomo izrek \ref{izrekEnaDelitevDaljiceVekt} iz razdelka \ref{odd5LinKombVekt} zapisali še v drugi obliki. + + + + \bizrek \label{HarmCetEnaSamaDelitev} + If $A$ and $B$ are different points on the line $p$ and $\lambda\neq -1$ + an arbitrary real number, then there is exactly one point $L$ on the line $p$ such that: + $$ \frac{\overrightarrow{AL}}{\overrightarrow{LB}}=\lambda.$$ + \eizrek + + +V razdelku \ref{odd5TalesVekt} smo ugotovili, kako lahko daljico $AB$ razdelimo v razmerju $m:n$ (izrek \ref{izrekEnaDelitevDaljice}). To idejo bomo sedaj uporabili za konstrukcijo točke $L$ iz prejšnjega izreka \ref{HarmCetEnaSamaDelitev}. + + + + \bzgled + The points $A$ and $B$ and the real number $\lambda\neq -1$ are given. + Construct a point $L$ on the line $AB$ such that: + $$\frac{\overrightarrow{AL}}{\overrightarrow{LB}}=\lambda.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.1.pic} +\caption{} \label{sl.pod.7.4.1.pic} +\end{figure} + + \textit{\textbf{Rešitev.}} (Figure \ref{sl.pod.7.4.1.pic}) + +V primeru $\lambda=0$ je dokaz neposreden - takrat je $L=A$. Naj bo $\lambda\neq 0$. +Konstruirajmo +poljubni vzporednici $a$ in $b$ skozi točki $A$ in $B$. S $P$ in $Q$ označimo takšne točke na premicah $a$ in $b$, +da velja $P,Q\div AB$ ter $|AP|=|\lambda|$ in $|BQ|=1$. Označimo še $Q'=\mathcal{S}_B(Q)$. +Obravnavali bomo dva primera. + +\textit{1)} Naj bo $\lambda>0$. Jasno je, da mora za iskano točko $L$ veljati $\mathcal{B}(A,L,B)$. Z $L_1$ označimo +presečišče premic $AB$ in $PQ$. Po Talesovem izreku \ref{TalesovIzrek} je: + $$\frac{\overrightarrow{AL_1}}{\overrightarrow{L_1B}}= +\frac{\overrightarrow{AP}}{\overrightarrow{QB}}=\lambda.$$ + + \textit{2)} Če je $\lambda<0$, iskana točka ne leži na daljici $AB$. Z $L_1$ označimo +presečišče premic $AB$ in $PQ'$. Ker je $\lambda\neq -1$, to presečišče obstaja. Po Talesovem izreku \ref{TalesovIzrek} je: + $$\frac{\overrightarrow{AL_2}}{\overrightarrow{L_2B}}= +\frac{\overrightarrow{AP}}{\overrightarrow{Q'B}} +=-\frac{\overrightarrow{AP}}{\overrightarrow{BQ'}}=-\left(-\lambda\right)=\lambda,$$ kar je bilo treba dokazati. \kdokaz + +Iz prejšnjega zgleda je torej jasno, da za vsakega od pogojev $\frac{\overrightarrow{AL}}{\overrightarrow{LB}}=\lambda$ ($\lambda>0$) oz. $\frac{\overrightarrow{AL}}{\overrightarrow{LB}}=\lambda$ ($\lambda<0$ in $\lambda\neq -1$) obstaja natanko ena rešitev za točko $L$ na premici $AB$. V prvem primeru točka $L$ leži na daljici $AB$, zato pravimo, da gre za \index{delitev daljice!notranja}\pojem{notranjo delitev} daljice $AB$ v razmerju $\lambda$ ($\lambda>0$). V drugem primeru točka $L$ ne leži na daljici $AB$, zato pravimo, da gre za \index{delitev daljice!zunanja}\pojem{zunanjo delitev} daljice $AB$ v razmerju $|\lambda|$ ($\lambda<0$ in $\lambda\neq -1$). + +Lahko rečemo tudi, da za pogoj $\frac{AL}{LB}=\lambda$ ($\lambda>0$, $\lambda\neq 1$) obstajata dve rešitvi za točko $L$ na premici - notranja oziroma zunanja delitev. + +Če sta $L_1$ in $L_2$ notranja in zunanja delitev daljice (za isto $\lambda$), velja: +$$\frac{\overrightarrow{AL_1}}{\overrightarrow{L_1B}}: +\frac{\overrightarrow{AL_2}}{\overrightarrow{L_2B}}=-1.$$ + Če izraz $\frac{\overrightarrow{AL_1}}{\overrightarrow{L_1B}}: +\frac{\overrightarrow{AL_2}}{\overrightarrow{L_2B}}$ definiramo kot \index{dvorazmerje parov točk}\pojem{dvorazmerje parov točk} $(A,B)$ in $(L_1,L_2)$ ter označimo +$$d(A,B;L_1,L_2)=\frac{\overrightarrow{AL_1}}{\overrightarrow{L_1B}}: +\frac{\overrightarrow{AL_2}}{\overrightarrow{L_2B}},$$ + vidimo, da je za notranjo in zunanjo delitev daljice $AB$ s točkama $L_1$ in $L_2$ ustrezno dvorazmerje enako $-1$, oz. $d(A,B;L_1,L_2)=-1$ +Tako dobimo idejo za novo definicijo. + +Pravimo, da različne kolinearne točke $A$, $B$, $C$ in $D$ določajo + \index{harmonična četverica točk} + \pojem{harmonično četverico točk} oz. da je par $(A, B)$, + \pojem{harmonično konjugiran} s parom $(C, D)$, oznaka + $\mathcal{H}(A,B;C,D)$, če je: + \begin{eqnarray}\label{formulaHarmEvkl} + \frac{\overrightarrow{AC}}{\overrightarrow{CB}}= + -\frac{\overrightarrow{AD}}{\overrightarrow{DB}}, + \end{eqnarray} oziroma: + \begin{eqnarray*} +d(A,B;C,D)=\frac{\overrightarrow{AC}}{\overrightarrow{CB}}: +\frac{\overrightarrow{AD}}{\overrightarrow{DB}}=-1. + \end{eqnarray*} + +Dokažimo osnovne lastnosti definirane relacije. + Kot smo že omenili, točki $C$ in $D$, za kateri velja $\mathcal{H}(A,B;C,D)$, predstavljata +notranjo in zunanjo delitev daljice $AB$ v nekemu razmerju $\lambda$ ($\lambda>0$ in $\lambda\neq 1$). Takšnih parov $C$, $D$ je torej +neskončno mnogo. Toda, če je dana ena od točk $C$ ali $D$, je druga enolično določena. To povejmo še z drugimi besedami v naslednji trditvi. + + + + \bizrek \label{HarmCetEnaSamaTockaD} + Let $C$ be a point on a line $AB$ different from the points $A$ and $B$ + and also from the midpoint of the line segment $AB$, + then there is exactly one point $D$ such that + $\mathcal{H}(A,B;C,D)$. + \eizrek + + + \textit{\textbf{Proof.}} Naj bo $\frac{\overrightarrow{AC}}{\overrightarrow{CB}}=\lambda$. Ker se točka $C$ razlikuje od točk $A$, $B$ in + središča daljice $AB$, je $\lambda\neq 0$ in $\lambda\neq 1$. Iščemo točko $D$, za katero je $\mathcal{H}(A,B;C,D)$ oz. $d(A,B;C,D)=-1$ ali ekvivalentno $\frac{\overrightarrow{AD}}{\overrightarrow{DB}}=-\lambda$. Ker je $-\lambda\neq -1$, po izreku \ref{HarmCetEnaSamaDelitev} obstaja ena sama točka $D$, za katero je to izpolnjeno. +\kdokaz + + + + \bizrek \label{HarmCetEF} + Let $A$, $B$, $C$ and $D$ be four different collinear points + on a line $p$ and $O$ a point not lying on this line. Suppose that + a line that is parallel to the line $OA$ through the point $B$ + intersects the lines $OC$ and $OD$ at the points $E$ and $F$. + Then: + $$\mathcal{H}(A,B;C,D) \hspace*{1mm} \Leftrightarrow \hspace*{1mm} \mathcal{S}_B(E)=F.$$ + \eizrek + + + \textit{\textbf{Proof.}} + (Figure \ref{sl.pod.7.4.2.pic}) + +($\Rightarrow$) Če velja $\mathcal{H}(A,B;C,D)$, je po Talesovemu izreku \ref{TalesovIzrek}: +$$\frac{\overrightarrow{AO}}{\overrightarrow{EB}}= +\frac{\overrightarrow{AC}}{\overrightarrow{CB}}= +-\frac{\overrightarrow{AD}}{\overrightarrow{DB}}= +-\frac{\overrightarrow{AO}}{\overrightarrow{FB}}= +\frac{\overrightarrow{AO}}{\overrightarrow{BF}},$$ + zato je +$\overrightarrow{EB}=\overrightarrow{BF}$ oz. $\mathcal{S}_B(E)=F$. + +($\Leftarrow$) Naj bo sedaj $\mathcal{S}_B(E)=F$. Iz tega sledi $\overrightarrow{EB}=\overrightarrow{BF}$, zato je (spet po Talesovemu izreku \ref{TalesovIzrek}): +$$\frac{\overrightarrow{AC}}{\overrightarrow{CB}}= +\frac{\overrightarrow{AO}}{\overrightarrow{EB}}= +\frac{\overrightarrow{AO}}{\overrightarrow{BF}}= +-\frac{\overrightarrow{AO}}{\overrightarrow{FB}}= +-\frac{\overrightarrow{AD}}{\overrightarrow{DB}},$$ +torej je po definiciji $\mathcal{H}(A,B;C,D)$. +\kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.2.pic} +\caption{} \label{sl.pod.7.4.2.pic} +\end{figure} + +Prejšnji izrek nam omogoča efektivno konstrukcijo četrte točke v harmonični četverici točk. + + + + \bzgled \label{HarmCetEnaSamaTockaDKonstr} + Let $C$ be a point that lies on a line $AB$. Suppose that $C$ is different from the points $A$, + $B$ and the midpoint of the line segment $AB$. Construct a point $D$ such that + $\mathcal{H}(A,B;C,D)$. + \ezgled + + + \textit{\textbf{Rešitev.}} Po izreku \ref{HarmCetEnaSamaTockaD} obstaja ena sama takšna točka $D$, za katero je $\mathcal{H}(A,B;C,D)$. Sedaj jo bomo tudi + konstruirali. Naj bo $O$ poljubna točka, ki ne leži na premici $AB$ (Figure \ref{sl.pod.7.4.2.pic}), in $l$ vzporednica premice $AO$ v točki $B$. Z $E$ označimo presečišče premic $OC$ in $l$ in $F=\mathcal{S}_B(E)$. Točko $D$ dobimo kot presečišče premic $OF$ in $AB$. + Po prejšnjem izreku \ref{HarmCetEF} je $\mathcal{H}(A,B;C,D)$. +\kdokaz + +Na podoben način kot v prejšnjem zgledu lahko za dane kolinearne točke $A$, $B$ in $D$ (točka $D$ ne leži na daljici $AB$) načrtamo takšno točko $C$, da velja $\mathcal{H}(A,B;C,D)$. + + Intuitivno je jasno, da iz $\mathcal{H}(A,B;C,D)$ sledi $\mathcal{H}(A,B;D,C)$, kajti če točki $C$ in $D$ delita daljico $AB$ v istem razmerju, enako velja tudi za točki $D$ in $C$. Zanimivo je, da iz $\mathcal{H}(A,B;C,D)$ sledi tudi $\mathcal{H}(C,D;A,B)$, kar pomeni, da če točki $C$ in $D$ delita daljico $AB$ v istem razmerju, potem tudi točki $A$ in $B$ delita daljico $CD$ v istem razmerju. Dokažimo formalno obe lastnosti. + + \bizrek + a) $\mathcal{H}(A,B;C,D) \hspace*{1mm} \Rightarrow + \hspace*{1mm} \mathcal{H}(A,B;D,C)$; \\ + \hspace*{22mm}b) $\mathcal{H}(A,B;C,D) + \hspace*{1mm} \Rightarrow + \hspace*{1mm} \mathcal{H}(C,D;A,B)$; + \eizrek + + + \textit{\textbf{Proof.}} + + $$a)\hspace*{1mm}\mathcal{H}(A,B;C,D) +\hspace*{1mm} \Rightarrow + \hspace*{1mm} + \frac{\overrightarrow{AC}}{\overrightarrow{CB}}= -\frac{\overrightarrow{AD}}{\overrightarrow{DB}} +\hspace*{1mm} \Rightarrow + \hspace*{1mm} +\frac{\overrightarrow{AD}}{\overrightarrow{DB}}= -\frac{\overrightarrow{AC}}{\overrightarrow{CB}} +\hspace*{1mm} \Rightarrow + \hspace*{1mm} +\mathcal{H}(A,B;D,C).$$ + + + $$b)\hspace*{1mm}\mathcal{H}(A,B;C,D) +\hspace*{1mm} \Rightarrow + \hspace*{1mm} + \frac{\overrightarrow{AC}}{\overrightarrow{CB}}= -\frac{\overrightarrow{AD}}{\overrightarrow{DB}} +\hspace*{1mm} \Rightarrow + \hspace*{1mm} +\frac{\overrightarrow{CA}}{\overrightarrow{AD}}= -\frac{\overrightarrow{CB}}{\overrightarrow{BD}} +\hspace*{1mm} \Rightarrow + \hspace*{1mm} +\mathcal{H}(C,D;A,B),$$ kar je bilo treba dokazati. \kdokaz + + + + \bizrek \label{izrek 1.2.1} + Let $A$, $B$, $C$ and $D$ be different collinear points. + Then $\mathcal{H}(A,B;C,D)$ if and only if there exists such a quadrilateral $PQRS$, + that: + $$A=PQ\cap RS, \hspace*{2mm}B=QR\cap PS, \hspace*{2mm} C\in PR \hspace*{2mm} + \textrm{and} + \hspace*{2mm} D\in QS.$$ + \eizrek + + + \textit{\textbf{Proof.}} ($\Rightarrow$) Naj bo $PQRS$ tak štirikotnik, + da je $A=PQ\cap RS$, $B=QR\cap PS$, $C\in PR$ in $D\in QS$ + (Figure \ref{sl.pod.7.4.7a.pic}). Po Menelajevem izreku + \ref{izrekMenelaj} za trikotnik $ABP$ in premico $QS$ dobimo: + $$\frac{\overrightarrow{AD}}{\overrightarrow{DB}}\cdot + \frac{\overrightarrow{BS}}{\overrightarrow{SP}}\cdot + \frac{\overrightarrow{PQ}}{\overrightarrow{QA}}=-1.$$ + Analogno z uporabo Cevovega izreka \ref{izrekCeva} za isti + trikotnik in točko $R$ dobimo: + $$\frac{\overrightarrow{AC}}{\overrightarrow{CB}}\cdot + \frac{\overrightarrow{BS}}{\overrightarrow{SP}}\cdot + \frac{\overrightarrow{PQ}}{\overrightarrow{QA}}=1.$$ +Iz teh dveh relacij sledi +$\frac{\overrightarrow{AC}}{\overrightarrow{CB}}= +-\frac{\overrightarrow{AD}}{\overrightarrow{DB}}$ oz. +$\mathcal{H}(A,B;C,D)$. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.7a.pic} +\caption{} \label{sl.pod.7.4.7a.pic} +\end{figure} + + ($\Leftarrow$) Predpostavimo sedaj, da velja + $\mathcal{H}(A,B;C,D)$. Brez škode za splošnost naj bo točka + $C$ med točkama $A$ in $B$; v drugih dveh primerih dokaz poteka + enako. Naj bo $P$ poljubna točka izven premice $AB$ in $Q$ + poljubna točka med točkama $A$ in $P$. Po Paschovem\footnote{\index{Pasch, M.} + \textit{M. Pasch} + (1843--1930), nemški matematik, ki je relacijo urejenosti točk vpeljal v svojem delu + \textit{Predavanja o novejši geometriji} iz leta 1882.} \ref{PaschIzrek} + aksiomu se premici $PC$ + in $QB$ sekata v neki točki $R$ ter premici $AR$ in $PB$ v neki + točki $S$, zato se tudi premici $AB$ in $QS$ sekata v neki točki $D_1$ (iz $AB\parallel QS$ + sledi, da je + $C$ središče daljice $AB$, kar ni možno zaradi + $\mathcal{H}(A,B;C,D)$). Sedaj iz prvega dela dokaza ($\Rightarrow$) + sledi $\mathcal{H}(A,B;C,D_1)$. Ker je po predpostavki še + $\mathcal{H}(A,B;C,D)$, iz enoličnosti četrte točke harmonične + četverice točk (izrek \ref{HarmCetEnaSamaTockaD}) + sledi $D=D_1$. Torej je $PQRS$ iskani + štirikotnik. + \kdokaz + + Opazimo, da je prejšnji izrek smiseln tudi v primeru, ko je $C$ + središče daljice $AB$ in $D$ točka v neskončnosti (Figure \ref{sl.pod.7.4.7a.pic}). + + Izjavo na desni strani ekvivalence iz prejšnjega izreka (obstoj + ustreznega štirikotnika) bi lahko privzeli kot definicijo + harmonične četverice točk v evklidski geometriji. V + projektivni geometriji je ta definicija še bolj naravna, saj v + takšni definiciji ne uporabljamo metrike. + + Iz prejšnjega izreka sledi tudi, da se relacija harmonične + četverice ohranja pri t. i. \index{središčna projekcija}\pojem{središčni projekciji} v prostoru. Res, če so $A$, + $B$, $C$ in $D$ točke, za katere velja $\mathcal{H}(A,B;C,D)$, + ter $A'$, $B'$, $C'$ in $D'$ središčne projekcije teh točk, + tedaj iz obstoja ustreznega štirikotnika za četverico $A$, + $B$, $C$, $D$ sledi obstoj ustreznega štirikotnika tudi za + četverico $A'$, $B'$, $C'$, $D'$. Drugi je središčna + projekcija prvega štirikotnika, saj središčna projekcija + ohranja kolinearnost. Seveda bi v prejšnjo obravnavo morali + vključiti tudi primer, ko so nekatere od središčnih projekcij + točke v neskončnosti. + + \index{geometrija!projektivna}V tem smislu lahko projektivno geometrijo opišemo kot + tisto geometrijo, ki se ukvarja z objekti in lastnostmi, ki se ohranjajo + pri središčni projekciji. Ker je 'biti harmonička četverica + točk' ena od takšnih lastnosti, je predmet preučevanja v projektivni + geometriji. Še več - v tej geometriji je ta relacija ena od + osnovnih pojmov (glej \cite{Mitrovic}). + + Ugotovili smo že, da na premici $AB$ obstajata natanko dve točki, ki delita daljico $AB$ v +razmerju $\lambda>0$, $\lambda\neq 1$ (v primeru $\lambda= 1$ je to le ena točka - središče daljice). Omenjeni točki predstavljata +notranjo in zunanjo delitev daljice v tem razmerju in skupaj s točkama $A$ in $B$ določata harmonijsko četverico točk. Zastavlja se vprašanje: Kaj predstavlja množico vseh +takšnih točk $X$ v ravnini, da velja $\frac{AX}{XB}=\lambda$ ($\lambda\neq 1$)? Odgovor nam da naslednji izrek. + + + + + + + + + \bizrek \label{ApolonijevaKroznica} + Suppose that $A$ and $B$ are points in the plane and $\lambda>0$, $\lambda\neq 1$ + an arbitrary real number. If $C$ and $D$ are points of the line $AB$ such that: + $$\frac{\overrightarrow{AC}}{\overrightarrow{CB}}= + -\frac{\overrightarrow{AD}}{\overrightarrow{DB}}=\lambda,$$ + i.e. $\mathcal{H}(A,B;C,D)$, then the set of all points $X$ of this plane such that: + $$\frac{AX}{XB}=\lambda,$$ + is a circle with the diameter $CD$ + (so-called Apollonius Circle\footnote{\index{Apolonij} + \textit{Apolonij iz Perge} (3.-- 2. st. pr. n. + š.), starogrški matematik.}). + \index{krožnica!Apolonijeva} + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.3.pic} +\caption{} \label{sl.pod.7.4.3.pic} +\end{figure} + + \textit{\textbf{Proof.}} +Naj bo $k$ krožnica nad premerom $CD$. Naj bo nato še +$X$ poljubna točka te ravnine in $p$ vzporednica premice $AX$ skozi točko $B$. Z $E$ in $F$ označimo presečišča premic $XC$ in $XD$ +s premico $p$ (Figure \ref{sl.pod.7.4.3.pic}). Po izreku \ref{HarmCetEF}, je točka $B$ središče +daljice $EF$, ker je $\mathcal{H}(A,B;C,D)$. Potrebno je dokazati, da velja: +$$\frac{AX}{XB}=\lambda \hspace*{1mm} \Leftrightarrow \hspace*{1mm} X\in k.$$ + +($\Leftarrow$) Naj bo $X\in k$. Potem je $\angle EXF=\angle CXD=90^0$ in velja $BE\cong BF\cong BX$ (izrek \ref{TalesovIzrKroz2}). Iz tega in iz Talesovega izreka \ref{TalesovIzrek} sledi: +$$\frac{AX}{XB}=\frac{AX}{EB}=\frac{AC}{CB}=\lambda.$$ + +($\Rightarrow$) Naj bo $\frac{AX}{XB}=\lambda$. Iz tega in iz Talesovega izreka \ref{TalesovIzrek} sledi: +$$\frac{AX}{XB}=\lambda=\frac{AC}{CB}=\frac{AX}{EB},$$ +zato je $XB\cong EB\cong BF$. Torej je $EXF$ +pravokotni trikotnik (izrek \ref{TalesovIzrKroz2}). Potem $\angle CXD=\angle EXF=90^0$, kar pomeni, da točka $X$ leži na krožnici $k$ nad +premerom $CD$ (izrek \ref{TalesovIzrKroz2}). +\kdokaz + + +Za dani točki $A$ in $B$ za različne vrednosti $\lambda\in \mathbb{R}$ ($\lambda>0$, $\lambda\neq 1$) imamo različne Apolonijeve krožnice (Figure \ref{sl.pod.7.4.4a.pic}). Označili jih bomo z $\mathcal{A}_{AB,\lambda}$, če pa vemo za katero daljico gre, lahko tudi krajše $k_{\lambda}$. V primeru $\lambda=1$ dejansko iščemo množico vseh takšnih točk $X$ te ravnine, za katere je: + $\frac{AX}{XB}=1$, oz. $AX\cong BX$. Takrat iskana množica ne predstavlja krožnice, ampak simetralo $s_{AB}$ daljice $AB$. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.4a.pic} +\caption{} \label{sl.pod.7.4.4a.pic} +\end{figure} + + + + \bzgled + Points $A$ and $B$ and a line $l$ in the plane are given. + Construct the point $L$ on the line $l$ such that $LA:LB=5:2$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.4.pic} +\caption{} \label{sl.pod.7.4.4.pic} +\end{figure} + + \textit{\textbf{Rešitev.}} + (Figure \ref{sl.pod.7.4.4.pic}) + +Načrtajmo najprej takšno točko $C$ na daljici $AB$, da velja $AC:CB=5:2$ + (glej zgled \ref{izrekEnaDelitevDaljice}), nato pa četrto točko harmonične četverice $\mathcal{H}(A,B;C,D)$ (glej zgled \ref{HarmCetEnaSamaTockaDKonstr}). Konstruirajmo Aplolonijevo krožnico $k_{\frac{5}{2}}$ nad premerom $CD$. Točka $L$ je potem tista, za katero velja $L\in k\cap l$. + +Po prejšnjem izreku (\ref{ApolonijevaKroznica}) je: +$$\frac{AL}{LB}=\frac{AC}{CB}=\frac{5}{2}.$$ + +Naloga ima 0, 1 ali 2 rešitvi, odvisno od števila presečišč krožnice $k$ in premice $l$. +\kdokaz + + + + \bizrek \label{HarmOhranjaVzporProj} + A parallel projection preserves the relation of a harmonic conjugate points, + i.e. if $A$, $B$, $C$, $D$ and $A'$, $B'$, $C'$, $D'$ are two quartets + of collinear points such that $AA'\parallel BB'\parallel CC'\parallel DD'$, then: + $$\mathcal{H}(A,B;C,D)\hspace*{1mm}\Rightarrow\hspace*{1mm} + \mathcal{H}(A',B';C',D').$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.5.pic} +\caption{} \label{sl.pod.7.4.5.pic} +\end{figure} + + \textit{\textbf{Rešitev.}} + (Figure \ref{sl.pod.7.4.5.pic}) + +Naj bo $\mathcal{H}(A,B;C,D)$ oz. $\frac{\overrightarrow{AC}}{\overrightarrow{CB}}: +\frac{\overrightarrow{AD}}{\overrightarrow{DB}}=-1$. Po Talesovem izreku \ref{TalesovIzrek} je: +$$\frac{\overrightarrow{A'C'}}{\overrightarrow{C'B'}}: +\frac{\overrightarrow{A'D'}}{\overrightarrow{D'B'}}= +\frac{\overrightarrow{AC}}{\overrightarrow{CB}}: +\frac{\overrightarrow{AD}}{\overrightarrow{DB}}=-1,$$ +kar pomeni, da velja tudi $\mathcal{H}(A',B';C',D')$. +\kdokaz + +Zanimivo je, da tudi središčna projekcija ohranja relacijo harmonične četverice točk. Zato je omenjena relacija predmet raziskovanja v projektivni geometriji (glej \cite{Mitrovic}). + + + + \bizrek \label{HarmCetSimKota} + Suppose that a line $BC$ intersects the bisectors of the interior and + the exterior angle at the vertex $A$ of a triangle $ABC$ at points $E$ and $F$. Then: + + a) $BE:CE=BF:CF=AB:AC$, + + b) $\mathcal{H}(B,C;E,F)$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.6.pic} +\caption{} \label{sl.pod.7.4.6.pic} +\end{figure} + + \textit{\textbf{Rešitev.}} Naj bo $L$ poljubna točka, za katero velja $\mathcal{B}(C,A,L)$. +Označimo z $M$ in $N$ točki, v katerih vzporednica premice $AC$ skozi točko $B$ seka po vrsti +simetrale $AE$ in $AF$ notranjega in zunanjega kota (Figure \ref{sl.pod.7.4.6.pic}). + +\textit{a)} Po izreku \ref{KotiTransverzala} je: + \begin{eqnarray*} +\angle BMA &\cong& \angle CAM=\angle CAE\cong\angle BAE=\angle BAM\\ +\angle BNA &\cong& \angle LAN=\angle LAF\cong\angle BAF=\angle BAN +\end{eqnarray*} + +Torej sta $AMB$ in $ANB$ enakokraka trikotnika z osnovnicama $AM$ oz. $AN$ (izrek \ref{enakokraki} in velja: + $$BM\cong BA\cong BN.$$ +Iz tega in iz posledice Talesovega izreka \ref{TalesovIzrekDolzine} (ker je $AC\parallel NM$) dobimo: +\begin{eqnarray*} +\frac{BE}{EC}&=& \frac{BM}{AC}=\frac{BA}{AC},\\ +\frac{BF}{FC}&=& \frac{BN}{AC}=\frac{BA}{AC}. +\end{eqnarray*} + +\textit{b)} Ker je $\mathcal{B}(B,E,C)$ in $\neg\mathcal{B}(B,F,C)$, je +\begin{eqnarray*} +\frac{\overrightarrow{BE}}{\overrightarrow{EC}}&=& \frac{BA}{AC},\\ +\frac{\overrightarrow{BF}}{\overrightarrow{FC}}&=& -\frac{BA}{AC}. +\end{eqnarray*} +Torej: +\begin{eqnarray*} +\frac{\overrightarrow{BE}}{\overrightarrow{EC}}= +-\frac{\overrightarrow{BF}}{\overrightarrow{FC}}, +\end{eqnarray*} +kar pomeni, da velja $\mathcal{H}(B,C;E,F)$. +\kdokaz + +Nadaljevali bomo z uporabo prejšnjega izreka. + + \bzgled \label{HarmTrikZgl1} + Construct a triangle $ABC$, with given: $a$, $t_a$, $b:c=2:3$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.8.pic} +\caption{} \label{sl.pod.7.4.8.pic} +\end{figure} + + \textit{\textbf{Rešitev.}} +Naj bo $ABC$ trikotnik, v katerem je stranica $BC\cong a$, težiščnica $AA_1\cong t_a$ in $AC:AB=2:3$ (Figure \ref{sl.pod.7.4.8.pic}). Torej lahko +konstruiramo stranico $BC$ in njeno središče $A_1$. +Iz danih pogojev oglišče $A$ leži na krožnici +$k(A_1,t_a)$ in Apolonijevi krožnici $\mathcal{A}_{BC,\frac{3}{2}}$, ker je $AB:AC=3:2$ (izrek \ref{ApolonijevaKroznica}). Torej oglišče $A$ dobimo kot eno od presečišč krožnic +$k$ in $\mathcal{A}_{BC,\frac{3}{2}}$. + +Čeprav tega dejstva ne potrebujemo pri konstrukciji, omenimo, da +Apolonijeva krožnica $\mathcal{A}_{BC,\frac{3}{2}}$ predstavlja krožnico s premerom $EF$, kjer sta $E$ in $F$ definirani kot v izreku \ref{HarmCetSimKota}. +\kdokaz + + + + \bzgled + Construct a parallelogram, with the sides congruent to + given line segments $a$ and $b$, and the diagonals in the ratio $3:7$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.9.pic} +\caption{} \label{sl.pod.7.4.9.pic} +\end{figure} + + \textit{\textbf{Rešitev.}} (Figure \ref{sl.pod.7.4.9.pic}) + +Naj bo $S$ presečišče diagonal iskanega paralelograma $ABCD$, v katerem je $AB\cong a$, $BC\cong b$ in $AC:BD=3:7$. Označimo še s $P$ središče starnice $AB$. +Ker se pri paralelogramu +diagonali razpolavljata, sta tudi daljici $SA$ in $SB$ v razmerju $2:3$. To pomeni, da lahko najprej konstruiramo +trikotnik $ASB$, podobno kot v prejšnjem primeru \ref{HarmTrikZgl1}: +$AB\cong a$, $SP=\frac{1}{2}b$ in $SA:SB=3:7$. + \kdokaz + + + + + + \bizrek \label{harmVelNal} + Suppose that $k(S, r)$ is the incircle and $k_a (S_a, r_a)$ the excircle of a triangle $ABC$ + and $P$ and $P_a$ the touching points of these circles with the aide $BC$. + Let $A'$ be the foot of the altitude $AA'$ and $E$ + intersection of the bisectors of the + interior angle at the vertex $A$ and the side $BC$. + If $L$ and $L_a$ are foots of the perpendiculars from the points $S$ and $S_a$ on the line $AA'$, then: + + \hspace*{2mm} (i) $\mathcal{H}(A,E;S,Sa)$ \hspace*{2mm} (ii) + $\mathcal{H}(A,A';L,La)$ \hspace*{2mm} (iii) + $\mathcal{H}(A',E;P,Pa)$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.3.7.pic} +\caption{} \label{sl.pod.7.3.7.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.3.7.pic}) + +\textit{(i)} Premici $CS$ in $CS_a$ sta simetrali +notranjega in zunanjega kota pri oglišču $C$ +trikotnika $ACE$, zato po izreku \ref{HarmCetSimKota} velja $\mathcal{H}(A,E;S,S_a)$. + +\textit{(ii)} Točke $A$, $A'$, $L$ in $L_a$ +so pravokotne projekcije točk $A$, $E$, $S$ in $S_a$ na premici $AA'$. Po izreku \ref{HarmOhranjaVzporProj} je $\mathcal{H}(A,A';L,La)$. + +\textit{(iii)} Analogno prejšnji trditvi, saj so +točke $A'$, $E$, $P$ in $P_a$ + pravokotne projekcije točk $A$, $E$, $S$ in $S_a$ na premici $BC$. +\kdokaz + + + \bzgled Construct a triangle $ABC$, with given: + + \hspace*{4mm} (i) $r$, $a$, $v_a$ \hspace*{5mm} (ii) $v_a$, + $r$, $b-c$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.10.pic} +\caption{} \label{sl.pod.7.4.10.pic} +\end{figure} + +\textbf{\textit{Solution.}} +Uporabimo oznake kot v veliki nalogi \ref{velikaNaloga} in izreku \ref{harmVelNal} (Figure \ref{sl.pod.7.4.10.pic}). V obeh primerih \textit{(i)} in \textit{(ii)} lahko uporabimo dejstvo $\mathcal{H}(A,A';L,La)$ iz izreka \ref{harmVelNal}, ker iz $AA'\cong v_a$ in $LA'\cong r_a$ lahko načrtamo četrto točko v harmonični četverici $\mathcal{H}(A,A';L,La)$. Tako dobimo $r_a\cong A'L_a$. + +\textit{(i)} Uporabimo relacijo $RR_a\cong a$ (velika naloga \ref{velikaNaloga}), narišemo krožnici $k(S,SR)$ in $k_a(S_a,R_a)$, nato pa še njune ustrezne tri skupne tangente (zgled \ref{tang2ehkroz}). + + +\textit{(ii)} Uporabimo relacijo $PP_a\cong b-c$ (velika naloga \ref{velikaNaloga}), narišemo krožnici $k(S,SP)$ in $k_a(S_a,P_a)$, nato pa še njune ustrezne tri skupne tangente. +\kdokaz + +V prejšnjem primeru smo videli, kako lahko s pomočjo dokazane relacije $\mathcal{H}(A,A';L,La)$ iz elementov trikotnika $v_a$ in $r$ dobimo $r_a$. Na podoben način iz vsakega znanega para trojice $(v_a,r,r_a)$ dobimo tretji element. To dejstvo bomo zapisali v obliki $\langle v_a,r,r_a\rangle$. + + + + + + \bzgled + Construct a cyclic quadrilateral $ABCD$, with the sides congruent to + given line segments $a$, $b$, $c$ in $d$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.4.11.pic} +\caption{} \label{sl.pod.7.4.11.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.4.11.pic}) + +Naj bo $ABCD$ tetivni štirikotnik s +stranicami, ki so skladne z danimi +daljicami, oz. $AB\cong a$, $BC\cong b$, $CD\cong c$ in $DA\cong d$, $h_{A,k}$ središčni razteg s središčem $A$ +in koeficientom $k=\frac{AD}{AB}=\frac{d}{a}$, $\mathcal{R}_{A,\alpha}$ rotacija s +središčem $A$ za orientirani kot $\alpha=\angle BAD$ ter kompozitum: + $$\rho_{A,k,\alpha}=\mathcal{R}_{A,\alpha}\circ h_{A,k}$$ +rotacijski razteg s središčem $A$, koeficientom $k$ in kotom $\alpha$. + +Naj bo $B'=h_{A,k}(B)$ in $C'=h_{A,k}(C)$. Ker je $|AB'|=k\cdot |AB|=\frac{|AD|}{|AB|}\cdot |AB|= |AD|$ in $\angle B'AD=\angle BAD= \alpha$, velja $R_{A,\alpha}(B')=D$, zato je tudi $\rho_{A,k,\alpha}(B)=D$. Naj bo $E=R_{A,\alpha}(C')$ +oz. $E=\rho_{A,k,\alpha}(C)$. Potem je trikotnik $ADE$ slika trikotnika $ABC$ pri rotacijskem raztegu $\rho_{A,k,\alpha}$ (ki je transformacija podobnosti), zato sta ta +dva trikotnika podobna s koeficientom podobnosti $k$. Iz tega in iz izreka \ref{TetivniPogoj} sledi: +$$\angle EDC=\angle EDA+\angle ADC=\angle CBA+\angle ADC=180^0,$$ +kar pomeni, da so točke $E$, $D$ in $C$ kolinearne. Daljico $ED$ lahko konstruiramo (z uporabo Talesovega izreka \ref{TalesovIzrek}), ker je: +$$ED\cong B'C'=k\cdot BC=\frac{AD}{AB}\cdot BC=\frac{b\cdot d}{a}.$$ +Po konstrukciji točk $C$, $D$ in $E$ lahko načrtamo tudi točko $A$, kajti: +$$\frac{AE}{AC}=\frac{AC'}{AC}=k,$$ +zato točka $A$ leži na Apolonijevi krožnici $\mathcal{A}_{EC,k}$, prav tako tudi na krožnici $k(D,d)$. +\kdokaz + + + \bnaloga\footnote{44. IMO Japan - 2003, Problem 4.} + $ABCD$ is cyclic. The foot of the perpendicular from $D$ to the + lines $AB$, $BC$, $CA$ are $P$, $Q$, $R$, respectively. Show that the angle bisectors of + $\angle ABC$ and $\angle CDA$ meet on the line $AC$ if and only if $RP \cong RQ$. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.3.IMO1.pic} +\caption{} \label{sl.pod.7.3.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Brez škode za splošnost predpostavimo, da +velja $\mathcal{B}(B,C,Q)$ in $\mathcal{B}(B,P,A)$. Po izreku +\ref{SimpsPrem} so točke $P$, $Q$ in $R$ kolinearne in ležijo na +\index{premica!Simsonova} Simsonovi premici (Figure +\ref{sl.pod.7.3.IMO1.pic}). Iz dokaza tega izreka sledi tudi + $\angle CDQ\cong \angle ADP$, + kar pomeni, da sta $CDQ$ in $ADP$ podobna pravokotna trikotnika, + zato je: $$CD:CQ=AD:AP \textrm{ oz. } CD:AD=CQ:AP.$$ + Če označimo $P_1=\mathcal{S}_A(P)$, iz prejšnje relacije dobimo: + \begin{eqnarray} \label{eqn.pod.7.3.IMO1} + CD:AD=CQ:AP_1. + \end{eqnarray} +($\Rightarrow$) Predpostavimo najprej, da se simetrali kotov $ABC$ +in $CDA$ sekata v točki $E$, ki leži na premici + $AC$. Po izreku \ref{HarmCetSimKota} je potem: + $BC:BA=CE:EA=DC:DA$. Iz tega in iz dokazane relacije + \ref{eqn.pod.7.3.IMO1} sledi $BC:BA=CQ:AP_1$ oz. $BC:CQ=BA:AP_1$. + Ker velja $\mathcal{B}(B,C,Q)$ + in $\mathcal{B}(B,A,P_1)$, iz prejšnje relacije dobimo + $BC:BQ=BA:BP_1$. Po obratnem Talesovem izreku + \ref{TalesovIzrekObr} je $CA\parallel QP_1$ oz. $AR\parallel QP_1$, + zato je po Talesovem + izreku \ref{TalesovIzrek} $PR:RQ=PA:AP_1=1:1$ oz. $PR\cong RQ$. + + ($\Leftarrow$) Naj bo sedaj $PR\cong RQ$. Ker je $A$ središče daljice $PP_1$, + je $RA$ srednjica + trikotnika $PQP_1$. Iz tega sledi $AR\parallel QP_1$ oz. + $CA\parallel QP_1$. Po Talesovem izreku je $BC:CQ=BA:AP_1$ + oz. $BC:BA=CQ:AP_1$. Iz dokazane relacije + \ref{eqn.pod.7.3.IMO1} sedaj dobimo $BC:BA=CD:AD$. Naj bosta $E_1$ + in $E_2$ presečišči simetral + kotov $ABC$ in $CDA$ s premico $AC$. Če uporabimo + prejšnjo relacijo in izrek \ref{HarmCetSimKota}, dobimo:\\ + $CE_1:AE_1=BC:BA=CD:AD=CE_2:AE_2$. Ker točki $E_1$ in $E_2$ obe ležita + na daljici $AC$, je $E_1=E_2$ (izrek \ref{HarmCetEnaSamaDelitev}), zato se simetrali + kotov $ABC$ in $CDA$ sekata na premici $AC$. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{The Right Triangle Altitude Theorem. Euclid's Theorems} \label{odd7VisinEvkl} + +V algebri in matematični analizi sta znana pojma aritmetične oz. geometrijske sredine dveh (ali več) števil. +Na tem mestu bomo najprej definirali t. i. aritmetično in geometrijsko sredino dveh daljic. +Pravimo, da je $x$ \index{aritmetična sredina daljic} \pojem{aritmetična sredina} daljic $a$ in $b$, če velja: + +\begin{eqnarray} \label{eqnVisEvkl1} +|x|=\frac{1}{2}\left( |a|+|b| \right). +\end{eqnarray} +Prav tako je $y$ \index{geometrijska sredina daljic}\pojem{geometrijska (tudi geometrična) sredina} daljic $a$ in $b$, če velja: + +\begin{eqnarray} \label{eqnVisEvkl2} +|y|= \sqrt{|a|\cdot |b|}. +\end{eqnarray} + +Torej aritmetična oz. geometrijska sredina dveh daljic je daljica z dolžino, ki je enaka aritmetični oz. geometrijski sredini dolžin teh dveh daljic. Relaciji \ref{eqnVisEvkl1} in \ref{eqnVisEvkl2} bomo pogosto pisali v krajši obliki: +\begin{eqnarray*} +x=\frac{1}{2}\left( a+b \right)\hspace*{1mm} \textrm{ oz. }\hspace*{1mm} +y=\sqrt{ab}. +\end{eqnarray*} + +Jasno je, da na zelo enostaven način lahko konstruiramo aritmetično sredino dveh danih daljic. V tem razdelku bomo izpeljali konstrukcijo geometrijske sredine dveh danih daljic. Dokažimo najprej glavna izreka, ki se nanašata na pravokotne trikotnike. + + + + + \bizrek \index{izrek!višinski}\label{izrekVisinski} + The altitude on the hypotenuse of a right-angled triangle + is the geometric mean + of the line segments into which it divides the hypotenuse.\\ + (The right triangle altitude theorem) + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6E.1.pic} +\caption{} \label{sl.pod.7.6E.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Naj bo $ABC$ pravokotni trikotnik s +hipotenuzo $AB$ in z višino $CC'$. Označimo $v_c=|CC'|$, $b_1=|AC'|$ in $a_1=|BC'|$ (Figure \ref{sl.pod.7.6E.1.pic}). + +Najprej je $\angle ACC'=90^0-\angle CAC'=\angle CBC'$ in + $\angle BCC'=90^0-\angle CBC'=\angle CAC'$. +Iz podobnosti pravokotnih trikotnikov $CC'A$ in $BC'C$ (izrek \ref{PodTrikKKK}) je: +$$\frac{CC'}{BC'}=\frac{C'A}{C'C},$$ +torej: +\begin{eqnarray} \label{eqnVisinski} +v_c^2=a_1\cdot b_1. +\end{eqnarray} +\kdokaz + + + + \bizrek \index{izrek!Evklidov}\label{izrekEvklidov} + Each leg of a right-angled triangle + is the geometric mean + of the hypotenuse and the line segment of the hypotenuse adjacent to the leg.\\ + (Euclid's\footnote{Starogrški filozof in matematik \index{Evklid}\textit{Evklid iz Aleksandrije} (3. st. pr. n. š.).} theorems) + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6E.2.pic} +\caption{} \label{sl.pod.7.6E.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Naj bo $ABC$ pravokotni trikotnik s +hipotenuzo $AB$ in višino $CC'$. Označimo $a=|BC|$, $b=|AC|$, $c=|AB|$, $b_1=|AC'|$ in $a_1=|BC'|$ (Figure \ref{sl.pod.7.6E.2.pic}). + +Enako kot v dokazu izreka \ref{izrekVisinski} je $\angle ACC'=90^0-\angle CAC'=\angle CBA$ in + $\angle CAC'=\angle CAB$. +Iz podobnosti pravokotnih trikotnikov $CC'A$ in $BCA$ (izrek \ref{PodTrikKKK}) je: +$$\frac{CA}{BA}=\frac{C'A}{CA},$$ +oziroma: +\begin{eqnarray} \label{eqnEvklidov1} +b^2=b_1\cdot c. +\end{eqnarray} +Analogno je tudi: +\begin{eqnarray} \label{eqnEvklidov2} +a^2=a_1\cdot c. +\end{eqnarray} +\kdokaz + + + + \bzgled \label{EvklVisPosl} + If $c$ is the length of the hypotenuse, $a$ and $b$ the lengths of the legs and $v_c$ + the length of the altitude on the hypotenuse of a right-angled triangle $ABC$, then + $$c\cdot v_c=a\cdot b.$$ + \ezgled + + +\textbf{\textit{Proof.}} + Po množenju relacij \ref{eqnEvklidov1} in \ref{eqnEvklidov2} iz evklidovega izreka \ref{izrekEvklidov} ter vstavljanju relacije \ref{eqnVisinski} iz višinskega izreka \ref{izrekVisinski} dobimo: + $$a^2b^2=a_1b_1 c^2=v^2c^2,$$ + iz tega pa sledi $ab=cv_c$. + \kdokaz + + Sedaj bomo izpeljali napovedano konstrukcijo. + + + + \bzgled + Construct a line segment $x$ that is the geometric mean of given line segments + $a$ and $b$, i.e. $x=\sqrt{ab}$. + \ezgled + + +\textbf{\textit{Proof.}} Konstrukcijo bomo izpeljali na dva načina. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6E.3.pic} +\caption{} \label{sl.pod.7.6E.3.pic} +\end{figure} + +\textit{1)} + Načrtajmo najprej takšne točke $P$, $Q$ in $R$, da velja $PQ\cong a$, $QR\cong b$ in $\mathcal{B}(P,Q,R)$ (Figure \ref{sl.pod.7.6E.3.pic}), nato polkrožnico $k$ nad premerom $PR$ in na koncu točko $X$ kot presečišče polkrožnice $k$ s pravokotnico premice $PR$ v točki $Q$. + + Dokažimo, da je $x=QX$ iskana daljica. Po izreku \ref{TalesovIzrKroz2} je $PXR$ pravokotni trikotnik s hipotenuzo $PR$, $XQ$ je po konstrukciji njegova višina na tej hipotenuzi. Po višinskem izreku \ref{izrekVisinski} je $x$ geometrijska sredina daljic $a$ in $b$. + +\textit{2)} Brez škode za splošnost naj bo $a>b$ (če je $a=b$ je tudi $x=a$). + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6E.4.pic} +\caption{} \label{sl.pod.7.6E.4.pic} +\end{figure} + + Načrtajmo najprej takšne točke $P$, $Q$ in $R$, da velja $PQ\cong a$, $QR\cong b$ in $\mathcal{B}(P,R,Q)$ (Figure \ref{sl.pod.7.6E.4.pic}), nato polkrožnico $k$ nad premerom $PQ$ in na koncu točko $X$ kot presečišče polkrožnice $k$ s pravokotnico premice $PQ$ v točki $R$. + + + Dokažimo, da je $x=QX$ iskana daljica. Po izreku \ref{TalesovIzrKroz2} je $PXQ$ pravokotni trikotnik s hipotenuzo $PQ$, $XR$ je po konstrukciji njegova višina na to hipotenuzo. Po Evklidovem izreku \ref{izrekEvklidov} je $x$ geometrijska sredina daljic $a$ in $b$. +\kdokaz + +V naslednjem zgledu bomo obravnavali še eno pomembno konstrukcijo. + + \bzgled \label{konstrKoren} + Construct a line segment $x=e\sqrt{6}$, for a given line segment $e$. + \ezgled + +\textbf{\textit{Proof.}} Podobno kot v prejšnjem zgledu bomo konstrukcijo izpeljali na dva načina. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6E.5.pic} +\caption{} \label{sl.pod.7.6E.5.pic} +\end{figure} + +\textit{1)} + Načrtajmo najprej takšne točke $P$, $Q$ in $R$, da velja: $PQ=3e$, $QR=2e$ in $\mathcal{B}(P,Q,R)$ (Figure \ref{sl.pod.7.6E.5.pic}), nato polkrožnico $k$ nad premerom $PR$ in na koncu točko $X$ kot presečišče polkrožnice $k$ s pravokotnico premice $PR$ v točki $Q$. + + Dokažimo, da je $x=QX$ iskana daljica. Po izreku \ref{TalesovIzrKroz2} je $PXR$ pravokotni trikotnik s hipotenuzo $PR$, $XQ$ je po konstrukciji njegova višina na tej hipotenuzi. Po višinskem izreku \ref{izrekVisinski} je $x$ geometrijska sredina daljic $PQ=3e$ in $QR=2e$, oz. $x=\sqrt{3e\cdot 2e}=e\sqrt{6}$. + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6E.6.pic} +\caption{} \label{sl.pod.7.6E.6.pic} +\end{figure} + +\textit{2)} + Načrtajmo najprej takšne točke $P$, $Q$ in $R$, da velja: $PQ=3e$, $QR=2e$ in $\mathcal{B}(P,R,Q)$ (Figure \ref{sl.pod.7.6E.6.pic}), nato polkrožnico $k$ nad premerom $PQ$ in na koncu točko $X$ kot presečišče polkrožnice $k$ s pravokotnico premice $PQ$ v točki $R$. + + + Dokažimo, da je $x=QX$ iskana daljica. Po izreku \ref{TalesovIzrKroz2} je $PXQ$ pravokotni trikotnik s hipotenuzo $PQ$, $XR$ je po konstrukciji njegova višina na to hipotenuzo. Po Evklidovem izreku \ref{izrekEvklidov} je $x$ geometrijska sredina daljic $PQ=3e$ in $QR=2e$, oz. $x=\sqrt{3e\cdot 2e}=e\sqrt{6}$. +\kdokaz + + V prejšnjem zgledu smo torej opisali postopek za konstrukcijo daljice $x=e\sqrt{n}$, kjer je $e$ dana daljica, $n\in \mathbb{N}$ pa dano naravno število. Vedno imamo dve možnosti za konstrukcijo - z uporabo višinskega izreka oz. Evklidovega izreka. Toda če je $n$ sestavljeno število, imamo več možnosti za izbiro daljic $PQ$ in $QR$. Npr. za $n=12$, lahko izberemo $PQ=3e$ in $QR=4e$, lahko pa tudi $PQ=6e$ in $QR=2e$. Če je $n$ praštevilo, npr. $n=7$, je najbolj ugodno, če izberemo $PQ=7e$ in $QR=e$. + + +%________________________________________________________________________________ + \poglavje{Pythagoras' Theorem} \label{odd7Pitagora} + +Dokažimo znameniti Pitagorov izrek. + + + + \bizrek \label{PitagorovIzrek} + \index{izrek!Pitagorov} + The square of the length of the hypotenuse of a right-angled triangle is + equal to the sum of the squares of the lengths of both legs, i.e. for a right-angled triangle + $ABC$ with the hypotenuse of length $c$ and the legs of lengths $a$ and $b$ is + $$a^2+b^2=c^2$$ + (Pythagoras'\footnote{Predpostavlja se, da je bil ta izrek znan že Egipčanom (pribl. 3000 let pr. n. š.) in Babiloncem (pribl. 2000 + let pr. n. š.), toda starogrški filozof in matematik \index{Pitagora}\textit{Pitagora z otoka Samosa} (582--497 pr. n. š.) ga je verjetno prvi dokazal. Prvi pisni dokument dokaza Pitagorovega izreka je dal \index{Evklid}\textit{Evklid iz Aleksandrije} (3. st. pr. n. š.) v svojem delu ‘‘Elementi’’. + Pri Starih Grkih se je Pitagorov izrek običajno nanašal na zvezo med ploščinami kvadratov nad stranicami pravokotnega trikotnika.} theorem) + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.0.pic} +\caption{} \label{sl.pod.7.6.0.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.6.0.pic}) + +Če uporabimo izrek \ref{izrekEvklidov} in oznake iz dokaza tega izreka, dobimo: +$$a^2 + b^2 = c\cdot a_1 + c\cdot b_1 = c\cdot (a_1 + b_1) = c^2,$$ kar je bilo treba dokazati. \kdokaz + +V prejšnji obliki se Pitagorov izrek nanaša na kvadrate dolžin stranic pravokotnega trikotnika. Drugo obliko, ki se nanaša na zvezo med ploščinami kvadratov nad stranicami pravokotnega trikotnika, bomo obravnavali v razdelku \ref{odd8PloTrik}. + +Pitagorov izrek nam omogoča, da izračunamo tretjo stranico, če sta dani dve stranici pravokotnega trikotnika. Če s $c$ označimo dolžino hipotenuze ter z $a$ in $b$ dolžini katet pravokotnega trikotnika, iz Pitagorovega izreka dobimo: +\begin{eqnarray*} +c&=&\sqrt{a^2+b^2}\\ +a&=&\sqrt{c^2-b^2}\\ +b&=&\sqrt{c^2-a^2} +\end{eqnarray*} + + + + \bizrek \label{PitagorovIzrekObrat} + \index{izrek!Obratni Pitagorov} + Let $ABC$ be an arbitrary triangle. If + $$|AC|^2+|BC|^2=|AB|^2,$$ + then $ABC$ is a right-angled triangle with the right angle at the vertex $C$.\\ + (Converse of Pythagorean Theorem) + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.0a.pic} +\caption{} \label{sl.pod.7.6.0a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Označimo: $c=|AB|$, $b=|AC|$ in $a=|BC|$ (Figure \ref{sl.pod.7.6.0a.pic}). Torej velja $a^2+b^2=c^2$ oz.: +\begin{eqnarray} \label{eqnPitagObrat1} +|AB|=c=\sqrt{a^2+b^2}. +\end{eqnarray} +Naj bo $A'B'C'$ takšen pravokotni trikotnik s pravim kotom pri oglišču $C'$, da velja +$A'C'\cong AC$ in $B'C'\cong BC$. Po Pitagorovem izreku je +$|A'B'|^2= |A'C'|^2+|B'C'|^2=b^2+a^2$ +oziroma: +\begin{eqnarray} \label{eqnPitagObrat2} +|A'B'|=\sqrt{a^2+b^2}, +\end{eqnarray} + zato je iz \ref{eqnPitagObrat1} in \ref{eqnPitagObrat2} $A'B'\cong AB$. +Trikotnika $ABC$ in $A'B'C'$ sta torej skladna (po izreku \textit{SSS} \ref{SSS}), iz tega pa sledi $\angle ACB\cong \angle A'C'B'=90^0$, kar pomeni, da je $ABC$ pravokotni trikotnik s pravim kotom pri oglišču $C$. +\kdokaz + + + +Če so dolžine stranic $a$, $b$ in $c$ pravokotnega trikotnika $ABC$ (kjer je $c$ dolžina hipotenuze) naravna števila, pravimo, da trojica $(a,b,c)$ predstavlja \index{pitagorejska trojica}\pojem{pitagorejsko trojico}. Po Pitagorovem \ref{PitagorovIzrek} izreku za pitagorejsko trojico $(a, b, c)$ velja $a^2+b^2=c^2$. Po obratnem Pitagorovem izreku pa, če za naravna števila $a$, $b$ in $c$ velja $a^2+b^2=c^2$, je $(a,b,c)$ pitagorejska trojica. Najbolj znana pitagorejska trojica\footnote{Ta trojica je bila znana že Starim Egipčanom in Babiloncem. Imenuje se tudi \index{trikotnik!egipčanski}\pojem{egipčanski trikotnik}, ker so s pomočjo njega Stari Egipčani določali pravi kot na terenu.} je $(3, 4, 5)$ (Figure \ref{sl.pod.7.6.0b.pic}), ker velja $3^2+4^2=5^2$. Pogosto srečamo tudi pitagorejski trojici $(5, 12, 13)$ in $(7, 24, 25)$. + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.0b.pic} +\caption{} \label{sl.pod.7.6.0b.pic} +\end{figure} + +Pravzaprav obstaja neskončno mnogo pitagorejskih trojic. Že iz ene takšne trojice $(a,b,c)$ jih lahko dobimo neskončno mnogo: $(ka, kb, kc)$ (za poljubno $k\in \mathbb{N}$). Očitno gre v tem primeru za podobne pravokotne trikotnike. +Za pitagorejsko trojico pravimo, da je \index{pitagorejska trojica!primitivna}\pojem{primitivna}, če $a$, $b$ in $c$ nimajo skupnega delitelja. Izkaže se, da obstaja tudi neskončno mnogo primitivnih pitagorejskih trojic. + +En postopek računanja novih pitagorejskih trojic dobimo na naslednji način. +Če sta $m$ in $n$ poljubni naravni števili ($m > n$), potem je: + \begin{eqnarray*} + a &=& m^2 - n^2,\\ + b &=& 2mn,\\ + c &=& m^2 + n^2. + \end{eqnarray*} + Z enostavnim računom se lahko prepričamo, da $(a, b, c)$ res predstavlja pitagorejsko trojico. + + V nadaljevanju bomo uporabili Pitagorov izrek še pri drugih likih. + + + + + \bzgled \label{PitagorovPravokotnik} + If $a$ and $b$ are the lengths of the sides and $d$ the lengths of the diagonal of + a rectangle, then: + $$d=\sqrt{a^2+b^2}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.0c.pic} +\caption{} \label{sl.pod.7.6.0c.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $ABCD$ poljubni pravokotnik. Označimo: $a=|AB|$, $b=|BC|$ in $d=|AC|$ (Figure \ref{sl.pod.7.6.0c.pic}). Ker je $ABC$ pravokotni trikotnik s hipotenuze $AC$, je po Pitagorovem izreku \ref{PitagorovIzrek} $d^2=a^2+b^2$ oz. $d=\sqrt{a^2+b^2}$. +\kdokaz + + Direktna posledica (za $a=b$) je naslednja trditev. + + + + + \bzgled \label{PitagorovKvadrat} + If $a$ is the length of the side and $d$ the length of the diagonal of a square, + then (Figure \ref{sl.pod.7.6.0d.pic}): + $$d=a\sqrt{2}.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.0d.pic} +\caption{} \label{sl.pod.7.6.0d.pic} +\end{figure} + +Dejstvo, da je $\sqrt{2}$ iracionalno število ($\sqrt{2}\notin \mathbb{Q}$) oz. da ga ni možno zapisati v obliki ulomka, ima za posledico, da diagonala kvadrata $d$ in stranica $a$ nista \index{sorazmerni daljici}\pojem{sorazmerni} oz. \index{primerljivi daljici}\pojem{primerljivi} daljici\footnote{Stari Grki niso še poznali iracionalnih števil. Pitagorejci - filozofska šola, ki jo je utemeljil starogrški matematik \index{Pitagora} + \textit{Pitagora z otoka Samosa} + (582--497 pr. n. š.) - so bili prepričanja, da je vse število. Pri tem so mislili na racionalna števila in verjeli, da sta vsaki dve daljici sorazmerni. Prvi, ki je ugotovil nasprotno, je bil Starogrški matematik in filozof iz pitagorejske šole \index{Hipas}\textit{Hipas iz Metaponta} (5. st. pr. n. š.). Za to odkritje, ki je bilo torej v popolnem nasprotju s pitagorejsko filozofijo, je vezano več legend - od njegovega samomora do tega, da so Hipasa utopili drugi pitagorejci ali pa so ga le izključili iz svojega kroga.}. To pomeni, da ne obstaja nobena daljica $e$ kot enota, da bi za neka naravna števila $n,m\in \mathbb{N}$ veljalo $d=n\cdot e$ in $a=m\cdot e$. + + + + \bzgled \label{PitagorovEnakostr} + Suppose that $a$ is the length of the side of an equilateral triangle $ABC$. + If $v$ is the length of the altitude, $R$ the circumradius and $r$ the inradius + of that triangle, then + $$v=\frac{a\sqrt{3}}{2},\hspace{2mm} R=\frac{a\sqrt{3}}{3}, + \hspace{2mm} r=\frac{a\sqrt{3}}{6}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.0e.pic} +\caption{} \label{sl.pod.7.6.0e.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $AA'$ višina trikotnika $ABC$ (Figure \ref{sl.pod.7.6.0e.pic}). + +V razdelku \ref{odd3ZnamTock} smo dokazali, da so pri enakostraničnem trikotniku vse štiri značilne točke enake ($O=S=T=V$). To pomeni, da je težišče $T$ trikotnika $ABC$ hkrati središče očrtane in včrtane krožnice tega trikotnika. Prav tako je $T$ hkrati višinska točka tega trikotnika, kar pomeni, da je $AA'$ tudi težiščnica, zato je točka $A'$ središče stranice $BC$, oz. $|BA'|=\frac{a}{2}$. Po izreku \ref{tezisce} je $AT:TA'=2:1$. Torej: + \begin{eqnarray} \label{eqnPitagEnakostr1} + R=|TA|=\frac{2}{3}|AA'|=\frac{2}{3}v \hspace*{1mm} \textrm{ in } \hspace*{1mm} + r=|TA'|=\frac{1}{3}|AA'|=\frac{1}{3}v. + \end{eqnarray} + +Ker je $ABA'$ pravokotni trikotnik s hipotenuzo $AB$, je po Pitagorovem izreku \ref{PitagorovIzrek} +\begin{eqnarray*} + v=\sqrt{a^2-\left(\frac{a}{2} \right)^2}=\sqrt{\frac{3a^2}{4}}=\frac{a\sqrt{3}}{2}. + \end{eqnarray*} + Če uporabimo še relaciji iz \ref{eqnPitagEnakostr1}, dobimo $R=\frac{a\sqrt{3}}{3}$ + in $r=\frac{a\sqrt{3}}{6}$. +\kdokaz + + + + \bzgled \label{PitagorovRomb} + If $a$ is the length of the side and $e$ and $f$ the lengths of the diagonals of a rhombus, then + $$\left(\frac{e}{2}\right)^2+\left(\frac{f}{2}\right)^2=a^2.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.0f.pic} +\caption{} \label{sl.pod.7.6.0f.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $S$ presečišče diagonal romba $ABCD$ (Figure \ref{sl.pod.7.6.0f.pic}). Po izreku \ref{paralelogram} je $S$ središče diagonal $AC$ in $BD$. Torej $|SA|=\frac{e}{2}$ in $|SB|=\frac{f}{2}$. Po izreku \ref{RombPravKvadr} sta diagonali $AC$ in $BD$ pravokotni, zato je $ASB$ pravokotni trikotnik s hipotenuzo dolžine $|AB|=c$ in katetama dolžin $|SA|=\frac{e}{2}$ in $|SB|=\frac{f}{2}$. Iskana relacija je sedaj direktna posledica Pitagorovega izreka \ref{PitagorovIzrek}. +\kdokaz + + + + \bzgled + If $c$ is the length of the hypotenuse, $a$ and $b$ the lengths of the legs + and $v_c$ the length of the altitude on the hypotenuse of a right-angled + triangle, then + $$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{v_c^2}.$$ + \ezgled + + +\textbf{\textit{Proof.}} + Če uporabimo Pitagorov izrek \ref{PitagorovIzrek} in trditev iz zgleda \ref{EvklVisPosl}, dobimo: + $$\frac{1}{a^2}+\frac{1}{b^2}=\frac{a^2+b^2}{a^2b^2} + =\frac{c^2}{a^2b^2}=\frac{c^2}{c^2v_c^2}=\frac{1}{v_c^2},$$ kar je bilo treba dokazati. \kdokaz + +V zgledu \ref{konstrKoren} smo obravnavali dva načina konstrukcije daljice $x=e\sqrt{n}$ ($n\in \mathbb{N}$) z uporabo višinskega in Evklidovega izreka. Na tem mestu bomo to nalogo rešili s pomočjo Pitagorovega izreka - ponovno na dva načina. + + + \bzgled \label{konstrKorenPit} + Construct a line segment $e\sqrt{7}$, for a given line segment $e$. + \ezgled + +\textbf{\textit{Solution.}} Konstrukcijo bomo izpeljali na dva načina. + + +\textit{1)} (Figure \ref{sl.pod.7.6.3.pic}). +Ideja je, da načrtamo pravokotni trikotnik $ABC$ s kateto $a=e\sqrt{7}$, kjer sta druga kateta $b=n\cdot e$ in hipotenuza $c=m\cdot e$ ($n$ in $m$ sta naravni ali vsaj racionalni števili). +Po Pitagorovem izreku \ref{PitagorovIzrek} je $\left(e\sqrt{7}\right)^2+b^2=c^2$ oz. $m^2 e^2-n^2e^2=7e^2$ ali ekvivalentno $(m+n)(m-n)=7$. Eno rešitev te enačbe po $m$ in $n$ dobimo, če rešimo sistem: +\begin{eqnarray*} +&& m+n=7\\ +&& m-n=1 +\end{eqnarray*} +S seštevanjem in odštevanjem enačb iz tega sistema dobimo torej eno možnost $m=4$ in $n=3$; iz tega je $c=4e$ in $b=3e$. Ta ideja nam omogoča konstrukcijo. + +Načrtajmo najprej pravokotni trikotnik $ABC$ s hipotenuzo $AB=4e$ in kateto $AC=3e$. Po Pitagorovem izreku \ref{PitagorovIzrek} je $BC^2=\left(4e\right)^2-\left(3e\right)^2=7e^2$ oz. $BC=e\sqrt{7}$. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.3.pic} +\caption{} \label{sl.pod.7.6.3.pic} +\end{figure} + +\textit{2)} (Figure \ref{sl.pod.7.6.3.pic}). +Konstruirajmo najprej pravokotni trikotnik $A_0A_1A_2$ s katetama $A_0A_1=A_1A_2=e$, nato pravokotni trikotnik $A_0A_2A_3$ s katetama $A_0A_2$ in $A_2A_3=e$. Postopek nadaljujemo in konstruiramo zaporedje pravokotnih trikotnikov $A_0A_{n-1}A_n$ s katetama $A_0A_{n-1}$ in $A_{n-1}A_n=e$. Po Pitagorovem izreku je: + \begin{eqnarray*} +A_0A_2&=&\sqrt{A_0A_1^2+A_1A_2^2}=\sqrt{e^2+e^2}=e\sqrt{2}\\ +A_0A_3&=&\sqrt{A_0A_2^2+A_2A_3^2}=\sqrt{2e^2+e^2}=e\sqrt{3}\\ +&\vdots&\\ +A_0A_n&=&\sqrt{A_0A_{n-1}^2+A_{n-1}A_n^2}=\sqrt{(n-1)e^2+e^2}=e\sqrt{n} +\end{eqnarray*} + Iskano daljico torej dobimo iz šestega pravokotnega trikotnika: $A_0A_7=e\sqrt{7}$. +\kdokaz + +Nekaj naslednjih primerov se nanaša na uporabo Pitagorovega izreka pri krožnici. + + + + \bzgled \label{PitagoraCofman} + (Example from the book \cite{Cofman}) + Let $a$ and $b$ be two circles of the same + radius $r=1$, touching each other externally at the point $P$, + and $t$ their common external tangent. + If $c_1$, $c_2$, $\ldots$, $c_n$,... is a sequence of circles touching + circles $a$ and $b$, the first of them touching the line $t$ and each circle from the sequence + touches the previous one. Calculate the diameter of the circle $c_n$.\\ + + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.4.pic} +\caption{} \label{sl.pod.7.6.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} + Označimo: z $A$ in $B$ središči krožnic $a$ in $b$, z $A'$ in $B'$ dotikališči teh krožnic s +tangento $t$, s $S_n$ središča krožnic $c_n$ ($n =1,2,\ldots$), z $r_n$ njihove polmere, z $d_n=2r_n$ njihove +premere, s $P_n$ dotikališča krožnic $c_n$ in $c_{n-1}$ (oz. krožnice $c_1$ in premice $t$) ter s $S_n'$ in $X_n$ pravokotne projekcije točk $S_n$ +oz. $P_n$ na premici $AA'$ (Figure \ref{sl.pod.7.6.4.pic}). Naj bo še $x_n=|AX_n|$. Če uporabimo Pitagorov izrek \ref{PitagorovIzrek} za trikotnik +$AS_nS_n'$, dobimo $|AS_n|^2=|S_nS_n'|^2+|S_n'A|^2$ oz. +$\left(1+r_n \right)^2=1+\left(x_n-r_n \right)^2$. Iz tega reševanjem po $r_n$ dobimo: + \begin{eqnarray} \label{eqnPitagoraCofman1} + r_n=\frac{x_n^2}{2\left(1+x_n \right)}. + \end{eqnarray} + Po drugi strani je jasno $x_n-x_{n+1}=d_n=2r_n$. Če to povežemo z relacijo \ref{eqnPitagoraCofman1}, dobimo: + \begin{eqnarray*} + x_{n+1}=x_n-2r_n=x_n-\frac{x_n^2}{1+x_n}=\frac{x_n}{1+x_n}. + \end{eqnarray*} + Ker pri tem velja $x_1=1$, z direktnim računanjem dobimo $x_2=\frac{1}{2}$, $x_3=\frac{1}{3}$, $\ldots$ Iz tega intuitivno sklepamo, da velja: + \begin{eqnarray} \label{eqnPitagoraCofman2} + x_n=\frac{1}{n}. + \end{eqnarray} +Dokažimo relacijo \ref{eqnPitagoraCofman2} formalno - z uporabo matematične indukcije. Relacija \ref{eqnPitagoraCofman2} jasno velja za $n=1$. Predpostavimo, da relacija \ref{eqnPitagoraCofman2} velja za $n=k$ ($x_k=\frac{1}{k}$) in dokažimo, da iz tega sledi, da velja za $n=k+1$ ($x_{k+1}=\frac{1}{k+1}$): + \begin{eqnarray*} + x_{k+1}=\frac{x_k}{1+x_k}=\frac{\frac{1}{k}}{1+\frac{1}{k}}=\frac{1}{k+1}. + \end{eqnarray*} +S tem smo dokazali, da relacija \ref{eqnPitagoraCofman2} velja za vsak $n\in \mathbb{N}$. + +Na koncu iz relacij \ref{eqnPitagoraCofman1} in \ref{eqnPitagoraCofman2} sledi: +\begin{eqnarray*} + d_n=2r_n=\frac{x_n^2}{1+x_n}= + \frac{\left(\frac{1}{n}\right)^2}{1+\frac{1}{n}}=\frac{1}{n(n+1)} + \end{eqnarray*} +oz. +\begin{eqnarray} \label{eqnPitagoraCofman3} + d_n=\frac{1}{n(n+1)}, + \end{eqnarray} + kar je bilo potrebno izračunati. +\kdokaz + +S pomočjo trditve iz prejšnjega zgleda oz. relacije \ref{eqnPitagoraCofman3} lahko pridemo do zanimive neskončne vrste. Namreč $c_1, c_2, \ldots +c_n,\ldots$ je neskončno zaporedje krožnic in pri tem velja: +\begin{eqnarray*} + d_1+d_2+\cdots + d_n+\cdots=|PP_1|=1. + \end{eqnarray*} +Sedaj iz relacije \ref{eqnPitagoraCofman3} dobimo: +\begin{eqnarray*} + \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{n\cdot (n+1)}+\cdots=1, + \end{eqnarray*} + oz. +\begin{eqnarray*} + \sum_{n=1}^{\infty}\frac{1}{n (n+1)}=1. + \end{eqnarray*} + + + + \bizrek \label{Jung} + Let $\mathcal{P}=\{A_1,A_2,\ldots A_n\}$ be a finite set of + points in the plane and $d=\max \{|A_iA_j|;\hspace*{1mm}1\leq i,j\leq n\}$. + Prove that there exists a circle with radius $\frac{d}{\sqrt{3}}$, + which contains all the points of this set (\index{izrek!Jungov}Jung's theorem\footnote{Nemški matematik + \index{Jung, H. W. E.}\textit{H. W. E. Jung} (1876--1953) + je leta 1901 dokazal + splošno trditev za $n$-razsežni primer.}). + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.1a.pic} +\caption{} \label{sl.pod.7.6.1a.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.6.1a.pic}) + +Po trditvi \ref{lemaJung} je dovolj dokazati, da za vsake tri +točke od teh $n$ točk obstaja ustrezen krog. Naj bodo to točke +$A$, $B$ in $C$. Po predpostavki nobena od stranic trikotnika $ABC$ +ne presega $d$. Brez škode za splošnost naj bo $BC$ največja +stranica tega trikotnika. Potem je $|AC|, |AB| \leq |BC| \leq d$. +Obravnavali bomo dva primera: + +\textit{1)} V primeru, ko je $ABC$ pravokotni ali topokotni trikotnik, +krog $\mathcal{K}'(S, \frac{|BC|}{2})$ (kjer je $S$ središče stranice $BC$) +vsebuje vse tri točke $A$, $B$ in $C$. Ker je $\frac{|BC|}{2}\leq +\frac{d}{2}<\frac{d}{\sqrt{3}}$, točke $A$, $B$ in $C$ ležijo tudi v +krogu $\mathcal{K}(S, \frac{d}{\sqrt{3}})$. + +\textit{2)} Če je trikotnik $ABC$ ostrokotni, mora biti $\angle BAC +\geq 60^0$, ker je $BC$ največja stranica tega trikotnika. Naj bo +$BA'C$ pravilen trikotnik, tako da sta točki $A$ in $A'$ na istem +bregu premice $BC$, in $\mathcal{K}'(S,R)$ krog, ki je določen z očrtano +krožnico tega trikotnika. Zaradi pogoja $\angle BAC \geq 60^0$ je +točka $A$ bodisi na robu kroga $\mathcal{K}'$ bodisi njegova notranja točka. +Polmer tega kroga je po izreku \ref{PitagorovEnakostr} enak +$R=\frac{|BC|\sqrt{3}}{3}=\frac{|BC|}{\sqrt{3}}$. Ker je $|BC|\leq d$, točki $A$, $B$ ležita tudi v krogu $\mathcal{K}(S, \frac{d}{\sqrt{3}})$. +\kdokaz + + + + \bizrek + Let $M$ and $N$ be common points of two congruent circles + $k$ and $l$ with a radius $r$. Points $P$ and $Q$ are the intersections of these two circles with + a line defined by their centres such that $P$ and $Q$ are on the same side of the line $MN$. Prove that + $$|MN|^2 + |PQ|^2 = 4r^2.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.1.pic} +\caption{} \label{sl.pod.7.6.1.pic} +\end{figure} + + +\textbf{\textit{Proof.}} +Naj bosta $O$ in $S$ središči teh dveh krožnic in $\overrightarrow{v} += \overrightarrow{OS}$ (Figure \ref{sl.pod.7.6.1.pic}). Translacija +$\mathcal{T}_{\overrightarrow{v}}$ +preslika krožnico $k$ v krožnico $l$ ter točko $M$ v neko točko $M'$. +Ker točka $M$ leži na krožnici +$k$, njena slika $M'$ leži na krožnici $l$. Iz istih razlogov je tudi + $\mathcal{T}_{\overrightarrow{v}}(P )=Q$. Zato je +$\overrightarrow{MM'} = \overrightarrow{v} = \overrightarrow{OS} = \overrightarrow{PQ}$, kar pomeni, + da sta premici $MM'$ in $OS$ vzporedni in $MM'\cong OS\cong PQ$. Ker je premica +$MN$ pravokotna na premici $OS$ ($OS$ je simetrala daljice $MN$), +je premica $MN$ pravokotna tudi na vzporednici $MM'$ premice $OS$. +Torej je $\angle NMM'$ pravi kot, zato je $NM'$ premer krožnice $l$. + Če uporabimo Pitagorov izrek \ref{PitagorovIzrek}, dobimo: +$$|MN|^2 + |PQ|^2 = |MN|^2 + |MM'|^2 = |NM'|^2 = 4r^2,$$ kar je bilo treba dokazati. \kdokaz + + + + \bzgled + Lines $b$ and $c$ and a point $A$ in the same plane are given. Construct + a square $ABCD$ such that the vertices $B$ and $C$ lie on the lines $b$ + and $c$, respectively. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.2.pic} +\caption{} \label{sl.pod.7.6.2.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.6.2.pic}) + +Uporabimo rotacijski razteg $\rho_{A,\sqrt{2},45^0}$. Ker je $\rho_{A,\sqrt{2},45^0}(B)=C$, lahko točko $C$ načrtamo iz pogoja +$C\in c\cap\rho_{A,\sqrt{2},45^0}(b)$. +\kdokaz + + + + \bnaloga\footnote{1. IMO, Romania - 1959, Problem 4.} + Construct a right triangle with given hypotenuse $c$ such that the median + drawn to the hypotenuse is the geometric mean of the two legs of the triangle. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.IMO1.pic} +\caption{} \label{sl.pod.7.6.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} + Predpostavimo, da je $ABC$ pravokotni trikotnik s hipotenuzo + $AB\cong c$ in je njegova težiščnica na hipotenuzo $CM=t_c$ + geometrijska sredina katet $AC=b$ in $BC=a$ oz. + $t_c=\sqrt{a b}$ (Figure \ref{sl.pod.7.6.IMO1.pic}). + + Iz izreka \ref{TalesovIzrKroz} sledi $t_c=\frac{1}{2}\cdot c$. Če + uporabimo še Pitagorov izrek (\ref{PitagorovIzrek}), dobimo: + $$(a+b)^2=a^2+b^2+2ab=c^2+2t^2_c=c^2+2\left( \frac{1}{2}\cdot c\right)^2= + \frac{6}{4}\cdot c^2$$ + oz. + $$a+b=\frac{\sqrt{6}}{2}\cdot c.$$ + Slednja relacija omogoča konstrukcijo. + + Opišimo najprej pomožno konstrukcijo daljice, ki ima dolžino + $\frac{\sqrt{6}}{2}\cdot c$. Načrtajmo najprej daljico $UV=c$ + in njeno središče $W$, nato pa pravokotni trikotnik $TUW$ + ($TU=c$ in $\angle TUW=90^0$) ter pravokotni trikotnik + $TWZ$ ($WZ=\frac{1}{2}\cdot c$ + in $\angle TWZ=90^0$). + + Sedaj bomo konstruirali trikotnik $ABC$. Načrtajmo daljico $DB\cong + TZ$ in $\angle BDX=45^0$. Točka $A$ naj bo eno od + presečišč krožnice $k(B,c)$ in kraka $DX$. Na koncu načrtajmo točko + $C$ kot presečišče daljice $BD$ s simetralo + $s_{DA}$ daljice $DA$. + + Dokažimo, da trikotnik $ABC$ izpolnjuje vse pogoje iz naloge. + Najprej iz konstrukcije sledi $A\in k(B,c)$, kar pomeni, da velja + $AB=c$. Točka $C$ po konstrukciji leži na simetrali daljice + $DA$, zato je $DC\cong AC$ oz. (izrek \ref{enakokraki}) $\angle + DAC\cong \angle ADC =45^0$. Iz trikotnika $ACD$ je po izreku \ref{VsotKotTrik} + $\angle ACD=90^0$. Ker je po konstrukciji $\mathcal{B}(D,C,B)$, je + tudi $\angle ACB=90^0$ in je $ABC$ pravokotni trikotnik s + hipotenuzo $AB=c$. + + Dokažimo še, da je njegova + težiščnica na hipotenuzo $CM=t_c$ + geometrijska sredina njegovih katet $AC=b$ in $BC=a$ oz. $t_c=\sqrt{a b}$. + Po konstrukciji sta trikotnika $TUW$ in $TWZ$ pravokotna in je + $UW=WZ=\frac{1}{2}\cdot c$ in $TW=c$, + zato iz Pitagorovega izreka sledi: + \begin{eqnarray*} + |TZ|^2&=&|TW|^2+|WZ|^2=\\ + &=& |TU|^2+|UW|^2+|WZ|^2=\\ + &=& + c^2+\left(\frac{c}{2}\right)^2+\left(\frac{c}{2}\right)^2=\\ + &=& \frac{6}{4}\cdot c^2 + \end{eqnarray*} + Po konstrukciji je $BD\cong TZ$, zato je tudi $|BD|^2=\frac{6}{4}\cdot + c^2$. Torej: + \begin{eqnarray*} + \frac{6}{4}\cdot c^2&=&|BD|^2=\left(|BC|+|CD|\right)^2=\\ + &=&\left(|BC|+|CA|\right)^2=(a+b)^2=\\ + &=&a^2+b^2+2ab=\\ + &=&c^2+2ab\\ + \end{eqnarray*} + Iz tega sledi $\left(\frac{c}{2} \right)^2=ab$. Ker je po izreku + \ref{TalesovIzrKroz} $t_c=\frac{c}{2}$, na koncu dobimo + $t_c^2=ab$. + + Število rešitev naloge je + odvisno od števila presečišč krožnice $k(B,c)$ in poltraka + $DX$. + \kdokaz + + + + \bnaloga\footnote{23. IMO, Hungary - 1982, Problem 5.} + The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are + divided by the inner points $M$ and $N$, respectively, so that + $$\frac{AM}{AC}=\frac{CN}{CE}=r.$$ + Determine $r$ if $B$, $M$ and $N$ are collinear. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.skk.4.2.IMO2.pic} +\caption{} \label{sl.skk.4.2.IMO2.pic} +\end{figure} + +\textbf{\textit{Solution.}} Označimo z $O$ središče pravilnega +šestkotnika $ABCDEF$ (Figure \ref{sl.skk.4.2.IMO2.pic}) in z $a$ +dolžino njegove stranice. + Iz predpostavke $\frac{AM}{AC}=\frac{CN}{CE}=r$ in iz dejstva $AC\cong CE$ + sledi $CM\cong EN$ in $AM\cong CN$. +Trikotnika $ACB$ in $CED$ sta skladna (izrek \textit{SSS} +\ref{SSS}), zato je $\angle ACB\cong\angle CED$. To pomeni, da sta +skladna tudi trikotnika $BCM$ in $DEN$ (izrek \textit{ASA} +\ref{KSK} oz. $\angle BMC\cong \angle DNE$. + Trikotnik $ACE$ je pravilen, zato je $\angle ACE=60^0$. Torej: + \begin{eqnarray*} + \angle BND&=&\angle BNC+\angle CND\\ + &=&\angle BMC- \angle ECA+180^0-\angle DNE=120^0 + \end{eqnarray*} + Ker je tudi $\angle BOD=120^0$ in $CO\cong CD\cong CB$, + točka $N$ leži na krožnici $k(C,CO)$. Zato je $AM\cong CN\cong + CB=a$. Če z $v$ označimo dolžino višine pravilnega trikotnika + $OCD$ in uporabimo relacijo iz zgleda \ref{PitagorovEnakostr}: + $$r=\frac{CN}{CE}=\frac{a}{2v}=\frac{\sqrt{3}}{3},$$ kar je bilo treba izračunati. \kdokaz + +Let $ABC$ be an acute-angled triangle with circumcentre $O$. Let $P$ on $BC$ be the foot of the altitude from $A$. +Suppose that $\angle BCA>\angle ABC+30^0$. +Prove that $\angle CAB+\angle COP<90^0$. + + \bnaloga\footnote{42. IMO, USA - 2001, Problem 1.} + Let $ABC$ be an acute-angled triangle with circumcentre $O$. Let $P$ on $BC$ be the foot of the altitude from $A$. + Suppose that $\angle BCA>\angle ABC+30^0$. + Prove that $\angle CAB+\angle COP<90^0$. + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.6.IMO3.pic} +\caption{} \label{sl.pod.7.6.IMO3.pic} +\end{figure} + +\textbf{\textit{Solution.}} Označimo z $A_1$ središče stranice +$BC$, s $P'$ in $A'$ drugi presečišči premic $AP$ in $AO$ z +očrtano krožnico $k(O,R)$ trikotnika $ABC$, ter s $Q$ pravokotno +projekcijo točke $A'$ na premici $BC$ (Figure +\ref{sl.pod.7.6.IMO3.pic}). Označimo še z $\alpha$, $\beta$ in +$\gamma$ notranje kote trikotnika $ABC$ ob ogliščih $A$, $B$ in +$C$. + +Ker je $AA'$ premer krožnice $k$, je $\angle AP'A'=90^0$ (izrek +\ref{TalesovIzrKroz}). To pomeni, da je $PP'A'Q$ pravokotnik in +velja: + \begin{eqnarray} +PQ\cong A'P'. \label{pod.7.6.IMO31} + \end{eqnarray} + Točka $O$ je središče premera $AA'$, $A_1$ pa je po Talesovem izreku +\ref{TalesovIzrek} središče dalice $PQ$. Točka $A_1$ je +torej skupno središče daljic $BC$ in $PQ$. + +Po izreku \ref{TockaNbetagama} je še $\angle A'AP'=\angle +OAP=\gamma-\beta$. Iz danega pogoja $\angle BCA>\angle ABC+30^0$ +oz. $\gamma-\beta>30^0$ torej sledi: + \begin{eqnarray} +\angle A'AP>30^0. \label{pod.7.6.IMO32} +\end{eqnarray} +Iz izreka \ref{SredObodKot} je $\angle BOC=2\alpha$. Ker je $BOC$ +enakokraki trikotnik z osnovnico $BC$, je $\angle +OCB=\frac{1}{2}\left( 180^0-2\alpha\right)=90^0-\alpha$ (izreka +\ref{enakokraki} in \ref{VsotKotTrik}). Pogoj $\angle CAB+\angle +COP<90^0$ oz. $\angle COP<90^0-\alpha$, ki ga želimo dokazati, je +zato ekvivalenten s pogojem $\angle COP<\angle OCB=\angle OCP$ +(velja $\angle OCB=\angle OCP$, ker je $ABC$ ostrokotni trikotnik +in je $\mathcal{B}(B,P,C)$). Slednji pa je po izreku +\ref{vecstrveckot} ekvivalenten s pogojem $CP|BC|^2. \label{pod.7.6.IMO3b} + \end{eqnarray} + +Toda iz neenakosti \ref{pod.7.6.IMO32} in iz izreka \ref{SredObodKot} +sledi $\angle A'OP'>60^0$, zato je v enakokrakem trikotniku $OA'P'$ +$\angle A'OP'$ največji kot (izreka \ref{enakokraki} in +\ref{VsotKotTrik}). To pomeni, da velja $A'P'>OA'=R$. Iz +\ref{pod.7.6.IMO31} je potem $PQ\cong A'P'>R$. Na koncu je: +\begin{eqnarray*} +\hspace*{-1.5mm} 4|OA_1|^2+2|BC|\cdot|PQ|>2|BC|\cdot|PQ|>2|BC|\cdot R=|BC|\cdot +2R>|BC|^2, + \end{eqnarray*} +kar je bilo še potrebno dokazati. + \kdokaz + + + +%_______________________________________________________________________________ + \poglavje{Napoleon Triangles} \label{odd7Napoleon} + +Definirajmo najprej dva trikotnika, ki ju priredimo poljubnemu trikotniku $ABC$. +Nad njegovimi stranicami z zunanje strani skonstruirajmo enakostranične trikotnike. Središča teh trikotnikov so oglišča t. i. \index{trikotnik!Napoleonov zunanji}\pojem{zunanjega Napoleonovega\footnote{Ta trditev je pripisana \textit{Napoleónu Bonapartu} (1769--1821), francoskemu cesaarju in vojskovodji, ki se je zanimal +za elementarno geometrijo, čeprav je vprašanje, ali je imel dovolj znanja, da bi to trditev tudi dokazal.} trikotnika} (Figure \ref{sl.pod.7.7n.1.pic}). Če enakostranične trikotnike konstuiramo z notranje strani, na enak način dobimo t. i. \index{trikotnik!Napoleonov notranji}\pojem{notranji Napoleonov trikotnik}. + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7n.1.pic} +\caption{} \label{sl.pod.7.7n.1.pic} +\end{figure} + + + + \bizrek + The outer and inner Napoleon triangles are equilateral. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7n.2.pic} +\caption{} \label{sl.pod.7.7n.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Naj bodo $P$,$ Q$ in $R$ središča pravilnih trikotnikov, ki so konstruirani z zunanje strani nad stranicami poljubnega trikotnika $ABC$, torej je $PQR$ zunanji Napoleonov trikotnik trikotnika $ABC$ (Figure \ref{sl.pod.7.7n.2.pic}). Dokažimo, da je $PQR$ enakostranični trikotnik. + +Naj bo: +$$\mathcal{I}=\mathcal{R}_{R,120^0}\circ +\mathcal{R}_{P,120^0}\circ\mathcal{R}_{Q,120^0}.$$ +Ker je $120^0+120^0+120^0=360^0$, je kompozitum $\mathcal{I}$ bodisi identiteta bodisi translacija (izrek \ref{rotacKomp2rotac}). Vendar druga možnost odpade, saj $\mathcal{I}(A)=A$. Torej $\mathcal{I}=\mathcal{E}$ oz.: + $$\mathcal{R}_{R,120^0}\circ +\mathcal{R}_{P,120^0}\circ\mathcal{R}_{Q,120^0}=\mathcal{E}.$$ +Iz tega sledi: + $$\mathcal{R}_{R,120^0}\circ\mathcal{R}_{P,120^0}= + \mathcal{R}^{-1}_{Q,120^0}=\mathcal{R}_{Q,-120^0}=\mathcal{R}_{Q,240^0}.$$ + Naj bo $Q'$ tretje oglišče enakostraničnega trikotnika $PQ'R$. Po izreku \ref{rotacKomp2rotac} je: + $$\mathcal{R}_{R,120^0}\circ\mathcal{R}_{P,120^0}=\mathcal{R}_{Q',240^0}.$$ + Torej velja $\mathcal{R}_{Q,240^0}=\mathcal{R}_{Q',240^0}$ oz. $Q=Q'$, kar pomeni, da je $PQR$ enakostranični trikotnik. + Na podoben način dokažemo (če uporabimo rotacije za kot $-120^0$), da je tudi notranji Napoleonov trikotnik enakostraničen. + \kdokaz + + +Dokažimo še dodatno lastnost dveh Napoleonovih trikotnikov. + + + + \bizrek + The outer and inner Napoleon triangles of an arbitrary + triangle $ABC$ have the same centre, which is at the same time the centroid of the triangle $ABC$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7n.3.pic} +\caption{} \label{sl.pod.7.7n.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Naj bosta $PQR$ in $P'Q'R'$ ustrezni zunanji in notranji Napoleonov trikotnik trikotnika $ABC$ (Figure \ref{sl.pod.7.7n.3.pic}). Označimo z $Y$ in $A_1$ središči daljic $PR$ in $BC$. Dokažimo najprej: + +\begin{enumerate} + \item Trikotnika $QP'C$ in $RBP'$ sta skladna in sta podobna trikotniku $ABC$. + \item Štirikotnika $ARP'Q$ in $CP'RQ'$ sta paralelograma. + \item $\overrightarrow{YA_1}=\frac{1}{2}\overrightarrow{AQ}$. +\end{enumerate} + +\textit{1}) Najprej iz $\angle BCP'\cong \angle ACQ=30^0$ sledi +$\angle P'CQ\cong\angle BCA$ (z rotacijo $\mathcal{R}_{C.30^0}$ se kot $\angle P'CQ$ preslika v kot $\angle BCA$). Velja tudi $\frac{CQ}{CA}=\frac{CP'}{CB}=\frac{\sqrt{3}}{3}$ (po izreku \ref{PitagorovEnakostr}, ker sta $CQ$ in $CP'$ polmera očrtanih krožnic dveh enakostraničnih trikotnikov s stranicama $CA$ oz. $CB$), zato sta si trikotnika $ABC$ in $QP'C$ podobna (izrek \ref{PodTrikSKS}) s +koeficientom podobnosti $k=\frac{\sqrt{3}}{3}$. + Analogno sta si tudi trikotnika $ABC$ in$ RBP'$ +podobna z enakim koeficientom podobnosti, kar pomeni, da sta trikotnika $QP'C$ in $RBP'$ skladna. + +\textit{2}) Iz skladnosti trikotnikov $QP'C$ in $RBP'$ in iz dejstva, da sta trikotnika $RAB$ in $QAC$ enakokraka, sledi: +$QP'\cong RB\cong RA$ in $P'R\cong CQ\cong QA$, zato je štirikotnik $ARP'Q$ paralelogram. Analogno dokažemo, da +je tudi štirikotnik $CP'RQ'$ paralelogram. + + +\textit{3}) Daljica $YA_1$ je srednjica trikotnika $RPP'$ z osnovnico $RP'$, zato je po izrekih \ref{srednjicaTrikVekt} in \ref{vektParalelogram}: + +$$\overrightarrow{YA_1}=\frac{1}{2}\overrightarrow{RP'}=\frac{1}{2}\overrightarrow{AQ}$$ + +Dokazati moramo le še začetno trditev. Ker je $\overrightarrow{YA_1}=\frac{1}{2}\overrightarrow{AQ}$, + se težiščnica $AA_1$ trikotnika $ABC$ in +težiščnica $QY$ trikotnika $PQR$ sekata v točki, ki ju deli v razmerju $2:1$. Torej gre za skupno težišče trikotnikov $ABC$ in $PQR$ (izrek \ref{tezisce}). Analogno se dokaže, da imata tudi trikotnika +$ABC$ in $P'Q'R'$ skupno težišče. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Ptolemy's Theorem} \label{odd7Ptolomej} + +Dokažimo zelo pomemben izrek, ki se nanaša na tetivne štirikotnike. + + + + \bizrek \label{izrekPtolomej} + \index{izrek!Ptolomejev}(Ptolemy’s\footnote{\index{Ptolomej Aleksandrijski}\textit{Ptolomej Aleksandrijski} (2. st.) je dokazal ta izrek v svojemu delu ‘‘Veliki zbornik’’.} theorem) + If $ABCD$ is a cyclic quadrilateral then + the product of the lengths of its diagonals + is equal to the sum of the products of the lengths of the pairs of opposite sides, + i.e. + $$|AC|\cdot |BD|=|AB|\cdot |CD|+|BC|\cdot |AD|.$$ + + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.1.pic} +\caption{} \label{sl.pod.7.7.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Brez škode za splošnost predpostavimo, da je $\angle CBD\leq\angle ABD$. +Naj bo $L$ takšna točka diagonale $AC$, da velja +$\angle ABL\cong \angle CBD$ (Figure \ref{sl.pod.7.7.2.pic}). Ker je še $\angle BAL=\angle BAC\cong\angle BDC$ (izrek \ref{ObodObodKot}), sta trikotnika +$ABL$ in $DBC$ podobna (izrek \ref{PodTrikKKK}), zato je $AB:DB=AL:DC$ oziroma: +\begin{eqnarray} \label{eqnPtolomej1} +|AB|\cdot |CD|=|AL|\cdot |BD|. +\end{eqnarray} +Ker je $\angle BCL=\angle BCA\cong\angle BDA$ (izrek \ref{ObodObodKot}) in $\angle LBC\cong\angle ABD$ (iz predpostavke $\angle CBD\cong\angle ABL$), velja +tudi $\triangle BCL\sim \triangle BDA$ (izrek \ref{PodTrikKKK}), zato je $BC:BD=CL:DA$ oziroma: +\begin{eqnarray} \label{eqnPtolomej2} +|BC|\cdot |AD|=|CL|\cdot |BD|. +\end{eqnarray} + Po seštevanju relacij \ref{eqnPtolomej1} in \ref{eqnPtolomej2} dobimo iskano enakost. +\kdokaz + + +Trditev iz zgleda \ref{zgledTrikABCocrkrozP} bomo na tem mestu dokazali na bolj enostaven način - z uporabo Ptolomejevega izreka. + + + + + \bzgled \label{zgledTrikABCocrkrozPPtol} + Let $k$ be the circumcircle of a regular triangle $ABC$. + If $P$ is an arbitrary point lying + on the shorter arc $BC$ of the circle $k$, then + $$|PA|=|PB|+|PC|.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.2a.pic} +\caption{} \label{sl.pod.7.7.2a.pic} +\end{figure} + +\textbf{\textit{Solution.}} +Štirikotnik +$ABPC$ je tetiven (Figure \ref{sl.pod.7.7.2a.pic}), zato iz Ptolomejevega izreka \ref{izrekPtolomej} sledi +$|PA|\cdot |BC|=|PB|\cdot |AC|+|PC|\cdot |AB|$. Ker je $ABC$ enakostranični trikotnik, oz. $|BC|=|AC|= |AB|$, iz prejšnje relacije sledi $|PA|=|PB|+|PC|$. + \kdokaz + + Še ena uporaba Ptolomejevega izreka se nanaša na enakokraki trapez. + + + + + \bzgled + Let $a$ and $c$ be the bases, $b$ the leg and $d$ the diagonal + isosceles trapezium. Prove that + $$ac=d^2-b^2.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.4.pic} +\caption{} \label{sl.pod.7.7.4.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.7.4.pic}) + +Po izreku \ref{trapezTetivEnakokr} je enakokraki trapez tetiven, zato lahko uporabimo Ptolomejev izrek \ref{izrekPtolomej} in dobimo $d^2=ac+b^2$ oz. $ac=d^2-b^2$. +\kdokaz + +Zanimivo je, da če v prejšnjem izreku vzamemo poseben primer $a=c$ (Figure \ref{sl.pod.7.7.4.pic}), ko je enakokraki trapez pravokotnik, dobimo formulo za diagonalo pravokotnika $d^2=a^2+b^2$, kar je sicer direktna posledica Pitagorovega izreka \ref{PitagorovIzrek}. V tem smislu lahko tudi rečemo, da je Pitagorov izrek posledica Ptolomejevega izreka, če pravokotni trikotnik dopolnimo do pravokotnika. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.5.pic} +\caption{} \label{sl.pod.7.7.5.pic} +\end{figure} + +Seveda nas uporaba Ptolomejevega izreka v kvadratu vodi v že znano zvezo (izrek \ref{PitagorovKvadrat}) med diagonalo $d$ in stranico $a$ kvadrata: $d^2=2a^2$ oz. $d=a\sqrt{2}$ (Figure \ref{sl.pod.7.7.5.pic}). + + Vemo, da za vsak pravilen $n$-kotnik obstaja očrtana +krožnica \ref{sredOcrtaneKrozVeck}. To nam omogoča uporabo Ptolomejevega izreka, če v pravilnem $n$-kotniku (v primeru $n>4$) izberemo ustrezna štiri oglišča. +Sedaj bomo to idejo uporabili v primerih $n=5$ in $n=7$. + + + + \bzgled \label{PtolomejPetkotnik} + Let $a$ be the side and $d$ the diagonal of a regular pentagon ($5$-gon). + Prove that + $$d=\frac{1+\sqrt{5}}{2}a.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.3.pic} +\caption{} \label{sl.pod.7.7.3.pic} +\end{figure} + +\textbf{\textit{Solution.}} + Naj bo $A_1A_2A_3A_4A_5$ pravilni petkotnik s stranico $a$ in diagonalo $d$ + (Figure \ref{sl.pod.7.7.3.pic}). Če uporabimo Ptolomejev izrek \ref{izrekPtolomej} za štirikotnik $A_1A_2A_3A_5$, dobimo: $d^2=ad+a^2$ oz. $d^2-ad-a^2=0$. Pozitivna rešitev te kvadratne enačbe po $d$ je ravno $d=\frac{1+\sqrt{5}}{2}a$. + \kdokaz + +Prejšnji primer nam omogoča konstrukcijo pravilnega petkotnika s stranico, ki je skladna daljici $a$. +Res, če je $ABCDE$ iskani pravilni petkotnik, lahko najprej +načrtamo trikotnik $ABC$, kjer je $AB\cong BC\cong a$ in $AC =\frac{1+\sqrt{5}}{2}a$ Za konstrukcijo $\frac{1+\sqrt{5}}{2}a$ uporabimo Pitagorov izrek (\ref{PitagorovIzrek}) za pravokotni trikotnik s katetama $a$ in $2a$ - njegova hipotenuza meri $a\sqrt{5}$. Tej hipotenuzi dodamo še daljico $a$ in dobljeno daljico s simetralo razdelimo na dve daljici. Središče očrtane krožnice pravilnega petkotnika $ABCDE$ je hkrati središče očrtane krožnice trikotnika $ABC$ (Figure \ref{sl.pod.7.7.3a.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.3a.pic} +\caption{} \label{sl.pod.7.7.3a.pic} +\end{figure} + + +Na podoben način lahko konstruiramo pravilni petkotnik, ki je včrtan dani krožnici. To naredimo tako, +da najprej konstruiramo pravilni petkotnik s poljubno stranico, nato pa uporabimo središčni razteg (Figure \ref{sl.pod.7.7.3a.pic}). + +Več o konstrukcijah pravilnih $n$-kotnikov bomo povedali v razdelku \ref{odd9LeSestilo}. + + + + \bzgled + Let $a$ be the side, $d$ the shorter and $D$ the longer diagonal + of a regular heptagon ($7$-gon). Prove that + $$\frac{1}{a}=\frac{1}{D}+\frac{1}{d}$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.2.pic} +\caption{} \label{sl.pod.7.7.2.pic} +\end{figure} + +\textbf{\textit{Solution.}} + +Naj bo $A_1A_2A_3A_4A_5A_6A_7$ pravilni sedemkotnik in $k$ očrtana +krožnica (Figure \ref{sl.pod.7.7.2.pic}). Štirikotnik +$A_1A_2A_3A_5$ je tetiven, zato iz Ptolomejevega izreka \ref{izrekPtolomej} sledi +$ad+aD=dD$ oz. $\frac{1}{a}=\frac{1}{D}+\frac{1}{d}$. + \kdokaz + +Če večkrat uporabimo Ptolomejev izrek, dobimo eno posplošitev naloge \ref{zgledTrikABCocrkrozPPtol}: + + + + \bzgled + Let $P$ be an arbitrary point of the shorter arc $A_1A_{2n+1}$ + of the circumcircle of a regular polygon $A_1A_2\cdots A_{2n+1}$. + If we denote $d_i=PA_i$ ($i\in \{1,2,\cdots , 2n+1 \}$), + prove that (Figure \ref{sl.pod.7.7.6.pic}) + $$d_1+d_3+\cdots +d_{2n+1}=d_2+d_4+\cdots +d_{2n}.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.7.6.pic} +\caption{} \label{sl.pod.7.7.6.pic} +\end{figure} + +%________________________________________________________________________________ + \poglavje{Stewart's Theorem} \label{odd7Stewart} + +Naslednji izrek se nanaša na zelo zanimivo in pomembno metrično lastnost trikotnika. + + + + \bizrek \label{StewartIzrek} + \index{izrek!Stewartov} (Stewart's\footnote{\index{Stewart, M.}\textit{M. Stewart} (1717--1785), + angleški matematik, ki je leta 1746 dokazal in objavil to trditev. S to trditvijo ga je seznanil + njegov učitelj - angleški matematik \index{Simson, R.}\textit{R. Simson} (1687--1768) - ki jo je objavil šele leta 1749. Predpostavlja se, + da je trditev morda bila znana že \index{Arhimed} \textit{Arhimedu iz Sirakuze} (3. st. pr. n. š.).} theorem) + If $X$ is an arbitrary point of the side + $BC$ of a triangle $ABC$, then + $$|AX|^2=\frac{|BX|}{|BC|}|AC|^2+\frac{|CX|}{|BC|}|AB|^2-|BX|\cdot |CX|.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.1.pic} +\caption{} \label{sl.pod.7.8.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Z $a$, $b$ in $c$ označimo dolžine ustreznih stranic $BC$, $AC$ in $AB$, s $p$, $q$ in $x$ pa po vrsti dolžine daljic $BX$, $CX$ in $AX$. Z $A'$ označimo še +nožišče višine $v_a$ iz oglišča $A$ trikotnika $ABC$ (Figure \ref{sl.pod.7.8.1.pic}). Predpostavimo, da je $\mathcal{B}(B,A',C)$. + +Dokažimo najprej trditev za primer, ko je $X=A'$. Iz Pitagorovega izreka \ref{PitagorovIzrek} sledi $v_a^2 =b^2- q^2$ +in $v_a^2 =c^2- p^2$. Z množenjem prve enakosti s $p$ in druge s $q$ ter po seštevanju dobljenih relacij in upoštevanju $p+q=a$ dobimo $av_a^2=pb^2+qc^2-apq$ oz.: +$$v_a^2=\frac{p}{a}b^2+\frac{q}{a}c^2-pq,$$ +kar pomeni, da v primeru višin trditev velja. + +Naj bo sedaj $X\neq A'$. Daljica $v_a$ je višina trikotnikov $ABX$ in $ABC$ iz oglišča $A$. Brez škode za splošnost predpostavimo, da je $\mathcal{B}(B,A',X)$. Če z $y$ označimo dolžino daljice $BA'$ in uporabimo dokazani del trditve za višine, +dobimo: +\begin{eqnarray*} +v_a^2&=&\frac{y}{p}x^2+\frac{p-y}{p}c^2-y(p-y),\\ +v_a^2&=&\frac{y}{a}b^2+\frac{a-y}{a}c^2-y(a-y). +\end{eqnarray*} +Če izenačimo desni strani teh dveh enakosti in poenostavimo, dobimo: +$$x^2=\frac{p}{a}b^2+\frac{q}{a}c^2-pq,$$ +kar pomeni, da trditev velja tudi v primeru, ko je $X\neq A'$. + +Dokaz za višino $v_a$ je podoben tudi v primeru, ko ni $\mathcal{B}(B,A',C)$, le da takrat dobimo $v_a^2=\frac{p}{a}b^2+\frac{q}{a}c^2+pq$. +\kdokaz + + +Stewartov izrek \ref{StewartIzrek} lahko zapišemo tudi v drugačni obliki: + + + \bizrek \label{StewartIzrek2} + Let $a=|BC|$, $b=|AC|$ and $c=|AB|$ be the length of + the sides of a triangle $ABC$. If $X$ is the point that divides the side $BC$ of this triangle + in the ratio $n:m$ and $x=|AX|$ then + $$x^2=\frac{n}{n+m}b^2+\frac{m}{n+m}c^2-\frac{mn}{(m+n)^2}bc.$$ + \eizrek + +Najbolj znana je uporaba Stewartovega izreka za težiščnice trikotnika. + + + + \bizrek \label{StwartTezisc} + If $a$, $b$ and $c$ are the sides and $t_a$ the triangle median + on the side $a$, then + $$t_a^2=\frac{1}{2}b^2+\frac{1}{2}c^2-\frac{1}{4}a^2.$$ + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.2.pic} +\caption{} \label{sl.pod.7.8.2.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Trditev je direktna posledica Stewartovega izreka \ref{StewartIzrek2}, saj je v primeru težiščnice $n=m=1$ (Figure \ref{sl.pod.7.8.2.pic}). +\kdokaz + + +Direktna posledica je naslednja trditev. + + + + \bzgled \label{StwartTezisc2} + If $a$, $b$ and $c$ are the sides and $t_a$, $t_b$ and $t_c$ the triangle medians + on those sides, respectively, then + $$t_a^2+t_b^2+t_c^2=\frac{3}{4}\left(a^2+ b^2+c^2\right).$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.3.pic} +\caption{} \label{sl.pod.7.8.3.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.8.3.pic}) + + Če trikrat uporabimo dokazano relacijo iz zgleda \ref{StwartTezisc}, dobimo: + \begin{eqnarray*} + t_a^2&=&\frac{1}{2}b^2+\frac{1}{2}c^2-\frac{1}{4}a^2\\ + t_b^2&=&\frac{1}{2}a^2+\frac{1}{2}c^2-\frac{1}{4}b^2\\ + t_c^2&=&\frac{1}{2}a^2+\frac{1}{2}b^2-\frac{1}{4}c^2 + \end{eqnarray*} + Po seštevanju vseh treh enakosti dobimo iskano relacijo. +\kdokaz + +V primeru enakokrakega trikotnika se enakost na desni strani Stewartovega izreka poenostavi. + + + + \bzgled \label{StewartEnakokraki} + If $ABC$ is an isosceles triangle with the base $BC$ and + $X$ an arbitrary point of this base, then + $$|AX|^2=|AB|^2-|BX|\cdot |CX|.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.8.pic} +\caption{} \label{sl.pod.7.8.8.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.8.8.pic}) + + Če je torej $AB\cong AC$, direktno iz Stewartovega izreka (\ref{StewartIzrek}) sledi: +\begin{eqnarray*} +|AX|^2 &=& \frac{|BX|}{|BC|}|AC|^2+\frac{|CX|}{|BC|}|AB|^2-|BX|\cdot |CX|=\\ +&=&\left(\frac{|BX|}{|BC|}+\frac{|CX|}{|BC|}\right)|AB|^2-|BX|\cdot |CX|=\\ +&=&\frac{|BX|+|CX|}{|BC|}|AB|^2-|BX|\cdot |CX|=\\ +&=&|AB|^2-|BX|\cdot |CX|, +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + +Z uporabo Stewartovega izreka bomo še malo nadaljevali. + + + + \bzgled + Let $E$ be the intersection of the side $BC$ with the bisector of the interior angle + $BAC$ of a triangle $ABC$. If we denote + $a=|BC|$, $b=|AC|$, $c=|AB|$, $l_a=|AE|$ and $s=\frac{a+b+c}{2}$, then + $$l_a=\frac{2\sqrt{bc}}{b+c}\sqrt{s(s-a)}.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.4.pic} +\caption{} \label{sl.pod.7.8.4.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.8.4.pic}) + + Po izreku \ref{HarmCetSimKota} je: $BE:CE=c:b$. Če uporabimo Stewartov izrek \ref{StewartIzrek2} za trikotnik $ABC$ in daljico $AE$, dobimo: +$$l_a^2=\frac{c}{b+c}b^2+\frac{b}{b+c}c^2-\frac{bc}{(b+c)^2}a^2.$$ +Po enostavnem preoblikovanju in poenostavljanju izraza na desni strani enakosti, dobimo iskano relacijo. +\kdokaz + + + \bizrek \label{izrekEulerStirik}\index{izrek!Eulerjev za štirikotnike} + (Euler's\footnote{Švicarski matematik \index{Euler, L.}\textit{L. Euler} (1707--1783).} theorem for quadrilaterals) + If $P$ and $Q$ are the midpoints of + the diagonals $e$ and $f$ of an arbitrary quadrilateral with sides $a$, $b$, $c$ and $d$, then + $$|PQ|^2=\frac{1}{4}\left(a^2+b^2+c^2+d^2-e^2-f^2 \right).$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.5.pic} +\caption{} \label{sl.pod.7.8.5.pic} +\end{figure} + + +\textbf{\textit{Proof.}} +Naj bosta $P$ in $Q$ središči diagonal $AC$ in $BD$ poljubnega štirikotnika $ABCD$ (Figure \ref{sl.pod.7.8.5.pic}). +Označimo še $a=|AB|$, $b=|BC|$, $c=|CD|$, $d=|DA|$, $e=|AC|$ in $f=|BD|$. Daljica $PQ$ je težiščnica trikotnika $AQC$, zato po izreku \ref{StwartTezisc} velja: + \begin{eqnarray} \label{eqnEulStirik} + |PQ|^2=\frac{1}{2}|AQ|^2+\frac{1}{2}|CQ|^2-\frac{1}{4}e^2. + \end{eqnarray} + Prav tako sta daljici $QA$ in $QC$ težiščnici trikotnikov $ABD$ in +$CBD$, zato je (izrek \ref{StwartTezisc}): + \begin{eqnarray*} + |AQ|^2&=&\frac{1}{2}d^2+\frac{1}{2}a^2-\frac{1}{4}f^2,\\ + |CQ|^2&=&\frac{1}{2}c^2+\frac{1}{2}b^2-\frac{1}{4}f^2. + \end{eqnarray*} + Če uvrstimo zadnji dve enakosti v \ref{eqnEulStirik}, dobimo iskano relacijo. + \kdokaz + + Direktno posledico prejšnjega izreka dobimo, če za štirikotnik izberemo paralelogram. + + \bizrek + A quadrilateral is a parallelogram if and only if the sum of + the squares of all of its sides is equal to the sum of the squares of its diagonals. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.6.pic} +\caption{} \label{sl.pod.7.8.6.pic} +\end{figure} + + +\textbf{\textit{Proof.}} +Naj bosta $P$ in $Q$ središči diagonal $AC$ in $BD$ poljubnega štirikotnika $ABCD$. Označimo še $a=|AB|$, $b=|BC|$, $c=|CD|$, $d=|DA|$, $e=|AC|$ in $f=|BD|$. Po izreku \ref{paralelogram} je $ABCD$ paralelogram natanko tedaj, ko je $P=Q$ (Figure \ref{sl.pod.7.8.6.pic}) oz.$|PQ|=0$. Slednje pa po prejšnjem izreku \ref{izrekEulerStirik} velja natanko tedaj, ko je +$\frac{1}{4}\left(a^2+b^2+c^2+d^2-e^2-f^2 \right)=0$ oz. $a^2+b^2+c^2+d^2=e^2+f^2$. +\kdokaz + + + + \bizrek \label{GMTmnl} + Let $A$ and $B$ be given points in the plane and + $m, n, l\in R^+\setminus \{0\}$. + Determine a set of all points of this plane such that + $$m|AX|^2 + n|BX|^2 = l^2.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.7.pic} +\caption{} \label{sl.pod.7.8.7.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.8.7.pic}) + + Po trditvi iz zgleda \ref{izrekEnaDelitevDaljice} obstaja ena sama takšna točka $S$ na daljici $AB$, da velja +$\overrightarrow{AS}:\overrightarrow{SB}=n:m$. Naj bo $X$ poljubna točka. Če uporabimo +Stewartov izrek \ref{StewartIzrek2} za trikotnik $AXB$ in daljico $XS$, dobimo: +\begin{eqnarray*} +|XS|^2=\frac{m}{n+m}|AX|^2+\frac{n}{n+m}|BX|^2-\frac{nm}{(n+m)^2}|AB|^2, \textrm{ oz.} +\end{eqnarray*} + \begin{eqnarray} \label{eqnStewMnozTock} +|XS|^2=\frac{1}{n+m}\left(m|AX|^2+n|BX|^2\right)-\frac{nm}{(n+m)^2}|AB|^2. +\end{eqnarray} +Točka $X$ leži na iskani množici točk natanko tedaj, ko je $m|AX|^2 + n|BX|^2 = l^2$. Zaradi \ref{eqnStewMnozTock} to velja natanko tedaj, ko je: + \begin{eqnarray*} +|XS|^2=\frac{1}{n+m}l^2-\frac{nm}{(n+m)^2}|AB|^2=c, +\end{eqnarray*} +kjer je $c$ konstanta, ki ni odvisna od točke $X$. Če je torej $c>0$, je iskana +množica točk krožnica $k(S, c)$. Če je $c=0$, je množica točk le $\{S\}$, če pa je $c<0$, je množica +točk prazna množica. +\kdokaz + + + + \bnaloga\footnote{50. IMO, Germany - 2009, Problem 2.} + Let $ABC$ be a triangle with circumcentre $O$. The points $P$ and $Q$ are interior points + of the sides $CA$ and $AB$, respectively. Let $K$, $L$ and $M$ be the midpoints of the segments $BP$, $CQ$ + and $PQ$, respectively, and let $l$ be the circle passing through $K$, $L$ and $M$. Suppose that the line + $PQ$ is tangent to the circle $l$. Prove that $|OP|=|OQ|$. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.8.IMO2.pic} +\caption{} \label{sl.pod.7.8.IMO2.pic} +\end{figure} + +\textbf{\textit{Solution.}} Označimo s $k(O,R)$ očrtano +krožnico trikotnika $ABC$ (Figure \ref{sl.pod.7.8.IMO2.pic}). + Daljici $MK$ in $ML$ sta srednjici trikotnikov $QPB$ in $PQC$, + zato je (izrek \ref{srednjicaTrikVekt}): + \begin{eqnarray} \label{eqn72} + \overrightarrow{MK}=\frac{1}{2}\overrightarrow{QB} \hspace*{3mm} + \textrm{in} \hspace*{3mm} \overrightarrow{ML}=\frac{1}{2}\overrightarrow{PC} + \end{eqnarray} + Iz izrekov \ref{KotiTransverzala} in \ref{KotaVzporKraki} potem sledi: + \begin{eqnarray} \label{eqn73} + \angle AQP \cong \angle QMK,\hspace*{3mm} + \angle APQ \cong \angle PML\hspace*{3mm} \textrm{in} \hspace*{3mm} + \angle BAC \cong \angle KML + \end{eqnarray} + Ker je po predpostavki premica $PQ$ tangenta krožnice $l$, je po + izreku \ref{ObodKotTang}: + $$\angle MLK\cong\angle QMK \hspace*{3mm} \textrm{in} \hspace*{3mm} + \angle MKL\cong\angle PML.$$ + Iz tega in \ref{eqn73} sledi: + \begin{eqnarray*} + \angle AQP \cong \angle MLK,\hspace*{3mm} + \angle APQ \cong \angle MKL\hspace*{3mm} \textrm{in} \hspace*{3mm} + \angle BAC \cong\angle KML, + \end{eqnarray*} + kar pomeni, da sta si trikotnika $AQP$ in $MLK$ podobna (po izreku + \ref{PodTrikKKK} je dovolj dokazati + skladnost dveh parov pripadajočih kotov). Zato iz definicije podobnosti + likov sledi: + $\frac{AQ}{ML}=\frac{AP}{MK}$. Če slednjo relacijo kombiniramo z relacijo + \ref{eqn72}, dobimo $\frac{AQ}{AP}=\frac{ML}{MK}= + \frac{\frac{1}{2}\cdot CP}{\frac{1}{2}\cdot + BQ}=\frac{CP}{BQ}$. Torej velja: + \begin{eqnarray} \label{eqn74} + |AQ|\cdot |BQ| = |AP|\cdot |CP| + \end{eqnarray} + Iz Stewartovega izreka \ref{StewartIzrek} za enakokraki trikotnik $AOB$ + ($|OA|=|OB|=R$) sledi: + $$|OQ|^2=|OA|^2\cdot \frac{QB}{AB}+|OB|^2\cdot \frac{QA}{AB} + -|AQ|\cdot |BQ|=R^2-|AQ|\cdot |BQ|.$$ + Analogno iz trikotnika $AOC$ po istem izreku dobimo: + $$|OP|^2=R^2-|AP|\cdot |CP|.$$ +Iz dokazane relacije \ref{eqn74} na koncu sledi $|OQ|^2=|OP|^2$ +oz. $|OQ|=|OP|$. + \kdokaz + + + +%________________________________________________________________________________ +\poglavje{Desargues' Theorem} \label{odd7Desargues} + +Tudi naslednji izrek je zgodovinsko povezan z razvojem projektivne geometrije. + + + + \bizrek \label{izrekDesarguesEvkl} \index{izrek!Desarguesov} + (Desargues’\footnote{ + \index{Desargues, G.} \textit{G. Desargues} (1591--1661), francoski arhitekt, ki je bil eden od + utemeljiteljev projektivne geometrije.} theorem) + Let $ABC$ and $A'B'C'$ be two triangles in the plane such that the lines $AA'$, $BB'$ and $CC'$ intersect + at a point $S$ + (i.e the triangles are \index{perspective triangles}\pojem{perspective with respect to the centre} + \color{blue} $S$). + If $P=BC\cap B'C'$, $Q=AC\cap A'C'$ and $R=AB\cap A'B'$, + then the points $P$, $Q$ and + $R$ are collinear + (i.e. triangles are \pojem{perspective with respect to the axis} \color{blue} $PQ$). + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10D.1.pic} +\caption{} \label{sl.pod.7.10D.1.pic} +\end{figure} + + +\textit{\textbf{Proof.}} (Figure +\ref{sl.pod.7.10D.1.pic}) + + Če uporabimo Menelajev izrek \ref{izrekMenelaj} za +trikotnike $SA'B'$, $SA'C'$ in $SB'C'$, dobimo: + \begin{eqnarray*} +\hspace*{-2mm} \frac{\overrightarrow{SA}}{\overrightarrow{AA'}}\cdot + \frac{\overrightarrow{A'R}}{\overrightarrow{RB'}}\cdot + \frac{\overrightarrow{B'B}}{\overrightarrow{BS}}=-1, \hspace*{1mm} +\frac{\overrightarrow{SA}}{\overrightarrow{AA'}}\cdot + \frac{\overrightarrow{A'Q}}{\overrightarrow{QC'}}\cdot + \frac{\overrightarrow{C'C}}{\overrightarrow{CS}}=-1, \hspace*{1mm} + \frac{\overrightarrow{SC}}{\overrightarrow{CC'}}\cdot + \frac{\overrightarrow{C'P}}{\overrightarrow{PB'}}\cdot + \frac{\overrightarrow{B'B}}{\overrightarrow{BS}}=-1. + \end{eqnarray*} + Iz teh treh relacij sledi: + \begin{eqnarray*} + \frac{\overrightarrow{A'Q}}{\overrightarrow{QC'}}\cdot + \frac{\overrightarrow{C'P}}{\overrightarrow{PB'}}\cdot + \frac{\overrightarrow{B'R}}{\overrightarrow{RA'}}=-1. + \end{eqnarray*} +Zato so po Menelajevem izreku \ref{izrekMenelaj} (obratna smer) za trikotnik $A'B'C'$ +točke $P$, $Q$ in $R$ kolinearne. + \kdokaz + +Na podoben način dokažemo tudi obratno trditev. + + + + \bizrek \label{izrekDesarguesObr} \index{izrek!Desarguesov obratni} + Let $ABC$ and $A'B'C'$ be two triangles in the + plane such that the lines $AA'$ and $BB'$ intersect at the point $S$. + If $P=BC\cap B'C'$, + $Q=AC\cap A'C'$ and $R=AB\cap A'B'$ are collinear points, then also + $S\in CC'$.\\ + (Converse of Desargues’ theorem) + \eizrek + + +Naslednji trditvi sta na določen način podobni Desarguesovem izreku \ref{izrekDesarguesEvkl}. + + + + \bizrek \label{izrekDesarguesZarkVzp} + Let $ABC$ and $A'B'C'$ be two triangles in the plane such that the lines $AA'$, $BB'$ and $CC'$ + are parallel to each other. + If $P=BC\cap B'C'$, $Q=AC\cap A'C'$ and $R=AB\cap A'B'$, + then the points $P$, $Q$ and + $R$ are collinear. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10D.2.pic} +\caption{} \label{sl.pod.7.10D.2.pic} +\end{figure} + +\textbf{\textit{Proof.}} + (Figure \ref{sl.pod.7.10D.2.pic}) + + Po trikratni uporabi Talesovega izreka \ref{TalesovIzrek} dobimo: + \begin{eqnarray*} + \frac{\overrightarrow{BP}}{\overrightarrow{PC}}= + \frac{\overrightarrow{BB'}}{\overrightarrow{C'C}},\hspace*{4mm} + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}= + \frac{\overrightarrow{CC'}}{\overrightarrow{A'A}},\hspace*{4mm} + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= + \frac{\overrightarrow{AA'}}{\overrightarrow{B'B}}. + \end{eqnarray*} + Po množenju teh treh relacij je naprej: + \begin{eqnarray*} + \frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot + \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot + \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=-1, + \end{eqnarray*} +in zato so po Menelajevem izreku \ref{izrekMenelaj} (obratna smer) za trikotnik $ABC$ +točke $P$, $Q$ in $R$ kolinearne. + \kdokaz + + + If $P=BC\cap B'C'$, $Q=AC\cap A'C'$ and $R=AB\cap A'B'$, + then the points $P$, $Q$ and + $R$ are collinear. + + \bizrek \label{izrekDesarguesOsNesk} + Let $ABC$ and $A'B'C'$ be two triangles in the plane such that + the lines $AA'$, $BB'$ and $CC'$ intersect + at a point $S$. + If + $BC\parallel B'C'$ and + $AC\parallel A'C'$, then also $AB\parallel A'B'$. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10D.3.pic} +\caption{} \label{sl.pod.7.10D.3.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.10D.3.pic}) + + Ker je $BC\parallel B'C'$ in + $AC\parallel A'C'$, iz Talesovega izreka \ref{TalesovIzrek} sledi + $\frac{\overrightarrow{SB}}{\overrightarrow{SB'}}= + \frac{\overrightarrow{SC}}{\overrightarrow{SC'}}$ in + $\frac{\overrightarrow{SA}}{\overrightarrow{SA'}}= + \frac{\overrightarrow{SC}}{\overrightarrow{SC'}}$. + Iz prejšnjih dveh relacij sledi najprej + $\frac{\overrightarrow{SB}}{\overrightarrow{SB'}}= + \frac{\overrightarrow{SA}}{\overrightarrow{SA'}}$, zato po + Talesovem ireku (obratna smer) \ref{TalesovIzrekObr} velja tudi $AB\parallel A'B'$. + \kdokaz + + Formulacije predhodnih treh izrekov so zelo podobne, čeprav so dokazi teh + izrekov bistveno različni. Formulacije se celo ne bi razlikovale, če bi + privzeli, da se vse vzporednice (v eni smeri) neke ravnine sekajo v + isti točki v neskončnosti ter da vse točke v neskončnosti neke ravnine + določajo natanko eno premico v neskončnosti. To je pravzaprav bila glavna motivacija za razvoj projektivne geometrije, + v kateri se vsaki dve premici v ravnini sekata (glej \cite{Mitrovic}). + + V nadaljevanju bomo videli nekaj posledic Desarguesovega izreka. + + + + \bzgled \label{zgled 3.2} + Let $p$, $q$ and $r$ be three lines in the plane which + intersect at the same point, and points $A$, $B$ and $C$ of this plane which do not belong + to these lines. Construct a triangle whose vertices belong to the given + lines, and the sides contain the given points. + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10D.4.pic} +\caption{} \label{sl.pod.7.10D.4.pic} +\end{figure} + + + \textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.10D.4.pic}) + + Naj bo $PQR$ tak trikotnik, da + njegova oglišča $P$, $Q$ in $R$ pripadajo premicam $p$, $q$ in + $r$, stranice $QR$, $PR$ in $PQ$ pa vsebujejo točke $A$, $B$ in + $C$. Naj bo $S$ skupna točka premic $p$, $q$ in $r$. + + Če je $P'Q'R'$ poljuben trikotnik, ki je perspektiven trikotniku + $PQR$ glede na središče $S$, katerega stranici $R'Q'$ in $R'P'$ + vsebujeta točki $A$ in $B$ (izpuščen je torej pogoj glede na točko + $C$), sta po Desarguesovem \ref{izrekDesarguesEvkl} izreku trikotnika $PQR$ in $P'Q'R'$ + perspektivna glede na neko os $s$. Torej os $s$, vsebuje točke + $A$, $B$ in $Z=PQ\cap P'Q'$. + + Trikotnik $PQR$ konstruiramo tako, da najprej + konstruiramo pomožni trikotnik $P'Q'R'$, pri katerem je točka + $R'\in r$ poljubna. Potem konstruiramo točko $Z$ kot + presečišče premic $AB$ in $P'Q'$. S točkama $Z$ in $C$ je + določena stranica $PQ$. + + V dokazu, da je $PQR$ iskani trikotnik, uporabimo obratni + Desargeusov izrek \ref{izrekDesarguesObr}. + \kdokaz + + + +%________________________________________________________________________________ +\poglavje{Power of a Point} \label{odd7Potenca} + +Med najbolj zanimive značilnosti krožnice, ki izpostavlja nekatere njene +metrične lastnosti, je potenca tačke\footnote{Besedo potenca je v tem pomenu prvi uporabil švicarski matematik \index{Steiner, J.}\textit{J. Steiner} (1769--1863).}. Pred prehodom na definicijo dokažimo naslednji izrek. + + + + \bizrek \label{izrekPotenca} + Suppose that $P$ is an arbitrary point in the plane of a circle $k(S,r)$. + For any line of this plane containing the point $P$ and intersecting the circle $k$ at points + $A$ and $B$, the expression $\overrightarrow{PA}\cdot \overrightarrow{PB}$ + (Figure \ref{sl.pod.7.12.1b.pic}) + is constant, furthermore: + $$\overrightarrow{PA}\cdot \overrightarrow{PB} = |PS|^2 - r^2.$$ + If $P$ is an exterior point of the circle $k$ and $PT$ its tangent at a point $T$, then: + $$\overrightarrow{PA}\cdot \overrightarrow{PB} = |PT|^2.$$ + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.1b.pic} +\caption{} \label{sl.pod.7.12.1b.pic} +\end{figure} + + +\textbf{\textit{Proof.}} Obravnavali bomo tri možne primere. + + +\textit{1)} (Figure \ref{sl.pod.7.12.1.pic}) + +Naj bo $P$ zunanja točka krožnice $k$. V tem primeru ni $\mathcal{B}(A,P,B)$, zato je (ekvivalenca \ref{eqnMnozVektRelacijaB} iz razdelka \ref{odd5DolzVekt}): +\begin{eqnarray} \label{eqnPotenIzr1} +\overrightarrow{PA}\cdot \overrightarrow{PB}>0. +\end{eqnarray} + Brez škode za splošnost predpostavimo, da velja $\mathcal{B}(P,A,B)$. Ker je $\angle PTA\cong\angle TBA=\angle TBP$ (izrek \ref{ObodKotTang}) in $\angle TPA=\angle BPT$, +sta si trikotnika $PAT$ in $PTB$ podobna (izrek \ref{PodTrikKKK}), zato je +$PA:PT=PT:PB$. Če uporabimo še Pitagorov izrek, dobimo: + $$|PA|\cdot |PB| = |PT|^2 = |PS|^2 - r^2.$$ + Iz tega zaradi relacije \ref{eqnPotenIzr1} sledi: +$$\overrightarrow{PA}\cdot \overrightarrow{PB} = |PS|^2 - r^2.$$ + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.1.pic} +\caption{} \label{sl.pod.7.12.1.pic} +\end{figure} + +\textit{2)} (Figure \ref{sl.pod.7.12.1a.pic}) + +Če točka $P$ leži na krožnici $k$, je $P=A$ ali $P=B$, torej: +$$\overrightarrow{PA}\cdot \overrightarrow{PB} = 0 = |PS|^2 - r^2.$$ + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.1a.pic} +\caption{} \label{sl.pod.7.12.1a.pic} +\end{figure} + + +\textit{3)} (Figure \ref{sl.pod.7.12.1a.pic}) + +Naj bo $P$ notranja točka krožnice $k$. V tem primeru je $\mathcal{B}(A,P,B)$, zato je (ekvivalenca \ref{eqnMnozVektRelacijaB} iz razdelka \ref{odd5DolzVekt}): +\begin{eqnarray} \label{eqnPotenIzr2} +\overrightarrow{PA}\cdot \overrightarrow{PB}<0. +\end{eqnarray} +Naj bosta +$A_1$ in $B_1$ presecišči premice $SP$ s krožnico $k$ (brez škode za splošnost naj bo $\mathcal{B}(A_1,S,P)$). Zaradi skladnosti ustreznih +obodnih kotov (izrek \ref{ObodObodKot}) je $\triangle APA_1\sim \triangle B_1PB$ (izrek \ref{PodTrikKKK}), zato je $AP:B_1P = PA_1:PB$, torej +$$|PA|\cdot |PB|=|PA_1|\cdot |PB_1|= +\left(r+|PS|\right)\cdot\left(r-|PS|\right)=r^2-|PS|^2.$$ + Iz tega zaradi relacije \ref{eqnPotenIzr1} sledi +$$\overrightarrow{PA}\cdot \overrightarrow{PB} = |PS|^2 - r^2,$$ kar je bilo treba dokazati. \kdokaz + +Konstanten produkt $\overrightarrow{PA}\cdot \overrightarrow{PB}$ iz prejšnjega izreka \ref{izrekPotenca} imenujemo +\index{potenca točke}\pojem{potenca točke} $P$ glede na krožnico $k$ in jo označimo $p(P,k)$. + +Po prejšnjem izreku (\ref{izrekPotenca}) je torej potenca točke $P$ glede na krožnico $k(S,r)$ število $|PS|^2 - r^2$. +To število je pozitivno, negativno ali nič, odvisno od tega, ali je $P$ zunanja ali notranja ali točka na kružnici $k$. Torej: + \begin{eqnarray*} + p(P,k)\hspace*{1mm}\left\{ + \begin{array}{ll} + >0, & \textrm{če je } OP>r; \\ + =0, & \textrm{če je } OP=r; \\ + <0, & \textrm{če je } OP0$, po prejšnjem izreku (\ref{PotencOsLema}) iskana množica +predstavlja pravokotnico premice $S_1S_2$. V primeru $r_1=r_2$ je jasno $|S_1P|=|S_2P|$, torej je iskana množica simetrala daljice $S_1S_2$. +Označimo to premico s $p$. + +V posebnem primeru, ko se krožnici sekata v točkah $A$ in $B$, je $p(A,k_1)=p(A,k_2)=0$ in $p(B,k_1)=p(B,k_2)=0$. To pomeni, da točki $A$ in $B$ ležita na iskani premici $p$, torej je to ravno premica $AB$. + +Če se krožnici dotikata v točki $T$, je $p(T,k_1)=p(T,k_2)=0$. To pomeni, da točka $T$ leži na premici $p$, ki je pravokotnica premice $S_1S_2$. Torej je premica $p$ skupna tangenta dveh danih krožnic skozi njuno dotikališče $T$. +\kdokaz + + Premica $p$ iz prejšnjega izreka se imenuje \index{potenčna!premica} + \pojem{potenčna premica} dveh + krožnic. Potenčno premico krožnic $k_1$ in $k_2$ bomo označili z $p(k_1,k_2)$. + +Zanimivo je ugotoviti kako efektivno konstruiramo potenčno premico dveh danih krožnic $k_1$ in $k_2$. + V primerih, ko se krožnici sekata ali dotikata, smo že dali odgovor v izreku \ref{PotencnaOs} (Figure \ref{sl.pod.7.12.4.pic}). + Ostane nam še konstrukcija potenčne premice v primeru, ko krožnici nimata skupnih točk. Ena možnost je, da direktno uporabimo izrek \ref{PotencOsLema}. Nekoliko hitrejši postopek je vezan za konstrukcijo +pomožne krožnice $l$, ki seka dani krožnici v točkah $A$ in $B$ oziroma $C$ in $D$. Potem presečišče +premic $AB$ in $CD$ - točka $X$ - leži na iskani potenčni premici $p(k_1,k_2)$. Res, iz $p(k_1,l)=AB$ in $p(k_2,l)=CD$ sledi $X\in p(k_1,l)$ in $X\in p(k_1,l)$ oz. $p(X,p_1)=p(X,l)=p(X,p_2)$. + +V izreku \ref{PotencnaOs} ni obravnavan en primer - ko + sta $k_1(S,r_1)$ in $k_2(s,r_2)$ koncentrični krožnici. V tem primeru je omenjena množica točk prazna množica. To dobimo iz pogoja +$|S_1P|^2-|S_2P|^2=r_1^2-r_2^2$. Če je namreč $S_1=S_2$ in $r_1\neq r_2$, dobimo pogoj $0=r_1^2-r_2^2\neq 0$, ki ni izpolnjen za nobeno točko $P$. + + +Definirajmo še nekatere pojme, ki so v zvezi s potenčno premico dveh krožnic. + +Množico vseh takšnih krožnic neke ravnine, da imata vsaki dve potenčno premico $p$, ki je enaka za vsak par krožnic te množice, imenujmo \index{šop krožnic}\pojem{šop krožnic}. Premica $p$ je \index{potenčna!os}\pojem{potenčna os} tega šopa (Figure \ref{sl.pod.7.12.5.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.5.pic} +\caption{} \label{sl.pod.7.12.5.pic} +\end{figure} + +Naj bo $p$ potenčna os nekega šopa krožnic. Iz izreka \ref{PotencnaOs} je jasno, da vsa središča krožnic tega šopa ležijo na isti premici $s$, ki je pravokotna na premico $p$. +Obravnavali bomo tri primere. + +\textit{1)} Če se vsaj dve krožnici šopa sekata v točkah $A$ in $B$, je $p=AB$, zato vse krožnice tega šopa potekajo skozi točki $A$ in $B$. V tem primeru pravimo, da gre za \index{šop krožnic!eliptični}\pojem{eliptični šop} krožnic. + +\textit{2)} Če se vsaj dve krožnici šopa dotikata v točki $T$, je $p$ pravokotnica premice $s$ v točki $T$, zato imajo vse krožnice skupno tangento $p$ v točki $T$. V tem primeru pravimo, da gre za \index{šop krožnic!parabolični}\pojem{parabolični šop} krožnic. + +\textit{3)} Če nobeni dve krožnici šopa nimata skupnih točk, pravimo, da gre za \index{šop krožnic!hiperbolični}\pojem{hiperbolični šop} krožnic. V tem primeru velja naslednja lastnost: če poljubna krožnica seka krožnice tega šopa (ni nujno, da seka vse) v točkah $A_i$ in $B_i$, potem vse premice $A_iB_i$ potekajo skozi eno točko, ki leži na premici $p$. + + + + + \bizrek \label{PotencnoSr} + Let $k$, $l$ and $j$ be three non-concentric circles with non-collinear centres. + Then there is exactly one point that has the same power with respect to all three circles. + This point is the intersection of their three radical axes $p(k,l)$, $p(l,j)$ and $p(k,j)$. + \eizrek + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.6.pic}) Ker so središča krožnic $k$, $l$ in $j$ tri nekolinearne točke, nobeni dve od premic $p(k,l)$, $p(l,j)$ in $p(k,j)$ nista vzporedni. Naj bo $P=p(k,l)\cap p(l,j)$. Potem je +$$p(P,k)=p(P,l)=p(P,j),$$ + oz. $P\in p(k,j)$, kar pomeni, da se potenčne premice $p(k,l)$, $p(l,j)$ in $p(k,j)$ sekajo v točki $P$. + +Če za neko drugo točko $\widehat{P}$ te ravnine velja $p(\widehat{P},k)=p(\widehat{P},l)=p(\widehat{P},j)$, je $\widehat{P}\in p(k,l),\hspace*{1mm}p(l,j),\hspace*{1mm}p(k,j)$, zato je $\widehat{P}=P$. +\kdokaz + + Točka iz prejšnjega izreka (\ref{PotencnoSr}) se imenuje \index{središče!potenčno} + \pojem{potenčno središče} treh krožnic. Potenčno središče krožnic $k$, $l$ in $j$ bomo označili s $p(k,l,j)$. + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.6.pic} +\caption{} \label{sl.pod.7.12.6.pic} +\end{figure} + + V primeru, ko so središča treh krožnic tri kolinearne točke, krožnice pa niso iz istega šopa, so vse tri potenčne premice vzporedne, ker so vse pravokotne na skupno centralo teh krožnic (izrek \ref{PotencnaOs}). + Iz tega in izreka \ref{PotencnoSr} direktno sledi naslednji izrek (Figure \ref{sl.pod.7.12.6.pic}). + + +Let $k$, $l$ and $j$ be three non-concentric circles with non-collinear centres. + Then there is exactly one point that has the same power with respect to all three circles. + This point is the intersection of their three radical axes $p(k,l)$, $p(l,j)$ and $p(k,j)$. + + Radical axes of three circles in the plane that are not + from the same pencil and no two of them are concentric, + belong to the same family of lines. + + + \bizrek \label{PotencnoSrSop} + Radical axes of three circles in the plane that are not + from the same pencil and no two of them are concentric, + belong to the same family of lines. + \eizrek + +Poseben primer izreka \ref{PotencnoSr}, ko se vsaki dve premici sekata, je naslednja trditev. + + + \bzgled + Let $k$, $l$, and $j$ be three circles of some plane with nonlinear + centres and: + \begin{itemize} + \item $A$ and $B$ intersections of the circles $k$ and $l$, + \item $C$ and $D$ intersections of the circles $l$ and $j$, + \item $E$ and $F$ intersections of the circles $j$ and $k$. + \end{itemize} + Prove that the lines $AB$, $CD$ and $EF$ intersect at a single point. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.7.pic} +\caption{} \label{sl.pod.7.12.7.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.7.pic}) + +Po izreku \ref{PotencnaOs} so $p(k,l)=AB$, $p(l,j)=CD$ in $p(j,k)=EF$ ustrezne potenčne osi. Po izreku \ref{PotencnoSr} se le-te sekajo v eni točki - potenčnem središču $P=p(k,l,j)$ teh treh krožnic. +\kdokaz + +Zelo podobna je tudi naslednja trditev. + + + + + \bzgled + Let $k$, $l$, and $j$ be three circles of some plane with nonlinear + centres and: + \begin{itemize} + \item $t_1$ the common tangent of the circles $k$ in $l$, + \item $t_2$ the common tangent of the circles $l$ in $j$, + \item $t_3$ the common tangent of the circles $j$ in $k$. + \end{itemize} + Prove that the lines $t_1$, $t_2$ and $t_3$ intersect at a single point. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.8.pic} +\caption{} \label{sl.pod.7.12.8.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.8.pic}) + +Po izreku \ref{PotencnaOs} so $p(k,l)=t_1$, $p(l,j)=t_2$ in $p(j,k)=t_3$ ustrezne potenčne osi. Po izreku \ref{PotencnoSr} se le-te sekajo v eni točki - potenčnem središču $P=p(k,l,j)$ teh treh krožnic. +\kdokaz + + + + \bzgled + a) Suppose that circles $k$ and $l$ are touching each other externally, and a line $t$ is + is the common tangent of these circles at their common point. Let $AB$ be a second + common tangent of these circles at touching points $A$ and $B$. Prove that the midpoint + the line segment $AB$ lies on the tangent $t$.\\ + b) Suppose that circles $k$ and $l$ intersect at points $P$ and $Q$. Let $AB$ be a common tangent + of these circles at touching points $A$ and $B$. Prove that the midpoint of the line segment $AB$ lies on + line $PQ$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.9.pic} +\caption{} \label{sl.pod.7.12.9.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.9.pic}) + +\textit{a)} Naj bo $S$ središče daljice $AB$. Potem je: +$p(S,k)= |SA|^2 = |SB|^2 = p(S,l)$, +zato točka $S$ leži na potenčni premici krožnic $k$ in $l$ oz. premici $t$ (izrek \ref{PotencnaOs}). + +\textit{b)} Enako kot v prejšnjem primeru, le da je potenčna premica krožnic $k$ in $l$ v tem primeru premica $AB$. +\kdokaz + + + + \bizrek \label{EulerjevaFormula} + \index{formula!Eulerjeva} + (Euler's\footnote{\index{Euler, L.} + \textit{L. Euler} + (1707--1783), švicarski matematik.} formula) If $k(S,r)$ is the incircle and $l(O,R)$ the circumcircle +of an arbitrary triangle, then + $$|OS|^2=R^2- 2Rr.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.10.pic} +\caption{} \label{sl.pod.7.12.10.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.10.pic}) + +Označimo z $A$, $B$ in $C$ oglišča trikotnika, $NM$ premer očrtane krožnice $l$, ki je pravokoten na +stranico $BC$ (in še $A,N\div BC$). Po izreku \ref{TockaN} leži točka $N$ na bisekstrisi notranjega kota pri oglišču $A$ oz. na poltraku +$AS$ (izrek \ref{SredVcrtaneKrozn}). Po izreku \ref{TockaN.NBNC} je $NS\cong NC$. Če uporabimo še potenco točke $S$ glede na krožnico $l$ (izrek \ref{PotencnoSr}) in dejstvo $\mathcal{B}(A,S,N)$, dobimo $p(S,l)= |SO|^2 - R^2 = \overrightarrow{SA}\cdot \overrightarrow{SN} + = -|SA|\cdot |SN| += -|SA|\cdot |CN|$, torej: + \begin{eqnarray} \label{eqnEulFormOS} + |SO|^2 - R^2 = -|SA|\cdot |CN|. +\end{eqnarray} + Označimo s $Q$ dotikališče včrtane krožnice $k$ s stranico $AC$ trikotnika $ABC$. + Po izrekih \ref{TangPogoj} in \ref{TalesovIzrKroz2} je $\angle AQS\cong\angle MCN=90^0$, iz izreka \ref{ObodObodKot} sledi še $\angle SAQ=\angle NAC\cong\angle NMC$. +Torej sta si trikotnika $AQS$ in $MCN$ podobna (izrek \ref{PodTrikKKK}), zato je: +$$\frac{AS}{MN}=\frac{SQ}{NC}$$ oz. $|AS|\cdot |NC|=|MN|\cdot |SQ|=2Rr$. Če to vstavimo v relacijo \ref{eqnEulFormOS}, dobimo: +$$|SO|^2 = R^2-|SA|\cdot |CN|=R^2-2Rr,$$ +kar je bilo treba dokazati. \kdokaz + + +Naslednja naloga je poseben primer prejšnjega izreka in je torej +njegova posledica (Figure \ref{sl.pod.7.12.10a.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.10a.pic} +\caption{} \label{sl.pod.7.12.10a.pic} +\end{figure} + + + \bnaloga\footnote{4. IMO, Czechoslovakia - 1962, Problem 6.} + Consider an isosceles triangle. Let $r$ be the radius of its circumscribed circle + and $\rho$ the radius of its inscribed circle. Prove that the distance $d$ between + the centres of these two circles is $$d = \sqrt{r(r-2\rho)}.$$ + \enaloga + +Tudi naslednja trditev je direktna posledica izreka \ref{EulerjevaFormula}. + + + \bizrek + If $k(S,r)$ is the incircle and $l(O,R)$ the circumcircle + of an arbitrary triangle, then + $$R\geq 2r.$$ + Equality is achieved for an equilateral triangle. + \eizrek + + Naslednja načrtovalna naloga je eden od desetih Apolonijevih problemov o dotiku krožnic, ki jih bomo bolj podrobno raziskovali v razdelku \ref{odd9ApolDotik}. + + + + \bzgled + Construct a circle through two given points $A$ and $B$ and tangent to a given line $t$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.11.pic} +\caption{} \label{sl.pod.7.12.11.pic} +\end{figure} + + +\textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.12.11.pic}) + +Naj bo $k$ iskana krožnica, ki poteka skozi točki $A$ in $B$ in se dotika premice $t$ v točki $T$. + +Če sta premici $AB$ in $t$ vzporedni, +tretjo točko $T$ krožnice $k$ dobimo kot presečišče premice +$t$ s simetralo daljice $AB$. + +Naj bo $P$ presečišče premic $AB$ in $t$. + +Uporabili bomo potenco točke $P$ glede na + krožnico $k$. Z $l$ označimo poljubno +krožnico, ki poteka skozi točki $A$ in $B$. Po izreku \ref{PotencnaOs} je $AB$ +potenčna premica krožnic $k$ in $l$, zato je $p(P,k)=p(P,l)$. Označimo s $PT$ in +$PT_1$ tangenti krožnic $k$ in $l$ v njunih točkah $P$ in $P_1$. Potem velja (izrek \ref{izrekPotenca}): + $$|PT|^2=p(P,k)=p(P,l)=|PT_1|^2$$ +oz. $|PT|=|PT_1|$. + +Zadnja relacija nam omogoča konstrukcijo tretje točke $T$ krožnice $k$. + \kdokaz + + + + \bzgled + Let $E$ be the intersection of the bisector of the interior angle at the vertex + $A$ with the side $BC$ of a triangle $ABC$ and $A_1$ the midpoint of this side. + Let $P$ and $Q$ + be intersections of the circumcircle of the triangle $AEA_1$ with the sides $AB$ + and $AC$ of the triangle $ABC$. Prove that: + $$BP\cong CQ.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.12.pic} +\caption{} \label{sl.pod.7.12.12.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.12.pic}) + +Označimo s $k$ očrtano krožnico trikotnika $AEA_1$. Če uporabimo potenco točk $B$ in $C$ glede na krožnico $k$, relaciji $\mathcal{B}(B,P,A)$ in $\mathcal{B}(C,Q,A)$ (po predpostavki točki $P$ in $Q$ ležita na stranicah $AB$ oz. $AC$ trikotnika $ABC$) ter +ekvivalenco \ref{eqnMnozVektRelacijaB} iz razdelka \ref{odd5DolzVekt}, dobimo: + \begin{eqnarray*} +p(B,k)&=&|BP|\cdot |BA|=|BE|\cdot |BA_1|,\\ +p(C,k)&=&|CP|\cdot |CA|=|CE|\cdot |CA_1|. +\end{eqnarray*} + Iz tega in relacije $BA_1\cong CA_1$ ter izreka \ref{HarmCetSimKota} dobimo: +\begin{eqnarray*} +\frac{|BP|\cdot |BA|}{|CP|\cdot |CA|} +=\frac{|BE|\cdot |BA_1|}{|CE|\cdot |CA_1|} +=\frac{|BE|}{|CE|} +=\frac{|BA|}{|CA|}. +\end{eqnarray*} +Torej velja $\frac{|BP|\cdot |BA|}{|CP|\cdot |CA|}=\frac{|BA|}{|CA|}$ oz. $|BP|=|CP|$. + \kdokaz + + \bzgled + Construct a circle that is perpendicular to three given circles + $k$, $l$ in $j$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.13.pic} +\caption{} \label{sl.pod.7.12.13.pic} +\end{figure} + + +\textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.12.13.pic}) + +Predpostavimo najprej, da so središča teh krožnic $k$, $l$ in $j$ nekolinearne +točke. Naj bo $x$ iskana krožnica s središčem $P$, ki je pravokotna na krožnice $k$, $l$ in $j$ ter $A\in x\cap k$, $B\in x\cap l$ in $C\in x\cap j$. Ker je $x\perp k,j,l$, so $PA$, $PB$ in $PC$ tangente teh krožnic iz točke $P$ (izrek \ref{pravokotniKroznici}). Po izreku \ref{izrekPotenca} je $p(P,k)=|PA|^2$, $p(P,l)=|PB|^2$ in $p(P,j)=|PC|^2$. Ker točke $A$, $B$ in $C$ ležijo na krožnici $x$ s središčem $P$, je $|PA|^2=|PB|^2=|PC|^2$ oz. $p(P,k)=p(P,l)= p(P,j)$. To pomeni, da je $P=p(k,l,j)$ potenčno središče krožnic $k$, $l$ in $j$. + +Iskano krožnico $x$ lahko torej načrtamo tako, da najprej narišemo njeno središče $P=p(k,l,j)$, nato polmer $PA$, kjer je premica $PA$ tangenta krožnice $k$ v točki $A$. +Jasno je, da v primeru, ko je $P$ notranja točka neke od krožnic $k$, $l$, $j$, naloga nima rešitev. + +Tudi v primeru, ko so središča +krožnic kolinearne točke, iskana krožnica ne +obstaja (le-ta je ‘‘degenerirana +krožnica’’ oz. premica, ki predstavlja njihovo skupno centralo). + \kdokaz + + + + \bzgled + Circles $k(O,r)$ and $l(S,\rho)$ and a point $P$ in the same + plane are given. + Construct a line passing through the point $P$, which determine congruent chords + on the circles $k$ and $l$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.14.pic} +\caption{} \label{sl.pod.7.12.14.pic} +\end{figure} + + +\textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.12.14.pic}) + +Naj bo $p$ premica, ki poteka skozi točko $P$, +krožnici $k$ in $l$ pa seka v takšnih točkah $A$ in $B$ oz. $C$ in $D$, da je +$AB\cong CD$. Naj bo +$\overrightarrow{v}$ vektor, ki je določen s +središčema daljic $AB$ in $CD$. +Potem je +$\mathcal{T}_{\overrightarrow{v}}:\hspace*{1mm}A,B\mapsto C,D$, +krožnica $k$ se s to translacijo preslika v +krožnico $k'$, ki poteka skozi točki $C$ in $D$. Torej se +krožnici $k'$ in $l$ sekata v točkah $C$ in $D$. +Tako se problem konstrukcije premice $p$ prevede na problem + konstrukcije vektorja + $\overrightarrow{v}$ oz. točke +$O'= \mathcal{T}_{\overrightarrow{v}}(O)$, ki predstavlja središče krožnice $k'$. + +Premica, ki +je določena s središčema $S$ in $O'$ krožnic $l$ in $k'$, je pravokotna na njuni skupni tetivi $CD$. +Ker je $\overrightarrow{OO'}= \overrightarrow{v} \parallel p$, sledi + $$\angle OO'S=90^0.$$ +Po izreku \ref{TalesovIzrKroz2} točka $O'$ leži na krožnici nad premerom $OS$. + +Točka $P$ leži na potenčni premici $p$ krožnic $k'$ in $l$ (izrek \ref{PotencnaOs}). Iz tega sledi $PL\cong PK$, kjer sta $PL$ in +$PK$ tangenti krožnic $l$ in $k'$ v točkah $L$ in $K$. +To pomeni, da lahko konstruiramo trikotnik, ki je skladen s pravokotnim trikotnikom $PKO'$ ($PK\cong PL$, $\angle PO'K=90^0$ in $KO'\cong r$), s tem pa tudi + $d$, ki je skladna z daljico $PO'$. Torej točka $O'$ pripada +preseku krožnice s središčem $P$ in polmerom $d$ ter krožnice nad premerom $OS$. + \kdokaz + + + + \bnaloga\footnote{36. IMO Canada - 1995, Problem 1.} + Let $A$, $B$, $C$, $D$ be four distinct points on a line, in that order $\mathcal{B}(A,B,C,D)$. The + circles with diameters $AC$ and $BD$ intersect at $X$ and $Y$. The line $XY$ + meets $BC$ at $Z$. Let $P$ be a point on the line $XY$ other than $Z$. The + line $CP$ intersects the circle with diameter $AC$ at $C$ and $M$, and the + line $BP$ intersects the circle with diameter $BD$ at $B$ and $N$. Prove + that the lines $AM$, $DN$, $XY$ are concurrent. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.IMO1.pic} +\caption{} \label{sl.pod.7.12.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Označimo s $k_1$ in $k_2$ krožnici +nad polmeroma $AC$ in $BD$ (Figure \ref{sl.pod.7.12.IMO1.pic}). Po +izreku \ref{KroznPresABpravokOS} je premica $XY$ pravokotna na +centrali $BC$ teh dveh krožnic in je hkrati njuna potenčna os +(izrek \ref{PotencnaOs}). Naj bo $S_1=AM\cap XY$ in $S_2=DN\cap +XY$. Dovolj je dokazati $S_1=S_2$. + + Ker $M\in k_1$, je po izreku \ref{TalesovIzrek} $\angle + AMC=90^0$ oz. $AS_1\perp MC$. Zaradi tega je + $\angle MCA\cong\angle AS_1Z$ (kota s pravokotnimi kraki + - izrek \ref{KotaPravokKraki}). Iz slednje skladnosti kotov + sledi, da sta si pravokotna trikotnika $AZS_1$ in $PZC$ podobna. + Zato je $\frac{AZ}{PZ}=\frac{ZS_1}{ZC}$ oz. + $|ZS_1|=\frac{|ZC|\cdot |ZA|}{|PZ|}$. Toda $|ZC|\cdot + |ZA|=p(Z,k_1)=|ZX|\cdot + |ZY|=|ZX|^2$. Iz tega sledi: + $$|ZS_1|=\frac{|ZX|^2}{|PZ|}.$$ + Na enak način se dokaže, da velja tudi: + $|ZS_2|=\frac{|ZX|^2}{|PZ|}$. Ker sta $S_1$ in $S_2$ na + istem poltraku $ZP$, je $S_1=S_2$. + \kdokaz + + + + \bzgled + Let $P$ be an arbitrary point in the plane of a triangle $ABC$ which + does not lie on any of lines containing altitudes of this triangle. Suppose $A_1$ is a point, + in which a perpendicular line of the line $AP$ at the point $P$ intersects the line $BC$. Analogously + we can also define points $B_1$ and $C_1$. + Prove that $A_1$, $B_1$ and $C_1$ are three collinear points. + \ezgled + +\textbf{\textit{Solution.}} Označimo z $A_C$ in $B_C$ pravokotni +projekciji oglišč $A$ in $B$ na premici $CP$. Analogno sta $A_B$ +in $C_B$ pravokotni projekciji oglišč $A$ in $C$ na premici $BP$ +ter $B_A$ in $C_A$ pravokotni projekciji oglišč $B$ in $C$ na +premici $AP$ (Figure \ref{sl.pd.7.4.6.pic}). Po Talesovem izreku \ref{TalesovIzrek} +je: + \begin{eqnarray} \label{4.1}\frac{AC_1}{C_1B} + \cdot \frac{BA_1}{A_1C} + \cdot \frac{CB_1}{B_1A}= +\frac{A_CP}{PB_C} + \cdot +\frac{B_AP}{PC_A} + \cdot +\frac{C_BP}{PA_B} + \end{eqnarray} + +\begin{figure}[!htb] +\centering +\input{sl.pd.7.4.6.pic} +\caption{} \label{sl.pd.7.4.6.pic} +\end{figure} + + + Iz $\angle AA_CC\cong\angle AC_AC=90^0$ sledi, + da točki $A_C$ in $C_A$ ležita na krožnici s premerom $AC$. Zato + je potenca točke $P$ na to krožnico enaka (izrek \ref{izrekPotenca}) + $\overrightarrow{PC}\cdot \overrightarrow{PA_C}= + \overrightarrow{PA}\cdot \overrightarrow{PC_A}$. + Analogno je $\overrightarrow{PB}\cdot \overrightarrow{PC_B}= + \overrightarrow{PC}\cdot \overrightarrow{PB_C}$ in + $\overrightarrow{PB}\cdot \overrightarrow{PA_B}= + \overrightarrow{PA}\cdot \overrightarrow{PB_A}$. Iz teh relacij + dobimo: $\frac{PA_C}{PC_A}=\frac{PA}{PC}$, + $\frac{PC_B}{PB_C}=\frac{PC}{PB}$ in + $\frac{PB_A}{PA_B}=\frac{PB}{PA}$. Če te relacije uvrstimo v + \ref{4.1}, dobimo $\frac{AC_1}{C_1B} + \cdot \frac{BA_1}{A_1C} + \cdot \frac{CB_1}{B_1A}=1$. + Ker je $\frac{\overrightarrow{AC_1}}{\overrightarrow{C_1B}} + \cdot \frac{\overrightarrow{BA_1}}{\overrightarrow{A_1C}} + \cdot \frac{\overrightarrow{CB_1}}{\overrightarrow{B_1A}}<0$, + velja + $\frac{\overrightarrow{AC_1}}{\overrightarrow{C_1B}} + \cdot \frac{\overrightarrow{BA_1}}{\overrightarrow{A_1C}} + \cdot \frac{\overrightarrow{CB_1}}{\overrightarrow{B_1A}}=-1$. Po + Menelajevem izreku so točke $A_1$, $B_1$ in $C_1$ kolinearne. +\kdokaz + + + + \bnaloga\footnote{41. IMO, S. Korea - 2000, Problem 1.} + $AB$ is tangent to the circles $CAMN$ and $NMBD$. $M$ lies + between $C$ and $D$ on the line $CD$, and $CD$ is parallel to $AB$. The chords + $NA$ and $CM$ meet at $P$; the chords $NB$ and $MD$ meet at $Q$. The rays $CA$ + and $DB$ meet at $E$. Prove that $PE\cong QE$. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.IMO3.pic} +\caption{} \label{sl.pod.7.12.IMO3.pic} +\end{figure} + +\textbf{\textit{Solution.}} Označimo z $L$ presečišče premic $MN$ +in $AB$ ter s $k$ in $l$ očrtani krožnici štirikotnikov + $CAMN$ in $NMBD$ (Figure \ref{sl.pod.7.12.IMO3.pic}). + +Po izreku \ref{PotencnaOs} je premica $MN$ potenčna os krožnic $k$ +in $l$, zato za njeno točko $L\in MN$ velja +$|LA|^2=p(L,k)=p(L,l)=|LB|^2$, kar pomeni, da je $L$ središče +daljice $AB$. Ker je po predpostavki $AB\parallel CD$, oz. +$AB\parallel PQ$, je po Talesovem izreku $MP:MQ=LA:LB=1$. Torej +je točka $M$ središče daljice $PQ$ oz. $MP\cong MQ$. + +Če uporabimo izreka \ref{ObodKotTang} in \ref{KotiTransverzala}, +dobimo: + \begin{eqnarray*} + \angle EAB &\cong& \angle AMC \cong\angle MAB\\ + \angle EBA &\cong& \angle BMQ \cong\angle MBA + \end{eqnarray*} +To pomeni, da sta trikotnika $AEB$ in $AMB$ skladna (izrek +\textit{ASA} \ref{KSK}), in sicer simetrična glede na os $AB$. To +pomeni, da je $EM\perp AB$. Ker je $AB\parallel PQ$, je tudi +$EM\perp PQ$ oz. $\angle PME\cong\angle QME =90^0$. +Da je $MP\cong MQ$, smo že dokazali, zato sta trikotnika $PME$ in $QME$ skladna +(izrek \textit{SAS} \ref{SKS}), iz tega pa sledi $PE\cong QE$. + \kdokaz + + + +\bnaloga\footnote{40. IMO, Romania - 1999, Problem 5.} + Two circles $k_1$ and $k_2$ are contained inside the circle $k$, and are tangent to $k$ + at the distinct points $M$ and $N$, respectively. $k_1$ passes through the center of + $k_2$. The line passing through the two points of intersection of $k_1$ and $k_2$ meets + $k$ at $A$ and $B$. The lines $MA$ and $MB$ meet $k_1$ at $C$ and $D$, respectively. + Prove that $CD$ is tangent to $k_2$. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.12.IMO4.pic} +\caption{} \label{sl.pod.7.12.IMO4.pic} +\end{figure} + +\textbf{\textit{Solution.}} Označimo z $O_1$ in $O_2$ središči +krožnic $k_1$ in $k_2$, z $r_1$ in $r_2$ njuna polmera ter z $E$ +drugo presečišče premice $AN$ s krožnico $k_2$ (Figure +\ref{sl.pod.7.12.IMO4.pic}). Brez škode za splošnost predpostavimo +$r_1\geq r_2$. + +Dokažimo najprej, da je premica $CE$ skupna tangenta krožnic $k_1$ +in $k_2$. Naj bo $\widehat{E}$ drugo presečišče očrtane krožnice +$k'$ trikotnika $CMN$ in krožnice $k_2$. Točka $A$ leži na +potenčnih oseh krožnic $k_1$ in $k_2$ oz. $k_1$ in $k'$, zato je +po izreku \ref{PotencnoSr} točka $A$ potenčno središče krožnic +$k_1$, $k_2$ in $k'$. To pomeni, da točka $A$ leži na potenčni +premici krožnic $k_2$ in $k'$ - premici $N\widehat{E}$. Iz tega +sledi, da je $\widehat{E}\in AN\cap k_2$ oz. $\widehat{E}=E$. +Torej so točke $M$, $C$, $E$ in $N$ konciklične in je po izreku +\ref{ObodKotTang} $\angle ACE \cong\angle ANM$. Označimo z $L$ +poljubno točko skupne tangente krožnic $k$ in $k_1$ v točki $M$, +ki leži v polravnini z robom $AC$, v kateri nista točki $B$ in +$D$. Po istem izreku \ref{ObodKotTang} (glede na krožnici $k$ in +$k_1$) je $\angle LMA \cong\angle MBA$ in $\angle LMC \cong\angle +MDC$. Iz izreka \ref{ObodObodKot} (za krožnico $k$ in tetivo $AM$) +dobimo še $\angle ANM \cong\angle ABM$. Če povežemo dokazane +relacije, dobimo: +$$\angle ACE \cong \angle ANM +\cong \angle ABM\cong \angle LMA \cong\ \angle CDM. $$ + Iz $\angle ACE \cong \angle CDM$ pa po izreku \ref{ObodKotTang} sledi, + da je $EC$ tangenta krožnice $k_1$. Ker v dokazu še nismo + uporabili dejstva $O_2\in k_1$, + analogno dokažemo tudi, da je $CE$ tangenta krožnice $k_2$. + + + Označimo s $T$ presečišče daljice $O_2O_1$ in krožnice $k_2$. Dovolj je še + dokazati $T\in CD$ in $\angle CTO_2=90^0$. Naj bo $O'_2$ + pravokotna projekcija točke $O_2$ na premici $O_1C$. Ker je $CE$ + skupna tangenta krožnic $k_1$ in $k_2$, sta polmera $O_1C$ in + $O_2E$ pravokotna na to tangento. Zatorej je $CEO_2O'_2$ + pravokotnik in velja $O'_2C\cong O_2E=r_2$. Iz tega sledi + $O_1O'_2=r_1-r_2=O_1T$. To pomeni, da sta trikotnika + $O_1O'_2O_2$ in $O_1TC$ skladna (izrek \textit{SAS} \ref{SKS}), zato je + $\angle CTO_1\cong\angle O_2O'_2O_1=90^0$ oz. $\angle + CTO_2=90^0$. Premici $CT$ in $AB$ sta pravokotni na + centralo $O_1O_2$ dveh krožnic. Torej je $CT\parallel AB$. Ker + je (zaradi že dokazane relacije $\angle ABM\cong \angle CDM$), je + tudi $CD\parallel AB$. Po Playfairovem aksiomu \ref{Playfair} + sta $CT$ in $CD$ ista premica oz. $T\in CD$. To pomeni, da se + premica $CD$ dotika krožnice $k_2$ v točki $T$. + \kdokaz + + + + +%________________________________________________________________________________ +\poglavje{The Theorems of Pappus and Pascal} \label{odd7PappusPascal} + +Izreka v tem razdelku sta zgodovinsko povezana z razvojem \index{geometrija!projektivna}projektivne geometrije. + + + + \bizrek \label{izrek Pappus} \index{izrek!Pappusov}(Pappus'\footnote{\index{Pappus} \textit{Pappus} iz Aleksandrije (3. st.), eden od zadnjih + velikih starogrških geometrov. Ta izrek je dokazal v evklidskem primeru, + uporabljajoč pri tem metriko. Toda fundamentalna vloga Pappusovega izreka + v projektivni geometriji je bila odkrita šele šestnajst stoletij + kasneje.} theorem) + Let $A$, $B$ and $C$ be three different + points of a line $p$ and $A'$, $B'$ and $C'$ three different points of another line + $p'$ in the same plane. Then the points + $X=BC'\cap B'C$, $Y=AC'\cap A'C$ and $Z=AB'\cap A'B$ are collinear. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10.1.pic} +\caption{} \label{sl.pod.7.10.1.pic} +\end{figure} + + + \textbf{\textit{Proof.}} +Naj bo $L=AB'\cap BC'$, $M=AB'\cap CA'$ in $N=CA'\cap BC'$ (Figure \ref{sl.pod.7.10.1.pic}). Uporabimo petkrat Menelajev izrek (\ref{izrekMenelaj}) +glede na trikotnik $LMN$ in premice $BA'$, $AC'$, $CB'$, $AB$, $A'B'$: + +\begin{eqnarray*} +& & \frac{\overrightarrow{LZ}}{\overrightarrow{ZM}}\cdot \frac{\overrightarrow{MA'}}{\overrightarrow{A'N}}\cdot \frac{\overrightarrow{NB}}{\overrightarrow{BL}}=-1,\\ +& & \frac{\overrightarrow{LA}}{\overrightarrow{AM}}\cdot \frac{\overrightarrow{MY}}{\overrightarrow{YN}}\cdot \frac{\overrightarrow{NC'}}{\overrightarrow{C'L}}=-1,\\ +& & \frac{\overrightarrow{LB'}}{\overrightarrow{B'M}}\cdot \frac{\overrightarrow{MC}}{\overrightarrow{CN}}\cdot \frac{\overrightarrow{NX}}{\overrightarrow{XL}}=-1,\\ +& & \frac{\overrightarrow{LA}}{\overrightarrow{AM}}\cdot \frac{\overrightarrow{MC}}{\overrightarrow{CN}}\cdot \frac{\overrightarrow{NB}}{\overrightarrow{BL}}=-1,\\ +& & \frac{\overrightarrow{LB'}}{\overrightarrow{B'M}}\cdot \frac{\overrightarrow{MA'}}{\overrightarrow{A'N}}\cdot \frac{\overrightarrow{NC'}}{\overrightarrow{C'L}}=-1. +\end{eqnarray*} +Iz teh petih relacij (če pomnožimo prve tri, nato pa v dobljeno relacijo vstavimo četrto in peto) dobimo: + $$\frac{\overrightarrow{LZ}}{\overrightarrow{ZM}}\cdot \frac{\overrightarrow{MY}}{\overrightarrow{YN}}\cdot \frac{\overrightarrow{NX}}{\overrightarrow{XL}}=-1.$$ + Po obratnem Menelajevem izreku (\ref{izrekMenelaj}) so točke $X$, $Y$ in $Z$ kolinearne. + \kdokaz + + Dokažimo sedaj še Pascalov\footnote{Ni točno znano, kako je to trditev za krožnico + dokazal + francoski matematik in filozof \index{Pascal, B.} + \textit{B. Pascal} (1623--1662), ker je + originalen dokaz izgubljen. Lahko pa predvidevamo, da je uprabljal rezultate + in metode tistega časa, kar pomeni, da je verjetno uporabljal Menelajev izrek.} + izrek za + krožnico. + + + + \bizrek \index{izrek!Pascalov} \label{izrekPascalEvkl} + Let $A$, $B$, $C$, $D$, $E$ and $F$ be arbitrary points on + some circle $k$. Then the points $X=AE\cap BD$, $Y=AF\cap CD$ and + $Z=BF\cap CE$ are collinear. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10.2.pic} +\caption{} \label{sl.pod.7.10.2.pic} +\end{figure} + + + \textbf{\textit{Proof.}} +Označimo $L=AE\cap BF$, $M=AE\cap CD$ in $N=CD\cap BF$. + Če trikrat uporabimo Menelajev izrek (\ref{izrekMenelaj}) za trikotnik $LMN$ in premice $BD$, +$AF$ in $CE$, dobimo (Figure \ref{sl.pod.7.10.2.pic}): + \begin{eqnarray*} + \hspace*{-2mm} \frac{\overrightarrow{LX}}{\overrightarrow{XM}}\cdot + \frac{\overrightarrow{MD}}{\overrightarrow{DN}}\cdot + \frac{\overrightarrow{NB}}{\overrightarrow{BL}}=-1,\hspace*{1mm} + \frac{\overrightarrow{LA}}{\overrightarrow{AM}}\cdot + \frac{\overrightarrow{MY}}{\overrightarrow{YN}}\cdot + \frac{\overrightarrow{NF}}{\overrightarrow{FL}}=-1,\hspace*{1mm} + \frac{\overrightarrow{LE}}{\overrightarrow{EM}}\cdot + \frac{\overrightarrow{MC}}{\overrightarrow{CN}}\cdot + \frac{\overrightarrow{NZ}}{\overrightarrow{ZL}}=-1. + \end{eqnarray*} + Če uporabimo potence točk $M$, $N$ in $L$ glede na krožnico + $k$ (izrek \ref{izrekPotenca}) ali pa upoštevamo podobnost ustreznih trikotnikov, sledi: + \begin{eqnarray*} + \overrightarrow{MC}\cdot\overrightarrow{MD}= + \overrightarrow{MA}\cdot\overrightarrow{ME},\hspace*{3mm} + \overrightarrow{NC}\cdot\overrightarrow{ND}= + \overrightarrow{NF}\cdot\overrightarrow{NB},\hspace*{3mm} + \overrightarrow{LA}\cdot\overrightarrow{LE}= + \overrightarrow{LB}\cdot\overrightarrow{LF}. + \end{eqnarray*} + Iz prejšnjih šestih relacij sledi: + \begin{eqnarray*} + \frac{\overrightarrow{LX}}{\overrightarrow{XM}}\cdot + \frac{\overrightarrow{MY}}{\overrightarrow{YN}}\cdot + \frac{\overrightarrow{NZ}}{\overrightarrow{ZL}}=-1. + \end{eqnarray*} +Torej so po Menelajevem izreku (\ref{izrekMenelaj}, obratna smer) točke $X$, $Y$ in +$Z$ kolinearne. + \kdokaz + + Prejšnji izrek, ki se nanaša na krožnico, + lahko posplošimo tudi na poljubno stožnico\footnote{Proučevanje stožnic se je +začelo že pri Starih Grkih. Termine elipsa, parabola, hiperbola + je prvi uporabil starogrški matematik \index{Apolonij} \textit{Apolonij} iz Perge (262--200 pr. n. š.) v + svojem +znanem delu \textit{Razprava o presekih stožca}, ki je +sestavljeno iz + osmih knjig, kjer stožnico definira kot presek ravnine s +krožnim stožcem. Komentar k temu Apolonijevemu delu je napisala starogrška filozofinja in matematičarka ter zadnja predstavnica antične znanosti \index{Hipatija} \textit{Hipatija} iz Aleksandrije (370--415). Ponovno zanimanje za stožnice sta oživela + nemški astronom \index{Kepler, J.} \textit{J. Kepler} (1571--1630) in +francoski matematik in filozof \index{Pascal, B.} \textit{B. +Pascal} (1623--1662) v 17. stoletju.} + oz. krivuljo drugega razreda. + V evklidskem primeru so to: elipsa, parabola in hiperbola. + To lahko ugotovimo na ta način, da stožnico definiramo + kot presek nosilk vseh + stranic stožca in neke ravnine. + Zaradi tega na stožnico $\mathcal{K}$ v splošnem primeru gledamo kot na + središčno + projekcijo neke krožnice $k$ na neko ravnino. Središče projiciranja je + vrh stožca $S$. Omenili smo že, da + središčna projekcija ohranja kolinearnost. Če so $A'$, $B'$, $C'$, + $D'$, $E'$ in $F'$ točke na stožnici $\mathcal{K}$ ter $X'$, $Y'$ in $Z'$ + ustrezne točke, definirane kot v izreku \ref{izrekPascalEvkl}, + so te točke slike nekih točk $A$, $B$, $C$, $D$, $E$, $F$, $X$, $Y$ in + $Z$. + Pri tem je prvih šest točk na krožnici $k$, zadnje tri pa so + po izreku \ref{izrekPascalEvkl} kolinearne. Iz tega sledi, da so tudi + točke $X'$, $Y'$ in $Z'$ kolinearne (Figure \ref{sl.pod.7.10.3.pic}). To pomeni, da izrek \ref{izrekPascalEvkl} + res velja v splošnem primeru za poljubne stožnice. Ta splošni izrek je + znan kot Pascalov izrek za stožnice\footnote{\index{Pascal, B.} \textit{B. Pascal} +(1623--1662), francoski matematik in filozof, ki je + že kot šestnajstletnik dokazal ta pomemben izrek o + stožnicah, objavil pa ga je +leta 1640, vendar se takrat trditev ni neposredno nanašala na +projektivno geometrijo.}. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.10.3.pic} +\caption{} \label{sl.pod.7.10.3.pic} +\end{figure} + + Ideje, ki smo jih obravnavali v tem razdelku, nas napeljujejo na ugotovitev, + da lahko Pascalov izrek izpeljemo tudi v projektivni geometriji. + Še več - v projektivni geometriji lahko definiramo in raziskujemo + tudi stožnice, vendar + ni možno razlikovati med elipso, hiperbolo ali + parabolo (glej \cite{Mitrovic}). + + +%________________________________________________________________________________ + \poglavje{The Golden Ratio} \label{odd7ZlatiRez} + + Pravimo, da točka $Z$ daljice $AB$ deli to daljico v razmerju \index{zlati!rez}\pojem{zlatega reza}\footnote{Takšno delitev je obravnaval že \index{Pitagora}\textit{Pitagora z otoka Samosa} (582--497 pr. n. š.), starogrški filozof in matematik. Prve znane zapise o zlatem rezu naj bi ustvaril starogrški matematik \index{Evklid} + \textit{Evklid iz Aleksandrije} (3. st. pr. n. š.). V svojem znamenitem delu \textit{Elementi} je zastavil problem: ‘‘\textit{Dano daljico razdeli na dva neenaka dela tako, da bo ploščina pravokotnika, ki ima dolžino enako celotni daljici, višino pa krajšemu delu daljice, enaka ploščini kvadrata, načrtanega nad daljšim delom daljice.}’’ +Termin zlatega reza, ki ga danes uporabljamo, je vpeljal \index{Leonardo da Vinci}\textit{Leonardo da Vinci} (1452--1519), +italijanski slikar, arhitekt in izumitelj. Zlati rez ljudje že tisočletja uporabljajo v slikarstvu in arhitekturi.}, če je razmerje dolžine celotne daljice proti daljšemu delu enako razmerju daljšega dela proti krajšemu (Figure \ref{sl.pod.7.15.1.pic}) oz: + + \begin{eqnarray} \label{eqnZlatiRez} + AB:AZ=AZ:ZB. + \end{eqnarray} + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.1.pic} +\caption{} \label{sl.pod.7.15.1.pic} +\end{figure} + +Če +sedaj krajši del ‘‘vstavimo’’ v daljšega, dobimo enako +razmerje. Res, ker je +$AZ:ZB=AB:AZ=(AZ+ZB):AZ$, velja tudi +$$ZB:(AZ-ZB)=AZ:ZB.$$ +Ta postopek lahko nadaljujemo (Figure \ref{sl.pod.7.15.2.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.2.pic} +\caption{} \label{sl.pod.7.15.2.pic} +\end{figure} + +Seveda se zastavlja vprašanje, kako konstruiramo takšno točko $Z$. To konstrukcijo bomo opisali v naslednjem zgledu. + + + + \bzgled + For a given line $AB$, construct a point $Z$ that divides the line segment into + the golden ratio. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.3.pic} +\caption{} \label{sl.pod.7.15.3.pic} +\end{figure} + + + \textbf{\textit{Solution.}} +Načrtajmo najprej krožnico $k(S,SB)$ s polmerom +$|SB|=\frac{1}{2}|AB|$, ki se dotika premice $AB$ v točki $B$ (Figure \ref{sl.pod.7.15.3.pic}). Konstruirajmo še presečišči +te krožnice s premico $AS$ - označimo ju z $X$ in $Y$ (naj bo pri tem $\mathcal{B}(A,X,S)$). Točko $Z$ sedaj dobimo kot + presečišče daljice $AB$ in krožnice +$l(A,AX)$. + +Dokažimo, da je $Z$ iskana točka. Če uporabimo potenco točke $A$ glede na krožnico $k$ (izrek \ref{izrekPotenca}), dobimo: +$$\hspace*{-1.5mm} |AB|^2=p(A,k)=|AX|\cdot |AY|=|AX|\cdot(|AX|+|XY|)=|AZ|\cdot(|AZ|+|AB|).$$ + Torej velja: +\begin{eqnarray} \label{eqnZlatiRez2} + |AB|^2=|AZ|\cdot(|AZ|+|AB|) + \end{eqnarray} + Naprej je: + \begin{eqnarray*} + |AB|^2=|AZ|\cdot(|AZ|+|AB|)\hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} + \frac{|AB|}{|AZ|}=\frac{|AZ|+|AB|}{|AB|}\\ + \hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} + \frac{|AB|}{|AZ|}=\frac{|AZ|}{|AB|}+1\\ + \hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} + \frac{|AB|}{|AZ|}-1=\frac{|AZ|}{|AB|}\\ + \hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} + \frac{|AB|-|AZ|}{|AZ|}=\frac{|AZ|}{|AB|}\\ + \hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} + \frac{|BZ|}{|AZ|}=\frac{|AZ|}{|AB|} + \end{eqnarray*} + kar je ekvivalentno z relacijo \ref{eqnZlatiRez}. To pomeni, da točka $Z$ deli daljico $AB$ v razmerju zlatega reza. + \kdokaz + + V naslednjem primeru bomo izračunali vrednost razmerja, ki ga določa zlati rez. + + + + \bizrek \label{zlatiRezStevilo} + If a point $Z$ divides a line segment $AB$ into the golden ratio + ($AZ$ is the longer part), then + \begin{eqnarray*} + && AZ:ZB=AB:AZ=\frac{\sqrt{5}+1}{2}, \hspace*{1mm}\textrm{ i.e.}\\ + && AZ=\frac{\sqrt{5}-1}{2}AB. + \end{eqnarray*} + \eizrek + + \textbf{\textit{Proof.}} + Relacija \ref{eqnZlatiRez2} iz prejšnjega izreka je ekvivalentna z relacijo: + $$|AB|^2-|AB|\cdot |AZ|-|AZ|^2=0.$$ + Če to kvadratno enačbo rešimo po $|AZ|$, dobimo: + $$|AZ|=\frac{\sqrt{5}+1}{2}|AB|,$$ + iz tega pa sledita iskani enakosti. + \kdokaz + + Število $$\Phi=\frac{\sqrt{5}+1}{2}$$ iz prejšnjega izreka, ki torej predstavlja vrednost razmerja, ki ga določa zlati rez, imenujemo \index{število!zlato}\pojem{zlato število}. Seveda gre za iracionalno število ($\Phi\notin \mathbb{Q}$). Njegova približna vrednost znaša: + $$\Phi=\frac{\sqrt{5}+1}{2}\doteq 1,62.$$ + + Prav tako je: + $$\frac{\sqrt{5}-1}{2}\doteq 0,62,$$ + kar pomeni, da je daljši del $AZ$ pri zlatem rezu približno $62\%$ celotne daljice $AB$. + + . + + \bzgled \label{zlatiRezKonstr} + Construct a golden rectangle $ABCD$ with two sides + $a$ and $b$ in the golden ratio - so-called \index{golden!rectangle}\pojem{golden rectangle}\color{green1}. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.4.pic} +\caption{} \label{sl.pod.7.15.4.pic} +\end{figure} + + + \textbf{\textit{Solution.}} Narišemo poljubno daljico $AB$, nato pa točko $Z$, ki deli daljico $AB$ v zlatem rezu tako, da je $AZ$ daljši del (glej prejšnji zgled \ref{zlatiRezKonstr}). Na koncu dobimo $D=\mathcal{R}_{A,90^0}(Z)$ in $C=\mathcal{T}_{\overrightarrow{AB}}(D)$ + (Figure \ref{sl.pod.7.15.4.pic}). + \kdokaz + + Če zlati pravokotnik po daljši stranici v razmerju zlatega reza razdelimo na dva pravokotnika, dobimo kvadrat in še en zlati pravokotnik. Novi zlati pravokotnik razdelimo na enak način in postopek nadaljujemo. Če povežemo ustrezna oglišča (krajišč diagonal kvadratov, ki določajo lomljenko), tako da narišemo krožne loke s središčnim kotom $90^0$, dobimo približno konstrukcijo t. i. \index{zlata spirala}\pojem{zlate spirale} ali \index{logaritemska spirala}\pojem{logaritemske spirale} (Figure \ref{sl.pod.7.15.5.pic}). + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.5.pic} +\caption{} \label{sl.pod.7.15.5.pic} +\end{figure} + +Logaritemska spirala\footnote{Predlog za ime logaritemska spirala je podal francoski matematik \index{Varignon, P.}\textit{P. Varignon} (1645--1722). Logaritemska spirala ima lastnost, da jo vsaka ravna premica iz središča spirale seka pod enakim kotom. Pojavlja se v različnih oblikah v naravi: od spiral v cvetu soncnič, polžjih hišic in pajkovih mrež pa vse do oddaljenih galaksij. Imenujemo jo tudi spirala mirabilis (čudežna spirala) - to ime je predlagal švicarski matematik \index{Bernoulli, J.}\textit{J. Bernoulli} (1667–-1748) zaradi njenih čudovitih lastnosti.} je krivulja, ki jo definiramo z enačbami v parametrični obliki: +\begin{eqnarray*} +x&=&a e^{bt}\cdot\cos t \\ +y&=&a e^{bt}\cdot\sin t, +\end{eqnarray*} +kjer je $t\in \mathbb{R}$ parameter, $a$ in $b$ poljubni realni konstanti ter $e\doteq 2,72$ t. i. \index{število!Eulerjevo }\pojem{Eulerjevo\footnote{Število $e$ je iracionalno število in predstavlja vrednost limite $\lim_{n\rightarrow \infty}\left(1+\frac{1}{n} \right)^n$ =e. Ime je dobilo po švicarskem matematiku \index{Euler, L.}\textit{L. Eulerju} (1707--1783).} število.}. +Prejšnja konstrukcija (s krožnimi loki) je, kot smo že omenili, približna, toda predstavlja zelo dobro aproksimacijo te krivulje (Figure \ref{sl.pod.7.15.5a.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.5a.pic} +\caption{} \label{sl.pod.7.15.5a.pic} +\end{figure} + +Nadaljevali bomo s pravilnim petkotnikom. + + + + \bzgled \label{zlatiRezPravPetk1} + A diagonal and a side of a regular pentagon are in the golden ratio. + \ezgled + + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.15.6.pic}) + + Označimo z $a$ stranico in $d$ diagonalo pravilnega petkotnika $ABCDE$. Iz \ref{PtolomejPetkotnik} izreka sledi: + $$d=\frac{1+\sqrt{5}}{2}a.$$ +Zato sta po izreku \ref{zlatiRezStevilo} daljici $d$ in $a$ v razmerju zlatega reza. +\kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.6.pic} +\caption{} \label{sl.pod.7.15.6.pic} +\end{figure} + + + + \bzgled + Two diagonals of a regular pentagon intersect at a point that + divides them in the golden ratio\footnote{Za to to lastnost petkotnika so vedeli že pitagorejci. Že omenjeno šolo pitagorejcev je ustanovil starogrški filozof in matematik \index{Pitagora}\textit{Pitagora z otoka Samosa} (582--497 pr. n. š.) v Crotoni v južni + Italiji. Njegovi učenci so se ukvarjali s filozofijo, z matematiko in naravoslovjem. Za svoj razpoznavni znak so + izbrali pentagram, ki je sestavljen iz petkotnikovih diagonal. Vpliv pitagorejske šole na matematiko Starih + Grkov je obstajal še več stoletij po Pitagorovi smrti.}. + \ezgled + + + \textbf{\textit{Proof.}} +S $P$ označimo presečišče diagonal $AC$ in $BD$ pravilnega petkotnika $ABCDE$ ter s $k$ +njegovo očrtano krožnico (Figure \ref{sl.pod.7.15.6.pic}). Koti $EAD$, $DAC$, $CAB$ in $DBC$, ki so prirejeni skladnim +tetivam $ED$, $CD$ in $CB$, so skladni (izrek \ref{SklTetSklObKot}). Ker je po izreku \ref{pravVeckNotrKot} notranji kot $EAB$ pravilnega petkotnika enak +$108^0$, so koti $EAD$, $DAC$, $CAB$ (oz. $PAB$) in +$DBC$ enaki $36^0$. Iz tega dobimo $\angle PBA=\angle DBA=\angle CBA-\angle CBD=108^0-36^0= 72^0$. Po izreku \ref{VsotKotTrik} iz vsote notranjih kotov trikotnika $ABP$ sledi +$\angle APB=180^0-\angle PBA-\angle PAB=72^0$. Torej je trikotnik $PAB$ enakokrak z osnovnico $PB$ (izrek \ref{enakokraki}), +oz. velja $AP\cong AB$. Iz tega sledi: +$$AC:AP=AC:AB.$$ +Če uporabimo še prejšnjo trditev \ref{zlatiRezPravPetk1}, ugotovimo, da točka $P$ deli diagonalo $AC$ v razmerju zlatega reza. +\kdokaz + + +Če v pravilnem petkotniku $ABCDE$ izberemo oglišča $A$, $B$ in $D$, dobimo enakokraki trikotnik $ABD$ z osnovnico, ki je enaka stranici $a$ petkotnika $ABCDE$, kraka pa sta enaka diagonali $d$ tega petkotnika. Krak in osnovnica tega trikotnika sta torej v razmerju zlatega reza (Figure \ref{sl.pod.7.15.7.pic}). Zato trikotnik $ABD$ imenujemo \index{zlati!trikotnik}\pojem{zlati trikotnik}. Koti zlatega trikotnika merijo $72^0$, $72^0$ in $36^0$. + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.15.7.pic} +\caption{} \label{sl.pod.7.15.7.pic} +\end{figure} + +Podobno kot pri zlatem pravokotniku, če uporabimo simetrale ustreznih notranjih kotov ob osnovnici zlatega trikotnika, dobimo zaporedje zlatih trikotnikov, ki so vsi podobni. S pomočjo ustreznih krožnih lokov lahko izpeljemo še eno približno konstrukcijo zlate (oz. logaritemske) spirale (Figure \ref{sl.pod.7.15.7.pic}). + +%________________________________________________________________________________ + \poglavje{Morley's Theorem and Some More Theorems} \label{odd7Morly} + +Pravimo, da sta poltraka $SP$ in $SQ$ \index{trisektrisa}\pojem{trisektrisi} kota $ASB$, če točki $P$ in $Q$ ležita v tem kotu in velja $$\angle ASP\cong\angle PSQ\cong\angle QSB,$$ +oz. gre za poltraka, ki delita kot na tri skladne kote +(Figure \ref{sl.pod.7.16.0.pic}). + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.0.pic} +\caption{} \label{sl.pod.7.16.0.pic} +\end{figure} + + Dokažimo najprej pomožno trditev - lemo. + + + + + \bizrek \label{izrekMorleyLema} + Let $Y'$, $Z$, $Y$ and $Z'$ be points in the plane such that + $Y'Y\cong ZY\cong ZZ'$ and + $$\angle Z'ZY\cong \angle ZYY'=180^0-2\alpha>60^0.$$ + If $A$ is a point in this plane that is on the different side + of the line $Z'Y'$ with respect to + the point $Z$ and also $Y'AZ'=3\alpha$, then the points $A$, $Y'$, $Z$, $Y$ and $Z'$ are + concyclic and also + $$\angle Z'AZ\cong \angle ZAY\cong \angle YAY'=\alpha.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.1a.pic} +\caption{} \label{sl.pod.7.16.1a.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Najprej je dani pogoj $180^0-2\alpha>60^0$ ekvivalenten s pogojem $\alpha<60^0$. + + Naj bo $s$ simetrala daljice $ZY$, $\mathcal{S}_s$ pa zrcaljenje čez premico $s$ (Figure \ref{sl.pod.7.16.1a.pic}). Torej $\mathcal{S}_s(Z)=Y$. Ker je $\mathcal{S}_s$ izometrija, ki ohranja velikost kotov in dolžine daljic, je tudi $\mathcal{S}_s(Z')=Y'$. Iz $ZY,Z'Y'\perp s$ sledi $ZY\parallel Z'Y'$, kar pomeni, da je $ZYY'Z'$ enakokraki trapez. + Po izreku \ref{trapezTetivEnakokr} je ta trapez tetiven - označimo s $k$ njegovo očrtano krožnico. Po izreku \ref{KotiTransverzala} sta kota $YZZ'$ in $Y'Z'Z$ suplementarna, zato je: + \begin{eqnarray} \label{eqnMorleyLema1} + \angle ZZ'Y'=180^0-\angle Z'ZY=2\alpha. + \end{eqnarray} + Iz $Z'Z\cong ZY$ po izreku \ref{SklTetSklObKot} sledi $\angle ZZ'Y\cong \angle YZ'Y'$, zato je po relaciji \ref{eqnMorleyLema1} $\angle ZZ'Y\cong \angle YZ'Y'=\alpha$. Trikotnik $Z'ZY$ je enakokrak z osnovnico $Z'Y$, zato je po izreku \ref{enakokraki} tudi $\angle ZYZ'\cong \angle ZZ'Y=\alpha$. Analogno je $\angle YY'Z\cong \angle ZY'Z'=\alpha$ in $\angle YZY'\cong \angle YY'Z=\alpha$. Če vse povežemo, dobimo $\angle ZZ'Y\cong \angle YZ'Y'\cong \angle ZYZ'\cong + \angle YY'Z\cong \angle ZY'Z'\cong\angle YZY'=\alpha$, torej: + \begin{eqnarray} \label{eqnMorleyLema2} + \angle Z'Y'Z \cong \angle Y'Z'Y= \alpha. + \end{eqnarray} + Ker je $\angle Z'ZY'=\angle Z'ZY-\angle Y'ZY=180^0-2\alpha-\alpha=180^0-3\alpha$ (ta kot obstaja, ker je $\alpha<60^0$) oz. $\angle Z'ZY'+\angle Z'AY'=180^0-3\alpha+3\alpha=180^0$, je po izreku \ref{TetivniPogoj} $Z'ZY'A$ tetivni štirikotnik, kar pomeni, da točka $A$ leži na krožnici $k$. + + Iz relacije \ref{eqnMorleyLema2} in izreka \ref{ObodObodKot} sledi: + \begin{eqnarray*} + &&\angle Z'AZ\cong \angle Z'Y'Z=\alpha,\\ + &&\angle Y'AY\cong \angle Y'Z'Y=\alpha. + \end{eqnarray*} + Na koncu je še: + \begin{eqnarray*} + \angle ZAY=\angle Z'AY'-\angle Z'AZ-\angle Y'AY =3\alpha-\alpha-\alpha=\alpha, + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + Sedaj smo pripravljeni za osnovni izrek. + + + + + + \bizrek \label{izrekMorley}\index{izrek!Morleyev} + If $X$, $Y$ and $Z$ are three points of intersection of the adjacent angle trisectors + of a triangle $ABC$, then $XYZ$ is an equilateral triangle.\\ + (Morley's\footnote{\index{Morley, F.}\textit{F. Morley} (1860--1937), angleški matematik, je odkril to lastnost trikotnika leta 1904, objavil pa jo je šele 20 let + pozneje. V tem času je bil izrek kot naloga objavljen v časopisu \textit{Educational Times}. Na tem mestu bomo podali eno od takrat predlaganih rešitev.} theorem) + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.1.pic} +\caption{} \label{sl.pod.7.16.1.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + (Figure \ref{sl.pod.7.16.1.pic}) + + Naj bosta $X$ in $V$ presečišči ustreznih trisektris notranjih kotov $ABC$ in $ACB$ trikotnika $ABC$, tako da je $X$ notranja točka trikotnika $BVC$. Označimo še z $Z_1$ in $Y_1$ točki trisektris $BV$ in $CV$, za kateri velja $\angle Z_1XV \cong \angle VXY_1=30^0$. Dokažimo najprej, da je $XY_1Z_1$ enakostranični trikotnik. +Ker sta premici $BX$ in $CX$ simetrali kotov $VBC$ in $VCB$, je točka $X$ središče včrtane krožnice trikotnika $BVC$, zato je tudi premica $VX$ simetrala kota $BVC$ (izrek \ref{SredVcrtaneKrozn}). Iz skladnosti trikotnikov $VXZ_1$ in +$VXY_1$ (izrek \textit{ASA} \ref{KSK}) sledi $XZ_1\cong XY_1$. Ker je še $\angle Z_1XY_1=60^0$, je $XY_1Z_1$ enakostranični trikotnik (izrek \ref{enakokraki}). Dovolj je, če še dokažemo, da velja $Y_1=Y$ in $Z_1=Z$ oz. $\angle BAZ_1\cong\angle Z_1AY_1\cong\angle Z_1AC$. + +Iz skladnosti trikotnikov $VXZ_1$ in +$VXY_1$ sledi še $Z_1V \cong Y_1V$. Torej je $VZ_1Y_1$ enakokraki trikotnik z osnovnico $Z_1Y_1$, zato je po izreku \ref{enakokraki}: +\begin{eqnarray} \label{eqnMorley1} +\angle VZ_1Y_1\cong\angle VY_1Z_1. + \end{eqnarray} +Naj bosta $Z'$ in $Y'$ takšni točki na stranicah $AB$ in $AC$ trikotnika $ABC$, da velja $BZ'\cong BX$ in $CY'\cong CX$. +Trikotnika $BZ'Z_1$ in $BXZ_1$ sta skladna (izrek \textit{SAS} \ref{SKS}), zato je $Z'Z_1\cong XZ_1$. Prav tako je $\angle Z'Z_1B\cong\angle XZ_1B$ oz. je premica $BV$ simetrala kota $Z'Z_1X$. +Analogno je tudi $Y'Y_1\cong XY_1$. Iz tega in iz dejstva, da je $XY_1Z_1$ enakostranični trikotnik, sledi: +\begin{eqnarray} \label{eqnMorley2} +Z'Z_1\cong Z_1Y_1\cong Y_1Y'. + \end{eqnarray} + Označimo s $3\alpha$, $3\beta$, $3\gamma$ mere +notranjih trikotnika $ABC$, kotov ob ogliščih $A$, $B$ in $C$. +Jasno je $3\alpha+3\beta+3\gamma=180^0$, torej: +\begin{eqnarray} \label{eqnMorley3} +2\alpha+2\beta+2\gamma=120^0. + \end{eqnarray} +Če uporabimo dejstvo, da je premica $BV$ simetrala konveksnega oz. nekonveksnega kota $Z'Z_1X$ ter relaciji \ref{eqnMorley1} in \ref{eqnMorley3}, po enostavnem računanju dobimo: +\begin{eqnarray*} + \angle Z'Z_1Y_1&=&\angle Z'Z_1V+\angle VZ_1Y_1=\\ + &=&\angle VZ_1X+\angle VZ_1Y_1=\\ + &=&60^0+2\angle VZ_1Y_1=\\ + &=&60^0+\angle VZ_1Y_1+\angle VY_1Z_1=\\ + &=&60^0+180^0-\angle Z_1VY_1=\\ + &=&60^0+2\beta+2\gamma=\\ + &=&60^0+120^0-2\alpha=\\ + &=&180^0-2\alpha. + \end{eqnarray*} + Analogno je tudi $\angle Y'Y_1Z_1=180^0-2\alpha$, zato je (če upoštevamo še relacijo \ref{eqnMorley2}) po lemi \ref{izrekMorleyLema} + $\angle Z'AZ_1\cong\angle Z_1AY_1\cong\angle Y_1AY'= \alpha$, kar je bilo potrebno dokazati. + \kdokaz + + + + \bizrek (Leibniz's\footnote{\index{Leibniz, G. W.}\textit{G. W. Leibniz} (1646--1716), nemški matematik.} theorem) + \label{izrekLeibniz}\index{theorem!Leibniz's} + If $T$ is the centroid and $X$ an arbitrary + point in the plane of a triangle $ABC$, then + $$|XA|^2 + |XB|^2 + |XC|^2 = \frac{1}{3}\left(|AB|^2 +|BC|^2 +|CA|^2\right) + 3|XT|^2 + ,\textrm{ i.e.}$$ + $$|XA|^2 + |XB|^2 + |XC|^2 = |TA|^2 +|TB|^2 +|TC|^2 + 3|XT|^2.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.2.pic} +\caption{} \label{sl.pod.7.16.2.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.16.2.pic}) + Naj bo $A_1$ središče +stranice $BC$ oz. $|AA_1|=t_a$ njegova +težiščnica. Ker je $AT:TA_1=2:1$ (izrek \ref{tezisce}), po +Stewartovem izreku \ref{StewartIzrek2} +glede na trikotnik +$AXA_1$ sledi: +\begin{eqnarray} \label{eqnLeibniz1} +|XT|^2=\frac{1}{3}|XA|^2+\frac{2}{3}|XA_1|^2-\frac{2}{9}t_a^2 +\end{eqnarray} +Če še enkrat uporabimo Stewartov izrek \ref{StewartIzrek2} (ali njegovo posledico za težiščnico \ref{StwartTezisc}) glede na trikotnik $BXC$, dobimo: + +\begin{eqnarray} \label{eqnLeibniz2} +|XA_1|^2=\frac{1}{2}|XB|^2+\frac{1}{2}|XC|^2-\frac{1}{4}|BC|^2 +\end{eqnarray} +in če vstavimo \ref{eqnLeibniz2} v \ref{eqnLeibniz1}: +\begin{eqnarray} \label{eqnLeibniz3} +|XT|^2=\frac{1}{3}\left(|XA|^2+|XB|^2+|XC|^2\right)-\frac{1}{6}|BC|^2-\frac{2}{9}t_a^2 +\end{eqnarray} +ter analogno še: +\begin{eqnarray} \label{eqnLeibniz4a} +\hspace*{-4mm} |XT|^2&=&\frac{1}{3}\left(|XA|^2+|XB|^2+|XC|^2\right)-\frac{1}{6}|AC|^2-\frac{2}{9}t_b^2\\ +\hspace*{-4mm} |XT|^2&=&\frac{1}{3}\left(|XA|^2+|XB|^2+|XC|^2\right)-\frac{1}{6}|AB|^2-\frac{2}{9}t_c^2\label{eqnLeibniz4} +\end{eqnarray} +S seštevanjem enakosti iz \ref{eqnLeibniz3} - \ref{eqnLeibniz4} dobimo: +\begin{eqnarray*} +3|XT|^2&=&|XA|^2+|XB|^2+|XC|^2-\frac{1}{6}\left(|BC|^2+|AC|^2+|AB|^2\right)-\\ +&-&\frac{2}{9}\left(t_a^2+t_b^2+t_c^2\right) +\end{eqnarray*} +Na koncu iz trditve \ref{StwartTezisc2} sledi: +\begin{eqnarray*} +3|XT|^2=|XA|^2+|XB|^2+|XC|^2-\frac{1}{3}\left(|BC|^2+|AC|^2+|AB|^2\right) +\end{eqnarray*} +oz. obe relaciji iz trditve. +\kdokaz + +Direktna posledica Leibnizovega izreka je naslednja trditev. + + + + + \bizrek + A point in the plane of a triangle for which the squared distances + from its vertices has a minimum value is its centroid. + \eizrek + + \textbf{\textit{Proof.}} Po Leibnitzovem izreku za poljubno točko $X$ v ravnini trikotnika $ABC$ s težiščem $T$ velja: +$$|XA|^2 + |XB|^2 + |XC|^2 = \frac{1}{3}\left(|AB|^2 +|BC|^2 +|CA|^2\right) + 3|XT|^2.$$ +Minimum vsote $|XA|^2 + |XB|^2 + |XC|^2$ po $X$ se torej doseže, ko je $|XT|$ najmanjše, to pa je, kadar je $X=T$. +\kdokaz + + + + + + + \bizrek (Carnot's\footnote{\index{Carnot, L. N. M.}\textit{L. N. M. Carnot} (1753--1823), + francoski matematik.} theorem)\index{theorem!Carnot's} + Let $P$, $Q$ and $R$ be points on the lines containing the + sides $BC$, $CA$ and $AB$ of a triangle $ABC$. Perpendicular lines on the lines $BC$, $CA$. + and $AB$ at the points $P$, $Q$ and $R$ intersect at one point if and only if + \begin{eqnarray} \label{eqnCarnotIzrek1} + |BP|^2 - |PC|^2 + |CQ|^2 - |QA|^2 + |AR|^2 - |RB|^2 = 0. + \end{eqnarray} + + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.3.pic} +\caption{} \label{sl.pod.7.16.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} Označimo s $p$, $q$ in $r$ pravokotnice premic $BC$, $CA$ in $AB$ skozi točke $P$, $Q$ in $R$ (Figure \ref{sl.pod.7.16.3.pic}). + +($\Rightarrow$) Predpostavimo najprej, da se premice $p$, $q$ in $r$ sekajo v neki točki $L$. Če šestkrat uporabimo Pitagorov izrek \ref{PitagorovIzrek}, dobimo: + \begin{eqnarray*} +\begin{array}{cc} + |AR|^2+|RL|^2=|AL|^2 & \hspace{6mm} -|AQ|^2-|QL|^2=-|AL|^2 \\ + |BP|^2+|PL|^2=|BL|^2 & \hspace{6mm} -|BR|^2-|RL|^2=-|BL|^2\\ + |CQ|^2+|QL|^2=|CL|^2 & \hspace{6mm} -|CP|^2-|PL|^2=-|CL|^2 +\end{array} +\end{eqnarray*} + Če seštejemo vseh šest enakosti, dobimo relacijo \ref{eqnCarnotIzrek1}. + +($\Leftarrow$) Predpostavimo sedaj, da velja +relacija \ref{eqnCarnotIzrek1}. +Pravokotnici $q$ in $r$ stranic $AC$ in $AB$ nista vzporedni (ker bi bile sicer po posledici Playfairjevega aksioma \ref{Playfair1} točke $A$, $B$ in $C$ kolinearne). Označimo z $\widehat{L}$ presečišče premic $q$ in $r$. Označimo s $\widehat{P}$ pravokotno projekcijo točke $\widehat{L}$ na premici $BC$. Ker se pravokotnice nosilk stranic trikotnika $ABC$ v točkah $\widehat{P}$, $Q$ in $R$ sekajo v točki $\widehat{L}$, iz prvega dela dokaza ($\Rightarrow$) sledi: + \begin{eqnarray} \label{eqnCarnotIzrek2} + |B\widehat{P}|^2 - |\widehat{P}C|^2 + |CQ|^2 - |QA|^2 + |AR|^2 - |RB|^2 = 0. + \end{eqnarray} + Iz \ref{eqnCarnotIzrek1} in \ref{eqnCarnotIzrek2} sledi: +$$|B\widehat{P}|^2 - |\widehat{P}C|^2=|BP|^2 - |PC|^2.$$ + Na enak način kot na koncu drugega dela dokaza izreka \ref{PotencOsLema} dobimo $P=\widehat{P}$, oz. premice $p$, $q$ in $r$ se sekajo v eni točki. +\kdokaz + + + + + \bizrek (Butterfly theorem\footnote{En dokaz tega izreka je objavil angleški matematik \index{Horner, W. J.}\textit{W. J. Horner} (1786--1837) leta 1815.}) + \index{theorem!butterfly} + Let $S$ be the midpoint of chord $PQ$ of a circle $k$. Suppose that $AB$ and $CD$ + are arbitrary chords of this circle passing through the point $S$. If $X$ and $Y$ + are the points of intersection of the chords $AD$ and $BC$ with the chord $PQ$, + then $S$ is the midpoint of the line segment $XY$. + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.4.pic} +\caption{} \label{sl.pod.7.16.4.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Označimo $x=|SX|$ in $y=|SY|$ + (Figure \ref{sl.pod.7.16.4.pic}). + Naj bodo +$X_1$ in $X_2$ oz. $Y_1$ in $Y_2$ +pravokotne projekcije točk $X$ oz. $Y$ na tetivah $AB$ +in $CD$. +Po Talesovem izreku (\ref{TalesovIzrekDolzine}) je: +\begin{eqnarray*} + \frac{x}{y}=\frac{XX_1}{YY_1}=\frac{XX_2}{YY_2} +\end{eqnarray*} + Če uporabimo slednje enakosti, nato +podobnost trikotnikov $AXX_1$ in $CYY_2$ oz. +trikotnikov $DXX_2$ in $BYY_1$ (izrek \ref{PodTrikKKK}) ter na koncu še potenco +točk $X$ in $Y$ glede na krožnico $k$ (izrek \ref{izrekPotenca}), dobimo: +\begin{eqnarray*} + \frac{x^2}{y^2}&=& \frac{|XX_1|}{|YY_1|}\cdot\frac{|XX_2|}{|YY_2|}= + \frac{|XX_1|}{|YY_2|}\cdot\frac{|XX_2|}{|YY_1|}=\\ + &=& \frac{|AX|}{|CY|}\cdot\frac{|DX|}{|BY|}=\frac{|AX|\cdot |DX|}{|CY|\cdot |BY|}=\\ + &=& \frac{|XP|\cdot |XQ|}{|YP|\cdot |YQ|}= + \frac{\left(|PS|-x\right)\cdot\left(|PS|+x\right)} + {\left(|PS|+y\right)\cdot\left(|PS|-y\right)}=\\ + &=& \frac{|PS|^2-x^2}{|PS|^2-y^2}. +\end{eqnarray*} + Iz $$\frac{x^2}{y^2}=\frac{|PS|^2-x^2}{|PS|^2-y^2}$$ na koncu sledi $x=y$. +\kdokaz + + + \bizrek \index{krožnica!Taylorjeva} \label{izrekTaylor} + Let $A'$, $B'$ and $C'$ be the foots of the altitudes of a triangle $ABC$. + The foot of the perpendiculars from the points $A'$, + $B'$ and $C'$ on the lines containing the adjacent sides + of that triangle lie on a circle, so-called \pojem{Taylor\footnote{\index{Taylor, B.} \textit{B. Taylor} (1685--1731), angleški matematik.} circle} \color{blue} of that triangle. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.5.pic} +\caption{} \label{sl.pod.7.16.5.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + Naj bosta $A_c$ in $A_b$ pravokotni projekciji točk $C'$ in $B'$ na premici $BC$, + $B_a$ in $B_c$ pravokotni projekciji točk $A'$ in $C'$ na premici $AC$ ter + $C_a$ in $C_b$ pravokotni projekciji točk $A'$ in $B'$ na premici $AB$ (Figure \ref{sl.pod.7.16.5.pic}). + + Trikotnik $A'B'C'$ je +pedalni trikotnik trikotnika $ABC$, zato iz dokaza izreka \ref{PedalniVS} sledi +$\angle B'A'C\cong\angle BAC=\alpha$ in $\angle C'B'A\cong\angle CBA=\beta$. Ker je $\angle B'B_aA'=\angle B'A_bA'=90^0$, je po izreku \ref{TalesovIzrKroz2} $A'A_bB_aB'$ tetivni štirikotnik, zato je tudi $\angle A_bB_aC\cong B'A'C=\alpha$ (izrek \ref{TetivniPogojZunanji}). Iz tega po izreku \ref{KotiTransverzala} sledi $A_bB_a \parallel BA$. + +Trikotnika $C'A_cB_c$ in $VA'B'$ sta perspektivna glede na točko $C$ (to pomeni, da se premice $C'V$, $A_cA'$ in $B_cB'$ sekajo v točki $C$). Ker je $C'A_c\parallel VA'$ in $C'B_c\parallel VB'$, je po posplošitvi Desarguesovega izreka \ref{izrekDesarguesOsNesk} tudi $A_cB_c\parallel A'B'$. Po izreku \ref{KotiTransverzala1} je potem $\angle B_cA_cA'\cong B'A'C=\alpha$. Ker je torej $\angle B_cA_cA'=\alpha=A_bB_aC$, je $A_cA_bB_aB_c$ tetivni štirikotnik (izrek \ref{TetivniPogojZunanji}); označimo njegovo očrtano krožnico s $k$. + +Ostane nam še dokaz, da tudi točki $C_a$ in $C_b$ ležita na +krožnici $k$. Analogno z dokazanim $A_cB_c\parallel A'B'$ je tudi $C_aB_a\parallel C'B'$ in analogno z dokazanim, da je $\angle A_bB_aC\cong B'A'C=\alpha$, je tudi $\angle C_aA_cB\cong\angle A'C'B=\gamma$. +Iz vzporednosti $C_aB_a\parallel C'B'$ sledi (izrek \ref{KotiTransverzala1}) +$\angle C_aB_aA \cong\angle C'B'A = \beta$, zato je: + $$\angle C_aB_aA_b=180^0-\angle A_bB_aC-\angle C_aB_aA= 180^0-\alpha-\beta=\gamma.$$ + Torej $\angle C_aA_cB\cong\angle C_aB_aA_b=\gamma$, kar pomeni (izrek \ref{TetivniPogojZunanji}), da je $C_aA_cA_bB_a$ tetivni štirikotnik, torej točka $C_a$ leži na krožnici $k$. Analogno dokažemo, da tudi točka $C_b$ leži na krožnici $k$. +\kdokaz + + A direct consequence is the following theorem. + + + + \bizrek + Let $A'$, $B'$ and $C'$ be the foots of the altitudes of a triangle $ABC$. + The foot of the perpendiculars from the points $A'$, + $B'$ and $C'$ on the lines containing the adjacent sides + of that triangle determine three congruent line segments. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.pod.7.16.6.pic} +\caption{} \label{sl.pod.7.16.6.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.16.6.pic}) + +V dokazu prejšnjega izreka \ref{izrekTaylor} smo ugotovili, da velja $A_bB_a\parallel BA$, zato je štirikotnik $C_bC_aA_bB_a$ trapez. Ta +trapez je hkrati tetiven (prejšnji izrek \ref{izrekTaylor}) in je zato tudi enakokrak (\ref{trapezTetivEnakokr}). Njegovi diagonali sta skladni (izrek \ref{trapezEnakokraki}), oz. velja +$A_bC_b\cong C_aB_a$. Analogno je tudi $A_bC_b\cong A_cB_c$. +\kdokaz + + + + \vspace*{12mm} + + +%________________________________________________________________________________ +\naloge{Exercises} + +\begin{enumerate} + + + +%Tales + +\item +Naj bo $S$ presečišče diagonal $AC$ in $BD$ trapeza $ABCD$. Naj bosta + $P$ in $Q$ presečišči vzporednice osnovnic $AB$ in $CD$ skozi točko $S$ s krakoma tega trapeza. Dokaži, da je $S$ središče daljice $PQ$. + +\item +Naj bo $ABCD$ trapez z osnovnico $AB$, točka $S$ presečišče njegovih +diagonal in $E$ presečišče nosilk krakov tega trapeza. Dokaži, da +premica $SE$ poteka skozi središči osnovnic $AB$ in $CD$. + +\item +Naj bodo $P$, $Q$ in $R$ točke, v katerih poljubna premica skozi točko $A$ seka +nosilke stranic $BC$ in $CD$ ter diagonalo $BD$ paralelograma +$ABCD$. Dokaži, da je $|AR|^2=|PR|\cdot |QR|$. + +\item + Točki $D$ in $K$ ležita na stranicah $BC$ in $AC$ trikotnika $ABC$ tako, da velja $BD:DC=2:5$ in $AK:KC=3:2$. Izračunaj razmerje, v katerem premica $BK$ deli daljico +$AD$. + +\item +Naj bo $P$ takšna točka stranice $AD$ paralelograma $ABCD$, da velja $\overrightarrow{AP}= +\frac{1}{n}\overrightarrow{AD}$, in $Q$ presečišče premic $AC$ in $BP$. Dokaži, da velja: + $$\overrightarrow{AQ}=\frac{1}{n + 1}\overrightarrow{AC}.$$ + + + +%Homotetija + +\item +Dani so: točka $A$, premici $p$ in $q$ ter daljici $m$ in $n$. Načrtaj premico $s$ skozi točko $A$, +ki naj seka premici $p$ in $q$ v takšnih točkah $X$ in $Y$, da velja $XA:AY= m:n$. + +\item Dani so: točka $S$, premice $p$, $q$ in $r$ ter daljici $m$ in $n$. Načrtaj premico $s$ skozi točko $S$, +ki naj seka premice $p$, $q$ in $r$ v takšnih točkah $X$, $Y$ in $Z$, da velja $XY:YZ=m:n$. + + \item V dani trikotnik $ABC$ včrtaj takšen pravokotnik $PQRS$, da stranica $PQ$ leži na stranici $BC$, +oglišči $R$ in $S$ ležita na stranicah $AB$ in $AC$, pri tem pa še velja $PQ=2QR$. + +\item +Načrtaj:\\ +(\textit{a}) romb, če sta dani stranica $a$ in razmerje diagonal $e:f$;\\ + (\textit{b}) trapez, če so dani: notranja kota $\alpha$ in $\beta$ ob eni osnovnici, razmerje te osnovnice in višine $a:v$ ter druga osnovnica $c$. + +\item +Naj bodo: $A$ točka v notranjosti kota $pSq$, $l$ premica in $\alpha$ kot v neki ravnini. Načrtaj takšen trikotnik +$APQ$, da velja: $P\in p$, $Q\in q$, $\angle PAQ\cong \alpha$ in +$PQ\parallel l$. + +\item +Načrtaj krožnico $l$, ki se dotika dane krožnice $k$ in dane premice $p$, če je dano še dotikališče:\\ +(\textit{a}) $l$ in $p$;\hspace*{3mm} +(\textit{b}) $l$ in $k$. + +\item +Načrtaj krožnico, ki se dotika krakov kota $pSq$ in:\\ +(\textit{a}) poteka skozi dano točko,\\ +(\textit{b}) se dotika dane krožnice. + +\item Naj bodo: $p$ in $q$ premici, $S$ točka ($S\notin p$ in $S\notin q$) ter daljici $m$ in $n$. +Načrtaj krožnici $k$ in $l$, ki se od zunaj dotikata v točki $S$, prva se dotika premice $p$, +druga premice $q$, razmerje polmerov pa je enako $m:n$. + + +\item +V ravnini so dani: premica $p$ ter točki $B$ in $C$, ki ležita na krožnici $k$. Načrtaj takšno točko $A$ +na krožnici $k$, da težišče trikotnika $ABC$ leži na premici $p$. + +\item +Naj bo $k$ krožnica s premerom $PQ$. Načrtaj kvadrat $ABCD$ tako, da velja $A,B\in PQ$ in $C,D\in k$. + +\item +V isti ravnini so dani: premici $p$ in $q$, točka $A$ ter daljici $m$ in $n$. Načrtaj takšen +pravokotnik $ABCD$, da velja $B\in p$, $D\in q$ +in $AB:AD=m:n$. + +\item V dani trikotnik $ABC$ včrtaj trikotnik tako, da bodo njegove stranice vzporedne z danimi premicami $p$, $q$ in $r$. + +\item Naj bodo $p$, $q$ in $r$ tri premice neke ravnine. Načrtaj premico $t$, ki je +pravokotna na premico $p$ in seka premice $p$, $q$ in $r$ po vrsti v takšnih točkah $P$, $Q$ in $R$, da velja $PQ\cong QR$. + + + + +%Podobnost trikotnikov + +\item Naj bo $P$ notranja točka trikotnika $ABC$ ter $A_1$, $B_1$ in $C_1$ pravokotne projekcije +točke $P$ na trikotnikovih stranicah $BC$, $AC$ in $AB$. Analogno so točke $A_2$, $B_2$ in $C_2$ določene s točko $P$ +in trikotnikom $A_1B_1C_1$,..., točke $A_{n+1}$, $B_{n+1}$ in $C_{n+1}$ s točko $P$ in trikotnikom $A_nB_nC_n$... Kateri od trikotnikov $A_1B_1C_1$, $A_2B_2C_2$, ... so podobni trikotniku $ABC$? + +\item +Naj bo $k$ očrtana krožnica štirikotnika $ABCD$, $E$ presečišče njegovih diagonal ter $CB\cong CD$. Dokaži, da je $\triangle ABC \sim\triangle BEC$. + +\item +Naj bo $ABCD$ paralelogram. V točkah $E$ in $F$ trikotniku $ABC$ očrtana krožnica seka premici $AD$ in $CD$. Dokaži, da je $\triangle EBC\sim\triangle EFD$. + +\item + Naj bosta $AA'$ in $BB'$ višini ostrokotnega trikotnika $ABC$. Dokaži, da je +$\triangle ABC\sim\triangle A'B'C$. + +\item +Naj bo nosilka višine $AD$ trikotnika $ABC$ hkrati tangenta očrtane krožnice tega trikotnika. Dokaži, da velja $|AD|^2=|BD|\cdot |CD|$. + +\item +V trikotniku $ABC$ naj bo notranji kot ob oglišču $A$ dvakrat večji +od notranjega kota ob oglišču $B$. Dokaži, da velja $|BC|^2= |AC|^2+|AC|\cdot |AB|$. + +\item Dokaži, da sta polmera očrtanih krožnic dveh podobnih trikotnikov sorazmerna z ustreznima stranicama teh dveh trikotnikov. + + + \item Krožnici s središčem $S$ je včrtan štirikotnik $ABCD$. Diagonali tega štirikotnika sta pravokotni in se sekata v točki $E$. Premica, ki poteka skozi točko $E$ in je pravokotna na stranici $AD$, seka stranico $BC$ v točki $M$. \\ +(\textit{a}) Dokaži, da je točka $M$ središče daljice $BC$.\\ +(\textit{b}) Določi množico vseh točk $M$, če se diagonala $BD$ spreminja in je vedno pravokotna na diagonalo $AC$. + +\item Naj bo $t$ tangenta očrtane krožnice $l$ trikotnika $ABC$ v oglišču $A$. +Naj bo $D$ takšna točka premice $AC$, da je $BD\parallel t$. Dokaži, da velja +$|AB|^2=|AC|\cdot |AD|$. + + \item Višinska točka ostrokotnega trikotnika naj deli njegovi višini v enakem razmerju (od oglišča do nožišča višine). Dokaži, da gre za enakokraki trikotnik. + +\item V trikotniku $ABC$ se višina $BD$ dotika očrtane krožnice tega trikotnika. +Dokaži:\\ +(\textit{a}) da je razlika kotov ob osnovnici $AC$ enaka $90^0$,\\ +(\textit{b}) da velja $|BD|^2=|AD|\cdot |CD|$. + + \item Krožnica s središčem na osnovnici $BC$ enakokrakega trikotnika $ABC$ se dotika +krakov $AB$ in $AC$. Točki $P$ in $Q$ sta presečišči teh krakov s poljubno +tangento te krožnice. Dokaži, da velja $4\cdot |PB|\cdot |CQ|=|BC|^2$. + +\item Naj bo $V$ višinska točka ostrokotnega trikotnika $ABC$, točka $V$ središče višine +$AD$, višino $BE$ pa točka $V$ deli v razmerju $3:2$. Izračunaj razmerje, v katerem $V$ deli višino $CF$. + +\item Naj bo $S$ zunanja točka krožnice $k$. $P$ in $Q$ +sta točki, v katerih se krožnica $k$ dotika svojih tangent iz točke $S$, $X$ in $Y$ pa presečišči te krožnice s poljubno premico, ki poteka skozi točko $S$. Dokaži da je $XP:YP=XQ:YQ$. + +\item Naj bo $D$ točka, ki leži na stranici $BC$ trikotnika $ABC$. Točki $S_1$ in $S_2$ naj bosta središči očrtanih krožnic trikotnikov $ABD$ in $ACD$. Dokaži, da velja + $\triangle ABC\sim\triangle AS_1S_2$. + +\item Točka $P$ leži na hipotenuzi $BC$ trikotnika $ABC$. Pravokotnica premice $BC$ v točki $P$ seka premici $AC$ in $AB$ v točkah $Q$ in $R$ ter očrtano krožnico trikotnika $ABC$ v točki $S$. Dokaži, da velja $|PS|^2=|PQ|\cdot |PR|$. + +\item Točka $A$ leži na kraku $OP$ pravega kota $POQ$. Naj bodo +$B$, $C$ in $D$ takšne točke kraka $OQ$, da velja $\mathcal{B}(O,B,C)$, $\mathcal{B}(B,C,D)$ in +$OA\cong OB\cong BC\cong CD$. Dokaži, da velja tudi $\triangle ABC\sim\triangle DBA$. + +\item Načrtaj trikotnik, če so znani naslednji podatki:\\ +(\textit{a}) $\alpha$, $\beta$, $R+r$, \hspace*{3mm} + (\textit{b}) $a$, $b:c$, $t_c-v_c$,\hspace*{3mm} + (\textit{c}) $v_a$, $v_b$, $v_c$. + +\item Naj bosta $AB$ in $CD$ osnovnici enakokrakega tangentnega trapeza $ABCD$, $r$ pa naj bo polmer včrtane krožnice. Dokaži, da je $|AB|\cdot |CD|=4r^2$. + + + + +%Harmon cetverica + +\item Dane so krožnica $k$ ter točki $A$ in $B$. Načrtaj takšno točko $X$ na krožnici $k$, da bo $AX:XB=2:5$. + +\item Načrtaj trikotnik z danimi podatki:\\ +(\textit{a}) $a$, $v_a$, $b:c$, \hspace*{3mm} + (\textit{b}) $a$, $t_a$, $b:c$,\hspace*{3mm} + (\textit{c}) $a$, $b$, $b:c$,\\ + (\textit{d}) $a$, $\alpha$, $b:c$,\hspace*{3mm} + (\textit{e}) $a$, $l_a$, $b:c$. + +\item Načrtaj trikotnik, če so dani naslednji podatki:\\ +(\textit{a}) $v_a$, $r$, $\alpha$, \hspace*{3mm} +(\textit{b}) $v_a$, $r_a$, $a$, \hspace*{3mm} +(\textit{c}) $v_a$, $t_a$, $b-c$. + +\item Načrtaj paralelogram, pri katerem sta ena stranica in ustrezna višina skladni z danima daljicama $a$ in $v_a$, diagonali pa sta v razmerju $3:5$. + +\item Točka $E$ naj bo presečišče simetrale notranjega kota $BAC$ trikotnika $ABC$ z njegovo stranico $BC$. Dokaži da velja: + $$\overrightarrow{AE}=\frac{|AC|}{|AB|+|AC|}\cdot\overrightarrow{AB}+ + \frac{|AB|}{|AB|+|AC|}\cdot\overrightarrow{AC}.$$ + +\item Dane so štiri kolinearne točke, za katere velja $\mathcal{H}(A,B;C,D)$. Načrtaj točko $L$, iz katere se daljice $AC$, $CB$ in $BD$ vidijo pod enakim kotom. + +\item Naj bo $AE$ ($E\in BC$) simetrala notranjega kota trikotnika $ABC$ ter $a=|BC|$, $b=|AC|$ in $c=|AB|$. Dokaži, da velja: +$$|BE|=\frac{ac}{b+c} \hspace*{1mm} \textrm{ in } \hspace*{1mm} |CE|=\frac{ab}{b+c}.$$ + +\item Naj bosta $AE$ ($E\in BC$) in $BF$ ($F\in AC$) simetrali notranjih kotov ter $S$ središče včrtane krožnice trikotnika $ABC$. Dokaži, da je $ABC$ enakokraki trikotnik (z osnovnico $AB$) natanko tedaj, ko je $AS:SE=BS:SF$. + + + + +%Menelaj Ceva + + +\item Dokaži, da simetrale zunanjih kotov poljubnega trikotnika sekajo nosilke nasprotnih stranic v treh kolinearnih točkah. + + + +%Ppitagorov iizrek + +\item Če so $a$, $b$ in $c$ ($a>b$) dane daljice, načrtaj takšno daljico $x$, da velja:\\ +(\textit{a}) $x=\sqrt{a^2+b^2}$, \hspace*{3mm} +(\textit{b}) $x=\sqrt{a^2-b^2}$, \hspace*{3mm} +(\textit{c}) $x=\sqrt{3ab}$,\\ +(\textit{d}) $x=\sqrt{a^2+bc}$, \hspace*{3mm} +(\textit{e}) $x=\sqrt{3ab-c^2}$, \hspace*{3mm} +(\textit{f}) $x=\frac{a\sqrt{ab+c^2}}{b+c}$. + + + +%Stewartov iizrek + +\item Naj bodo $a$, $b$ in $c$ stranice nekega trikotnika in velja $a^2+b^2=5c^2$. Dokaži, da sta težiščnici, +ki ustrezata stranicama $a$ in $b$, med seboj pravokotni. + +\item Naj bodo $a$, $b$, $c$ in $d$ stranice, $e$ in $f$ diagonali ter $x$ daljica, ki je določena s središčema + stranic $b$ in $d$ nekega štirikotnika. Dokaži: +$$x^2 = \frac{1}{4} \left(a^2 +c^2 -b^2 -d^2 +e^2 +f^2 \right).$$ + +\item Naj bodo $a$, $b$ in $c$ stranice trikotnika $ABC$. Dokaži, da je razdalja središča $A_1$ stranice $a$ od nožišča $A'$ višine na +to stranico enaka: +$$|A_1A'|=\frac{|b^2-c^2|}{2a}.$$ + + + +%Pappus in Ppascal + +\item Naj bosta ($A$, $B$, $C$) ter ($A_1$, $B_1$, $C_1$) trojici kolinearnih točk neke ravnine, ki nista na +isti premici. Če je $AB_1\parallel A_1B$ in $AC_1\parallel A_1C$, tedaj je tudi $CB_1\parallel C_1B$. (\textit{Pappusov izrek}\footnote{Pappus iz Aleksandrije\index{Pappus} (3. st.), starogrški matematik. Gre za posplošitev Pappusovega izreka (glej izrek \ref{izrek Pappus}), če za $X$, $Y$ in $Z$ izberemo točke v neskončnosti.}) + + + +%Desarguesov iizrek + + +\item Naj bodo $P$, $Q$ in $R$ takšne točke stranic $BC$, $AC$ in $AB$ +trikotnika $ABC$, da so premice $AP$, $BQ$ in +$CR$ iz istega šopa. Dokaži: Če je $X=BC\cap QR$, $Y=AC\cap PR$ in $Z=AB\cap PQ$, so točke +$X$, $Y$ in $Z$ +kolinearne. + + +\item Naj bodo $AA'$, $BB'$ in $CC'$ višine trikotnika $ABC$ ter $X=B'C'\cap BC$, $Y=A'C'\cap AC$ in $Z=A'B'\cap AB$. Dokaži, da so $X$, $Y$ in $Z$ kolinearne točke. + +\item Naj bosta $A$ in $B$ točki izven premice $p$. Načrtaj presečišče premic $p$ in $AB$ brez direktnega risanja premice +$AB$. + +\item Naj bosta $p$ in $q$ premici neke ravnine, ki se sekata v točki $S$, ki je ‘‘izven papirja’’, in $A$ +točka te ravnine. Načrtaj premico, ki poteka skozi točki $A$ in $S$. + +\item Načrtaj trikotnik tako, da njegova oglišča ležijo na treh danih vzporednih premicah, nosilke njegovih stranic pa gredo skozi tri dane +točke. + + + + +%Ppotenca + +\item +Dana je krožnica $k(S,r)$.\\ +(\textit{a}) Katere vrednosti vse lahko ima potenca točke glede na krožnico $k$?\\ + (\textit{b}) Katera je najmanjša vrednost te potence in za katero točko se ta minimalna vrednost doseže?\\ +(\textit{c}) Določi množico vseh točk, za katere je potenca glede na krožnico enaka $\lambda\in \mathbb{R}$. + +\item Naj bosta $k_a(S_a,r_a)$ in $l(O,R)$ pričrtana in očrtana krožnica nekega trikotnika. Dokaži enakost\footnote{Trditev je posplošitev Eulerjeve formule za krožnico (glej izrek \ref{EulerjevaFormula}). \index{Euler, L.} + \textit{L. Euler} + (1707--1783), švicarski matematik.}: + $$S_aO^2=R^2+2r_aR.$$ + + +\item Načrtaj krožnico, ki poteka skozi dani točki $A$ in $B$ in se dotika dane krožnice $k$. + +\item Dokaži, da so središča daljic, ki so določena s skupnimi tangentami dveh krožnic, kolinearne točke. + +\item Načrtaj krožnico, ki je pravokotna na dve dani krožnici, tretjo dano krožnico +pa seka v točkah, ki določata premer te tretje krožnice. + + + +%Raazno + +\item Naj bosta $M$ in $N$ presečišči stranic $AB$ in $AC$ trikotnika $ABC$ s premico, +ki poteka skozi središče včrtane krožnice tega trikotnika in je vzporedna z njegovo +stranico $BC$. Izrazi dolžino daljice $MN$ kot funkcijo dolžin stranic trikotnika $ABC$. + +\item Naj bo $AA_1$ težiščnica trikotnika $ABC$. Točki $P$ in $Q$ naj bosta presečišči +simetral kotov $AA_1B$ in $AA_1C$ s stranicami $AB$ in $AC$. Dokaži, da je +$PQ\parallel BC$. + +\item V trikotniku $ABC$ naj bo vsota (ali razlika) notranjih kotov $ABC$ in $ACB$ enaka pravemu +kotu. Dokaži, da je $|AB|^2+|AC|^2=4r^2$, kjer je $r$ polmer očrtane krožnice tega trikotnika. + +\item Naj bo $AD$ višina trikotnika $ABC$. Dokaži, da je +vsota (ali razlika) notranjih kotov $ABC$ in $ACB$ enaka pravemu kotu natanko tedaj, ko je: +$$\frac{1}{|AB|^2}+\frac{1}{|AC|^2}=\frac{1}{|AD|^2}.$$ + +\item Izrazi razdaljo med težiščem in središčem očrtane krožnice trikotnika kot +funkcijo dolžin njegovih stranic in polmera očrtane krožnice. + +\item Dokaži, da pri trikotniku $ABC$ simetrala zunanjega kota ob oglišču $A$ in simetrali +notranjih kotov ob ogliščih $B$ in $C$ sekajo nosilke nasprotnih stranic v treh kolinearnih točkah. + +\item Dokaži, da so pri trikotniku $ABC$ središče višine $AD$, središče včrtane krožnice in točka, v kateri se stranica $BC$ dotika pričrtane krožnice tega trikotnika, tri +kolinearne točke. + +\item Dokaži Simsonov izrek \ref{SimpsPrem} z uporabo Menelajevega izreka \ref{izrekMenelaj}. + +\item Skozi točko $M$ stranice $AB$ trikotnika $ABC$ je konstruirana premica, ki seka +premico $AC$ v točki $K$. Izračunaj razmerje, v katerem premica $MK$ deli stranico $BC$, +če je $AM:MB=1:2$ in $AK:AC=3:2$. + +\item Naj bo $A_1$ središče stranice $BC$ trikotnika $ABC$ in naj bosta $P$ in $Q$ takšni točki stranic +$AB$ in $AC$, da velja $BP:PA=2:5$ in $AQ:QC=6:1$. Izračunaj razmerje, v katerem premica $PQ$ deli težiščnico $AA_1$. + +\item Dokaži, da se pri poljubnem trikotniku premice, ki so določene z oglišči in +dotikališči ene pričrtane krožnice z nosilkami nasprotnih stranic, sekajo v skupni točki. + +\item Kaj predstavlja množica vseh točk, iz katerih odseka tangent glede na dve dani krožnici predstavljata dve skladni daljici? + +\item Naj bosta $PP_1$ in $QQ_1$ zunanji tangenti krožnic $k(O,r)$ in $k_1(O_1,r_1)$ (točke $P$, $P_1$, $Q$ in $Q_1$ so ustrezna dotikališča). Točka $S$ naj bo presečišče teh dveh tangent, $A$ eno od presečišč krožnic $k$ in $k_1$ ter $L$ in $L_1$ presečišči +premice $SO$ s premicama $PQ$ in $P_1Q_1$. Dokaži, da velja $\angle LAO\cong\angle L_1AO_1$. + +\item Dokaži, da je stranica pravilnega desetkotnika enaka +večjem delu delitve polmera očrtane krožnice tega desetkotnika v razmerju zlatega reza. + +\item Naj bodo $a_5$, $a_6$ in $a_{10}$ stranice pravilnega petkotnika, šestkotnika in desetkotnika, ki so včrtani isti krožnici. Dokaži, da velja: + $$a_5^2=a_6^2+a_{10}^2.$$ + + +\item Naj bodo $t_a$, $t_b$ in $t_c$ težiščnice in $s$ polobseg nekega trikotnika. Dokaži, da velja: + $$t_a^2+t_b^2+t_c^2\geq s^2.$$ % zvezek - dodatni MG + + +\end{enumerate} + + + + +% DEL 8 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% PLOŠČINA +%________________________________________________________________________________ + + \del{Area of Figures} \label{pogPLO} + + +%________________________________________________________________________________ +\poglavje{Area of Figures. Definition} \label{odd8PloLik} + + V tem razdelku bomo definirali pojem ploščine določenega razreda likov. Ploščina temelji na pojmu dolžine daljice v nekem sistemu merjenja in na teoriji infinitezimalnega računa\footnote{Prve korake v tej smeri sta naredila starogrška matematika \index{Evdoks}\textit{Evdoks} (408--355 pr. n. š.) in \index{Arhimed}\textit{Arhimed} (287--212 pr. n. š.) pri računanju prostornine teles.} + +Naj bo $\Omega_0$ kvadrat s stranico dolžine $1$ v določenem sistemu merjenja. Omenjeni kvadrat predstavlja \index{enota merjenja ploščine}\pojem{enoto merjenja ploščine} - imenujemo ga tudi \index{enotski kvadrat}\pojem{enotski kvadrat}. Intuitivno ploščina nekega lika $\Phi$ predstavlja število enotskih kvadratov, ki so skladni kvadratu $\Omega_0$ in s katerimi lahko ''prekrijemo'' lik $\Phi$. + +Sedaj bomo bolj formalno pristopili k definiranju ploščine. Naj bo $\Phi$ poljubni lik v ravnini. Enotski kvadrat $\Omega_0$ določa tlakovanje $(4,4)$ (glej razdelek \ref{odd3Tlakovanja}) - označimo ga s $\mathcal{T}_0$. S $\underline{S}_0$ označimo število kvadratov tlakovanja $\mathcal{T}_0$, ki ležijo v liku $\Phi$, oz. so njegova podmnožica. +S $\overline{S}_0$ pa označimo število kvadratov tlakovanja $\mathcal{T}_0$, ki imajo z likom $\Phi$ vsaj eno skupno točko (Figure \ref{sl.plo.8.1.1.pic}). Jasno je, da velja: + $$0\leq\underline{S}_0\leq\overline{S}_0.$$ + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.1.1.pic} +\caption{} \label{sl.plo.8.1.1.pic} +\end{figure} + + +Če vsako stranico kvadrata razdelimo na $10$ daljic, lahko enotski kvadrat $\Omega_0$ razdelimo na $10^2$ skladnih kvadratov, ki so vsi skladni enemu od teh kvadratov $\Omega_1$, ki ima stranico dolžine $\frac{1}{10}$. Kvadrat $\Omega_1$ določa novo tlakovanje $\mathcal{T}_1$. Podobno kot v prejšnjem primeru naj bo $\underline{S}_1$ število kvadratov tlakovanja $\mathcal{T}_1$, ki ležijo v liku $\Phi$, in $\overline{S}_1$ število kvadratov tlakovanja $\mathcal{T}_1$, ki imajo z likom $\Phi$ vsaj eno skupno točko. Jasno je, da velja $\underline{S}_1\leq\overline{S}_1$ in še: + $$0\leq\underline{S}_0\leq\frac{\underline{S}_1}{10^2} +\leq\frac{\overline{S}_1}{10^2}\leq\overline{S}_0.$$ + +Če postopek nadaljujemo, dobimo zaporedje kvadratov $\Omega_n$, ki imajo stranico dolžine $\frac{1}{10^n}$, tlakovanja $\mathcal{T}_n$ in pare števil $\underline{S}_n$ in $\overline{S}_n$, za katere velja: +$$0\leq\underline{S}_0\leq\frac{\underline{S}_1}{10^2}\leq \cdots \leq \frac{\underline{S}_n}{10^{2n}}\leq\cdots\leq \frac{\overline{S}_n}{10^{2n}}\leq\cdots +\leq\frac{\overline{S}_1}{10^2}\leq\overline{S}_0.$$ + +Zaporedje $\frac{\underline{S}_n}{10^{2n}}$ je naraščajoče in navzgor omejeno, zato je po znanem izreku matematične analize konvergentno in ima svojo limito: +$$\underline{S}=\lim_{n\rightarrow\infty}\frac{\underline{S}_n}{10^{2n}}.$$ + +Podobno je zaporedje $\frac{\overline{S}_n}{10^{2n}}$ padajoče in navzdol omejeno, zato je konvergentno in ima svojo limito: +$$\overline{S}=\lim_{n\rightarrow\infty}\frac{\overline{S}_n}{10^{2n}}.$$ + +Če je $\underline{S}=\overline{S}$, pravimo, da je lik $\Phi$ merljiv. Število $S=\underline{S}=\overline{S}$ pa je njegova \index{ploščina lika}\pojem{ploščina}. Označimo jo z $p(\Phi)$ oz. $p_{\Phi}$. + +Intuitivno je jasno, da je ploščina lika $\Phi$ ni odvisna od lege enotskega kvadrata - torej je odvisna le od sistema merjenja daljic, v katerem ima stranica enotskega kvadrata dolžino 1. Na tem mestu tega dejstva ne bomo formalno dokazovali. + +Dokažimo prvo pomembno lastnost ploščine. + + + + \bizrek \label{ploscDaljice} + The area of an arbitrary point is 0. + The area of an arbitrary line segment is 0. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.1.2.pic} +\caption{} \label{sl.plo.8.1.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} + +\textit{1)} Naj bo $A$ poljubna točka. Jasno je, da za vsak $n\in \mathbb{N}$ velja: $\underline{S}_n=0$ in $0\leq\overline{S}_n\leq4$ (točka leži v največ štirih kvadratih tlakovanja $T_n$). Torej velja: +$$\underline{S}=\lim_{n\rightarrow\infty}\frac{\underline{S}_n}{10^{2n}} +=\lim_{n\rightarrow\infty}\frac{0}{10^{2n}}=0$$ +in +$$0\leq\overline{S}=\lim_{n\rightarrow\infty}\frac{\overline{S}_n}{10^{2n}}\leq +\lim_{n\rightarrow\infty}\frac{4}{10^{2n}}=0,$$ +zato je $p_A=S=\underline{S}=\overline{S}=0$ + +\textit{2)} Naj bo $AB$ poljubna daljica (Figure \ref{sl.plo.8.1.2.pic}) in $|AB|=d$. S $k$ označimo celi del števila $d$, oz. $k=[d]$. Izberimo enotski kvadrat $\Omega_0$, tako da je eno njegovo oglišče točka $A$, ena njegova stranica pa leži na daljici $AB$. V tem primeru je $\underline{S}_n=0$ in $\overline{S}_n\leq (k+1)\cdot 10^n$. Tedaj je: +$$\underline{S}=\lim_{n\rightarrow\infty}\frac{\underline{S}_n}{10^{2n}} +=\lim_{n\rightarrow\infty}\frac{0}{10^{2n}}=0$$ +in +$$0\leq\overline{S}=\lim_{n\rightarrow\infty}\frac{\overline{S}_n}{10^{2n}}\leq +\lim_{n\rightarrow\infty}\frac{(k+1)10^n}{10^{2n}}= +\lim_{n\rightarrow\infty}\frac{k+1}{10^n}=0,$$ +zato je $p_{AB}=S=\underline{S}=\overline{S}=0.$ +\kdokaz + +Naslednji izrek, ki se nanaša na osnovne lastnosti ploščine, bomo podali brez dokaza (glej \cite{Lucic}). + + \bizrek \label{ploscGlavniIzrek} + Naj bo $p$ ploščina, definirana na množici $\mu$ merljivih likov + ravnine. + Potem velja: + \begin{enumerate} + \item $p(\Omega_0)=1$; + \item $(\forall \Phi\in\mu)\hspace*{1mm}p(\Phi)\geq 0$; + \item $(\forall \Phi_1,\Phi_2\in\mu)\hspace*{1mm} + \left(\Phi_1\cong\Phi_2 + \hspace*{1mm}\Rightarrow\hspace*{1mm}p(\Phi_1)=p(\Phi_2)\right)$; + \item $(\forall \Phi_1,\Phi_2\in\mu)\hspace*{1mm} + \left(p(\Phi_1\cap\Phi_2)=0 + \hspace*{1mm}\Rightarrow\hspace*{1mm}p(\Phi_1\cup\Phi_2)= + p(\Phi_1)+p(\Phi_2)\right)$. + \end{enumerate} + \eizrek + +Zelo koristen je naslednji izrek. + + + \bizrek \label{ploscLomljenke} + Ploščina poljubne lomljenke je enaka 0. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.1.3.pic} +\caption{} \label{sl.plo.8.1.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.1.3.pic}) + +Trditev je direktna posledica izrekov \ref{ploscDaljice} in \ref{ploscGlavniIzrek} (\textit{4}). +\kdokaz + + + +%________________________________________________________________________________ + \poglavje{Area of Rectangles, Parallelogram, and Trapezoids} \label{odd8PloParalel} + + + Na tem mestu bomo izpeljali formule za ploščino določenih štirikotnikov. + + \bizrek \label{ploscPravok} + Če je $p$ ploščina pravokotnika s stranicama dolžin $a$ in $b$, potem + je: + $$p=ab.$$ + \eizrek + + + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.1a.pic} +\caption{} \label{sl.plo.8.2.1a.pic} +\end{figure} + + + \textbf{\textit{Proof.}} +Naj bo $ABCD$ pravokotnik s stranicama $AB$ in $AD$ dolžin $|AB|=a$ in $|AD|=b$ (Figure \ref{sl.plo.8.2.1a.pic}). +Izberimo enotski kvadrat $\Omega_0$, tako da je eno njegovo oglišče točka $A$, ena njegova stranica pa leži na daljici $AB$. Naj bodo $\mathcal{T}_n$ ($n\in \mathbb{N}\cup\{0\}$) ustrezna tlakovanja, ki jih določa enotni kvadrat $\Omega_0$. + +Označimo z $a_n=[a\cdot 10^n]$ in $b_n=[b\cdot 10^n]$ (kjer $[x]$ predstavlja celi del števila $x$). Najprej je: + \begin{eqnarray} \label{eqnPloscPrav1a} + \frac{a_n}{10^n}\leq a\leq \frac{a_n+1}{10^n} \hspace*{2mm} \textrm{ in } \hspace*{2mm} +\frac{b_n}{10^n}\leq b\leq \frac{b_n+1}{10^n}, + \end{eqnarray} +nato pa še: + \begin{eqnarray} \label{eqnPloscPrav2a} + \underline{S}_n=\frac{a_n\cdot b_n}{10^{2n}} \hspace*{2mm} \textrm{ in } \hspace*{2mm} +\overline{S}_n=\frac{\left(a_n+1\right)\cdot \left(b_n+1\right)}{10^{2n}}. + \end{eqnarray} +Iz \ref{eqnPloscPrav2a} in \ref{eqnPloscPrav1a} dobimo: +\begin{eqnarray*} +0&\leq&\lim_{n\rightarrow\infty}\left(\overline{S}_n-\underline{S}_n\right)=\\ +&=&\lim_{n\rightarrow\infty}\left(\frac{\left(a_n+1\right)\cdot \left(b_n+1\right)}{10^{2n}}-\frac{a_n\cdot b_n}{10^{2n}}\right)=\\ +&=& \lim_{n\rightarrow\infty}\frac{a_n+ b_n+1}{10^{2n}}\leq\\ +&\leq& \lim_{n\rightarrow\infty}\frac{a+b+1}{10^n}=\\ +&=& 0. + \end{eqnarray*} +Torej: +\begin{eqnarray*} +\lim_{n\rightarrow\infty}\left(\overline{S}_n-\underline{S}_n\right)=0, + \end{eqnarray*} +zato je $\underline{S}=\lim_{n\rightarrow\infty}\underline{S}_n +=\lim_{n\rightarrow\infty}\overline{S}_n=\overline{S}$, kar pomeni, da je pravokotnik $ABCD$ merljiv lik s ploščino $p=S=\underline{S}=\overline{S}$. + +Dokažimo še $p=ab$. Iz \ref{eqnPloscPrav1a} dobimo: + \begin{eqnarray*} + \frac{a_n\cdot b_n}{10^{2n}} \leq ab\leq \frac{\left(a_n+1\right)\cdot \left(b_n+1\right)}{10^{2n}}, + \end{eqnarray*} +Ker to velja za vsak $n\in \mathbb{N}$, sledi: + \begin{eqnarray*} + \hspace*{-2mm} \underline{S}=\lim_{n\rightarrow\infty}\underline{S}_n= +\lim_{n\rightarrow\infty}\frac{a_n\cdot b_n}{10^{2n}} \leq ab\leq \lim_{n\rightarrow\infty}\frac{\left(a_n+1\right)\cdot \left(b_n+1\right)}{10^{2n}}=\lim_{n\rightarrow\infty}\overline{S}_n=\overline{S}. + \end{eqnarray*} +Iz $p=S=\underline{S}=\overline{S}$ na koncu sledi $p=ab$. +\kdokaz + +Direktna posledica prejšnjega izreka \ref{ploscPravok} je naslednja trditev (Figure \ref{sl.plo.8.2.2.pic}). + + \bizrek \label{ploscKvadr} + Če je $p_{\square}$ ploščina kvadrata s stranico dolžine $a$, potem je: + $$p_{\square}=a^2.$$ + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.2.pic} +\caption{} \label{sl.plo.8.2.2.pic} +\end{figure} + +Izpeljimo formule za ploščino še za nekatere štirikotnike. + + + \bizrek \label{ploscParal} + Če je $p$ ploščina paralelograma s stranicama dolžin $a$ in $b$ ter + pripadajočima višinama dolžin $v_a$ in $v_b$, potem + je: + $$p=av_a=bv_b.$$ + \eizrek + + + \textbf{\textit{Proof.}} + Naj bo $ABCD$ paralelogram s stranico $AB$ dolžine $|AB|=a$ in pripadajočo višino $v_a$. Označimo z $E$ in $F$ pravokotni projekciji oglišč $D$ in $C$ na premici $AB$. Štirikotnik $EFCD$ je pravokotnik s stranicama $|CD|=a$ in $|FC|=v_a$, zato je po izreku \ref{ploscPravok}: +\begin{eqnarray} +p_{EFCD}=av_a. \label{eqnPloscPrav2} +\end{eqnarray} +Pravokotna trikotnika $AED$ in $BFC$ sta skladna (izrek \textit{ASA} \ref{KSK}) zato je po izreku \ref{ploscGlavniIzrek} \textit{3)}: +\begin{eqnarray} +p_{AED}=p_{BFC}. \label{eqnPloscPrav1} +\end{eqnarray} +Brez škode za splošnost predpostavimo, da je $\angle BAD\leq 90^0$. V tem primeru sta točki $B$ in $E$ na isti strani točke $A$ (v primeru $E=A$ gre za pravokotnik $ABCD$ in trditev sledi direktno iz izreka \ref{ploscPravok}). Obravnavali bomo več možnih primerov: $\mathcal{B}(A,E,B)$, $E=B$ in $\mathcal{B}(A,B,E)$. + +\textit{1)} Naj bo $\mathcal{B}(A,E,B)$ (Figure \ref{sl.plo.8.2.3.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.3.pic} +\caption{} \label{sl.plo.8.2.3.pic} +\end{figure} + +Če uporabimo relaciji \ref{eqnPloscPrav2} in \ref{eqnPloscPrav1} ter izreka \ref{ploscGlavniIzrek} \textit{4)} in \ref{ploscDaljice}, dobimo: + \begin{eqnarray*} + p_{ABCD}=p_{AED}+p_{EBCD}=p_{BFC}+p_{EBCD}=p_{EFCD}=av_a. +\end{eqnarray*} + +\textit{2)} V primeru $E=B$ (Figure \ref{sl.plo.8.2.3a.pic}) podobno iz relacij \ref{eqnPloscPrav1} in \ref{eqnPloscPrav1} ter izrekov \ref{ploscGlavniIzrek} \textit{4)} in \ref{ploscDaljice} dobimo: + \begin{eqnarray*} + p_{ABCD}=p_{AED}+p_{BCD}=p_{BFC}+p_{BCD}=p_{EFCD}=av_a. +\end{eqnarray*} + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.3a.pic} +\caption{} \label{sl.plo.8.2.3a.pic} +\end{figure} + + + +\textit{3)} Predpostavimo, da velja $\mathcal{B}(A,B,E)$ (Figure \ref{sl.plo.8.2.3b.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.3b.pic} +\caption{} \label{sl.plo.8.2.3b.pic} +\end{figure} + +Daljici $BC$ in $DE$ se potem sekata v neki točki $L$. Po izrekih \ref{ploscGlavniIzrek} \textit{4)} in \ref{ploscDaljice} je: +$p_{AED}=p_{ABLD}+p_{BEL}$ in $p_{BFC}=p_{EFCL}+p_{BEL}$. Ker je po relaciji \ref{eqnPloscPrav1} $p_{AED}=p_{BFC}$, je tudi: + \begin{eqnarray*} + p_{ABLD}=p_{EFCL}. +\end{eqnarray*} +Iz tega in iz izrekov \ref{ploscGlavniIzrek} \textit{4)} in \ref{ploscDaljice} sledi: + \begin{eqnarray*} + p_{ABCD}=p_{ABLD}+p_{DLC}=p_{EFCL}+p_{DLC}=p_{EFCD}=av_a, +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + \bizrek \label{ploscTrapez} + Če je $p$ ploščina trapeza z osnovnicama dolžin $a$ in $c$ ter + višino dolžine $v$, potem + je: + $$p=\frac{a+c}{2}\cdot v.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.4.pic} +\caption{} \label{sl.plo.8.2.4.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Naj bo $ABCD$ trapez z osnovnicama $AB$ in $CD$ dolžin $|AB|=a$ in $|CD|=c$ ter višino dolžine $v$ + (Figure \ref{sl.plo.8.2.4.pic}). + S $S$ označimo središče kraka $BC$ in $\mathcal{S}_S:\hspace*{1mm}A,D\mapsto A', D'$. Ker je $S$ skupno središče daljic $AA'$ in $DD'$, je po izreku \ref{paralelogram} $AD'A'D$ paralelogram. Ker je še $\mathcal{S}_S(B)=C$ in $\mathcal{B}(A,B,D')$, je tudi $\mathcal{B}(A',C,D)$. Izometrija $\mathcal{S}$ preslika trapez $ABCD$ v skladni trapez $A'CBD'$, zato je $|BD'|=|CD|=c$ oz. $|AD'|=a+c$, in po izreku \ref{ploscGlavniIzrek} \textit{3)} še $p_{ABCD}=p_{A'CBD'}$. Torej je paralelogram $AD'A'D$ z osnovnico $AD'$ dolžine $|AD'|=a+c$ in višino, ki je enaka višini trapeza $ABCD$ dolžine $v$, razdeljen na dva skladna trapeza $ABCD$ in $A'CBD'$ z enakima ploščinama. + +Iz tega in izrekov \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscDaljice} in \ref{ploscParal} sledi: + \begin{eqnarray*} + 2\cdot p_{ABCD}=p_{ABCD}+p_{A'CBD'}=p_{AD'A'D}=\left(a+c\right)v, +\end{eqnarray*} +oz. iskana relacija. +\kdokaz + + +Če uporabimo izrek \ref{srednjTrapez}, vidimo, da izraz $\frac{a+c}{2}$ predstavlja dolžino srednjice trapeza, zato za ploščino trapeza velja tudi formula: + $$p=mv,$$ +kjer je $m$ dolžina srednjice trapeza in $v$ njegova višina + (Figure \ref{sl.plo.8.2.4a.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.4a.pic} +\caption{} \label{sl.plo.8.2.4a.pic} +\end{figure} + + + + + \bzgled \label{ploscStirikPravok} + Če je $p$ ploščina štirikotnika s pravokotnima diagonalama dolžin $e$ in + $f$, potem je: + $$p=\frac{ef}{2}.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.5.pic} +\caption{} \label{sl.plo.8.2.5.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Naj bo $ABCD$ štirikotnik s pravokotnima diagonalama $AC$ in $BD$ dolžin $|AC|=e$ in $|BD|=f$ (Figure \ref{sl.plo.8.2.5.pic}). Vzporednice teh diagonal skozi oglišča štirikotnika $ABCD$ določajo pravokotnik $A'B'C'D'$ s stranicama dolžin $|A'B'|=f$ in $|B'C'|=e$. Iz skladnosti trikotnikov $ASD$ in $DA'A$ (izrek \textit{SAS} \ref{SKS}) sledi $p_{ASD}=p_{DA'A}$ (izrek \ref{ploscGlavniIzrek} \textit{3)}). Analogno je tudi $p_{ASB}=p_{BB'A}$, $p_{CSB}=p_{BC'C}$ in $p_{CSD}=p_{DD'C}$. Iz tega in iz izrekov \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscDaljice} in \ref{ploscPravok} sledi: + \begin{eqnarray*} + 2\cdot p_{ABCD}&=&2\cdot\left(p_{ASD}+p_{ASB}+p_{CSB}+p_{CSD}\right)=\\ +&=&2\cdot p_{ASD}+2\cdot p_{ASB}+2\cdot p_{CSB}+2\cdot p_{CSD}=\\ +&=& p_{ASD}+ p_{DA'A}+ p_{ASB}+p_{BB'A}+\\ + && + p_{CSB}+p_{BC'C} +p_{CSD}+p_{DD'C}=\\ +&=& p_{A'B'C'D'}= ef, +\end{eqnarray*} + +oz. iskana relacija. +\kdokaz + + +Kot posledico imamo naslednji trditvi (Figure \ref{sl.plo.8.2.5a.pic}). + + + \bzgled \label{ploscDeltoid} + Če je $p$ ploščina deltoida z diagonalama dolžin $e$ in + $f$, potem je: + $$p=\frac{ef}{2}.$$ + \ezgled + + \textbf{\textit{Proof.}} + + Trditev je direktna posledica izreka \ref{ploscStirikPravok} in definicije deltoida. +\kdokaz + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.5a.pic} +\caption{} \label{sl.plo.8.2.5a.pic} +\end{figure} + + + \bzgled \label{ploscRomb} + Če je $p$ ploščina romba z diagonalama dolžin $e$ in + $f$, potem je: + $$p=\frac{ef}{2}.$$ + \ezgled + + \textbf{\textit{Proof.}} + Trditev je direktna posledica izrekov \ref{ploscStirikPravok} in \ref{RombPravKvadr}. +\kdokaz + + + + \bzgled \label{ploscRomb1} + Če so: $a$ dolžina stranice, $v$ višina ter $e$ in + $f$ dolžini diagonal romba, potem je: + $$av=\frac{ef}{2}.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.2.6.pic} +\caption{} \label{sl.plo.8.2.6.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.2.5a.pic}) + + Trditev je direktna posledica izrekov \ref{ploscRomb1} in \ref{ploscParal}. +\kdokaz + + +%________________________________________________________________________________ + \poglavje{Area of Triangles} \label{odd8PloTrik} + +V tem razdelku bomo izpeljali več formul za ploščino trikotnika. + + \bizrek \label{PloscTrik} Če je $p_\triangle$ ploščina + trikotnika $ABC$, $v_a$, $v_b$ in $v_c$ dolžine višin, ki ustrezajo stranicam $BC$, $AC$ in $AB$ z dolžinami $a$, $b$ in $c$, potem velja: + $$p_\triangle=\frac{a\cdot v_a}{2}=\frac{b\cdot v_b}{2} + =\frac{c\cdot v_c}{2}.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.1a.pic} +\caption{} \label{sl.plo.8.3.1a.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $D$ četrto oglišče paralelograma $ABCD$ (Figure \ref{sl.plo.8.3.1a.pic}). Trikotnika $ABC$ in $ADC$ sta skladna (izreka \ref{paralelogram} in \ref{SSS}), zato je $p_{ABC}=p_{ADC}$ (izrek \ref{ploscGlavniIzrek} \textit{3)}). Paralelogram $ABCD$ ima stranico $BC$ in pripadajočo višino dolžin $a$ in $v_a$, zato je po izreku \ref{ploscParal} $p_{ABCD}=av_a$. Če uporabimo še izreka \ref{ploscGlavniIzrek} \textit{4)} in \ref{ploscDaljice}, dobimo: + \begin{eqnarray*} + 2\cdot p_{ABC}=p_{ABC}+p_{ADC}=p_{ABCD}=av_a, +\end{eqnarray*} +oz. iskano relacijo $p_\triangle=\frac{a\cdot v_a}{2}$. Analogno je tudi $p_\triangle=\frac{b\cdot v_b}{2}$ oz. $p_\triangle=\frac{c\cdot v_c}{2}$. +\kdokaz + + \bizrek \label{PloscTrikVcrt} Če je $p_\triangle$ ploščina + trikotnika $ABC$ s + polobsegom $s=\frac{a+b+c}{2}$ in polmerom včrtane krožnice + $r$, potem velja: + $$p_\triangle=sr.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.1.pic} +\caption{} \label{sl.plo.8.3.1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Naj bo $S$ središče včrtane krožnice trikotnika $ABC$ (Figure \ref{sl.plo.8.3.1.pic}). + +Po izrekih \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscDaljice} in \ref{PloscTrik} je: +\begin{eqnarray*} + p_{ABC}=p_{SBC}+p_{ASC}+p_{ABS}&=& + \frac{a\cdot r}{2}+\frac{b\cdot r}{2}+\frac{c\cdot r}{2}=\\ +&=& + \frac{a+b+c}{2}\cdot r=sr, +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + + \bizrek \label{PloscTrikPricrt} Če je $p_\triangle$ + ploščina trikotnika $ABC$ s + polobsegom $s=\frac{a+b+c}{2}$ in polmerom pričrtane krožnice + $r_a$, potem velja: + $$p_\triangle=(s-a)r_a.$$ + \eizrek + + + + \textbf{\textit{Proof.}} Uporabimo oznake iz velike naloge \ref{velikaNaloga} (Figure \ref{sl.plo.8.3.2.pic}). Iz iste naloge je: $AR=s-a$ in $AR_a=s$. + +Pravokotna trikotnika $ARS$ in $AR_aS_a$ sta si podobna (izrek \ref{PodTrikKKK}), zato je $\frac{SR}{S_aR_a}=\frac{AR}{AR_a}$, oz. $\frac{r}{r_a}=\frac{s-a}{s}$. Če uporabimo še prejšnji izrek \ref{PloscTrikVcrt}, dobimo: +\begin{eqnarray*} + p_{ABC}=sr=(s-a)r_a, +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.2.pic} +\caption{} \label{sl.plo.8.3.2.pic} +\end{figure} + + + \bizrek \label{PloscTrikHeron} Če je $p_\triangle$ + ploščina trikotnika $ABC$ s + polobsegom $s=\frac{a+b+c}{2}$, potem velja + \index{formula!Heronova} + (Heronova\footnote{\index{Heron} + \textit{Heron iz Aleksandije}(20--100), + starogrški matemetik.} formula): + $$p_\triangle=\sqrt{s(s-a)(s-b)(s-c)}.$$ + \eizrek + + + \textbf{\textit{Proof.}} Tudi v tem primeru uporabimo oznake iz velike naloge \ref{velikaNaloga} (Figure \ref{sl.plo.8.3.2.pic}). Iz iste naloge je: $BP=s-b$ in $BP_a=s-c$. + +Kota $SBP$ in $BS_aP_a$ sta skladna, ker imata paroma pravokotna kraka (izrek \ref{KotaPravokKraki}). Torej sta si pravokotna trikotnika $SBP$ in $BS_aP_a$ podobna (izrek \ref{PodTrikKKK}), zato je $\frac{SP}{BP_a}=\frac{BP}{S_aP_a}$ oz. $\frac{r}{s-c}=\frac{s-b}{r_a}$ oz. $rr_a=(s-b)(s-c)$. + Če uporabimo še izreka \ref{PloscTrikVcrt} in \ref{PloscTrikPricrt}, dobimo: +\begin{eqnarray*} +p_{ABC}^2=sr(s-a)r_a=s(s-a)rr_a=s(s-a)(s-b)(s-c), +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + + \bizrek \label{PloscTrikOcrt} Če je $p_\triangle$ + ploščina trikotnika $ABC$ z + dolžinami stranic $a$, $b$ in $c$ ter $R$ polmer očrtane krožnice, + potem velja: + $$p_\triangle=\frac{abc}{4R}.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.3.pic} +\caption{} \label{sl.plo.8.3.3.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.3.3.pic}) + + Trditev je direktna posledica izrekov \ref{PloscTrik} in \ref{izrekSinusni}: +\begin{eqnarray*} +p_{ABC}=\frac{av_a}{2}=\frac{a}{2}\cdot v_a=\frac{a}{2}\cdot \frac{bc}{2R}=\frac{abc}{4R}, +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + +Prejšnje formule za ploščino trikotnika lahko uporabimo za računanje polmerov očrtane, včrtane in pričrtanih krožnic trikotnika kot funkcije njegovih stranic. + + \bzgled \label{PloscTrikOcrtVcrt} Če sta $R$ in $r$ polmera + očrtane in včrtane krožnice, $r_a$, $r_b$ in $r_c$ polmeri pričrtanih krožnic ter + $s=\frac{a+b+c}{2}$ polobseg + trikotnika $ABC$, + potem velja: + + (i) $R=\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$,\hspace*{7mm} + (ii) $r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$,\\ + (iii) $r_a=\sqrt{\frac{s(s-b)(s-c)}{s-a}}$,\hspace*{2.7mm} + (iv) $r_b=\sqrt{\frac{(s-a)s(s-c)}{s-b}}$,\hspace*{2.7mm} + (v) $r_c=\sqrt{\frac{(s-a)(s-b)s}{s-c}}$. + + \ezgled + + + + \textbf{\textit{Proof.}} + Trditev je direktna posledica izrekov \ref{PloscTrikOcrt}, \ref{PloscTrikVcrt}, \ref{PloscTrikPricrt} in \ref{PloscTrikHeron}. +\kdokaz + + + \bzgled \label{PloscTrikVcrtPricrt} Če je $r$ polmer + včrtane krožnice, $r_a$, $r_b$ in $r_c$ polmeri pričrtanih krožnic ter $p_{\triangle}$ ploščina trikotnika $ABC$, + potem velja: + $$p_{\triangle}=\sqrt{rr_ar_br_c}.$$ + \ezgled + + + \textbf{\textit{Proof.}} + Če uporabimo trditve iz prejšnjega zgleda \ref{PloscTrikOcrtVcrt}, dobimo: + \begin{eqnarray*} +rr_ar_br_c=s(s-a)(s-b)(s-c)=p_{\triangle}^2, +\end{eqnarray*} +kar je bilo treba dokazati. \kdokaz + + \bizrek \label{ploscTrikPedalni} + Naj bo $A'B'C'$ pedalni trikotnik ostrokotnega trikotnika $ABC$, $s'$ polobseg + trikotnika $A'B'C'$, $R$ polmer očrtane krožnice in $p_{\triangle}$ ploščina + trikotnika $ABC$, potem je: + $$p_{\triangle} = s'R.$$ + \eizrek + + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.3a.pic} +\caption{} \label{sl.plo.8.3.3a.pic} +\end{figure} + + + \textbf{\textit{Proof.}} Označimo z $O$ središče očrtane krožnice trikotnika $ABC$ (Figure \ref{sl.plo.8.3.3a.pic}). Po izreku \ref{PedalniLemaOcrtana} je $OA\perp B'C'$, zato iz izreka \ref{ploscStirikPravok} sledi $p_{AC'OB'}=\frac{|AO|\cdot |B'C'|}{2}=\frac{R\cdot |B'C'|}{2}$. Analogno je tudi $p_{BA'OC'}=\frac{R\cdot |A'C'|}{2}$ in $p_{CB'OA'}=\frac{R\cdot |A'B'|}{2}$. Ker gre za ostrokotni trikotnik $ABC$, je točka $O$ v njegovi notranjosti (glej razdelek \ref{odd3ZnamTock}), zato po izrekih \ref{ploscGlavniIzrek} \textit{4)} in \ref{ploscDaljice} velja: +\begin{eqnarray*} +p_{\triangle}&=&p_{AC'OB'}+p_{BA'OC'}+p_{CB'OA'}=\\ +&=&\frac{R}{2}\left(|B'C'|+|A'C'|+|A'B'| \right)=s'R, +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + \bzgled \label{ploscTrikPedalni1} + Naj bosta $R$ in $r$ polmera očrtane in včrtane krožnice ostrokotnega + trikotnika $ABC$ ter $s$ in $s'$ polobsega tega trikotnika in njegovega + pedalniga trikotnika. + Dokaži, da velja: + $$\frac{R}{r}=\frac{s}{s'}.$$ + \ezgled + +\textbf{\textit{Proof.}} Po izrekih \ref{PloscTrikVcrt} in \ref{ploscTrikPedalni} je: +\begin{eqnarray*} +p_{ABC}=sr=s'R. +\end{eqnarray*} +Iz tega pa dobimo iskano relacijo. +\kdokaz + +V posebnem primeru dobimo formuli za ploščino pravokotnega in enakostraničnega trikotnika. + + \bzgled \label{PloscTrikPravokotni} Če je $p_\triangle$ ploščina + pravokotnega trikotnika $ABC$ s katetama dolžin $a$ in $b$, potem velja: + $$p_\triangle=\frac{ab}{2}.$$ + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.4a.pic} +\caption{} \label{sl.plo.8.3.4a.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.3.4a.pic}) + Trditev je direktna posledica izreka \ref{PloscTrik}, saj je v tem primeru $v_a=b$. +\kdokaz + + \bizrek \label{PloscTrikEnakostr} Če je $p_\triangle$ ploščina + enakostraničnega trikotnika $ABC$ s stranico dolžine $a$, potem velja: + $$p_\triangle=\frac{a^2\sqrt{3}}{4}.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.4.pic} +\caption{} \label{sl.plo.8.3.4.pic} +\end{figure} + + + \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.3.4.pic}) + Trditev je direktna posledica izrekov \ref{PloscTrik} in \ref{PitagorovEnakostr}: +\begin{eqnarray*} +p_{ABC}=\frac{av_a}{2}=\frac{a}{2}\cdot v_a=\frac{a}{2}\cdot \frac{a\sqrt{3}}{2}=\frac{a^2\sqrt{3}}{4}, +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + Nadaljevali bomo z uporabo formul za ploščino trikotnika. + + + + + + \bzgled \label{CarnotOcrtLema} + Naj bo $P$ notranja točka trikotnika $ABC$ ter $x$, $y$ in $z$ razdalje te točke od njegovih stranic $BC$, $AC$ in $AB$ z dolžinami $a$, $b$ in $c$. Če je $r$ polmer včrtane krožnice tega trikotnika, potem je: + $$xa+yb+zc=2\cdot p_{\triangle ABC}=r(a+b+c).$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.6.pic} +\caption{} \label{sl.plo.8.3.6.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.3.6.pic}) + +$x$, $y$ in $z$ so dolžine višin trikotnikov $PBC$, $APC$ in $ABP$, zato +po izrekih \ref{PloscTrik} in \ref{PloscTrikVcrt} sledi: + \begin{eqnarray*} + xa+yb+zc &=& 2\cdot p_{\triangle PBC}+2\cdot p_{\triangle APC}+2\cdot p_{\triangle ABP}\\ + &=& 2\cdot p_{\triangle ABC}\\ + &=& r(a+b+c), + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + + \bzgled \index{izrek!Vivianijev} (Vivianijev\footnote{\index{Viviani, V.}\textit{V. Viviani} (1622--1703), + italijanski matematik in fizik.} izrek) + Naj bo $P$ notranja točka enakostraničnega trikotnika $ABC$ ter $x$, $y$ in $z$ razdalje te točke od njegovih stranic $BC$, $AC$ in $AB$. Če je $v$ višina tega trikotnika, potem je: + $$x+y+z=v.$$ + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.6a.pic} +\caption{} \label{sl.plo.8.3.6a.pic} +\end{figure} + +\textbf{\textit{Proof.}} +Po trditvi iz prejšnjega zgleda \ref{CarnotOcrtLema} in izreka \ref{PloscTrik}: +$xa+ya+za=2\cdot p_{\triangle ABC}=v_aa$. +\kdokaz + + \bizrek \label{CarnotOcrt}\index{izrek!Carnotov o očrtani krožnici}(Carnotov\footnote{\index{Carnot, L. N. M.}\textit{L. N. M. Carnot} (1753--1823), francoski matematik.} izrek o očrtani krožnici.) + Vsota razdalj središča očrtane krožnice trikotnika od njegovih stranic je enaka vsoti polmerov očrtane in včrtane krožnice tega trikotnika. Če so torej $l(O,R)$ očrtana in $k(S,r)$ včrtana krožnica ter $A_1$, $B_1$ in $C_1$ središča stranic trikotnika $ABC$, velja: + $$|OA_1|+|OB_1|+|OC_1|=R+r.$$ + + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.7.pic} +\caption{} \label{sl.plo.8.3.7.pic} +\end{figure} + +\textbf{\textit{Proof.}} Označimo $x=|OA_1|$, $y=|OB_1|$ in $z=|OC_1|$ (Figure \ref{sl.plo.8.3.7.pic}). Po trditvi iz zgleda \ref{CarnotOcrtLema} je: +\begin{eqnarray} \label{eqnCarnotOcrt1} +xa+yb+zc=r(a+b+c) +\end{eqnarray} + Iz skladnosti trikotnikov $BA_1O$ in $CA_1O$ (izrek \textit{SAS} \ref{SKS}) in izreka \ref{SredObodKot} sledi: $\angle BAC=\frac{1}{2}\angle BOC=\angle BOA_1$. To pomeni, da velja $\triangle OA_1B\sim \triangle AB'B \sim \triangle AC'C$ (izrek \ref{PodTrikKKK}), zato je: + \begin{eqnarray*} + \frac{OA_1}{AB'}=\frac{OB}{AB} \hspace*{2mm} \textrm{ in } \hspace*{2mm} + \frac{OA_1}{AC'}=\frac{OB}{AC} +\end{eqnarray*} +oziroma: +\begin{eqnarray*} + cx=R\cdot |AB'| \hspace*{2mm} \textrm{ in } \hspace*{2mm} + bx=R\cdot |AC'| +\end{eqnarray*} +Po seštevanju zadnjih dveh enakosti dobimo: +\begin{eqnarray*} + (b+c)x=R(|AB'|+|AC'|) +\end{eqnarray*} +in analogno še: +\begin{eqnarray*} + (a+c)y=R\cdot (|BA'|+|BC'|)\\ + (a+b)z=R\cdot (|CA'|+|CB'|). +\end{eqnarray*} + +Po seštevanju zadnjih treh relacij dobimo: +\begin{eqnarray} \label{eqnCarnotOcrt2} +(b+c)x+(a+c)y+(a+b)z=R(a+b+c). +\end{eqnarray} +Če na koncu seštejemo enakosti \ref{eqnCarnotOcrt1} in \ref{eqnCarnotOcrt2} in dobljeno enakost delimo z $a+b+c$, dobimo: +\begin{eqnarray*} +x+y+z=r+R, +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + + \bnaloga\footnote{12. IMO, Hungary - 1970, Problem 1.} + Let $M$ be a point on the side $AB$ of triangle $ABC$. Let $r_1$, $r_2$ and $r$ be the radii + of the inscribed circles of triangles $AMC$, $BMC$ and $ABC$. Let $q_1$, $q_2$ and $q$ + be the radii of the escribed circles of the same triangles that lie in the angle + $ACB$. Prove that + $$\frac{r_1}{q_1}\cdot\frac{r_2}{q_2}=\frac{r}{q}.$$ + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.IMO1.pic} +\caption{} \label{sl.plo.8.3.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Označimo najprej $a=|BC|$, $b=|AC|$, +$c=|AB|$, $x=|AM|$, $y=|BM|$ in $m=|CM|$ (Figure +\ref{sl.plo.8.3.IMO1.pic}). + + Če uporabimo izreka \ref{PloscTrikVcrt} in + \ref{PloscTrikPricrt}, dobimo: + \begin{eqnarray*} + && p_{\triangle ABC}=\frac{a+b+x+y}{2}\cdot r=\frac{a+b-x-y}{2}\cdot q\\ + && p_{\triangle AMC}=\frac{b+m+x}{2}\cdot r_1=\frac{b+m-x}{2}\cdot q_1\\ + && p_{\triangle ABC}=\frac{a+m+y}{2}\cdot r_2=\frac{a+m-y}{2}\cdot q_2 + \end{eqnarray*} +Iz tega sledi: + \begin{eqnarray*} + \frac{r}{q}=\frac{a+b+x+y}{a+b-x-y},\hspace*{6mm} + \frac{r_1}{q_1}=\frac{b+m+x}{b+m-x},\hspace*{6mm} + \frac{r_2}{q_2}=\frac{a+m+y}{a+m-y} + \end{eqnarray*} +Zato je relacija +$\frac{r_1}{q_1}\cdot\frac{r_2}{q_2}=\frac{r}{q}$, ki jo želimo +dokazati, ekvivalentna z: +$$(a+b+x+y)(a+m-y)(b+m-x)=(a+b-x-y)(a+m+y)(b+m+x),$$ +slednja pa je po preoblikovanju ekvivalentna z: +$$(a+b)(a+m)x+(a+b)(b+m)y=c(a+m)(b+m)+(x+y)xy,$$ +oziroma, če upoštevamo $x+y=c$ in dodatno preoblikujemo, z: +$$m^2=a^2\frac{x}{c}+b^2\frac{y}{c}-xy.$$ + Zadnja relacija velja, ker predstavlja Stewartov izrek + \ref{StewartIzrek} za trikotnik $ABC$ in daljico $CM$. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Area of Polygons} \label{odd8PloVeck} + +Ploščino poljubnega večkotnika dobimo, če ga z njegovimi diagonalami razdelimo na unijo trikotnikov (Figure \ref{sl.plo.8.4.1.pic}). Po izrekih \ref{ploscGlavniIzrek} in \ref{ploscDaljice} je ploščina tega večkotnika enaka vsoti ploščin teh trikotnikov. + + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.4.1.pic} +\caption{} \label{sl.plo.8.4.1.pic} +\end{figure} + +Posebej za pravilni šestkotnik velja naslednja trditev. + + \bizrek + Če je $p$ ploščina in $a$ dolžina stranice pravilnega šestkotnika, potem je: + $$p=\frac{3\sqrt{3}\cdot a^2}{2}.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.4.2.pic} +\caption{} \label{sl.plo.8.4.2.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.4.2.pic}) + +Trditev je direktna posledica izrekov \ref{ploscGlavniIzrek}, \ref{ploscDaljice} \ref{PloscTrikEnakostr} in dejstva, da pravilni šestkotnik lahko razdelimo na šest pravilnih trikotnikov (razdelek \ref{odd3PravilniVeck}). +\kdokaz + +Za ploščino tangentnih večkotnikov velja podobna formula kot v primeru trikotnikov (izrek \ref{PloscTrikVcrt}). + + \bizrek \label{ploscTetVec} + Če so: $s$ polobseg, $k(S,r)$ včrtana krožnica tangentnega večkotnika + $A_1A_2\ldots A_n$ in $p$ + njegova ploščina, tedaj je $$p = sr.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.4.3.pic} +\caption{} \label{sl.plo.8.4.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.4.3.pic}) + +Trikotniki $A_1SA_2$, $A_2SA_3$, ... , $A_{n-1}SA_n$ in $A_nSA_1$ imajo vsi enako dolžino višine iz oglišča $S$, ki je enaka $r$. Če uporabimo izreke \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscDaljice} in \ref{PloscTrik}, dobimo: +\begin{eqnarray*} + p &=& p_{A_1SA_2}+p_{A_2SA_3}+\cdots +p_{A_{n-1}SA_n} +p_{A_nSA_1}=\\ + &=& \frac{|A_1A_2|\cdot r}{2}+\frac{|A_2A_3|\cdot r}{2}+\cdots+ +\frac{|A_{n-1}A_n|\cdot r}{2}+\frac{|A_n A_1|\cdot r}{2}=\\ +&=&sr, +\end{eqnarray*} +kar je bilo treba dokazati. \kdokaz + + + + \bnaloga\footnote{30. IMO, Germany - 1989, Problem 2.} + Let $S$, $S_a$, $S_b$ and $S_c$ be incentre and excentres of an acute-angled triangle $ABC$. + $N_a$, $N_b$ and $N_c$ are midpoints of line segments $SS_a$, $SS_b$ and $SS_c$, respectively. + Let $\mathcal{S}_{X_1X_2\ldots X_n}$ be the area of a polygon $X_1X_2\ldots X_n$. + Prove that: + $$\mathcal{S}_{S_aS_bS_c} = + 2\mathcal{S}_{AN_cBN_aCN_b} \geq 4\mathcal{S}_{ABC}.$$ + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.3.IMO2.pic} +\caption{} \label{sl.plo.8.3.IMO2.pic} +\end{figure} + +\textbf{\textit{Solution.}} Naj bo $l$ očrtana krožnica +trikotnika $ABC$ (Figure \ref{sl.plo.8.3.IMO2.pic}). Po veliki +nalogi (\ref{velikaNaloga}) je točka $N_a$ središče tistega loka +$BC$ krožnice $l$, na katerem točka $A$ ne leži - označimo ta lok +z $l_{BC}$. Podobno sta $N_b$ in $N_c$ središči ustreznih lokov +$AC$ in $AB$ na krožnici $l$. + + Najprej iz $SN_a\cong N_aS_a$ sledi + $\mathcal{S}_{SCN_a}=\mathcal{S}_{N_aCS_a}$ oz. + $\mathcal{S}_{SCS_a}=2\cdot\mathcal{S}_{SCN_a}$. Če uporabimo + analogne enakosti in upoštevamo, da je $S$ notranja točka trikotnika + $S_aS_bS_c$, dobimo: + \begin{eqnarray*} +\mathcal{S}_{S_aS_bS_c} &=& +\mathcal{S}_{SCS_a}+\mathcal{S}_{SBS_a}+ +\mathcal{S}_{SCS_b}+\mathcal{S}_{SAS_b}+ +\mathcal{S}_{SAS_c}+\mathcal{S}_{SBS_c}=\\ + &=& +2\cdot\left(\mathcal{S}_{SCN_a}+\mathcal{S}_{SBN_a}+ +\mathcal{S}_{SCN_b}+\mathcal{S}_{SAN_b}+ +\mathcal{S}_{SAN_c}+\mathcal{S}_{SBN_c}\right)=\\ +&=& 2\mathcal{S}_{AN_cBN_aCN_b}. + \end{eqnarray*} + + Dokažimo še $2\mathcal{S}_{AN_cBN_aCN_b} \geq 4\mathcal{S}_{ABC}$, + oz. $\mathcal{S}_{AN_cBN_aCN_b} \geq 2\mathcal{S}_{ABC}$. + Naj bo $V$ višinska točka trikotnika $ABC$ in $V_a=S_{BC}(V)$. + Trikotnik $ABC$ je po predpostavki ostrokotni, zato točka + $V$ leži v njegovi notranjosti. + Po izreku \ref{TockaV'} točka $V_a$ leži na loku $l_{BC}$. Omenili smo že, + da je $N_a$ središče tega loka, zato je višina trikotnika + $BN_aC$ daljša od višine trikotnika $BV_aC$ za isto osnovnico $BC$. + Iz tega sledi $\mathcal{S}_{BN_aC}\geq \mathcal{S}_{BV_aC}$. Iz + skladnosti trikotnikov $BV_aC$ in $BVC$ dobimo: + $$\mathcal{S}_{BN_aC}\geq \mathcal{S}_{BV_aC}=\mathcal{S}_{BVC}.$$ + Podobno je tudi + $\mathcal{S}_{AN_bC}\geq \mathcal{S}_{AVC}$ in + $\mathcal{S}_{AN_cB}\geq \mathcal{S}_{AVB}$. Ker je $V$ + notranja točka trikotnika $ABC$, sledi: + \begin{eqnarray*} +\mathcal{S}_{AN_cBN_aCN_b}&=&\mathcal{S}_{ABC}+ +\mathcal{S}_{BN_aC}+\mathcal{S}_{AN_bC}+\mathcal{S}_{AN_cB}\geq\\ + &\geq& \mathcal{S}_{ABC}+ +\mathcal{S}_{BVC}+\mathcal{S}_{AVC}+\mathcal{S}_{AVB}=\\ +&=& 2\mathcal{S}_{ABC}, + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + + +%________________________________________________________________________________ +\poglavje{Koch Snowflake} \label{odd8PloKoch} + +V tem razdelku bomo obravnavali lik, ki je omejen (podmnožica nekega kroga), ima končno ploščino, njegov obseg pa je neskončen. + +Najprej bomo definirali posebno vrsto večkotnikov. Enakostranični trikotnik $ABC$ s stranico dolžine $a=a_0$ označimo s $\mathcal{K}_0(a)$. Večkotnik $\mathcal{K}_1(a)$ dobimo, če vsako stranico trikotnika $\mathcal{K}_0(a)$ razdelimo na tri enake dele in nad srednjim narišemo enakostranični trikotnik z dolžino stranice $a_1=\frac{1}{3}\cdot a_0$. Postopek nadaljujemo in večkotnik $\mathcal{K}_n(a)$ dobimo, če vsako stranico trikotnika $\mathcal{K}_{n-1}(a)$ razdelimo na tri enake dele in nad srednjim narišemo enakostranični trikotnik z dolžino stranice $a_n=\frac{1}{3}\cdot a_{n-1}$ (Figure \ref{sl.plo.8.5.Koch1.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.Koch1.pic} +\caption{} \label{sl.plo.8.5.Koch1.pic} +\end{figure} + + Če opisani postopek nadaljujemo do neskončnosti, dobimo t. i. +\index{Kochova snežinka} \pojem{Kochovo\footnote{Ta lik je leta 1904 definiral švedski matematik \index{Koch, H.}\textit{Helge von Koch} (1870--1924). Danes Kochovo snežinko uvršcamo +med t. i. \textit{fraktale}.} snežinko} $\mathcal{K}(a)$. + +Izračunajmo najprej obseg večkotnika $\mathcal{K}_n(a)$. + +Iz postopka, po katerem smo dobili večkotnik $\mathcal{K}_n(a)$, je jasno, da ima večkotnik $\mathcal{K}_n(a)$ $4$-krat več stranic kot $\mathcal{K}_{n-1}(a)$ (vsaka stranica večkotnika $\mathcal{K}_{n-1}(a)$ je zamenjana s štirimi stranicami večkotnika $\mathcal{K}_n(a)$). Označimo s $s_n$ število stranic večkotnika $\mathcal{K}_n(a)$. Očitno je $s_n$ geometrijsko zaporedje s kvocientom $q=4$ in začetnim členom $s_0=3$, zato je: +\begin{eqnarray} \label{eqnKoch1} +s_n=3\cdot 4^n +\end{eqnarray} +Vsaka stranica $a_n$ večkotnika $\mathcal{K}_n(a)$ je po konstrukciji $3$-krat krajša od stranice $a_{n-1}$ večkotnika $\mathcal{K}_{n-1}(a)$. Tako imamo spet geometrijsko zaporedje $a_n$ s kvocientom $q=\frac{1}{3}$ in začetnim členom $a_0=a$, zato je: +\begin{eqnarray} \label{eqnKoch2} +a_n=\left( \frac{1}{3} \right)^n\cdot a +\end{eqnarray} +Ker za obseg $o_n$ večkotnika $\mathcal{K}_n(a)$ velja $o_n=s_n\cdot a_n$, iz \ref{eqnKoch1} in \ref{eqnKoch2} dobimo: +\begin{eqnarray} \label{eqnKoch3} +o_n=3\cdot \left( \frac{4}{3} \right)^n\cdot a +\end{eqnarray} +Če želimo dobiti obseg $o$ Kochove snežinke $\mathcal{K}(a)$ ga računamo kot $o=\lim_{n\rightarrow\infty}o_n$, toda iz \ref{eqnKoch3} dobimo (ker je $\frac{4}{3}>1$): +\begin{eqnarray} \label{eqnKoch4} +o=\lim_{n\rightarrow\infty}o_n=\lim_{n\rightarrow\infty}3\cdot \left( \frac{4}{3} \right)^n\cdot a=3a\cdot\lim_{n\rightarrow\infty}\left( \frac{4}{3} \right)^n=\infty +\end{eqnarray} +kar pomeni, da Kochova snežinka $\mathcal{K}(a)$ nima končnega obsega, torej obseg $o_n$ večkotnika $\mathcal{K}_n(a)$ neomejeno raste, ko se število korakov $n$ povečuje. + +Izračunajmo sedaj ploščino $p$ Kochove snežinke $\mathcal{K}(a)$. Označimo s $p_n$ ploščino večkotnika $\mathcal{K}_n(a)$. + Ploščino $p_n$ dobimo, če ploščini $p_{n-1}$ dodamo določeno število ploščin manjših trikotnikov +stranice $a_n$. Koliko je teh trikotnikov? Po konstrukciji večkotnik $\mathcal{K}_n(a)$ dobimo iz večkotnika $\mathcal{K}_{n-1}(a)$, če pri vsaki stranici večkotnika $\mathcal{K}_{n-1}(a)$ pričrtamo manjši trikotnik. Torej je število teh trikotnikov enako $s_{n-1}$. Zaradi tega in izreka \ref{PloscTrikEnakostr} je: +\begin{eqnarray*} +p_n=p_{n-1}+s_{n-1}\cdot \frac{a_n^2\cdot \sqrt{3}}{4}=p_{n-1} ++\frac{3}{16}\cdot \left(\frac{4}{9} \right)^n\cdot a^2\sqrt{3}. +\end{eqnarray*} +Ploščina $p$ Kochove snežinke $\mathcal{K}(a)$ je torej neskončna vsota: +\begin{eqnarray*} +p&=&p_0+ \sum_{n=1}^{\infty}\frac{3}{16}\cdot \left(\frac{4}{9} \right)^n\cdot a^2\sqrt{3}=\\ +&=&\frac{a^2\cdot \sqrt{3}}{4}+ \frac{3}{16} a^2\sqrt{3}\cdot\sum_{n=1}^{\infty} \left(\frac{4}{9}\right)^n=\\ +&=&\frac{a^2\cdot \sqrt{3}}{4}+ \frac{3}{16} a^2\sqrt{3}\cdot\left(\frac{4}{9}+\left(\frac{4}{9}\right)^2+\cdots + \left(\frac{4}{9}\right)^n+\cdots \right). +\end{eqnarray*} +Če izračunamo vsoto neskončnega geometrijskega zaporedja in poenostavimo, dobimo: +\begin{eqnarray} \label{eqnKoch5} +p=\frac{2a^2\sqrt{3}}{5}. +\end{eqnarray} + + +%________________________________________________________________________________ +\poglavje{Circumference and Area of a Circle} \label{odd8PloKrog} + +%OBSEG + +V razdelku \ref{odd3NeenTrik} smo že definirali obseg večkotnika kot vsoto vseh njegovih stranic. V tem smislu je obseg predstavljal daljico. V nadaljevanju bomo pod obsegom razumeli tudi dolžino te daljice oz. vsoto dolžin vseh stranic večkotnika. Na tem mestu bomo obravnavali obseg kroga. Čeprav je intuitivno jasen, je ta nov pojem potrebno najprej definirati. + +Naj bo $k(S,r)$ poljubna krožnica. Pripadajoči krog označimo s $\mathcal{K}(S,r)$. Obseg tega kroga intuitivno predstavlja dolžino krožnice $k$. Naj bo $A_1A_2\ldots A_n$ ($n\in \mathbb{N}$ in $n\geq 3$) pravilni večkotnik, ki je včrtan krožnici $k$. Tangente te krožnice v ogliščih večkotnika $A_1A_2\ldots A_n$ določajo nosilke stranic pravilnega večkotnika $B_1B_2\ldots B_n$, ki je krožnici $k$ očrtan (Figure \ref{sl.plo.8.5.1.pic}). + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.1.pic} +\caption{} \label{sl.plo.8.5.1.pic} +\end{figure} + +Označimo z $\underline{o}_n$ obseg večkotnika $A_1A_2\ldots A_n$ in $\overline{o}_n$ obseg večkotnika $B_1B_2\ldots B_n$. Po trikotniški neenakosti (izrek \ref{neenaktrik}) je $|A_1B_1|+|B_1A_2|>|A_1A_2|$, $|A_2B_2|+|B_2A_3|>|A_2A_3|$,..., $|A_nB_n|+|B_nA_1|>|A_nA_1|$ (Figure \ref{sl.plo.8.5.1a.pic}). Če seštejemo vse te neenakosti dobimo $\overline{o}_n>\underline{o}_n$. Podobno velja $\underline{o}_n<\underline{o}_{n+1}$ in $\overline{o}_n>\overline{o}_{n+1}$. Torej za vsak $n\in \mathbb{N}$, $n\geq 3$ velja: +\begin{eqnarray*} +\underline{o}_1<\underline{o}_2<\cdots< +\underline{o}_n<\overline{o}_n<\cdots<\overline{o}_2<\overline{o}_1. +\end{eqnarray*} + To pomeni, da je $\underline{o}_n$ naraščajoče zaporedje navzgor omejeno z $\overline{o}_1$, zato je po znanem izreku matematične analize konvergentno in ima svojo limito: +\begin{eqnarray*} +\underline{o}=\lim_{n\rightarrow\infty}\underline{o}_n. +\end{eqnarray*} + Podobno je $\overline{o}_n$ padajoče zaporedje navzdol omejeno z $\underline{o}_1$, zato ima svojo limito: +\begin{eqnarray*} +\overline{o}=\lim_{n\rightarrow\infty}\overline{o}_n. +\end{eqnarray*} + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.1a.pic} +\caption{} \label{sl.plo.8.5.1a.pic} +\end{figure} + +Na tem mestu ne bomo formalno dokazovali intuitivno jasnega dejstva, da je za dovolj veliko naravno število $n$ razlika $\overline{o}_n-\underline{o}_n$ poljubno majhna, oz. $\underline{o}=\overline{o}$. \index{obseg!kroga}\pojem{Obseg kroga} $\mathcal{K}$ z oznako $o$ potem predstavlja $o=\underline{o}=\overline{o}$. + +Dokažimo naslednji pomemben izrek. + + \bizrek + Če je $o$ obseg kroga s polmerom $r$, potem je razmerje + $o:2r$ + konstantno in tako neodvisno od izbire kroga. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.1b.pic} +\caption{} \label{sl.plo.8.5.1b.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.1b.pic}) + +Naj bosta $\mathcal{K}(S,r)$ in $\mathcal{K}'(S',r')$ poljubna kroga z obsegoma $o$ in $o'$ ter $k(S,r)$ in $k'(S',r')$ pripadajoči krožnici. Po izreku \ref{RaztKroznKrozn1} obstaja središčni razteg, ki preslika krožnico $k$ v krožnico $k'$. Po izreku \ref{RaztTransPod} je ta središčni razteg tranformacija podobnosti (označimo jo s $f$) z nekim koeficientom $\lambda$. Najprej je $r':r=\lambda$ oziroma: +\begin{eqnarray} + r'=\lambda r. \label{eqnPloscKrogrr'oo'} +\end{eqnarray} +Naj bosta $A_1A_2\ldots A_n$ in $B_1B_2\ldots B_n$ ($n\in \mathbb{N}$ in $n\geq 3$) včrtani in očrtani pravilni večkotnik krožnice $k$ z obsegoma $\underline{o}_n$ in $\overline{o}_n$. Označimo: +$$f:S,A_1,A_2,\ldots,A_n,B_1,B_2,\ldots,B_n\mapsto S',A_1',A_2',\ldots,A_n',B_1',B_2',\ldots,B_n'.$$ +Večkotnika $A_1'A_2'\ldots A_n'$ in $B_1'B_2'\ldots B_n'$ ($n\in \mathbb{N}$ in $n\geq 3$) sta včrtani in očrtani pravilni večkotnik krožnice $k$ in za njuna obsega $\underline{o'}_n$ in $\overline{o'}_n$ velja $\underline{o'}_n=\lambda\underline{o}_n$ in $\overline{o'}_n=\lambda\overline{o}_n$. Iz tega in iz definicije obsega kroga sledi: +\begin{eqnarray}\label{eqnPloscKrogrr'oo'1} + o'=\underline{o'}=\lim_{n\rightarrow\infty}\underline{o'}_n= +\lim_{n\rightarrow\infty}\lambda\underline{o}_n= +\lambda\lim_{n\rightarrow\infty}\underline{o}_n=\lambda \underline{o}=\lambda o. +\end{eqnarray} +Iz \ref{eqnPloscKrogrr'oo'} in \ref{eqnPloscKrogrr'oo'1} sledi: +$$\frac{o'}{2r'}=\frac{\lambda o}{2\lambda r}=\frac{o}{2r},$$ kar je bilo treba dokazati. \kdokaz + +Konstanto iz prejšnjega izreka imenujemo +\index{število!$\pi$}\pojem{število $\pi$} (\index{Arhimedova konstanta}\pojem{Arhimedova konstanta} oz. \index{število!Ludolfovo}\pojem{Ludolfovo število}) in ga tudi označimo s $\pi$. + +Iz prejšnjega izreka in iz definicije števila $\pi$ dobimo naslednji izrek. + + + \bizrek \label{obsegKtoznice} + Če je $o$ obseg kroga s polmerom $r$, potem velja: + $$o=2r\pi.$$ + \eizrek + +najprej bomo podali prvo grobo oceno števila $\pi$. + + \bizrek \label{stevPiOcena} + Za število $\pi$ velja: + $$3<\pi<4.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.2.pic} +\caption{} \label{sl.plo.8.5.2.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.2.pic}) + +Naj bo $\mathcal{K}$ poljuben krog s polmerom $r$ in z obsegom $o$ ter $k$ pripadajoča krožnica. Po izreku \ref{obsegKtoznice} je $o=2r\pi$. +Izberimo za pravilni včrtani večkotnik $n=6$ in pravilni očrtani večkotnik $n=4$. Iz definicije obsega kroga je: +$$\underline{o}_6r_1$) dva kroga z istim središčem (pripadajoči krožnici $k_1$ in $k_2$ sta koncentrični). Množico $\mathcal{K}_2\setminus \mathcal{K}_1$ imenujemo +\index{krožni!kolobar}\pojem{krožni kolobar}. + + + \bizrek \label{ploscKrozKolob} + Če je $p_{kl}$ ploščina krožnega kolobarja, ki ga določata krožnici $k_1(S,r_1)$ + in $k_2(S,r_2)$ ($r_2>r_1$), potem je: + $$p_{kl}=\left( r_2^2-r_1^2 \right)\cdot\pi.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.4.pic} +\caption{} \label{sl.plo.8.5.4.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.4.pic}) + +Označimo s $\mathcal{K}_1(S,r_1)$ in $\mathcal{K}_2(S,r_2)$ pripadajoča kroga krožnic $k_1(S,r_1)$ in $k_2(S,r_2)$. + +Iz $\mathcal{K}_2=\mathcal{K}_2\setminus \mathcal{K}_1\cup \mathcal{K}_1 $ in $\left( \mathcal{K}_2\setminus \mathcal{K}_1 \right) \cap \mathcal{K}_1=\emptyset$ po izrekih \ref{ploscGlavniIzrek} \textit{4)} in \ref{ploscKrog} sledi + $p_{\mathcal{K}_2}=p_{\mathcal{K}_2\setminus \mathcal{K}_1} + p_{\mathcal{K}_1}$ oz.: +$$p_{kl}=p_{\mathcal{K}_2\setminus \mathcal{K}_1}= p_{\mathcal{K}_2}- p_{\mathcal{K}_1}=r_2^2\pi-r_1^2\pi +=\left( r_2^2-r_1^2 \right)\cdot\pi,$$ kar je bilo treba dokazati. \kdokaz + +Na podoben način kot pri dolžini krožnega loka (izrek \ref{obsegKrozLok}) dobimo naslednjo trditev. + + + \bizrek \label{ploscKrozIzsek} + Če je $p_i$ ploščina krožnega izseka z ustreznim središčnim kotom z mero + (v kotnih stopinjah) $\alpha$ in s + polmerom $r$, potem je: + $$p_i=\frac{r^2\pi\cdot\alpha}{360^0}.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.5.pic} +\caption{} \label{sl.plo.8.5.5.pic} +\end{figure} + + + + \bzgled + Če v formuli za ploščino kroga število $\pi$ zamenjamo z njegovo približno + vrednostjo $3$, dobimo formulo za ploščino v ustrezno krožnico včrtanega + pravilnega dvanajstkotnika\footnote{To nalogo je rešil kitajski matematik + \index{Liu, H.}\textit{H. Liu} (3. st.)}. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.6.pic} +\caption{} \label{sl.plo.8.5.6.pic} +\end{figure} + + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.6.pic}) + +Naj bo $A_1A_2\ldots A_{12}$ pravilni dvanajstkotnik, ki je +včrtan krožnici $k(S,r)$. Potem je $A_1A_3\ldots A_{11}$ +pravilni šestkotnik, ki je včrtan isti krožnici. Označimo s $p_{12}$ ploščino + omenjenega dvanajstkotnika. Po izreku \ref{ploscGlavniIzrek} \textit{4)} je $p_{12}=12\cdot p_{SA_1A_2}$. + +Ker je še trikotnik $A_1SA_3$ pravilen, je višina $A_1P$ + trikotnika $SA_1A_2$ enaka polovici stranice $A_1A_3$ +šestkotnika $A_1A_3\ldots A_{11}$, torej: + $$|A_1P|=\frac{1}{2}\cdot |A_1A_3|=\frac{r}{2}.$$ +Po izreku \ref{PloscTrik} je potem: + $$p_{12}=12\cdot p_{SA_1A_2}=12\cdot\frac{|SA_2|\cdot |A_1P|}{2}=12\cdot\frac{r^2}{4}=3\cdot r^2,$$ kar je bilo treba dokazati. \kdokaz + + + + \bzgled \label{HipokratoviLuni} + Nad katetama pravokotnega trikotnika konstruirajmo polkroga + in od tako povečanega trikotnika odrežimo polkrog nad hipotenuzo + (Figure \ref{sl.plo.8.5.7.pic}). Pokaži, da je ploščina ostanka + enaka ploščini prvotnega + pravokotnega trikotnika\footnote{Specialno trditev (naslednji zgled v tem razdelku), če je dani trikotnik enakokrak, je dokazal \index{Hipokrat}\textit{Hipokrat iz Kiosa} (5. st. pr. n. š.), starogrški +matematik. Omenjeni figuri imenujemo \index{Hipokratovi luni}\pojem{Hipokratovi luni}. Hipokrat je prvič odkril, da lahko nekatere like s krivim robom z ravnilom in s šestilom pretvorimo v ploščinsko enak kvadrat.}. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.7.pic} +\caption{} \label{sl.plo.8.5.7.pic} +\end{figure} + +\textbf{\textit{Solution.}} +Naj bo $ABC$ omenjeni trikotnik s hipotenuzo $c$ ter katetama $a$ in $b$. +Označimo s $p$ iskano ploščino, $p_{\triangle}$ ploščino trikotnika $ABC$ ter $p_a$, $p_b$ in $p_c$ ploščine ustreznih polkrogov. Po izrekih \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscKrozIzsek} in \ref{PloscTrik} ter Pitagorovem izreku \ref{PitagorovIzrek} je: + \begin{eqnarray*} + p&=& p_a+p_b-p_c+p_{\triangle}=\\ + &=& \left(\frac{a}{2} \right)^2\cdot \pi ++\left(\frac{a}{2} \right)^2\cdot \pi-\left(\frac{a}{2} \right)^2\cdot \pi+p_{\triangle}=\\ + &=& \frac{\pi}{4}\cdot \left(a^2+b^2-c^2 \right)+p_{\triangle}=\\ + &=& \frac{\pi}{4}\cdot 0+p_{\triangle}=\\ + &=& p_{\triangle}, + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + \bzgled \label{HipokratoviLuni2} + Dan je kvadrat $PQRS$. + Načrtaj kvadrat, ki ima ploščino enako ploščini lika, ki nastane, ko + krogu, ki je očrtan kvadratu $PQRS$, odrežemo + presek s krogom s + polmerom $PQ$ + (Figure \ref{sl.plo.8.5.8.pic}). + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.8.pic} +\caption{} \label{sl.plo.8.5.8.pic} +\end{figure} + +\textbf{\textit{Solution.}} +Če izberemo enakokraki pravokotni trikotnik $QSQ'$ (kjer je $Q'=\mathcal{S}_p(Q)$), je po prejšnjem izreku (\ref{HipokratoviLuni}) ploščina iskanega lika enaka ploščini enakokrakega pravokotnega trikotnika $SPQ$ (Figure \ref{sl.plo.8.5.8.pic}), njegova ploščina + pa je enaka ploščini kvadrata $PLSN$, kjer je $L$ središče daljice $SQ$ in $N=\mathcal{S}_{SP}(L)$ + (Figure \ref{sl.plo.8.5.8a.pic}). + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.8a.pic} +\caption{} \label{sl.plo.8.5.8a.pic} +\end{figure} +\kdokaz + + + \bzgled + Kvadrat $ABCD$ povečajmo za štiri polkroge nad stranicami, od + dobljenega lika pa odrežimo kvadratu očrtan krog (Figure \ref{sl.plo.8.5.9.pic}). Dokaži, da je + ploščina ostanka enaka ploščini prvotnega kvadrata. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.9.pic} +\caption{} \label{sl.plo.8.5.9.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $S$ središče kvadrata $ABCD$. +Trditev je direktna posledica trditve iz dokaza prejšnjega izreka (\ref{HipokratoviLuni2}) - uporabimo ustrezno relacijo za trikotnike $ASB$, $BSC$, $CSD$ in $DSA$ ter vse seštejemo. +\kdokaz + + + \bzgled \label{arbelos} + Naj bo $C$ poljubna točka na premeru $AB$ polkrožnice $k$ ter $D$ presečišče te + polkrožnice s pravokotnico premera $AB$ v točki $C$. Naj bosta $j$ in $l$ + polkrožnici s premeroma $AC$ in $BC$. Dokaži, da je ploščina lika, ki ga + omejujejo polkrožnice $k$, $l$ in $j$, enaka ploščini kroga s premerom + $CD$\footnote{To nalogo je v svoji ‘‘Knjigi lem’’ objavil starogrški matematik \index{Arhimed}\textit{Arhimed iz Sirakuze} (3. st. pr. n. š.). +Omenjeno figuro je imenoval \index{arbelos}\textit{arbelos} (kar v grščini pomeni čevljarski nož).}. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.10.pic} +\caption{} \label{sl.plo.8.5.10.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.10.pic}) + +Označimo z $a=|AC|$ in $b=|BC|$ ustrezna premera, $p$ ploščino omenjenega lika in $p_0$ ploščino kroga s premerom $|CD|$. Po izreku \ref{izrekVisinski} + je $|CD|^2 = ab$. Če uporabimo še izreka \ref{ploscKrozIzsek} in \ref{ploscKrog}, +dobimo: +\begin{eqnarray*} +p&=&\frac{1}{2}\left(\frac{a+b}{2}\right)^2\cdot\pi- +\frac{1}{2}\left(\frac{a}{2}\right)^2\cdot\pi- +\frac{1}{2}\left(\frac{b}{2}\right)^2\cdot\pi=\\ +&=& \frac{ab}{4}\cdot\pi=\\ +&=& \left(\frac{|CD|}{2}\right)^2\cdot\pi=p_0, +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + + \bzgled + Naj bo $k$ včrtana krožnica pravilnega trikotnika in $\mathcal{K}$ pripadajoči krog. Včrtajmo potem tri + manjše krožnice, ki se dotikajo krožnice $k$ in dveh stranic tega + trikotnika, nato še tri manjše krožnice, ki se dotikajo prejšnjih treh + krožnic in dveh stranic trikotnika, in tako naprej postopek nadaljujemo do + neskončnosti. Ali je ploščina vseh pripadajočih krogov, razen kroga $\mathcal{K}$, + manjša + od polovice ploščine kroga $\mathcal{K}$? + \ezgled + + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.11.pic} +\caption{} \label{sl.plo.8.5.11.pic} +\end{figure} + +\textbf{\textit{Solution.}} (Figure \ref{sl.plo.8.5.11.pic}) + +Naj bo $ABC$ dani trikotnik in $k(S,r)$ včrtana krožnica tega trikotnika. S +$k_i(S_i ,r_i)$ ($i\in\mathbb{N}$) označimo krožnice iz danega zaporedja krožnic pri oglišču $A$ ter s $p_i$ ($i \in \mathbb{N}$) + ploščine pripadajočih krogov. Krožnico $k$ označimo še s $k_0(S_0 ,r_0)$ in ploščino pripadajočega kroga $\mathcal{K}$ s $p=p_0$. Naj bosta +$k_n(S_n ,r_n)$ in $k_{n+1}(S_{n+1} ,r_{n+1})$ zaporedni krožnici omenjenega zaporedja. Njuni dotikališči s stranico $AB$ trikotnika $ABC$ +označimo s $T_n$ in $T_{n+1}$, pravokotno projekcijo središča $S_{n+1}$ na premico $S_nT_n$ pa z $L_n$. V +pravokotnem trikotniku $S_{n+1}L_nS_n$ je dolžina katete $|S_nL_n| = r_n- r_{n+1}$ in hipotenuze $|S_{n+1}S_n| = r_n+ r_{n+1}$. Kot +$L_nS_{n+1}S_n$ je enak polovici notranjega kota pri oglišču $A$, in sicer $30°$. Iz tega sledi $\angle S_{n+1}S_nL_n=60^0$. Naj bo $S_n'=\mathcal{S}_{S_{n+1}L_n}(S_n)$. Trikotnika $S_{n+1}L_nS_n$ in $S_{n+1}L_nS_n'$ sta skladna (izrek \textit{SAS} \ref{SKS}), zato je tudi $\angle S_{n+1}S_n'L_n=60^0$. Torej je $S_{n+1}S_n'S_n$ enakostranični trikotnik, kar pomeni, da velja $|S_{n+1}S_n| =|S_n'S_n| =2\cdot|S_nL_n|$ oz. +$$r_n+ r_{n+1}=2\cdot(r_n-r_{n+1}).$$ +Če iz zadnje enakosti izrazimo $r_{n+1}$, dobimo: + \begin{eqnarray*} + r_{n+1}=\frac{1}{3}\cdot r_n. +\end{eqnarray*} +Iz tega pa po izreku \ref{ploscKrog} sledi: + \begin{eqnarray*} + p_{n+1}=\frac{1}{9}\cdot p_n. +\end{eqnarray*} +Zaporedje $p_n$ je torej geometrijsko zaporedje s koeficientom $q=\frac{1}{9}$ in začetno vrednostjo $p_0=p$, zato je: + \begin{eqnarray*} + p_n=\left( \frac{1}{9} \right)^n \cdot p. +\end{eqnarray*} +To pomeni, da skupna ploščina $p_A$ krožnic iz zaporedja pri oglišču $A$ predstavlja vsoto neskončnega geometrijskega zaporedja $p_n$ ($n\in \mathbb{N}$). Torej: +\begin{eqnarray*} + p_A&=&p_1+p_2+p_3+\cdots=\\ +&=&\frac{1}{9}\cdot p+\left(\frac{1}{9}\right)^2\cdot p+ +\left(\frac{1}{9}\right)^3\cdot p+\cdots=\\ +&=&\frac{1}{9}p\cdot \frac{1}{1-\frac{1}{9}}=\\ +&=& \frac{1}{8}\cdot p. +\end{eqnarray*} +Vsota ploščin vseh krožnic pri ogliščih $A$, $B$ in $C$ je potem enaka $\frac{3}{8}\cdot p$ in ni večja od polovice $p$. +\kdokaz + + + + \bnaloga\footnote{6. IMO, USSR - 1964, Problem 3.} + A circle is inscribed in triangle $ABC$ with sides $a$, $b$, $c$. Tangents to the circle + parallel to the sides of the triangle are constructed. Each of these tangents + cuts off a triangle from triangle $ABC$. In each of these triangles, a circle is inscribed. + Find the sum of the areas of all four inscribed circles (in terms of $a$, $b$, $c$). + \enaloga + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.5.IMO1.pic} +\caption{} \label{sl.plo.8.5.IMO1.pic} +\end{figure} + +\textbf{\textit{Solution.}} Naj bo $k(S,\rho)$ včrtana krožnica +trikotnika $ABC$ z višinami $v_a$, $v_b$ in $v_c$ in $p$ ploščina +ustreznega kroga. Označimo z $AB_aC_a$, $A_bBC_b$ in $A_cB_cC$ +nove trikotnike, $k_a$, $k_b$ in $k_c$ včrtane krožnice teh +trikotnikov s polmeri $\rho_a$, $\rho_b$ in $\rho_c$ ter $p_a$, +$p_b$ in $p_c$ ploščine ustreznih krogov. (Figure +\ref{sl.plo.8.5.IMO1.pic}). + +Trikotnika $ABC$ in $AB_aC_a$ sta si podobna s koeficientom +$\frac{v_a-2\rho}{v_a}$. Zato je: + $$\rho_a=\frac{v_a-2\rho}{v_a}\cdot \rho.$$ + Ker po izreku \ref{PloscTrikVcrt} za ploščino trikotnika $ABC$ + velja $p_\triangle=s\cdot \varrho$ + (kjer je $s=\frac{a+b+c}{2}$ polobseg trikotnika $ABC$), sledi: +$$\rho_a^2=\left(1-\frac{2\rho}{v_a}\right)^2\cdot \rho^2= + \left(1-\frac{2\frac{P_\triangle}{s}}{2\frac{P_\triangle}{a}}\right)^2\cdot \rho^2= + \frac{(s-a)^2}{s^2}\cdot \rho^2.$$ +Podobno dobimo $\rho_b^2= + \frac{(s-b)^2}{s^2}\cdot \rho^2$ in $\rho_c^2= + \frac{(s-c)^2}{s^2}\cdot \rho^2$. Iz trditve \ref{PloscTrikOcrtVcrt} + sledi + $\rho^2=\frac{(s-a)(s-b)(s-c)}{s}$, zato je: +\begin{eqnarray*} + && p+p_a+p_b+p_c= \pi (\rho^2+\rho_a^2+\rho_b^2+\rho_c^2)=\\ + &&=\pi + \rho^2(1+\frac{(s-a)^2}{s^2}+\frac{(s-b)^2}{s^2}+\frac{(s-c)^2}{s^2})=\\ + &&= + \frac{(s-a)(s-b)(s-c)(s^2+(s-a)^2+(s-b)^2+(s-c)^2)}{s^3}\cdot \pi, +\end{eqnarray*} + kar je bilo treba izraziti. \kdokaz + + +%________________________________________________________________________________ +\poglavje{Pythagoras' Theorem and Area} \label{odd8PloPitagora} + + V razdelku \ref{odd7Pitagora} smo dokazali Pitagorov izrek \ref{PitagorovIzrek} in napovedali, da ga bomo izrazili v obliki, ki se nanaša na ploščine. + + \bizrek \index{izrek!Pitagorov za ploščine} \label{PitagorovIzrekPlosc}(Pitagorov izrek v obliki ploščin)\\ + Ploščina kvadrata nad hipotenuzo pravokotnega trikotnika je enaka vsoti ploščin kvadratov nad obema katetama tega trikotnika. + + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.6.1.pic} +\caption{} \label{sl.plo.8.6.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} Na tem mestu bomo podali dva dokaza in eno idejo dokaza tega izreka. + +\textit{1)} Izrek je direktna posledica Pitagorovega izreka \ref{PitagorovIzrek} in formule za ploščino kvadrata \ref{ploscKvadr} (Figure +\ref{sl.plo.8.6.1.pic}). + + +\textit{2)\footnote{Ta dokaz je objavil \index{Evklid}\textit{Evklid iz Aleksandrije} (3. st. pr. n. š.) v svojem znamenitem delu ‘‘Elementi’’, ki je sestavljeno iz 13 knjig.}} + Naj bodo $AMNB$, $BPQC$ in $CKLA$ kvadrati, ki so po vrsti konstruirani nad hipotenuzo in +katetama pravokotnega trikotnika $ABC$. Naj bosta $D$ in $E$ projekciji točke $C$ na premicah $AB$ in $MN$ (Figure +\ref{sl.plo.8.6.1.pic}). + +Ker je ploščina kvadrata $AMNB$ enaka vsoti ploščin pravokotnikov $NBDE$ in $EDAM$, je dovolj dokazati, da sta ploščini kvadratov $BPQC$ in +$CKLA$ enaki ploščinam pravokotnikov $NBDE$ in $EDAM$. Dokazali bomo le prvo enakost $p_{BPQC}=p_{NBDE}$ (druga +enakost $p_{CKLA}=p_{EDAM}$ je potem analogna). + +Za to relacijo $p_{BPQC}=p_{NBDE}$ je dovolj dokazati enakost ploščin trikotnikov $BPQ$ in +$NBD$, saj je $p_{BPQC}=2\cdot p_{BPQ}$ in $p_{NBDE}=2\cdot p_{NBD}$. + Tedaj: + \begin{eqnarray*} + p_{BPQ}&=&p_{BPA}=\hspace*{3mm} \textrm{(trikotnika imata enako osnovnico in višino)}\\ + &=&p_{NBC}=\hspace*{3mm} \textrm{(trikotnika sta skladna zaradi } \mathcal{R}_{B,90^0+\angle CBA}\textrm{)}\\ + &=&p_{NBD}\hspace*{4mm} \textrm{(trikotnika imata enako osnovnico in višino)} + \end{eqnarray*} +Torej imata trikotnika $BPQ$ in $NBD$ enaki ploščini, kar je bilo dovolj dokazati. + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.6.2.pic} +\caption{} \label{sl.plo.8.6.2.pic} +\end{figure} + +\textit{3)\footnote{Ta dokaz (na podlagi ponazoritve) pripisujejo Starim Indijcem. Lahko rečemo, da mu zadostuje le komentar: ‘‘Poglej!’’ - tako je preprost in eleganten.}} Ideja dokaza je podana s sliko +\ref{sl.plo.8.6.2.pic}. Formalni dokaz bomo prepustili bralcu. +\kdokaz + + \bzgled + Ploščina krožnega kolobarja, ki ga določata očrtana in včrtana krožnica pravilnega $n$-kotnika s stranico dolžine $a$ ni odvisna od števila $n$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.6.3.pic} +\caption{} \label{sl.plo.8.6.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $AB$ ena stranica, $S$ središče te stranice in $O$ središče očrtane oz. včrtane krožnice poljubnega pravilnega $n$-kotnika $\mathcal{V}_n(a)$ (Figure \ref{sl.plo.8.6.3.pic}). Po predpostavki je $|AB|=a$, zato je $|AS|=\frac{a}{2}$. Označimo z $R$ in $r$ polmera očrtane in včrtane krožnice $n$-kotnika $\mathcal{V}_n(a)$. Iz skladnosti trikotnikov $OSA$ in $OSB$ (izrek \textit{SSS} \ref{SSS}) sledi $\angle OSA\cong \angle OSB=90^0$. Torej $OSA$ je pravokotni trikotnik s hipotenuzo $OA$, zato je po Pitagorovem izreku \ref{PitagorovIzrek} $|OA|^2-|OS|^2=|AS|^2$ oz. $R^2-r^2=\left(\frac{a}{2}\right)^2$. Iz te relacije in formule za ploščino krožnega kolobarja \ref{ploscKrozKolob} dobimo: + $$p_k=\left(R^2-r^2\right)\cdot \pi=\left(\frac{a}{2}\right)^2\cdot \pi=\frac{a^2\pi}{4},$$ +kar pomeni, da je ploščina kolobarja odvisna le od dolžine stranice pravilnega $n$-kotnika $\mathcal{V}_n(a)$. +\kdokaz + + \bzgled + S pomočjo trditve iz zgleda \ref{PitagoraCofman} dokaži neenakost: + \begin{eqnarray*} + \frac{1}{1^2\cdot 2^2}+\frac{1}{2^2\cdot 3^2}+\cdots +\frac{1}{n^2 \left(n+1\right)^2}+\cdots<\frac{2}{3} + \end{eqnarray*} + oziroma + \begin{eqnarray*} + \sum_{n=1}^{\infty}\frac{1}{n^2 \left(n+1\right)^2}<\frac{2}{3}. + \end{eqnarray*} + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.plo.8.6.4.pic} +\caption{} \label{sl.plo.8.6.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.6.4.pic}) + +Uporabimo oznake iz zgleda \ref{PitagoraCofman}. Po relaciji \ref{eqnPitagoraCofman3} iz tega zgleda je $d_n=\frac{1}{n(n+1)}$, zato je $r_n=\frac{1}{2n(n+1)}$. Če s $p_n$ označimo ploščino pripadajočega kroga krožnice $c_n$, je (izrek \ref{ploscKrog}): +\begin{eqnarray*} +p_n=\pi r_n^2=\frac{\pi}{4}\frac{1}{n^2\left(n+1\right)^2}. +\end{eqnarray*} +Torej je vsota ploščin vseh pripadajočih krogov enaka: +\begin{eqnarray} \label{eqnPitagoraCofmanPlo2} +\sum_{n=1}^{\infty} p_n=\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{1}{n^2\left(n+1\right)^2}. +\end{eqnarray} +Toda ta vsota je manjša od ploščine $p$ lika, ki ga omejujejo daljica +$A'B'$ ter krožna loka $A'P$ in $B'P$ krožnic $a$ in $b$ s središčnim kotom $90^0$. Ploščina $p$ je enaka razliki ploščine pravokotnika $A'ABB'$ in ploščine dveh ustreznih krožnih izsekov s središčnim kotom $90^0$. Torej (izreka \ref{ploscPravok} in \ref{ploscKrozIzsek}): + \begin{eqnarray} \label{eqnPitagoraCofmanPlo3} +p=2-2\cdot\frac{\pi}{4}=2-\frac{\pi}{2}. +\end{eqnarray} +Ker je $\sum_{n=1}^{\infty} p_n3$ (izrek \ref{stevPiOcena}), dobimo: +\begin{eqnarray*} +\sum_{n=1}^{\infty}\frac{1}{n^2\left(n+1\right)^2}<\frac{8}{\pi}-2<\frac{8}{3}-2=\frac{2}{3}, +\end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + + +%________________________________________________________________________________ +\naloge{Exercises} + + +\begin{enumerate} + +% Ttrikotniki + +\item Naj bo $T$ težišče trikotnika $ABC$. Dokaži: + $$p_{TBC}=p_{ATC}=p_{ABT}.$$ + +\item Naj bodo $c$ hipotenuza, $v_c$ pripadajoča višina ter $a$ in $b$ kateti pravokotnega trikotnika. Dokaži, da velja $c+v_c>a+b$. +% zvezek - dodatni MG + + + \item Načrtaj kvadrat, ki ima enako ploščino kot pravokotni trikotnik s katetama, ki sta skladni z danima daljicama $a$ in $b$. % (Hipokratovi luni) + +\item Naj bo $ABC$ enakokraki pravokotni trikotnik z dolžino katet $|CB|=|CA|=a$. Označimo z $A_1$, $B_1$, $C_1$ in $C_2$ točke, za katere velja: $\overrightarrow{CA_1}=\frac{n-1}{n}\cdot \overrightarrow{CA}$, $\overrightarrow{CB_1}=\frac{n-1}{n}\cdot \overrightarrow{CB}$, $\overrightarrow{AC_1}=\frac{n-1}{n}\cdot \overrightarrow{AB}$ in $\overrightarrow{AC_2}=\frac{n-2}{n}\cdot \overrightarrow{AB}$ za nek $n\in \mathbb{N}$. Izrazi ploščino štirikotnika, ki ga določajo premice $AB$, $A_1B_1$, $CC_1$ in $CC_2$, kot funkcijo $a$ in $n$. + +\item Dan je pravokotni trikotnik $ABC$ s hipotenuzo $AB$ in ploščino $p$. Naj bo $C'=\mathcal{S}_{AB}(C)$, $B'=\mathcal{S}_{AC}(B)$ in $A'=\mathcal{S}_{BC}(A)$. Izrazi ploščino trikotnika $A'B'C'$ kot funkcijo $p$. + +\item Naj bosta $R$ in $Q$ točki, v katerih se trikotniku $ABC$ včrtana krožnica dotika njegovih stranic $AB$ in $AC$. Naj simetrala notranjega kota $ABC$ seka premico $QR$ v točki $L$. Določi razmerje ploščin trikotnikov $ABC$ in $ABL$. + % pripremni zadaci - naloga 193 + + +\item Določi točko v notranjosti trikotnika $ABC$, za katero je produkt njenih razdalj od stranic tega trikotnika maksimalen. % zvezek - dodatni MG + +\item Trikotniku + $ABC$ s stranicami dolžin $a$, $b$ in $c$ je včrtana krožnica. Načrtane so + tangente te krožnice, ki so vzporedne s stranicami + trikotnika. + Vsaka od tangent v notranjosti trikotnika določa ustrezne daljice dolžin $a_1$, $b_1$ in $c_1$. Dokaži, da velja: + $$\frac{a_1}{a}+\frac{b_1}{b}+\frac{b_1}{b}=1.$$ % zvezek - dodatni MG + +\item Naj bosta $T$ in $S$ težišče in središče včrtane krožnice trikotnika $ABC$. Naj bo tudi $|AB|+|AC|=2\cdot |BC|$. Dokaži, da je $ST\parallel BC$. % zvezek - dodatni MG + +\item Naj bo $p$ ploščina trikotnika $ABC$, $R$ polmer očrtane krožnice in $s'$ obseg pedalnega trikotnika. Dokaži, da velja $p=Rs'$. %Lopandic - nal 918 + +% Sstirikotniki + + +\item Naj bo $L$ poljubna točka v notranjosti paralelograma $ABCD$. Dokaži, da velja: + $$p_{LAB}+p_{LCD}=p_{LBC}+p_{LAD}.$$ %Lopandic - nal 890 + +\item Naj bo $P$ središče kraka $BC$ trapeza $ABCD$. Dokaži: + $$p_{APD}=\frac{1}{2}\cdot p_{ABCD}.$$ + +\item Naj bodo $o$ obseg, $v$ višina in $p$ ploščina tangentnega trapeza. Dokaži, da velja: $p=\frac{o\cdot v}{4}$. + +\item Načrtaj premico $p$, ki poteka skozi oglišče $D$ trapeza $ABCD$ ($AB>CD$), tako da razdeli ta trapez na dva ploščinsko enaka lika. + + +\item Načrtaj premici $p$ in $q$, ki potekata skozi oglišče $D$ kvadrata $ABCD$ in ga razdelita na ploščinsko enake like. + +\item Naj bo $ABCD$ kvadrat, $E$ središče njegove stranice $BC$ in $F$ točka, za katero je $\overrightarrow{AF}=\frac{1}{3}\cdot \overrightarrow{AB}$. Točka $G$ je četrto oglišče pravokotnika $FBEG$. Kolikšen del ploščine kvadrata $ABCD$ predstavlja ploščina trikotnika $BDG$? + +\item Naj bosta $\mathcal{V}$ in $\mathcal{V}'$ podobna večkotnika s koeficientom podobnosti $k$. Potem je: + $$p_{\mathcal{V}'}=k^2\cdot p_{\mathcal{V}}.$$ Dokaži. + +\item Naj bodo $a$, $b$, $c$ in $d$ dolžine stranic, $s$ polobseg in $p$ ploščina poljubnega štirikotnika. Dokaži, da velja: + $$p=\sqrt{(s-a)(s-b)(s-c)(s-d)}.$$ %Lopandic - nal 924 + +\item Naj bodo $a$, $b$, $c$ in $d$ dolžine stranic in $p$ ploščina tetivnotangentnega štirikotnika. Dokaži, da velja: + $$p=\sqrt{abcd}.$$ %Lopandic - nal 925 + +% Veckotniki + +\item Dan je pravokotnik $ABCD$ s stranicama dolžin $a=|AB|$ in $b=|BC|$. Izračunaj ploščino lika, ki predstavlja unijo pravokotnika $ABCD$ in njegove slike pri zrcaljenju čez premico $AC$. + +\item Naj bo $ABCDEF$ pravilni šestkotnik, točki $P$ in $Q$ pa središči njegovih stranic $BC$ in $FA$. Kolikšen del ploščine tega šestkotnika predstavlja ploščina trikotnika $PQD$? + + +% KKrog + +\item V kvadratu so včrtani štirje skladni krogi, tako da se vsak krog dotika dveh stranic in dveh krogov. Dokaži, da je vsota ploščin teh krogov enaka ploščini temu kvadratu včrtanega kroga. + +\item Izračunaj ploščino kroga, ki je včrtan trikotniku s stranicami dolžin 9, 12 in 15. %resitev 54 + +\item Naj bo $P$ središče osnovnice $AB$ trapeza $ABCD$, za katerega velja $|BC|=|CD|=|AD|=\frac{1}{2}\cdot |AB|=a$. Izrazi ploščino lika, ki ga določajo osnovnica $CD$ ter krajša krožna loka $PD$ in $PC$ krožnic s središčema $A$ in $B$, kot funkcijo osnovnice $a$. + +\item Tetiva $PQ$ ($|PQ|=d$) krožnice $k$ se dotika njene konciklične krožnice $k'$. Izrazi ploščino kolobarja, ki ga določata krožnici $k$ in $k'$, kot funkcijo tetive $d$. + +\item Naj bo $r$ polmer včrtanega kroga tetivnega večkotnika $\mathcal{V}$, ki je razdeljen na trikotnike $\triangle_1,\triangle_2,\ldots,\triangle_n$, tako da nobena dva trikotnika nimata skupnih notranjih točk. Naj bodo $r_1,r_2,\ldots , r_n$ polmeri včrtanih krožnic teh trikotnikov. Dokaži, da je: + $$\sum_{i=1}^n r_i\geq r.$$ + +\end{enumerate} + + +% DEL 3 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% INVERZIJA +%________________________________________________________________________________ + + + \del{Inversion} \label{pogINV} +\poglavje{Definition and Basic Properties of Inversion} +\label{odd9DefInv} + +V tem poglavju bomo definirali inverzijo - novo preslikavo, ki na +določen način predstavlja ‘‘zrcaljenje’’ čez krožnico. +Pričakujemo, da ima ta nova preslikava podobne lastnosti kot +zrcaljenje čez premico. Jasno je, da inverzija ne more biti izometrija, +toda kljub temu bomo navedli naslednje željene skupne lastnosti ‘‘dveh +zrcaljenj’’ (Figure \ref{sl.inv.9.1.1.pic}): + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.1.1.pic} +\caption{} \label{sl.inv.9.1.1.pic} +\end{figure} + +\begin{itemize} + \item zrcaljenje je bijektivna preslikava, + \item zrcaljenje je involucija, + \item zrcaljenje preslika eno od +območij ravnine, ki ga os (krožnica) zrcaljenja določa, v drugo +območje, + \item vse +fiksne točke zrcaljenja ležijo na osi (krožnici) tega +zrcaljenja, + \item če je $X'$ slika točke $X$ pri zrcaljenju, potem je premica + $XX'$ +pravokotna na os (krožnico) tega zrcaljenja. +\end{itemize} + + Preslikava, ki jo bomo sedaj +definirali, naj bi torej izpolnjevala vse te lastnosti. Namesto imena + ‘‘zrcaljenje čez krožnico’’ bomo raje uporabljali + termin ‘‘inverzija’’\index{inverzija}\footnote{Inverzijo je prvi vpeljal + nemški matematik +\index{Magnus, L. I.} \textit{L. I. Magnus} (1790--1861) leta 1831. +Prve ideje inverzije so se pojavile že pri starogrškem matematiku +\index{Apolonij} \textit{Apoloniju iz Perge} (3.--2. st. pr. n. +š.) in švicarskem matematiku \index{Steiner, J.} \textit{J. +Steinerju} (1796--1863)}. + + + V evklidski ravnini bomo inverzijo definirali na naslednji način. +Naj bo $k(O,r)$ krožnica v evklidski ravnini $\mathbb{E}^2$. Točka $X'$ je slika neke točke $X\in \mathbb{E}^2\setminus +\{O\}$ v \index{inverzija} \pojem{inverziji} $\psi_k$ glede na +krožnico $k$ (njena \index{inverzivna slika} \pojem{inverzivna slika}), če pripada poltraku $OX$ in velja $|OX|\cdot +|OX'|=r^2$, $k$ je \index{krožnica!inverzije}\pojem{krožnica +inverzije}, $O$ pa \index{središče!inverzije} \pojem{središče +inverzije}. + + Inverzijo $\psi_k$ glede na krožnico $k(O,r)$ bomo pogosto označevali + tudi $\psi_{O,r}$. + +Iz same definicije neposredno sledi, da so edine negibne točke +inverzije ravno točke na krožnici inverzije. In vse točke +krožnice inverzije so negibne. Velja torej ekvivalenca: +$$\psi_k(X)=X\hspace*{1mm}\Leftrightarrow\hspace*{1mm}X\in k.$$ + +Inverzija $\psi_k$ je bijektivna preslikava na množici +$\mathbb{E}^2\setminus \{O\}$. Velja tudi $\psi_k^{-1}=\psi_k$, torej je +$\psi_k^2$ je identična preslikava. Obe lastnosti sledita +neposredno iz definicije. Direktno sledi tudi, da se zunanje +točke krožnice $k$ z inverzijo $\psi_k$ preslikajo v notranje +točke in obratno (brez točke O). + + Iz same definicije sledi, da je premica $XX'$ pravokotna na + krožnici inverzije, kar pomeni, da je izpolnjena tudi zadnja + željena lastnost. + +Sedaj bomo konstruirali sliko $X'=\psi_k(X)$ poljubne točke $X$ +pri inverziji. Iz vsega povedanega (posebej iz dejstva +$\psi_k^{-1}=\psi_k$) sledi, da zadošča opis konstrukcije, ko je $X$ v zunanjosti krožnice $k$. + Naj bo v tem primeru $OX$ tangenta +krožnice $k$ v njeni točki $T$, potem iz $|OX|\cdot +|OX'|=r^2=|OT|^2$ sledi, da sta trikotnika $OTX$ in $OX'T$ podobna +in je zato $\angle TX'O=90^0$ (Figure \ref{sl.inv.9.1.2.pic}). Ker +je točka $X'$ tudi na poltraku $OX$, to točko dobimo kot +pravokotno projekcijo točke $T$ na poltrak $OX$. + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.1.2.pic} +\caption{} \label{sl.inv.9.1.2.pic} +\end{figure} + +%slikaNova7-4 +%\includegraphics[width=50mm]{slikaNova7-4.pdf} + + + Iz relacije v definiciji inverzije $|OX|\cdot +|OX'|=r^2$ sledi, da če se točka $X$ približuje središču $O$, + se njena slika $X'$ ‘‘oddaljuje proti neskončnosti’’. Velja tudi - če je +točka $X$ v bližini krožnice $k$, je tudi njena slika v bližini +te krožnice, seveda na drugi strani. Vse te ugotovitve lahko +zapišemo v bolj formalni obliki. + +\bizrek \label{invUrejenost} Če velja $\psi_k:A,B \mapsto A',B'$ in +$\mathcal{B}(O,A,B)$, + potem je tudi $\mathcal{B}(O,B',A')$. + \eizrek + + Točka $O$ nima svoje slike. Intuitivno je njena slika točka v +neskončnosti. Več o tem bomo povedali v poglavju \ref{odd9InvRavn}. + +Dokažimo naslednjo pomembno lastnost inverzije. + +\bizrek \label{invPodTrik} Naj bodo $O$, $A$ in $B$ tri +nekolinearne točke in $\psi_k$ inverzija glede na krožnico +$k(O,r)$. Če sta $A'$ in $B'$ sliki točk $A$ in $B$ v tej +inverziji, sta si trikotnika $OAB$ in $OB'A'$ podobna, torej: +$$\psi_{O,r}:A,B\mapsto A',B' \Rightarrow \triangle OAB \sim \triangle +OB'A'.$$ + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.1.3.pic} +\caption{} \label{sl.inv.9.1.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} Ker je $\psi_{O,r}:A,B\mapsto A',B'$, + je tudi +$OA\cdot OA'=OB\cdot OB'=r^2$ (Figure \ref{sl.inv.9.1.3.pic}). +Podobnost trikotnikov sedaj sledi iz $OA:OB'=OB:OA'$ in $\angle AOB +\cong \angle B'OA'$. + \kdokaz + + Naslednja trditev je direktna posledica prejšnjega izreka. + +\bizrek Naj bodo $O$, $A$ in $B$ tri nekolinearne točke in +$\psi_k$ inverzija glede na krožnico $k(O,r)$. Če sta $A'$ in +$B'$ slike točk $A$ in $B$ v tej inverziji, potem velja: + + (i) $\angle +OAB \cong \angle OB'A'$ in $\angle OBA \cong \angle OA'B'$, + +(ii) točke $A$, $B$, $B'$ in $A'$ so konciklične, + +(iii) krožnica, ki poteka skozi točke $A$, $B$, $B'$ in $A'$, je +pravokotna na krožnico inverzije. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.1.4.pic} +\caption{} \label{sl.inv.9.1.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} + \textit{(i)} Skladnost kotov sledi iz podobnosti trikotnikov $OAB$ in + $OB'A'$(prejšnji izrek \ref{invPodTrik}). + + + \textit{(ii)} Naj bo $l$ očrtana krožnica trikotnika + $ABA'$ (Figure \ref{sl.inv.9.1.4.pic}). Iz definicije inverzije sledi + $OA\cdot OA'=OB\cdot + OB'=r^2$. To pomeni, da je potenca točke $O$ na krožnico $l$ + enaka $OA\cdot OA'=OB\cdot + OB'$, torej točka $B'$ leži na tej krožnici. + + Omenimo, da tetivnost štirikotnika $AA'B'B$ sledi tudi direktno + iz $\angle +OAB \cong \angle OB'A'$, ker je potem $\angle A'AB + \angle BB'A' += 180^0 $. + + \textit{(iii)} Iz $\psi_{O,r}(A)=A'$ sledi, da je ena izmed točk + $A$ in $A'$ v notranjosti, druga pa v zunanjosti krožnice + inverzije $k$. Po posledici Dedekindovega aksioma + (\ref{DedPoslKrozKroz}) imata krožnici $k$ in $l$ dve skupni + točki; eno od njiju označimo s $T$. Ker je $OA\cdot OA'=OB\cdot + OB'=r^2=OT^2$, sledi, da je $OT$ tangenta krožnice $k$, + kar pomeni, da sta krožnici $k$ in $l$ pravokotni (izrek + \ref{pravokotniKroznici}). + \kdokaz + +Na enak način kot v delu (\textit{iii}) prejšnjega izreka, +dokazujemo tudi naslednjo trditev. + + \bizrek \label{invPravKrozn} Naj +bo $X'$ slika točke $X$ ($X\neq X'$) pri inverziji $\psi_k$ +glede na krožnico $k(O,r)$. Potem je vsaka krožnica, ki gre +skozi točki $X$ in $X'$ (tudi premica $XX'$) pravokotna na +krožnici inverzije $k$. + \eizrek + + V naslednjem primeru bomo videli, kako lahko inverzijo s pomočjo + določenih raztegov prevedemo v inverzijo s koncentrično + krožnico inverzije. + +\bzgled \label{invRazteg} Kompozitum inverzije $\psi_{S,r}$ in +raztega $h_{S,k}$ z istim središčem $S$ in pozitivnim +koeficientom ($k>0$) je inverzija s središčem $S$ in +koeficientom $r\sqrt{k}$. + \ezgled +\textbf{\textit{Proof.}} Naj bo $f=h_{S,k}\circ \psi_{S,r}$. Če za +poljubno točko $X$ označimo $X'=f(X)$, je najprej jasno, da točka +$X'$ leži na poltraku $SX$. Označimo še $X_1=\psi_{S,k}(X)$. Iz +definicij inverzije in raztega sledi $|SX_1| \cdot |SX| =r^2$ in +$|SX'| = k\cdot |SX_1| $. Zato je $|SX'| \cdot |SX| =k\cdot +r^2=\left(r\sqrt{k}\right)^2$ oz. $\psi_{S,r\sqrt{k}}(X)=X' =f(X)$. +Ker to velja za poljubno točko $X$, je $f=h_{S,k}\circ +\psi_{S,r}=\psi_{S,r\sqrt{k}}$. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Image of a Circle or Line Under an Inversion} \label{odd9SlokaKrozPrem} + +V tem razdelku bomo ugotovili, da inverzija ni kolineacija, kar pomeni, da +ne ohranja relacije kolinearnosti točk. Obravnavali bomo slike + premic in krožnic pri inverziji. Ker je definicijsko območje + inverzije evklidska ravnina brez središča inverzije, je smiselno + vpeljati naslednjo oznako: če je $\Phi$ poljubni lik +evklidske ravnine + in $S$ točka te ravnine, je + $$\Phi^S= \Phi \setminus + \{S\}.$$ + +\bizrek \label{InverzKroznVkrozn} + Naj bo $\psi_i$ inverzija glede na krožnico $i(S,r)$ evklidske ravnine. + Če sta $p$ premica in $k$ +krožnica te ravnine, velja (Figure \ref{sl.inv.9.2.1.pic}): + +(i) če $S\in p$, je $\psi_i(p^S)=p^S$, + +(ii) če $S\notin p$, je $\psi_i(p)=j^S$, kjer je $j$ krožnica, ki +poteka skozi točko $S$, + +(iii) če $S\in k$, je y $\psi_i(k^S)=q$, kjer je $q$ premica, ki +ne poteka skozi točko $S$; + +(iv) če $S\notin k$, je $\psi_i(k)=k'$, kjer je k' krožnica, ki +ne poteka skozi točko $S$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.2.1.pic} +\caption{} \label{sl.inv.9.2.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} + + (\textit{i}) Iz definicije inverzije sledi, da slika poljubne + točke + $X \in p^S$ leži na odprtem poltraku $SX$, zato velja $\psi_i(X)\in + p^S$. Ker je še $\psi_i^{-1}=\psi_i$, je poljubna točka $Y\in + p^S$ slika točke $\psi_i^{-1}(Y)=\psi_i(Y)\in + p^S$. Torej velja $\psi_i(p^S)=p^S$. + + (\textit{ii}) Naj bo $P$ pravokotna projekcija točke + $S$ na premici $p$ (Figure \ref{sl.inv.9.2.2.pic}). + Ker $S\notin p$, je potem $P\neq S$, zato +obstaja slika točke $P$ pri inverziji $\psi_i$ -- označimo jo s $P'$. +Naj bo $X$ poljubna točka premice $p$ ($X\neq P$) in $X'=\psi_i(X)$. +Po izreku \ref{invPodTrik} sta si trikotnika $SPX$ in $SX'P'$ podobna, +zato je $\angle SX'P' \cong \angle SPX=90^0$. Torej točka $X'$ leži +na krožnici nad premerom $SP'$ - označimo ga z $j$. Ker je $X'\neq +S$, velja $X'\in j^S$ oz. $\psi_i(p)\subseteq j^S$. Z obratnim +postopkom dokažemo, da je vsaka točka množice $j^S$ slika neke točke +premice $p$, zato je + $\psi_i(p)= j^S$. + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.2.2.pic} +\caption{} \label{sl.inv.9.2.2.pic} +\end{figure} + + (\textit{iii}) Direktna posledica (\textit{ii}), ker je $\psi_i$ involucija, + oz. $\psi_i^{-1}=\psi_i$. + + (\textit{iv}) (Figure \ref{sl.inv.9.2.2.pic}) Naj bosta $P$ in $Q$ + presečišči krožnice $k$ s premico, ki poteka skozi točko $S$, + in središče krožnice + $k$ (primer, ko sta krožnici $i$ in $k$ koncentrični, je enostaven). + Brez škode za splošnost predpostavimo, da velja $\mathcal{B}(S,P,Q)$. + Naj bo $P'=\psi_i(P)$, +$Q'=\psi_i(Q)$ in $X$ poljubna točka krožnice $k$, ki je +različna od točk $P$ in $Q$ ter $X'=\psi_i(X)$. Po izreku +\ref{invPodTrik} je $\triangle SPX \sim \triangle SX'P'$ in $\triangle SQX +\sim \triangle SX'Q'$. Zaradi tega je + $\angle SX'P' \cong \angle SPX$ in $\angle SX'Q'\cong \angle SQX$, torej: +$$\angle P'X'Q'=\angle SX'P'-\angle SX'Q' +=\angle SPX-\angle SQX=\angle PXQ=90^0.$$ + Točka $X'$ potem leži na +krožnici nad premerom $P'Q'$ - označimo jo s $k'$. Z obratnim +postopkom lahko dokažemo, da je tudi vsaka točka krožnice $k'$ +slika neke točke krožnice $k$, zato je $\psi_i(k)=k'$ + +Trditev analogno dokažemo tudi v primeru, kadar velja +$\mathcal{B}(P,S,Q)$. + \kdokaz + +Prejšnji izrek nam poda tudi efektiven način konstrukcije slike +premice oz. krožnice v različnih primerih. Uporabili bomo +oznake iz tega izreka (\ref{InverzKroznVkrozn}). + +V primeru (\textit{ii}) - slika \ref{sl.inv.9.2.2a.pic} - je dovolj +določiti sliko pravokotne projekcije $P$ središča inverzije $S$ +na premici $p$, ki jo preslikamo. Točka $P'=\psi_i(P)$ s +središčem inverzije določa premer krožnice $j$. Če pa premica +$p$ seka krožnico inverzije $i(S,r)$ npr. v točkah $M$ in $N$, +sta tidve točki fiksni in je slika premice $p$ očrtana krožnica +trikotnika $SMN$ (brez točke $S$). Tangenta krožnice inverzije +$i$ pa se preslika v krožnico (brez točke $S$), ki se od znotraj +dotika krožnice $i$. + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.2.2a.pic} +\caption{} \label{sl.inv.9.2.2a.pic} +\end{figure} + + +V primeru (\textit{iii}) naredimo obraten postopek konstrukciji iz +(\textit{ii}) in dobimo pravokotno projekcijo na premico, ki je +slika dane krožnice. Tudi ostali posebni primeri (ko krožnica +seka ali se dotika krožnice inverzije) so obratni primerom iz +(\textit{ii}). + + Tudi v primeru (\textit{iv}), + kadar se krožnica preslika v krožnico, nam je postopek + konstrukcije jasen iz samega izreka - slika $k'$ je določena s + točkama $P'$ in $Q'$. + Moramo pa biti pozorni, da se pri tem središče krožnice $k$ ne + preslika v središče nove krožnice $k'$! + Kje v tem primeru leži slika središča + te krožnice? Označimo z $O'$ sliko središča $O$ krožnice + $k$ pri inverziji $\psi_i$. + Naj bosta $t_1=ST_1$ in $t_2=ST_2$ tangenti krožnice $k$ + v točkah $T_1$ in $T_2$ ter $T'_1=\psi_i(T_1)$ in + $T'_2=\psi_i(T_2)$ +(Figure \ref{sl.inv.9.2.3.pic}). Jasno je, da velja $T'_1, T'_2 \in +k'$, toda $t_1$ in $t_2$ sta tudi tangenti krožnice $k'$ (če je +denimo $Y\in k'\cap t_1$, je tudi $X=\psi_i(Y)\in k\cap +t_1=\{T_1\}$, oz. $X=T'_1$). Po izreku \ref{invPodTrik} je + $\triangle ST_1O \sim \triangle SO'T'_1$ oz. $\angle SO'T'_1 \cong \angle + ST_1O=90^0$. Iz istih razlogov je tudi $\angle SO'T'_2 \cong \angle + ST_2O=90^0$. To pomeni, da so točke $T'_1$, $O'$ in $T'_2$ + kolinearne. Torej točko $O'$ lahko dobimo kot presečišče daljice + $T'_1T'_2$ in poltraka $SO$. + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.2.3.pic} +\caption{} \label{sl.inv.9.2.3.pic} +\end{figure} + +Še eno vprašanje se ponuja glede primera (\textit{iv}). +Katere krožnice so fiksne pri inverziji? Ena takšna je seveda +sama krožnica inverzije $i$, saj velja, da so vse njene točke +fiksne. Naslednji izrek bo odkril tudi vse druge možnosti. + + \bizrek \label{InverzKroznFiks} + Naj bo $\psi_i$ inverzija glede na krožnico $i(S,r)$. + Edine fiksne krožnice te inverzije so krožnica $i$ in krožnice, + ki so pravokotne na to krožnico, torej: + $$\psi_i(k)=k \Leftrightarrow k=i \vee k\perp i.$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.2.4.pic} +\caption{} \label{sl.inv.9.2.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} + + ($\Leftarrow$) Če je $k=i$, je trditev trivialna. Naj bosta $k$ in + $i$ +pravokotni krožnici (Figure \ref{sl.inv.9.2.4.pic}). S $T$ +označimo eno njuno presečišče. Premica $ST$ je tangenta +krožnice $k$ (izrek \ref{TangPogoj}). Naj bo $X$ poljubna točka +krožnice $k$ ($X\notin i$) in $X'$ drugo presečišče te +krožnice s poltrakom $SX$. Tedaj je: + $$p(S,k)= |SX|\cdot |SX'| = |OT|^2 = r^2,$$ +zato je $\psi_i(X)=X'\in k$. Analogno je vsaka točka $Y$ +krožnice $k$ slika neke točke te krožnice, zato velja +$\psi_i(k)=k$. + + ($\Rightarrow$) Naj bo $\psi_i(k)=k$. Če je $k\neq i$, potem + obstaja točka $X$ na +krožnici $k$, ki ne leži na krožnici $i$. Naj bo $X'=\psi_i(X)$. +Jasno je, da velja tudi $X'\notin i$. Iz definicije inverzije sledi, +da sta točki $X$ in $X'$ na poltraku z začetno točko $S$, kar +pomeni, da je $S$ zunanja točka krožnice $k$ (ker ni +$\mathcal{B}(X,S,X')$). Torej obstaja tangenta iz točke $S$ na +krožnico $k$ -- s $T$ označimo dotikališče. Tedaj je: + $$p(S,k)=|ST|^2 = |SX| \cdot |SX'| = r^2,$$ + zato točka $T$ leži na +krožnici $i$. To pomeni, da sta krožnici $k$ in $i$ pravokotni +(izrek \ref{TangPogoj}). + \kdokaz + + Omenimo, da analogno velja tudi za fiksne premice, le da + pravokotnost premice s krožnico inverzije pomeni, da premica poteka skozi + središče inverzije. Torej so edine fiksne premice inverzije + tiste, ki potekajo skozi središče inverzije (del (\textit{i}) izreka + \ref{InverzKroznVkrozn}). + + Prejšnji izrek velja tudi, če namesto inverzije govorimo o + zrcaljenju čez premico. Edine fiksne premice (krožnice) + tega zrcaljenja so namreč os zrcaljenja in tiste premice (krožnice), + ki so pravokotne na to os. To bo motiv, da na določen način + izenačimo premice in krožnice, s tem pa tudi osno zrcaljenje in + inverzijo. O tem bomo več izvedeli v naslednjem razdelku. + + \bzgled Dani so kot $\gamma$ ter daljici $c$ in $p$. +Načrtaj trikotnik $ABC$, če je $AB\cong c$, $\angle BCA\cong \gamma$ +in $|BA'|\cdot|BC|=p^2$ (daljica $AA'$ je višina trikotnika $ABC$). + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.2.5.pic} +\caption{} \label{sl.inv.9.2.5.pic} +\end{figure} + +\textbf{\textit{Solution.}} Naj bo $\triangle ABC$ iskani trikotnik, +ki izpolnjuje pogoje iz naloge (Figure \ref{sl.inv.9.2.5.pic}). +Najprej iz pogoja $\angle BCA\cong \gamma$ sledi, da točka $C$ +leži na loku $l$, ki je določen s tetivo $AB$ in obodnim kotom +$\gamma$. Ker je $A'$ nožišče višine, točka $A'$ leži na +krožnici $k$ nad premerom $AB$. Naj bo $\psi_i$ inverzija glede +na krožnico $i(B,p)$. Iz pogoja + $|BA'|\cdot|BC|=p^2$ sledi $\psi_i(A')=C$. Ker je še $A'\in k$, + velja tudi $C\in k'$, kjer je $k'=\psi_i(k)$ premica (izrek + \ref{InverzKroznVkrozn}). + + Če torej najprej načrtamo daljico $AB\cong c$, dobimo točko + $C$ kot presečišče premice $k'=\psi_i(k)$ z lokom $l$. +\kdokaz + +\bzgled \label{MiquelKroznice}(Miquelov\footnote{\index{Miquel, A.}\textit{A. Miquel} (1816–-1851), francoski +matematik, ki je ta izrek objavil leta 1840.} izrek o šestih +krožnicah) + Naj bodo $k_1$, $k_2$, $k_3$ in $k_4$ takšne krožnice, da + se krožnici $k_1$ in $k_2$ sekata v točkah $A$ in +$A_0$, krožnici $k_2$ in $k_3$ v točkah $B$ in $B_0$, +krožnici $k_3$ in $k_4$ v točkah $C$ in $C_0$, krožnici +$k_4$ in $k_1$ pa v točkah $D$ in $D_0$. Če so točke $A$, $B$, $C$ +in $D$ konciklične, so točke $A_0$, $B_0$, $C_0$ in $D_0$ +konciklične ali kolinearne. \index{izrek!Miquelov.} + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.7a.pic} +\caption{} \label{sl.inv.9.3.7a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Označimo s $k$ krožnico, ki jo +določajo točke $A$, $B$, $C$ in $D$. Naj bo $\psi_i$ inverzija +glede na poljubno krožnico $i$ s središčem v točki $A$ (Figure +\ref{sl.inv.9.3.7a.pic}) in + $$\psi_i:\hspace*{1mm}B,C,D,A_0,B_0,C_0,D_0 \mapsto + B',C',D',A'_0,B'_0,C'_0,D'_0.$$ + Ker gredo krožnice $k$, $k_1$ in +$k_2$ skozi točko $A$, so po izreku \ref{InverzKroznVkrozn} +njihove slike $k'$, $k'_1$ in $k'_2$ pri inverziji $\psi_i$ +premice, ki določajo oglišča trikotnika $A_0'B'D'$. Krožnici +$k_3$ in $k_4$ pa ne gresta skozi točko $A$, zato sta njuni sliki +krožnici $k'_3$ in $k'_4$, ki se sekata v točkah $C'\in B'D'$ in +$C'_0$. Naj bo $k'_0$ očrtana krožnica trikotnika +$A'_0D'_0B'_0$. Iz zgleda \ref{Miquelova točka} sledi, da se +krožnice $k'_3$, $k'_4$ in $k'_0$ sekajo v eni točki - $C'_0$, +oz. $A'_0,B'_0,C'_0,D'_0\in k'_0$. Zaradi tega je tudi +$A_0,B_0,C_0,D_0\in k_0=\psi_i(k'_0)$. Po izreku +\ref{InverzKroznVkrozn} pa je $k_0$ krožnica ali premica. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{The Inversive Plane} +\label{odd9InvRavn} + +Motiv za naslednjo obravnavo je dejstvo, da pri inverziji +definicijsko območje ni cela evklidska ravnina - središče inverzije +nima svoje slike. To nam pogosto oteži delo, kot je to npr. pri +formulaciji izreka \ref{InverzKroznVkrozn}, kjer moramo vedno paziti, +kdaj premica oz. krožnica poteka skozi središče inverzije. Tudi slike +premice in krožnice niso vedno cela premica oz. krožnica, ker moramo v +določenih primerih izločiti središče inverzije. + +Zaradi vsega navedenega se ponuja druga rešitev; namesto da iz +definicijskega območja izključimo središče inverzije, lahko +evklidski ravnini formalno dodamo eno točko, ki bo slika središča inverzije +pri tej inverziji. Iz definicije inverzije za sliko $X'$ točke $X$ +pri inverziji $\psi_{S,k}$ velja: + $|SX'|\cdot |SX|=r^2$. Če je $S=X$, potem formalno dobilmo + $|SX'|\cdot 0=r^2$, kar pomeni, da mora biti + $|SX'|=\infty$. Tako novo točko intuitivno vidimo kot + ‘‘neskončno oddaljeno točko’’, zato jo bomo formalno + označili s simbolom $\infty$ in imenovali \index{točka!v neskončnosti}\pojem{točka v neskončnosti}. Množico, ki nastane z dodajanjem + te nove točke evklidski ravnini, imenujemo \index{inverzivna + ravnina} \pojem{inverzivna ravnina}, + ki jo bomo označili z $\widehat{E}^2$. Torej: + $$\widehat{E}^2=E^2 \cup \{\infty\}.$$ + Vendar moramo biti pozorni - ne smemo namreč mešati +inverzivne ravnine z \index{razširjena evklidska ravnina} +\pojem{razširjeno evklidsko ravnino}, ki jo dobimo, če evklidski +ravnini dodamo eno (neskončno oddaljeno) premico in opredelimo, da se +vzporedni premici na njej sekata. Razširjena evklidska ravnina je +model t. i. projektivne geometrije, ki smo jo že omenjali v uvodu. + +V primeru inverzivne ravnine smo torej dodali le eno točko +$\infty$ in pri tem zahtevali, da za vsako inverzijo $\psi_{S,k}$ +velja $\psi_{S,k}(S)=\infty$. Ker je $\psi_{S,k}$ involucija +(velja $\psi_{S,k}^{-1}=\psi_{S,k}$), naj po dogovoru velja tudi +$\psi_{S,k}(\infty)=S$. + +V delu (\textit{iii}) izreka \ref{InverzKroznVkrozn} smo videli, da +je slika lika $k^S$ (krožnice brez točke $S$) pri inverziji +$\psi_{S,k}$ premica $q$, ki ne poteka skozi točko $S$. Če pa liku +$k^S$ dodamo točko $S$, bomo za sliko dobili množico $q \cup +\{\infty\}$. Zato bomo v inverzivni ravnini na naraven način +premicam dodelili točko $\infty$. Če takšne premice z dodano +točko $\infty$ in krožnice skupaj imenujemo $i$-krožnice, lahko +prevedemo izrek \ref{InverzKroznVkrozn} in rečemo, da se z +inverzijo $i$-krožnica preslika v $i$-krožnico. Toda videli smo +tudi določene analogije zrcaljenja čez premico in inverzijo. +Tako bomo obe preslikavi v inverzivni ravnini imenovali +$i$-inverziji. Ni težko videti, da lahko potem posplošimo izrek +\ref{InverzKroznVkrozn} in ga zapišemo v enostavnejši +obliki. + + + \bizrek \label{InverzInvRavKvK} + Z $i$-inverzijo se $i$-krožnica preslika v $i$-krožnico. + \eizrek + + Dejstvo, da v inverzivni ravnini vsaka premica vsebuje točko + $\infty$, pomeni, da se dve vzporednici sekata v tej točki. Premici, + ki se sekata (v navadni točki evklidske ravnine) imata v + inverzivni ravnini dve skupni točki. + Ker imata vzporednici + le eno skupno točko, rečemo tudi, da se dotikata v + točki $\infty$. Slika dveh vzporednic pri $i$-inverziji bosta + bodisi vzporednici (če gre za osno zrcaljenje) bodisi + dve krožnici, ki se dotikata v središču inverzije (če gre + za inverzijo). Dve krožnici, ki se dotikata (ker imata le eno + skupno točko), se preslikata bodisi v vzporednici bodisi v + krožnici, ki se dotikata. Podobno velja tudi v primeru dotika + premice in krožnice. Vse te primere v inverzivni ravnini lahko + veliko krajše formuliramo v naslednji obliki. + + \bizrek \label{InverzDotik} + Naj bo $\psi_i$ poljubna inverzija. + Če se $i$-krožnici $k$ in $l$ dotikata, se dotikata tudi + $i$-krožnici $\psi_i(k)$ in $\psi_i(l)$. + \eizrek + +S pomočjo tega izreka bomo najprej dokazali naslednji pomemben +izrek. + + \bizrek \label{InverzKonf} + Inverzija je \index{konformna preslikava}konformna preslikava, kar pomeni: + + Kot, pod katerim se premici $p$ in $q$ sekata v presečišču +$A$, je enak kotu, pod katerim se sekata njuni inverzni sliki $p'$ +in $q'$ v pripadajoči točki $A'$. + \eizrek + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.1.pic} +\caption{} \label{sl.inv.9.3.1.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $\psi_i$ inverzija glede na +krožnico $i(S,r)$, pri kateri se premici $p$ in $q$ in njuno +presečišče $A$ preslikajo v $p'$, $q'$ in $A'$ (Figure +\ref{sl.inv.9.3.1.pic}). + +Primer, če je $A=S$, je trivialen, ker sta tedaj premici $p$ in +$q$ nepremični. + + Predpostavimo najprej, da $A\notin i$. V tem primeru je jasno $A\neq + A'$. + + Naj bo točka +$P$ presečišče pravokotnice premice $p$ v točki $A$ in +simetrale daljice $AA'$ ter $k_p$ krožnica s središčem $P$, ki +gre skozi točko $A$. Iz konstrukcije krožnice $k_p$ je jasno, da +tudi točka $A'$ leži na tej krožnici in je premica $p$ njena +tangenta v točki $A$. Če na podoben način s $Q$ označimo +presečišče pravokotnice premice $q$ v točki $A$ in simetrale +daljice $AA'$ ter $k_q$ krožnico s središčem $Q$, ki gre skozi +točko $A$, sledi, da točka $A'$ leži na krožnici $k_q$ in je +premica $q$ njena tangenta v točki $A$. + +Ker gresta obe krožnici $k_p$ in $k_q$ skozi par točk $A$, $A'$ +inverzije $\psi_i$, sta $k_p$ in $k_q$ pravokotni na krožnico +inverzije $i$ (izrek \ref{invPravKrozn}). Po izreku +\ref{InverzKroznFiks} je potem $\psi_i(k_p)=k_p$ in +$\psi_i(k_q)=k_q$. + +Premici $p$ in $q$ se po vrsti dotikata krožnic $k_p$ in $k_q$. +Po izreku \ref{InverzDotik} se dotikata $i$-krožnici $p'$ in +$\psi_i(k_p)=k_p$, oz. $q'$ in $\psi_i(k_q)=k_q$. Zaradi vsega +tega velja: + $$\angle p,q \cong \angle k_p,k_q \cong \angle p',q'.$$ + + Kadar je $A\in i$, primer prevedemo na prejšnjega + (z inverzijo glede na koncentrično krožnico) z uporabo izreka + \ref{invRazteg}, ker razteg + ohranja velikost kotov. + \kdokaz + + Jasno je, da bi bil dokaz praktično enak, če bi bili $p$ in $q$ + krožnici oz. krožnica in premica. Zato lahko prejšnji izrek + zapišemo v bolj splošni obliki. + + \bizrek + Vsaka $i$-inverzija ohranja kote med dvema $i$-krožnicama. + \eizrek + + Poseben primer prejšnjega izreka se nanaša na prave kote. Iz + tega sledi, da je pravokotnost invarianta inverzij in velja + naslednji izrek. + + \bizrek \label{InverzKonfPrav} + Vsaka $i$-inverzija preslika dve pravokotni $i$-krožnici v + pravokotni $i$-krožnici. + \eizrek + + Sedaj bomo uporabili dokazane trditve, ki veljajo + v evklidski ravnini. + Dejstvo, da se krožnica, ki poteka skozi središče + inverzije, s to inverzijo preslika v premico, nam omogoča + reševanje različnih načrtovalnih nalog, kjer bomo namesto + iskane krožnice najprej konstruirali njeno sliko - premico. To + se nanaša tudi na druge naloge, kjer bomo neko trditev, ki se + nanaša na krožnico, z inverzijo prevedli v ekvivalentno trditev, + ki se nanaša na premico. V obeh primerih je zaželeno, da imamo + vsaj eno točko krožnice, ki jo potem izberemo za središče inverzije. + + Omenimo še, da bomo zaradi vsega navedenega pogosto rekli, + da je slika premice $p$ ($S\notin p$) pri inverziji $\psi_{S,r}$ kar + krožnica $j$ ($j\ni S$) namesto $j^S$. In obratno - v istem + primeru bomo pisali $\psi_{S,r}(j)=p$. + + \bzgled Dane so krožnica $k$ ter točki $A$ in $B$. Načrtaj + krožnico (premico) $x$, ki je pravokotna na krožnico $k$ ter + poteka skozi točki $A$ in $B$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.2.pic} +\caption{} \label{sl.inv.9.3.2.pic} +\end{figure} + +\textbf{\textit{Solution.}} Naj bo $x$ krožnica, ki je +pravokotna na krožnico $k$ in + poteka skozi točki $A$ in $B$, ter $\psi_i$ +inverzija glede na poljubno krožnico $i$ s središčem $A$ (Figure +\ref{sl.inv.9.3.2.pic}). + +Predpostavimo najprej, da $A\notin k$. V tem primeru je +$k'=\psi_i(k)$ krožnica in $x'=\psi_i(x)$ premica, ki gre skozi +točko $B'=\psi_i(B)$ (izrek \ref{InverzKroznVkrozn}). Po izreku +\ref{InverzKonfPrav} iz $x \perp k$ sledi $x' \perp k'$. Zato je +$x'$ premica, ki poteka skozi točko $B'$ in je pravokotna na +krožnico $k'$ oz. poteka skozi njeno središče. + +Premico $x'$ torej lahko načrtamo kot premico, ki gre skozi +točki $B'=\psi_i(B)$ in središče krožnice $k'=\psi_i(k)$. Na +koncu je $x=\psi_i(x')$. + +Če pa velja $A\in k$, je $k'$ premica (izrek +\ref{InverzKroznVkrozn}) in premico $x'$ načrtamo kot njeno +pravokotnico skozi točko $B'$. + +Ker vedno lahko narišemo eno samo pravokotnico $x'$, obstaja +natanko ena rešitev tudi za $i$-krožnico $x$. Toda $x$ je +krožnica natanko tedaj, ko $S\notin x'$ oz. kadar točke $A$, +$B$ in središče krožnice $k$ niso kolinearne. V primeru +kolinearnosti teh točk pa je rešitev $x$ premica. + \kdokaz + + \bzgled \label{TriKroznInv} + Če se med tremi krožnicami $k_1$, $k_2$ in $k_3$ + neke ravnine po dve od zunaj dotikata, + je krožnica $k$, ki jo določajo dotikališča, + pravokotna na vsako od krožnic $k_1$, $k_2$ in $k_3$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.3.pic} +\caption{} \label{sl.inv.9.3.3.pic} +\end{figure} + +\textbf{\textit{Proof.}} Označimo z $A$ dotikališče krožnic +$k_2$ in $k_3$, z $B$ dotikališče krožnic $k_1$ in $k_3$ ter s +$C$ dotikališče krožnic $k_2$ in $k_1$ (Figure +\ref{sl.inv.9.3.3.pic}). Krožnica $k$ je potem očrtana krožnica +trikotnika $ABC$. + +Naj bo $\psi_{A,r}$ inverzija s središčem $A$ in poljubnim +polmerom $r$. Krožnici $k_2$ in $k_3$ se dotikata v središču +inverzije $A$, zato se s to inverzijo preslikata v premici $k'_2$ +in $k'_3$ (izrek \ref{InverzKroznVkrozn}), ki nimata skupnih točk +oz. $k'_2\parallel k'_3$. Krožnica $k_1$, ki ne poteka skozi točko +$A$ in se krožnic $k_2$ in $k_3$ dotika v točkah $C$ in $B$, se +preslika v krožnico $k'_1$ (izrek \ref{InverzKroznVkrozn}), ki se +dotika premic $k'_2$ in $k'_3$ (izrek \ref{InverzDotik}) v točkah +$C'=\psi_{A,r}(C)$ in $B'=\psi_{A,r}(B)$. Torej sta premici $k'_2$ +in $k'_3$ vzporedni tangenti krožnice $k'_1$ v točkah $C'$ in +$B'$, zato je $B'C'$ njuna skupna pravokotnica, daljica $B'C'$ +pa je premer krožnice $k'_1$. Premica $B'C'$ je slika krožnice $k$, +ki poteka skozi središče inverzije $A$ in skozi točki $B$ in $C$ +(izrek \ref{InverzKroznVkrozn}), oz. $\psi_{A,r}(k)=k'=B'C'$. Ker +je premica $k'=B'C'$ pravokotna na premici $k'_2$ in $k'_3$ ter +na krožnico $k'_1$, je po izreku \ref{InverzKonfPrav} krožnica +$k$ pravokotna na vsako od krožnic $k_1$, $k_2$ in $k_3$. + \kdokaz + + \bzgled Po dve od štirih krožnic se od zunaj dotikata v + točkah $A$, $B$, $C$ in $D$. Dokaži, da so točke $A$, $B$, $C$ + in $D$ bodisi konciklične bodisi kolinearne. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.3c.pic} +\caption{} \label{sl.inv.9.3.3c.pic} +\end{figure} + +\textbf{\textit{Proof.}} Označimo s $k$ krožnico (oz. premico), +ki poteka skozi točke $B$, $C$ in $D$. Naj bodo $k_1$, $k_2$, $k_3$ +in $k_4$ takšne krožnice, da se $k_1$ in $k_2$ dotikata v točki +$A$, $k_1$ in $k_4$ v točki $B$, $k_3$ in $k_4$ v točki $C$ ter +$k_3$ in $k_2$ v točki $D$. Označimo s $\psi_i$ inverzijo glede +na poljubno krožnico $i$ s središčem $A$ (Figure +\ref{sl.inv.9.3.3c.pic}). Naj bodo $B'$, $C'$ in $D'$ slike točk +$B$, $C$ in $D$ ter $k'_1$, $k'_2$, $k'_3$ in $k'_4$ slike +krožnic $k_1$, $k_2$, $k_3$ in $k_4$ v tej inverziji. Iz izrekov +\ref{InverzKroznVkrozn} in \ref{InverzDotik} sledi, da sta $k'_3$ +in $k'_4$ krožnici, ki se dotikata v točki $C'$, $k'_1$ in +$k'_2$ pa vzporedni premici, ki sta tangenti krožnic $k'_4$ in +$k'_3$ v točkah $B'$ in $D'$. Iz vzporednosti premic $k'_1$ in +$k'_2$ sledi + $\angle k'_2,D'C' \cong \angle k'_1,B'C'$, zato sta po izreku + \ref{ObodKotTang} + skladna tudi središčna kota $D'S_2C'$ in $B'S_1C'$. Zaradi tega + je $\angle D'C'S_2 \cong \angle B'C'S_1$, kar pomeni, da so + točke $B'$, $C'$ in $D'$ kolinearne oz. je $k'$ premica. Po + izreku \ref{InverzKroznVkrozn} njena slika glede na inverzijo + $k=\psi_i(k')$ poteka skozi središče inverzije, zato velja + $A,B,C,D\in k$. + \kdokaz + + \bzgled Naj bodo $k$, $l$ in $j$ tri medsebojno pravokotne + krožnice s skupnimi tetivami $AB$, $CD$ in $EF$. Dokaži da + se očrtani krožnici trikotnikov $ACE$ in $ADF$ dotikata v + točki $A$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.3a.pic} +\caption{} \label{sl.inv.9.3.3a.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $AB$ skupna tetiva krožnic $k$ in +$l$, $CD$ skupna tetiva krožnic $l$ in $j$ ter $EF$ skupna +tetiva krožnic $j$ in $k$ (Figure \ref{sl.inv.9.3.3a.pic}). +Označimo z $x$ in $y$ očrtani krožnici trikotnikov $ACE$ in +$ADF$ in s $\psi_i$ inverzijo glede na poljubno krožnico $i$ s +središčem $A$. Naj bo: + \begin{eqnarray*} + && \psi_i:\hspace*{1mm}B,C,D,E,F \mapsto B',C',D',E',F' \hspace*{2mm} + \textrm{ in + }\\ + && \psi_i:\hspace*{1mm}k,l,j,x,y \rightarrow k',l',j',x',y'. + \end{eqnarray*} + Ker velja $A\in k, l, x, y$ in $A\notin j$, + so po izreku \ref{InverzKroznVkrozn} + $k'=E'F'$, $l'=C'D'$, $x'=E'C'$ in + $y'=D'F'$ premice, $j'\ni C',D',E',F'$ pa krožnica. Iz + medsebojne pravokotnosti krožnic $k$, $l$ in $j$ ter + dejstva $B\in k\cap l$ sledi, da sta premici $k'$ in $l'$ + pravokotni v točki $B'$, krožnica $j'$ pa pravokotna na + obe premici $k'$ in $l'$ (izrek \ref{InverzKonfPrav}). To + pomeni, da sta $C'D'$ in $E'F'$ pravokotna premera krožnice + $j'$, zato je štirikotnik $E'C'F'D'$ kvadrat in je $E'C' \parallel + D'F'$. Iz vzporednosti premic $x'$ in $y'$ sledi, da se njuni + inverzni sliki (krožnici $x$ in $y$) dotikata v središču + inverzije $A$. + \kdokaz + + \bzgled Naj bosta $k$ in $l$ krožnici neke ravnine s središčema + $O$ in $S$. Naj bodo $t_i$ tangente krožnice $k$, + ki sekajo krožnico $l$ v točkah $A_i$ in $B_i$. Dokaži, da + obstaja krožnica, ki se dotika vseh očrtanih krožnic trikotnikov + $SA_iB_i$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.4.pic} +\caption{} \label{sl.inv.9.3.4.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $\psi_l$ inverzija glede na +krožnico $l$ (Figure \ref{sl.inv.9.3.4.pic}). Premice $t_i$ se s +to inverzijo preslikajo v krožnice $t'_i$, ki so očrtane +krožnice trikotnikov $SA_iB_i$ (izrek \ref{InverzKroznVkrozn}). +Ker se premice $t_i$ dotikajo krožnice $k$, se vse krožnice $t'_i$ +dotikajo krožnice $k'=\psi_l(k)$ (izrek \ref{InverzDotik}). + \kdokaz + + \bzgled + Naj bo $ST$ premer krožnice $k$, $t$ tangenta te krožnice v + točki $T$ ter $PQ$ in $PR$ njeni tangenti v točkah $Q$ in $R$. + Dokaži, da če so $L$, $Q'$ in $R'$ presečišča poltrakov $SP$, $SQ$ in + $SR$ s premico $t$, je točka $L$ središče daljice + $Q'R'$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.5.pic} +\caption{} \label{sl.inv.9.3.5.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $\psi_i$ inverzija glede na +krožnico $i(S,ST)$ (Figure \ref{sl.inv.9.3.5.pic}). Točka $T$ je +pravokotna projekcija središča inverzije $S$ na premici $t$. Ker +je še $\psi_i(T)=T$, iz izreka \ref{InverzKroznVkrozn} sledi +$\psi_i(t)=k$ oz. $k'=\psi_i(k)=t$. Zato je tudi $\psi_i(Q)=Q'$ +in $\psi_i(R)=R'$. Označimo $P'=\psi_i(P)$. Iz $L\in SP$ sledi, +da so točke $S$, $P$, $L$ in $P'$ kolinearne. Tangenti $q=PQ$ in +$r=PR$ krožnice $k$ se preslikata v krožnici $q'$ in $r'$, ki +potekata skozi točko $S$ in se dotikata premice $k'$ v točkah $Q'$ in $R'$ (izreka +\ref{InverzKroznVkrozn} in \ref{InverzDotik}). Torej je premica +$t=k'$ skupna tangenta krožnic $q'$ in $r'$, ki se sekata v +točkah $S$ in $P'$. Točka $L$ pa leži na njuni potenčni osi +$SP'$, zato je + $|LQ'|^2=|LR'|^2$, torej je točka $L$ središče daljice $Q'R'$. +\kdokaz + + + \bzgled Naj bodo $a$, $b$ in $c_0$ krožnice + s premeri $PQ$, $PR$ in $RQ$, kjer je $\mathcal{B}(P,R,Q)$. + Naj bo $c_0$, $c_1$, $c_2$, + ... $c_n$, ... zaporedje krožnic na istem bregu premice $PQ$, ki + se dotikajo krožnic $a$, $b$ in se vsaka krožnica v zaporedju + dotika prejšnje. Dokaži, da je oddaljenost središča krožnice + $c_n$ od premice $PQ$ $n$-krat večja od premera te + krožnice\footnote{Ta problem + je obravnaval \index{Pappus} \textit{Pappus iz Aleksandrije} (4. st.). + Vzorec, ki nastane s + polkrožnicami $a$, $b$, $c_0$, je raziskoval že \index{Arhimed} + \textit{Arhimed + iz Sirakuze} (3. st. pr. n. š.) - glej zgled \ref{arbelos}.}\\ + (primer iz knjige \cite{Cofman}). + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.6.pic} +\caption{} \label{sl.inv.9.3.6.pic} +\end{figure} + +\textbf{\textit{Proof.}} (Figure \ref{sl.inv.9.3.6.pic}). Naj bo +$i$ krožnica s središčem v točki $P$, ki je pravokotna na +krožnico $c_n$ (krožnica $i$ poteka skozi dotikališči tangent iz +točke $P$ na krožnico $c_n$). Z inverzijo $\psi_i$ glede na to +krožnico se premica $l=PQ$ in krožnici $a$ in $b$ preslikajo v +premice $l$, $a'$ in $b'$ (izrek \ref{InverzKroznVkrozn}); pri tem +sta premici $a'$ in $b'$ pravokotni na premico $l$ (izrek +\ref{InverzKonfPrav}). Krožnice $c_0$, $c_1$, $c_2$, ..., $c_n$, +... se z isto inverzijo preslikajo v skladne krožnice $c'_0$, +$c'_1$, $c'_2$, ..., $c'_n$, ... , ki se vse dotikajo vzporednic + $a'$, $b'$ (izrek \ref{InverzDotik}) + in so vse skladne krožnici $c_n$, ker je + $c'_n=\psi_i(c_n)=c_n$ (izrek \ref{InverzKroznFiks}). +Ker je tudi $c_0'\perp l$, središče krožnice $c'_0$ leži na +premici $l$, zato je oddaljenost središča krožnice $c_n=c'_n$ +od te premice $n$-krat večje od premera te krožnice. + \kdokaz + + + + \bzgled + Naj bo $t$ skupna zunanja tangentna krožnica + krožnic $k$ in $l$, ki se od zunaj dotikata v + točki $A$, ter $c_0$, $c_1$, $c_2$, ... $c_n$, ... zaporedje + krožnic, ki se dotikajo krožnic $k$ in $l$, vsaka krožnica v + zaporedju se dotika prejšnje, krožnica $c_0$ pa se dotika tudi + krožnice $t$. Dokaži, da obstaja krožnica (ali premica), ki je + pravokotna na vsako krožnico iz danega zaporedja. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.8.pic} +\caption{} \label{sl.inv.9.3.8.pic} +\end{figure} + +\textbf{\textit{Proof.}} Naj bo $i$ poljubna krožnica s središčem +$A$ in $\psi_i$ inverzija glede na to krožnico (Figure +\ref{sl.inv.9.3.8.pic}). Krožnici $k$ in $l$ se pri tej inverziji +preslikata v premici $k'$ in $l'$ (izrek \ref{InverzKroznVkrozn}), ki +sta vzporedni, ker se krožnici $k$ in $l$ dotikata v središču +inverzije $A$. Krožnice iz zaporedja $c_0$, $c_1$, $c_2$, ..., +$c_n$, ... pa se preslikajo v krožnice $c'_0$, $c'_1$, $c'_2$, ..., +$c'_n$, ..., kjer se vsaki dve sosednji dotikata in se vse krožnice +dotikajo vzporednic $k'$ in $l'$ (izrek \ref{InverzDotik}). Zato so +vse krožnice iz tega + zaporedja medsebojno skladne. Premica $n'$, ki jo določajo + njihova +središča, je somernica premic $k'$ in $l'$. Premica $n'$ + je pravokotna na krožnice $c'_0$, $c'_1$, $c'_2$, ..., $c'_n$, ..., + zato je njena slika $n=\psi_i(n')$ pravokotna na krožnice + $c_0$, $c_1$, $c_2$, ..., $c_n$, ... (izrek \ref{InverzKonf}). + Na koncu lahko ugotovimo, da je $n$ krožnica natanko tedaj, + ko premica $n'$ ne +gre skozi točko $A$ oz. kadar krožnici $k$ in $l$ nista skladni, +sicer je $n$ premica. \kdokaz + + + \bzgled Naj bodo $A$, $B$, $C$ in $D$ štiri poljubne komplanarne + točke. Dokaži, da je kot, pod katerim se sekata očrtani + krožnici trikotnikov $ABC$ in $ABD$, enak kotu, pod katerim se + sekata očrtani + krožnici trikotnikov $ACD$ in $BCD$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.9.pic} +\caption{} \label{sl.inv.9.3.9pic} +\end{figure} + +\textbf{\textit{Proof.}} Označimo s $k_A$, $k_B$, $k_C$ in $k_D$ +očrtane krožnice trikotnikov $BCD$, $ACD$, $ABD$ in $ABC$ (Figure +\ref{sl.inv.9.3.9pic}). Z inverzijo $\psi_i$ glede na poljubno +krožnico $i$ s središčem $A$ se naša trditev prevede v +ekvivalenten izrek \ref{ObodKotTang}. + \kdokaz + + + + \bzgled Naj bodo $A$, $B$ in $C$ točke, ki ležijo na premici + $l$, + in $P$ točka izven te premice. Dokaži, da so točka $P$ in + središča očrtanih krožnic trikotnikov $APB$, $BPC$ in $APC$ + štiri konciklične točke. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.3.10.pic} +\caption{} \label{sl.inv.9.3.10pic} +\end{figure} + +\textbf{\textit{Proof.}} Označimo s $K_A$, $K_B$, $K_C$ +središča očrtanih krožnic trikotnikov $BPC$, $APC$ in $APB$ +(Figure \ref{sl.inv.9.3.10pic}). Če je $X=\mathcal{S}_{K_A}(P)$, +$Y=\mathcal{S}_{K_B}(P)$ in $Z=\mathcal{S}_{K_C}(P)$ oz. +$h_{P,2}(K_A)=X$, $h_{P,2}(K_B)=Y$ in $h_{P,2}(K_C)=Z$, iz +koncikličnosti točk $X$, $Y$, $Z$ in $P$ sledi koncikličnost +točk $K_A$, $K_B$, $K_C$ in $P$ (iz $X,Y,Z,P\in k$ sledi +$K_A,K_B,K_C,P\in h^{-1}_{P,2}$). Dokažimo torej, da so točke +$X$, $Y$, $Z$ in $P$ konciklične. + +Naj bo $\psi_{P,r}$ inverzija s središčem $P$ in poljubnim +polmerom $r$ ter + $$\psi_{P,r}:\hspace*{1mm} A,B,C,X,Y,Z\mapsto A',B',C',X',Y',Z'.$$ +Inverzija $\psi_{P,r}$ preslika očrtane krožnice trikotnikov $BPC$, +$APC$ in $APB$ v premice $B'C'$, $A'C'$ in $A'B'$ (izrek +\ref{InverzKroznVkrozn}). Ker so $PX$, $PY$ in $PZ$ premeri teh +krožnic, so točke $X'$, $Y'$ in $Z'$ pravokotne projekcije središča +inverzije $P$ na premice $B'C'$, $A'C'$ in $A'B'$. Premica $l$ se +po izreku \ref{InverzKroznVkrozn} preslika v očrtano krožnico +trikotnika $A'B'C'$, ki gre tudi skozi točko $P$. Po izreku +\ref{SimpsPrem} točke $X'$, $Y'$ in $Z'$ ležijo na +\index{premica!Simsonova} Simsonovi premici $s$, kar pomeni (izrek +\ref{InverzKroznVkrozn}), da točke $X$, $Y$ in $Z$ ležijo na +krožnici $\psi_{P,r}(s)$, ki poteka skozi točko $P$, zato so točke + $X$, $Y$, $Z$ in $P$ konciklične. + \kdokaz + +%________________________________________________________________________________ + \poglavje{Metric Properties of Inversion} \label{odd9MetrInv} + +Iz same definicije je jasno, da inverzija ni izometrija. V +prejšnjem poglavju smo ugotovili, da inverzija ne ohranja +relacije kolinearnosti točk, kar pomeni, da ni niti +transformacija podobnosti. Vendar nas zanima, kako se spreminja +razdalja med točkami, čeprav se z inverzijo daljica $AB$ v splošnem +primeru preslika v lok $A'B'$. Odgovor bo dal +naslednji pomemben izrek. + + \bizrek \label{invMetr} Če sta $A'$ in $B'$ ($A',B'\neq S$) sliki točk $A$ in + $B$ pri inverziji $\psi_{S,r}$, potem velja: + $$|A'B'|=\frac{r^2\cdot |AB|}{|SA|\cdot |SB|}$$ + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.4.1.pic} +\caption{} \label{sl.inv.9.4.1.pic} +\end{figure} + + \textbf{\textit{Proof.}} Obravnavali bomo dva možna +primera. + +\textit{(i)} Naj bodo točke $S$, $A$ in $B$ nekolinearne (Figure +\ref{sl.inv.9.4.1.pic}). Po izreku \ref{invPodTrik} sta trikotnika +$ASB$ in $B'SA'$ podobna in je $A'B':AB=SB':SA$. Ker je $B'$ +slika točke $B$ pri inverziji $\psi_{S,r}$, velja tudi $|SB|\cdot +|SB'|=r^2$ oz. $|SB'|=\frac{r^2}{|SB|}$. Iz teh dveh relacij +sledi: + $$|A'B'|=\frac{|SB'|\cdot |AB|}{|SA|}= + \frac{r^2\cdot |AB|}{|SA|\cdot |SB|}.$$ + + +\textit{(ii)} Naj bodo $S$, $A$ in $B$ kolinearne točke. Brez +škode za splošnost predpostavimo, da velja $\mathcal{B}(S,A,B)$. +Tedaj je $\mathcal{B}(S,B',A')$ (izrek \ref{invUrejenost}), zato +velja: + $$|A'B'|=|SA'|-|SB'|= +\frac{r^2}{|SA|}-\frac{r^2}{|SB|}= \frac{(|SB|- |SA|)\cdot +r^2}{|SA|\cdot |SB|} + = \frac{r^2\cdot |AB|}{|SA|\cdot |SB|} ,$$ kar je bilo treba dokazati. \kdokaz + + Iz prejšnjega izreka vidimo, da razdalja med slikama točk $A'$ in $B'$ + narašča, če se vsaj ena izmed originalov $A$ in $B$ + približa središču inverzije $S$, kar je logično, kajti videli + smo, da je intuitivno slika tega središča točka v neskončnosti. + + V razdelku \ref{odd7Ptolomej} smo že dokazali Ptolomejev \index{izrek!Ptolomejev splošni} izrek (\ref{izrekPtolomej}), ki se nanaša na + tetivne + štirikotnike. Sedaj bomo to trditev posplošili. + +\bizrek + Če je $ABCD$ poljubni konveksni štirikotnik, potem je: + $$|AB|\cdot |CD|+|BC|\cdot |AD|\geq |AC|\cdot |BD|.$$ + Enakost velja natanko tedaj, ko je $ABCD$ tetivni +štirikotnik. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.4.2a.pic} +\caption{} \label{sl.inv.9.4.2a.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $\psi_{A,r}$ inverzija s središčem v +točki $A$ (Figure \ref{sl.inv.9.4.2a.pic}). S $k$ označimo očrtano +krožnico trikotnika $ABD$. Naj bodo $B'$, $C'$ in $D'$ slike +točk $B$, $C$ in $D$ ter $k'$ slika krožnice $k$ pri inverziji +$\psi_{A,r}$. Po izreku \ref{InverzKroznVkrozn} je $k'$ premica, +ki vsebuje točke $B'$ in $D'$. + Iz prejšnjega izreka \ref{invMetr} sledi: + $$|B'C'|=\frac{r^2\cdot |BC|}{|AB|\cdot |AC|}, \hspace*{2mm} + |C'D'|=\frac{r^2\cdot |CD|}{|AC|\cdot |AD|} \hspace*{1mm} \textrm{ in } + \hspace*{1mm} |B'D'|=\frac{r^2\cdot |BD|}{|AB|\cdot |AD|}.$$ + Za točke $B'$, $C'$ in $D'$ velja trikotniška neenakost + \ref{neenaktrik}: + $$B'C'+C'D'\geq B'D',$$ +kjer enakost velja natanko tedaj, ko so točke $B'$, $C'$ in $D'$ +kolinearne (in $\mathcal{B}(B',C',D')$) oziroma ko $C'\in k'$ (in +$\mathcal{B}(B',C',D')$). To pa velja natanko tedaj, ko krožnica $k$ +vsebuje točko $C$ oziroma ko je štirikotnik $ABCD$ tetiven (in +konveksen). + Prejšnjo neenakost +lahko zapišemo tudi v obliki: + $$\frac{r^2\cdot |BC|}{|AB|\cdot |AC|}+\frac{r^2\cdot |CD|}{|AC|\cdot |AD|} + \geq\frac{r^2\cdot |BD|}{|AB|\cdot |AD|}, \hspace*{2mm} \textrm{oz.}$$ + $$|AB|\cdot |CD|+|BC|\cdot |AD|\geq |AC|\cdot |BD|,$$ kar je bilo treba dokazati. \kdokaz + + Naslednja trditev bo posplošitev zgleda \ref{zgledTrikABCocrkrozP} oz. + \ref{zgledTrikABCocrkrozPPtol}. + +\bzgled \label{zgledABCPinv} Naj bo $k$ očrtana krožnica pravilnega +trikotnika $ABC$. Če je $P$ poljubna točka v ravnini tega +trikotnika, tedaj velja ekvivalenca: točka $P$ ne leži na krožnici $k$, +natanko tedaj, ko obstaja trikotnik s stranicami, ki so skladne z +daljicami $PA$, $PB$ in $PC$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.4.3.pic} +\caption{} \label{sl.inv.9.4.3.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $\psi_{P,r}$ inverzija s središčem v +točki $P$ (Figure \ref{sl.inv.9.4.3.pic}). Naj bodo $A'$, $B'$ in +$C'$ slike točk $A$, $B$ in $C$ ter $k'$ slika krožnice $k$ pri +inverziji $\psi_{P,r}$. + Iz izreka \ref{invMetr} sledi: + $$|A'B'|=\frac{r^2\cdot |AB|}{|PA|\cdot |PB|}, \hspace*{2mm} + |B'C'|=\frac{r^2\cdot |BC|}{|PB|\cdot |PC|} \hspace*{1mm} \textrm{ in } + \hspace*{1mm} |A'C'|=\frac{r^2\cdot |AC|}{|PA|\cdot |PC|}.$$ +Ker je trikotnik $ABC$ pravilen, iz prejšnjih treh relacij sledi: +$$A'B':B'C':A'C'=PC:PA:PB.$$ + +Sedaj lahko začnemo z dokazovanjem iskane ekvivalence: + \begin{eqnarray*} + P \notin k & \Leftrightarrow& \psi_{P,r}(k) \textrm{ predstavlja krožnico + \hspace*{2mm}(izrek \ref{InverzKroznVkrozn})}\\ + & \Leftrightarrow& \textrm{točke }A', B', C' \textrm{ so nekolinearne}\\ + & \Leftrightarrow& \textrm{daljice }A'B', B'C', A'C' \textrm{ so stranice + nekega trikotnika}\\ + & \Leftrightarrow& \textrm{daljice }PC, PA, PB \textrm{ so stranice + nekega trikotnika,} + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + Še ena posplošitev bo povezana s Torricellijevo točko (izrek + \ref{izrekTorichelijev}). + + \bizrek \label{izrekToricheliFerma} + Točka, za katero je vsota razdalj od + oglišč nekega trikotnika minimalna, je +Torricellijeva\footnote{\index{Torricelli, E.} \textit{E. +Torricelli} (1608--1647), italijanski matematik in fizik. Lastnost iz +izreka je dokazal francoski matematik \index{Fermat, P.} \textit{P. +Fermat} (1601--1665). Zaradi tega se ta točka imenuje tudi +\index{točka!Torricellijeva}\index{točka!Fermatova}\pojem{Fermatova +točka}.} točka tega trikotnika. + \eizrek + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.4.4.pic} +\caption{} \label{sl.inv.9.4.4.pic} +\end{figure} + + \textbf{\textit{Proof.}} Uporabimo enake oznake kot pri izreku + \ref{izrekTorichelijev} (Figure \ref{sl.inv.9.4.4.pic}). Dokazali + smo, +da se krožnice $k$, $l$ in $j$ sekajo v Torricellijevi točki $P$ +trikotnika $ABC$. Ker je $BEC$ pravilni trikotnik, iz izreka +\ref{zgledTrikABCocrkrozP} sledi $|PB|+|PC|=|PE|$ oz.: + $$|PA|+|PB|+|PC|=|AE|.$$ + Dokažimo, da je ta vsota minimalna ravno za Torricellijevo točko + $P$. Naj bo $P'\neq P$. Točka $P'$ potem +ne leži na nobeni od krožnic $k$, $l$ in $j$. Brez škode za +splošnost naj bo $P' \notin k$. Iz prejšnje trditve +\ref{zgledABCPinv} sledi, da obstaja trikotnik s stranicami $P'E$, +$P'B$ in $P'C$. Zaradi tega in trikotniške neenakosti (izrek +\ref{neenaktrik}) je $|P'B| + |P'C| +> |P'E|$ oziroma: + $$|P'A| + |P'B| + |P'C| > |P'A| + |P'E| \geq |AE|=|PA|+|PB|+|PC|,$$ kar je bilo treba dokazati. \kdokaz + + + + + \bnaloga\footnote{37. IMO, India - 1996, Problem 2.} + Let $P$ be a point inside triangle $ABC$ such that + $$\angle APB -\angle ACB = \angle APC -\angle ABC.$$ + Let $D$, $E$ be the incentres of triangles $APB$, $APC$, respectively. Show +that $AP$, $BD$, $CE$ meet at a point. + \enaloga + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.4.5.pic} +\caption{} \label{sl.inv.9.4.5.pic} +\end{figure} + + \textbf{\textit{Proof.}} +Naj bo $r>\max\{|AB|, |AC|\}$ poljuben in $\psi_{A,r}$ inverzija s +središčem v točki $A$ in polmerom $r$ ter $B'$, $C'$ in $P'$ +slike točk $B$, $C$ in $P$ pri inverziji $\psi_{A,r}$ (Figure +\ref{sl.inv.9.4.5.pic}). + +Po izreku \ref{invPodTrik} sta trikotnika $ABC$ in $AC'B'$ +podobna, zato je: + $$\angle AC'B'\cong \angle ABC \textrm{ in } \angle +AB'C'\cong \angle ACB.$$ + Iz podobnosti $\triangle ABP \sim \triangle +AP'B'$ in $\triangle ACP \sim \triangle AP'C'$ pa sledi: + $$\angle AB'P'\cong \angle APB \textrm{ in } \angle +AC'P'\cong \angle APC.$$ + Če uporabimo dobljene štiri relacije skladnosti kotov in + začetni pogoj iz naloge ($\angle APB -\angle ACB = \angle APC -\angle + ABC$), dobimo: + \begin{eqnarray*} + \angle C'B'P' &=& \angle AB'P'-\angle AB'C' = \angle APB - \angle + ACB = \\ &=& \angle APC -\angle ABC = \angle AC'P' -\angle AC'B' + =\\ + &=& \angle B'C'P' + \end{eqnarray*} +Sedaj iz $\angle C'B'P'\cong \angle B'C'P'$ sledi $P'B'\cong +P'C'$. Če uporabimo relacijo iz izreka \ref{invMetr}, dobimo + $\frac{|PB| \cdot r^2}{|AP|\cdot |AB|}=\frac{|PC| \cdot r^2}{|AP|\cdot + |AC|}$ oz.: + $$\frac{|PB| }{ |AB|}=\frac{|PC| }{ |AC|}.$$ + Naj bo $BD \cap AP =X$ in $CE \cap AP =Y$. Potrebno je še + dokazati, da velja $X=Y$. Ker sta $D$ in $E$ središči + včrtanih krožnic trikotnikov $APB$ +in $APC$, sta premici $BD$ in $CE$ simetrali (notranjih) kotov +$ABP$ in $ACP$. Zato je (izrek \ref{HarmCetSimKota}) + $$ \frac{\overrightarrow{PX}}{\overrightarrow{XA}}=\frac{PB}{BA} + =\frac{PC}{CA}=\frac{\overrightarrow{PY}}{\overrightarrow{YA}}.$$ + Iz $ \frac{\overrightarrow{PX}}{\overrightarrow{XA}} + =\frac{\overrightarrow{PY}}{\overrightarrow{YA}}$ sledi (izrek + \ref{izrekEnaDelitevDaljice}) $X=Y$, kar pomeni, da se premice $AP$, $BD$ in $CE$ sekajo v isti +točki $X=Y$. + \kdokaz + +%________________________________________________________________________________ +\poglavje{Inversion and Harmonic Conjugate Points} +\label{odd9InvHarm} + +V razdelku \ref{odd7Harm} smo videli, da harmonično četverico točk +definiramo na dva ekvivalentna načina. Za štiri kolinearne točke +$A$, $B$, $C$ in $D$ je $\mathcal{H}(A,B;C,D)$, če je izpolnjen en +izmed dveh (ekvivalentnih) pogojev: +\begin{itemize} + \item $\frac{\overrightarrow{AC}}{\overrightarrow{CB}}=- + \frac{\overrightarrow{AD}}{\overrightarrow{DB}}$, + \item obstaja tak štirikotnik $ABCD$, da velja + $PQ \cap RS=A$, $PS \cap QR=B$, $PR \cap AB=C$ in $QS \cap AB=D$. +\end{itemize} + + +Sedaj bomo raziskovali še nekatere možnosti za ekvivalentne definicije + relacije harmonične četverice točk - najprej s pomočjo inverzije. + + \bizrek \label{invHarm} Naj bo $AB$ premer krožnice $i$ ter $C$ in $D$ točki + premice $AB$, ki sta različni od $A$ in $B$. Če je $\psi_i$ + inverzija glede na krožnico $i$, potem velja: + $$\mathcal{H}(A,B;C,D) \hspace*{1mm} \Leftrightarrow \hspace*{1mm} + \psi_i(C)=D.$$ + \eizrek +\textbf{\textit{Proof.}} Naj bo $O$ središče daljice $AB$ (Figure +\ref{sl.inv.9.6.1.pic}). Tedaj je: + \begin{eqnarray*} + \mathcal{H}(A,B;C,D) &\hspace*{1mm} \Leftrightarrow + \hspace*{1mm}& + \frac{\overrightarrow{AC}}{\overrightarrow{CB}}=- + \frac{\overrightarrow{AD}}{\overrightarrow{DB}}\\ + &\hspace*{1mm} \Leftrightarrow + \hspace*{1mm}& + \frac{\overrightarrow{OC}-\overrightarrow{OA}} + {\overrightarrow{OB}-\overrightarrow{OC}}=- + \frac{\overrightarrow{OD}-\overrightarrow{OA}} + {\overrightarrow{OB}-\overrightarrow{OD}}\\ + &\hspace*{1mm} \Leftrightarrow + \hspace*{1mm}& + \frac{\overrightarrow{OC}+\overrightarrow{OB}} + {\overrightarrow{OB}-\overrightarrow{OC}}=- + \frac{\overrightarrow{OD}+\overrightarrow{OB}} + {\overrightarrow{OB}-\overrightarrow{OD}}\\ + &\hspace*{1mm} \Leftrightarrow + \hspace*{1mm}& \overrightarrow{OC} \cdot \overrightarrow{OD} = + OB^2\\ + &\hspace*{1mm} \Leftrightarrow + \hspace*{1mm}& \psi_i(C)=D, + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.6.1.pic} +\caption{} \label{sl.inv.9.6.1.pic} +\end{figure} + +Naslednji izrek bo dal četrto možnost ekvivalentne definicije + relacije harmonične četverice točk. + + \bizrek \label{harmPravKrozn} + Naj bodo $A$, $B$, $C$ in $D$ različne kolinearne točke ter $k$ in $l$ + krožnici nad premeroma $AB$ in +$CD$. Tedaj velja ekvivalenca: +$$\mathcal{H}(A,B;C,D) \Leftrightarrow k\perp l.$$ + \eizrek + + \textbf{\textit{Proof.}} Naj bosta $O$ in $S$ središči krožnic $k$ in $l$ + (Figure \ref{sl.inv.9.6.1.pic}). Iz obeh strani ekvivalence + sledi, da se krožnici +sekata. Eno njuno presečišče označimo s $T$. Iz prejšnjega +izreka \ref{invHarm} je: + $$\mathcal{H}(A,B;C,D) \hspace*{1mm} \Leftrightarrow \hspace*{1mm} + \psi_k(C)=D.$$ + Torej moramo dokazati le še: +$$k\perp l \hspace*{1mm} \Leftrightarrow \hspace*{1mm} + \psi_k(C)=D.$$ + Toda sedaj imamo: + \begin{eqnarray*} + k\perp l \hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} OT\perp TS \hspace*{2mm} + \textrm{ (izrek \ref{pravokotniKroznici})}\\ + &\Leftrightarrow& \hspace*{1mm} OT \textrm{ je tangenta krožnice + }l \hspace*{2mm}\textrm{ (izrek \ref{TangPogoj})}\\ +&\Leftrightarrow& \hspace*{1mm} \overrightarrow{OC}\cdot \overrightarrow{OD} += OT^2\hspace*{2mm}\textrm{ (izrek \ref{izrekPotenca})}\\ +&\Leftrightarrow& \hspace*{1mm} \psi_k(C)=D, + \end{eqnarray*} + kar je bilo treba dokazati. \kdokaz + + Iz prejšnjega izreka \ref{harmPravKrozn} direktno sledita (že prej dokazani) dejstvi, da + iz $\mathcal{H}(A,B;C,D)$ sledi $\mathcal{H}(C,D;A,B)$ oz. + $\mathcal{H}(B,A;C,D)$. + + Že od prej nam je znano, da za tri kolinearne točke $A$, $B$ in $C$, + kjer točka $C$ ni središče daljice $AB$, + obstaja ena sama točka $D$, tako da velja $\mathcal{H}(A,B;C,D)$. Torej je za tri +dane točke v harmonični četverici točk četrta enolično določena. +Eno od možnih konstrukcij te točke nam da prejšnji izrek +\ref{invHarm} ($D=\psi_k(C)$). Samo dve točki $A$ in $B$ pa nista +dovolj, da bi določili drugi par točk $C$ in $D$. Takšnih parov +$(C,D)$, za katere velja $\mathcal{H}(A,B;C,D)$, je neskončno mnogo (Figure +\ref{sl.inv.9.6.2.pic}). Za njuno določitev + je potreben še en pogoj. Takšne pogoje bomo obravnavali v + naslednjih dveh primerih. + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.6.2.pic} +\caption{} \label{sl.inv.9.6.2.pic} +\end{figure} + + \bzgled + Dane so točke $A$, $B$ in $S$. Konstruiraj takšni točki $C$ in $D$, + da je $S$ središče +daljice $CD$ in velja $\mathcal{H}(A,B;C,D)$. + \ezgled + + \textbf{\textit{Solution.}} +Naj bo $k$ krožnica nad premerom $AB$. Če je $l$ krožnica nad +premerom $CD$, je $S$ središče te krožnice. Iz izreka +\ref{harmPravKrozn} sledi $k \perp l$ (Figure +\ref{sl.inv.9.6.3.pic}). Torej je dovolj načrtati krožnico $l$, ker +sta potem točki $C$ in $D$ presečišči te krožnice s premico $AB$. Krožnico $l$ pa lahko načrtamo, če najprej konstruiramo tangento +krožnice $k$ iz točke $S$ v točki $T$. + \kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.6.3.pic} +\caption{} \label{sl.inv.9.6.3.pic} +\end{figure} + + + \bzgled \label{harmDaljicad} + Dani so daljica $d$ in točki $A$ in $B$. Konstruiraj takšni točki $C$ + in $D$, da velja $\mathcal{H}(A,B;C,D)$ in $CD\cong d$. + \ezgled + \textbf{\textit{Solution.}} + + Dovolj je konstruirati središče $S$ daljice $CD$, nato pa nadaljujemo enako + kot v + prejšnjem zgledu. Pravokotni trikotnik $OTS$ lahko +konstruiramo, ker sta znani obe kateti $|OT| = \frac{1}{2}\cdot +|AB|$ in $|ST| = \frac{1}{2}\cdot |CD|= \frac{1}{2}\cdot |d|$. Iz +njega pa dobimo hipotenuzo $OS$ (Figure \ref{sl.inv.9.6.3.pic}). + \kdokaz + +V naslednjih primerih si bomo ogledali uporabo prejšnjih dveh +konstrukcij. + + + +\bzgled + Načrtaj trikotnik s podatki $v_a$, $l_a$ in $b-c$. + \ezgled + +\textbf{\textit{Solution.}} + Uporabili bomo oznake iz velike naloge \ref{velikaNaloga} +(Figure \ref{sl.inv.9.6.4.pic}). Najprej lahko načrtamo pravokotni +trikotnik $AA'E$, ker je $AA'\cong v_a$ in $AE\cong l_a$. Po +zgledu \ref{harmVelNal} je $\mathcal{H}(A',E;P,Pa)$. Iz velike +naloge pa sledi $PP_a=b -c$, zato lahko po prejšnjem zgledu +\ref{harmDaljicad} konstruiramo točki $P$ in $P_a$. Nato +narišemo središče $S$ včrtane krožnice trikotnika $ABC$, na +koncu še včrtano krožnico, njuni tangenti iz točke $A$ ter +oglišči $B$ in $C$. + \kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.6.4.pic} +\caption{} \label{sl.inv.9.6.4.pic} +\end{figure} + + + +\bzgled + Načrtaj trikotnik s podatki $v_a$, $a$ in $r+r_a$. + \ezgled + +\textbf{\textit{Solution.}} Tudi v tem primeru bomo uporabili +oznake iz velike naloge \ref{velikaNaloga} (Figure +\ref{sl.inv.9.6.4.pic}). Po zgledu \ref{harmVelNal} je +$\mathcal{H}(A,L;A',La)$. Ker je $AA'\cong v_a$ in $LL_a =r+r_a$, +lahko po zgledu \ref{harmDaljicad} narišemo najprej daljico +$AA'\cong v_a$, nato pa še točki $L$ in $L_a$. Tako dobimo +$r\cong LA'$ in $r_a\cong L_aA'$. + +Iz velike naloge sledi $RR_a = a$. To pomeni, da lahko narišemo +pravokotni trapez $SRR_aS_a$ ($RR_a=a$, $SR=r$ in $S_aR_a=r_a$). +Nato konstruiramo včrtano krožnico $k(S,r)$ in pričrtano +krožnico $k_a(S_a,r_a)$ ter na koncu njune skupne tangente (dve +zunanji in še eno notranjo), ki so nosilke stranic trikotnika +$ABC$. + \kdokaz + +%________________________________________________________________________________ + \poglavje{Feuerbach Points} +\label{odd9Feuerbach} + +V tem razdelku bomo obravnavali še nekatere lastnosti včrtane in +pričrtanih krožnic trikotnika. + + \bzgled Naj bo $ABC$ trikotnik s polobsegom $s$ ter $D$ in $E$ takšni + točki + na + premici $BC$, da velja $|AD|=|AE|=s$. Dokaži, da se očrtana + krožnica trikotnika $ADE$ in stranici $BC$ pričrtana krožnica + trikotnika $ABC$ dotikata. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.5.0.pic} +\caption{} \label{sl.inv.9.5.0.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bo $l$ očrtana krožnica + trikotnika $AED$ in $P_a$, $Q_a$ ter $R_a$ točke, v katerih +se pričrtana krožnica $k_a$ trikotnika $ABC$ dotika stranice + $BC$ ter nosilk $AC$ in $AB$ (Figure \ref{sl.inv.9.5.0.pic}). Z $i$ + označimo krožnico s središčem $A$ in polmerom $AE$ (oz. + $AD$). +Iz velike naloge \ref{velikaNaloga} sledi $|AR_a|=|AQ_a|=s=|AE|=|AD|$, +kar + pomeni, da točki $R_a$ in $Q_a$ ležita na krožnici $i$. + Po izreku \ref{pravokotniKroznici} sta krožnici $i$ in $k_a$ + pravokotni. Zato se krožnica $k_a$ pri inverziji $\psi_i$ preslika + vase (izrek \ref{InverzKroznFiks}), premica $BC$ pa v krožnico $l$ + (izrek \ref{InverzKroznVkrozn}). + Ker se premica $BC$ dotika + krožnice $k_a$ v točki $P_a$, se njuni sliki (krožnici $l$ in $k_a$) + dotikata v + točki $T=\psi_i(P_a)$ (izrek \ref{InverzDotik}). + \kdokaz + +\bzgled \label{InvOcrtVcrt} Naj bodo $P$, $Q$ in $R$ točke, v +katerih se včrtana krožnica + dotika stranic trikotnika $ABC$. Dokaži, da so višinska točka trikotnika + $PQR$ in središči + očrtane ter včrtane krožnice trikotnika $ABC$ kolinearne + točke. + \ezgled + + \begin{figure}[!htb] +\centering +\input{sl.inv.9.5.0a.pic} +\caption{} \label{sl.inv.9.5.0a.pic} +\end{figure} + + \textbf{\textit{Proof.}} Naj bosta $l(O,R)$ in $k(S,r)$ očrtana in + včrtana krožnica + trikotnika $ABC$ ter $\psi_k$ inverzija glede na krožnico $k$ + (Figure \ref{sl.inv.9.5.0a.pic}). + +Označimo še $A'=SA \cap QR$, $B'=SB \cap PR$ in $C'=SC \cap PQ$. +Iz skladnosti trikotnikov $ARA'$ in $AQA'$ (izrek \textit{SAS} \ref{SKS}) +sledi, da je točka $A'$ središče stranice $QR$ in velja $AS +\perp RQ$. Analogno sledi tudi, da sta točki $B'$ in $C'$ +središči daljic $PR$ in $PQ$ ter velja $BS \perp PR$ in $CS +\perp PQ$. + +Iz postopka konstrukcije slike točke pri inverziji sledi: + $$\psi_k:A,B,C \mapsto A',B',C'.$$ + Torej se očrtana krožnica $l$ z inverzijo $\psi_k$ preslika v + očrtano krožnico trikotnika $A'B'C'$ (izrek \ref{InverzKroznVkrozn}). + Ta pa je Eulerjeva krožnica $e_1$ trikotnika $PQR$, ker gre skozi + središča njegovih stranic. Središče te krožnice - točka + $E_1$ - leži na Eulerjevi premici $e_p$ trikotnika $PQR$, ki je + določena z njegovo višinsko točko $V_1$ in s središčem + očrtane krožnice $S$. + + Dokažimo še, da točka $O$ leži na premici $e_p$. Čeprav se z + inverzijo središče + krožnice $l$ (točka $O$) ne preslika v središče krožnice + (točko $E_1$), leži točka $E_1$ na premici $SO$. To + pomeni tudi, da točka $O$ leži na premici $SE_1=e_p$, zato so + točke $V_1$, $O$ in $S$ kolinearne. + \kdokaz + + Naslednja enostavna trditev je le uvod v izrek o t. i. Feuerbachovih + točkah. + +\bizrek Naj bo $k$ včrtana krožnica trikotnika $ABC$ in $k_a$ njegovi stranici $BC$ pričrtana krožnica. Če je $A_1$ +središče stranice $BC$ ter $i$ krožnica s središčem $A_1$ in +polmerom $\frac{1}{2}|b-c|$, je +$$\psi_i:k,k_a\rightarrow k, k_a.$$ + \eizrek + \textbf{\textit{Proof.}} Naj bosta $P$ in $P_a$ dotikališči + krožnic +$k$ in $k_a$ z njegovo stranico $BC$ (Figure +\ref{sl.inv.9.5.1.pic}). Iz velike naloge \ref{velikaNaloga} +sledi, da točki $P$ in $P_a$ ležita na krožnici $i$. Torej sta po +izreku \ref{pravokotniKroznici} krožnici $k$ in $k_a$ +pravokotni na krožnico $i$, zato je $\psi_i:k,k_a\rightarrow k, +k_a$ (izrek \ref{InverzKroznFiks}). + \kdokaz + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.5.1.pic} +\caption{} \label{sl.inv.9.5.1.pic} +\end{figure} + + + \bizrek \index{krožnica!Eulerjeva} + Eulerjeva krožnica trikotnika se dotika + včrtane + krožnice in vseh treh pričrtanih + krožnic tega trikotnika\footnote{Točke dotika so t. i. + \index{točka!Feuerbachova} \pojem{Feuerbachove točke} tega + trikotnika. \index{Feuerbach, K. W.} \textit{K. W. Feuerbach} + (1800--1834), nemški matematik, ki je to trditev dokazal leta 1822.}. + \eizrek + +\textbf{\textit{Proof.}} + Uporabili bomo oznake kot pri veliki nalogi + \ref{velikaNaloga}. Naj bo še $e$ Eulerjeva krožnica tega + trikotnika (Figure \ref{sl.inv.9.5.1.pic}). Iz prejšnje trditve inverzija $\psi_i$ glede na + krožnico $i$ s središčem $A_1$ in polmerom + $\frac{1}{2}|b-c|$ (oz. vsebuje točki $P$ in $P_a$) preslika krožnici + $k$ in $k_a$ vsako vase. Določimo sliko Eulerjeve krožnice $e$ + pri inverziji $\psi_i$. + +Krožnica $e$ vsebuje središče inverzije $A_1$, zato se z +inverzijo preslika v neko premico $e'$ (izrek +\ref{InverzKroznVkrozn}). Ostane nam še dokaz, da se +premica $e'$ dotika krožnic $k$ in $k_a$ (izrek +\ref{InverzDotik}). Najprej premica $e'$ vsebuje točko +$\psi_i(A')=E$, ki je presečišče simetrale notranjega kota pri +oglišču $A$ s stranico $BC$. (izreka \ref{harmVelNal} in +\ref{invHarm}). Toda premica $e'$ vsebuje tudi točki +$\psi_i(B_1)=B'_1$ in $\psi_i(C_1)=C'_1$. Po izreku +\ref{invPodTrik} sta si trikotnika $A_1B_1C_1$ in $A_1C'_1B'_1$ +podobna. Iz tega sledi, da premica $e'$ s premico $AB$ določa kot +$ACB$. Ker premica $e'$ seka premico $BC$ v točki $E$, ki leži +na simetrali kota $BAC$, $e'$ predstavlja drugo skupno tangento +krožnic $k$ in $k_a$. Označimo s $F'$ in $F'_a$ dotikališči +premice $e'$ s krožnicama $k$ in $k_a$. Potem se krožnica $e$ +dotika krožnic $k$ in $k_a$ v točkah $F=\psi_i(F')$ in +$F_a=\psi_i(F'_a)$. Na enak način dokažemo, da se $e$ dotika +tudi krožnic $k_b$ in $k_c$. + \kdokaz + + +%________________________________________________________________________________ + \poglavje{Stainer's Theorem} +\label{odd9Stainer} + +V tem razdelku bomo na eleganten način (s pomočjo inverzije) +dokazali Stainerjev izrek, ki je povezan s problemom obstoja +zaporedja krožnic, ki se zaporedoma ciklično dotikajo, hkrati pa se +dotikajo tudi dveh danih, od znotraj mimobežnih, krožnic. Jasno +je, da takšno zaporedje v splošnem primeru (za poljubni od znotraj +mimobežni krožnici) ne obstaja. Toda dokazali bomo, da če za +dve krožnici $k$ in $l$ obstaja vsaj eno +takšno zaporedje, potem začetni člen tega zaporedja lahko +izberemo kot poljubno krožnico, ki se dotika krožnic $k$ in $l$. + + Najprej bomo rešili dve +pomožni nalogi. + + + + \bzgled \label{StainerjevLema1} + Naj bosta $k_1(S_1,r_1)$ in $k_2(S_2,r_2)$ dve krožnici neke ravnine. Načrtaj krožnico s +središčem na premici $S_1S_2$, ki je pravokotna na obe krožnici. + \ezgled + + + \begin{figure}[!htb] +\centering +\input{sl.inv.9.7.0.pic} +\caption{} \label{sl.inv.9.7.0.pic} +\end{figure} + + + \textbf{\textit{Solution.}} + Naj bosta $k_1(S_1,r_1)$ in $k_2(S_2,r_2)$ poljubni + krožnici. V primeru, če rešitev obstaja, središče + iskane krožnice pripada +potenčni premici $l=p(k_1,k_2)$ dveh krožnic (Figure +\ref{sl.inv.9.7.0.pic}). To je primer, kadar je presečišče $S$ +potenčne premice $l$ in centrale $S_1S_2$ zunanja točka obeh +krožnic oz. ko sta krožnici $k_1$ in $k_2$ mimobežni (od znotraj +ali od zunaj). Če iz točke $S$ načrtamo tangento na eno od krožnic +(npr. $k_1$), dobimo tudi polmer $r=ST_1$ iskane krožnice +$k(S,r)$, kjer je točka $T_1$ dotikališče tangente iz $S$ na +krožnico $k_1$. Naj bo točka $T_2$ dotikališče tangente iz $S$ +na krožnico $k_2$. Ker točka $S$ leži na potenčni premici dveh +krožnic, je: + $$|ST_1|=p(S,k_1)=p(S,k_2)=|ST_2|$$ + krožnica $k(S,r)$ pravokotna na krožnici $k_1(S_1,r_1)$ in + $k_2(S_2,r_2)$ po izreku \ref{pravokotniKroznici}. + \kdokaz + + + %slika + + \bzgled \label{StainerjevLema2} + Če je $k$ krožnica v notranjosti krožnice $l$, obstaja inverzija, + ki krožnici +$k$ in $l$ preslika v dve koncentrični krožnici. + \ezgled + + + \begin{figure}[!htb] +\centering +\input{sl.inv.9.7.1.pic} +\caption{} \label{sl.inv.9.7.1.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + Naj bo $p$ premica, določena s središčema krožnic $k$ in $l$; + hkrati je tudi pravokotna na +obe krožnici. Iz prejšnjega zgleda \ref{StainerjevLema1} sledi, +da obstaja krožnica $n$ s središčem na premici $p$, ki je +pravokotna na krožnici $k$ in $l$ (Figure +\ref{sl.inv.9.7.1.pic}). + Ker središče krožnice $n$ leži na +premici $p$, je krožnica $n$ pravokotna tudi na premico $p$. Z $I$ +in $J$ označimo presečišči krožnice $n$ s premico $p$. Naj bo $i$ +poljubna krožnica s središčem $I$ in $\psi_i$ inverzija glede na to +krožnico. Ker inverzija ohranja kote, se premica $p$ in +krožnica $n$ (brez točke $I$) s $\psi_i$ preslikata v pravokotni premici $p'=p$ +in $n'$, krožnici $k$ in $l$ pa v krožnici $k'$ in $l'$, ki sta +pravokotni na ti dve premici (izreka \ref{InverzKroznVkrozn} in +\ref{InverzKonf}). Torej sta krožnici $k'$ in $l'$ koncentrični. + \kdokaz + + + \bzgled + (Stainerjev izrek \index{izrek!Steinerjev} + \footnote{\index{Steiner, J.} + \textit{J. Steiner} (1769--1863), švicarski matematik.}.) + Naj bo $l$ krožnica v notranjosti krožnice $k$ in $a_1$, $a_2$, +..., $a_n$ zaporedje krožnic, ki se dotikajo krožnic $k$ in $l$, +vsaka pa se dotika tudi sosednje krožnice v zaporedju. + Če se dotikata še krožnici $a_n$ +in $a_1$, velja ta lastnost neodvisno od izbire prve krožnice +$a_1$ tega zaporedja. + \ezgled + + \begin{figure}[!htb] +\centering +\input{sl.inv.9.7.2.pic} +\caption{} \label{sl.inv.9.7.2.pic} +\end{figure} + + + \textbf{\textit{Proof.}} + Če uporabimo prejšnji izrek \ref{StainerjevLema2}, se trditev + prevede v primer, ko sta krožnici koncentrični (dotik krožnic je + invarianta inverzije \ref{InverzDotik}). Toda v tem primeru se novo zaporedje krožnic + enostavno dobi iz prvega z rotacijo s središčem v skupnem središču krožnic + $\psi_i(k)$ in $\psi_i(l)$ (Figure +\ref{sl.inv.9.7.2.pic}). + \kdokaz + + +%________________________________________________________________________________ +\poglavje{Problem of Apollonius} +\label{odd9ApolDotik} + + Sedaj se bomo ukvarjali s t. i. \index{problem!Apolonijev} + Apolonijevimi\footnote{Starogrški + matematik + \index{Apolonij} + \textit{Apolonij iz Perge} (3.--2. st. pr. n. š.), je reševal te probleme.} + problemi o dotiku + krožnic, ki jih lahko na +eleganten način rešimo z uporabo inverzije. Gre za probleme, ki +so naslednje oblike: + +\textit{ + \vspace*{2mm} + Načrtaj krožnico $k$, ki izpolnjuje tri +pogoje, od katerih ima vsak eno izmed naslednjih oblik: +\vspace*{2mm} +\begin{itemize} + \item vsebuje dano točko, + \item se dotika dane premice, + \item se dotika dane krožnico. +\end{itemize}} + Jasno je, da vse točke, premice in krožnice iz omenjenih pogojev ležijo + v isti ravnini. Nekatere +od teh problemov smo že srečali. Npr. konstruirati krožnico, ki +vsebuje dani točki in se dotika dane premice. Ni težko +ugotoviti, da obstaja deset Apolonijevih problemov. Po navadi jih +navajamo v naslednjem zaporedju: +\begin{enumerate} + \item načrtaj krožnico, ki vsebuje tri dane točke. $(A,B,C)$, +\item načrtaj krožnico, ki vsebuje dani točki in se dotika dane +premice $(A,B,p)$, + \item načrtaj krožnico, ki vsebuje dani točki in se dotika + dane +krožnice $(A,B,k)$, + \item načrtaj krožnico, ki vsebuje dano točko in se +dotika dveh danih premic $(A,p,q)$, + \item načrtaj krožnico, ki vsebuje dano točko ter se + dotika dane premice in dane +krožnice $(A,p,k)$, + \item načrtaj krožnico, ki vsebuje dano točko in +se dotika dveh danih krožnic $(A,k_1,k_2)$, + \item načrtaj krožnico, ki se dotika treh danih premic + $(p,q,r)$, + \item načrtaj krožnico, ki se dotika + dveh danih premic in dane krožnice $(p,q,k)$, + \item načrtaj krožnico, ki se dotika dane premice in dveh danih krožnic +$(p,k_1,k_2)$, + \item načrtaj krožnico, ki se dotika +treh danih krožnic $(k_1,k_2,k_3)$. +\end{enumerate} + + Takoj vidimo, da sta prvi in sedmi problem trivialna, + pa tudi vse ostale probleme lahko +rešimo brez uporabe inverzije. Inverzija pa nam da splošno +metodo za njihovo reševanje. Ta metoda temelji na dejstvu, da se v +določenem primeru krožnica z inverzijo preslika v premico (izrek +\ref{InverzKroznVkrozn}). Ilustrirali jo bomo na primeru +petega Apolonijevega problema: + +\bzgled + Načrtaj krožnico, ki vsebuje dano točko $A$ ter se + dotika dane premice $p$ in dane +krožnice $k$. + \ezgled + + + \begin{figure}[!htb] +\centering +\input{sl.inv.9.8.1.pic} +\caption{} \label{sl.inv.9.8.1.pic} +\end{figure} + + + \textbf{\textit{Solution.}} + +Predpostavimo, da je $x$ krožnica, ki vsebuje točko $A$ ter se +dotika premice $p$ in krožnice $k$ (Figure +\ref{sl.inv.9.8.1.pic}). Obravnavali bomo splošni primer, tako da +točka $A$ ne leži niti na premici $p$ niti na krožnici $k$. Z +$i$ označimo krožnico s središčem $A$ in poljubnim polmerom +$r$. Naj bo $\psi_i$ inverzija glede na to krožnico ter $p'$, +$k'$ in $x'$ slike premice $p$ ter krožnic $k$ in $x$ pri tej +inverziji. Ker velja $A\notin p,k$ in $A\in x$, sta $p'$ in $k'$ +krožnici, $x'$ je pa premica (izrek \ref{InverzKroznVkrozn}). +Premica $p$ in krožnica $x$ imata natanko eno skupno točko, zato +to velja tudi za sliki $p'$ in $x'$, torej je premica $x'$ tangenta +krožnice $p'$. Analogno je premica $x'$ tudi tangenta krožnice +$k'$. Torej se problem prevede na konstrukcijo premice $x'$, ki je +skupna tangenta krožnic $p'$ in $k'$ (zgled \ref{tang2ehkroz}). +Potem je $x= \psi_i^{-1}(x')= \psi_i(x')$. + +Naloga ima nič, eno, dve, tri ali štiri rešitve, odvisno od +medsebojne lege krožnic $p'$ in $k'$ oz. od števila njunih +skupnih tangent. + +V primeru, ko velja ali $A\in p$ ali $A\in k$, je $x'$ premica, ki +se dotika ene krožnice ($k'$ ali $p'$) in je vzporedna z eno +premico ($p'$ ali $k'$). + +Če je $A\in p \cap k$, naloga nima rešitev ali pa jih je neskončno mnogo, +odvisno od tega, ali se krožnici sekata ali dotikata. V obeh primerih +je namreč premica $x'$ vzporedna z dvema premicama $p'$ in $k'$, +toda v prvem primeru se premici $p'$ in $k'$ sekata, v drugem sta pa +vzporedni. + \kdokaz + +Takoj ugotovimo, da pri reševanju tega problema ni bilo preveč +pomembno, ali sta $p$ in $k$ ravno premica in krožnica, pomembno pa je +bilo, da sta sliki $p'$ in $k'$ krožnici. Toda $p'$ in +$k'$ sta krožnici tudi v primeru, kadar sta npr. $p$ in $k$ dve +premici in $A\notin p,k$. Torej četrti in šesti Apolonijev problem +se rešita na popolnoma enak način kot peti problem, ki smo ga pravkar +rešili. + +Tudi drugi in tretji problem lahko rešimo z uporabo inverzije +glede na poljubno krožnico s središčem $A$. Oba problema se +prevedeta na konstrukcijo tangente iz točke $B'$ na krožnico $p'$ (oz. $k'$). + + Problemi 8, 9 in +10 se po vrsti prevedejo na probleme 4, 5 in 6. Ideja je v tem, da +najprej načrtamo krožnico, ki je koncentrična z iskano krožnico +in vsebuje središče ene od danih krožnic - tiste z najmanjšim +polmerom. + +V nadaljevanju bomo rešili primer, ki je podoben Apolonijevim +problemom o dotiku krožnic. + +\bzgled + Dani so točka $A$, krožnici $k$ in $l$ ter kota $\alpha$ in $\beta$. + Načrtaj + krožnico, ki poteka skozi točko $A$, +krožnici $k$ in $l$ pa seka pod kotoma $\alpha$ in $\beta$. + \ezgled + + + \begin{figure}[!htb] +\centering +\input{sl.inv.9.8.2.pic} +\caption{} \label{sl.inv.9.8.2.pic} +\end{figure} + +\textbf{\textit{Solution.}} + +Predpostavimo, da je $x$ krožnica, ki vsebuje točko $A$ in s +krožnicama $k$ in $l$ določa kota $\alpha$ in $\beta$ (Figure +\ref{sl.inv.9.8.2.pic}). Spet bomo obravnavali splošni primer, +ko točka $A$ ne leži na nobeni izmed krožnic $k$ in $l$. Z +$i$ označimo krožnico s središčem $A$ in poljubnim polmerom +$r$ ter s $\psi_i$ inverzijo glede na to krožnico. Naj bodo $k'$, +$l'$ in $x'$ slike krožnic $k$, $l$ in $x$ pri tej inverziji. Ker +velja $A\notin k,l$ in $A\in x$, sta $k'$ in $l'$ krožnici, $x'$ +pa je premica (izrek \ref{InverzKroznVkrozn}). Ker je inverzija +konformna preslikava (izrek \ref{InverzKonf}), premica $x'$ seka +krožnici $k'$ in $l'$ pod kotoma $\alpha$ in $\beta$. + +Premico $x'$ lahko načrtamo kot skupno tangento dveh krožnic +$k'_1$ in $l'_1$, ki sta koncentrični s krožnicama $k'$ in $l'$. +Pri risanju krožnic $k'_1$ in $l'_1$ upoštevamo dejstvo, da +premica $x'$ s krožnicama $k'$ in $l'$ določa tetivi s +središčnima kotoma $2\alpha$ in $2\beta$. Na koncu je še $x= +\psi_i^{-1}(x')= \psi_i(x')$. + +Tudi v tem primeru je število rešitev odvisno od medsebojne lege +krožnic $k'_1$ in $l'_1$ oz. števila njunih skupnih tangent. + +V primeru, ko velja ali $A\in k$ ali $A\in l$, je $x'$ premica, ki +s premico $k'$ (ali $l'$) določa kot $\alpha$, s krožnico $p'$ +(ali $k'$) pa kot $\beta$. + +Če je $A\in p \cap k$, naloga v splošnem nima rešitev, +kajti gre za premico $x'$, ki z danima premicama $k'$ in $l'$ +določa kota $\alpha$ in $\beta$. Naloga ima neskončno mnogo rešitev, +kadar premici $k'$ in $l'$ določata kot $|\beta \pm \alpha|$. + \kdokaz + + +%________________________________________________________________________________ +\poglavje{Constructions With Compass Alone} \label{odd9LeSestilo} + + Pri konstrukcijah v evklidski geometriji smo vedno uporabljali + ravnilo in šestilo, kar pomeni, da smo uporabljali + elementarne konstrukcije\index{konstrukcije!z ravnilom in šestilom} (glej razdelek \ref{elementarneKonstrukcije}). + + Toda z uporabo ravnila in šestila, oz. omenjenih elementarnih konstrukcij, v evklidski geometriji ni možno izpeljati vseh konstrukcij. Najbolj znani so naslednji primeri\footnote{Vsa tri probleme so zastavili že Stari Grki. Kasneje je te probleme poskušalo rešiti veliko znanih matematikov kot tudi laikov. Dejstvo, da omenjenih konstrukcij ni mogoče izpeljati le z uporabo ravnila in šestila, je bilo dokazano šele v 19. stoletju, ko je francoski matematik \index{Galois, E.} \textit{E. Galois} (1811--1832) razvil \index{grupa}teorijo grup. Dokaza za trisekcijo kota in podvojitev kocke je leta 1837 prvi podal francoski matematik \index{Wantzel, P. L.}\textit{P. L. Wantzel} (1814--1848). Dejstvo nerešljivosti kvadrature kroga pa je posledica transcendentnosti števila $\pi$, kar je leta 1882 dokazal nemški matematik \index{Lindemann, C. L. F.}\textit{C. L. F. Lindemann} (1852–-1939).}: + \begin{itemize} + \item razdelitev danega kota z dvema poltrakoma na tri skladne dele (\pojem{trisekcija kota})\index{trisekcija kota}; + \item konstrukcija kvadrata, ki ima enako ploščino kot dani krog (\pojem{kvadratura kroga})\index{kvadratura kroga}; + \item konstrukcija roba kocke, ki ima dvakrat tolikšno prostornino kot kocka z danim robom $a$ (\pojem{podvojitev kocke})\index{podvojitev kocke}. + \end{itemize} + +\index{konstrukcije!pravilnih $n$-kotnikov}Razen tega je znan tudi problem konstrukcije pravilnih $n$-kotnikov z ravnilom in s šestilom, ki ga ni možno rešiti za vsak $n\in\{3,4,\ldots\}$\footnote{Znani nemški matematik \index{Gauss, C. F.}\textit{C. F. Gauss} (1777--1855) je leta 1796 dokazal, da lahko z ravnilom in šestilom konstruiramo pravilen $n$-kotnik natanko tedaj, ko je $n=2^k\cdot p$, kjer je $p$ bodisi enak 1 bodisi praštevilo, ki ga lahko zapišemo v obliki $2^{2^l}+1$, za $l\in\{0,1,2,\ldots\}$ (t. i. \pojem{Fermatova števila}\normalcolor, ki sicer niso vsa prosta - npr. za $n=5$, se imenujejo po francoskem matematiku \index{Fermat, P.}\textit{P. Fermatu} (1601--1665)). Pravilni $n$-kotnik torej lahko konstruiramo z ravnilom in s šestilom, če je $n\in\{3,4,5,6,8,10,12,16,17,\ldots\}$, če pa je $n\in\{7,9,11,13,14,15,18,19,\ldots\}$, ta konstrukcija ni možna.}. + + V projektivni geometriji, v kateri ni niti vzporednosti niti metrike, + pri načrtovanju uporabljamo le ravnilo\index{konstrukcije!z ravnilom} oz. samo tiste + elementarne konstrukcije, + ki jih lahko naredimo le z ravnilom. Jasno je, da v tej geometriji ni možno + načrtati (niti definirati) likov, kot so npr. kvadrat, + paralelogram, krožnica, ... + + Zastavlja se vprašanje, kaj lahko konstruiramo, če uporabljamo + le šestilo oz. samo tiste + elementarne konstrukcije, + ki jih lahko naredimo samo s šestilom. Pri tem štejemo, da je + premica načrtana, če sta načrtani dve njeni točki, ne smemo + pa uporabiti direktne konstrukcije presečišča dveh premic (z + ravnilom). Presenetljivo je, da na ta način - le s šestilom - + lahko izpeljemo vse + konstrukcije, ki jih lahko naredimo s šestilom in z + ravnilom\footnote{Konstrukcije le s pomočjo šestila sta raziskovala + italijanski matematik \index{Maskeroni, L.} + \textit{L. Maskeroni} (1750--1800) (po njem + jih imenujemo \index{konstrukcije!Maskeronijeve}\index{konstrukcije!s šestilom} + \pojem{Maskeronijeve konstrukcije}) in že sto +let pred njim danski matematik \index{Mor, G.} \textit{G. Mor} +(1640--1697) v svoji knjigi ‘‘Danski Evklid’’ iz leta 1672. Teoretično +osnovo teh konstrukcij je podal avstrijski matematik \index{Adler, +A.} \textit{A. Adler}, ki je leta 1890 dokazal, da se vsaka +načrtovalna naloga, ki jo lahko rešimo s pomočjo ravnila in šestila, da +rešiti tudi samo z uporabo šestila.}! + +Nadaljevali bomo z naslednjimi konstrukcijami, pri katerih bomo +uporabljali le šestilo. V želji, da pokažemo, da lahko ravnilo +‘‘nadomestimo’’ s šestilom, bo glavno vlogo imela inverzija. V +dosedanji uporabi inverzije pri konstrukcijah smo najpogosteje +uporabljali dejstvo, da inverzija v določenem primeru krožnico +preslika v premico. Tako smo problem konstrukcije iskane krožnice s +pomočjo inverzije prevedli v problem konstrukcije iskane premice, +kar je v večini primerov bolj enostavno. Sedaj bomo poskusili ravno +obratno - probleme povezane s konstrukcijo premice (in uporabe +ravnila) bomo z inverzijo prevedli v probleme konstrukcije krožnice +(in uporabe šestila). + + \bzgled \label{MaskeroniNAB} + Dani sta točki $A$ in $B$ ter $n\in \mathbb{N}$. + Načrtaj takšno točko $P_n$, da velja + $\overrightarrow{AP_n}=n\cdot \overrightarrow{AB}$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.9.1.pic} +\caption{} \label{sl.inv.9.9.1.pic} +\end{figure} + +%PAZI NA BARVO! \normalcolor + + \textbf{\textit{Solution.}} Konstrukcijo bomo izpeljali + induktivno. Za $n=1$ je jasno $P_1=B$. Načrtajmo najprej točko + $P_2$, za katero velja $\overrightarrow{AP_2}=2\cdot \overrightarrow{AB}$ + (Figure +\ref{sl.inv.9.9.1.pic}). Načrtajmo sedaj krožnici $k_1(A,AB)$ +in $k_2(B,BA)$. Eno od njunih presečišč označimo s $Q_1$. +Načrtajmo nato krožnico $l_1(Q_1,Q_1B)$. Presečišče krožnic +$l_1$ in $k_2$, ki ni točka $A$, označimo s $Q_2$. Točko $P_2$ +dobimo kot eno od presečišč krožnic $l_2(Q_2,Q_2B)$ in $k_2$ (tisto, ki +ni $Q_1$). \normalcolor Relacija $\overrightarrow{AP_2}=2\cdot +\overrightarrow{AB}$ sledi iz dejstva, da so trikotniki $ABQ_1$, +$Q_1BQ_2$ in $Q_2BP_2$ vsi pravilni. + + Predpostavimo, da smo z opisanim postopkom + načrtali točke $P_k$ ($k\leq n-1$), za katere velja + $\overrightarrow{AP_k}=k\cdot \overrightarrow{AB}$. Točko $P_n$ + lahko narišemo na enak način. + Načrtamo najprej krožnico $k_n(P_{n-1},P_{n-1}P_{n-2})$. Presečišče krožnic $l_{n-1}$ in $k_n$, ki ni točka $P_{n-2}$, +označimo s $Q_n$. Točko $P_n$ dobimo kot eno od presečišč +krožnic $l_n$ in $k_n$ (tisto, ki ni $Q_{n-1}$). Trikotniki +$P_{n-2}P_{n-1}Q_{n-1}$, $Q_{n-1}P_{n-1}Q_n$ in $Q_nP_{n-1}P_n$ so +vsi pravilni. Zato je +$\overrightarrow{P_{n-1}P_n}=\overrightarrow{P_{n-2}P_{n-1}}$. Če +uporabimo še indukcijsko predpostavko +$\overrightarrow{AP_k}=k\cdot \overrightarrow{AB}$ za $k=n-1$, +dobimo $\overrightarrow{AP_n}=n\cdot \overrightarrow{AB}$, kar +pomeni, da je $P_n$ iskana točka. + \kdokaz + + Omenili smo že pomembnost inverzije v naših konstrukcijah. + Sedaj smo pripravljeni dokazati postopek konstrukcije slike + točke pri inverziji le s pomočjo šestila. + + \bzgled Dani sta krožnica $i(S,r)$ in točka $X$. Načrtaj točko + $X'=\psi_i(X)$. \label{MaskeroniInv} + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.9.2.pic} +\caption{} \label{sl.inv.9.9.2.pic} +\end{figure} + + \textbf{\textit{Solution.}} Če $X\in i$ je trivialno $X'=X$. + + Predpostavimo najprej, da je $X$ zunanja točka krožnice + inverzije $i$ (Figure \ref{sl.inv.9.9.2.pic}). Ker je središče $S$ njena notranja točka, + krožnica $k(X,XS)$ seka krožnico inverzije v dveh točkah + (izrek \ref{DedPoslKrozKroz}) - denimo $P$ in $Q$, ki ju lahko + narišemo. Nato točko $X'$ dobimo kot drugo presečišče + krožnic $l_P(P,PS)$ in $l_Q(Q,QS)$ (prvo je točka $S$). + Dokažimo še, da je $X'=\psi_i(X)$. Ker je po konstrukciji $XS\cong + XP$ in $PX'\cong PS$, je tudi $\angle PX'S\cong \angle PSX'= + \angle PSX\cong\angle SPX$. To pomeni, da sta trikotnika $PSX'$ + in $XPS$ podobna, zato je $\frac{|PX'|}{|XS|}=\frac{|SX'|}{|PS|}$ oz. + $|SX|\cdot |SX'|=|PX'|\cdot |PS|=r^2$, torej $X'=\psi_i(X)$. + + Če je $X$ notranja točka krožnice + inverzije, potem obstaja takšno naravno število $n$, da velja + $|n\cdot \overrightarrow{SX}|>r$. Naj bo $X_n$ točka, za katero + velja $\overrightarrow{SX_n}=n\cdot \overrightarrow{SX}$ in + $X'_n=\psi_i(X_n)$. Iz prejšnjega zgleda in prvega dela dokaza + točki $X_n$ in $X'_n$ lahko narišemo. + Za točko $X'=\psi_i(X)$ velja: + $$|SX'|=\frac{r^2}{|SX|}=\frac{n\cdot r^2}{|SX_n|}=n\cdot |SX'_n|,$$ + kar pomeni, da lahko tudi točko $X'$ narišemo s pomočjo + prejšnjega zgleda. + \kdokaz + + \bzgled Dani sta točki $A$ in $B$. Načrtaj središče daljice + $AB$. \label{MaskeroniSred} + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.9.3.pic} +\caption{} \label{sl.inv.9.9.3.pic} +\end{figure} + + \textbf{\textit{Solution.}} Naj bo $\psi_i$ inverzija + glede na krožnico $i(A,AB)$ (Figure \ref{sl.inv.9.9.3.pic}). + Najprej načrtamo takšno točko + $X$, da velja $\overrightarrow{AX}=2\cdot\overrightarrow{AB}$ + (zgled \ref{MaskeroniNAB}), nato pa $X'=\psi_i(X)$ (zgled \ref{MaskeroniInv}). + Ker je $X$ zunanja točka krožnice $i$, je $X'$ njena notranja točka. + Zato je $\mathcal{B}(A,X',B)$. Za točko $X'$ velja tudi + $|AX'|=\frac{|AB|^2}{|AX|}=\frac{|AB|^2}{2\cdot |AB|}=\frac{1}{2}\cdot |AB|,$ + kar pomeni, da je $X'$ središče daljice $AB$. + \kdokaz + + \bzgled \label{MaskeroniProj} Dane so tri nekolinearne točke $P$, $Q$ in $R$. Načrtaj + pravokotno projekcijo točke $P$ na premico $QR$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.9.4.pic} +\caption{} \label{sl.inv.9.9.4.pic} +\end{figure} + + \textbf{\textit{Solution.}} Načrtajmo najprej središči daljic + $PQ$ in $PR$ (zgled \ref{MaskeroniSred}) in ju označimo po vrsti s $Q_1$ + in $R_1$ (Figure \ref{sl.inv.9.9.4.pic}). Točko $P'$ načrtamo + kot drugo presečišče krožnic $k(Q_1,Q_1P)$ in $l(R_1,R_1P)$. Točka $P'$ leži na krožnicah s premeroma $PQ$ in $PR$. Zato + je $\angle PP'Q=\angle PP'R=90^0$. Torej točka $R$ leži na + premici $P'Q$, kar pomeni, da so točke $P'$, $Q$ in $R$ + kolinearne. Zato je točka $P'$ pravokotna projekcija točke $P$ + na premici $QR$. + \kdokaz + + \bzgled \label{MaskeroniOcrt} Dane so tri nekolinearne točke $A$, $B$ in $C$. + Načrtaj + središče očrtane krožnice in očrtano krožnico trikotnika + $ABC$. + \ezgled + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.9.5.pic} +\caption{} \label{sl.inv.9.9.5.pic} +\end{figure} + + \textbf{\textit{Solution.}} + Naj bo $l$ očrtana krožnica trikotnika $ABC$ s središčem v + točki $O$ ter $\psi_i$ inverzija s središčem $A$ in poljubnim + polmerom (Figure \ref{sl.inv.9.9.5.pic}). Ker $A\in l$, iz izreka + \ref{InverzKroznVkrozn} sledi, + da je $l'=\psi_i(l)$ premica. Če je $P'$ pravokotna projekcija + središča inverzije $A$ na premici $l'$ in $P=\psi_i(P')$, je + $AP$ premer krožnice $l$ (posledica konstrukcije v dokazu + (\textit{ii}) izreka + \ref{InverzKroznVkrozn}). + + Dokazana dejstva nam omogočajo konstrukcijo. Najprej načrtamo + poljubno krožnico $i$ s središčem v točki $A$ in + točki $B'=\psi_i(B)$ in $C'=\psi_i(C)$ (zgled + \ref{MaskeroniInv}), nato pravokotno projekcijo $P'$ točke $A$ + na premici $B'C'$ (zgled \ref{MaskeroniProj}) in $P=\psi_i(P')$. + Središče očrtane krožnice dobimo kot središče daljice $AP$ + (zgled \ref{MaskeroniSred}). Krožnica $l(O,A)$ je očrtana + krožnica trikotnika $ABC$. + \kdokaz + + Eno od elementarnih konstrukcij samo z uporabo šestila predstavlja konstrukcija + presečišča dveh krožnic. S pomočjo inverzije jo bomo + uporabili za naslednji elementarni konstrukciji, ki ju štejemo za + konstrukciji z ravnilom in s + šestilom. + + \bzgled \label{MaskeroniKrPr} Dane so štiri točke $A$, $B$, $C$ in $D$. + Načrtaj presečišče premic $AB$ in $CD$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.9.6.pic} +\caption{} \label{sl.inv.9.9.6.pic} +\end{figure} + + \textbf{\textit{Solution.}} Predpostavimo, da premici $AB$ in $CD$ + nista vzporedni. + Naj bo $S$ poljubna točka, ki ne leži na + nobeni od premic $AB$ in $CD$. Jasno je, da takšna točka + obstaja, toda efektivno konstrukcijo takšne točke lahko dobimo, + če uporabimo postopek konstrukcij točk $Q_1$ in $Q_2 $ iz + zgleda \ref{MaskeroniNAB} glede na daljico $AB$. + Naj bo $\psi_i$ inverzija glede na poljubno krožnico $i$ s + središčem v točki $S$ (Figure \ref{sl.inv.9.9.6.pic}). + Načrtajmo točke $A'=\psi_i(A)$, $B'=\psi_i(B)$, $C'=\psi_i(C)$ in + $D'=\psi_i(D)$ (zgled \ref{MaskeroniInv}) ter očrtane krožnice + trikotnikov $SA'B'$ in $SC'D'$ (zgled \ref{MaskeroniOcrt}) oz. + slike premic $AB$ in $CD$ pri inverziji $\psi_i$. Točka $L'$ je + drugo presečišče dveh očrtanih krožnic (prvo presečišče je točka $S$) + Le-to obstaja v primeru, kadar premici $AB$ in $CD$ + nista vzporedni. Presečišče $L$ premic $AB$ in $CD$ je slika + presečišča $L'$ njunih slik - torej $L=\psi_i(L')$ (zgled \ref{MaskeroniInv}). + \kdokaz + + + \bzgled \label{MaskeroniKtKr}Dane so krožnica $k$ ter točki $A$ in $B$. + Načrtaj presečišča premice $AB$ in krožnice $k$. + \ezgled + + +\begin{figure}[!htb] +\centering +\input{sl.inv.9.9.7.pic} +\caption{} \label{sl.inv.9.9.7.pic} +\end{figure} + + \textbf{\textit{Solution.}} Konstrukcijo lahko izpeljemo na enak + način kot v prejšnjem primeru, le da je $k'$ očrtana krožnica + trikotnika, ki ga določajo točke $C'=\psi_i(C)$, $D'=\psi_i(D)$ + in $E'=\psi_i(E)$, kjer so $C$, $D$ in $E$ poljubne točke + krožnice $k$ + (Figure \ref{sl.inv.9.9.7.pic}). +\kdokaz + + Z zadnjima dvema konstrukcijama smo dokazali, da se vse + elementarne konstrukcije, ki se nanašajo na ravnilo in šestilo (glej razdelek \ref{elementarneKonstrukcije}), + lahko naredijo samo s šestilom. Kot smo že omenili, pri tem štejemo, da je + premica načrtana, če sta načrtani dve njeni točki (podobno velja za daljico in poltrak), ne smemo + pa uporabiti direktne konstrukcije presečišča dveh premic oz. premice in krožnice (z uporabo + ravnila). Iz zadnjih dveh konstrukcij (zgleda \ref{MaskeroniKrPr} in \ref{MaskeroniKtKr}) pa sledi, da je slednje možno tudi samo s šestilom. + + Ker vsako načrtovalno nalogo, ki jo lahko rešimo z ravnilom in s šestilom, + izpeljemo z elementarnimi konstrukcijami, ki se + nanašajo na ravnilo in šestilo, iz prejšnjega sledi, + da to nalogo lahko rešimo le s šestilom. + + +%NALOGE +%________________________________________________________________________________ + +\naloge{Exercises} + + +\begin{enumerate} + + \item +Dokaži, da kompozitum dveh inverzij $\psi_{S,r_1}$ in $\psi_{S,r_2}$ glede na koncentrični + krožnici predstavlja razteg. Določi središče in koeficient + tega raztega. + + \item Naj bodo $A$, $B$, $C$ in $D$ štiri kolinearne točke. + Konstruiraj takšni točki $E$ in $F$, da velja $\mathcal{H}(A,B;E,F)$ +in $\mathcal{H}(C,D;E,F)$. + + \item V ravnini so dani točka $A$, premica $p$ in krožnica $k$. + Načrtaj krožnico, ki poteka skozi točko $A$ in je + pravokotna na premico $p$ in krožnico $k$. + + \item Reši tretji, četrti, deveti in deseti Apolonijev problem. + + \item Naj bodo $A$ točka, $p$ premica, $k$ krožnica + in $\omega$ kot v neki ravnini. Načrtaj krožnico, +ki poteka skozi točko $A$, se dotika premicp $p$ in s krožnico $k$ +določa kot $\omega$. + + \item Določi geometrijsko mesto točk dotika dveh krožnic, ki + se dotikata krakov danega kota v dveh danih točkah $A$ in $B$. + + \item Načrtaj trikotnik, če so znani naslednji podatki: +\begin{enumerate} + \item $a$, $l_a$, $v_a$ + \item $v_a$, $t_a$, $b-c$ + \item $b+c$, $v_a$, $r_b-r_c$ + \end{enumerate} + + + \item Naj bosta $c(S,r)$ in $l$ krožnica in premica v isti ravnini, ki +nimata skupnih točk. Naj bodo še $c_1$, $c_2$ in $c_3$ krožnice +te ravnine, ki se medsebojno (po dve) dotikajo in se vsaka od +njih dotika še $c$ in $l$. Izrazi razdaljo točke $S$ od premice +$l$ s $r$\footnote{Predlog za MMO 1982. (SL 12.)}. + +\item Naj bo $ABCD$ pravilni tetraeder. Poljubni točki +$M$, ki leži na robu $CD$, pridružimo točko $P = f(M)$, ki +je presečišče pravokotnice skozi točko $A$ na premico $BM$ in +pravokotnice skozi točko $B$ na premico $AM$. Določi geometrijsko +mesto vseh točk $P$, če točka $M$ zavzame vse vrednosti na +robu $CD$. + +\item Naj bo $ABCD$ tetivnotangentni štirikotnik in $P$, $Q$, +$R$ in $S$ dotikališča stranic $AB$, $BC$, $CD$ in $AD$ z včrtano +krožnico tega štirikotnika. Dokaži, da velja $PR\perp QS$. + +\item Dokaži, da sta središči tetivnotangentnemu štirikotniku včrtane in očrtane +krožnice ter presečišče njegovih diagonal kolinearne točke +(\index{izrek!Newtonov}Newtonov izrek\footnote{\index{Newton, +I.}\textit{I. Newton} (1643--1727), angleški fizik in matematik}). + +\item Naj bosta $p$ in $q$ vzporedni tangenti krožnice $k$. +Krožnica $c_1$ se dotika premice $p$ v točki $P$ in krožnice $k$ v +točki $A$, krožnica $k_2$ pa se dotika premice $q$ ter krožnic $k$ +in $k_1$ v točkah $Q$, $B$ in $C$. Dokaži, da je presečišče premic +$PB$ in $AQ$ središče trikotniku $ABC$ očrtane krožnice. + + \item Krožnici $k_1$ in $k_3$ se od zunaj dotikata v točki $P$. Prav + tako se tudi krožnici $k_2$ in $k_2$ od zunaj dotikata v isti + točki. Krožnica $k_1$ seka krožnici $k_2$ in $k_4$ še v točkah + $A$ in $D$, krožnica $k_3$ pa seka krožnici $k_2$ in $k_4$ še v točkah + $B$ in $C$. Dokaži, da velja\footnote{Predlog za MMO 2003. (SL 16.)}: + $$\frac{|AB|\cdot|BC|}{|AD|\cdot|DC|}=\frac{|PB|^2}{|PD|^2}.$$ + +\item Naj bo $A$ točka, ki leži na krožnici $k$. Samo s + šestilom načrtaj kvadrat $ABCD$ (oz. njegova oglišča), ki je včrtan dani +krožnici. + +\item Dani sta točki $A$ in $B$. Le z uporabo + šestila načrtaj takšno točko $C$, da velja + $\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}$. + + \item Samo s pomočjo + šestila razdeli dano daljico v razmerju $2:3$. +\end{enumerate} + + + + + + + + + +% DEL 10 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +% REŠITVE IN NAPOTKI +%________________________________________________________________________________ + +\del{Solutions and Hints} +\footnotesize +%REŠITVE - Aksiomi evklidske geometrije ravnine +%________________________________________________________________________________ + +\poglavje{Introduction} + +There are no exercises in the first chapter. \color{viol4}$\ddot\smile$ \normalcolor %smile + +\poglavje{Axioms of Planar Euclidean Geometry} + +\begin{enumerate} + +\item \res{Naj bodo $P$, $Q$ in $R$ notranje točke stranic trikotnika $ABC$. Dokaži, da so $P$, $Q$ in $R$ nekolinearne.} + +Predpostavimo nasprotno, da točke $P$, $Q$ in $R$ ležijo na + neki premici $l$. To pomeni, da premica $l$ seka vse tri +stranice trikotnika $ABC$ in po predpostavki ne poteka skozi nobeno +njegovih oglišč. To je v nasprotju s posledico \ref{PaschIzrek} + Paschevega aksioma \ref{AksPascheva}, zatorej so $P$, $Q$ in + $R$ nekolinearne točke. + +\item \res{Naj bosta $P$ in $Q$ točki stranic $BC$ in $AC$ trikotnika $ABC$ in hkrati +različni od njegovih oglišč. Dokaži, da se daljici $AP$ in $BQ$ +sekata v eni točki.} + + Iz predpostavke, da točka $P$ leži na stranici $BC$ trikotnika + $ABC$, po definiciji daljice $BC$ velja $\mathcal{B}(B,P,C)$. + Iz aksioma \ref{AksII2} sledi, da ne velja + $\mathcal{B}(B,C,P)$, kar pomeni, da točka $B$ ne leži na + daljici $PC$. V trikotniku $APC$ premica $BQ$ seka stranico $AC$, + ne seka pa stranice $PC$, zato po posledici \ref{PaschIzrek} + Paschevega aksioma \ref{AksPascheva} premica $BQ$ seka + stranico $AP$ tega trikotnika. Torej premica $BQ$ seka daljico + $AP$ v neki točki $X$ oz. $BQ\cap [AP]=\{X\}$. Analogno je + tudi $[BQ]\cap AP=\{\widehat{X}\}$. Dokažimo, da je + $X=\widehat{X}$. Predpostavimo nasprotno, da velja $X\neq + \widehat{X}$. Ker je $[AP]\subset AP$ in $[BQ]\subset BQ$, + velja $X,\widehat{X}\in AP\cap BQ$. Po aksiomu \ref{AksI1} predstavljata + $AP$ in $BQ$ isto premico oz. točke $A$, $B$, + $P$ in $Q$ ležijo na isti premici $l$. Toda na tej premici, ki + je določena s točkama $B$ in $P$ (aksiom \ref{AksI1}), leži po + predpostavki tudi točka $C$, kar pomeni, da so točke $A$, $B$ + in $C$ kolinearne. To pa je v nasprotju s predpostavko, da je + $ABC$ trikotnik, zato je $X=\widehat{X}$. To pomeni $X\in + [AP]\cap[BQ]$. Če bi daljici $[AP]$ in $[BQ]$ imeli še eno + skupno točko, bi ta bila druga skupna točka tudi premic $AP$ + in $BQ$. Dokazali smo, da to ni možno, zatorej je + $[AP]\cap[BQ]=\{X\}$. + +\item \res{Točke $P$, $Q$ in $R$ ležijo po vrsti na stranicah $BC$, $AC$ in $AB$ trikotnika + $ABC$ in so različne od njegovih oglišč. Dokaži, da se daljici $AP$ +in $QR$ sekata v eni točki.} + +Uporabimo prejšnjo nalogo najprej za daljici $AP$ in $BQ$ v +trikotniku $ABC$, nato pa še enkrat za daljici $AX$ +($\{X\}=[AP]\cap[BQ]$) in $QR$ v trikotniku $ABQ$. + + + +\item \res{Premica $p$, ki leži v ravnini štirikotnika, seka njegovo +diagonalo $AC$ in ne poteka skozi nobeno oglišče tega štirikotnika. +Dokaži, da premica $p$ seka natanko dve stranici tega +štirikotnika.} + +Dvakrat uporabimo posledico \ref{PaschIzrek} Paschevega aksioma + \ref{AksPascheva} za premico $p$ in trikotnika $ABC$ in $ADC$. + +\item \label{nalAks3}\res{Dokaži, da je polravnina konveksni lik.} + +Naj bosta $X$ in $Y$ poljubni točki polravnine $pA$. Potrebno je dokazati, da tudi cela daljica $XY$ leži v tej polravnini. Po definiciji polravnine iz $X,Y\in pA$ sledi $X,A\ddot{-} p$ in $Y,A\ddot{-} p$. Ker je $\ddot{-} p$ ekvivalenčna relacija, je tudi $X,Y\ddot{-} p$. Naj bo $Z$ poljubna točka daljice $XY$. Predpostavimo, da ni $Z,A\ddot{-} p$ oz. da velja $Z,A\div p$ oz. $Z,X\div p$ (ali $Z \in p$). V tem primeru daljica $ZX$ seka premico $p$ v neki točki $T$. Točka $T$ je potem skupna točka premice $p$ in daljice $XY$ (izrek \ref{izrekAksIIDaljica}), kar je v nasprotju z dokazanim $X,Y\ddot{-} p$. Torej velja $Z,A\ddot{-} p$, zato je tudi $Z\in pA$, kar pomeni, da je polravnina $pA$ konveksni lik. + +\item \label{nalAks4} \res{Dokaži, da je presek dveh konveksnih likov +konveksni lik.} + +Naj bosta $\Phi_1$ in $\Phi_2$ konveksna lika. Dokažimo, da je +potem tudi $\Phi_1\cap\Phi_2$ konveksni lik. Naj bosta $A, B\in +\Phi_1\cap\Phi_2$ poljubni točki. V tem primeru je $A, B\in +\Phi_1$ in $A,B\in\Phi_2$. Ker sta lika $\Phi_1$ in $\Phi_2$ +konveksna, je $[AB]\subseteq\Phi_1$ in $[AB]\subseteq\Phi_2$ +oz. $[AB]\subseteq\Phi_1\cap\Phi_2$, kar pomeni, da je tudi +$\Phi_1\cap\Phi_2$ konveksni lik. + +\item \res{Dokaži, da je poljubni trikotnik konveksni lik.} + +Trikotnik $ABC$ je presek polravnin $ABC$, $ACB$ in +$BCA$, zato je po prejšnjih dveh nalogah \ref{nalAks3} in +\ref{nalAks4} konveksni lik. + +\item \res{Če je $\mathcal{B}(A,B,C)$ in $\mathcal{B}(D,A,C)$, je tudi +$\mathcal{B}(B,A,D)$. Dokaži.} + +Po izreku \ref{izrekAksIIPoltrak} določa točka $A$ na premici $AB$ dva poltraka. +Označimo s $p$ poltrak $AC$ in $p'$ njegov komplementarni (dopolnilni) poltrak. Iz $\mathcal{B}(A,B,C)$ sledi $B,C\ddot{-} A$ iz $\mathcal{B}(D,A,C)$ pa $D,C\div A$, zato velja $B\in p$ in $D\in p'$. Torej točki $B$ in $D$ ne ležita na istem poltraku z začetno točko $A$, zato ni $B,D\ddot{-} A$, kar pomeni, da velja $B,D\div A$ oz. $\mathcal{B}(B,A,D)$. + + + +\item \res{Naj bodo $A$, $B$, $C$ in $D$ takšne kolinearne točke, da je +$\neg\mathcal{B}(B,A,C)$ in $\neg\mathcal{B}(B,A,D)$. Dokaži, da +velja $\neg\mathcal{B}(C,A,D)$.} + +Iz $\neg\mathcal{B}(B,A,C)$ in $\neg\mathcal{B}(B,A,D)$ sledi $B,C\ddot{-} A$ oz. $B,D\ddot{-} A$. Ker je $\ddot{-} A$ ekvivalenčna relacija, je tudi tranzitivna, zato velja $C,D\ddot{-} A$ oz. $\neg\mathcal{B}(C,A,D)$. + +\item \res{Naj bo $A_1A_2\ldots,A_{2k+1}$ poljubni večkotnik z lihim +številom oglišč. Dokaži, da ne obstaja premica, ki seka vse +njegove stranice.} + +Predpostavimo nasprotno, da neka premica $p$ seka vse stranice +tega večkotnika. Dokaži, da bi bile v tem primeru točke +$A_1,A_3,\ldots,A_{2k+1}$ na istem bregu premice $p$. + +\item \label{nalAks11}\res{Če izometrija $\mathcal{I}$ preslika lika $\Phi_1$ in $\Phi_2$ +v lika $\Phi'_1$ in $\Phi'_2$, potem se presek +$\Phi_1\cap\Phi_2$ s to izometrijo preslika v presek +$\Phi'_1\cap\Phi'_2$. Dokaži.} + +Če za poljubno točko $X$ velja $X\in \Phi_1\cap\Phi_2$, je $X\in \Phi_1$ in $X\in\Phi_2$. Iz tega sledi $\mathcal{I}(X)\in \mathcal{I}(\Phi_1)$ in $\mathcal{I}(X)\in \mathcal{I}(\Phi_2)$ oz. $X'\in \Phi'_1$ in $X'\in \Phi'_2$ (kjer je $X'=\mathcal{I}(X)$). Torej velja $X'\in\Phi'_1\cap\Phi'_2$. Na ta način smo dokazali $\mathcal{I}(\Phi_1\cap\Phi_2)\subseteq \Phi'_1\cap\Phi'_2$. Drugo inkluzijo $\mathcal{I}(\Phi_1\cap\Phi_2)\supseteq \Phi'_1\cap\Phi'_2$ dokažemo na enak način z uporabo izometrije $\mathcal{I}^{-1}$. + +\item \res{Dokaži, da sta poljubna poltraka neke +ravnine med seboj skladna.} + +Uporabimo aksiom \ref{aksIII2} in izrek \ref{izrekIzoB}. + +\item \res{Dokaži, da sta poljubni premici neke +ravnine med seboj skladni.} + +Uporabimo aksioma \ref{aksIII2} in \ref{aksIII1}. + + +\item \res{Naj bosta $k$ in $k'$ dve krožnici +neke ravnine s središčema $O$ in $O'$ ter polmeroma $AB$ in $A'B'$. +Dokaži ekvivalenco: $k\cong k' \Leftrightarrow AB\cong A'B'$.} + +($\Leftarrow$) Naj bo $AB\cong A'B'$. Označimo s $P$ in $P'$ poljubni točki krožnic $k$ oz. $k'$. Ker sta $OP$ in $O'P'$ polmera teh krožnic, je $OP\cong O'P'$. Po izreku \ref{izrekAB} obstaja izometrija $\mathcal{I}$, ki preslika točki $O$ in $P$ v točki $O'$ in $P'$. Naj bo $X$ poljubna točka krožnice $k$ in $X'=\mathcal{I}(X)$. Iz $\mathcal{I}:O,X,P\rightarrow O',X',P'$ sledi $O'X'\cong OX$. Ker $X,P\in k$, je tudi $OX\cong OP$. Iz prejšnjih relacij sledi $O'X'\cong O'P'$ oz. $X'\in k'$. Torej velja $\mathcal{I}(k)\subseteq k'$. Podobno je $\mathcal{I}(k)\supseteq k'$ (uporabimo $\mathcal{I}^{-1}$), zato je $\mathcal{I}(k)= k'$. + +($\Rightarrow$) Naj b sedaj $k\cong k'$. Predpostavimo nasprotno - da ni $AB\cong A'B'$. Brez škode za splošnost naj bo $AB>A'B'$. + Iz $k\cong k'$ sledi, da obstaja izometrija $\mathcal{I}$, ki preslika krožnico $k$ v krožnico $k'$. Naj bo $PQ$ poljubni premer krožnice $k$. Po izreku \ref{premerInS} je $PQ=2\cdot AB$. Naj bo $P'=\mathcal{I}(P)$ in $Q'=\mathcal{I}(Q)$. Torej $P'Q'$ je tetiva krožnice, za katero velja $P'Q'\cong PQ =2\cdot AB>2\cdot A'B'$. Toda relacija $P'Q'>2\cdot A'B'$ ni možna (izrek \ref{premerNajdTetiva}), zato je $AB\cong A'B'$. + +\item \res{Naj bo $\mathcal{I}$ +neidentična izometrija ravnine z dvema negibnima točkama $A$ in +$B$. Naj bo $p$ premica te ravnine, ki je vzporedna s premico +$AB$, in $A\notin p$. Dokaži, da na premici $p$ ni negibnih točk +izometrije $\mathcal{I}$.} + +Predpostavimo nasprotno - da obstaja fiksna točka $C$ izmetrije $\mathcal{I}$, ki leži na premici $p$. Točka $C$ ne leži na premici $AB$, ker bi bili v nasprotnem vzporednici $AB$ in $p$ enaki, kar pa je v nasprotju s predpostavko $A\notin p$. Torej imamo tri nekolinearne fiksne točke izometrije $\mathcal{I}$, zato je po izreku \ref{IizrekABC2} $\mathcal{I}$ identična preslikava. To je v nasprotju s predpostavko, zato na premici $p$ ni fiksnih točk izometrije $\mathcal{I}$. + +\item \res{ Naj bo $S$ edina negibna točka +izometrije $\mathcal{I}$ v neki ravnini. Dokaži, da če ta izometrija +preslika premico $p$ vase, je $S\in p$.} + +Predpostavimo, da je $S\notin p$. Označimo z $N$ pravokotno projekcijo točke $S$ na premici $p$ in $N'=\mathcal{I}(N)$. Iz $S\notin p$ sledi $N\neq S$. Ker je $S$ edina fiksna točka te izometrije, je $N'\neq N$. Po predpostavki $\mathcal{I}:p\rightarrow p$ je tudi $N'\in p$. Iz $\mathcal{I}:SN,p\rightarrow SN', p$ sledi $\angle SN',p=\angle SN,p=90^0$. V tem primeru bi iz točke $S$ obstajali dve različni pravokotnici na premici $p$, kar po izreku \ref{enaSamaPravokotnica} ni možno. Torej predpostavka $S\notin p$ odpade, kar pomeni, da velja $S\in p$. + +\item \label{nalAks17}\res{Dokaži, da se poljubni dve premici +neke ravnine ali sekata ali sta vzporedni. } + +Po aksiomu \ref{AksI1} imata dve različni premici $p$ in $q$ neke ravnine kvečjemu eno skupno točko. Če imata eno skupno točko, se po definiciji premici sekata, če pa nimata skupnih točk, sta po definiciji vzporedni. + +\item \label{nalAks18}\res{Če neka +premica v ravnini seka eno od dveh vzporednic iste ravnine, + potem seka tudi drugo vzporednico. Dokaži.} + +Naj bo $p\parallel q$ in $l\cap p =\{A\}$. Če je $p=q$, je jasno tudi $l\cap q =\{A\}$. Druga možnost je pa $p\cap q =\emptyset$. Po Playfairjevem aksiomu je potem $l\cap q \neq\emptyset$. Ker ne more biti niti $l=q$, se po prejšnji nalogi \ref{nalAks17} premici $l$ in $q$ sekata. + +\item \res{Dokaži, da vsaka izometrija preslika vzporednici v vzporednici.} + +Uporabimo nalogo \ref{nalAks11}. + +\item \res{Naj bodo $p$, $q$ in $r$ takšne premice neke ravnine, tako da velja +$p\parallel q$ in $r\perp p$. Dokaži, da je $r\perp q$.} + +Trditev je direktna posledica naloge \ref{nalAks18} in izreka \ref{KotiTransverzala}. + +\item \res{Dokaži, da konveksen $n$-kotnik ne more imeti več kot treh +ostrih kotov.} + +Uporabimo dejstvo, da je vsota zunanjih kotov konveksnega $n$-kotnika +enaka $360^0$ (izrek \ref{VsotKotVeckZuna}). + + +\end{enumerate} + + +%REŠITVE - Skladnost trikotnikov +%________________________________________________________________________________ + +\poglavje{Congruence. Triangles and Polygons} + +\begin{enumerate} + + + \item \res{Naj bo $S$ točka, ki leži v kotu $pOq$, točki $A$ in $B$ pa + pravokotni projekciji točke $S$ na krakih $p$ in $q$ tega +kota. Dokaži, da je $SA\cong SB$ natanko tedaj, ko je premica +$OS$ simetrala kota $pOq$.} + +V obeh smereh ekvivalence dokaži skladnost trikotnikov $OSA$ in +$OSB$. V dokazu za ($\Rightarrow$) izrek \textit{SSA} +\ref{SSK}, v dokazu za ($\Leftarrow$) pa izrek \textit{ASA} +\ref{KSK}. + +\item \res{Dokaži, da je vsota diagonal konveksnega štirikotnika večja od +vsote dveh njegovih nasprotnih stranic.} + +Uporabimo trikotniško neenakost za trikotnika $ASB$ in $CSD$, kjer +je $S$ presečišče diagonal $AC$ in $BD$ konveksnega štirikotnika +$ABCD$. + + \item \res{Dokaži, da je v vsakemu trikotniku + največ ena stranica krajša od pripadajoče višine.} + + Označimo z $a$, $b$ in $c$ stranice in $v_a$, $v_b$, $v_c$ + pripadajoče višine trikotnika $ABC$. Predpostavimo nasprotno, da + je npr. $a \frac{3}{4}(a + b + c)$ + \end{enumerate}} + + Uporabimo zgled \ref{neenTezisZgl}. + +\item \res{Naj bo premica $p$ mimobežnica krožnice $k$. Dokaži, da so +vse točke te krožnice na istem bregu premice $p$.} + +Predpostavimo nasprotno, da sta točki $X,Y\in k$ na različnih +bregovih premice $p$. V tem primeru tetiva $XY$ seka premico $p$ +v neki točki $N$, ki je po zgledu \ref{tetivaNotrTocke} notranja +točka krožnice $k$. Premica $p$, ki vsebuje notranjo točko $N$ +krožnice $k$, je po izreku \ref{DedPoslKrozPrem} njena sekanta, +kar je v nasprotju s predpostavko, da je $p$ mimobežnica. + +\item \res{Če krožnica $k$ leži v nekem konveksnem liku $\Phi$, potem tudi +krog, ki je določen s to krožnico, leži v tem liku. Dokaži.} + +Dovolj je dokazati, da poljubna notranja točka $X$ krožnice $k$ +leži v liku $\Phi$. Naj bo $AB$ poljubna tetiva, ki vsebuje +točko $X$ (obstoj takšne tetive sledi iz izreka +\ref{DedPoslKrozPrem}). Ker je $\Phi$ konveksni lik, iz $A,B\in +k\subset \Phi$ sledi $X\in [AB]\subset \Phi$. + +\item \res{Naj bosta $p$ in $q$ različni tangenti krožnice $k$, ki se jo dotikata v +točkah $P$ in $Q$. Dokaži ekvivalenco: $p \parallel q$ natanko +tedaj, ko je $AB$ premer krožnice $k$.} + +Uporabimo dejstvo $SP\perp p$ in $SQ\perp q$, kjer je $S$ središče +krožnice $k$. + +\item \res{Če je $AB$ tetiva krožnice $k$, potem je presek premice $AB$ in +kroga, ki ga krožnica $k$ določa, enak tej tetivi. Dokaži.} + +Označimo s $\mathcal{K}$ omenjeni krog. Potrebno je dokazati +$[AB]=AB\cap \mathcal{K}$. Inkluzija $[AB]\subseteq AB\cap \mathcal{K}$ sledi +direktno iz aksioma \ref{AksII1} in zgleda +\ref{tetivaNotrTocke}. Dokažimo še inkluzijo $AB\cap +\mathcal{K}\subseteq [AB]$. Naj bo $X\in AB\cap \mathcal{K}$. Predpostavimo, da + $X\notin [AB]$. V tem primeru bi bilo $\mathcal{B}(X,A,B)$ ali + $\mathcal{B}(A,B,X)$. Ni težko dokazati, da bi v vsakem od teh + dveh primerov dobilo $SX>SA$, kar ni možno, ker je $X\in \mathcal{K}$. + Torej velja $X\in[AB]$, oz. $AB\cap +\mathcal{K}\subseteq [AB]$. + +\item \res{Naj bo $S'$ pravokotna projekcija središča $S$ krožnice $k$ na +premici $p$. Dokaži, da je $S'$ zunanja točka te krožnice +natanko tedaj, ko premica $p$ krožnice ne seka.} + +Uporabimo izrek \ref{TangSekMimobKrit}. + +\item \res{Naj bo $V$ višinska točka trikotnika $ABC$, pri katerem velja +$CV \cong AB$. Določi velikost kota $ACB$.} + +Označimo z $A'$, $B'$ in $C'$ nožišča višin iz oglišč $A$, $B$ +in $C$ trikotnika $ABC$. Najprej sta skladna kota $C'CB$ in +$A'AB$ (kota s pravokotnima krakoma - izrek +\ref{KotaPravokKraki}). Iz skladnosti trikotnikov $CVA'$ in +$ABA'$ (izrek \textit{ASA} \ref{KSK}) sledi $CA'\cong AA'$. To +pomeni, da je $CAA'$ enakokraki pravokotni trikotnik s +hipotenuzo $AC$, zato je po izreku \ref{enakokraki} $\angle +ACB=\angle ACA'=45^0$. + +\item \res{Naj bo $CC'$ višina pravokotnega trikotnika $ABC$ ($\angle ACB = +90^0$). Če sta $O$ in $S$ središči včrtanih krožnic trikotnikov +$ACC'$ in $BCC'$, je simetrala notranjega kota $ACB$ +pravokotna na premici $OS$. Dokaži.} + +Naj bo $I$ središče včrtane krožnice trikotnika $ABC$. Dokažemo, da +je $I$ višinska točka trikotnika $COS$. + +\item \res{Naj bo $ABC$ trikotnik, v katerem je $\angle ABC = 15^0$ in +$\angle ACB = 30^0$. Naj bo $D$ takšna točka stranice $BC$, da je + $\angle BAD=90^0$. Dokaži, da je $BD = 2AC$.} + +Označimo s $S$ središče daljice $BD$. Točka $S$ je središče +očrtane krožnice pravokotnega trikotnika $BAD$ (izrek +\ref{TalesovIzrKroz2}). Torej $SA\cong SB\cong SD$. Ker je $BSA$ +enakokraki trikotnik z osnovnico $AB$, je po izreku +\ref{enakokraki} $\angle BAS\cong \angle ABS=15^0$. Za zunanji +kot $ASD$ trikotnika $BSA$ potem velja $\angle ASD=\angle BAS + +\angle ABS=30^0$ (izrek \ref{zunanjiNotrNotr}). Torej velja +$\angle ASC=30^0=\angle ACS$, zato je $ASC$ enakokraki trikotnik +z osnovnico $SC$ oz. $AS\cong AC$ (izrek \ref{enakokraki}). Iz +tega sledi $BD = 2SB=2AS=2AC$. + + +\item \res{Dokaži, da obstaja takšen petkotnik, s katerim je mogoče tlakovati ravnino.} + +Lahko izberemo takšen petkotnik $ABCDE$, da je $ABCE$ +kvadrat, $ECD$ enakokraki pravokotni trikotnik z osnovnico $CE$ +in $A,D\div CE$. + +\item \res{Dokaži, da obstaja takšen desetkotnik, s katerim je mogoče tlakovati ravnino.} + +Naredimo unijo po dveh ustreznih pravilnih šestkotnikov - celic + tlakovanja $(6,3)$. + +\item \res{V neki ravnini je vsaka točka pobarvana rdeče ali črno. + Dokaži, da obstaja pravilni trikotnik, ki ima vsa + oglišča enake barve.} + +Uporabimo tlakovanje $(3,6)$. + +\item \res{Naj bodo $l_1,l_2,\ldots, l_n$ ($n > 3$) loki, ki vsi ležijo na isti +krožnici. Središčni kot vsakega loka je kvečjemu enak $180^0$. + Dokaži, da če ima vsaka trojica lokov vsaj eno skupno točko, + obstaja točka, ki leži na vsakem loku.} + + Označimo s $k(S,r)$ dano krožnico. Naj bodo $I_i$ ($i\in + \{1,2,\ldots,n\}$) krožni izseki, ki jih določajo loki $l_k$, + in $J_i=I_i\setminus\{S\}$. Ker je za vsak $l_i$ središčni kot + kvečjemu enak $180^0$, so $J_i$ konveksni liki (dokaži). Po + predpostavki ima vsaka trojica lokov $l_{i_1}$, $l_{i_2}$ in $l_{i_2}$ + ($i_1,i_2,i_3\in\{1,2,\ldots,n\}$) neko skupno točko + $X_{i_1i_2i_3}$. Ta točka pripada tudi vsakemu od likov + $J_{i_1}$, $J_{i_2}$ in $J_{i_2}$. Po Hellyjevem izreku + \ref{Helly} obstaja točka $Y$, ki pripada vsakemu od likov + $J_1,J_2,\ldots, J_n$. Če z $X$ označimo presečišče poltraka + $SY$ s krožnico $k$, sledi, da tudi odprta daljica $(SX]$ + pripada vsakem od teh likov. Ker za vsak $i$ velja $l_i=J_i\cap k$, točka $X$ + pripada vsakem od lokov $l_1,l_2,\ldots, l_n$. + +%drugi del + +\item +\res{Naj bosta $p$ in $q$ pravokotnici, ki se sekata v točki $A$. Če +je $B, B'\in p$, $C, C'\in q$, $AB\cong AC'$, $AB'\cong AC$, +$\mathcal{B}(B,A,B')$ in $\mathcal{B}(C,A,C')$, potem +pravokotnica na premico $BC$ skozi točko $A$ poteka skozi središče +daljice $B'C'$. Dokaži.} + +Najprej sta po izreku \textit{SAS} \ref{SKS} trikotnika $BAC$ in +$C'AB'$ skladna. Iz tega sledi: $\angle AC'B'\cong\angle +ABC=\beta$ in $\angle AB'C'\cong\angle ACB=90^0-\beta$. Označimo +s $S$ središče daljice $B'C'$, s $P$ pa presečišče premic $BC$ +in $AS$. Po izreku \ref{TalesovIzrKroz2} je $SA\cong SB'\cong +SC'$, zato je $\angle CAP=\angle C'AS\cong\angle AC'S=\angle +AC'B'=\beta$ (izrek \ref{enakokraki}). Ker je še $\angle +ACP=\angle ACB=90^0-\beta$, je iz trikotnika $ACP$ po izreku +\ref{VsotKotTrik} $\angle APC=90^0$ oz. $AS\perp BC$. To pomeni, +da pravokotnica $AP$ premice $BC$ poteka skozi središče daljice +$B'C'$. + +\item +\res{Dokaži, da se simetrale notranjih kotov pravokotnika, ki ni +kvadrat, sekajo v točkah, ki so oglišča kvadrata.} + +Označimo s $s_{\alpha}$, $s_{\beta}$, $s_{\gamma}$ in +$s_{\delta}$ simetrale notranjih kotov ob ogliščih $A$, $B$, $C$ +in $D$ pravokotnika $ABCD$ ter $P=s_{\alpha}\cap s_{\beta}$, +$Q=s_{\gamma}\cap s_{\delta}$, $L=s_{\beta}\cap s_{\gamma}$ in +$K=s_{\alpha}\cap s_{\delta}$. Dokažimo, da je $PKQL$ kvadrat. +Najprej iz $\angle PAB=45^0$ in $\angle PBA=45^0$ sledi, da je +$ABP$ enakokraki pravokotni trikotnik z osnovnico $AB$ (izreka +\ref{enakokraki} in \ref{VsotKotTrik}). Torej $\angle APB=90^0$ +in $AP\cong BP$. Analogno so tudi ostali notranji koti +štirikotnika $PKQL$ pravi koti. Dovolj je dokazati še $PK\cong +PL$. Iz skladnosti trikotnikov $AKD$ in $BLC$ (izrek \ref{KSK}) +je $AK\cong BL$. Če slednje povežemo z že dokazanim $AP\cong +BP$, dobimo $PK\cong PL$. + +\item + \res{Dokaži, da se simetrale notranjih kotov paralelograma, ki ni +romb, sekajo v točkah, ki so oglišča pravokotnika. Dokaži še, da so diagonale +tega pravokotnika vzporedne s stranicami paralelograma in so +enake razliki sosednjih stranic tega paralelograma.} + +Podobno kot pri prejšnji nalogi označimo s $s_{\alpha}$, +$s_{\beta}$, $s_{\gamma}$ in $s_{\delta}$ simetrale notranjih +kotov ob ogliščih $A$, $B$, $C$ in $D$ pravokotnika $ABCD$ ter +$P=s_{\alpha}\cap s_{\beta}$, $Q=s_{\gamma}\cap s_{\delta}$, +$L=s_{\beta}\cap s_{\gamma}$ in $K=s_{\alpha}\cap s_{\delta}$. +Dokažimo, da je $PKQL$ pravokotnik. Naj bosta $E$ in $F$ +središči stranic $AD$ in $BC$ paralelograma $ABCD$. V trikotniku +$PAB$ je + $\angle APB=180^0-\angle PAB-\angle +PBA=180^0-\frac{1}{2}\left(\angle DAB+\angle CBA +\right)=180^0-\frac{1}{2}\cdot 180^0=90^0$ (izreka + \ref{VsotKotTrik} in \ref{paralelogram}). Analogno so tudi + ostali notranji koti štirikotnika $PKQL$ pravi koti. Po izreku + \ref{TalesovIzrKroz2} je $E$ središče očrtane krožnice + pravokotnega trikotnika $AKD$ s hipotenuzo $AD$ in velja + $EK\cong EA$. Torej $\angle EKA\cong \angle EAK \cong\angle + KAB$ (izrek \ref{enakokraki}) oz. po izreku + \ref{KotiTransverzala} je $EK\parallel AB$). Analogno je tudi $FL\parallel AB$. Ker je še $EF\parallel AB$ (izrek \ref{srednjTrapez}) je tudi + $KL\parallel AB$. Brez škode za splošnost pedpostavimo, da je $AB>AD$. Označimo s $T$ presečišče simetrale + $s_{\alpha}$ s stranico $CD$. Po izreku \ref{KotiTransverzala} + je $\angle DTA\cong\angle TAB\cong\angle DAT$. To pomeni, da je + $ADT$ enakokraki trikotnik, zato je $AD\cong DT$ (izrek + \ref{enakokraki}). Ker je $KL\parallel AB\parallel CT$ in + $s_{\alpha}\parallel s_{\gamma}$, je štirikotnik $KLCT$ + paralelogram. Torej $PQ\cong KL\cong CT=CD-DT=CD-AB$. + + +\item +\res{Dokaži, da sta simetrali dveh sokotov med seboj +pravokotni.} + +Simetrali določata kot, ki je enak polovici ustreznega +iztegnjenega kota. + +\item \res{Naj bosta $B'$ in $C'$ nožišči višin iz oglišč $B$ in $C$ trikotnika +$ABC$. Dokaži ekvivalenco $AB\cong AC \Leftrightarrow BB'\cong +CC'$.} + +V obe smeri dokaži ekvivalenco $\triangle BC'C\cong\triangle CB'B$. V +smeri ($\Rightarrow$) uporabi izrek \textit{ASA} \ref{KSK}, v +smeri ($\Leftarrow$) pa \textit{SSA} \ref{SSK}. + +\item \res{Dokaži, da je trikotnik pravilen, +če središče trikotniku očrtane krožnice in njegova višinska točka sovpadata. +Ali podobna trditev velja za poljubni dve značilni +točki tega trikotnika?} + +Uporabimi dejstvo, da so simetrale stranic v tem primeru nosilke +višin tega trikotnika. Na podoben način bi dokazali tudi, da +podobna trditev velja za poljubni dve značilni točki tega +trikotnika. + +\item \res{Dokaži, da sta ostrokotna trikotnika $ABC$ in $A'B'C'$ skladna natanko +tedaj, ko imata skladni višini $CD$ in $C'D'$, stranici $AB$ in +$A'B'$ ter kota $ACD$ in $A'C'D'$.} + +Iz skladnosti trikotnikov $ABC$ in $A'B'C'$ direktno sledi +najprej $AB\cong A'B'$, $AC\cong A'C'$ in $\angle BAC\cong \angle B'A'C'$, nato +pa še $\triangle ACD\cong\triangle A'C'D'$ (izrek \textit{ASA} +\ref{KSK}) oz. $CD\cong C'D'$ in $\angle ACD\cong \angle A'C'D'$. + +Predpostavimo, da velja $CD\cong C'D'$, $AB\cong A'B'$ in $\angle ACD\cong \angle A'C'D'$. Iz skladnosti +trikotnikov $ACD$ in $A'C'D'$ (izrek \textit{ASA} \ref{KSK}) +sledi, da obstaja izometrija $\mathcal{I}$, za katero velja +$\mathcal{I}:A,C,D \mapsto A',C',D'$ (izrek \ref{IizrekABC}. +Le-ta preslika poltrak $AD$ v poltrak $A'D'$. Ker ležita $B$ in +$B'$ na poltrakih $AD$ in $A'D'$ in velja $AB\cong A'B'$, je +tudi $\mathcal{I}(B)=B'$. Torej $\mathcal{I}:A,B,C \mapsto +A',B',C'$, kar pomeni, da je $\triangle ABC\cong\triangle A'B'C'$. + +\item \res{Če je $ABCD$ pravokotnik ter $AQB$ in $APD$ pravilna + trikotnika z enako orientacijo, je daljica $PQ$ +skladna z diagonalo tega pravokotnika. Dokaži.} + +Dokažemo, da sta $PAQ$ in $DAB$ skladna trikotnika. + +\item \res{Naj bosta $BB'$ in $CC'$ višini trikotnika $ABC$ ($AC>AB$) ter + $D$ takšna točka poltraka $AB$, da velja $AD\cong AC$. Točka +$E$ je presečišče premice $BB'$ s premico, ki poteka skozi točko $D$ in je +vzporedna s premico $AC$. Dokaži, da je $BE=CC'-BB'$.} + +Naj bo $F$ presečišče premice $ED$ s premico, ki je v točki $B'$ +vzporedna s premico $AB$. Štirikotnik $ADFB'$ je paralelogram, +zato je $FB'\cong DA$ in $\angle DFB' \cong\angle DAB'$ (izrek +\ref{paralelogram}). Ker je še po predpostavki $AD\cong AC$, +velja tudi $FB'\cong AC$. To pomeni, da sta pravokotna +trikotnika $FEB'$ in $AC'C$ skladna (izrek \textit{ASA} \ref{KSK}), zato +je $EB'\cong C'C$ oz. $BE=EB'-BB'=CC'-BB'$. + +\item \res{Naj bo $ABCD$ konveksni štirikotnik, pri katerem velja + $AB\cong BC\cong CD$ in $AC\perp BD$. Dokaži, da je $ABCD$ + romb.} + + Ker je $ABCD$ konveksni štirikotnik, se njegovi diagonali + sekata v neki točki $S$. Dokaži $\triangle ABS\cong\triangle CBS$ in + $\triangle CBS\cong\triangle CDS$. + +\item \res{Naj bo $BC$ osnovnica enakokrakega trikotnika $ABC$. Če sta $K$ in +$L$ takšni točki, da je $\mathcal{B}(A,K,B)$, $\mathcal{B}(A,C,L)$ in $KB\cong LC$, potem +središče daljice $KL$ leži na osnovnici $BC$. Dokaži.} + +Naj bo $O$ središče daljice $KL$. Označimo z $M$ četrto oglišče +paralelograma $CKBM$, skupno središče njunih diagonal $BC$ in +$KM$ (izrek \ref{paralelogram}) pa s $S$. Če uporabimo dejstvo, +da sta trikotnika $ABC$ in $MLC$ enakokraka z osnovnicama $BC$ +in $ML$, iz izreka \ref{enakokraki} dobimo $\angle ABC\cong\angle +ACB$ in $\angle CML\cong\angle CLM$. Iz skladnosti trikotnikov +$BKC$ in $CMB$ (izreka \ref{paralelogram} in \textit{SSS} +\ref{SSS}) pa sledi $\angle ABC=\angle KBC\cong\angle MCB$. Kot +$ACM$ je zunanji kot trikotnika $MCL$, zato je po izreku +\ref{zunanjiNotrNotr} $\angle ACM =\angle CML+\angle CLM=2\angle +CML$. Torej $\angle CML=\frac{1}{2}\angle +ACM=\frac{1}{2}\left(\angle ACB+\angle +BCM\right)\frac{1}{2}\left(\angle ACB+\angle ABC\right)=\angle +ABC$. To pomeni, da sta premici $ML$ in $BC$ vzporedni (izrek +\ref{KotiTransverzala}). Daljica $OS$ je srednjica trikotnika +$KML$ z osnovnico $ML$, zato je po izreku \ref{srednjicaTrik} +$SO\parallel MN$. Po Playfairovem aksiomu \ref{Playfair} sta +$SO$ in $BC$ ista premica (sovpadata), zato točka $O$ leži na premici $BC$. +Po Paschevem aksiomu \ref{AksPascheva} za trikotnik $ABC$ in +premico $KL$ točka $O$ leži na stranici $BC$. + +\item \res{Naj bo $S$ središče trikotniku $ABC$ včrtane krožnice. +Premica, ki poteka skozi točko $S$ in je vzporedna s stranico $BC$ +tega trikotnika, seka stranici $AB$ in $AC$ po vrsti v točkah +$M$ in $N$. Dokaži, da je $BM+NC=NM$.} + +Dokažemo, da sta $BSM$ in $SCN$ enakokraka trikotnika z +osnovnicama $BS$ in $SC$. + +\item \res{Naj bo $ABCDEFG$ konveksni sedemkotnik. Izračunaj vsoto +konveksnih kotov, ki jih določa lomljenka $ACEGBDFA$.} + +Uporabimo dejstvo, da je dvakratna vsota zunanjih kotov +sedemkotnika, ki jih določajo presečišča diagonal sedemkotnika +$ABCDEFG$, enaka $720^0$. Rezultat: $540^0$. + +\item \label{nalSkl34} +\res{Dokaži, da so središča stranic in nožišče poljubne višine trikotnika, +v katerem nobeni dve stranici nista skladni, oglišča +enakokrakega trapeza.} + +Označimo z $A_1$, $B_1$ in $C_1$ središča stranic $BC$, $CA$ in +$BA$ trikotnika $ABC$ ter $A'$ nožišče višine tega trikotnika iz +oglišča $A$. Iz $|AB|\neq |AC|$ sledi $A'\neq A_1$. Dokažimo, da +je štirikotnik $A'A_1B_1C_1$ enakokraki trapez. Daljica $B_1C_1$ +je srednjica trikotnika $ABC$ z osnovnico $BC$, zato je po +izreku \ref{srednjicaTrik} $B_1C_1\parallel BC$. Ker $A',A_1\in +BC$, je tudi $B_1C_1\parallel A',A_1$. Torej je štirikotnik +$A'A_1B_1C_1$ trapez. Dokažimo še, da je enakokrak oz. +$C_1A'\cong B_1A_1$. Toda to dejstvo sledi iz izrekov +\ref{TalesovIzrKroz2} in \ref{srednjicaTrik}, ker je: +$C_1A'=\frac{1}{2}AB= B_1A_1$. + +Omenimo še, da lahko z uporabo trditve iz te naloge še na drug +način dokažemo izrek o Eulerjevi krožnici \ref{EulerKroznica}. + + \item \res{Naj bo $ABC$ pravokotni trikotnik s pravim kotom pri oglišču $C$. +Točki $E$ in $F$ naj bosta presečišči simetral notranjih kotov pri +ogliščih $A$ in $B$ z nasprotnima katetama, $K$ in $L$ pa +pravokotni projekciji točk $E$ in $F$ na hipotenuzo tega +trikotnika. Dokaži, da je $\angle LCK=45^0$.} + +Označimo z $\alpha$ in $\beta$ notranja kota pri ogliščih $A$ in +$B$ trikotnika $ABC$. Po izreku \ref{VsotKotTrik} je +$\alpha+\beta=90^0$. Po izreku \textit{ASA} \ref{KSK} velja +$\triangle ACE\cong\triangle AKE$ in $\triangle BCF\cong\triangle BLF$, zato +je $EC\cong EK$ in $FC\cong FL$. Torej sta trikotnika $CEK$ in $CFL$ + enakokraka z osnovnicama $CK$ in $CL$, kar pomeni, da je $\angle ECK\cong\angle EKC$ in +$\angle FCL\cong\angle FLC$. Če +uporabimo izrek \ref{zunanjiNotrNotr} za trikotnik $CFL$ in +izrek \ref{VsotKotTrik} za trikotnik $ALF$, dobimo: $\angle +FCL=\frac{1}{2}\angle LFA=\frac{1}{2}\left(90^0-\alpha \right)$. +Analogno je tudi $\angle ECK=\frac{1}{2}\left(90^0-\beta +\right)$. Iz tega sledi $\angle FCL+\angle +ECK=\frac{1}{2}\left(90^0-\alpha +90^0-\beta\right)=45^0$. Torej +$\angle LCK=90^0-(\angle FCL+\angle ECK)=90^0-45^0=45^0$. + +\item \res{Naj bo $M$ središče stranice $CD$ kvadrata $ABCD$ in $P$ takšna točka + diagonale $AC$, da velja $3AP=PC$. Dokaži, da je $\angle BPM$ +pravi kot.} + +Naj bo $S$ središče diagonale $AC$ in $V$ središče daljice $SB$. Dokažemo, da je $V$ +višinska točka trikotnika $PBC$ (glej še zgled +\ref{zgledPravokotnik}). + + \item \res{Naj bodo $P$, $Q$ in $R$ središča stranic $AB$, $BC$ in $CD$ + paralelograma $ABCD$. Premici $DP$ in $BR$ naj sekata daljico +$AQ$ v točkah $K$ in $L$. Dokaži, da je $KL= \frac{2}{5} AQ$.} + +Naj bo $S$ središče daljice $AD$ in $M$ presečišče daljic $SC$ +in $BR$. Dokaži najprej, da je $CM\cong AK$, daljica $LQ$ +srednjica trikotnika $CBM$ in daljica $PK$ srednjica trikotnika +$LAB$. + + \item \res{Naj bo $D$ središče hipotenuze $AB$ pravokotnega +trikotnika $ABC$ ($AC>BC$). Točki $E$ in $F$ naj bosta presečišči +poltrakov $CA$ in $CB$ s premico, ki poteka skozi $D$ in je pravokotna +na premico $CD$. Točka $M$ naj bo središče daljice $EF$. Dokaži, da +je $CM\perp AB$.} + +Označimo s $T$ presečišče premic $AB$ in $CM$ ter $\angle +CAB=\alpha$ in $\angle CBA=\beta$. Po predpostavki je +$\alpha+\beta=180^0$. Ker je $D$ središče hipotenuze $AB$ +pravokotnega trikotnika $ABC$, je po Talesovem izreku +\ref{TalesovIzrKroz2} $DC\cong DA$. To pomeni, da je $\triangle +CDA$ enakokraki trikotnik in velja $\angle DCE=\angle +DCA\cong\angle DAC=\alpha$. Iz $FE\perp CD$ in $FC\perp CE$ +sledi $\angle CFE\cong\angle DCE=\alpha$ (izrek +\ref{KotaPravokKraki}). Točka $M$ je središče hipotenuze $FE$ +pravokotnega trikotnika $FCE$, zato je (podobno kot pri +trikotniku $ABC$) $\angle FCM\cong\angle CFM=\angle CFE=\alpha$. +V trikotniku $BCT$ je vsota dveh notranjih kotov $\angle +CBT+\angle BCT=\angle CBA+\angle FCM=\alpha+\beta=90^0$. Po +izreku \ref{VsotKotTrik} je torej $\angle CTB=90^0$ oz. +$CM\perp AB$. + +\item \res{Naj bosta $A_1$ in $C_1$ središči stranic $BC$ in $AB$ trikotnika $ABC$. + Simetrala notranjega kota pri oglišču $A$ seka daljico +$A_1C_1$ v točki $P$. Dokaži, da je $\angle APB$ pravi kot.} + +Po izreku \ref{KotiTransverzala}, je $\angle C_1PA\cong\angle +PAC$. Torej $\angle C_1PA\cong\angle PAC\cong\angle PAC_1$, kar +pomeni, da je $PAC_1$ enakokraki trikotnik z osnovnico $AP$ oz. +$C_1A\cong C_1P$ (izrek \ref{enakokraki}). Ker je $C_1$ središče +stranice $AB$, je $C_1B\cong C_1A\cong C_1P$. Torej +točka $P$ leži na krožnici s premerom $AB$, zato je po Talesovem +izreku \ref{TalesovIzrKroz2} $\angle APB=90^0$. + + \item \res{Naj bosta $P$ in $Q$ takšni točki stranic $BC$ in $CD$ kvadrata $ABCD$, + da je premica $PA$ simetrala kota $BPQ$. Določi velikost kota + $PAQ$.} + + Naj bo $A'=pr_{\perp PQ}(A)$. Dokaži najprej, da velja $\triangle + ABP\cong\triangle AA'P$ in $\triangle ADQ\cong\triangle AA'Q$. + Rezultat: $\angle PAQ=45^0$. + +\item \res{Dokaži, da središče očrtane krožnice leži najbliže najdaljši stranici +trikotnika.} + +Naj bo $O$ središče očrtane krožnice trikotnika $ABC$ ter $A_1$, +$B_1$ in $C_1$ središča njegovih stranic $BC$, $AC$ in $AB$. +Dovolj je dokazati, da npr. iz $AC>AB$ sledi $OB_1AB$. Po izreku \ref{vecstrveckot} je +v tem primeru $\angle ABC>\angle ACB$. Ker je $B_1C_1$ srednjica +trikotnika $ABC$, je $B_1C_1\parallel BC$ (izrek +\ref{srednjicaTrik}). Zato je po izreku \ref{KotiTransverzala} +$\angle AC_1B_1\cong \angle ABC$ in $\angle AB_1C_1\cong \angle +ACB$. Iz $\angle ABC>\angle ACB$ sedaj sledi: $\angle +B_1C_1O=90^0-\angle AC_1B_1= 90^0-\angle ABC<90^0-\angle +ACB=90^0-\angle AB_1C_1=\angle C_1B_1O$ oz. $\angle +B_1C_1O<\angle C_1B_1O$. Po izreku \ref{vecstrveckot} za +trikotnik $B_1OC_1$ je potem $OB_1\gamma$ sledi $SB\gamma$ +sledi najprej $\angle +SBP=\frac{1}{2}\beta>\frac{1}{2}\gamma=\angle SCP$. Naj bo $B'$ +točka na poltraku $PC$, za katero velja $PB'\cong PB$. Iz +skladnosti trikotnikov $SPB'$ in $SPB$ (izrek \textit{SAS} +\ref{SKS}) sledi $SB'\cong SB$ in $\angle SB'P\cong \angle +SBP>\angle SCP$. Dokažimo, da velja $\mathcal{B}(P,B',C)$ oz. +$PB'\angle SB'P\cong\angle SBP$ - izrek +\ref{zunanjiNotrNotrVecji} za trikotnik $SCB'$). Ker je torej +$\mathcal{B}(P,B',C)$, je $\angle SB'C=180^0-\angle +SB'P=180^0-\frac{1}{2}\beta>90^0$. V topokotnem trikotniku +$SB'C$ je potem $SC\gamma$) označimo središčna kota. Ker je po izreku \ref{TangPogoj} $\angle ST_1P=\angle ST_2P=90^0$ dobimo (iz štirikotnika $PT_1ST_2$ - izrek \ref{VsotKotVeck}): $\angle T_1PT_2=180^0-\gamma= + \frac{\beta+\gamma}{2}-\gamma= + \frac{\beta-\gamma}{2}$. + +\item \res{Naj bo $L$ pravokotna projekcija poljubne točke $K$ krožnice $k$ + na njeni tangenti $t$ skozi točko $T\in k$ in $X$ točka, ki je +simetrična točki $L$ glede na premico $KT$. Določi geometrijsko +mesto točk $X$.} + +Označimo s $S$ središče krožnice $k$ in $\alpha=\angle LTK$. Dokažimo najprej, da so $S$, $X$ in $K$ kolinearne točke. Iz skladnosti trikotnikov $TLK$ in $TXK$ je $\angle XTK\cong \angle LTK=\alpha$, $\angle TXK\cong \angle TLK=90^0$ in $\angle TKX\cong \angle TKL=90^0-\alpha$. Po izrekih \ref{ObodKotTang} in \ref{SredObodKot} je $\angle TSK=2\angle LTK=2\alpha$. Torej iz enakokrakega trikotnika $TSK$ (izreka \ref{enakokraki} in \ref{VsotKotTrik}) dobimo: $\angle TKS=\frac{180^0-2\alpha}{2}=90^0-\alpha=\angle TKX$ oz. $\angle TKS\cong\angle TKX$, zato $X\in SK$. Na koncu je $\angle SXT\cong\angle KXT=90^0$, kar pomeni, da po izreku \ref{TalesovIzrKroz2} točka $X$ leži na krožnici $l$ s premerom $TS$. Ni težko dokazati, da velja tudi obratno - vsako točko $X\in l$ dobimo iz neke točke $K\in k$ po opisanem postopku. + +\item \res{Naj bosta $BB'$ in $CC'$ višini trikotnika $ABC$ ter $t$ +tangenta očrtane krožnice tega trikotnika v točki $A$. Dokaži, da +je $B'C'\parallel t$.} + +Označimo s $P$ poljubno točko tangente $t$, tako da sta $P$ in $C$ na različnih bregovih te tangente. Po izreku \ref{ObodKotTang} je $\angle PAC'=\angle PAB\cong\angle ACB$. Ker je $BCC'B'$ tetivni štirikotnik (izrek \ref{TalesovIzrKroz2}), je po izreku \ref{TetivniPogojZunanji} $\angle AC'B'\cong\angle ACB$. Torej velja $\angle AC'B'\cong\angle PAC'$, zato je po izreku \ref{KotiTransverzala} $PA\parallel B'C'$ oz. $B'C'\parallel t$. + +\item \res{V pravokotnem trikotniku $ABC$ je nad kateto $AC$ kot premerom +načrtana krožnica, ki seka hipotenuzo $AB$ v točki $E$. Tangenta te +krožnice v točki $E$ seka drugo kateto $BC$ v točki $D$. Dokaži, da je $BDE$ +enakokraki trikotnik.} + +Če uporabimo izrek \ref{ObodKotTang}, dobimo $\angle DEC\cong\angle EAC$. Torej: +$\angle BED=90^0-\angle DEC=90^0-\angle EAC=90^0-\angle BAC=\angle ABC=\angle EBD$. Iz +$\angle BED\cong\angle EBD$ po izreku \ref{enakokraki} sledi $BD\cong ED$, kar pomeni, da je $BDE$ enakokraki trikotnik. + + +\item \res{V pravi kot z vrhom $A$ je včrtana krožnica, ki se dotika krakov tega kota v +točkah $B$ in $C$. Poljubna tangenta te krožnice seka premici $AB$ in $AC$ po vrsti +v točkah $M$ in $N$ (tako, da je $\mathcal{B}(A,M,B)$). Dokaži, da velja: +$$\frac{1}{3}\left(|AB|+|AC|\right) < |MB|+|NC| < +\frac{1}{2}\left(|AB|+|AC|\right).$$} + +Označimo s $T$ dotikališče omenjene krožnice s premico $MN$ in $s=\frac{1}{2}(|MN|+|NA|+|AM|)$ polobseg trikotnika $MAN$. Dana krožnica je trikotniku $MAN$ pričrtana krožnica, zato po veliki nalogi velja: +\begin{eqnarray} \label{nalSkk8Eqn1} +|AB|=|AC|=s. +\end{eqnarray} +Dokažimo najprej neenakost $|MB|+|NC| <\frac{1}{2}\left(|AB|+|AC|\right)$. +Če uporabimo izrek \ref{TangOdsek} in trikotniško neenakost \ref{neenaktrik} ter relacijo \ref{nalSkk8Eqn1}, dobimo: + \begin{eqnarray*} +|MB|+|NC| &=& |MT|+|NT| =|MN|=\\ + &=& \frac{1}{2}\left( |MN|+|MN| \right)\\ + &<& \frac{1}{2}\left( |MN|+|NA|+|AM| \right)=\\ + &=& s=|AB|=\\ + &=& \frac{1}{2}\left(|AB|+|AC|\right). +\end{eqnarray*} +Dokažimo še drugo neenakost: $|MB|+|NC| >\frac{1}{3}\left(|AB|+|AC|\right)$. +Ker je $MAN$ pravokotni trikotnik s hipotenuzo $MN$, po izreku \ref{vecstrveckotHipot} sledi: +\begin{eqnarray} \label{nalSkk8Eqn2} +|MN|>|MA| \hspace*{1mm} \textrm{ in } \hspace*{1mm} |MN|>|NA|. +\end{eqnarray} +Če uporabimo izrek \ref{TangOdsek} ter relaciji \ref{nalSkk8Eqn2} in \ref{nalSkk8Eqn1}, dobimo: + \begin{eqnarray*} +|MB|+|NC| &=& |MT|+|NT| =|MN|=\\ + &=& \frac{1}{3}\left( |MN|+|MN|+|MN| \right)\\ + &>& \frac{1}{3}\left( |MN|+|NA|+|AM| \right)=\\ + &=& \frac{1}{3}\cdot2s=\frac{1}{3}\cdot2\cdot|AB|=\\ + &=& \frac{1}{3}\left(|AB|+|AC|\right). +\end{eqnarray*} + +\item \res{Dokaži, da je v pravokotnem trikotniku vsota katet enaka vsoti +premerov očrtane in včrtane krožnice.} + +Naj bo $ABC$ pravokotni trikotnik s pravim kotom ob oglišču $C$. Uporabimo oznake kot pri veliki nalogi \ref{velikaNaloga}. Ker je $\angle PCQ=\angle BCA=90^0$, je štirikotnik $PCQS$ kvadrat, zato po veliki nalogi (\ref{velikaNaloga}) sledi: + \begin{eqnarray} \label{nalSkk9Eqn} +r=|SQ|=|CP|=s-c. +\end{eqnarray} + Iz tega in Talesovega izreka (\ref{TalesovIzrKroz2}) sledi: + $$R+r=\frac{c}{2}+s-c=b+c.$$ + +\item \res{Simetrale notranjih kotov konveksnega štirikotnika naj se sekajo v šestih različnih točkah. +Dokaži, da so štiri od teh točk oglišča tetivnega štirikotnika.} + +Uporabi kriterij za tetivnost štirikotnika - izrek \ref{TetivniPogoj}. + +\item \res{Naj bodo: $c$ dolžina hipotenuze, $a$ in $b$ dolžini +katet ter $r$ polmer včrtane krožnice pravokotnega trikotnika. Dokaži, da velja:} +\begin{enumerate} + \item \res{$2r + c \geq 2 \sqrt{ab}$} + +Uporabi relacijo \ref{nalSkk9Eqn} ter neenakost aritmetične in geometrijske sredine. + + \item \res{$a + b + c > 8r$} + +Uporabi relacijo \ref{nalSkk9Eqn} in Pitagorov izrek \ref{PitagorovIzrek}. +\end{enumerate} + +\item \res{Naj bosta $P$ in $Q$ središči krajših lokov $AB$ in $AC$ +pravilnemu trikotniku $ABC$ očrtane krožnice. Dokaži, da stranici $AB$ in $AC$ tega trikotnika razdelita +tetivo $PQ$ na tri skladne daljice.} + +Označimo z $X$ in $Y$ presečišči tetive $PQ$ s stranicama $AB$ in $AC$ trikotnika $ABC$. Dokažimo, da velja $PX\cong XY\cong YQ$. Po izreku \ref{TockaN} sta $BQ$ in $CP$ simetrali notranjih kotov $ABC$ in $ACB$ trikotnika $ABC$. Po izreku \ref{ObodObodKot} je: + \begin{eqnarray*} +\angle PAX &=& \angle PAB\cong\angle PCB=\frac{1}{2}\angle ACB=30^0\\ +\angle APX &=& \angle APQ\cong\angle ABQ=\frac{1}{2}\angle ABC=30^0. +\end{eqnarray*} +Torej: + $\angle PAX = \angle APX=30^0$, kar pomeni, da je $APX$ enakokraki trikotnik z osnovnico $AP$ (izrek \ref{enakokraki}) oz. $PX\cong XA$. Po izreku \ref{zunanjiNotrNotr} je še $\angle AXY=\angle APX+\angle PAX=60^0$. Ker je še $\angle XAY=\angle BAC= 60^0$, je $AXY$ pravilni trikotnik, zato je $XA\cong XY\cong YA$. Analogno je tudi $AQY$ enakokraki trikotnik z osnovnico $AQ$ oz. $QY\cong YA$. Če povežemo dokazane relacije skladnosti daljic, dobimo: $PX\cong XA\cong XY\cong AY\cong YQ$. + +\item \res{Naj bodo $k_1$, $k_2$, $k_3$, $k_4$ štiri krožnice, od katerih se vsaka od zunaj dotika ene stranice in dveh nosilk stranic poljubnega konveksnega +štirikotnika. Dokaži, da so središča teh krožnic konciklične točke.} + +Dokaži najprej, da dva nasprotna kota dobljenega štirikotnika merita $\frac{\alpha+\beta}{2}$ in $\frac{\gamma+\delta}{2}$, kjer so $\alpha$, $\beta$, $\gamma$ in $\delta$ notranji koti začetnega štirikotnika, nato pa uporabi izreka \ref{TetivniPogoj} in \ref{VsotKotVeck}. + +\item \res{Krožnici $k$ in $l$ se od zunaj dotikata v točki $A$. Točki $B$ in $C$ sta dotikališči +skupne zunanje tangente teh dveh krožnic. Dokaži, da je $\angle BAC$ pravi kot.} + +Naj bo $L$ presečišče skupne tangente krožnic $k$ in $l$ v točki $A$ z daljico $BC$. Ker sta $XA$ in $XB$ tangenti krožnice $k$ v točkah $A$ in $B$, je $XA\cong XB$ (izrek \ref{TangOdsek}). Analogno sta $XA$ in $XC$ tangenti krožnice $l$ v točkah $A$ in $C$ oz. je $XA\cong XC$. Torej je $X$ središče krožnice s premerom $BC$ in točka $A$ leži na tej krožnici. Po Talesovem izreku \ref{TalesovIzrKroz2} je $\angle BAC=90^0$. + +\item \res{Naj bo $ABCD$ deltoid ($AB\cong AD$ in $CB\cong CD$). Dokaži:} +\begin{enumerate} + \item \res{$ABCD$ je tangentni štirikotnik.} + +Uporabi izrek \ref{TangentniPogoj}. + + \item \res{$ABCD$ je tetivni štirikotnik natanko tedaj, ko je $AB\perp BC$.} + +Uporabi izrek \ref{TetivniPogoj}. +\end{enumerate} + + + +\item \res{Krožnici $k$ in $k_1$ se od zunaj dotikata v točki $T$, v kateri se sekata premici $p$ in $q$. Premica $p$ ima +s krožnicama še presečišči $P$ in $P_1$, premica $q$ pa $Q$ in $Q_1$. Dokaži, da je $PQ\parallel P_1Q_1$.} + +Naj bo $t$ skupna tangenta krožnic $k$ in $k_1$, $X\in t$ poljubna točka te tangente, za katero velja $P,X\div QQ_1$ in $Y$ točka, ki je simetrična točki $X$ glede na $T$. Po izrekih \ref{ObodKotTang} in \ref{sovrsnaSkladna} velja: + \begin{eqnarray*} + \angle QPP_1= \angle QPT \cong\angle QTX\cong\angle Q_1TY\cong\angle Q_1P_1T + =\angle Q_1P_1P + \end{eqnarray*} +Torej $\angle QPP_1 \cong\angle Q_1P_1P$, zato je po izreku \ref{KotiTransverzala} $PQ\parallel P_1Q_1$. + +\item \res{Naj bo $MN$ skupna tangenta krožnic $k$ in $l$ ($M$ in $N$ sta dotikališči), +ki se sekata v točkah $A$ in $B$. Izračunaj vrednost vsote $\angle MAN+\angle MBN$.} + +Brez škode za splošnost predpostavimo, da je $d(A,MN) ||AE|-|BE||= |\left(|AC|-|EC|\right)-\left(|BD|-|ED|\right)| = ||AC|-|BD||. + \end{eqnarray*} + + +\item \res{Naj bo $S$ presečišče nosilk krakov $AD$ in $BC$ +trapeza $ABCD$ z osnovnico $AB$. Dokaži, da se očrtani krožnici trikotnikov $SAB$ in +$SCD$ dotikata v točki $S$.} + +Uporabi izrek \ref{TangOdsek}. + +\item \res{Premici $PB$ in $PD$ se dotikata krožnice $k(O,r)$ v točkah $B$ in $D$. +Premica $PO$ seka krožnico $k$ v točkah $A$ in $C$ ($\mathcal{B}(P,A,C)$). Dokaži, da je +premica $BA$ simetrala kota $PBD$.} + + +Uporabi izreka \ref{enakokraki} in \ref{TangOdsek}. + +\item \res{Štirikotnik $ABCD$ je včrtan krožnici s središčem $O$. Diagonali $AC$ in +$BD$ sta pravokotni. Naj bo $M$ pravokotna projekcija središča $O$ +na premici $AD$. Dokaži, da je + $$|OM|=\frac{1}{2}|BC|.$$} + +Iz skladnosti trikotnikov $OMA$ in $OMD$ (izrek \textit{SSA} \ref{SSK}) je $MA\cong MD$ oz. točka $M$ je središče daljice $AD$. +Označimo z $E$ presečišče diagonal $AC$ in $BD$. Naj bo $D'$ točka, ki je simetrična točki $D$ glede na središče $O$. Jasno je, da velja $D'\in k$. Po Talesovem izreku \ref{TalesovIzrKroz2} je $\angle DAD'=90^0$. Daljica $MO$ je srednjica trikotnika $DAD'$ za stranico $AD'$, zato je $|OM|=\frac{1}{2}|AD'|$. Dovolj je dokazati še $AD'\cong BC$ oz. (po izreku \ref{SklTetSklObKot}) $\angle ADD'\cong \angle BDC$. Če sdaj uporabimo izreka \ref{VsotKotTrik} in \ref{ObodObodKot}, dobimo: +\begin{eqnarray*} + \angle ADD' &=& 90^0-\angle AD'D =90^0- \angle ACD =\\ +&=& 90^0-\angle ECD =\angle EDC =\\ +&=& \angle BDC. + \end{eqnarray*} + + +\item \res{Daljici $AB$ in $BC$ sta sosednji stranici pravilnega devetkotnika, ki je včrtan krožnici $k$ s središčem $O$. +Točka $M$ je središče stranice $AB$, točka $N$ pa središče +polmera $OX$ krožnice $k$, ki je pravokoten na premico $BC$. Dokaži, da je +$\angle OMN=30^0$.} + +Iz $\angle AOB\cong\angle BOC=\frac{360^0}{9}=40^0$ sledi: + $$\angle AOX=\angle AOB + \angle BOX=\angle AOB + \frac{1}{2}\angle BOC=60^0.$$ +Ker je še $OA\cong OB$, je $AOB$ pravilni trikotnik (izreka \ref{enakokraki} in \ref{VsotKotTrik}). Njegova težiščnica $AN$ je hkrati višina oz. $\angle ONA=90^0$. Ker je še $OMA=90^0$, po Talesovem izreku točki $N$ in $M$ ležita na krožnici s premerom $OA$ oz. $ONMA$ je tetiven štirikotnik. Če na koncu uporabimo še izrek \ref{ObodObodKot}, dobimo: +\begin{eqnarray*} + \angle OMN=\angle OAN=\frac{1}{2}\angle OAX=30^0. + \end{eqnarray*} + +\item \res{Krožnici $k_1$ in $k_2$ se sekata v točkah $A$ in $B$. Naj bo $p$ premica, ki poteka skozi točko $A$, krožnico $k_1$ seka še v točki $C$, krožnico $k_2$ pa še v točki $D$, ter $q$ premica, ki poteka skozi točko $B$, krožnico $k_1$ seka še v točki $E$, krožnico $k_2$ pa še v točki $F$. Dokaži, da +je $\angle CBD\cong\angle EAF$.} + +Če uporabimo izreka \ref{VsotKotTrik} in \ref{ObodObodKot}, dobimo: +\begin{eqnarray*} + \angle CBD &=& 180^0-\angle BCD - \angle BDC =\\ + &=& 180^0-\angle BCA - \angle BDA =\\ + &=& 180^0-\angle BEA - \angle BFA =\\ + &=& 180^0-\angle FEA - \angle EFA =\\ + &=& \angle EAF =\\ + \end{eqnarray*} + + +\item \res{Krožnici $k_1$ in $k_2$ se sekata v točkah $A$ in $B$. Načrtaj premico $p$, ki poteka skozi točko $A$, tako da bo +dolžina daljice $MN$, kjer sta $M$ in $N$ presečišči premice $p$ +s krožnicama $k_1$ in $k_2$, maksimalna.} + +Naj bo $p$ poljubna premica, ki poteka skozi točko $A$ in krožnici $k_1$ in $k_2$ seka v točkah $M$ in $N$. +Označimo s $S_1$ in $S_2$ središči krožnic $k_1$ in $k_2$ ter s $P_1$ in $P_2$ središči daljic $MA$ in $NA$. Ker je $MN=2\cdot P_1P_2$, se problem maksimuma daljice $MN$ prevede na maksimum daljice $P_1P_2$. Iz skladnosti trikotnikov $MS_1P_1$ in $AS_1P_1$ (izrek \textit{SSS} \ref{SSS}) sledi $\angle S_1P_1M\cong\angle S_1P_1A=90^0$ oz. $\angle P_2P_1S_1=90^0$. Analogno je tudi $\angle P_2P_1S_1=90^0$. Torej je $S_1S_2P_2P_1$ pravokotni trapez z višino $P_1P_2$, zato je $P_1P_2\leq S_1S_2$. Enakost se doseže, ko je $S_1S_2P_2P_1$ pravokotnik oz. $P_1P_2\parallel S_1S_2$. To pomeni, da premico $p$ (za katero je dolžina $|MN|$ maksimalna) narišemo kot vzporednico premice $S_1S_2$ skozi točko $A$. + +\item \res{Naj bo $L$ pravokotna projekcija poljubne točke $K$ krožnice $k$ na njeni +tangenti skozi točko $T\in k$ ter $X$ točka, ki je simetrična točki $L$ glede na premico +$KT$. Določi geometrijsko mesto točk $X$.} + + +Označimo s $S$ središče krožnice $k$ in $k_1$ krožnico nad premerom $ST$. +Dokažimo, da je geometrijsko mesto točk $X$ (označimo ga z $\mathcal{G}$) enako krožnici $k_1$. Potrebno je dokazati ekvivalenco: +\begin{eqnarray*} + X\in \mathcal{G} \hspace*{1mm} \Leftrightarrow \hspace*{1mm} X\in k_1 + \end{eqnarray*} + +Za točki $T$ in $S$ je po definiciji jasno, da velja hkrati $T\in \mathcal{G}$ in $T\in k_1$ oz. $S\in \mathcal{G}$ in $S\in k_1$. Predpostavimo naprej, da je $X\neq T$ in $X\neq S$. + +Predpostavimo najprej $X\in \mathcal{G}$. Označimo $\angle LTK=\alpha$. + Trikotnika $XKT$ in $LKT$ sta simetrična glede na premico $KT$ (glej razdelek \ref{odd6OsnZrc}), zato je $\triangle XKT\cong \triangle LKT$ oz. $\angle XTK\cong \angle LTK=\alpha$ in $\angle KXT\cong \angle KLT=90^0$. Dokažimo, da je tudi $\angle SXT=90^0$. Dovolj je dokazati, da so točke $S$, $X$ in $K$ kolinearne oz. da velja $\angle TKX=\angle TKS$. Najprej je (iz $\triangle XKT\cong \triangle LKT$): + $\angle TKX\cong\angle TKL =90^0-\alpha$. Če uporabimo izrek \ref{enakokraki} enakokrakega trikotnika $KST$ z osnovnico $KT$ in izreka \ref{ObodKotTang} ter \ref{SredObodKot}, je: + $\angle TKS=\frac{1}{2}(180^0-\angle KST)=\frac{1}{2}(180^0-2\angle LTK)=90^0-\alpha$. +Torej $\angle TKX=\angle TKS=90^0-\alpha$, kar pomeni, da so točke $S$, $X$ in $K$ kolinearne oz. $\angle SXT=90^0$. Iz slednjega je $X\in k_1$. + +Predpostavimo sedaj, da je $X\in k_1$. Označimo s $K$ presečišče poltraka $SX$ s krožnico $k$ in z $L$ točko, ki je simetrična točki $X$ glede na premico $TK$. + Iz $X\in k_1$ po Talesovem izreku \ref{TalesovIzrKroz2} sledi $\angle SXT=90^0$, zato je tudi $\angle KXT=90^0$. +Da dokažemo $X\in \mathcal{G}$, je dovolj dokazati, da $X$ leži na tangenti $t$ krožnice $k$ skozi točko $T$. Naj bo $L'$ poljubna točka na tangenti $t$, za katero velja $L,X\ddot{-} ST$. Dovolj je dokazati $\angle LTK\cong\angle L'TK$. +Označimo $\angle LTK=\alpha$. + Trikotnika $XKT$ in $LKT$ sta simetrična glede na premico $KT$, zato je $\triangle XKT\cong \triangle LKT$ oz. $\angle XTK\cong \angle LTK=\alpha$ in $\angle KLT\angle KXT\cong =90^0$. Če uporabimo izreka \ref{ObodKotTang} in \ref{SredObodKot}, dobimo: +$\angle L'TK=\frac{1}{2}\angle TSK=\frac{1}{2}(180^0-2\angle SKT)=90^0-\angle SKT=90^0-\angle XKT=\angle XTK=\alpha=\angle LTK$. +Torej $L'\in t$ oz. $X\in \mathcal{G}$. + +\item \res{Dokaži, da je tetivni večkotnik z lihim številom oglišč, ki ima vse notranje kote skladne, pravilni večkotnik.} + +Označimo z $O$ središče očrtane krožnice tega večkotnika in z $B_i$ središča stranic $A_iB_{i+1}$ ($i\in \{1,2,\ldots , 2n-1 \}$, $A_0=A_{2n-1}$ in $A_{2n}=A_1$). Uporabimo dejstvo, da je $A_iOA_{i+1}$ enakokraki trikotnik z osnovnico $A_iA_{i+1}$ in dokažimo najprej $\triangle OB_{i-1}A_i\cong \triangle OB_{i+1}A_{i+1}$. + +\item \res{Dve krožnici se dotikata od znotraj v točki $A$. Daljica $AB$ je premer večje +krožnice, tetiva $BK$ večje krožnice pa se dotika manjše krožnice v točki $C$. Dokaži +da je premica $AC$ simetrala kota $BAK$.} + +Označimo z $EA$ skupno tangento krožnic $k$ (večje) in $l$ (manjše) ter z $D$ drugo presečišče premice $AB$ s krožnico $l$. Naj bo še $L$ drugo presečišče poltraka $AK$ s krožnico $l$, $\alpha=\angle DCB$ in $\beta=\angle DBC$. + Po izreku \ref{ObodKotTang} je: + \begin{eqnarray} \label{nalSkk36Eqn1} + \angle CAB=\angle CAD\cong \angle DCB=\alpha +\end{eqnarray} +in + \begin{eqnarray} \label{nalSkk36Eqn2} + \angle LDA\cong \angle EAL=\angle EAK\cong \angle ABK=\angle DBC=\beta. +\end{eqnarray} +Kot $CDA$ je zunanji kot trikotnika $CDB$, zato je po izreku \ref{zunanjiNotrNotr} + $\angle CDA=\angle DCB+\angle CBD=\alpha+\beta$. Iz tega in \ref{nalSkk36Eqn2} sledi: + \begin{eqnarray} \label{nalSkk36Eqn3} + \angle LDC= \angle CDA-\angle LDA=\alpha+\beta-\beta=\alpha. +\end{eqnarray} +Iz relacij \ref{nalSkk36Eqn3} in \ref{nalSkk36Eqn1} ter izreka \ref{ObodObodKot}: +\begin{eqnarray*} + \angle KAC= \angle LAC\cong \angle LDC=\alpha=\angle CAB. +\end{eqnarray*} +To pomeni, da je premica $AC$ simetrala kota $BAK$. + +Naloga je poseben primer splošnejše naloge \ref{nalSkk47}. + +\item \res{Naj bo $BC$ tetiva krožnice $k$. Določi geometrijsko mesto višinskih točk +vseh trikotnikov $ABC$, kjer je $A$ poljubna točka, ki leži na krožnici $k$.} + +Iskano geometrijsko mesto točk je krožnica, ki poteka skozi točki $B$ in $C$. Izračunaj mero kota $BVC$, kjer je $V$ višinska točka trikotnika $ABC$, in uporabi izrek \ref{ObodKotGMT}. Glej tudi izrek \ref{TockaV'a}. + +\item \res{Imejmo štirikotnik s tremi topimi notranjimi koti. Dokaži, da daljša +diagonala poteka iz oglišča, ki pripada ostrem kotu.} + +Označimo z $ABCD$ štirikotnik, pri katerem so notranji koti pri ogliščih $A$, $B$ in $C$ topi. +Naj bo $k$ krožnica s premerom $BD$. Ker je $\angle BAD>90^0$ in $\angle BCD>90^0$, sta po izreku \ref{obodKotGMTZunNotr} $A$ in $C$ notranji točki krožnice $k$. To pomeni, da je $AC90^0$. To pomeni $\mathcal{B}(A',A,V)$ (kjer je $A'$ višina tega trikotnika) oz. $V$ je zunanja točka krožnice $k$ in velja $|\overrightarrow{OV}|>1$. + Lahko pogledamo tudi nalogo \ref{OlimpVekt15} (razdelek \ref{odd5DolzVekt}). + +\item \res{Izračunaj kote, ki jih določajo vektorji $\overrightarrow{OA}$, $\overrightarrow{OB}$ in $\overrightarrow{OC}$, če točke $A$, $B$ in $C$ ležijo na krožnici s središčem +$O$ in dodatno velja še: +$$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{0}.$$} + +Po Hamiltonovem izreku \ref{Hamilton} je $\overrightarrow{0}=\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}=\overrightarrow{OV}$, kjer je $V$ višinska točka trikotnika $ABC$. Torej $V=O$, kar pomeni, da je $ABC$ enakostranični trikotnik so vsi iskani koti enaki $120^0$. + + \item \res{Naj bodo $A$, $B$, $C$ in $D$ točke, ki ležijo na krožnici s središčem +$O$, in velja +$$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = \overrightarrow{0}.$$ +Dokaži, da je $ABCD$ pravokotnik.} + +Uporabimo Hamiltonov izrek \ref{Hamilton}. + +\item \res{Naj bodo $\overrightarrow{a}$, $\overrightarrow{b}$ in $\overrightarrow{c}$ vektorji neke ravnine, za katere velja $|\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{c}| =x$. Razišči, v katerem primeru velja tudi $|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}| = x$.} + +Predpostavimo najprej, da so $\overrightarrow{a}$, $\overrightarrow{b}$ in $\overrightarrow{c}$ nekolinearni vektorji. +Naj bo $O$ poljubna točka in $A$, $B$ in $C$ takšne točke, da velja $\overrightarrow{OA}=\overrightarrow{a}$, $\overrightarrow{OB}=\overrightarrow{b}$ in $\overrightarrow{OC}=\overrightarrow{c}$ (izrek \ref{vektAvObst1TockaB}). Iz nekolinearnosti vektorjev $\overrightarrow{a}$, $\overrightarrow{b}$ in $\overrightarrow{c}$ sledi nekolinearnost točk $A$, $B$ in $C$. Iz pogoja $|\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{c}| =x$ sledi $|\overrightarrow{OA}| = |\overrightarrow{OB}| = |\overrightarrow{OC}| =x$, kar pomeni, da je točka $O$ središče očrtane krožnice trikotnika $ABC$. Po Hamiltonovem izreku \ref{Hamilton} je: + \begin{eqnarray*} +x + &=& +|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}|=\\ + &=& +|\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}|=\\ + &=& |\overrightarrow{OV}|, +\end{eqnarray*} + kjer je $V$ višinska točka, ki torej leži na očrtani krožnici trikotnika $ABC$. To je možno natanko tedaj, ko je $ABC$ pravokotni trikotnik. + +Če so $\overrightarrow{a}$, $\overrightarrow{b}$ in $\overrightarrow{c}$ kolinearni vektorji, je dani pogoj izpolnjen natanko tedaj, ko $\overrightarrow{a}$, $\overrightarrow{b}$ in $\overrightarrow{c}$ niso vsi iste orientacije. + + + %Tales + %______________________________________________________ + + +\item \res{Dano daljico $AB$ razdeli:} + + (\textit{a}) \res{na pet enakih daljic,}\\ + (\textit{b}) \res{v razmerju $2:5$,}\\ + (\textit{c}) \res{na tri daljice, ki so v razmerju $2:\frac{1}{2}:1$.} + +Uporabimo zgled \ref{izrekEnaDelitevDaljice}. + + + \item \res{Dana je daljica $AB$. Samo z uporabo ravnila z možnostjo risanja vzporednic (konstrukcije v afini geometriji) načrtaj točko $C$, če je:} + +(\textit{a}) \res{$\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}$,} \hspace*{3mm} + (\textit{b}) \res{$\overrightarrow{AC}=\frac{3}{5}\overrightarrow{AB}$,} \hspace*{3mm} + (\textit{c}) \res{$\overrightarrow{AC}=-\frac{4}{7}\overrightarrow{AB}$.} + +Uporabimo nalogo \ref{nalVekt4} in zgled \ref{izrekEnaDelitevDaljiceNan}. + + + \item \res{Dane so daljice $a$, $b$ in $c$. Načrtaj daljico $x$, tako da bo:} + + (\textit{a}) \res{$a:b=c:x$;} \hspace*{3mm} + (\textit{b}) \res{$x=\frac{a\cdot b}{c}$,} \hspace*{3mm} + (\textit{c}) \res{$x=\frac{a^2}{c}$,}\hspace*{3mm}\\ + (\textit{č}) \res{$x=\frac{2ab}{3c}$,}\hspace*{3mm} + (\textit{č}) \res{$(x+c):(x-c)=7:2$.} + +Uporabimo talesov izrek \ref{TalesovIzrek}. + + \item \res{Naj bosta $M$ in $N$ točki na kraku $OX$, $P$ točka na kraku $OY$ kota +$XOY$ ter $NQ\parallel MP$ in $PN\parallel QS$ ($Q\in OY$, $S\in OX$). Dokaži, da je +$|ON|^2=|OM|\cdot |OS|$ (za daljico $ON$ v tem primeru pravimo, da je \index{geometrijska sredina daljic}\pojem{geometrijska sredina} \color{green1} daljic $OM$ +in $OS$).} + +Po Talesovem izreku \ref{TalesovIzrek} je: + $$\frac{|ON|}{|OM|}=\frac{|OQ|}{|OP|}=\frac{|OS|}{|ON|},$$ iz tega sledi iskana relacija. + +\item \res{Naj bo $ABC$ trikotnik ter $Q$, $K$, $L$, $M$, $N$ in $P$ takšne točke poltrakov $AB$, $AC$, $BC$, +$BA$, $CA$ in $CB$, da velja $AQ\cong CP\cong AC$, $AK\cong BL\cong AB$ in $BM\cong CN\cong BC$. +Dokaži, da so $MN$, $PQ$ in $LK$ tri vzporednice.} + +Uporabimo izreka \ref{vektParamPremica} in \ref{vektDelitDaljice}. + +\item \res{Naj bo $P$ središče težiščnice $AA_1$ trikotnika $ABC$. Točka $Q$ je presečišče premice $BP$ +s stranico $AC$. Izračunaj razmerji $AQ:QC$ in $BP:PQ$.} + +Naj bo $R$ središče daljice $QC$. Po izreku \ref{srednjicaTrikVekt} (za trikotnika $BQC$ in $AA_aR$) je: + $$\overrightarrow{PQ}=\frac{1}{2}\overrightarrow{A_1R}=\frac{1}{4}\overrightarrow{BQ},$$ + zato je $BP:PQ=3:1$. Iz $\overrightarrow{PQ}=\frac{1}{2}\overrightarrow{A_1R}$ sledi $PQ\parallel A_1R$, zato je po Talesovem izreku $AQ:QR=AP:PA_1=1:1$. Torej $AQ\cong QR \cong RC$ oz. $AQ:QC=1:2$. + +Glej tudi zgled \ref{TezisceSredisceZgled}. + +\item \res{Točki $P$ in $Q$ ležita na stranicah $AB$ in $AC$ trikotnika $ABC$, pri tem pa velja $\frac{|\overrightarrow{PB}|}{|\overrightarrow{AP}|} + +\frac{|\overrightarrow{QC}|}{|\overrightarrow{AQ}|}=1$. Dokaži, da težišče tega trikotnika leži na daljici $PQ$.} + +Naj bo $\widehat{T}$ presečišče premice $PQ$ s težiščnico $AA_1$ tega trikotnika. Dovolj je dokazati $\widehat{T}=T$. +Ker so točke $A$, $\widehat{T}$ in $A_1$ kolinearne, je $\overrightarrow{A\widehat{T}}=\lambda\overrightarrow{AA_1}$ za nek $\lambda\in R$. Za $\widehat{T}=T$ je dovolj dokazati, da je $\lambda=\frac{2}{3}$. + +Označimo $\alpha=\frac{|\overrightarrow{PB}|}{|\overrightarrow{AP}|}$ in $\beta=\frac{|\overrightarrow{QC}|}{|\overrightarrow{AQ}|}$. Po predpostavki je $\alpha+\beta=1$. Iz danih pogojev je + $\overrightarrow{AB}=\left(1+\alpha\right)\cdot\overrightarrow{AP}$ in + $\overrightarrow{AC}=\left(1+\beta\right)\cdot\overrightarrow{AQ}$. Če uporabimo slednji relaciji in izrek \ref{vektSredOSOAOB}, dobimo: + \begin{eqnarray*} + \overrightarrow{A\widehat{T}} &=& \lambda\overrightarrow{AA_1}=\\ + &=& +\frac{\lambda}{2}\cdot \left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\\ + &=& +\frac{\lambda}{2}\cdot \left(\left(1+\alpha\right)\cdot\overrightarrow{AP} ++\left(1+\beta\right)\cdot\overrightarrow{AQ}\right)=\\ + &=& +\frac{\lambda}{2}\cdot \left(1+\alpha\right)\cdot\overrightarrow{AP} ++\frac{\lambda}{2}\left(1+\beta\right)\cdot\overrightarrow{AQ}. + \end{eqnarray*} +Ker točka $T$ leži na premici $PQ$ in ker je $\alpha+\beta=1$, je po izreku \ref{vektParamPremica}: + \begin{eqnarray*} +1&=&\frac{\lambda}{2}\cdot \left(1+\alpha\right) ++\frac{\lambda}{2}\left(1+\beta\right)=\\ +&=&\frac{\lambda}{2}\cdot \left(1+\alpha+1+\beta\right)=\\ + &=&\frac{3\lambda}{2}\\ + \end{eqnarray*} +oz. $\lambda=\frac{2}{3}$. + + + \item \res{Naj bodo $a$, $b$ in $c$ trije poltraki s skupnim izhodiščem $S$ ter $M$ +točka na poltraku $a$. Če se točka $M$ “giblje” po poltraku $a$, + je razmerje razdalj te točke od premic $b$ in $c$ konstantno. Dokaži.} + +Uporabimo Talesov izrek \ref{TalesovIzrek}. + + \item \res{Naj bo $D$ točka, ki leži na stranici $BC$ trikotnika $ABC$ ter + $F$ in $G$ točki, v katerih premica, ki poteka skozi točko $D$ in je vzporedna s težiščnico $AA_1$, seka premici $AB$ in $AC$. +Dokaži, da je vsota $|DF|+|DG|$ konstantna, če se točka $D$ “giblje” po stranici +$BC$.} + +Dokažemo, da je $|DF|+|DG|=|AA_1|$. Glej nalogo \ref{nalVekt29}. + + + \item + \res{Načrtaj trikotnik s podatki:} + +V obeh primerih uporabi iste oznake kot v veliki nalogi \ref{velikaNaloga} in izreku \ref{velNalTockP'}. + + (\textit{a}) \res{$v_a$, $r$, $b-c$} + +Najprej narišemo pravokotni trikotnik $SPP_a$ ($SP=r$, $\angle P=90^0$ in $PP_a=b-c$), točko $P'$ in na koncu uporabimo dejstvo, da oglišče $A$ leži na poltraku $P_aP'$. + + (\textit{b}) \res{$\beta$, $r$, $b-c$} + +Podobno kot v prejšnjem primeru. + + +\end{enumerate} + + + + +%REŠITVE - Izometrije +%________________________________________________________________________________ + +\poglavje{Isometries} + +\begin{enumerate} + \item + \res{Dana je premica $p$ ter točki $A$ in $B$, ki ležita na + nasprotnih straneh premice $p$. Konstruiraj točko + $X$, ki leži na +premici $p$, tako da bo razlika $|AX|-|XB|$ maksimalna.} + +Naj bo $A'=\mathcal{S}_p(A)$. Dokažemo, da je $X=A'B\cap p$ iskana +točka (glej tudi zgled \ref{HeronProbl}). + + \item \res{V ravnini so dane premice $p$, $q$ in $r$. Konstruiraj + enakostranični trikotnik $ABC$, tako da + oglišče $B$ leži na premici $p$, $C$ na $q$, +višina iz oglišča $A$ pa na premici $r$.} + +Ker je $\mathcal{S}_r(B)=C$ in $B\in p$, oglišče $C$ dobimo iz +pogoja $C\in \mathcal{S}_r(p)\cap q$. + +\item \res{Dan je štirikotnik $ABCD$ in točka $S$. Načrtaj paralelogram +s središčem v točki $S$, tako +da njegova oglišča ležijo na nosilkah +stranic danega štirikotnika.} + +Uporabimo središčno zrcaljenje $\mathcal{S}_S$. + +\item \res{Naj bo $\mathcal{I}$ indirektna izometrija ravnine, ki + preslika točko $A$ v točko $B$, +$B$ pa v $A$. Dokaži, da je $\mathcal{I}$ osno zrcaljenje.} + +Naj bo $s$ simetrala daljice $AB$. Kompozitum +$\mathcal{I}\circ\mathcal{S}_s$ je direktna izometrija z dvema +fiksnima točkama $A$ in $B$, zato po izreku \ref{izo2ftIdent} +predstavlja identiteto oz. +$\mathcal{I}\circ\mathcal{S}_s=\mathcal{E}$. Če množimo enakost +z desne s $\mathcal{S}_s$, dobimo $\mathcal{I}=\mathcal{S}_s$. + +\item \res{Naj bosta $K$ in $L$ točki, ki sta simetrični z ogliščem +$A$ trikotnika $ABC$ glede na +simetrali notranjih kotov ob ogliščih $B$ in $C$. Točka $P$ naj bo +dotikališče včrtane krožnice tega trikotnika in stranice $BC$. +Dokaži, da je $P$ središče daljice $KL$.} + +Označimo s $s_\beta$ in $s_\gamma$ omenjene simetrale notranjih +kotov ob ogliščih $B$ in $C$ ter s $Q$ in z $R$ dotikališči +včrtane krožnice trikotnika $ABC$ s stranicama $AC$ in $AB$. Po +izreku \ref{TangOdsek} je $AQ\cong AR$, $BR\cong BP$ in $CP\cong +CQ$. Torej $\mathcal{S}_{s_\beta}:A,R\mapsto K,P$ in +$\mathcal{S}_{s_\gamma}:A,Q\mapsto L,P$. Iz tega sledi $KP\cong +AR\cong AQ\cong LP$, kar pomeni, da je $P$ središče daljice +$KL$. + +\item \res{Naj bosta $k$ in $l$ krožnici na različnih bregovih premice +$p$. Načrtaj enakostranični trikotnik $ABC$, tako da njegova višina +$AA'$ leži na premici $p$, oglišče $B$ na krožnici $k$, oglišče +$C$ pa na krožnici $l$.} + +Ker velja $\mathcal{S}_p(B)=C$ in $B\in k$, točko $C$ dobimo iz +pogoja $C\in \mathcal{S}_p(k)\cap l$. + +\item \res{Naj bo $k$ krožnica ter $a$, $b$ in $c$ premice v isti ravnini. +Načrtaj trikotnik $ABC$, ki je +včrtan krožnici $k$, tako da bodo njegove stranice $BC$, $AC$ in + $AB$ vzporedne po vrsti s premicami $a$, $b$ in $c$.} + + Najprej lahko načrtamo simetrale stranic $BC$, $AC$ in $AB$ kot + premice, ki gredo skozi središče $O$ krožnice $k$ in so + pravokotne po vrsti s premicami $a$, $b$ in $c$. Označimo te + simetrale s $p$, $q$ in $r$. Ker ima kompozitum + $\mathcal{I}=\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p$ + fiksni točki $O$ in $A$, je po izreku \ref{izo1ftIndZrc} + $\mathcal{I}=\mathcal{S}_{OA}$. Torej lahko premico $OA$ + narišemo kot simetralo daljice $XX'$, kjer je $X$ poljubna + točka. Oglišče $A$ dobimo kot presečišče premice $OA$ in + krožnice $k$. Nato je $B=\mathcal{S}_p(A)$ in + $C=\mathcal{S}_r(A)$. + +\item \res{Naj bo $ABCDE$ tetivni petkotnik, v katerem je $BC\parallel +DE$ in $CD\parallel EA$. +Dokaži, da oglišče $D$ leži na simetrali daljice $AB$.} + +Označimo z $O$ središče očrtane krožnice petkotnika $ABCDE$. Ker +je $BC\parallel DE$, imata premici $BC$ in $DE$ isto pravokotnico +iz središča $S$, ki je pravzaprav skupna simetrala stranic $BC$ +in $DE$. Označimo jo s $p$. Podobno imata stranici $CD$ in $EA$ +skupno simetralo $q$. Označimo še z $r$ simetralo daljice $AB$. +Dokažemo, da velja $D\in r$. Kompozitum + $\mathcal{S}_{OD}\circ\mathcal{S}_q +\circ\mathcal{S}_p\circ\mathcal{S}_r +\circ\mathcal{S}_q\circ\mathcal{S}_p$ je direktna izometrija, ki +ima fiksni točki $O$ in $D$, zato po izreku \ref{izo2ftIdent} +predstavlja identiteto. Torej: +$$\mathcal{S}_{OD}\circ\mathcal{S}_q +\circ\mathcal{S}_p\circ\mathcal{S}_r +\circ\mathcal{S}_q\circ\mathcal{S}_p=\mathcal{E}.$$ + +Premice $p$, $q$ in $r$ so iz istega šopa $\mathcal{X}_O$, zato +kompozitum $\mathcal{S}_q \circ\mathcal{S}_p\circ\mathcal{S}_r$ +predstavlja osno zrcaljenje (izrek \ref{izoSop}) oz. +$$\mathcal{S}_q +\circ\mathcal{S}_p\circ\mathcal{S}_r=\mathcal{S}_l +=\mathcal{S}^{-1}_l=\left(\mathcal{S}_q +\circ\mathcal{S}_p\circ\mathcal{S}_r\right)^{-1}=\mathcal{S}_r +\circ\mathcal{S}_p\circ\mathcal{S}_q.$$ + Torej je: + \begin{eqnarray*} +\mathcal{E}&=&\mathcal{S}_{OD}\circ\mathcal{S}_q +\circ\mathcal{S}_p\circ\mathcal{S}_r +\circ\mathcal{S}_q\circ\mathcal{S}_p =\\ +&=&\mathcal{S}_{OD}\circ\mathcal{S}_r +\circ\mathcal{S}_p\circ\mathcal{S}_q +\circ\mathcal{S}_q\circ\mathcal{S}_p = +\mathcal{S}_{OD}\circ\mathcal{S}_r. +\end{eqnarray*} + Iz +$\mathcal{E}=\mathcal{S}_{OD}\circ\mathcal{S}_r$ pa sledi +$\mathcal{S}_{OD}=\mathcal{S}_r$ oz. $OD=r$ in $D\in r$. + + +\item \res{Premice $p$, $q$ in $r$ ležijo v isti ravnini. Dokaži +ekvivalenco $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p + =\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_r$ natanko + tedaj, ko premice $p$, $q$ in $r$ pripadajo istem šopu.} + +Označimo +$\mathcal{I}=\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p$. + +($\Leftarrow$) Predpostavimo, da premice $p$, $q$ in $r$ +pripadajo istemu šopu. Po izreku \ref{izoSop} izometrija +$\mathcal{I}$ predstavlja osmo zrcaljenje. Iz tega sledi +$\mathcal{I}=\mathcal{I}^{-1}$ oz. +$\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p + =\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_r$ + +($\Rightarrow$) Predpostavimo, da velja + $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p + =\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_r$. Iz danega + pogoja sledi $\mathcal{I}=\mathcal{I}^{-1}$ oz. +$\mathcal{I}^2=\mathcal{E}$. Predpostavimo nasprotno - da +premice $p$, $q$ in $r$ niso iz istega šopa. Po izreku +\ref{izoZrcdrsprq} je +$\mathcal{I}=\mathcal{G}_{2\overrightarrow{AB}}$ +($\overrightarrow{AB}\neq \overrightarrow{0}$). Toda v tem +primeru bi bilo +$\mathcal{E}=\mathcal{I}^2=\mathcal{G}^2_{2\overrightarrow{AB}} +=\mathcal{T}_{4\overrightarrow{AB}}$ (izrek +\ref{izoZrcdrsZrcdrs}), kar ni možno, ker je po predpostavki +$\overrightarrow{AB}\neq \overrightarrow{0}$. Zatorej premice +$p$, $q$ in $r$ pripadajo istemu šopu. + +\item \res{Naj bodo $O$, $P$ in $Q$ tri nekolinearne točke. Konstruiraj +kvadrat $ABCD$ (v ravnini $OPQ$) s središčem v točki $O$, tako +da točki $P$ in $Q$ po vrsti +ležita na premicah $AB$ in $BC$.} + +Uporabimo dejstvo, da $Q,\mathcal{R}_{O,90^0}(P)\in BC$. + +\item \res{Naj bosta $\mathcal{R}_{S,\alpha}$ rotacija in $\mathcal{S}_p$ +osno zrcaljenje v isti ravnini ter $S\in p$. +Dokaži, da kompozituma $\mathcal{R}_{S,\alpha}\circ\mathcal{S}_p$ +in $\mathcal{S}_p\circ\mathcal{R}_{S,\alpha}$ predstavljata osno +zrcaljenje.} + +Uporabimo izreka \ref{rotacKom2Zrc} in \ref{izoSop}. + +\item \res{Dani sta točka $A$ in krožnica $k$ v isti ravnini. Načrtaj +kvadrat $ABCD$, tako da krajišči diagonale $BD$ ležita na krožnici $k$.} + +Iz $B,D\in k$ in $\mathcal{R}_{A,90^0}(B)=D$ sledi $D\in k\cap +\mathcal{R}_{A,90^0}(k)$, kar omogoča konstrukcijo oglišča $D$. +Nato je $B=\mathcal{R}_{A,-90^0}(D)$ in +$C=\mathcal{R}_{D,90^0}(A)$. + +\item \res{Naj bo $ABC$ poljubni trikotnik. Dokaži: + $$\mathcal{R}_{C,2\measuredangle BCA}\circ + \mathcal{R}_{B,2\measuredangle ABC}\circ + \mathcal{R}_{A,2\measuredangle CAB}=\mathcal{E}.$$} + + Po izreku \ref{rotacKom2Zrc} je: + \begin{eqnarray*} + &&\mathcal{R}_{C,2\measuredangle BCA}\circ + \mathcal{R}_{B,2\measuredangle ABC}\circ + \mathcal{R}_{A,2\measuredangle CAB}=\\ &&= + \mathcal{S}_{CA}\circ\mathcal{S}_{CB}\circ\mathcal{S}_{BC} + \circ\mathcal{S}_{BA}\circ\mathcal{S}_{AB}\circ\mathcal{S}_{AC}= + \mathcal{E}. + \end{eqnarray*} + +\item \res{Dokaži, da kompozitum osnega zrcaljenja $\mathcal{S}_p$ +in središčnega zrcaljenja $\mathcal{S}_S$ ($S\in p$) +predstavlja osno zrcaljenje.} + + Naj bo $q$ pravokotnica premice $p$ skozi točko $S$. Po izreku + \ref{izoSrZrcKom2Zrc} je: + $$\mathcal{S}_S\circ\mathcal{S}_p= + \mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_p=\mathcal{S}_q.$$ + +\item \res{Naj bodo $O$, $P$ in $Q$ tri nekolinearne točke. Konstruiraj +kvadrat $ABCD$ (v ravnini $OPQ$) s središčem v točki $O$, tako + da točki $P$ in $Q$ +ležita po vrsti na premicah $AB$ in $CD$.} + +Uporabimo dejstvo, da $Q,\mathcal{S}_O(P)\in CD$. + +\item \res{Kaj predstavlja kompozitum translacije in središčnega zrcaljenja?} + +Uporabimo izreka \ref{transl2sred} in \ref{izoKomp3SredZrc}. + +\item \res{Dani so premica $p$ ter krožnici $k$ in $l$, ki +ležijo v isti ravnini. Načrtaj premico, ki je vzporedna s +premico $p$, tako da na krožnicah $k$ in $l$ določa skladni tetivi.} + +Označimo s $K$ in $L$ središči krožnic $k$ in $l$. Naj bo $q$ +iskana premica, ki je vzporedna s premico $p$, krožnici $k$ in +$l$ pa seka po vrsti v takšnih točkah $A$, $B$, $C$ in $D$, da +velja $AB\cong CD$. Naj bo še $\mathcal{B}(A,B,C,D)$. +Označimo s $K_q$ in $L_q$ pravokotni projekciji središč $K$ in +$L$ na premici $q$ ter $K_p$ in $L_p$ pravokotni projekciji +središč $K$ in $L$ na premici $p$. Iz skladnosti trikotnikov +$KAK_q$ in $KBK_q$ (izrek \textit{SSA} \ref{SSK}) sledi, da je +$K_q$ središče tetive $AB$. Analogno je $L_q$ središče tetive +$CD$. Iz $AB\cong CD$ potem sledi $K_qB\cong L_qD$ oz. +$\overrightarrow{K_qB} = \overrightarrow{L_qD}$. Torej: +$$\overrightarrow{BD}=\overrightarrow{BD}+\overrightarrow{0}= +\overrightarrow{BD}+\overrightarrow{K_qB}+\overrightarrow{DL_q}= +\overrightarrow{K_qL_q}=\overrightarrow{K_pL_p}.$$ Vektor +$\overrightarrow{v}=\overrightarrow{K_pL_p}$ lahko konstruiramo. +To pomeni, da prejšnja analiza omogoča konstrukcijo, ker je +$\mathcal{T}_{\overrightarrow{v}}(B)=D$ oz. $D\in +\mathcal{T}_{\overrightarrow{v}}(k)\cap l$. + +\item \res{Naj bo $c$ premica, ki seka vzporednici $a$ in $b$, ter $l$ + daljica. Načrtaj enakostranični trikotnik $ABC$, tako da velja + $A\in a$, $B\in b$, $C\in c$ in $AB\cong l$.} + + Najprej konstruiramo poljubni enakostranični trikotnik + $A_1B_1C_1$, ki izpolnjuje pogoje $A_1\in a$, $B_1\in b$ in + $A_1B_1\cong l$, nato uporabimo ustrezno translacijo. + +\item \res{Dokaži, da kompozitum rotacije in osnega zrcaljenja +neke ravnine +predstavlja zrcalni zdrs natanko tedaj, ko središče +rotacije ne leži na osi osnega zrcaljenja.} + +Uporabimo izreke \ref{rotacKom2Zrc}, \ref{izoZrcdrsprq} in +\ref{izoSop}. + +\item \res{Naj bo $ABC$ enakostranični trikotnik. Dokaži, +da kompozitum $\mathcal{S}_{AB} + \circ\mathcal{S}_{CA} + \circ\mathcal{S}_{BC}$ +predstavlja zrcalni zdrs. Določi še vektor in os tega zdrsa.} + +Označimo $\mathcal{I}=\mathcal{S}_{AB} \circ\mathcal{S}_{CA} + \circ\mathcal{S}_{BC}$. Po izreku \ref{izoZrcdrsprq} je + $\mathcal{I}$ zrcalni zdrs. Naj bodo $A_1$, $B_1$ in $C_1$ po + vrsti središča daljic $BC$, $AC$ in $AB$ trikotnika $ABC$. + Ker je le-ta pravilen je $\mathcal{I}(A_1C_1)=A_1C_1$, kar + pomeni, da je premica $A_1C_1$ os tega zrcalnega zdrsa. Ni + težko dokazati, da za točko $A'_1=\mathcal{I}(A_1)$ (obe + ležita na osi $A_1C_1$) velja + $\overrightarrow{A_1A'_1}=3\overrightarrow{A_1C_1}$, zato je + $\mathcal{I}=\mathcal{G}_{3\overrightarrow{A_1C_1}}$. + +\item \res{Dani sta točki $A$ in $B$ na istem bregu premice +$p$. +Načrtaj daljico $XY$, ki leži na premici $p$ in je skladna +z dano daljico $l$, tako da bo vsota +$|AX|+|XY|+|YB|$ minimalna.} + +Naj bo $A'=\mathcal{G}_{\overrightarrow{MN}}(A)$ (kjer $M,N\in +p$ in $MN\cong l$). Točko $Y$ dobimo kot presečišče premic $p$ +in $X'Y$ (glej tudi zgled \ref{HeronProbl}). + +\item \res{Naj bo $ABC$ enakokraki pravokotni trikotnik s pravim +kotom pri oglišču $A$. Kaj predstavlja kompozitum +$\mathcal{G}_{\overrightarrow{AB}}\circ \mathcal{G}_{\overrightarrow{CA}}$?} + +Naj bosta $p$ in $q$ simetrali stranic $CA$ in $AB$ trikotnika +$ABC$. Po izreku \ref{izoZrcDrsKompSrOsn} je: + $$\mathcal{G}_{\overrightarrow{AB}}\circ + \mathcal{G}_{\overrightarrow{CA}}= + \mathcal{S}_q\circ\mathcal{S}_A\circ\mathcal{S}_A\circ\mathcal{S}_p= + \mathcal{S}_q\circ\mathcal{S}_p.$$ Ker je $ABC$ enakokraki + pravokotni trikotnik s pravim kotom pri oglišču $A$, sta + premici $p$ in $q$ pravokotni in se sekata v središču $S$ + hipotenuze $BC$. Zatorej je + $\mathcal{G}_{\overrightarrow{AB}}\circ + \mathcal{G}_{\overrightarrow{CA}}=\mathcal{S}_q + \circ\mathcal{S}_p=\mathcal{S}_S$. + +\item \res{V isti ravnini so dane premice $a$, $b$ in $c$. +Načrtaj točki $A\in a$ in $B\in b$ +tako, da bo $\mathcal{S}_c(A)=B$.} + +Iz $A\in a$ sledi $\mathcal{S}_c(A)\in \mathcal{S}_c(a)$ oz. +$B\in \mathcal{S}_c(a)$. Ker je še $B\in b$, dobimo točko $B$ iz +pogoja $B\in \mathcal{S}_c(a)\cap b$. Nato je še +$A=\mathcal{S}_c(B)$. + +\item \res{Dani sta premici $p$ in $q$ ter točka $A$ v isti ravnini. +Načrtaj točki $B$ in $C$ tako, +da bosta premici $p$ in $q$ simetrali notranjih kotov pri +ogliščih $B$ in $C$ trikotnika $ABC$.} + +Uporabimo dejstvo, da je premica $BC$ določena s točkama +$\mathcal{S}_p(A)$ in $\mathcal{S}_q(A)$. + +\item \res{Naj bodo $p$, $q$ in $r$ premice ter $K$ in $L$ točki v +isti ravnini. Načrtaj +premici $s$ in $s'$, ki gresta po vrsti skozi točki $K$ in $L$, +tako da velja $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p(s)=s'$.} + +Označimo +$\mathcal{I}=\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p$. +Premica $s'$ je določena s točkama $L$ in $\mathcal{I}(K)$, +premica $s$ pa s točkama $K$ in $\mathcal{I}^{-1}(K)$. Če je +$\mathcal{I}(K)=L$, ima naloga neskončno mnogo rešitev - premica $s$ +je poljubna premica, ki poteka skozi točko $K$. + + +\item \res{Naj bo $s$ simetrala enega od kotov, ki jih določata premici $p$ +in $q$. Dokaži, da je $\mathcal{S}_s\circ\mathcal{S}_p = +\mathcal{S}_q\circ\mathcal{S}_s$.} + +Dokažemo lahko celo več - da velja ekvivalenca (pod +predpostavko, da se premici $p$ in $q$ sekata). Če dano enakost +množimo z desne s $\mathcal{S}_s$, dobimo ekvivalentno enakost +$\mathcal{S}_s\circ\mathcal{S}_p\circ\mathcal{S}_s = +\mathcal{S}_q$. Po izreku \ref{izoTransmutacija} je +ekvivalentna z enakostjo $\mathcal{S}_{\mathcal{S}_s(p)} = +\mathcal{S}_q$ oz. $\mathcal{S}_s(p) = q$, kar je ekvivalentno s +tem, da je $s$ simetrala enega od kotov, ki ga določata premici +$p$ in $q$. + +\item \label{nalIzo27} +\res{Naj bo $S$ središče trikotniku $ABC$ včrtane krožnice in $P$ točka, + v kateri se ta krožnica dotika stranice $BC$. Dokaži: $$\mathcal{S}_{SC} + \circ\mathcal{S}_{SA}\circ\mathcal{S}_{SB} =\mathcal{S}_{SP}.$$} + + Enakost je direktna posledica izreka \ref{izo1ftIndZrc}, ker + je\\ + $\mathcal{S}_{SC}\circ\mathcal{S}_{SA}\circ\mathcal{S}_{SB}:S,P\mapsto + S,P$. + +\item \res{Premice $p$, $q$ in $r$ neke ravnine potekajo skozi središče +$S$ krožnice $k$. +Načrtaj trikotnik $ABC$, ki je očrtan tej krožnici, tako da +bodo premice $p$, $q$ in $r$ simetrale notranjih kotov pri +ogliščih $A$, $B$ in $C$ tega trikotnika.} + +Uporabimo prejšnjo nalogo \ref{nalIzo27}. + +\item \res{Premice $p$, $q$, $r$, $s$ in $t$ neke ravnine se sekajo +v točki $O$, točka $M$ pa leži na premici $p$. +Načrtaj tak petkotnik, da bo $M$ središče ene njegove stranice, +premice $p$, $q$, $r$, $s$ in $t$ pa simetrale stranic.} + +Točka $M$ leži na eni od simetral stranic. Brez škode za +splošnost naj bo $M\in p$. Naj bo $ABCDE$ takšen petkotnik, da +so premice $p$, $q$, $r$, $s$ in $t$ po vrsti simetrale +njegovih stranic $AB$, $BC$, $CD$, $DE$ in $EA$. Naj bo +$\mathcal{I}=\mathcal{S}_t\circ\mathcal{S}_s\circ\mathcal{S}_r +\circ\mathcal{S}_q\circ\mathcal{S}_p$. Ker +$\mathcal{I}:O,A\mapsto O,A$, je po izreku \ref{izo1ftIndZrc} +$\mathcal{I}=\mathcal{S}_{OA}$. Premico $OA$ dobimo kot +simetralo daljice $XX'$, kjer je $X$ poljubna točka in +$X'=\mathcal{I}(X)$, nato pa oglišče $A$ kot presečišče premice +$OA$ in pravokotnice premice $p$ skozi točko $M$. + +\item \res{Točka $P$ leži v ravnini trikotnika $ABC$. Dokaži, da +premice, ki so simetrične s +premicami $AP$, $BP$ in $CP$ glede na simetrale notranjih kotov + ob ogliščih $A$, $B$ in $C$ tega trikotnika, pripadajo istem šopu.} + +Označimo z $s_{\alpha}$, $s_{\alpha}$ in $s_{\alpha}$ simetrale notranjih +kotov ob ogliščih $A$, $B$ in $C$ trikotnika $ABC$ +ter $a=\mathcal{S}_{s_{\alpha}}(AP)$, $b=\mathcal{S}_{s_{\beta}}(BP)$ in $c=\mathcal{S}_{s_{\gamma}}(CP)$. +Dokažimo, da premice $a$, $b$ in $c$ pripadajo istemu šopu. Ker $\mathcal{S}_{s_{\alpha}}:AC, p\rightarrow AB, + a$ je $\measuredangle CAP=\measuredangle a,AP$. Torej $\mathcal{R}_{A,2\measuredangle CAP} + =\mathcal{R}_{A,2\measuredangle a,AP}$ oz. $\mathcal{S}_{AC}\circ \mathcal{S}_{AP} + =\mathcal{S}_a\circ \mathcal{S}_{AB}$. Iz tega sledi $\mathcal{S}_a=\mathcal{S}_{AC} + \circ\mathcal{S}_{AP}\circ \mathcal{S}_{AB}$. Analogno je: $\mathcal{S}_b=\mathcal{S}_{BA} + \circ\mathcal{S}_{BP}\circ \mathcal{S}_{BC}$ in $\mathcal{S}_c=\mathcal{S}_{CB}\circ\mathcal{S}_{CP} + \circ \mathcal{S}_{CA}$. Zatorej je (izreka \ref{izoSop} in \ref{izoTransmutacija}): + \begin{eqnarray*} + \mathcal{I}&=&\mathcal{S}_a\circ\mathcal{S}_b\circ\mathcal{S}_c=\\ + &=& + \mathcal{S}_{AC}\circ\mathcal{S}_{AP}\circ \mathcal{S}_{AB}\circ + \mathcal{S}_{BA}\circ\mathcal{S}_{BP}\circ \mathcal{S}_{BC}\circ + \mathcal{S}_{CB}\circ\mathcal{S}_{CP}\circ \mathcal{S}_{CA}=\\ + &=& + \mathcal{S}_{AC}\circ\mathcal{S}_{AP}\circ \mathcal{S}_{BP}\circ\mathcal{S}_{CP}\circ \mathcal{S}_{CA}=\\ + &=& + \mathcal{S}_{AC}\circ\mathcal{S}_{AX}\circ \mathcal{S}_{CA}=\\ + &=& + \mathcal{S}_{\mathcal{S}_{AC}(AX)}. + \end{eqnarray*} +Po izreku \ref{izoSop} premice $a$, $b$ in $c$ pripadajo istem šopu. + +\item \res{Izračunaj kot, ki ga določata premici $p$ in $q$, če velja: +$\mathcal{S}_p\circ\mathcal{S}_q\circ\mathcal{S}_p = +\mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_q$.} + + Če dano enakost množimo z leve po vrsti z $\mathcal{S}_q$, +$\mathcal{S}_p$ in $\mathcal{S}_q$, dobimo ekvivalentno enakost: +$\mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_q\circ +\mathcal{S}_p\circ\mathcal{S}_q\circ\mathcal{S}_p = +\mathcal{E}$ oz. +$\left(\mathcal{S}_q\circ\mathcal{S}_p\right)^3 = \mathcal{E}$. + +Če je $p=q$, je zadnja enakost jasno izpolnjena. + +V primeru $p\parallel q$ je $\mathcal{S}_q\circ\mathcal{S}_p$ +translacija, zato je vedno +$\left(\mathcal{S}_q\circ\mathcal{S}_p\right)^3 +=\mathcal{T}^3_{\overrightarrow{v}}=\mathcal{T}_{3\overrightarrow{v}}\neq +\mathcal{E}$. + +Če se premici $p$ in $q$ sekata, predstavlja kompozitum +$\mathcal{S}_q\circ\mathcal{S}_p$ rotacijo +$\mathcal{R}_{S,\omega}$ ($p\cap q=\{S\}$ in +$\omega=2\measuredangle p,q$). V tem primeru je torej +$\left(\mathcal{S}_q\circ\mathcal{S}_p\right)^3 +=\mathcal{R}^3_{S,\omega}=\mathcal{R}_{S,3\omega}= \mathcal{E}$. +To pomeni $3\omega=360^0$ oz. $\omega=120^0$, zato premici $p$ +in $q$ določata kot $60^0$. + +\item \res{Naj bosta $\mathcal{R}_{A,\alpha}$ in $\mathcal{R}_{B,\beta}$ +rotaciji v isti ravnini. Določi vse točke $X$ v tej ravnini, za +katere velja +$\mathcal{R}_{A,\alpha}(X)=\mathcal{R}_{B,\beta}(X)$.} + +Pogoj $\mathcal{R}_{A,\alpha}(X)=\mathcal{R}_{B,\beta}(X)$ je +ekvivalenten s pogojem +$\mathcal{R}_{A,\alpha}\circ\mathcal{R}^{-1}_{B,\beta}(X)=X$ +oz. $\mathcal{R}_{A,\alpha}\circ\mathcal{R}_{B,-\beta}(X)=X$. +Potrebno je torej določiti fiksne točke izometrije +$\mathcal{I}=\mathcal{R}_{A,\alpha}\circ\mathcal{R}_{B,-\beta}$. +Uporabimo izrek \ref{rotacKomp2rotac}. Obravnavali bomo več +primerov. + + \textit{1}) Če je $A=B$ in $\alpha-\beta=k\cdot360^0$ (za nek + $k\in \mathbb{Z}$), je $\mathcal{I}=\mathcal{E}$, zato pogoj + jasno velja za vsako točko te ravnine. + + \textit{2}) Če je $A=B$ in $\alpha-\beta\neq k\cdot360^0$ (za + vsak $k\in \mathbb{Z}$), je + $\mathcal{I}=\mathcal{R}_{A,\alpha-\beta}$ in je pogoj + izpolnjen le za točko $X=A$. + + \textit{3}) Če je $A\neq B$ in $\alpha-\beta=k\cdot360^0$ (za + nek $k\in \mathbb{Z}$), predstavlja $\mathcal{I}$ translacijo, + zato pogoj ne ustreza nobeni točki te ravnine. + + \textit{4}) Če je $A\neq B$ in $\alpha-\beta\neq k\cdot360^0$ + (za vsak $k\in \mathbb{Z}$), je $\mathcal{R}_{C,\alpha-\beta}$ + (kjer velja $\measuredangle CAB=\frac{\alpha}{2}$ in + $\measuredangle BAC=\frac{\beta}{2}$), kar pomeni, da pogoj + velja le za točko $C$. + +\item \res{Premici $p$ in $q$ se pod kotom $60^0$ sekata v središču $O$ enakostraničnega + trikotnika $ABC$. Dokaži, da + sta odseka, ki jih na premicah določata stranici + trikotnika $ABC$, skladni daljici.} + +Uporabimo rotacijo $\mathcal{R}_{O,120^0}$. + +\item \res{Točka $S$ naj bo središče pravilnega petkotnika $ABCDE$. +Dokaži, da velja: + $$\overrightarrow{SA} + \overrightarrow{SB} + \overrightarrow{SC} + + \overrightarrow{SD} + \overrightarrow{SE} = \overrightarrow{0}.$$} + +Naj bo $\overrightarrow{SA} + \overrightarrow{SB} + + \overrightarrow{SC} + \overrightarrow{SD} + \overrightarrow{SE} + = \overrightarrow{SX}$. Ker je $ABCDE$ pravilni petkotnik, + velja $\mathcal{R}_{S,72^0}:A,B,C,D,E,S,X\mapsto + B,C,D,E,A,S,X'$. Zato je $ \overrightarrow{SB} + + \overrightarrow{SC} + \overrightarrow{SD} + \overrightarrow{SE} + +\overrightarrow{SA} = \overrightarrow{SX'}$. To pomeni, da je + $\overrightarrow{SX}= \overrightarrow{SX'}$ oz. + $X=X'=\mathcal{R}_{S,72^0}(X)$. Ker je $S$ edina fiksna točka + rotacije $\mathcal{R}_{S,72^0}$, je $X=S$. Zatorej je: + $\overrightarrow{SA} + \overrightarrow{SB} + + \overrightarrow{SC} + \overrightarrow{SD} + \overrightarrow{SE} + = \overrightarrow{SS}=\overrightarrow{0}$. + +\item \res{Dokaži, da se diagonale pravilnega petkotnika +sekajo v točkah, ki so tudi +oglišča pravilnega petkotnika.} + +Uporabimo rotacijo s središčem v središču pravilnega petkotnika za +kot $72^0$. + +\item \res{Naj bosta $ABP$ in $BCQ$ pravilna trikotnika z +isto orientacijo in $\mathcal{B}(A,B,C)$. Točki $K$ in $L$ sta +središči daljic $AQ$ in $PC$. Dokaži, da je tudi $BLK$ pravilni +trikotnik.} + +Z rotacijo $\mathcal{R}_{B,-60^0}$ se točki $A$ in $Q$ +preslikata v točki $P$ in $C$. Torej se daljica $AQ$ in njeno +središče $K$ preslikata v daljico $PC$ in njeno središče $L$. Iz +$\mathcal{R}_{B,-60^0}(K)=L$ pa sledi, da je $BLK$ pravilni +trikotnik. + +\item \res{Dane so tri koncentrične krožnice in premica v isti ravnini. +Načrtaj pravilni trikotnik tako, da njegova oglišča +po vrsti ležijo na teh krožnicah, ena stranica pa bo vzporedna +dani premici.} + +Najprej načrtamo poljubni tak trikotnik brez pogoja, da je ena +njegova stranica vzporedna dani premici. To lahko naredimo tako, +da izberemo poljubno oglišče $A$ na eni od krožnic in uporabimo +rotacijo $\mathcal{R}_{A,60^0}$. Nato pa uporabimo ustrezno +rotacijo s središčem v središču koncentričnih krožnic, ki +načrtani trikotnik preslika v trikotnik, pri katerem je ena +njegova stranica vzporedna dani premici. + +\item \res{Točka $P$ je notranja točka pravilnega trikotnika +$ABC$, tako da velja +$\angle APB=113^0$ in $\angle BPC=123^0$. Izračunaj velikosti +kotov trikotnika, čigar stranice so skladne z daljicami +$PA$, $PB$ in $PC$.} + +Najprej je jasno $\angle APC=360^0-113^0-123^0=124^0$. Označimo +$P'=\mathcal{R}_{A,60^0}(P)$. $APP'$ je pravilni trikotnik, zato +je $PA\cong P'P$. Ker je še $C=\mathcal{R}_{A,60^0}(B)$, velja +$\triangle ABP\cong\triangle ACP'$, zato je $PB\cong P'C$ in $\angle +APB\cong\angle AP'C$. To pomeni, da so stranice $P'P$, $P'C$ in +$PC$ trikotnika $PCP'$ po vrsti skladne z daljicami $PA$, $PB$ +in $PC$. Pri tem je $\angle P'PC=124^0-60^0=64^0$, $\angle +PP'C=113^0-60^0=53^0$ in $\angle PP'C=180^0-64^0-53^0=63^0$. + +\item \res{Dane so nekolinearne točke $P$, $Q$ in $R$. +Načrtaj trikotnik $ABC$, tako +da bodo $P$, $Q$ in $R$ središča kvadratov, ki so konstruirani +nad stranicami $BC$, $CA$ in $AB$ tega trikotnika.} + +Poiščemo fiksno točko kompozituma +$\mathcal{R}_{Q,90^0}\circ\mathcal{R}_{R,90^0}\circ\mathcal{R}_{P,90^0}$. + +\item \res{Naj bosta $A$ in $B$ točki ter $p$ premica v isti +ravnini. Dokaži, da je +kompozitum $\mathcal{S}_B\circ\mathcal{S}_p\circ\mathcal{S}_A$ + osno zrcaljenje natanko tedaj, ko je $AB\perp p$.} + +Središčni zrcaljenji $\mathcal{S}_A$ in $\mathcal{S}_B$ +predstavimo kot kompozitume +$\mathcal{S}_A=\mathcal{S}_a\circ\mathcal{S}_{a_1}$ in +$\mathcal{S}_B=\mathcal{S}_{b_1}\circ\mathcal{S}_b$, kjer sta +$a$ in $b$ pravokotni na premico $p$. + +\item \res{Naj bodo $p$, $q$ in $r$ tangente trikotniku $ABC$ včrtane krožnice, +ki so vzporedne z njegovimi stranicami +$BC$, $AC$ in $AB$. Dokaži, +da premice $p$, $q$, $r$, $BC$, $AC$ in $AB$ določajo tak + šestkotnik, v katerem so pari nasprotnih stranic skladne daljice.} + +Uporabimo središčno zrcaljenje $\mathcal{S}_S$, kjer je $S$ +središče včrtane krožnice trikotnika $ABC$. + +\item \res{Načrtaj trikotnik s podatki: $\alpha$, $t_b$, $t_c$.} + +Najprej lahko načrtamo težiščnico $BB_1\cong t_b$ in težišče +$T$. Ker oglišče $A$ leži na loku $l$ nad tetivo $BB_1$ in +obodnim kotom $\alpha$, oglišče $B$ na krožnici +$k(T,\frac{2}{3}t_c)$, pri tem pa velja +$C=\mathcal{S}_{B_1}(A)$, dobimo oglišče $C$ iz pogoja $C\in +k(T,\frac{2}{3}t_c)\cap\mathcal{S}_{l}(A)$. + +\item \res{Naj bosta $ALKB$ in $ACPQ$ kvadrata, ki sta načrtana zunaj trikotnika $ABC$ +nad stranicama $AB$ in +$AC$, ter $X$ središče stranice $BC$. Dokaži, da +je $AX\perp LQ$ in +$|AX|=\frac{1}{2}|QL|$.} + +Naj bo $Q'=\mathcal{S}_A(C)$. V tem primeru velja +$R_{A,90^0}:Q,L\mapsto Q',B$. Torej $QL\cong Q'B$ in $QL\perp +Q'B$ (izrek \ref{rotacPremPremKot}). Ker je $AX$ srednjica +trikotnika $BCQ$ za osnovnico $BQ'$, je +$\overrightarrow{AX}=\frac{1}{2}\overrightarrow{BQ'}$, zatorej +je $AX\perp LQ$ in $|AX|=\frac{1}{2}|QL|$. + +\item \res{Naj bo $O$ središče pravilnega trikotnika $ABC$ ter $D$ in +$E$ točki na stranicah $CA$ in $CB$, tako + da velja $CD\cong CE$. Točka $F$ je četrto oglišče paralelograma + $BODF$. Dokaži, +da je trikotnik $OEF$ pravilen.} + +Naj bo $\mathcal{I}=\mathcal{T}_{\overrightarrow{OB}}\circ +\mathcal{R}_{C,-60^0}$. Po izreku \ref{izoKompTranslRot} je +$\mathcal{I}$ rotacija za isti kot $-60^0$ s središčem v neki +točki $\widehat{O}$. Torej +$\mathcal{T}_{\overrightarrow{OB}}\circ +\mathcal{R}_{C,-60^0}=\mathcal{R}_{\widehat{O},-60^0}$. Pri tem +je +$\mathcal{R}_{\widehat{O},-60^0}(E)=\mathcal{T}_{\overrightarrow{OB}}\circ +\mathcal{R}_{C,-60^0}(E)=F$, zato je $\widehat{O}EF$ pravilni +trikotnik. Potrebno je še dokazati, da je $\widehat{O}=O$ oz. +da je $O$ fiksna točka rotacije +$\mathcal{R}_{\widehat{O},-60^0}$. Če je +$O'=\mathcal{R}_{C,-60^0}(O)$, je $O'$ središče simetričnega +enakostraničnega trikotnika $AB'C$ (kjer je +$C'=\mathcal{S}_{AC}(B)$). Zato je +$\overrightarrow{O'O}=\overrightarrow{OB}$ oz. +$\mathcal{T}_{\overrightarrow{OB}}(O')=O$. Iz tega sledi +$\mathcal{R}_{\widehat{O},-60^0}(O)=O$ oz. $\widehat{O}=O$, kar +pomeni, da je $OEF$ + pravilni trikotnik. + +\item \res{Naj bo $L$ točka, v kateri se trikotniku +$ABC$ včrtana krožnica dotika njegove stranice $BC$. +Dokaži: $$\mathcal{R}_{C,\measuredangle ACB}\circ\mathcal{R}_{A,\measuredangle BAC} +\circ\mathcal{R}_{B,\measuredangle CBA} =\mathcal{S}_L.$$} + +Označimo z $M$ in $N$ točki, v katerih se včrtana krožnica +trikotnika $ABC$ dotika njegovih stranic $AC$ in $AB$. Po izreku +\ref{rotacKomp2rotac} je $\mathcal{R}_{C,\measuredangle ACB} +\circ\mathcal{R}_{A,\measuredangle BAC}\circ +\mathcal{R}_{B,\measuredangle CBA}=\mathcal{S}_{\widehat{L}}$. +Ker je: +\begin{eqnarray*} + \mathcal{S}_{\widehat{L}}(L)&=& + \mathcal{R}_{C,\measuredangle +ACB} \circ\mathcal{R}_{A,\measuredangle BAC}\circ +\mathcal{R}_{B,\measuredangle CBA}(L)=\\ +&=& + \mathcal{R}_{C,\measuredangle +ACB} \circ\mathcal{R}_{A,\measuredangle BAC}(N)= + \mathcal{R}_{C,\measuredangle ACB}(M)= L, + \end{eqnarray*} + je $\widehat{L}=L$ oz. $\mathcal{R}_{C,\measuredangle ACB} +\circ\mathcal{R}_{A,\measuredangle BAC}\circ +\mathcal{R}_{B,\measuredangle CBA}=\mathcal{S}_L$. + +\item \res{Točke $P$ in $Q$ ter $M$ in $N$ so središča po dveh kvadratov, +ki so načrtani zunaj nad +nasprotnimi stranicami poljubnega štirikotnika. Dokaži, da je +$PQ\perp MN$ in $PQ\cong MN$.} + + Naj bo $ABCD$ dani štirikotnik, $P$ in $Q$ središči kvadratov, + ki sta konstruirani nad stranicama $AB$ in $CD$ ter $M$ in $N$ + središči kvadratov, ki sta konstruirani nad stranicama $BC$ in + $AD$. Kompozitum $\mathcal{I}=\mathcal{R}_{N,90^0}\circ +\mathcal{R}_{P,90^0}$ po izreku \ref{rotacKomp2rotac} +predstavlja središčno zrcaljenje. Ker je pri tem +$\mathcal{I}(B)=D$, je središče tega zrcaljenja pravzaprav +središče diagonale $BD$; označimo ga s $S$. Torej +$\mathcal{R}_{N,90^0}\circ \mathcal{R}_{P,90^0}=\mathcal{S}_S$. +Po istem izreku \ref{rotacKomp2rotac} za točko $S$ velja $\angle +NPS=\frac{1}{2}90^0=45^0$ in $\angle PNS=\frac{1}{2}90^0=45^0$. +To pomeni, da je $PNS$ enakokraki pravokotni trikotnik z +osnovnico $NP$ in pravim kotom ob oglišču $S$. Analogno je tudi +$MSQ$ enakokraki pravokotni trikotnik z osnovnico $MQ$ in s pravim +kotom ob oglišču $S$. Iz teh dveh dejstev sledi $R_{S,90^0}: +M,N\mapsto Q,P$, zato je $MN\cong QP$ in $MN\perp QP$ (izrek +\ref{rotacPremPremKot}). + + +\item \res{Naj bosta $APB$ in $ACQ$ pravilna trikotnika, ki sta zunaj trikotnika $ABC$ +načrtana nad +stranicama $AB$ in $AC$. Točka $S$ je središče +stranice $BC$ in $O$ središče trikotnika $ACQ$. Dokaži, da je +$|OP|=2|OS|$.} + +Naj bo $\mathcal{I}=\mathcal{R}_{O,120^0}\circ +\mathcal{R}_{P,60^0}$. Po izreku \ref{rotacKomp2rotac} je: + $$\mathcal{I}=\mathcal{R}_{O,120^0}\circ +\mathcal{R}_{P,60^0}=\mathcal{R}_{\widehat{S},180^0} +=\mathcal{S}_{\widehat{S}},$$ kjer je $\widehat{S}$ oglišče + trikotnika $OP\widehat{S}$ in $\angle + \widehat{S}PO=\frac{1}{2}60^0=30^0$ ter $\angle + PO\widehat{S}=\frac{1}{2}120^0=60^0$. Ker je + $\mathcal{S}_{\widehat{S}}(S)=\mathcal{I}(S)=S$ oz. + $\widehat{S}=S$. + + Če označimo $O'=\mathcal{S}_S$, je $POO'$ pravilni trikotnik, + zato je $|OP|=|OO'|=2|OS|$. + +\item \res{Dokaži, da osno zrcaljenje in translacija neke ravnine +komutirata natanko tedaj, ko je os tega zrcaljenja vzporedna z +vektorjem translacije.} + +Če uporabimo izrek \ref{izoTransmutacija}, dobimo: + \begin{eqnarray*} +\mathcal{T}_{\overrightarrow{v}}\circ \mathcal{S}_p= +\mathcal{S}_p\circ\mathcal{T}_{\overrightarrow{v}} +&\Leftrightarrow& +\mathcal{T}_{\overrightarrow{v}}\circ \mathcal{T}_p\circ\mathcal{T}^{-1}_{\overrightarrow{v}}= + \mathcal{S}_p\\ +&\Leftrightarrow& +\mathcal{S}_{\mathcal{T}_{\overrightarrow{v}}(p)}= + \mathcal{S}_p +\Leftrightarrow +\mathcal{T}_{\overrightarrow{v}}(p)= + p +\Leftrightarrow +\overrightarrow{v}\parallel + p. +\end{eqnarray*} + +\item \res{V isti ravnini so dani premica $p$, krožnici $k$ in $l$ ter daljica $d$. +Načrtaj romb $ABCD$ s stranico, ki je skladna daljici $d$, stranica $AB$ leži na +premici $p$, oglišči $C$ in $D$ pa po vrsti ležita na krožnicah $k$ in $l$.} + +Uporabi translacijo za vektor $\overrightarrow{v}$, ki je +vzporeden s premico $p$ in $|\overrightarrow{v}|=|d|$. V tem +primeru je $D\in \mathcal{T}_{\overrightarrow{v}}(k)\cap l$. + +\item \res{Naj bo $p$ premica, $A$ in $B$ pa točki, ki ležita na istem bregu +premice $p$, ter $d$ daljica v isti ravnini. +Načrtaj točki $X$ in $Y$ na premici $p$ tako, da bo $AX\cong BY$ in $XY\cong d$.} + +Naj bo $\overrightarrow{v}$ vektor, ki je vzporeden s premico +$p$ in $|\overrightarrow{v}|=|d|$. Naj bo +$A'=\mathcal{T}_{\overrightarrow{v}}(A)$, $Y$ presečišče +simetrale daljice $A'B$ s premico $p$ in +$X=\mathcal{T}^{-1}_{\overrightarrow{v}}(Y)$. + +Točka $Y$ leži na simetrali daljice $A'B$, zato je $A'Y\cong YB$. Ker +$\mathcal{T}_{\overrightarrow{v}}:A,X\mapsto A',Y$ je štirikotnik $AYYA'$ +paralelogram, zatorej je $AX\cong A'Y\cong BY$. Iz $\mathcal{T}_{\overrightarrow{v}}(X)=Y$ +pa sledi $\overrightarrow{XY}=\overrightarrow{v}$ oz. $|XY|=|\overrightarrow{v}|=|d|$. + +\item \res{Naj bo $H$ višinska točka trikotnika $ABC$ in $R$ polmer očrtane krožnice tega +trikotnika. Dokaži, da je $|AB|^2+|CH|^2=4R^2$.} + +Naj bo $O$ središče očrtane krožnice $k(O,R)$ trikotnika $ABC$ +in $A'=\mathcal{S}_O(A)$. Daljica $AA'$ je premer krožnice $k$, +zato je $\angle ACA'=90^0$ oz. $A'C\perp AC$. Ker je še +$BH\perp AC$, je $A'C\parallel BH$. Analogno je tudi +$A'B\parallel CH$, kar pomeni, da je štirikotnik $BA'CH$ +paralelogram. Torej velja $CH\cong A'B$. Ker je $A'BA$ +pravokotni trikotnik, je po Pitagorovem izreku +\ref{PitagorovIzrek}: +$|AB|^2+|CH|^2=|AB|^2+|A'B|^2=|AA'|^2=4R^2$. + +\item \res{Naj bo $EAB$ trikotnik, ki je načrtan nad stranico $AB$ kvadrata +$ABCD$. Naj bo tudi $M=pr_{\perp AE}(C)$ in $N=pr_{\perp BE}(D)$ ter točka $P$ +presečišče premic $CM$ in $DN$. Dokaži, da je $PE\perp AB$.} + +Uporabimo translacijo za vektor $\overrightarrow{CB}$. + +\item \res{Načrtaj enakostranični trikotnik $ABC$ tako, da njegova oglišča po vrsti +ležijo na treh vzporednicah $a$, $b$ in $c$ v isti ravnini, +središče tega trikotnika pa leži na premici $s$, ki seka +premice $a$, $b$ in $c$.} + + Najprej narišmo poljubni pravilni trikotnik $A_1B_1C_1$, kjer $A_1\in a$, $B_1\in b$ in $C_1\in c$, + nato pa uporabimo translacijo za vektor $\overrightarrow{S_1S}$, kjer je $S_1$ središče trikotnika + $A_1B_1C_1$, točka $S$ presečišče premice $s$ z vzporednico premic $a$, $b$ in $c$ skozi točko $S_1$. + +\item \res{Če ima petkotnik vsaj dve osi simetrije, je pravilen. Dokaži.} + + Naj bo $\mathfrak{G}(\mathcal{V}_5)$ grupa simetrij + našega petkotnika $\mathcal{V}_5$ ter $p$ in $q$ njegovi + osi simetrije. Jasno je, da je ta grupa končna, zato po + izreku \ref{GrupaLeonardo} predstavlja bodisi ciklično grupo + $\mathfrak{C}_n$ bodisi diedrsko grupo $\mathfrak{D}_n$. Ker + vsebuje osne simetrije, je + $\mathfrak{G}(\mathcal{V}_5)=\mathfrak{D}_n$. Pri tem je + jasno $n\leq 5$. Ker $\mathcal{S}_p, \mathcal{S}_q \in + \mathfrak{G}(\mathcal{V}_5)=\mathfrak{D}_n$, se osi $p$ in + $q$ sekata v neki točki $S$, kompozitum $\mathcal{S}_q\circ + \mathcal{S}_p$ predstavlja rotacijo $\mathcal{R}_{S, + \alpha}$. Ker grupa + $\mathfrak{G}(\mathcal{V}_5)=\mathfrak{D}_n$ vsebuje vsaj + dve osni simetriji, je $n\geq 2$. Torej $n\in \{2,3,4,5\}$. + Pri tem je osnovna rotacija te grupe (za najmanjši kot) + $\mathcal{R}_{S, \frac{360^0}{n}}$. Ker ima petkotnik pet + oglišč, je število $n$ delitelj števila 5, kar pomeni + $n=5$. Torej $\mathfrak{G}(\mathcal{V}_5)=\mathfrak{D}_5$, + torej je $\mathcal{V}_5$ pravilni petkotnik. + +\item \res{Naj bodo $A$, $B$ in $C$ tri kolinearne točke. Kaj predstavlja +kompozitum $\mathcal{G}_{\overrightarrow{BC}}\circ \mathcal{S}_A$?} + +Označimo s $p$ premico, ki poteka skozi točke $A$, $B$ in $C$. +Naj bosta $b$ in $a$ pravokotnici premice $p$ v točkah $A$ in $B$ ter $s$ simetrala daljice $BC$. Tedaj je: + $$\mathcal{G}_{\overrightarrow{BC}}\circ \mathcal{S}_A= + \mathcal{S}_s\circ \mathcal{S}_b\circ \mathcal{S}_p + \circ \mathcal{S}_p\circ \mathcal{S}_a= + \mathcal{S}_s\circ \mathcal{S}_b\circ \mathcal{S}_a.$$ + Ker so premice $a$, $b$ in $s$ pravokotne na premico $p$, so iz šopa vzporednic, + zato je po izreku \ref{izoSop} kompozitum $\mathcal{S}_s + \circ \mathcal{S}_b\circ \mathcal{S}_a$ (oz. $\mathcal{G}_{\overrightarrow{BC}} + \circ \mathcal{S}_A$) osno zrcaljenje. + +\item \res{Naj bodo $p$, $q$ in $r$ premice, ki niso iz istega šopa, ter $A$ točka v isti +ravnini. Načrtaj premico $s$, ki poteka skozi točko $A$, tako da velja +$\mathcal{S}_r\circ \mathcal{S}_q\circ \mathcal{S}_p(s)=s'$ in $s\parallel s'$.} + +Po izreku \ref{izoZrcdrsprq} je kompozitum $\mathcal{S}_r\circ \mathcal{S}_q\circ +\mathcal{S}_p(s)=s'$ zrcalni zdrs - označimo ga z $\mathcal{G}_{2\overrightarrow{PQ}}$. +Premica $s$ je vzporedna ali pravokotna na os zrcalnega zdrsa. Le-ta je določena s +središčema daljic $XX'$ in $YY'$, kjer sta $X$ in $Y$ poljubni točki ter $\mathcal{S}_r +\circ \mathcal{S}_q\circ \mathcal{S}_p: X, Y\mapsto X', Y'$. + +%nove naloge +%___________________________________ + +\item \res{Naj bosta $Z$ in $K$ notranji točki pravokotnika $ABCD$. +Načrtaj točke $A_1$, $B_1$, $C_1$ in $D_1$, ki po vrsti ležijo na +stranicah $AB$, $BC$, $CD$ in $DA$ tega pravokotnika, tako da velja +$\angle ZA_1A\cong\angle B_1A_1B$, $\angle A_1B_1B\cong\angle C_1B_1C$, +$\angle B_1C_1C\cong\angle D_1C_1D$ in $\angle C_1D_1D\cong\angle KD_1A$.} + + Najprej načrtamo točki $Z'=S_{CB}\circ S_{AB}(Z)=S_B(Z)$ in $K'=S_{CD}\circ S_{AD}(Z)=S_D(K)$. + Nato dokažemo in uporabimo dejstvo, da so točke $Z'$, $B_1$, $C_1$ in $K'$ kolinearne. + + +\item \res{Točka $A$ leži na premici $a$, točka $B$ pa na premici $b$. +Določi rotacijo, ki preslika premico $a$ v premico $b$ in točko $A$ v točko $B$.} + + Če sta premici $a$ in $b$ vzporedni, je iskana rotacija + središčno zrcaljenje s središčem, ki je središče daljice + $AB$. Če pa se premici $a$ in $b$ sekata v točki + $O$, je središče rotacije presečišče simetrale daljice $AB$ + in simetrale kota $aOb$, kot rotacije pa enak kotu $aOb$. + +\item \res{V središču kvadrata se sekata dve pravokotnici. +Dokaži, da ti pravokotnici sekata stranice kvadrata v točkah, ki so +oglišča novega kvadrata.} + + Uporabimo rotacijo $\mathcal{R}_{S,90^0}$, kjer je $S$ središče kvadrata. + +\item \res{Dana je krožnica $k$ ter premice $a$, $b$, $c$, $d$ in $e$, ki ležijo +v isti ravnini. Krožnici $k$ včrtaj petkotnik s stranicami, ki +so po vrsti vzporedne s premicami $a$, $b$, $c$, $d$ in $e$.} + + + Najprej načrtamo simetrale $p$, $q$, $r$, $s$ in $t$ stranic + iskanega petkotnika $ABCDE$, ki so pravzaprav pravokotnice na + premice $a$, $b$, $c$, $d$ in $e$ iz središča $O$ + krožnice $k$. Kompozitum $\mathcal{I}= \mathcal{S}_t\circ + \mathcal{S}_s\circ \mathcal{S}_r\circ \mathcal{S}_q\circ + \mathcal{S}_p$ je indirektna izometrija s fiksnima točkama + $O$ in $A$, zato je po izreku \ref{izo1ftIndZrc} + $\mathcal{I}=\mathcal{S}_{OA}$. Os $OA$ lahko načrtamo kot + simetralo daljice $XX'$, kjer je $X$ poljubna točka in + $X'=\mathcal{I}(X)$. To omogoča konstrukcijo oglišča $A$, + nato pa ostalih oglišč petkotnika $ABCDE$. + +\item \res{Točka $P$ leži v notranjosti kota $aOb$. Načrtaj premico $p$ skozi točko $P$, + ki s krakoma $a$ in $b$ določa trikotnik z najmanjšo ploščino.} + + Dokažimo, da ima trikotnik najmanjšo ploščino, + če je daljica $OP$ njegova težiščnica. Označimo ta + trikotnik z $OAB$ ($A\in a$ in $B\in b$). Naj bosta $X$ in + $Y$ presečišča poljubne premice (različne od $AB$) skozi + točko $P$. Dokažimo, da je ploščina trikotnika $XOY$ večja + od ploščine trikotnika $AOB$. Brez škode za splošnost naj bo + $\mathcal{B}(O,X,A)$ oz. $\mathcal{B}(O,B,Y)$. V tem primeru + točki $A$ in $B$ načrtamo na naslednji način. Naj bo + $a'=\mathcal{S}_P(a)$, $\{B\}=b\cap a'$ in + $A=\mathcal{S}_P(B)$. Označimo z $X'=\mathcal{S}_P$. Jasno + je $X'\in a'$. Ker $\mathcal{S}_P:X,A,P\mapsto X',B,P$, sta + trikotnika $XAP$ in $X'BP$ skladna, zato imata enako + ploščino. Če z $p_{V_1V_2\cdots V_n}$ označimo ploščino + poljubnega večkotnika $V_1V_2\cdots V_n$, je: + \begin{eqnarray*} + p_{AOB}&=&p_{AXPB}+p_{XAP}=\\ + &=&p_{AXPB}+p_{X'BP}>p_{AXPB}+p_{X'BP}+ + p_{X'BY}=\\ + &=&p_{XOY}. + \end{eqnarray*} + +\item \res{Paralelogram $PQKL$ naj bo včrtan v paralelogram $ABCD$ (oglišča prvega ležijo na stranicah +drugega). Dokaži, da imata paralelograma skupno središče.} + + Uporabimo središčno zrcaljenje $\mathcal{S}_S$, kjer je $S$ presečišče diagonal paralelograma $ABCD$. + +\item \res{Loki $l_1, l_2,\cdots , l_n$ ležijo na krožnici $k$ in je vsota njihovih +dolžin manjša od polobsega te krožnice. Dokaži, da obstaja tak premer +$PQ$ krožnice $k$, da nobeno od njegovih krajišč ne leži na katerem od +lokov $l_1, l_2,\cdots , l_n$.} + + Naj bo $S$ središče krožnice $k$. Naj bo + $\mathcal{S}_S:l_1, l_2,\cdots , l_n\rightarrow l'_1, l'_2,\cdots , l'_n$. + Predpostavimo nasprotno, da za vsak premer $PQ$ eno od njegovih krajišč leži + na katerem od lokov $l_1, l_2,\cdots , l_n$. Ker je $\mathcal{S}_S(P)=Q$, + ležita obe krajišči na nekem loku $l_1, l_2,\cdots , l_n$, $l'_1, l'_2,\cdots , l'_n$. + To pomeni, da tudi vsaka točka krožnice $k$ leži na katerem od lokov $l_1, l_2,\cdots , + l_n$, $l'_1, l'_2,\cdots , l'_n$, kar ni možno, saj je skupna dolžina teh lokov manjša od obsega krožnice $k$. + +\item \res{Dan je krog $K(S,20)$. Igralca $\mathcal{A}$ in $\mathcal{B}$ +izmenično rišeta kroge s polmeri $x_i$ ($1b$) dane daljice, načrtaj takšno daljico $x$, da velja:}\\ +(\textit{a}) \res{$x=\sqrt{a^2+b^2}$,} \hspace*{3mm} +(\textit{b}) \res{$x=\sqrt{a^2-b^2}$,} \hspace*{3mm} +(\textit{c}) \res{$x=\sqrt{3ab}$,}\\ +(\textit{d}) \res{$x=\sqrt{a^2+bc}$,} \hspace*{3mm} +(\textit{e}) \res{$x=\sqrt{3ab-c^2}$,} \hspace*{3mm} +(\textit{f}) \res{$x=\frac{a\sqrt{ab+c^2}}{b+c}$.} + +Uporabi Pitagorov izrek \ref{PitagorovIzrek} in višinski izrek \ref{izrekVisinski}. + +%Stewartov iizrek + +\item +\res{Naj bodo $a$, $b$ in $c$ stranice nekega trikotnika in velja $a^2+b^2=5c^2$. Dokaži, da sta težiščnici + $t_a$ in $t_b$ med seboj pravokotni.} + +Označimo s $T$ težišče danega trikotnika $ABC$ ter $t_a$ in $t_b$ dolžini ustreznih težiščnic. Dokažimo, da je $ATB$ pravokotni trikotnik. +Po posledici \ref{StwartTezisc} Stewartovega izreka \ref{StewartIzrek} je: + \begin{eqnarray*} +t_a^2=\frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4}\\ +t_b^2=\frac{a^2}{2}+\frac{c^2}{2}-\frac{b^2}{4}. + \end{eqnarray*} +Iz tega in iz predpostavke $a^2+b^2=5c^2$ sledi: +\begin{eqnarray*} +|AT|^2+|BT|^2 + &=& +\left(\frac{2}{3}t_a\right)^2+\left(\frac{2}{3}t_b\right)^2=\\ + &=& +\frac{4}{9}\left(\frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4} ++\frac{a^2}{2}+\frac{c^2}{2}-\frac{b^2}{4}\right)=\\ + &=& +\frac{4}{9}\left(\frac{a^2+b^2}{4}+c^2\right)=\\ + &=& c^2=|AB|^2. + \end{eqnarray*} +Po obratnem Pitagorovem izreku \ref{PitagorovIzrekObrat} je $ATB$ pravokotni trikotnik s hipotenuzo $AB$, kar pomeni, da sta ustrezni težiščnici pravokotni. + +\item +\res{Naj bodo $a$, $b$, $c$ in $d$ stranice, $e$ in $f$ diagonali ter $x$ daljica, ki je določena s središčema + stranic $b$ in $d$ nekega štirikotnika. Dokaži: +$$x^2 = \frac{1}{4} \left(a^2 +c^2 -b^2 -d^2 +e^2 +f^2 \right).$$} + +Uporabimo posledico \ref{StwartTezisc} Stewartovega izreka \ref{StewartIzrek}. + +\item +\res{Naj bodo $a$, $b$ in $c$ stranice trikotnika $ABC$. Dokaži, da je razdalja središča $A_1$ stranice $a$ od nožišča $A'$ višine na +to stranico enaka: +$$|A_1A'|=\frac{|b^2-c^2|}{2a}.$$} + +Brez škode za splošnost predpostavimo, da velja $b\geq c$. +Označimo $x=|A_1A'|$, $v_a=|AA'|$ in $t_a=|AA_1|$. Če uporabimo Pitagorov izrek \ref{PitagorovIzrek} za trikotnika $AA'A_1$ in $AA'B$, dobimo: + \begin{eqnarray*} + v_a^2 &=& t_a^2-x^2;\\ + v_a^2 &=& c^2- \left( \frac{a}{2} -x \right)^2. + \end{eqnarray*} + Po odštevanju enačb in reševanju dobljene enačbe po $x$ dobimo: + \begin{eqnarray*} +x=\frac{1}{a}\left( t_a^2-c^2+ \frac{a^2}{2} \right). + \end{eqnarray*} +Na koncu uporabimo še relacijo za $t_a^2$ iz izreka \ref{StwartTezisc}: + \begin{eqnarray*} +x &=& \frac{1}{a}\left( t_a^2-c^2+ \frac{a^2}{4} \right)=\\ + &=& \frac{1}{a}\left( \frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4}-c^2+ \frac{a^2}{4} \right)=\\ + &=& \frac{b^2-c^2}{2a}. + \end{eqnarray*} + +%Pappus in Ppascal + +\item +\res{Naj bosta ($A$, $B$, $C$) in ($A_1$, $B_1$, $C_1$) trojici kolinearnih točk neke ravnine, ki nista na +isti premici. Če je $AB_1\parallel A_1B$ in $AC_1\parallel A_1C$, tedaj je tudi $CB_1\parallel C_1B$. (\textit{Pappusov izrek}\footnote{Pappus iz Aleksandrije\index{Pappus} (3. st.), starogrški matematik. Gre za posplošitev Pappusovega izreka (glej izrek \ref{izrek Pappus}), če za $X$, $Y$ in $Z$ izberemo točke v neskončnosti.})} + +Uporabimo Talesov izrek \ref{TalesovIzrek} in obratni Talesov izrek \ref{TalesovIzrekObr}. + +%Desarguesov iizrek + + +\item \label{nalPodDesarg1} +\res{Naj bodo $P$, $Q$ in $R$ takšne točke stranic $BC$, $AC$ in $AB$ +trikotnika $ABC$, da so premice $AP$, $BQ$ in +$CR$ iz istega šopa. Dokaži: Če je $X=BC\cap QR$, $Y=AC\cap PR$ in $Z=AB\cap PQ$, so točke +$X$, $Y$ in $Z$ +kolinearne.} + +Uporabimo Desarguesov izrek \ref{izrekDesarguesEvkl} za trikotnika $ABC$ in $PQR$. + + +\item +\res{Naj bodo $AA'$, $BB'$ in $CC'$ višine trikotnika $ABC$ ter $X=B'C'\cap BC$, $Y=A'C'\cap AC$ in $Z=A'B'\cap AB$. Dokaži, da so $X$, $Y$ in $Z$ kolinearne točke.} + +Direktna posledica izreka \ref{VisinskaTocka} in prejšnje naloge \ref{nalPodDesarg1}. + +\item +\res{Naj bosta $A$ in $B$ točki izven premice $p$. Načrtaj presečišče premic $p$ in $AB$ brez direktnega risanja premice +$AB$.} + +Načrtaj poljubne kolinearne točke $C$, $C'$ in $S$, nato pa: $Y=p\cap AC$, $X=p\cap BC$, $A'=SA\cap C'Y$ in $B'=SB\cap C'X$. Po Desarguesovem izreku \ref{izrekDesarguesEvkl} se premici $AB$ in $A'B'$ sekata v točki $Z$, ki je kolinearna s točkama $X$ in $Y$, torej leži na premici $p$. To pomeni, da presečišče premic $p$ in $AB$ dobimo kot presečišče premic $p$ in $A'B'$, torej brez direktnega risanja premice +$AB$. + + +\item +\res{Naj bosta $p$ in $q$ premici neke ravnine, ki se sekata v točki $S$, ki je ‘‘izven papirja’’, in $A$ +točka te ravnine. Načrtaj premico, ki poteka skozi točki $A$ in $S$.} + +Uporabimo izrek \ref{izrekDesarguesOsNesk}. + +\item +\res{Načrtaj trikotnik tako, da njegova oglišča ležijo na treh danih vzporednih premicah, nosilke njegovih stranic pa potekajoo skozi tri dane +točke.} + +Uporabimo Desarguesov izrek \ref{izrekDesarguesEvkl}. Glej zgled \ref{zgled 3.2}. + + +%Ppotenca + +\item +\res{Dana je krožnica $k(S,r)$.} + +(\textit{a}) \res{Katere vrednosti vse lahko ima potenca točke glede na krožnico $k$?} + +Ker po izreku \ref{izrekPotenca} potenco neke točke $P$ glede na krožnico $k$ dobimo kot: $p(S,k)=|PS|^2-r^2$, se vrednost potence giblje na intervalu $[-r^2,\infty)$. + + (\textit{b}) \res{Katera je najmanjša vrednost te potence in za katero točko se ta minimalna vrednost doseže?} + +Iz prejšnjega primera je jasno, da je minimalna vrednost potence $-r^2$ in se doseže za središče $S$ krožnice $k$. + +(\textit{c}) \res{Določi množico vseh točk, za katere je potenca glede na krožnico enaka $\lambda\in \mathbb{R}$.} + +Pogoja $p(S,k)=|PS|^2-r^2=\lambda$ in $|PS|^2=r^2+\lambda$ sta si ekvivalentna. + +V primeru, ko je $\lambda>-r^2$, gre za krožnico $k(S,r^2+\lambda)$. Če je $\lambda=-r^2$, je iskana množica enaka $\{S \}$, v primeru $\lambda<-r^2$ pa prazni množici $\emptyset$. + +\item \res{Naj bo $k_a(S_a,r_a)$ pričrtana in $l(O,R)$ očrtana krožnica nekega trikotnika. Dokaži enakost\footnote{Trditev je posplošitev Eulerjeve formule za krožnico (glej izrek \ref{EulerjevaFormula}). \index{Euler, L.} + \textit{L. Euler} + (1707--1783), švicarski matematik.}: + $$S_aO^2=R^2+2r_aR.$$} + +Dokaz je podoben dokazu izreka \ref{EulerjevaFormula}. + + +\item +\res{Načrtaj krožnico, ki poteka skozi dani točki $A$ in $B$ in se dotika dane krožnice $k$.} + +Iskano krožnico označimo z $x$. +Načrtamo poljubno krožnico $j$, ki poteka skozi točki $A$ in $B$, krožnico $k$ pa seka v točkah $C$ in $D$. Uporabimo dejstvo, da je presečišče premic $AB$ in $CD$ potenčno središče krožnic $k$, $j$ in $x$. + +Naloga predstavlja enega od desetih Apolonijevih problemov o dotiku krožnic (glej razdelek \ref{odd9ApolDotik}). + +\item +\res{Dokaži, da so središča daljic, ki so določena s skupnimi tangentami dveh krožnic, kolinearne točke.} + +Omenjene točke ležijo na potenčni premici dveh krožnic. + +\item +\res{Načrtaj krožnico, ki je pravokotna na dve dani krožnici, tretjo dano krožnico +pa seka v točkah, ki določata premer te tretje krožnice.} + +Označimo dane krožnice po vrsti z $k(K, r_k)$, $j(J,r_j)$ in $l(L,r_l)$, iskano krožnico pa z $x(X,r_x)$. + +Naj bo $P\in x\cap k$, $Q\in x\cap l$ in $R, R_1\in x\cap j$. +Po predpostavki je $x\perp k, l$. Po izreku \ref{TangPogoj} sta $XP$ in $XQ$ tangenti krožnic $k$ in $l$ v točkah $P$ in $Q$. Ker je $XP\cong XQ$, je $p(X,k)=p(X,l)$ oz. središče $X$ krožnice $X$ leži na potenčni premici $p(k,l)$ krožnic $k$ in $l$. + +Poiščemo še eno geometrijsko mesto točk za $X$, tako da vključimo pogoj za krožnico $j$. +Po predpostavki je $RR_1$ premer krožnice $j$. To pomeni, da je $J$ središče daljice $RR_1$, iz skladnosti trikotnikov $XJR$ in $XJR_1$ (izrek \textit{SSS} \ref{SSS}) pa sledi $\angle XJR=90^0$. Po Pitagorovem izreku \ref{PitagorovIzrek} (za $\triangle XJR$) je potem: + \begin{eqnarray*} + |XR|^2=|XJ|^2+r_j^2. + \end{eqnarray*} +Iz pravokotnega trikotnika $XQL$ po istem izreku dobimo: + \begin{eqnarray*} + |XQ|^2=r_l^2-|XL|^2. + \end{eqnarray*} +Ker je $XR\cong XQ$, iz prejšnjih dveh relacij sledi $|XJ|^2+r_j^2=r_l^2-|XL|^2$ oz.: +\begin{eqnarray*} + |XJ|^2+|XL|^2=r_l^2-r_j^2. + \end{eqnarray*} +Torej točka $X$ leži na neki krožnici $g$, ki jo lahko načrtamo (glej izrek \ref{GMTmnl}). To pomeni, da točko $S$ dobimo kot presečišče krožnice $g$ in potenčne premice $p(k,l)$ krožnic $k$ in $l$. + + + +%Razno + +\item +\res{Naj bosta $M$ in $N$ presečišči stranic $AB$ in $AC$ trikotnika $ABC$ s premico, +ki poteka skozi središče včrtane krožnice tega trikotnika in je vzporedna z njegovo +stranico $BC$. Izrazi dolžino daljice $MN$ kot funkcijo dolžin stranic trikotnika $ABC$.} + +Uporabimo izrek \ref{HarmCetSimKota} in dokažemo $\frac{AS}{SE}=\frac{b+c}{a}$. +Nato uporabimo \ref{TalesovIzrek} in dokažemo $\frac{MN}{BC}=\frac{AS}{AE}$. Rezultat: $|MN|=\frac{a(b+c)}{a+b+c}$. + +\item +\res{Naj bo $AA_1$ težiščnica trikotnika $ABC$. Točki $P$ in $Q$ naj bosta presečišči +simetral kotov $AA_1B$ in $AA_1C$ s stranicama $AB$ in $AC$. Dokaži, da je +$PQ\parallel BC$.} + +Po izreku \ref{HarmCetSimKota} je: +$$\frac{AP}{PB}=\frac{AA_1}{A_1B}=\frac{AA1}{A_1C}=\frac{AQ}{QC}.$$ +Iz obratnega Talesovega izreka \ref{TalesovIzrekObr} sledi $PQ\parallel BC$. + +\item +\res{V trikotniku $ABC$ naj bo vsota (ali razlika) notranjih kotov $ABC$ in $ACB$ enaka pravemu +kotu. Dokaži, da je $|AB|^2+|AC|^2=4r^2$, kjer je $r$ polmer očrtane krožnice tega trikotnika.} + +Naj bo $\alpha=\angle BAC$, $\beta=\angle ABC$ in $\gamma=\angle ACB$ ter $k(O,r)$ očrtana krožnica trikotnika $ABC$. + +Če je $\beta+\gamma=90^0$, je $\alpha=90^0$ (izrek \ref{VsotKotTrik}), kar pomeni, da je $ABC$ pravokotni trikotnik s hipotenuzo $BC$; trditev je v tem primeru trivialna posledica izreka \ref{TalesovIzrKroz2} in Pitagorovega izreka \ref{PitagorovIzrek}. + +Naj bo $\beta-\gamma=90^0$ oz. $\gamma=\beta-90^0$. Označimo $C'=\mathcal{S}_O(C)$. Po izreku \ref{TalesovIzrKroz2} je najprej $\angle CAC'=90^0$. Ker je $BCC'A$ tetivni štirikotnik, je po izreku \ref{TetivniPogoj}: + \begin{eqnarray*} +\angle C'CA &=& 90^0-\angle AC'C=\\ + &=& 90^0-(180^0-\beta)=\beta-90^0=\\ + &=& \gamma =\angle BCA. + \end{eqnarray*} + +Po izreku \ref{ObodObodKot} je $AC'\cong AB$, zato iz Pitagorovega izreka \ref{PitagorovIzrek} dobimo: + \begin{eqnarray*} +|AB|^2+|AC|^2=|AC'|^2+|AC|^2=|C'C|^2=4r^2. + \end{eqnarray*} + +\item +\res{Naj bo $AD$ višina trikotnika $ABC$. Dokaži, da je +vsota (ali razlika) notranjih kotov $ABC$ in $ACB$ enaka pravemu kotu natanko tedaj, ko je: +$$\frac{1}{|AB|^2}+\frac{1}{|AC|^2}=\frac{1}{|AD|^2}.$$} + +Obravnavamo trikotnika $ADB$ in $CDA$. + +\item +\res{Izrazi razdaljo med težiščem in središčem očrtane krožnice trikotnika kot +funkcijo dolžin njegovih stranic in polmera očrtane krožnice.} + +Uporabimo Stewartov izrek \ref{StewartIzrek} za trikotnik $OAA_1$ in izrek \ref{StwartTezisc}; $k(O,R)$ je očrtana krožnica, $T$ težišče in $A_1$ središče stranice $BC$ trikotnika $ABC$. + + Rezultat: +$|OT|=\sqrt{R^2-\frac{1}{9} \left(a^2+b^2+c^2 \right)}$. + +\item +\res{Dokaži, da pri trikotniku $ABC$ simetrala zunanjega kota ob oglišču $A$ in simetrali +notranjih kotov ob ogliščih $B$ in $C$ sekajo nosilke nasprotnih stranic v treh kolinearnih točkah.} + +Uporabimo Menelajev izrek \ref{izrekMenelaj} in izrek \ref{HarmCetSimKota}. + +\item +\res{Dokaži, da so pri trikotniku $ABC$ središče višine $AD$, središče včrtane krožnice in točka, v kateri se stranica $BC$ dotika pričrtane krožnice tega trikotnika, tri +kolinearne točke.} + +Uporabimo izrek \ref{velNalTockP'}. + +\item +\res{Dokaži Simsonov izrek \ref{SimpsPrem} z uporabo Menelajevega izreka \ref{izrekMenelaj}.} + +Naj bo $S$ poljubna točka očrtane krožnice trikotnika $ABC$ ter $P$, $Q$ in $R$ pravokotne projekcije te točke na nosilkah $BC$, $AC$ in $BC$. Uporabi: +$\triangle SRA\sim\triangle SPC$, $\triangle SPB\sim\triangle SQA$ in $\triangle SQC\sim\triangle SRB$. + + +\item +\res{Skozi točko $M$ stranice $AB$ trikotnika $ABC$ je konstruirana premica, ki seka +premico $AC$ v točki $K$. Izračunaj razmerje, v katerem premica $MK$ deli stranico $BC$, +če je $AM:MB=1:2$ in $AK:AC=3:2$.} + +Naj bo $P$ presečišče premic $MK$ in $BC$. Po Menelajevem izreku je +$$-1=\frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot +\frac{\overrightarrow{CK}}{\overrightarrow{KA}}\cdot +\frac{\overrightarrow{AM}}{\overrightarrow{MB}}= +\frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot +\frac{-1}{3}\cdot\frac{1}{2},$$ zato je $\frac{\overrightarrow{BP}}{\overrightarrow{PC}}=6:1$. + +\item +\res{Naj bo $A_1$ središče stranice $BC$ trikotnika $ABC$ in naj bosta $P$ in $Q$ takšni točki stranic +$AB$ in $AC$, da velja $BP:PA=2:5$ in $AQ:QC=6:1$. Izračunaj razmerje, v katerem premica $PQ$ deli težiščnico $AA_1$.} + +Naj bo $R$ presečišče premic $PQ$ in $BC$. Uporabimo Menelajev izrek \ref{izrekMenelaj} najprej za trikotnik $ABC$ in premico $PQ$, nato pa za trikotnik $AA_1C$ in isto premico. Rezultat: $17:60$. + +\item +\res{Dokaži, da se pri poljubnem trikotniku premice, ki so določene z oglišči in +dotikališči ene pričrtane krožnice z nosilkami nasprotnih stranic, sekajo v skupni točki.} + +Uporabimo Cevov izrek \ref{izrekCeva} in veliko nalogo \ref{velikaNaloga}. + +\item +\res{Kaj predstavlja množica vseh točk, iz katerih odseka tangent glede na dve dani krožnici predstavljata dve skladni daljici?} + +Ker je v tem primeru potenca iz teh točk glede na krožnici enaka, je iskana množica del njune potenčne premice, ki je v zunanjosti obeh krožnic. + +\item +\res{Naj bosta $PP_1$ in $QQ_1$ zunanji tangenti krožnic $k(O,r)$ in $k_1(O_1,r_1)$ (točke $P$, $P_1$, $Q$ in $Q_1$ so ustrezna dotikališča). Točka $S$ naj bo presečišče teh dveh tangent, $A$ eno od presečišč krožnic $k$ in $k_1$ ter $L$ in $L_1$ presečišči +premice $SO$ s premicama $PQ$ in $P_1Q_1$. Dokaži, da velja $\angle LAO\cong\angle L_1AO_1$.} + +Brez škode za splošnost naj bo $\mathcal{B}(S,O,O_1)$. +Naj bo še $h_{S,\lambda}$ središčni razteg s koeficientom $\lambda=\frac{\overrightarrow{SO_1}}{\overrightarrow{SO}}$. Potem je $h_{S,\lambda}(k)=k_1$ in: + $$h_{S,\lambda}:\hspace*{1mm} O,P,Q,L \mapsto O_1,P_1,Q_1,L_1.$$ +Naj bo še $A_1=h_{S,\lambda}(A)$. Iz $h_{S,\lambda}(k)=k_1$ in $A\in k$ sledi + $A_1\in k_1$. Po izreku \ref{homotOhranjaKote} je $\angle L_1A_1O_1\cong\angle LAO$, torej je dovolj dokazati $\angle L_1A_1O_1\cong\angle L_1AO_1$. + +Iz podobnosti pravokotnih trikotnikov $SP_1O_1$ in $SL_1P_1$ (izrek \ref{PodTrikKKK}) in izreka \ref{izrekPotenca} sledi: + \begin{eqnarray*} +\overrightarrow{SO_1}\cdot \overrightarrow{SL_1}=|SP_1|^2=p(S.k_1)= +\overrightarrow{SA}\cdot \overrightarrow{SA_1}. + \end{eqnarray*} +To pomeni, da je $O_1L_1AA_1$ tetivni štirikotnik, zato je $\angle L_1A_1O_1\cong\angle L_1AO_1$. + + +\item +\res{Dokaži, da je stranica pravilnega desetkotnika enaka +večjemu delu delitve polmera očrtane krožnice tega desetkotnika v razmerju zlatega reza.} + +Naj bo $k(S,r)$ središče očrtane krožnice in $AB$ ($a=|AB|$) ena stranica pravilnega desetkotnika. V trikotniku $ASB$ merijo notranji koti: + $\angle ASB=36^0$ in $\angle SAB=\angle SBA=72^0$. + +Označimo s $P$ takšno točko, da velja $\mathcal{B}(B,A,P)$ in $AP\cong AS$. Ker je $SAP$ enakokraki trikotnik z osnovnico $PS$, je po izreku \ref{enakokraki} $\angle SPA\cong\angle PSA$. Iz izreka \ref{zunanjiNotrNotr} (za trikotnik $SAP$) sledi + $\angle APS=\angle PSA=\frac{1}{2}72^0=36^0$. Torej velja $\triangle SPB\sim\triangle ASB$ (izrek \ref{PodTrikKKK}), zato je $\frac{PB}{SB}=\frac{SB}{AB}$ oz. $\frac{a+r}{r}=\frac{r}{a}$, iz tega pa direktno sledi naša trditev. + + +\item \label{nalPod75} +\res{Naj bodo $a_5$, $a_6$ in $a_{10}$ stranice pravilnega petkotnika, šestkotnika in desetkotnika, ki so včrtani isti krožnici. Dokaži, da velja: + $$a_5^2=a_6^2+a_{10}^2.$$} + +Uporabimo podobnost trikotnikov iz prejšnje naloge \ref{nalPod75}, pri tem pa upoštevamo ustrezni višini teh dveh trikotnikov. + +\item + \res{Naj bodo $t_a$, $t_b$ in $t_c$ težiščnice in $s$ polobseg nekega trikotnika. Dokaži, da velja: + $$t_a^2+t_b^2+t_c^2\geq s^2.$$} % zvezek - dodatni MG + +Če uporabimo izrek \ref{StwartTezisc2}: + \begin{eqnarray*} +t_a^2+t_b^2+t_c^2 &=& \frac{3}{4}\left(a^2+ b^2+c^2\right)=\\ + &\geq& \frac{3}{4}\cdot 3\cdot\left(\frac{a+ b+c}{3}\right)^2=\\ + &=& s^2. + \end{eqnarray*} + +\end{enumerate} + + + + + +%REŠITVE - Ploščina +%________________________________________________________________________________ + +\poglavje{Area of Figures} + + +\begin{enumerate} + +% Ttrikotniki + +\item \res{Naj bo $T$ težišče trikotnika $ABC$. Dokaži: + $$p_{TBC}=p_{ATC}=p_{ABT}.$$} + +Označimo z $A_1$ središče stranice $BC$. Potem je po izreku \ref{PloscTrik}: $p_{AA_1B}=p_{AA_1C}$ in $p_{TA_1B}=p_{TA_1C}$. Če odštejemo enakosti, dobimo $p_{AA_1B}-p_{TA_1B}$ in $p_{AA_1C}-p_{TA_1C}$ oz. po izreku \ref{ploscGlavniIzrek} \textit{4)}: $p_{ABT}=p_{ATC}$. Analogno dobimo tudi $p_{TBC}=p_{ATC}$. + + +\item \res{Naj bodo $c$ hipotenuza, $v_c$ pripadajoča višina ter $a$ in $b$ kateti pravokotnega trikotnika. Dokaži, da velja $c+v_c>a+b$.} % zvezek - dodatni MG + +Po izreku \ref{PloscTrik} za ploščino $p$ tega trikotnika velja $p=\frac{1}{2}ab=\frac{1}{2}cv_c$, zato je $ab=cv_c$, če pa še uporabimo Pitagorov izrek \ref{PitagorovIzrek}: + \begin{eqnarray*} + (a+b)^2=a^2+2ab+b^2= +c^2+2cv_cCD$), tako da razdeli ta trapez na dva ploščinsko enaka lika.} + +Naj bodo: $v$ višina trapeza ter $a$ in $c$ dolžini njegovih osnovnic $AB$ in $CD$. +Če je $X$ točka, v kateri premica $p$ seka osnovnico $AB$, iz danega pogoja velja: + $p_{AXD}=\frac{1}{2}p_{ABCD}$ oz.: +\begin{eqnarray*} + \frac{|AX|\cdot v}{2} =\frac{1}{2}\cdot \frac{a+c}{2}\cdot v. + \end{eqnarray*} + +Torej je daljica $AX$ enaka srednjici trapeza oz. $|AX|=\frac{a+c}{2}$, kar omogoča konstrukcijo točke $X$ in premice $p$. + + +\item \res{Načrtaj premici $p$ in $q$, ki potekata skozi oglišče $D$ kvadrata $ABCD$ in ga razdelita na ploščinsko enake like.} + +Če sta $P$ in $Q$ točki, v katerih premica $p$ oz. $q$ seka stranico $AB$ oz. $BC$ kvadrata $ABCD$, dokažemo, da je $AP:PB=2:1$ in $BQ:QC=1:2$. + +\item \res{Naj bo $ABCD$ kvadrat, $E$ središče njegove stranice $BC$ in $F$ točka, za katero je $\overrightarrow{AF}=\frac{1}{3}\cdot \overrightarrow{AB}$. Točka $G$ je četrto oglišče pravokotnika $FBEG$. Kolikšen del ploščine kvadrata $ABCD$ predstavlja ploščina trikotnika $BDG$?} + +Označimo z $a$ dolžino stranice kvadrata $ABCD$. +Naj bo $S$ središče tega kvadrata in $H=\mathcal{T}_{\overrightarrow{AF}}(D)$. Po izrekih \ref{PloscTrik}, \ref{ploscKvadr} in \ref{ploscGlavniIzrek} \textit{4)} je: +\begin{eqnarray*} + p_{BDG} &=& p_{BSG}+p_{SDG}= p_{FSG}+p_{SHG}=\\ + &=& p_{FSH}=\frac{1}{2}\cdot |FH| \cdot |GS|= \\ + &=& \frac{1}{2}\cdot a \cdot \left( \frac{a}{2}-\frac{a}{3} \right)=\\ + &=& \frac{a^2}{12} = \frac{1}{12}\cdot p_{ABCD}. + \end{eqnarray*} + + +\item \label{nalPloKoef} +\res{Naj bosta $\mathcal{V}$ in $\mathcal{V}'$ podobna večkotnika s koeficientom podobnosti $k$. Dokaži, da je: + $$p_{\mathcal{V}'}=k^2\cdot p_{\mathcal{V}}.$$} + +najprej dokažemo, da trditev velja za trikotnike (uporabimo izrek \ref{PloscTrik}), nato pa uporabimo izrek \ref{ploscGlavniIzrek} \textit{4)}. + + +\item \label{nalPloHeronStirik} + \res{Naj bodo $a$, $b$, $c$ in $d$ dolžine stranic, $s$ polobseg in $p$ ploščina poljubnega štirikotnika. Dokaži, da velja: + $$p=\sqrt{(s-a)(s-b)(s-c)(s-d)}.$$} %Lopandic - nal 924 + +Označimo z $A$, $B$, $C$ in $D$ oglišča danega štirikotnika, tako da je: $|AB|=a$, $|BC|=b$, $|CD|=c$ in $|DA|=d$. Če je $ABCD$ paralelogram, je zaradi njegove tetivnosti pravokotnik (izreka \ref{paralelogram} in \ref{TetivniPogoj}) in je trditev trivialna. + +Brez škode za splošnost torej predpostavimo, da se nosilki $BC$ in $AD$ sekata v neki točki $P$ in pri tem velja še $a>c$. Označimo še $|PC|=x$, $|PD|=y$ in $p'=p_{PCD}$. +Po izreku \ref{TetivniPogojZunanji} je $\angle PCD\cong \angle BAD$, zato je $\triangle CDP \cong \triangle ABP$ (izrek \ref{PodTrikKKK}). Koeficient podobnosti teh dveh trikotnikov je $k=\frac{AB}{CD}=\frac{a}{c}$. Torej po trditvi iz prejšnje naloge \ref{nalPloKoef} velja: + $$\frac{p_{ABP}}{p_{CDP}}=\frac{a^2}{c^2}$$ +oz. po izreku \ref{ploscGlavniIzrek} \textit{4)} +$$\frac{p_{ABCD}+p_{CDP}}{p_{CDP}}=\frac{a^2}{c^2}.$$ +Iz tega dobimo: +\begin{eqnarray} \label{nalPloEqnHeronStirik1} + p=\frac{a^2-c^2}{c^2}\cdot p'. + \end{eqnarray} +Izračunajmo najprej $p'=p_{PCD}$. Iz omenjene podobnosti $\triangle CDP \cong \triangle ABP$ sledi še: +\begin{eqnarray*} + \frac{x+b}{y}=\frac{y+d}{x}=\frac{a}{c}. + \end{eqnarray*} + Če preoblikujemo enakosti, dobimo sistem enačb za $x$ in $y$: + \begin{eqnarray*} + && ax-cy=cd\\ + && cx-ay=-bc + \end{eqnarray*} +in njegove rešitve: +\begin{eqnarray*} + x &=& \frac{c(ad+bc)}{a^2-c^2}\\ + y &=& \frac{c(ab+cd)}{a^2-c^2} + \end{eqnarray*} +Iz tega dobimo: +\begin{eqnarray} \label{nalPloEqnHeronStirik2} + x+y &=& \frac{c(b+d)}{a+c}\\ + x-y &=& \frac{c(d-b)}{a-c} + \end{eqnarray} +Naj bo $s'$ polobseg trikotnika $PCD$. Z uporabo relacij \ref{nalPloEqnHeronStirik2} dobimo: + \begin{eqnarray*} + s' &=& \frac{x+y+c}{2}=\frac{c}{a-c}\cdot (s-c)\\ + s'-c &=& \frac{x+y-c}{2}=\frac{c}{a-c}\cdot (s-a)\\ + s'-x &=& \frac{c+x+y}{2}=\frac{c}{a+c}\cdot (s-d)\\ + s'-y &=& \frac{c+x+y}{2}=\frac{c}{a+c}\cdot (s-b) + \end{eqnarray*} +Iz prejšnjih relacij po Heronovi formuli za ploščino trikotnika (izrek \ref{PloscTrikHeron}) velja: +\begin{eqnarray*} + p' &=& p_{PCD}=\sqrt{s'(s'-c)(s'-x)(s'-y)}=\\ + &=& \frac{c^2}{a^2-c^2}\sqrt{(s-a)(s-b)(s-c)(s-d)}. + \end{eqnarray*} +Če to uvrstimo v relacijo \ref{nalPloEqnHeronStirik1}, dobimo: +\begin{eqnarray*} + p = \frac{a^2-c^2}{c^2}\cdot p'=\sqrt{(s-a)(s-b)(s-c)(s-d)}. + \end{eqnarray*} + + +\item \res{Naj bodo $a$, $b$, $c$ in $d$ dolžine stranic in $p$ ploščina tetivnotangentnega štirikotnika. Dokaži, da velja: + $$p=\sqrt{abcd}.$$} %Lopandic - nal 925 + +Naj bo $s$ polobseg danega štirikotnika. +Ker gre za tangentni štirikotnik, za njegove stranice po izreku \ref{TangentniPogoj} velja $a+c=b+d$. Zato je: +\begin{eqnarray*} + s-a &=& \frac{b+c+d-a}{2}=c\\ + s-b &=& \frac{a+c+d-b}{2}=d\\ + s-c &=& \frac{a+b+d-c}{2}=a\\ + s-d &=& \frac{a+b+c-d}{2}=b. + \end{eqnarray*} +Ker je štirikotnik še tetiven, lahko uporabimo trditev iz prejšnje naloge \ref{nalPloHeronStirik} in dobimo: +$$p=\sqrt{(s-a)(s-b)(s-c)(s-d)}=\sqrt{abcd}.$$ + +% Veckotniki + +\item \res{Dan je pravokotnik $ABCD$ s stranicama $a=|AB|$ in $b=|BC|$. Izračunaj ploščino lika, ki predstavlja unijo pravokotnika $ABCD$ in njegove slike pri zrcaljenju čez premico $AC$.} + +Naj bo $B'=\mathcal{S}_{AC}(B)$ in $D'=\mathcal{S}_{AC}(D)$, $P$ presečišče daljic $AB'$ in $CD$ ter $S$ presečišče diagonal $AC$ in $BD$ pravokotnika $ABCD$. Označimo še $d=|AC|=|BD|$ in $v=|SP|$. Iz podobnosti trikotnikov $AB'C$ in $ASP$ (izrek \ref{PodTrikKKK}) dobimo $v:b=\frac{d}{2}:a$ oz. $v=\frac{bd}{2a}$. Če uporabimo dobljeno relacijo, izreke \ref{PloscTrik}, \ref{ploscPravok} in \ref{ploscGlavniIzrek} ter Pitagorov izrek \ref{PitagorovIzrek}, za ploščino $p$ iskanega lika velja: + +\begin{eqnarray*} + p &=& 2\cdot \left(p_{AB'C}+p_{ADC}-p_{APC}\right)= +2\cdot \left(p_{ABC}+p_{ADC}-p_{APC}\right)=\\ +&=& 2\cdot \left(p_{ABCD}-p_{APC}\right)=2\cdot \left( ab-\frac{dv}{2} \right)=\\ +&=& 2\cdot \left( ab-\frac{bd^2}{4a} \right)=2ab-\frac{b(a^2+b^2)}{2a}=\\ +&=& \frac{b(3a^2-b^2)}{2a}. + \end{eqnarray*} + +\item \res{Naj bo $ABCDEF$ pravilni šestkotnik, točki $P$ in $Q$ pa središči njegovih stranic $BC$ in $FA$. Kolikšen del ploščine tega šestkotnika predstavlja ploščina trikotnika $PQD$?} + +Uporabimo izrek \ref{srednjTrapez}. Rezultat: $p_{PQD}=\frac{3}{8}p_{ABCDEF}$. + + +% KKrog + +\item \res{V kvadratu so včrtani štirje skladni krogi, tako da se vsak krog dotika dveh stranic in dveh krogov. Dokaži, da je vsota ploščin teh krogov enaka ploščini temu kvadratu včrtanega kroga.} + +Če z $a$ označimo dolžino stranice kvadrata, sta $r=\frac{a}{2}$ oz. $r_1=\frac{a}{4}$ polmera včrtanega kroga oz. vsakega od manjših včrtanih krogov. Potem je omenjena vsota ploščin enaka: +$$4\cdot r_1^2\pi=4\left(\frac{a}{4} \right)^2\cdot \pi=\left(\frac{a}{2} \right)^2\cdot \pi=r^2\pi.$$ + +\item \res{Izračunaj ploščino kroga, ki je včrtan trikotniku s stranicami dolžin 9, 12 in 15.} + +Uporabimo Heronovo formulo \ref{PloscTrikHeron} in izrek \ref{PloscTrikVcrt}. Rezultat: ploščina trikotnika $p_{\triangle}=54$; ploščina kroga $p=9\pi$. + +\item \res{Naj bo $P$ središče osnovnice $AB$ trapeza $ABCD$, za katerega velja $|BC|=|CD|=|AD|=\frac{1}{2}\cdot |AB|=a$. Izrazi ploščino lika, ki ga določajo osnovnica $CD$ ter krajša krožna loka $PD$ in $PC$ krožnic s središčema $A$ in $B$, kot funkcijo osnovnice $a$.} + +Ker je po predpostavki $\overrightarrow{PB}=\overrightarrow{DC}$, je štirikotnik $PBDC$ paralelogram, zato je $BC\cong PD$. Ker je po predpostavki $AD\cong BC$ in $AD\cong AP$, je $PD\cong AD\cong AP$ in $APD$ je enakostranični trikotnik. Torej $\angle PAD= 60^0$. Analogno je tudi $\angle CBP=60^0$. Iskano ploščino $p_0$ dobimo kot razliko ploščine trapeza $p$ in dvakratne ploščine krožnega izseka s središčnim kotom $60^0$: +$$p_0=\frac{a^2}{12}\left(9\sqrt{3}-4\pi \right).$$ + + +\item \res{Tetiva $PQ$ ($|PQ|=d$) krožnice $k$ se dotika njene konciklične krožnice $k'$. Izrazi ploščino kolobarja, ki ga določata krožnici $k$ in $k'$, kot funkcijo tetive $d$.} + + Uporabimo Pitagorov izrek \ref{PitagorovIzrek}. Rezultat: $\frac{d^2\pi}{4}$. + +\item \res{Naj bo $r$ polmer včrtanega kroga tetivnega večkotnika $\mathcal{V}$, ki je razdeljen na trikotnike $\triangle_1,\triangle_2,\ldots,\triangle_n$, tako da nobena dva trikotnika nimata skupnih notranjih točk. Naj bodo $r_1,r_2,\ldots , r_n$ polmeri včrtanih krožnic teh trikotnikov. Dokaži, da je: + $$\sum_{i=1}^n r_i\geq r.$$} + +Naj bo $s$ polobseg in $p$ ploščino večkotnika $\mathcal{V}$ ter $s_i$ polobseg in $p_i$ ($i\in \{1,2,\ldots , n\}$) ploščina trikotnika $\triangle_i$. Če uporabimo izreka \ref{ploscTetVec} in \ref{PloscTrikVcrt} ter dejstvo, da za vsak $i\in \{1,2,\ldots , n\}$ velja $s\geq s_i$, dobimo: + $$sr=p= \sum_{i=1}^n p_i = \sum_{i=1}^n s_ir_i\leq \sum_{i=1}^n sr_i=s\cdot\sum_{i=1}^n r_i.$$ + + +\end{enumerate} + + + + +%REŠITVE - Inverzija +%________________________________________________________________________________ + +\poglavje{Inversion} + + +\begin{enumerate} + + \item \res{Dokaži da kompozitum dveh inverzij $\psi_{S,r_1}$ in $\psi_{S,r_2}$ glede na koncentrični + krožnici predstavlja razteg. Določi središče in koeficient + tega raztega.} + + Iz definicije inverzije sledi, da je kompozitum inverzij + $\psi_{S,r_2}\circ\psi_{S,r_1}$ + razteg s središčem $S$ in koeficientom $\frac{r_2^2}{r_1^2}$. + Trditev je tudi direktna + posledica naloge \ref{invRazteg}. + + + \item \res{Naj bodo $A$, $B$, $C$ in $D$ štiri kolinearne točke. + Konstruiraj takšni točki $E$ in $F$, da velja $\mathcal{H}(A,B;E,F)$ +in $\mathcal{H}(C,D;E,F)$.} + + Uporabimo izrek \ref{harmPravKrozn} - najprej narišemo krožnico, ki + je pravokotna na krožnici s premeroma $AB$ in $CD$, + + + \item \res{V ravnini so dani točka $A$, premica $p$ in krožnica $k$. + Načrtaj krožnico, ki poteka skozi točko $A$ in je + pravokotna na premico $p$ in krožnico $k$.} + + Uporabi inverzijo s središčem $A$. + + \item \res{Reši tretji, četrti, deveti in deseti Apolonijev problem.} + + Pri tretjem in četrtem Apolonijevem problemu uporabimo inverzijo + s središčem v eni od danih točk. Pri devetem in desetem + Apolonijevem problemu najprej načrtamo krožnico, ki je + koncentrična z iskano krožnico in poteka skozi središče ene od + krožnic. Na ta način problema prevedemo na peti oz. šesti + Apolonijev problem. + + \item \res{Naj bodo: $A$ točka, $p$ premica, $k$ krožnica + in $\omega$ kot v neki ravnini. Načrtaj krožnico, +ki poteka skozi točko $A$, se dotika premice $p$ in s krožnico $k$ +določa kot $\omega$.} + + Uporabimo inverzijo s središčem $A$. + + + \item \res{Določi geometrijsko mesto točk dotika dveh krožnic, ki + se dotikata krakov danega kota v dveh danih točkah $A$ in $B$.} + + Uporabimo inverzijo s središčem $B$. + + \item \res{Načrtaj trikotnik, če so znani naslednji podatki: +\begin{enumerate} + \item $a$, $l_a$, $v_a$ + \item $v_a$, $t_a$, $b-c$ + \item $b+c$, $v_a$, $r_b-r_c$ + \end{enumerate}} + + Uporabimo veliko nalogo (glej izrek \ref{velikaNaloga}) in + ustrezne harmonične četverice točk. + + +\item \res{Naj bosta $c(S,r)$ in $l$ krožnica in premica v isti ravnini, ki +nimata skupnih točk. Naj bodo še $c_1$, $c_2$ in $c_3$ krožnice te +ravnine, ki se medseboj (po dve) dotikajo in se vsaka od njih +dotika še $c$ in $l$. Izrazi razdaljo točke $S$ od premice $l$ s +$r$\footnote{Predlog za MMO 1982. (SL 12.)}.} + + Najprej dokažemo, da obstaja krožnica $n$, ki je pravokotna na + premico $l$ in krožnico $c$. Naj bo $Y$ eno od presečišč + krožnice $n$ in pravokotnice na premico $l$ iz središča krožnice + $c$. Uporabimo kompozitum $f=\psi_i\circ \mathcal{R}\circ \psi_i$, + kjer je $i$ poljubna krožnica s središčem $Y$, $\mathcal{R}$ + pa rotacija s središčem v središču krožnice $l'=\psi_i(l)$, + ki preslika dotikališče krožnic $l'$ in $c'_3=\psi_i(c_3$ v točko + $Y$. Če je: + $f:\hspace*{1mm}l, c, c_1, c_2, c_3\mapsto + \widehat{l}, \widehat{c}, \widehat{c_1}, \widehat{c_2}, \widehat{c_3},$ + dokažemo, da je $\widehat{l}=l$, $\widehat{c}=c$, $\widehat{c_3}$ + premica, ki je vzporedna s premico $l$, ter $\widehat{c_1}$ in + $\widehat{c_2}$ skladni krožnici, ki se med seboj dotikata in + se hkrati dotikata tudi vzporednic $l$ in $\widehat{c_3}$. + Na koncu sledi, da je razdalja središča krožnice $c$ od premice $l$ + enaka $7r$. + + +\item \res{Naj bo $ABCD$ pravilni tetraeder. Poljubni točki +$M$, ki leži na robu $CD$, pridružimo točko $P = f(M)$, ki je +presečišče pravokotnice skozi točko $A$ na premico $BM$ in +pravokotnice skozi točko $B$ na premico $AM$. Določi geometrijsko +mesto vseh točk $P$, če točka $M$ zavzame vse vrednosti na robu +$CD$.} + +Točka $P$ je višinska točka trikotnika $ABM$. Če je $S$ središče roba +$AB$, najprej dokažemo, da velja $\overrightarrow{SP}\cdot +\overrightarrow{SM}=\frac{a^2}{4}$, kjer je $a$ rob +pravilnega tetraedra. Nato uporabimo inverzijo +$\psi_{S,\frac{a}{2}}$ (v ravnini $SCD$). Geometrijsko mesto točk +je potem slika daljice $CD$ pri tej inverziji oz. ustrezni krožni +lok s središčem $S$, s krajišči v višinskih točkah trikotnikov +$ACD$ in $BCD$. + + +\item \res{Naj bo $ABCD$ tetivnotangentni štirikotnik ter $P$, $Q$, +$R$ in $S$ dotikališča stranic $AB$, $BC$, $CD$ in $AD$ z včrtano +krožnico tega štirikotnika. Dokaži, da velja $PR\perp QS$.} + +Uporabimo inverzijo glede na včrtano krožnico. Dokažemo, da so +slike oglišč štirikotnika $ABCD$ oglišča pravokotnika, stranice +tega pravokotnika pa vzporedne z daljicama $PR$ in $QS$. + + +\item \res{Dokaži, da sta središči tetivnotangentnemu štirikotniku včrtane in očrtane +krožnice ter presečišče njegovih diagonal kolinearne točke +(\index{izrek!Newtonov}Newtonov izrek\footnote{\index{Newton, +I.}\textit{I. Newton} (1643--1727), angleški fizik in matematik}).} + +Uporabimo prejšnjo nalogo in dokažemo, da je središče $G$ +pravokotnika iz te naloge hkrati središče daljice, ki jo določata +središče včrtane krožnice štirikotnika $ABCD$ (središče inverzije) +in presečišče daljic $PR$ in $QS$. Točka $G$ je namreč središče +slike očrtane krožnice pri omenjeni inverziji. + + +\item \res{Naj bosta $p$ in $q$ vzporedni tangenti krožnice $k$. +Krožnica $c_1$ se dotika premice $p$ v točki $P$ in krožnice $k$ v +točki $A$, krožnica $k_2$ pa se dotika premice $q$ in krožnic $k$ +in $k_1$ v točkah $Q$, $B$ in $C$. Dokaži, da je presečišče premic +$PB$ in $AQ$ središče trikotniku $ABC$ očrtane krožnice.} + +Uporabimo inverzijo s središčem $B$ in najprej dokažemo, da je $PB$ +skupna tangenta krožnic $k$ in $k_2$, nato pa, da je presečišče +premic $PB$ in $AQ$ potenčno središče krožnic $k$, $k_1$ in $k_2$. + +\item \res{Krožnici $k_1$ in $k_3$ se od zunaj dotikata v točki $P$. Prav + tako se tudi krožnici $k_2$ in $k_2$ od zunaj dotikata v isti + točki. Krožnica $k_1$ seka krožnici $k_2$ in $k_4$ še v točkah + $A$ in $D$, krožnica $k_3$ pa seka krožnici $k_2$ in $k_4$ še v točkah + $B$ in $C$. Dokaži, da velja\footnote{Predlog za MMO 2003. (SL 16.)}: + $$\frac{|AB|\cdot|BC|}{|AD|\cdot|DC|}=\frac{|PB|^2}{|PD|^2}.$$} + +Naj bo $\psi_P$ inverzija s poljubnim premerom $r$. Ta +preslika štirikotnik $A'B'C'D'$ v paralelogram (izrek +\ref{InverzDotik}). Torej velja $A'B'\cong C'D'$. Če uporabimo +izrek \ref{invMetr}, dobimo + $\frac{|AB|\cdot r^2}{|PA|\cdot|PB|}=\frac{|CD|\cdot + r^2}{|PC|\cdot|PD|}$ oz. + $\frac{|AB|}{|CD|}=\frac{|PA|\cdot|PB|}{|PC|\cdot|PD|}$. Na + podoben način iz $C'B'\cong A'D'$ sledi + $\frac{|CB|}{|AD|}=\frac{|PC|\cdot|PB|}{|PA|\cdot|PD|}$. Z + množenjem dveh relacij dobimo: + $\frac{|AB|\cdot|BC|}{|AD|\cdot|DC|}=\frac{|PB|^2}{|PD|^2}$. + + \item \res{Naj bo $A$ točka, ki leži na krožnici $k$. Samo + s šestilom načrtaj kvadrat $ABCD$ (oz. njegova oglišča), ki je včrtan dani +krožnici.} + + Najprej načrtamo pravilni šestkotnik $AB_1B_2CD_1D_2$, ki je + včrtan dani krožnici. + + \item \res{Dani sta točki $A$ in $B$. Samo z uporabo + šestila načrtaj takšno točko $C$, da velja + $\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}$.} + + Uporabimo podoben postopek kot v nalogi \ref{MaskeroniSred}. + Najprej narišemo točko $X$, za katero velja + $\overrightarrow{AX}=3\cdot \overrightarrow{AB}$, nato pa iskano + točko + $X'=\psi_k(X)$, kjer je $k$ krožnica s središčem $A$ in polmerom $AB$. + + \item \res{Samo s pomočjo + šestila razdeli dano daljico v razmerju $2:3$.} + + Za dano daljico $AB$ najprej narišemo točko $X$, za katero je + $\overrightarrow{AX}=2\cdot \overrightarrow{AB}$ (podobno kot + pri prejšnji nalogi), nato pa + točko $Y$, za katero velja + $\overrightarrow{AY}=\frac{1}{5}\cdot \overrightarrow{AX}$. $Y$ + je iskana točka, ker velja + $\overrightarrow{AY}=\frac{2}{5}\cdot \overrightarrow{AB}$. +\end{enumerate} +\newpage + + +\normalsize + +%________________________________________________________________________________ +% LITERATURA - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - +%________________________________________________________________________________ +\begin{thebibliography}{1} + + + \bibitem{Berger} Berger, M. +\emph{Geometry}, Springer-Verlag, Berlin 1987. + + \bibitem{Cofman} Cofman, J. +\emph{What to solve?}, Oxford University Press, Oxford, 1990. + + \bibitem{CoxeterRevisited} Coxeter, H. S. M.; Greitzer, S. L. +\emph{Geometry Revisited}, Random House, New York, 1976. + + \bibitem{Djerasimovic} Djerasimovi\'c, B. +\emph{Zbirka zadataka iz geometrije}, Stručna knjiga, Beograd, 1987. + + \bibitem{MMO} Djuki\'c, D.; Jankovi\'c, V.; + Mati\'c, I.; Petrovi\'c, N. +\emph{The IMO Compendium}, Springer, New York, 2006. + + \bibitem{Efimov} Efimov, N. V. +\emph{Višja geometrija}, Nauka, Moskva, 1978. + + \bibitem{Evklid} Evklid, \emph{Elementi}, Naučna knjiga, Beograd, 1949. + + \bibitem{Fetisov} Fetisov, A. I. +\emph{O euklidskoj i neeuklidskim geometrijama}, Školska knjiga, Zagreb, 1981. + + \bibitem{KratkaZgodCasa} Hawking, S. W. +\emph{Kratka zgodovina časa}, DMFA, Ljubljana, 2003. + + \bibitem{Hilbert} Hilbert, D. \emph{Osnove geometrije}, Naučno delo, Beograd, 1957. + + \bibitem{ZutaKnjiga} Lopandi\'c, D. +\emph{Geometrija}, Naučna kniga, Beograd, 1979. + + \bibitem{Lopandic} Lopandi\'c, D. +\emph{Zbirka zadataka iz osnova geometrije}, PMF, Beograd, 1971. + + \bibitem{Lucic} Luči\'c, Z. +\emph{Euklidska i hiperbolička geometrija}, Matematički +fakultet, Beograd, 1994. + + \bibitem{Martin} \emph{Martin, G.}, Transformation Geometry, Springer-Verlang, New York, 1982. + + \bibitem{Mitrovic} Mitrovi\'c, M. +\emph{Projektivna geometrija}, DMFA, Ljubljana, 2009. + + \bibitem{MitrovicMG} Mitrovi\'c, M.; Ognjanovi\'c, S.; Veljkovi\'c, M.; Petkovi\'c, L.; Lazarevi\'c, N. +\emph{Geometrija za prvi razred Matematičke gimnazije}, Krug, Beograd, 1996. + + \bibitem{Nice} Niče, V. +\emph{Uvod u sintetičku geometriju}, Školska knjiga, Zagreb, +1956. + + \bibitem{Prasalov} Prasalov, V. V. \emph{Zadači po geometriji}, Nauka, Moskva, 1986. + + \bibitem{Prvanovic} Prvanovi\'c, M. \emph{Osnove geometrije}, Gradjevinska knjiga, Beograd, 1987. + + \bibitem{Tosic} Toši\'c, R.; Petrovi\'c, V. +\emph{Zbirka zadataka iz osnova geometrije}, Gradjevinska +knjiga, Novi Sad, 1982. + + \bibitem{Stojanovic} Stojanović, V. \emph{Matematiskop III}, Nauka, Beograd, 1988. + + \bibitem{Struik} Struik, D. J. +\emph{Kratka zgodovina matematike}, Državna založba Slovenije, +Ljubljana, 1978. + + \bibitem{Oblika} Weeks, J. R. +\emph{Oblika prostora}, DMFA, Ljubljana, 1998. + +%%%% Pregledano v celoti! Roman + +\end{thebibliography} + +\footnotesize + + \printindex + +\end{document} diff --git a/examples/book_translation/translate_latex_book.ipynb b/examples/book_translation/translate_latex_book.ipynb new file mode 100644 index 0000000..8f27b32 --- /dev/null +++ b/examples/book_translation/translate_latex_book.ipynb @@ -0,0 +1,1163 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Translate a book writen in LaTeX from Slovenian into English\n", + "\n", + "With permission of the author, we will demonstrate how to translate the book [Euclidean Plane Geometry](https://sites.google.com/site/projektivna/), written by Milan Mitrović from Slovenian into English, without modifying any of the LaTeX commands.\n", + "\n", + "To achieve this, we will first split the book into chunks, each roughly a page long, then translate each chunk into English, and finally stitch them back together." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 1. Read in the data" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "1485565" + ] + }, + "execution_count": 19, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "import openai\n", + "from transformers import GPT2Tokenizer\n", + "\n", + "# OpenAI GPT-2 tokenizer is the same as GPT-3 tokenizer\n", + "# we use it to count the number of tokens in the text\n", + "tokenizer = GPT2Tokenizer.from_pretrained(\"gpt2\")\n", + "\n", + "with open(\"data/geometry_slovenian.tex\", \"r\") as f:\n", + " text = f.read()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### 1.1 Count the tokens in each chunk" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": {}, + "outputs": [ + { + "name": "stderr", + "output_type": "stream", + "text": [ + "Token indices sequence length is longer than the specified maximum sequence length for this model (1327 > 1024). Running this sequence through the model will result in indexing errors\n" + ] + }, + { + "data": { + "text/plain": [ + "1473" + ] + }, + "execution_count": 20, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "chunks = text.split('\\n\\n')\n", + "ntokens = []\n", + "for chunk in chunks:\n", + " ntokens.append(len(tokenizer.encode(chunk)))\n", + "max(ntokens)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "It turns out that a double newline is a good separator in this case, in order not to break the flow of the text. Also no individual chunk is larger than 1500 tokens. The model we will use is text-davinci-002, which has a limit of 4096 tokens, so we don't need to worry about breaking the chunks down further.\n", + "\n", + "We will group the shorter chunks into chunks of around 1000 tokens, to increase the coherence of the text, and the frequency of breaks within the text." + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "869" + ] + }, + "execution_count": 21, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "def group_chunks(chunks, ntokens, max_len=1000):\n", + " \"\"\"\n", + " Group very short chunks, to form approximately a page long chunks.\n", + " \"\"\"\n", + " batches = []\n", + " cur_batch = \"\"\n", + " cur_tokens = 0\n", + "\n", + " # iterate over chunks, and group the short ones together\n", + " for chunk, ntoken in zip(chunks, ntokens):\n", + " cur_tokens += ntoken + 2 # +2 for the newlines between chunks\n", + "\n", + " # if adding this chunk would exceed the max length, finalize the current batch and start a new one\n", + " if ntoken + cur_tokens > max_len:\n", + " batches.append(cur_batch)\n", + " cur_batch = chunk\n", + " else:\n", + " cur_batch += \"\\n\\n\" + chunk\n", + " batches.append(cur_batch)\n", + " return batches\n", + "\n", + "chunks = group_chunks(chunks, ntokens)\n", + "len(chunks)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Notice that adding a sample untranslated and translated first command, where only the content of the chapter name needs to be translated, helps to get more consistent results.\n", + "\n", + "The format of the prompt sent to the model consists of:\n", + "1. A high level instruction to translate only the text, but not commands into the desired language\n", + "2. A sample untranslated command, where only the content of the chapter name needs to be translated\n", + "3. The chunk of text to be translated\n", + "4. The translated sample command from 2, which shows the model the beginning of the translation process\n", + "\n", + "The expected output is the translated chunk of text." + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Let $\\mathcal{I}=\\mathcal{S}_{AB} \\circ\\mathcal{S}_{CA}\n", + " \\circ\\mathcal{S}_{BC}$. By \\ref{izoZrcdrsprq} is\n", + " $\\mathcal{I}$ a mirror reflection. Let $A_1$, $B_1$ and $C_1$ be in order the center points of the lines $BC$, $AC$ and $AB$ of the triangle $ABC$.\n", + " Because it is a right triangle is $\\mathcal{I}(A_1C_1)=A_1C_1$, which\n", + " means that the line $A_1C_1$ is of this mirror reflection. It is not\n", + " difficult to prove that for the point $A'_1=\\mathcal{I}(A_1)$ (both\n", + " lie on the axis $A_1C_1$) is\n", + " $\\overrightarrow{A_1A'_1}=3\\overrightarrow{A_1C_1}$, so\n", + " $\\mathcal{I}=\\mathcal{G}_{3\\overrightarrow{A_1C_1}}$.\n", + "\n", + "\\item \\res{Given are the points $A$ and $B$ on the same side of the line\n", + "$p$.\n", + "Draw the line $XY$, which lies on the line $p$ and is consistent\n", + "with the given line $l$, so that the sum\n", + "$|AX|+|XY|+|YB|$ is minimal.}\n", + "\n", + "Let $A'=\\mathcal{G}_{\\overrightarrow{MN}}(A)$ (where $M,N\\in\n", + "p$ and $MN\\cong l$). The point $Y$ is obtained as the intersection of the lines $p$\n", + "and $X'Y$ (see also example \\ref{HeronProbl}).\n", + "\n", + "\\item \\res{Let $ABC$ be an isosceles right triangle with a right angle at the vertex $A$. What does the composite\n", + "$\\mathcal{G}_{\\overrightarrow{AB}}\\circ \\mathcal{G}_{\\overrightarrow{CA}}$ represent?}\n", + "\n", + "Let $p$ and $q$ be the simetrali of the sides $CA$ and $AB$ of the triangle\n", + "$ABC$. By \\ref{izoZrcDrsKompSrOsn} is:\n", + " $$\\mathcal{G}_{\\overrightarrow{AB}}\\circ\n", + " \\mathcal{G}_{\\overrightarrow{CA}}=\n", + " \\mathcal{S}_q\\circ\\mathcal{S}_A\\circ\\mathcal{S}_A\\circ\\mathcal{S}_p=\n", + " \\mathcal{S}_q\\circ\\mathcal{S}_p.$$ Because $ABC$ is an isosceles\n", + " right triangle with a right angle at the vertex $A$, the lines $p$ and $q$ are perpendicular and intersect at the center $S$\n", + " of the hypotenuse $BC$. Therefore\n", + " $\\mathcal{G}_{\\overrightarrow{AB}}\\circ\n", + " \\mathcal{G}_{\\overrightarrow{CA}}=\\mathcal{S}_q\n", + " \\circ\\mathcal{S}_p=\\mathcal{S}_S$.\n", + "\n", + "\\item \\res{In the same plane are given the lines\n", + "$a$, $b$ and $c$.\n", + "Draw the points $A\\in a$ and $B\\in b$\n", + "so that $\\mathcal{S}_c(A)=B$.}\n" + ] + } + ], + "source": [ + "def translate_chunk(chunk, engine='text-davinci-002',\n", + " dest_language='English',\n", + " sample_translation=(\"\\poglavje{Osnove Geometrije} \\label{osn9Geom}\", \"\\poglavje{The basics of Geometry} \\label{osn9Geom}\")\n", + " ):\n", + " prompt = f'''Translate only the text from the following LaTeX document into {dest_language}. Leave all LaTeX commands unchanged\n", + " \n", + "\"\"\"\n", + "{sample_translation[0]}\n", + "{chunk}\"\"\"\n", + "\n", + "{sample_translation[1]}\n", + "'''\n", + " response = openai.Completion.create(\n", + " prompt=prompt,\n", + " engine=engine,\n", + " temperature=0,\n", + " top_p=1,\n", + " max_tokens=1500,\n", + " )\n", + " result = response['choices'][0]['text'].strip()\n", + " result = result.replace('\"\"\"', '') # remove the double quotes, as we used them to surround the text\n", + " return result\n", + "print(translate_chunk(chunks[800], engine='text-davinci-002', dest_language='English'))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can see here that this one chunk in particular translates only the text, but leaves LaTeX commands intact.\n", + "\n", + "Let's now translate all the chunks in the book - this will take 2-3 hours, as we're processing requests sequentially." + ] + }, + { + "cell_type": "code", + "execution_count": 39, + 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" translated_chunks.append(translate_chunk(chunk, engine='text-davinci-002', dest_language=dest_language))\n", + "\n", + "# join the chunks together\n", + "result = '\\n\\n'.join(translated_chunks)\n", + "\n", + "# save the final result\n", + "with open(f\"data/geometry_{dest_language}.tex\", \"w\") as f:\n", + " f.write(result)" + ] + } + ], + "metadata": { + "interpreter": { + "hash": "aee8b7b246df8f9039afb4144a1f6fd8d2ca17a180786b69acc140d282b71a49" + }, + "kernelspec": { + "display_name": "Python 3.9.10 64-bit", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.9.10" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/dbpedia_samples.jsonl b/examples/dbpedia_samples.jsonl new file mode 100644 index 0000000..0e022fa --- /dev/null +++ b/examples/dbpedia_samples.jsonl @@ -0,0 +1,200 @@ +{"text": " Morada Limited is a textile company based in Altham Lancashire. Morada specializes in curtains.", "category": "Company"} +{"text": " The Armenian Mirror-Spectator is a newspaper published by the Baikar Association in Watertown Massachusetts.", "category": "WrittenWork"} +{"text": " Mt. Kinka (\u91d1\u83ef\u5c71 Kinka-zan) also known as Kinkazan is located in the heart of the city of Gifu Gifu Prefecture Japan and rises to a height of 329 m (1079 ft). Previously called Mt. Inaba (\u7a32\u8449\u5c71 Inaba-yama) it has long served as the representative symbol of Gifu. It stands along the Nagara River creating bountiful nature within the city. Though it is the most famous mountain in the city Mount Dodo to the north is the tallest.", "category": "NaturalPlace"} +{"text": " Planning the Play of a Bridge Hand is a book on contract bridge co-written by Canadian teacher and author Barbara Seagram and British author David Bird. It was published by Master Point Press in 2009.The book teaches novice bridge players some basic techniques of declarer play including suit establishment ruffing losers and the finesse.", "category": "WrittenWork"} +{"text": " Wang Yuanping (born 8 December 1976) is a retired Chinese athlete who specialised in the 800 metres. She won several medals at the regional level.Her personal bests in the event are 2:00.63 seconds outdoors (Jinzhou 2000) and 2:03.41 seconds indoors (Yokohama 2004).", "category": "Athlete"} +{"text": " The Incorporated VIllage of Westhampton Beach is an incorporated village in the Town of Southampton Suffolk County New York United States. As of the 2010 census the village population was 1721.", "category": "Village"} +{"text": " Andersons Creek is a creek in Warrandyte and Park Orchards east of Melbourne Victoria Australia. It is a tributary of the Yarra River.", "category": "NaturalPlace"} +{"text": " The Three Horseshoes is a public house in Drybridge Street in the Overmonnow area of Monmouth Wales. The pub has also been used as an Inn and also known as The Three Horse Shoes Inn. The building has been a Grade II Listed building since 15 August 1974. 19th century 2 storeys roughcast as stone with a hooded doorway", "category": "Building"} +{"text": " The Brewer's Art is a Baltimore Maryland brewpub and restaurant. Opened on Friday September 13 1996. In 2008 it was named by Esquire magazine as the #1 Best Bar in America.", "category": "Company"} +{"text": " The P\u00e2r\u00e2ul S\u0103r\u0103\u021bii is a tributary of the Cibin River in Romania.", "category": "NaturalPlace"} +{"text": " Jean-Fran\u00e7ois Imbernon (born October 17 1951 in Perpignan France is a retired French international rugby union player.He played as a Lock for USA Perpignan. He earned his first cap with the French national team on 7 February 1976 against Ireland at Parc des Princes.", "category": "Athlete"} +{"text": " Le Cadeau released in Italy as Il regalo is a 1982 French and Italian film. It stars Claudia Cardinale.", "category": "Film"} +{"text": " Mykola Kanevets (Ukrainian: \u041c\u0438\u043a\u043e\u043b\u0430 \u041a\u0430\u043d\u0456\u0432\u0435\u0446\u044c) is the Artistic Director and Ballet Master of the Cheremosh Ukrainian Dance Company in Edmonton Alberta Canada.A native of Kiev Ukraine Mykola attended the National University of Culture and Performing Arts in Kiev Ukraine where he graduated from the Faculty of Choreography with the distinction of Ballet Master and Choreographer.", "category": "Artist"} +{"text": " Jenna Rose Swerdlow (born September 28 1998) is an American teenage singer who gained media attention as a pre-teen with her single My Jeans. After the video went viral on YouTube and received 14 million views Swerdlow is considered a semi-viral star.", "category": "Artist"} +{"text": " The Spice of Life is a smooth jazz studio album by Earl Klugh released in April 2008. The album received a Grammy nomination for Best Pop Instrumental Album at the 51st Grammy Awards in 2009.", "category": "Album"} +{"text": " Lomatium macrocarpum is a perennial flowering plant in the carrot family known by the common names bigseed lomatium biscuit root or even bigseed biscuitroot. It is native to much of western North America where it can be found in various types of habitat including the grasslands of the Great Plains. It is spreading or erect perennial herb growing up to about half a meter long with hairy gray-green herbage.", "category": "Plant"} +{"text": " Physena is the sole genus of the flowering plant family Physenaceae. It contains two species of shrubs and small trees which are endemic to Madagascar. The APG II system of 2003 (unchanged from the APG system of 1998) does recognize this family and assigns it to the order Caryophyllales in the clade core eudicots.", "category": "Plant"} +{"text": " David John Weatherley (born 1 March 1939) is a New Zealander actor known for his roles as Spencer the butler and the voice of Benglo the Fearcat in Power Rangers Operation Overdrive and Barliman Butterbur in The Lord of the Rings: The Fellowship of the Ring.Weatherley was born in London England and moved to Canada for a military career. He eventually moved to New Zealand to engage in a theatre acting career.", "category": "Artist"} +{"text": " Draba incrassata is an uncommon species of flowering plant in the mustard family known by the common name Sweetwater Mountains draba. It is endemic to California where it is known mainly from the Sweetwater Mountains of Mono County. It grows in alpine rock fields on the barren high mountain peaks. Draba incrassata is a small perennial herb forming mats of thick oval-shaped leaves.", "category": "Plant"} +{"text": " Pimelea ferruginea is a small shrub native to southwest Western Australia. It was described by Labillardiere in 1805.", "category": "Plant"} +{"text": " Lindsay Ell is a country music singer songwriter and guitarist from Calgary Alberta.She performed at the South by Southwest music festival held in Austin Texas in March 2009 the welcome reception of the 2009 Juno Awards held in Vancouver British Columbia and was also a featured artist at the 2010 Winter Olympics.", "category": "Artist"} +{"text": " Scopula fuscata is a moth of the Geometridae family. It is found from south-western Saskatchewan west to British Columbia and south to California and Arizona. The habitat consists of montane areas including foothills.The wingspan is 24-28 mm. The wings and body are light tan sprinkled with darker yellow-brown or grey-brown scales. There is one generation per year with adults on wing in late June and early July in the northern part of the range.", "category": "Animal"} +{"text": " Oxmoor Center is a Louisville Kentucky shopping mall located at 7900 Shelbyville Road in eastern Louisville.", "category": "Building"} +{"text": " Ghostquake (also known as Haunted High) is a 2012 American made-for-television horror film produced by Syfy. The film was directed by Jeffrey Lando and written by Paul A. Birkett and Anthony C. Ferrante. The film stars Danny Trejo and MC Gainey. It follows a group of high school students trying to escape the wrath of a few ghastly spirits following an earthquake at their school Holloman High School.", "category": "Film"} +{"text": " The Masonic Temple in Great Falls Montana is a building from 1914. It was listed on the National Register of Historic Places in 2000.Address is 821 Central Avenue Great Falls Motana 59401 Phone number is 453-9080.Cascade No. 34 meets 2nd and 4th Tuesdays at 7:30pm Sept-June.Euclid No. 58 meets year-round 1st and 3rd Tuesdays at 7:30pm Sept-May 3rd Tuesdays at 7:30pm June-Aug. Delta No. 128 meets 2nd Wednesdays at 7:30pm Sept-June.", "category": "Building"} +{"text": " Harold Frederick Weaver Hawkins (1893-1977) was an English painter who specialized in ambitious sometimes mural-sized modernist allegories of morality for an age of atomic warfare and global over-population.", "category": "Artist"} +{"text": " Robert Murray Waddington (24 October 1927 \u2013 15 March 2007) was Dean of Manchester in the last quarter of the 20th century.Born in Bognor Regis on 24 October 1927 he was educated at Dulwich College Selwyn College Cambridge and Ely Theological College. Ordained in 1954 he began his career at St John\u2019s Bethnal Green. Afterwards he was Chaplain at Slade School in Warwick Queensland. He returned to England in 1959 to join the Oratory of the Good Shepherd an order of celibate priests.", "category": "OfficeHolder"} +{"text": " Jason Gary King (born 13 April 1985 in Maidstone England) is a speedway rider who was formerly the club captain of Newcastle Diamonds in the British Premier League. His brother Daniel is also a speedway rider.", "category": "Athlete"} +{"text": " The African Queen is a 1951 adventure film adapted from the 1935 novel of the same name by C. S. Forester. The film was directed by John Huston and produced by Sam Spiegel and John Woolf. The screenplay was adapted by James Agee John Huston John Collier and Peter Viertel. It was photographed in Technicolor by Jack Cardiff and had a music score by Allan Gray.", "category": "Film"} +{"text": " The Fiat Barchetta (Italian pronunciation: [\u02c8fiat bar\u02c8ketta]) (Type 183) is a roadster produced by the Italian manufacturer Fiat from 1995 to 2005 (though production was paused between May 2002 and 2004). Barchetta in Italian means 'little boat'.", "category": "MeanOfTransportation"} +{"text": " Sardar Vallabhbhai Patel National Memorial is a museum and exhibition centre dedicated to Sardar Vallabhbhai Patel at Moti Shahi Mahal located in Shahibaug Ahmedabad Gujarat. Moti Shahi Mahal was constructed by Mughal emperor Shahjahan between 1618 and 1622. It is surrounded by a garden.", "category": "Building"} +{"text": " Under Cover 2 is the 5th solo album of Joe Lynn Turner released in 1999. Just like Under Cover the album consists mainly of covers of Turner's favourite artists.", "category": "Album"} +{"text": " The Atakora River is a tributary of Lake Volta in Ghana it flows about 60 km east to the Lake Volta. Its entire course is in south Ghana.", "category": "NaturalPlace"} +{"text": " Death from Above is a 2011 horror film by director Bruce Koehler. The film features professional wrestling stars Kurt Angle Sid Eudy James Storm Matt Morgan Terry Gerin and Jessica Kresa.", "category": "Film"} +{"text": " Portraits of Cuba is an album by Cuban musician Paquito D'Rivera released through Chesky Records in 1996. In 1997 the album won D'Rivera the Grammy Award for Best Latin Jazz Performance.", "category": "Album"} +{"text": " Jimmy Cross (17 November 1938 - 8 October 1978) was an American radio producer and singer who attained a minor Billboard Hot 100 hit with the novelty song I Want My Baby Back in 1965. He was born in Dothan Alabama[citation needed] and became the producer of the syndicated radio series Country Concert.I Want My Baby Back was originally issued on the Tollie label and reached #92 on the Billboard Hot 100 in February 1965.", "category": "Artist"} +{"text": " Timothy Floyd Tim Burchett (born August 25 1964) is an American Republican politician currently the mayor of Knox County Tennessee. He previously served in Tennessee General Assembly first in the Tennessee House of Representatives and later in the Tennessee State Senate in which he represented Tennessee's District 7 part of Knox County. On August 5 2010 Burchett was elected mayor of Knox County replacing Mike Ragsdale.", "category": "OfficeHolder"} +{"text": " Daniel Lawrence Dan Whitney (born February 17 1963) best known by his stage name and character Larry the Cable Guy is an American stand-up comedian actor voice actor and former radio personality.", "category": "Artist"} +{"text": " Renealmia is a plant genus in the family Zingiberaceae. Species include: Renealmia alpinia Renealmia aurantifera Renealmia cernua Renealmia dolichocalyx Renealmia oligotricha Renealmia sessilifolia Renealmia thrysoidesE.g. Alpinia nutans was formerly placed herein too.", "category": "Plant"} +{"text": " Jeff Chapman (born July 17 1969 in Brunswick Georgia) is the bass singer for the Kingdom Heirs. He has been a member of the group since 2002. He has previously traveled with Bob Wills The Sound The Anchormen and The Blackwoods.He has twice been nominated as favorite bass in the Singing News fan awards.Chapman has a wife Angie two sons Justin and Sean and daughter Taylor.", "category": "Artist"} +{"text": " Arenaria ursina is a species of flowering plant in the pink family known by the common name Bear Valley sandwort.", "category": "Plant"} +{"text": " Living Fossil is a classic science fiction story on the concepts of human extinction and future evolution by L. Sprague de Camp. It was first published in the magazine Astounding Science-Fiction for February 1939. It first appeared in book form in the anthology A Treasury of Science Fiction (Crown Publishers 1948); it later appeared in the anthologies Gates to Tomorrow (Atheneum 1973) and The SFWA Grand Masters Volume 1 (Tor Books 1999).", "category": "WrittenWork"} +{"text": " Brachyglottis huntii commonly called rautini or Chatham Island Christmas tree is a species in the Asteraceae family and is found only on the Chatham Islands in New Zealand.", "category": "Plant"} +{"text": " Luktvatnet is a lake that lies in the northern part of the municipality of Vefsn in Nordland county Norway. The lake lies between the mountains Korgfjellet and Lukttinden about 5 kilometres (3.1 mi) south of Elsfjord. The European route E06 highway passes along the northern shore of the lake.", "category": "NaturalPlace"} +{"text": " The IAR 79 is a bi-engine bomber military reconnaissance aircraft with a wood and metal structure that saw service in World War II built under licence in Brasov Romania by Industria Aeronautic\u0103 Rom\u00e2n\u0103", "category": "MeanOfTransportation"} +{"text": " Enrico Perucconi (born 4 January 1925 in Morazzone Varese Italy) was an Italian athlete who competed mainly in the 100 metres.", "category": "Athlete"} +{"text": " Central National-Gottesman Inc. is one of the world's largest distributors of pulp paper paperboard and newsprint. The firm's products are sold in over 75 countries through a network of 43 offices located in the United States and abroad. With annual revenues exceeding $3 billion Forbes ranked Central National-Gottesman 137th in its annual list of The Largest Private Companies.", "category": "Company"} +{"text": " The Kout Food Group is a Kuwaiti-based conglomerate founded in 1982.In Kuwait it operates franchises of Burger King Pizza Hut and Taco Bell.Its UK arm Kout Food Group Restaurants UK it operates under brands such as Burger King KFC and Maison Blanc. In August 2013 it acquired the Little Chef chain for \u00a315 million.", "category": "Company"} +{"text": " Fab Five: The Texas Cheerleader Scandal is a Lifetime Television made-for-TV drama film starring Jenna Dewan Ashley Benson and Tatum O'Neal and directed by Tom McLoughlin. The film premiered on August 2 2008. It is based on a true story which occurred at McKinney North High School in McKinney Texas in 2006 in which five teenage cheerleaders became notorious for bullying truancies violations of the school dress code and general disrespect to the school community and authority.", "category": "Film"} +{"text": " Qadi Mahalleh (Persian: \u0642\u0627\u062f\u064a \u0645\u062d\u0644\u0647\u200e also Romanized as Q\u0101d\u012b Ma\u1e29alleh) is a village in Pazevar Rural District Rudbast District Babolsar County Mazandaran Province Iran. At the 2006 census its population was 228 in 59 families.", "category": "Village"} +{"text": " Eungella Dam is one of Queensland's more established freshwater fisheries. Eungella has made a name for producing extra oversized Sooty grunter and more recently Barramundi.Eungella Dam was constructed in 1969 to meet the requirements of a thermal power station at Collinsville and the town water requirement of Collinsville and Scottsville.", "category": "NaturalPlace"} +{"text": " The American Motor Car Company was a short-lived company in the automotive industry founded in 1906 lasting until 1913. It was based in Indianapolis Indiana United States. The American Motor Car Company pioneered the underslung design.", "category": "Company"} +{"text": " Hawkeye & Mockingbird was a comic book ongoing series published by Marvel Comics starring superheroes Hawkeye and Mockingbird.", "category": "WrittenWork"} +{"text": " Margaret Anderson Kelliher (born March 11 1968) is a Minnesota politician and a former member of the Minnesota House of Representatives. A member of the Minnesota Democratic\u2013Farmer\u2013Labor Party she represented District 60A which includes portions of the city of Minneapolis in Hennepin County located in the Twin Cities metropolitan area. First elected in 1999 she served until 2011 also serving as the Speaker from 2007 to 2011.", "category": "OfficeHolder"} +{"text": " John Whitlow Wyatt (September 27 1907 \u2013 July 16 1999) was a professional baseball pitcher. He played all or part of sixteen seasons in Major League Baseball for the Detroit Tigers (1929\u201333) Chicago White Sox (1933\u201336) Cleveland Indians (1937) Brooklyn Dodgers (1939\u201344) and Philadelphia Phillies (1945). While injuries sidetracked much of Wyatt's early career he is most famous for his performance in 1941 when his team (the Dodgers) won the National League pennant.", "category": "Athlete"} +{"text": " William Thomas Burton (31 January 1878 in Black Rock St Michael Barbados \u2013 22 August 1946 St Michael Barbados) was a coloured West Indian cricketer best known as a member of the 1900 and 1906 West Indian tourists to England. He is generally known as Tommie Burton.He was the son of a black mother and a white father. He was brought up in Barbados and served for some years there as a practice bowler and in trial matches.", "category": "Athlete"} +{"text": " Tulemalu Lake is a lake in Kivalliq Region Nunavut Canada.", "category": "NaturalPlace"} +{"text": " Sten Stjernqvist is a Swedish former footballer who played as a forward.", "category": "Athlete"} +{"text": " David Parlett (born 1939) is a games scholar from South London who has studied both card games and board games. His published works include many popular books on games and the more academic volumes The Oxford Guide to Card Games and The Oxford History of Board Games both now out of print. Parlett also invented a number of board games the most successful of which is Hare and Tortoise (1974). The German edition was awarded Spiel des Jahres (Game of the Year) in 1979.", "category": "Artist"} +{"text": " Karl Nabersberg (sometimes written as Carl Nabersberg) was a German youth leader.Nabersberg was the son of a Crefeld shopkeeper. In 1923 he joined the Jugendorganisation the forerunner of the Hitler Youth in his home town. On 28 December 1925 he was admitted as a member of the National Socialist German Workers' Party (member number 26269) and as a member of the Sturmabteilung.", "category": "OfficeHolder"} +{"text": " \u0160etonje is a village situated in Petrovac na Mlavi municipality in Serbia.", "category": "Village"} +{"text": " Dr. Joseph de Graft-Johnson (1933\u20131999) was an engineer academic and politician. He became the Vice-President of Ghana between 1979 and 1981.", "category": "OfficeHolder"} +{"text": " Patties Foods (previously Patties Bakery) is an Australian food manufacturing company that produces meat pies baked goods frozen fruits and pre-made desserts. Headquartered in Bairnsdale Victoria Australia Patties Foods is represented in the Australian market by the Four'N Twenty Patties Herbert Adams Creative Gourmet Nanna's and Chefs Pride brands. Patties is the largest meat pie producing company in Australia and the world.", "category": "Company"} +{"text": " Double Butte is the 2579-foot (786 m) mountain summit distinguished by two buttes (the other at abou 2480 feet or 756 metres) in Riverside County California. It is the western most summit of a mountain range north of Winchester California east of Perris Valley and west of the San Jacinto Valley. The eastern ridge is composed primarily of metamorphic rock of the Triassic - Jurassic French Valley formation.", "category": "NaturalPlace"} +{"text": " Mount Carmel \u2013 Blytheswood Public School is an elementary school in the north end of Leamington Ontario Canada. It is part of the Greater Essex County District School Board and serves students from JK to Grade 8 from the communities of Blytheswood and Mount Carmel and surrounding areas.", "category": "EducationalInstitution"} +{"text": " La combi asesina (The Killer Combination) is a 1982 Mexican film. It was directed by Gustavo Alatriste.", "category": "Film"} +{"text": " Halimium ocymoides (basil-leaved rock rose) syn. Cistus algarvensis is a species of flowering plant in the family Cistaceae native to southern Portugal and southern Spain. It is an erect evergreen shrub growing to 60 cm (24 in) tall by 100 cm (3 ft) wide with woolly grey-green leaves and bright yellow flowers in spring. The flowers may have a dark brown blotch at the base of each petal.In cultivation this plant requires a sandy soil and full sun.", "category": "Plant"} +{"text": " Kaala Patthar (English: Black Stone) is a 1979 Indian Bollywood action/drama film. It was produced and directed by Yash Chopra. The story was written by Salim-Javed. This film is the fourth collaboration between Amitabh Bachchan Shashi Kapoor and director Yash Chopra after the hugely successful Deewaar (1975) Kabhie Kabhie (1976) and Trishul (1978). However this film did average business at the box office. It was nominated for Filmfare awards.", "category": "Film"} +{"text": " Martin G.S. Mansergh (born 31 December 1946) is a former Irish Fianna F\u00e1il politician and historian. He was a Teachta D\u00e1la (TD) for the Tipperary South constituency from 2007 until 2011. He was previously a Senator from 2002 to 2007. He played a leading role in formulating Fianna F\u00e1il policy on Northern Ireland.", "category": "OfficeHolder"} +{"text": " Shriniwas Ganesh Sardesai (1907-1996) popularly known as S.G. Sardesai was an Indian freedom fighter from Maharashtra and one of the great communist leaders produced by the communist movement in India. He is author of the book Progress and conservatism in ancient India famous for his profound theoretical analysis. He was the Central Executive Committee of pre-split Communist Party of India during the Indo-china conflict.", "category": "OfficeHolder"} +{"text": " USS Tuluran (AG-46) \u2013 also known as USS Lake Superior (ID-2995) \u2013 was a commercial cargo ship acquired by the U.S. Navy for service during both World War I when she was known as USS Lake Superior and also during World War II when she was known as USS Tuluran.", "category": "MeanOfTransportation"} +{"text": " The American Journal of Gastroenterology is a peer-reviewed medical journal published for the American College of Gastroenterology by the Nature Publishing Group.", "category": "WrittenWork"} +{"text": " William Lindsay (September 4 1835 \u2013 October 15 1909) was a Democratic U.S. Senator from Kentucky from 1893 to 1901.Born near Lexington Virginia Lindsay attended the common schools and settled in Clinton Kentucky in 1854. There he taught school and studied law. He was admitted to the bar and commenced practice in Clinton in 1858.", "category": "OfficeHolder"} +{"text": " Brian Schroeder (a.k.a. Pushead) is an artist record label owner and writer within the hardcore punk and heavy metal field. He has created artwork for many bands artists and athletes including Metallica The Misfits Dr. Dre Travis Barker Craig Johnson and Kool Keith. He has designed many record covers T-shirts skateboards and a pair of Nike SB Dunks. His record label Pusmort Records has released albums by Negative Gain Poison Idea and Final Conflict.", "category": "Artist"} +{"text": " Panicum anceps is a species of grass known by the common name beaked panicgrass. It is native to the southeastern United States where it occurs as far north as New Jersey and as far west as Kansas and Texas.This species is a rhizomatous perennial grass with stems growing up to 1.3 meters tall. The leaves have erect blades up to half a meter tall. The inflorescence is a panicle up to 40 centimeters long bearing pale green or yellowish spikelets. The grass produces an abundance of seed.", "category": "Plant"} +{"text": " Shukan ST is a weekly newspaper published by The Japan Times for learners of English language. It is originally titled as Student Times but changed to Shukan ST since a significant portion of its readers are not students. It has articles on news movie lifestyle in English-speaking countries opinions and other kinds attracting learners of English and helping them with notes on terms.", "category": "Company"} +{"text": " The Tiger Hotel is a hotel in Columbia Missouri. Built as a hotel in 1928 the building later housed a retirement home and banquet center. In 2012 the building was fully restored and reopened as a boutique hotel. It was listed on the National Register of Historic Places in 1980.", "category": "Building"} +{"text": " Emi Motoi (\u672c\u4e95 \u3048\u307f Motoi Emi born October 11 in Kanagawa) is a Japanese voice actress.", "category": "Artist"} +{"text": " The Hudson River is a 49.5-mile-long (79.7 km) tributary of the Broad River in the U.S. state of Georgia. Via the Broad River it is part of the Savannah River watershed.The headwaters are in Banks County near the city of Homer. Grove Creek feeds into the Hudson near the Franklin County line. The river then constitutes most of the southern border of Franklin County separating it from Madison County.", "category": "NaturalPlace"} +{"text": " This article details Car Nos. 10\u201313 of the Manx Electric Railway on the Isle of Man.This was the third batch of motorcars delivered to the railway in 1895 at the same time as the cars for the new Snaefell Mountain Railway were delivered. They were constructed to a very similar design to those provided for the mountain line.", "category": "MeanOfTransportation"} +{"text": " Catharanthus roseus commonly known as the Madagascar rosy periwinkle is a species of Catharanthus native and endemic to Madagascar. Other English names occasionally used include Cape periwinkle rose periwinkle rosy periwinkle and old-maid.", "category": "Plant"} +{"text": " Thapanzeik is a village in Homalin Township Hkamti District in the Sagaing Region of northwestern Burma.", "category": "Village"} +{"text": " USS Spiegel Grove (LSD-32) was a Thomaston-class dock landing ship of the United States Navy. She was named for Spiegel Grove the home and estate in Fremont Ohio of Rutherford B. Hayes the 19th President of the United States.", "category": "MeanOfTransportation"} +{"text": " Acmella is a genus of thirty species of plants in the aster family Asteraceae. It is native to the Americas and has been introduced to Asia Africa the Pacific islands and Australia.One familiar species is Acmella oleracea which has been widely cultivated for centuries. It is used for food and medicine and as an insecticide and an ornamental plant.", "category": "Plant"} +{"text": " Mirbelia is a plant genus belonging to the Fabaceae family. It is endemic to Australia occurring in every mainland state except South Australia.", "category": "Plant"} +{"text": " Nigma puella is a species of spider belonging to the family Dictynidae. It is found in Europe Azores Madeira Canary Islands and parts of North Africa.Like most members of the family this is a small spider but the female is striking with a light green abdomen marked with a bold maroon blotch and a variable amount of barring in the same colour. The male is reddish-brown. This species makes a horizontal web over the top surface of a leaf.", "category": "Animal"} +{"text": " The Madrisa (or Madrisahorn) is a mountain in the R\u00e4tikon mountain range overlooking Klosters in the Swiss canton of Graub\u00fcnden. Its summit (2826 metres) is located near the Austrian border.The Madrisa is constituted by several secondary summits notably the Gargeller Madrisa (2770 metres) overlooking Gargellen in Austria.Ski lifts up to 2600 metres are located on the Klosters side.", "category": "NaturalPlace"} +{"text": " Temporary Temple is a live album by Psychic TV. The album was recorded on July 28 1984 in London and released on 12 vinyl. It was later coupled with another concert and released on CD as Temporary Temple & Atonal.", "category": "Album"} +{"text": " La hija de Juan Sim\u00f3n (Juan Sim\u00f3n's Daughter) is a musical play by Nemesio M. Sobrevila which has been made into two Spanish films. It is also the name of the title track and the song has been recorded by numerous artists such as Leonardo Favio.The first film directed by Jos\u00e9 Luis S\u00e1enz de Heredia was released in 1935 and starred Angelillo Pilar Mu\u00f1oz and Manuel Arb\u00f3. Luis Bu\u00f1uel was the executive producer for Film\u00f3fono and had a small role as an actor.", "category": "Film"} +{"text": " Book Of Matches is a poetry book written by Simon Armitage first published in 1993 by Faber and Faber. Several poems featured in the book are studied as part of the GCSE English Literature examination in the UK.The book is written in three sections the first (Book of Matches) containing 30 short sonnets. Each is meant to be read within 20 seconds the amount of time it would take for a match to be lit and burn out.", "category": "WrittenWork"} +{"text": " The Last Supper is the fourth album released by American stand-up comedian Jim Gaffigan. It focuses largely on his love of food.", "category": "Album"} +{"text": " The Miami Center is a skyscraper in downtown Miami Florida. Although Miami Center is not the city's tallest building it is a symbol of early downtown. Built in 1983 it is older compared with most of the taller buildings in Miami which have been built in the last decade. In addition the Miami Center is immediately adjacent to Bayfront Park and is unobstructed when looking at the skyline from Miami Beach to the east. The building is 484 ft (148 m) tall and has 34 floors.", "category": "Building"} +{"text": " Duboisia hopwoodii is a shrub native to the arid interior region of Australia. Common names include pituri pitchuri thornapple or pitcheri. It has an erect habit usually growing to between 1 and 3 metres in height and has long narrow leaves. Flowers are white and bell-shaped with violet-striped throats. These appear between June and November in the species native range followed by purple-black rounded berries which are 3 to 6 mm in diameter.", "category": "Plant"} +{"text": " Jelenin svet (Jelena's World) is a 2008 independent documentary film written and directed by Tanja Brzakovi\u0107 about former World No. 1 female tennis player Jelena Jankovi\u0107.", "category": "Film"} +{"text": " Jay Cashman Inc. is an American heavy-construction company based in Quincy Massachusetts with satellite offices in Boston Jupiter Florida and Staten Island New York. As of 2006 the company has about 1000 employees. The company was one of the major contractors on the Boston's Central Artery/Tunnel Project. In 2004 Jay Cashman Inc.", "category": "Company"} +{"text": " Hashemanli (Persian: \u0647\u0627\u0634\u0645\u0646\u0644\u064a\u200e also Romanized as H\u0101shemanl\u012b; also known as H\u0101shem El\u00e1) is a village in Jafarbay-ye Jonubi Rural District in the Central District of Torkaman County Golestan Province Iran. At the 2006 census its population was 135 in 27 families.", "category": "Village"} +{"text": " Rani Kasula Rangamma is a Telugu film starring Chiranjeevi.", "category": "Film"} +{"text": " The 20/20 Experience \u2013 The Complete Experience is a compilation album by American singer-songwriter Justin Timberlake. It was released on September 27 2013 by RCA Records.", "category": "Album"} +{"text": " R.C. Bigelow Inc better known as the Bigelow Tea Company is an American tea company based in Fairfield Connecticut. The company was founded by Ruth C. Bigelow in the late 1940s based on a recipe she marketed as Constant Comment tea. Bigelow is still a 100% family-owned business that markets over 50 varieties of tea including black and green as well as herbal teas all of which are still blended in Fairfield. They also own America's only tea plantation in Charleston South Carolina.", "category": "Company"} +{"text": " Thomas Eyre(fl. 1890s) was a footballer who made 65 appearances in the Football League playing for Lincoln City. He played at left back. Either side of Lincoln he played for Ashfield and Hamilton Academical in Scotland.", "category": "Athlete"} +{"text": " Malleable Iron Range Company was a company that existed from 1896 to 1985 and primarily produced kitchen ranges made of malleable iron but also produced a variety of other related products. The company's primary trademark was 'Monarch' and was colloquially often referred to as the Monarch Company or just Monarch.", "category": "Company"} +{"text": " The Chiltern School is a coeducational special school located over two sites in Dunstable and Houghton Regis in Bedfordshire England. The school accepts pupils from all over the Central Bedfordshire area.The school was formed in 2012 from the merger of Glenwood School in Dunstable and Hillcrest School in Houghton Regis.", "category": "EducationalInstitution"} +{"text": " Kim Dae-Eun (born September 17 1984) is a South Korean gymnast. He is the 2004 Olympic All-around silver medalist. He won the gold medal on the parallel bars at the 2007 World Artistic Gymnastics Championships.Kim was part of the South Korean team that won the bronze medal in the team event at the 2006 Asian Games.", "category": "Athlete"} +{"text": " Arayik Vladimirovich Harutyunyan (Armenian: \u0531\u0580\u0561\u0575\u056b\u056f \u0540\u0561\u0580\u0578\u0582\u0569\u0575\u0578\u0582\u0576\u0575\u0561\u0576 Russian: \u0410\u0440\u0430\u0438\u043a \u0410\u0440\u0443\u0442\u044e\u043d\u044f\u043d) (born 14 December 1973) is the current Prime Minister of the Nagorno-Karabakh Republic. He was suggested by the President of Nagorno-Karabakh Bako Sahakyan and was unanimously approved by the Parliament of Karabakh on 14 September 2007 by 32 out of 32 present parliamentarians.", "category": "OfficeHolder"} +{"text": " Shelton Hank Williams also known as Hank Williams III and Hank 3 (born December 12 1972) is an American musician singer and multi-instrumentalist including guitar bass drums banjo and vocals. In addition to his honky tonk recordings Williams' style alternates between country punk and metal.", "category": "Artist"} +{"text": " Helicella orzai is a species of air-breathing land snails terrestrial pulmonate gastropod mollusks in the family Hygromiidae the hairy snails and their allies. This species is endemic to Spain.", "category": "Animal"} +{"text": " Gro\u00dfe Schmalenau is a river of North Rhine-Westphalia Germany.", "category": "NaturalPlace"} +{"text": " The Tupolev ANT-29 (military designation DIP \u2013 Dvukhmotorny istrebitel pushechny twin-engined cannon fighter) was a 1930s twin-engined cannon-armed fighter designed by Alexander Arkhangelsky and built by Tupolev.Design work started in 1932 on a twin-engined aircraft capable of carrying two APK-100 cannons. The resulting design was the ANT-29 and it first flew in February 1935. A monoplane with a tall and narrow fuselage and powered by two Hispano-Suiza 12Ybrs engines.", "category": "MeanOfTransportation"} +{"text": " Charles Corm (1894-1963) was a Lebanese writer businessman and philanthropist. He is considered to be the leader of the Phoenicianism movement in Lebanon which ignited a surge of nationalism that led to Lebanon's independence. In a country torn by sectarian conflicts Corm's intention was to find a common root shared by all Lebanese beyond their religious beliefs (the Phoenicians were pagans).", "category": "Artist"} +{"text": " Joseph Hubert Ruetz (October 21 1916 \u2013 January 2 2003) was a professional football player in the All-America Football Conference for the Chicago Rockets in 1946 and 1948. Prior to that he played at the collegiate level while attending the University of Notre Dame. He played guard for the Irish with the exception of playing one season at quarterback. In 1938 he graduated from Notre Dame with cum laude honors.", "category": "Athlete"} +{"text": " The Reef House is a historic house located at 411 S. Poplar St. in Carbondale Illinois. William A. Reef built the house for his family circa 1892. The Queen Anne-style cottage may have been designed by local carpenter A. M. Etherton though records of its designer do not exist. The house features fishscale shingle siding on its second floor and clapboard siding on its first; the clapboard siding is adorned with stickwork.", "category": "Building"} +{"text": " MAKO Surgical Corp. (Stryker Medical) is a publicly traded medical device company based in Florida. On September 25 2013 the Board of Directors of Mako Surgical accepted a deal to merge with Stryker Medical for $1.65B subject to shareholder approval.", "category": "Company"} +{"text": " Pop Carn is a 2003 Indian Tamil film written and directed by actor-cum-director Nassar and starring Mohanlal and Simran Bagga in lead roles and introducing newcomers Kunal Shah and Jyothi Nawal. The film which had music scored by Yuvan Shankar Raja was released on 30 January 2003 but flopped at the box office. Nonetheless the film was dubbed into Malayalam and released in 2007 under the same name.", "category": "Film"} +{"text": " USNS Mount Baker (T-AE-34) is the seventh of eight Kilauea-class ammunition ships to serve with the Military Sealift Command. She is the second U.S. Navy ship to bear the name and is named for Mount Baker a 10781-foot volcano in the Cascade Range of Washington. Ammunition ships operated by Military Sealift Command provide logistic support to US Navy ships at sea.Mount Baker was built by Ingalls Shipbuilding Pascagoula Mississippi.", "category": "MeanOfTransportation"} +{"text": " Dansere is an album by Jan Garbarek. The album was recorded in November 1975 and features the Bobo Stenson Quartet.", "category": "Album"} +{"text": " Divraz (Persian: \u062f\u064a\u0648\u0631\u0632\u200e also Romanized as D\u012bvraz) is a village in Bala Khiyaban-e Litkuh Rural District in the Central District of Amol County Mazandaran Province Iran. At the 2006 census its population was 393 in 95 families.", "category": "Village"} +{"text": " The D\u0103ih\u0103\u021ba\u0219u River is a tributary of the Dumbr\u0103vanu River in Romania.", "category": "NaturalPlace"} +{"text": " Zeisters also known as Fat Guy Goes Nutzoid is a 1986 comedy film produced by Troma Entertainment. Troma was originally set to title the film Fat Boy Goes Nutzoid but at the request of the lawyers of the hip-hop group The Fat Boys it was changed to Fat Guy.", "category": "Film"} +{"text": " Paul Gobeil (born March 1 1942 in Saint-R\u00e9mi-de-Tingwick Quebec) is a former Canadian politician and businessman.From 1985 to 1989 Mr. Gobeil was a Liberal member of the National Assembly for the riding of Verdun and served as Minister assigned to Administration President of the Treasury Board and as Minister of International Affairs for the Government of Quebec.", "category": "OfficeHolder"} +{"text": " Ruff Ryders: Past Present Future is the fifth compilation album from American hip hop record label Ruff Ryders Entertainment released on November 21 2011.", "category": "Album"} +{"text": " Ridi Viharaya (Sinhala: \u0dbb\u0dd2\u0daf\u0dd3 \u0dc0\u0dd2\u0dc4\u0dcf\u0dbb\u0dba) or Silver Temple is a 2nd-century BCE Theravada Buddhist temple in the village of Ridigama Sri Lanka. Built during the reign of Dutthagamani of Anuradhapura the temple is considered as the place where the silver ore which provided silver to complete Ruwanwelisaya; one of the largest stupa in Sri Lanka was discovered.", "category": "Building"} +{"text": " Grand Canyon Preparatory Academy is a public charter college preparatory school in Tempe Arizona.", "category": "EducationalInstitution"} +{"text": " Aricoceras is an extinct genus of the Adrianitidae family. They are an extinct group of ammonoid which are shelled cephalopods related to squids belemnites octopuses and cuttlefish and more distantly to the nautiloids.", "category": "Animal"} +{"text": " Blackburn High School is a public secondary school for girls and boys in years 7 to 12 in Blackburn a suburb of Melbourne Victoria Australia. Blackburn High School is an outstanding secondary school for aspiring young men and women. It aims to educate tomorrow's minds today. Started in 1956 the school has a proud tradition of academic excellence and exceptional music achievement.The school is nationally recognised as a leading educational institution for music education.", "category": "EducationalInstitution"} +{"text": " Chris Nieratko (born February 19 1976) is an American humorist and author. Nieratko is a past editor of Big Brother Magazine and currently reviews pornographic films for Vice magazine as well as being the author of the related Skinema book. He also appeared on MTV's Jackass.", "category": "Artist"} +{"text": " Warlock is a 1959 film released by Twentieth Century Fox and shot in DeLuxe Color and CinemaScope. It is a Western adapted from the novel by Oakley Hall (screenplay written by Robert Alan Aurthur).", "category": "Film"} +{"text": " Sieniawa [\u0255e\u02c8\u0272ava] is a village in the administrative district of Gmina Raba Wy\u017cna within Nowy Targ County Lesser Poland Voivodeship in southern Poland. It lies approximately 5 kilometres (3 mi) south-east of Raba Wy\u017cna 10 km (6 mi) north-west of Nowy Targ and 59 km (37 mi) south of the regional capital Krak\u00f3w.The village has a population of 1900.", "category": "Village"} +{"text": " Michael Adam (born 9 December 1984) is a German politician. He has been District Administrator (Landrat) of Regen since 2011.", "category": "OfficeHolder"} +{"text": " Thunderbird High School is a public high school located in northwestern Phoenix Arizona. The school is a part of the Glendale Union High School District.", "category": "EducationalInstitution"} +{"text": " Nayef Al Khater (born May 10 1978) is a Qatari football player. He currently plays for Al Wakrah as a defender.", "category": "Athlete"} +{"text": " Black Cobra Woman (Italian: Eva nera) also known as Black Cobra is an Italian 1976 exploitation movie written and directed by Joe D'Amato.", "category": "Film"} +{"text": " Joe Cuba a.k.a Sonny (April 22 1931 \u2013 February 15 2009) was a musician of Puerto Rican descent who was known as the Father of Latin Boogaloo.", "category": "Artist"} +{"text": " Jacob LeBlanc (born February 2 1981 in Auburn California) is an American retired professional soccer player.", "category": "Athlete"} +{"text": " Kevin B. Kamenetz is the 12th and current County Executive of Baltimore County Maryland serving since 2010. He is a member of the Democratic Party. He previously served as a four-term County Councilman representing the Second District of Baltimore County.", "category": "OfficeHolder"} +{"text": " Thomas Frederick Fred Peart Baron Peart PC (30 April 1914 \u2013 26 August 1988) was a British Labour politician who served in the Labour governments of the 1960s and 1970s and was a candidate for Deputy Leader of the Party.", "category": "OfficeHolder"} +{"text": " Grand Lake is a large lake in the interior of Newfoundland of the Canadian province of Newfoundland and Labrador. It has an area of 534 km\u00b2 making it the largest lake on Newfoundland. Consequently it is one of if not the deepest.", "category": "NaturalPlace"} +{"text": " A Colossal Failure of Common Sense: The Inside Story of the Collapse of Lehman Brothers is a 2009 non-fiction book written by Lawrence G. McDonald and Patrick Robinson which chronicles the events surrounding the bankruptcy of Lehman Brothers in the context of the financial crisis of 2007\u20132010 and the subprime mortgage crisis.", "category": "WrittenWork"} +{"text": " Thatching (31 May 1975 \u2013 1999) was an Irish Thoroughbred racehorse and sire. The horse's early career was delayed and disrupted by injury and he did not show his best form until switched to sprinting distances in the spring of 1979 when he won the Duke of York Stakes. He improved further when equipped with blinkers that summer recording impressive victories in both the Cork and Orrery Stakes and the July Cup.", "category": "Animal"} +{"text": " Ya\u015far Kurt (Armenian: \u0545\u0561\u0577\u0561\u0580 \u053f\u0578\u0582\u0580\u0569 b.August 16 1968 in Istanbul Turkey) is a Turkish-Armenian rock artist.", "category": "Artist"} +{"text": " Then and Now is a historical novel by W. Somerset Maugham. Set in Florence Italy during the Renaissance the story focuses on three months in the life of Niccolo Machiavelli the Florentine politician diplomat philosopher and writer in the early years of the 16th century. The book was first published by Heinemann in 1946.", "category": "WrittenWork"} +{"text": " Abdollah Masud-e Sofla (Persian: \u0639\u0628\u062f\u0627\u0644\u0647 \u0645\u0633\u0639\u0648\u062f\u0633\u0641\u0644\u064a\u200e also Romanized as \u2018Abdoll\u0101h Mas\u2018\u016bd-e Sofl\u00e1; also known as Abdollah Mas\u2019ood \u2018Abdoll\u0101h Mas\u2018\u016bd and Abdull\u0101h Mas\u016bd) is a village in Hesar-e Valiyeasr Rural District Avaj District Buin Zahra County Qazvin Province Iran. At the 2006 census its population was 72 in 15 families.", "category": "Village"} +{"text": " Springhill High School (SHS) is a secondary school in Springhill Nova Scotia Canada. SHS is part of the Chignecto-Central Regional School Board and is the only high school in the town of Springhill. The school is home to many sports teams and clubs. These include: basketball soccer badminton track and field softball Students Against Destructive Decisions drama club homework club and book club.", "category": "EducationalInstitution"} +{"text": " Charniele L. Herring (/\u0283\u0251r\u02c8n\u025bl \u02c8h\u025br\u026a\u014b/ shar-NEL HERR-ing; born September 25 1969) is an American politician. She has served in the Virginia House of Delegates since 2009 representing the 46th district made up the city of Alexandria and part of Fairfax County near Washington D.C. Herring is a member of the Democratic Party. She has been the House minority whip since 2012 and in December 2012 she was the first African-American to be elected chair of the Democratic Party of Virginia.", "category": "OfficeHolder"} +{"text": " Symmoca dodecatella is a moth of the Symmocidae family. It is found in Portugal and Spain.The wingspan is about 18\u201319 mm. The forewings are grey sprinkled with black mainly along the margin. The hindwings are grey.", "category": "Animal"} +{"text": " Ali Abbasov Mammad oglu (Azerbaijani: \u018fli Abbasov M\u0259mm\u0259d o\u011flu) (born 1953 in Azerbaijan) is the current Minister of Communications and Information Technologies of the Republic of Azerbaijan.", "category": "OfficeHolder"} +{"text": " Worlds Beyond was an American digest magazine of science fiction and fantasy fiction in 1950 and 1951. The magazine only issued three monthly issues from December 1950 to February 1951 but is notable for having printed stories by Cyril M. Kornbluth Jack Vance Mack Reynolds Graham Greene John Christopher Lester del Rey Judith Merril and others.Worlds Beyond was published by Hillman Periodicals and was edited by Damon Knight.", "category": "Company"} +{"text": " The Daily News Journal commonly abbreviated to DNJ is a newspaper serving Murfreesboro Tennessee Rutherford County and surrounding communities. Published in Murfreesboro it serves as the primary local newspaper with competition from The Murfreesboro Post and other publications. The newspaper is not in competition with The Tennessean of Nashville as both are owned by Gannett Company.The roots the DNJ date back to 1849 and the founding of Murfreesboro News.", "category": "WrittenWork"} +{"text": " Echinocereus fendleri is a species of cactus known by the common names pinkflower hedgehog cactus and Fendler's hedgehog cactus. It grows in deserts and woodlands in the Southwestern United States and Northeastern Mexico. It is most common in New Mexico.The taxonomy of the species is uncertain with authors recognizing up to eight varieties.", "category": "Plant"} +{"text": " Michael F. Kitt Snr (13 September 1914 \u2013 24 December 1974) was an Irish Fianna F\u00e1il politician and long-serving Teachta D\u00e1la (TD).He was elected to D\u00e1il \u00c9ireann for the first time at the 1948 general election for the Galway North constituency but lost his seat at the 1951 general election and failed to be elected again at the 1954 general election.", "category": "OfficeHolder"} +{"text": " Epidendrum mancum is an epiphytic orchid that grows in the tropical low elfin cloud forests of Ecuador and Amazonas Peru at altitudes of 2\u20143 km .", "category": "Plant"} +{"text": " Salempur Masanda is a village in Jalandhar District near the Jalandhar Cantonment in Punjab India.", "category": "Village"} +{"text": " Yaleh Gonbad (Persian: \u064a\u0644\u0647 \u06af\u0646\u0628\u062f\u200e; also known as Em\u0101mz\u0101deh Imamzade-Ele-Geumbez and Im\u0101mz\u0101deh) is a village in Ilat-e Qaqazan-e Gharbi Rural District Kuhin District Qazvin County Qazvin Province Iran. At the 2006 census its population was 429 in 105 families.", "category": "Village"} +{"text": " Popeyes Louisiana Kitchen is an American chain of fried chicken fast food restaurants founded in 1972 in New Orleans Louisiana. Often referred to as Popeyes and sometimes as Popeyes Chicken & Biscuits or Popeyes Chicken & Seafood[citation needed] it was acquired by Sandy Springs Georgia-based AFC Enterprises originally America's Favorite Chicken Company in 1993.", "category": "Company"} +{"text": " The White Umfolozi River originates just south of Vryheid KwaZulu-Natal South Africa and joins the Black Umfolozi River at 28\u00b020\u203258\u2033S 31\u00b058\u203246\u2033E to form the Umfolozi River before it flows east towards the Indian Ocean.", "category": "NaturalPlace"} +{"text": " The Albatros L 74 was a two-seated German training biplane produced by Albatros Flugzeugwerke. Only two were produced.", "category": "MeanOfTransportation"} +{"text": " The University of Nevada School of Medicine is an academic division of the University of Nevada Reno and grants the Doctor of Medicine (MD) degree. The School of Medicine was founded in 1969 as the first medical school in the state of Nevada. More than 1500 MDs have graduated from the School of Medicine. The pre-clinical campus is located in Reno but the third and fourth years can be spent in hospitals and clinics throughout Nevada.", "category": "EducationalInstitution"} +{"text": " Leon Kroll (December 6 1884 \u2013 October 25 1974) was an American painter and lithographer. Known as a figurative artist Life Magazine described him as the dean of U.S. nude painters yet he was an exceptional landscape painter and also produced an exceptional body of still life compositions.Born into a musical family on lower Second Avenue in New York City Kroll's father was a violinist and his cousin was William Kroll.", "category": "Artist"} +{"text": " Michael E. DeBakey High School for Health Professions at Qatar (DHSHP@Q in short) is a private international middle and secondary school in Doha Qatar. The school is a branch campus of Michael E. DeBakey High School for Health Professions of Houston Texas United States. Charlesetta Deason is the CEO and President.Named after Michael E. DeBakey the school opened in September 2008 with grades 8 through 10 with 100 students per grade; the school will ultimately cover grades 7-12.", "category": "EducationalInstitution"} +{"text": " The Richleighs of Tantamount is a children\u2019s historical novel written by British author Barbara Willard. It was originally published in the United Kingdom in 1966 by the publishers Constable before being published in the United States by Harcourt Brace & World in June 1967. C. Walter Hodges drew the line illustrations and painted the cover portrait for the original edition.", "category": "WrittenWork"} +{"text": " Ennea is a genus of air-breathing land snails terrestrial pulmonate gastropod mollusks in the family Streptaxidae.Ennea is the type genus of the subfamily Enneinae.", "category": "Animal"} +{"text": " Come Live with Me is a 1941 American romantic comedy film produced and directed by Clarence Brown and starring James Stewart and Hedy Lamarr. Based on a story by Virginia Van Upp the film is about a beautiful Viennese refugee seeking United States citizenship who arranges a marriage of convenience with a struggling writer.", "category": "Film"} +{"text": " St. Thomas Episcopal Church is a parish church in the Episcopal Diocese of Iowa. The church is located in Sioux City Iowa United States at 1200 Douglas Street. The church building is listed on the National Register of Historic Places.", "category": "Building"} +{"text": " Nuno Daniel Costeira Valente (born 22 November 1991 in Ada\u00fafe - Braga ) is a Portuguese footballer who plays for Vizela on loan from S.C. Braga as a midfielder.", "category": "Athlete"} +{"text": " Jaagoo (Bengali: \u099c\u09be\u0997\u09cb) is a Bangladeshi sports based romantic Movies. Its writer and director Khijir Hayat Khan. Adnan Karim sponsored youth football game built this film was released in 2010. The film is produced by Sharjeel Karim and Adnan Karim and directed by Khijir Hayat Khan who has written the story screenplay and dialogues of the film. The film features Ferdous Ahmed and Afsana Ara Bindu in lead roles and with supporting Arefin Shuvo Tariq Anam Ronok Hasaan and many more.", "category": "Film"} +{"text": " John Edward Hatton AO (born 29 May 1933) is former Australian politician and an National Trust of Australia nominated Australian Living Treasure. He was the independent member of the Legislative Assembly of the New South Wales parliament for the seat of South Coast from 1973 to 1995. Notably the allegations about police corruption Hatton raised in Parliament resulted in the Wood Royal Commission. He is currently a social activist in his local community.", "category": "OfficeHolder"} +{"text": " Trichoptilus subtilis is a moth of the Pterophoridae family that is known from South Africa.", "category": "Animal"} +{"text": " Sin\u00e9ad Madden (born in Galway Ireland) is an Irish singer-songwriter and fiddle player best known as a member of the Moya Brennan band. She also teaches at Waltons New School of Music in Dublin.", "category": "Artist"} +{"text": " Philip Sprint is a German footballer who currently plays for Hertha BSC.Sprint made his professional debut for Hertha BSC on 12 August 2012 in a 2. Bundesliga match against FSV Frankfurt coming on in the 50th minute for Marvin Knoll after starting goalkeeper Sascha Burchert had been sent off.", "category": "Athlete"} +{"text": " River Roads Mall was an enclosed shopping mall located in the city of Jennings a suburb of St. Louis Missouri United States. Opened in 1962 as one of the nation's first shopping malls the mall declined in the 1990s becoming a dead mall and eventually being shuttered in 1995. Demolition of the long-vacant mall began in 2006.", "category": "Building"} +{"text": " The Brown-patched Kangaroo lizard (Otocryptis wiegmanni) also called Wiegmann's Agama or Sri Lankan Kangaroo Lizard is a small ground dwelling agamid lizard endemic to the wet zone forests and lower mountain forests (up to 1300 metres) of Sri Lanka. It is commonly seen in the leaf litter of shady rain forests. When perceiving danger it spurts away quickly on its large hind legs and might eventually climb up a sapling or tree.", "category": "Animal"} +{"text": " Shiho Kawaragi (\u6cb3\u539f\u6728 \u5fd7\u7a42 Kawaragi Shiho born April 29 1976 in Tokyo) is a Japanese voice actress who works for Kenyu-Office. When voicing adult games and hentai OVAs she is also known as Kaname Yuzuki (\u67da\u6728\u304b\u306a\u3081 Yuzuki Kaname) She is currently married since March 2012.", "category": "Artist"} +{"text": " Down in the Shacks Where the Satellite Dishes Grow is the second album by the Judybats released in 1992.", "category": "Album"} +{"text": " Turn of Faith is a 2001 film directed by Charles Jarrott. It stars Ray Mancini and Mia Sara.", "category": "Film"} +{"text": " Frederick William Seward (July 8 1830 \u2013 April 25 1915) was the Assistant Secretary of State during the American Civil War serving in Abraham Lincoln's administration as well as under Andrew Johnson during Reconstruction and for over two years under Rutherford B. Hayes.", "category": "OfficeHolder"} +{"text": " Ivoprop Corporation founded in 1984 by Ivo Zdarsky is an American manufacturer of composite propellers for homebuilt and ultralight aircraft as well as airboats. The company headquarters is located in Bellflower California.Zdarsky started the company after carving his own propeller for a homebuilt ultralight trike that he flew from Cold War Czechoslovakia over the Iron Curtain to Vienna in 1984.", "category": "Company"} +{"text": " Wave Broadband is a provider of residential business and enterprise class cable TV broadband internet and telephone services on the West Coast currently serving about 400000 customers within communities in western Washington state Oregon Sacramento California and the San Francisco Bay Area. Wave Broadband provides services via their fiber-optic network and uses Northwest Open Access Network as the backbone for most of their service areas in Washington.", "category": "Company"} +{"text": " Andie Tong is a comic book artist known for his work on books such as Spectacular Spider-Man UK The Batman Strikes! and Tangent: Superman's Reign.", "category": "Artist"} +{"text": " Merdani is a village in the municipality of Busova\u010da Bosnia and Herzegovina.", "category": "Village"} +{"text": " Kamam (Persian: \u0643\u0627\u0645\u0645\u200e also Romanized as K\u0101mam) is a village in Mangur-e Sharqi Rural District Khalifan District Mahabad County West Azerbaijan Province Iran. At the 2006 census its population was 98 in 16 families.", "category": "Village"} +{"text": " Ficus greiffiana is a species of plant in the Moraceae family. It is found in Argentina Brazil Colombia and Guyana.", "category": "Plant"} +{"text": " Toni Amboaje is a Spanish singer who currently works for metal band Sauze which formed on early 2008.", "category": "Artist"} +{"text": " Mount Whittier is a mountain in Carroll County New Hampshire in the northern Ossipee Mountains. Named after John Greenleaf Whittier the peak is not to be confused with nearby Nickerson Mountain which was once known as Mount Whittier.There are no hiking trails on Mount Whittier. There was once a CCC alpine ski trail on the northern face.", "category": "NaturalPlace"} +{"text": " El Rompe Discoteka: The Mix Album is a 2007 album by Hector El Father.", "category": "Album"} +{"text": " e-Spirit is a commercial software company that develops and markets the FirstSpirit CMS Web content management system. The company was founded in 1999 in Dortmund Germany and established a US presence in 2011. The company's FirstSpirit CMS is a Java-based offering now in its fifth major release.[citation needed]", "category": "Company"} +{"text": " The Valley is the first novel by Barry Pilton published in 2005 by Bloomsbury. It is a humorous account of the effect of outsiders on the rural status quo in a fictional mid-Wales valley during the 1980s and is being adapted for television.", "category": "WrittenWork"} +{"text": " Sema Group plc was an Anglo-French IT services company. It was listed on the London Stock Exchange and was a constituent of the FTSE 100 Index but was acquired by Schlumberger in 2001.", "category": "Company"} +{"text": " Bent Hansen (born 1954) is a retired Danish ice hockey forward. He played for 18 years in Denmark for the R\u00f8dovre SIK and KSF. He also competed for the Danish national team. His son Jannik Hansen also played for the R\u00f8dovre team and was drafted into the NHL by the Vancouver Canucks in 2004. During his hockey career Hansen also worked as a carpenter.", "category": "Athlete"} +{"text": " Behind the Sun is a 2004 album by Dive.", "category": "Album"} +{"text": " Mungaru Male (English: Pre Monsoon Rain) is a 2006 Kannada language movie directed by Yograj Bhat and produced by E Krishnappa. The film stars Ganesh Pooja Gandhi Anant Nag Padmaja Rao in lead roles.", "category": "Film"} +{"text": " Megachile perihirta commonly known as the Western leafcutting bee is a bee in the genus Megachile. The bee is native to western North America ranging from Nebraska to Texas and Mexico west to California and north to British Columbia and Alberta and often inhabits meadows and orchards. The bee is black with long whitish-yellow hair more so below the thorax and abdomen. The abdomen however is mostly bare although each segment has scattered whitish hair.", "category": "Animal"} +{"text": " Sukeban Deka The Movie (\u30b9\u30b1\u30d0\u30f3\u5211\u4e8b) is a live action Japanese film that was released in 1987. The movie closely follows a TV and manga series Sukeban Deka written and illustrated by Shinji Wada. The movie stars Yoko Minamino and Yui Asaka who were also in the TV series. The movie was followed by Sukeban Deka II in 1988.", "category": "Film"} +{"text": " The Maple School District is a public school district in Douglas County Wisconsin United States based in Maple Wisconsin.", "category": "EducationalInstitution"} +{"text": " Mount Waverley Secondary College is a public secondary school located in the Melbourne suburb of Mount Waverley. The school consists of roughly 1900 students and is one of the largest in the state.The school consists of two campuses (Junior & Senior) both situated on Stephensons Road in Mount Waverley. The Junior site holds years 7 and 8 with year levels 9 to 12 at the Senior Campus. The campuses are a short walking distance apart.", "category": "EducationalInstitution"} +{"text": " Jon-Paul Roger JP Pietersen (born 12 July 1986 in Stellenbosch South Africa) is a South African rugby union footballer. He generally plays fullback or wing for the Sharks (in the Super Rugby competition) and the Natal Sharks in the Currie Cup. He played in more than 50 tests for the Springboks.", "category": "Athlete"} +{"text": " Deltocolpodes is a genus of beetles in the family Carabidae containing the following species: Deltocolpodes brendelli Morvan 1992 Deltocolpodes championi Morvan 1992 Deltocolpodes duluchus Morvan 1992 Deltocolpodes heinigeri Morvan 1992 Deltocolpodes jalepensis Morvan 1992 Deltocolpodes kirschenhoferi Morvan 1992 Deltocolpodes nepalensis Morvan 1992 Deltocolpodes perreaui Deuve 1985 Deltocolpodes rectangulus Morvan 1992 Deltocolpodes rolex Morvan 1992 Deltocolpodes salpensis Deuve 1985 Deltocolpodes sikkimensis Morvan 1992\u2191", "category": "Animal"} +{"text": " Stanhopea martiana is a species of orchid endemic to southwestern Mexico.", "category": "Plant"} +{"text": " Yawarmayu (Quechua yawar blood mayu river blood river hispanicized spelling Yahuarmayo) is a river in Peru located in the Puno Region Carabaya Province Ayapata District. It originates near the border of the districts Ayapata and Coasa. Its direction is mainly to the northwest where it meets Inambari River as a right affluent. The confluence is north of the village Yawarmayu (Yahuarmayo).", "category": "NaturalPlace"} +{"text": " The Charles Granke House at 406 S. Seventh St. in Hamilton Montana is a historic house that was built in 1906. It includes Colonial Revival and Queen Anne architecture. It was listed on the National Register of Historic Places in 1988. The listing included two contributing buildings.It was built in approximately 1906 by the Anaconda Copper Mining Company as a worker cottage for workers at the sawmill that operated in Hamilton until 1915. Charles W.", "category": "Building"} +{"text": " Passiflora monadelpha is a species of plant in the Passifloraceae family. It is endemic to Ecuador.", "category": "Plant"} +{"text": " Mangifera persiciformis or Peach Mango is a species of plant in the Anacardiaceae family. It is endemic to China.", "category": "Plant"} diff --git a/examples/fine-tuned_qa/answers_with_ft.py b/examples/fine-tuned_qa/answers_with_ft.py new file mode 100644 index 0000000..32507e8 --- /dev/null +++ b/examples/fine-tuned_qa/answers_with_ft.py @@ -0,0 +1,150 @@ +import argparse + +import openai + + +def create_context( + question, search_file_id, max_len=1800, search_model="ada", max_rerank=10 +): + """ + Create a context for a question by finding the most similar context from the search file. + :param question: The question + :param search_file_id: The file id of the search file + :param max_len: The maximum length of the returned context (in tokens) + :param search_model: The search model to use + :param max_rerank: The maximum number of reranking + :return: The context + """ + results = openai.Engine(search_model).search( + search_model=search_model, + query=question, + max_rerank=max_rerank, + file=search_file_id, + return_metadata=True, + ) + returns = [] + cur_len = 0 + for result in results["data"]: + cur_len += int(result["metadata"]) + 4 + if cur_len > max_len: + break + returns.append(result["text"]) + return "\n\n###\n\n".join(returns) + + +def answer_question( + search_file_id="", + fine_tuned_qa_model="", + question="Which country won the European Football championship in 2021?", + max_len=1800, + search_model="ada", + max_rerank=10, + debug=False, + stop_sequence=["\n", "."], + max_tokens=100, +): + """ + Answer a question based on the most similar context from the search file, using your fine-tuned model. + :param question: The question + :param fine_tuned_qa_model: The fine tuned QA model + :param search_file_id: The file id of the search file + :param max_len: The maximum length of the returned context (in tokens) + :param search_model: The search model to use + :param max_rerank: The maximum number of reranking + :param debug: Whether to output debug information + :param stop_sequence: The stop sequence for Q&A model + :param max_tokens: The maximum number of tokens to return + :return: The answer + """ + context = create_context( + question, + search_file_id, + max_len=max_len, + search_model=search_model, + max_rerank=max_rerank, + ) + if debug: + print("Context:\n" + context) + print("\n\n") + try: + # fine-tuned models requires model parameter, whereas other models require engine parameter + model_param = ( + {"model": fine_tuned_qa_model} + if ":" in fine_tuned_qa_model + and fine_tuned_qa_model.split(":")[1].startswith("ft") + else {"engine": fine_tuned_qa_model} + ) + response = openai.Completion.create( + prompt=f"Answer the question based on the context below\n\nText: {context}\n\n---\n\nQuestion: {question}\nAnswer:", + temperature=0, + max_tokens=max_tokens, + top_p=1, + frequency_penalty=0, + presence_penalty=0, + stop=stop_sequence, + **model_param, + ) + return response["choices"][0]["text"] + except Exception as e: + print(e) + return "" + + +if __name__ == "__main__": + parser = argparse.ArgumentParser( + description="Rudimentary functionality of the answers endpoint with a fine-tuned Q&A model.", + formatter_class=argparse.ArgumentDefaultsHelpFormatter, + ) + parser.add_argument( + "--search_file_id", help="Search file id", required=True, type=str + ) + parser.add_argument( + "--fine_tuned_qa_model", help="Fine-tuned QA model id", required=True, type=str + ) + parser.add_argument( + "--question", help="Question to answer", required=True, type=str + ) + parser.add_argument( + "--max_len", + help="Maximum length of the returned context (in tokens)", + default=1800, + type=int, + ) + parser.add_argument( + "--search_model", help="Search model to use", default="ada", type=str + ) + parser.add_argument( + "--max_rerank", + help="Maximum number of reranking for the search", + default=10, + type=int, + ) + parser.add_argument( + "--debug", help="Print debug information (context used)", action="store_true" + ) + parser.add_argument( + "--stop_sequence", + help="Stop sequences for the Q&A model", + default=["\n", "."], + nargs="+", + type=str, + ) + parser.add_argument( + "--max_tokens", + help="Maximum number of tokens to return", + default=100, + type=int, + ) + args = parser.parse_args() + response = answer_question( + search_file_id=args.search_file_id, + fine_tuned_qa_model=args.fine_tuned_qa_model, + question=args.question, + max_len=args.max_len, + search_model=args.search_model, + max_rerank=args.max_rerank, + debug=args.debug, + stop_sequence=args.stop_sequence, + max_tokens=args.max_tokens, + ) + print(f"Answer:{response}") diff --git a/examples/fine-tuned_qa/olympics-1-collect-data.ipynb b/examples/fine-tuned_qa/olympics-1-collect-data.ipynb new file mode 100644 index 0000000..7a88051 --- /dev/null +++ b/examples/fine-tuned_qa/olympics-1-collect-data.ipynb @@ -0,0 +1,513 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# 1. Collect Wikipedia data about Olympic Games 2020\n", + "\n", + "The idea of this project is to create a question answering model, based on a few paragraphs of provided text. Base GPT-3 models do a good job at answering questions when the answer is contained within the paragraph, however if the answer isn't contained, the base models tend to try their best to answer anyway, often leading to confabulated answers. \n", + "\n", + "To create a model which answers questions only if there is sufficient context for doing so, we first create a dataset of questions and answers based on paragraphs of text. In order to train the model to answer only when the answer is present, we also add adversarial examples, where the question doesn't match the context. In those cases, we ask the model to output \"No sufficient context for answering the question\". \n", + "\n", + "We will perform this task in three notebooks:\n", + "1. The first (this) notebook focuses on collecting recent data, which GPT-3 didn't see during it's pre-training. We picked the topic of Olympic Games 2020 (which actually took place in the summer of 2021), and downloaded 713 unique pages. We organized the dataset by individual sections, which will serve as context for asking and answering the questions.\n", + "2. The [second notebook](olympics-2-create-qa.ipynb) will utilize Davinci-instruct to ask a few questions based on a Wikipedia section, as well as answer those questions, based on that section.\n", + "3. The [third notebook](olympics-3-train-qa.ipynb) will utilize the dataset of context, question and answer pairs to additionally create adversarial questions and context pairs, where the question was not generated on that context. In those cases the model will be prompted to answer \"No sufficient context for answering the question\". We will also train a discriminator model, which predicts whether the question can be answered based on the context or not." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 1.1 Data extraction using the wikipedia API\n", + "Extracting the data will take about half an hour, and processing will likely take about as much." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "909" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "import pandas as pd\n", + "import wikipedia\n", + "\n", + "\n", + "def filter_olympic_2020_titles(titles):\n", + " \"\"\"\n", + " Get the titles which are related to Olympic games hosted in 2020, given a list of titles\n", + " \"\"\"\n", + " titles = [title for title in titles if '2020' in title and 'olympi' in title.lower()]\n", + " \n", + " return titles\n", + "\n", + "def get_wiki_page(title):\n", + " \"\"\"\n", + " Get the wikipedia page given a title\n", + " \"\"\"\n", + " try:\n", + " return wikipedia.page(title)\n", + " except wikipedia.exceptions.DisambiguationError as e:\n", + " return wikipedia.page(e.options[0])\n", + " except wikipedia.exceptions.PageError as e:\n", + " return None\n", + "\n", + "def recursively_find_all_pages(titles, titles_so_far=set()):\n", + " \"\"\"\n", + " Recursively find all the pages that are linked to the Wikipedia titles in the list\n", + " \"\"\"\n", + " all_pages = []\n", + " \n", + " titles = list(set(titles) - titles_so_far)\n", + " titles = filter_olympic_2020_titles(titles)\n", + " titles_so_far.update(titles)\n", + " for title in titles:\n", + " page = get_wiki_page(title)\n", + " if page is None:\n", + " continue\n", + " all_pages.append(page)\n", + "\n", + " new_pages = recursively_find_all_pages(page.links, titles_so_far)\n", + " for pg in new_pages:\n", + " if pg.title not in [p.title for p in all_pages]:\n", + " all_pages.append(pg)\n", + " titles_so_far.update(page.links)\n", + " return all_pages\n", + "\n", + "\n", + "pages = recursively_find_all_pages([\"2020 Summer Olympics\"])\n", + "len(pages)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 1.2 Filtering the Wikipedia pages and splitting them into sections by headings\n", + "We remove sections unlikely to contain textual information, and ensure that each section is not longer than the token limit" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "('Bermuda at the 2020 Summer Olympics',\n", + " 'Equestrian',\n", + " \"Bermuda entered one dressage rider into the Olympic competition by finishing in the top four, outside the group selection, of the individual FEI Olympic Rankings for Groups D and E (North, Central, and South America), marking the country's recurrence to the sport after an eight-year absence. The quota was later withdrawn, following an injury of Annabelle Collins' main horse Joyero and a failure to obtain minimum eligibility requirements (MER) aboard a new horse Chuppy Checker.\",\n", + " 104)" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "\n", + "import re\n", + "from typing import Set\n", + "from transformers import GPT2TokenizerFast\n", + "\n", + "import numpy as np\n", + "from nltk.tokenize import sent_tokenize\n", + "\n", + "tokenizer = GPT2TokenizerFast.from_pretrained(\"gpt2\")\n", + "\n", + "def count_tokens(text: str) -> int:\n", + " \"\"\"count the number of tokens in a string\"\"\"\n", + " return len(tokenizer.encode(text))\n", + "\n", + "def reduce_long(\n", + " long_text: str, long_text_tokens: bool = False, max_len: int = 590\n", + ") -> str:\n", + " \"\"\"\n", + " Reduce a long text to a maximum of `max_len` tokens by potentially cutting at a sentence end\n", + " \"\"\"\n", + " if not long_text_tokens:\n", + " long_text_tokens = count_tokens(long_text)\n", + " if long_text_tokens > max_len:\n", + " sentences = sent_tokenize(long_text.replace(\"\\n\", \" \"))\n", + " ntokens = 0\n", + " for i, sentence in enumerate(sentences):\n", + " ntokens += 1 + count_tokens(sentence)\n", + " if ntokens > max_len:\n", + " return \". \".join(sentences[:i][:-1]) + \".\"\n", + "\n", + " return long_text\n", + "\n", + "discard_categories = ['See also', 'References', 'External links', 'Further reading', \"Footnotes\",\n", + " \"Bibliography\", \"Sources\", \"Citations\", \"Literature\", \"Footnotes\", \"Notes and references\",\n", + " \"Photo gallery\", \"Works cited\", \"Photos\", \"Gallery\", \"Notes\", \"References and sources\",\n", + " \"References and notes\",]\n", + "\n", + "\n", + "def extract_sections(\n", + " wiki_text: str,\n", + " title: str,\n", + " max_len: int = 1500,\n", + " discard_categories: Set[str] = discard_categories,\n", + ") -> str:\n", + " \"\"\"\n", + " Extract the sections of a Wikipedia page, discarding the the references and other low information sections\n", + " \"\"\"\n", + " if len(wiki_text) == 0:\n", + " return []\n", + "\n", + " # find all headings and the coresponding contents\n", + " headings = re.findall(\"==+ .* ==+\", wiki_text)\n", + " for heading in headings:\n", + " wiki_text = wiki_text.replace(heading, \"==+ !! ==+\")\n", + " contents = wiki_text.split(\"==+ !! ==+\")\n", + " contents = [c.strip() for c in contents]\n", + " assert len(headings) == len(contents) - 1\n", + "\n", + " cont = contents.pop(0).strip()\n", + " outputs = [(title, \"Summary\", cont, count_tokens(cont)+4)]\n", + " \n", + " # discard the discard categories, accounting for a tree structure\n", + " max_level = 100\n", + " keep_group_level = max_level\n", + " remove_group_level = max_level\n", + " nheadings, ncontents = [], []\n", + " for heading, content in zip(headings, contents):\n", + " plain_heading = \" \".join(heading.split(\" \")[1:-1])\n", + " num_equals = len(heading.split(\" \")[0])\n", + " if num_equals <= keep_group_level:\n", + " keep_group_level = max_level\n", + "\n", + " if num_equals > remove_group_level:\n", + " if (\n", + " num_equals <= keep_group_level\n", + " ):\n", + " continue\n", + " keep_group_level = max_level\n", + " if plain_heading in discard_categories:\n", + " remove_group_level = num_equals\n", + " keep_group_level = max_level\n", + " continue\n", + " nheadings.append(heading.replace(\"=\", \"\").strip())\n", + " ncontents.append(content)\n", + " remove_group_level = max_level\n", + "\n", + " # count the tokens of each section\n", + " ncontent_ntokens = [\n", + " count_tokens(c)\n", + " + 3\n", + " + count_tokens(\" \".join(h.split(\" \")[1:-1]))\n", + " - (1 if len(c) == 0 else 0)\n", + " for h, c in zip(nheadings, ncontents)\n", + " ]\n", + "\n", + " # Create a tuple of (title, section_name, content, number of tokens)\n", + " outputs += [(title, h, c, t) if t 1024). Running this sequence through the model will result in indexing errors\n" + ] + }, + { + "data": { + "text/html": [ + "
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    titleheadingcontenttokens
    02020 Summer OlympicsSummaryThe 2020 Summer Olympics (Japanese: 2020年夏季オリン...713
    12020 Summer OlympicsHost city selectionThe International Olympic Committee (IOC) vote...126
    22020 Summer OlympicsImpact of the COVID-19 pandemicIn January 2020, concerns were raised about th...369
    32020 Summer OlympicsQualifying event cancellation and postponementConcerns about the pandemic began to affect qu...298
    42020 Summer OlympicsEffect on doping testsMandatory doping tests were being severely res...163
    \n", + "
    " + ], + "text/plain": [ + " title heading \\\n", + "0 2020 Summer Olympics Summary \n", + "1 2020 Summer Olympics Host city selection \n", + "2 2020 Summer Olympics Impact of the COVID-19 pandemic \n", + "3 2020 Summer Olympics Qualifying event cancellation and postponement \n", + "4 2020 Summer Olympics Effect on doping tests \n", + "\n", + " content tokens \n", + "0 The 2020 Summer Olympics (Japanese: 2020年夏季オリン... 713 \n", + "1 The International Olympic Committee (IOC) vote... 126 \n", + "2 In January 2020, concerns were raised about th... 369 \n", + "3 Concerns about the pandemic began to affect qu... 298 \n", + "4 Mandatory doping tests were being severely res... 163 " + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "res = []\n", + "for page in pages:\n", + " res += extract_sections(page.content, page.title)\n", + "df = pd.DataFrame(res, columns=[\"title\", \"heading\", \"content\", \"tokens\"])\n", + "df = df[df.tokens>40]\n", + "df = df.drop_duplicates(['title','heading'])\n", + "df = df.reset_index().drop('index',axis=1) # reset index\n", + "df.head()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Save the section dataset\n", + "We will save the section dataset, for the [next notebook](olympics-2-create-qa.ipynb)" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": {}, + "outputs": [], + "source": [ + "df.to_csv('olympics-data/olympics_sections.csv', index=False)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 1.3 (Optional) Exploring the data " + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "Concerns and controversies at the 2020 Summer Olympics 51\n", + "United States at the 2020 Summer Olympics 46\n", + "Great Britain at the 2020 Summer Olympics 42\n", + "Canada at the 2020 Summer Olympics 39\n", + "Olympic Games 39\n", + "Name: title, dtype: int64" + ] + }, + "execution_count": 5, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "df.title.value_counts().head()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "There appear to be winter and summer Olympics 2020. We chose to leave a little ambiguity and noise in the dataset, even though we were interested in only Summer Olympics 2020." + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "True 3567\n", + "False 305\n", + "Name: title, dtype: int64" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "df.title.str.contains('Summer').value_counts()" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "False 3774\n", + "True 98\n", + "Name: title, dtype: int64" + ] + }, + "execution_count": 7, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "df.title.str.contains('Winter').value_counts()" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": {}, + "outputs": [ + { + "data": { + "image/png": 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    " + ] + }, + "metadata": { + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "import pandas as pd\n", + "from matplotlib import pyplot as plt\n", + "\n", + "df = pd.read_csv('olympics-data/olympics_sections.csv')\n", + "df[['tokens']].hist()\n", + "# add axis descriptions and title\n", + "plt.xlabel('Number of tokens')\n", + "plt.ylabel('Number of Wikipedia sections')\n", + "plt.title('Distribution of number of tokens in Wikipedia sections')\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can see that the majority of section are fairly short (less than 500 tokens)." + ] + } + ], + "metadata": { + "interpreter": { + "hash": "be4b5d5b73a21c599de40d6deb1129796d12dc1cc33a738f7bac13269cfcafe8" + }, + "kernelspec": { + "display_name": "Python 3.7.3 64-bit ('base': conda)", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.7.3" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/fine-tuned_qa/olympics-2-create-qa.ipynb b/examples/fine-tuned_qa/olympics-2-create-qa.ipynb new file mode 100644 index 0000000..9834cec --- /dev/null +++ b/examples/fine-tuned_qa/olympics-2-create-qa.ipynb @@ -0,0 +1,751 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# 2. Creating a synthetic Q&A dataset\n", + "We use [`davinci-instruct-beta-v2`](https://beta.openai.com/docs/engines/instruct-series-beta), a model specialized in following instructions, to create questions based on the given context. Then we also use [`davinci-instruct-beta-v2`](https://beta.openai.com/docs/engines/instruct-series-beta) to answer those questions, given the same context. \n", + "\n", + "This is expensive, and will also take a long time, as we call the davinci engine for each section. You can simply download the final dataset instead.\n", + "\n", + "We're using the dataset created using the [previous notebook](olympics-1-collect-data.ipynb)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 2.1 Read in the data, and create a context\n", + "Create a context by concatenating the title, the heading and the content of that section" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": {}, + "outputs": [ + { + "data": { + "text/html": [ + "
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    02020 Summer OlympicsSummaryThe 2020 Summer Olympics (Japanese: 2020年夏季オリン...7132020 Summer Olympics\\nSummary\\n\\nThe 2020 Summ...
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    " + ], + "text/plain": [ + " title heading \\\n", + "0 2020 Summer Olympics Summary \n", + "1 2020 Summer Olympics Host city selection \n", + "2 2020 Summer Olympics Impact of the COVID-19 pandemic \n", + "3 2020 Summer Olympics Qualifying event cancellation and postponement \n", + "4 2020 Summer Olympics Effect on doping tests \n", + "\n", + " content tokens \\\n", + "0 The 2020 Summer Olympics (Japanese: 2020年夏季オリン... 713 \n", + "1 The International Olympic Committee (IOC) vote... 126 \n", + "2 In January 2020, concerns were raised about th... 369 \n", + "3 Concerns about the pandemic began to affect qu... 298 \n", + "4 Mandatory doping tests were being severely res... 163 \n", + "\n", + " context \n", + "0 2020 Summer Olympics\\nSummary\\n\\nThe 2020 Summ... \n", + "1 2020 Summer Olympics\\nHost city selection\\n\\nT... \n", + "2 2020 Summer Olympics\\nImpact of the COVID-19 p... \n", + "3 2020 Summer Olympics\\nQualifying event cancell... \n", + "4 2020 Summer Olympics\\nEffect on doping tests\\n... " + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "import pandas as pd\n", + "df = pd.read_csv('olympics-data/olympics_sections.csv')\n", + "df['context'] = df.title + \"\\n\" + df.heading + \"\\n\\n\" + df.content\n", + "df.head()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 2.2 Create questions based on the context\n", + "Use davinci-instruct to generate a number of plausible questions relating to the Wikipedia section contents.\n", + "\n", + "Note: We have used temperature=0, but it may be beneficial to experiment with a higher temperature to get a higher diversity of questions.\n", + "\n", + "**WARNING: This step will last a long time, and consume a lot of tokens, as it calls davinci-instruct for every section to generate a number of questions.**" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "1. What is the 2020 Summer Olympics?\n", + "2. When did the 2020 Summer Olympics take place?\n", + "3. Who won the most medals at the 2020 Summer Olympics?\n", + "4. Who won the most gold medals at the 2020 Summer Olympics?\n", + "5. Who won the most medals at the 2020 Summer Olympics?\n" + ] + } + ], + "source": [ + "import openai\n", + "\n", + "def get_questions(context):\n", + " try:\n", + " response = openai.Completion.create(\n", + " engine=\"davinci-instruct-beta-v2\",\n", + " prompt=f\"Write questions based on the text below\\n\\nText: {context}\\n\\nQuestions:\\n1.\",\n", + " temperature=0,\n", + " max_tokens=257,\n", + " top_p=1,\n", + " frequency_penalty=0,\n", + " presence_penalty=0,\n", + " stop=[\"\\n\\n\"]\n", + " )\n", + " return response['choices'][0]['text']\n", + " except:\n", + " return \"\"\n", + "\n", + "\n", + "df['questions']= df.context.apply(get_questions)\n", + "df['questions'] = \"1.\" + df.questions\n", + "print(df[['questions']].values[0][0])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The prompt is designed to generate a number of questions. Example questions above were generated based on the summary section of the 2020 Summer Olympics page.\n", + "\n", + "We can observe that the questions 3 and 5 above repeat. Sometimes the generated questions could be ambiguous without the context. We will show that even despite these limitations we can create a successful model." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The 2020 Summer Olympics (Japanese: 2020年夏季オリンピック, Hepburn: Nisen Nijū-nen Kaki Orinpikku), officially the Games of the XXXII Olympiad (第三十二回オリンピック競技大会, Dai Sanjūni-kai Orinpikku Kyōgi Taikai) and branded as Tokyo 2020 (東京2020, Tōkyō Nii Zero Nii Zero), was an international multi-sport event held from 23 July to 8 August 2021 in Tokyo, Japan, with some preliminary events that began on 21 July.\n", + "Tokyo was selected as the host city during the 125th IOC Session in Buenos Aires, Argentina, on 7 September 2013. Originally scheduled to take place from 24 July to 9 August 2020, the event was postponed to 2021 in March 2020 as a result of the COVID-19 pandemic, the first such instance in the history of the Olympic Games (previous games had been cancelled but not rescheduled). However, the event retained the Tokyo 2020 name for marketing and branding purposes. It was largely held behind closed doors with no public spectators permitted due to the declaration of a state of emergency in the Greater Tokyo Area in response to the pandemic. The Summer Paralympics were held between 24 August and 5 September 2021, 16 days after the completion of the Olympics.The 2020 Games were the fourth Olympic Games to be held in Japan, following the Tokyo 1964 (Summer), Sapporo 1972 (Winter) and Nagano 1998 (Winter) games. Tokyo is the first city in Asia to hold the Summer Games twice. The 2020 Games were the second of three consecutive Olympics to be held in East Asia, following the 2018 Winter Olympics in Pyeongchang, South Korea and preceding the 2022 Winter Olympics in Beijing, China.\n", + "New events were introduced in existing sports for 2020, including 3x3 basketball, freestyle BMX and mixed gender team events in a number of existing sports, as well as the return of madison cycling for men and an introduction of the same event for women. New IOC policies also allowed the host organizing committee to add new sports to the Olympic program for just one Games. The disciplines added by the Japanese Olympic Committee were baseball and softball, karate, sport climbing, surfing and skateboarding, the last four of which made their Olympic debuts, and the last three of which will remain on the Olympic program.The United States topped the medal count by both total golds (39) and total medals (113), with China finishing second by both respects (38 and 88). Host nation Japan finished third, setting a record for the most gold medals and total medals ever won by their delegation at an Olympic Games with 27 and 58. Great Britain finished fourth, with a total of 22 gold and 65 medals, becoming the first nation at the Summer Olympics to increase or equal their total medals won in the two Games subsequent to hosting them. The Russian delegation competing as the ROC (not to be confused with the Republic of China (Taiwan) which competed as Chinese Taipei, not ROC) finished fifth with 20 gold medals and third in the overall medal count, with 71 medals. Bermuda, the Philippines and Qatar won their first-ever Olympic gold medals. Burkina Faso, San Marino and Turkmenistan won their first-ever Olympic medals.\n" + ] + } + ], + "source": [ + "print(df.content.values[0])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 2.3 Create answers based on the context\n", + "Use davinci-instruct to answer the questions given the relevant Wikipedia section contents\n", + "\n", + "Note: We have used temperature=0, but it may be beneficial to experiment with a higher temperature to get a higher diversity of questions.\n", + "\n", + "**WARNING: This step will last a long time, and consume a lot of tokens, as it calls davinci-instruct for every section to answer all the questions.**" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "1. The 2020 Summer Olympics is an international multi-sport event held from 23 July to 8 August 2021 in Tokyo, Japan.\n", + "2. The 2020 Summer Olympics took place from 23 July to 8 August 2021.\n", + "3. The United States topped the medal count by both total golds (39) and total medals (113), with China finishing second by both respects (38 and 88).\n", + "4. The United States topped the medal count by both total golds (39) and total medals (113), with China finishing second by both respects (38 and 88).\n", + "5. The United States topped the medal count by both total golds (39) and total medals (113), with China finishing second by both respects (38 and 88).\n" + ] + } + ], + "source": [ + "def get_answers(row):\n", + " try:\n", + " response = openai.Completion.create(\n", + " engine=\"davinci-instruct-beta-v2\",\n", + " prompt=f\"Write questions based on the text below\\n\\nText: {row.context}\\n\\nQuestions:\\n{row.questions}\\n\\nAnswers:\\n1.\",\n", + " temperature=0,\n", + " max_tokens=257,\n", + " top_p=1,\n", + " frequency_penalty=0,\n", + " presence_penalty=0\n", + " )\n", + " return response['choices'][0]['text']\n", + " except Exception as e:\n", + " print (e)\n", + " return \"\"\n", + "\n", + "\n", + "df['answers']= df.apply(get_answers, axis=1)\n", + "df['answers'] = \"1.\" + df.answers\n", + "df = df.dropna().reset_index().drop('index',axis=1)\n", + "print(df[['answers']].values[0][0])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "These are the answers to the questions above based on the context around the host city selection. \n", + "\n", + "We can see that answers 3-5 contain the correct answer, but instead of answering the question directly, the answer is a verbatim extraction. Despite these occasional lower quality answers, we will show that the model can learn the task reasonably well, given a high number of examples." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 2.4 Save the Olympics Q&A dataset based on Wikipedia sections\n", + "We save the file for use in the [next notebook](olympics-3-train-qa.ipynb)" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": {}, + "outputs": [], + "source": [ + "df.to_csv('olympics-data/olympics_qa.csv', index=False)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 2.5 Search file\n", + "We create a search file ([API reference](https://beta.openai.com/docs/api-reference/files/list)), which can be used to retrieve the relevant context when a question is asked.\n" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": {}, + "outputs": [], + "source": [ + "df = df[df.tokens<2000]\n", + "df[['context', 'tokens']].rename(columns={'context':'text','tokens':'metadata'}).to_json('olympics-data/olympics_search.jsonl', orient='records', lines=True)\n", + "\n", + "search_file = openai.File.create(\n", + " file=open(\"olympics-data/olympics_search.jsonl\"),\n", + " purpose='search'\n", + ")\n", + "olympics_search_fileid = search_file['id']" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 2.6 Answer questions based on the context provided\n", + "\n", + "We will use a simple implementation of the answers endpoint. This works by simply using the [/search endpoint](https://beta.openai.com/docs/api-reference/searches), which searches over an indexed file to obtain the relevant sections which can be included in the context, following by a question and answering prompt given a specified model." + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Athletics at the 2020 Summer Olympics – Women's 4 × 100 metres relay\n", + "Summary\n", + "\n", + "The women's 4 × 100 metres relay event at the 2020 Summer Olympics took place on 5 and 6 August 2021 at the Japan National Stadium. There were 16 competing relay teams, with each team having 5 members from which 4 were selected in each round.\n", + "\n", + "###\n", + "\n", + "Athletics at the 2020 Summer Olympics – Men's 4 × 100 metres relay\n", + "Qualification\n", + "\n", + "National Olympic Committees (NOCs) could qualify one relay team in one of three following ways:\n", + "The top 8 NOCs at the 2019 World Athletics Championships qualified a relay team.\n", + "The top 8 NOCs at the 2021 World Athletics Relays qualified a relay team.\n", + "Where an NOC placed in the top 8 at both the 2019 World Championships and the 2021 World Relays, the quota place was allocated to the world top list as of 29 June 2021. In this case, 4 teams did so, so there are 4 places available through the world rankings.A total of five athletes may be entered for a relay team. Should a NOC have also entered individual athletes in the corresponding individual event (100 m), the entered individual athletes must be included in the total of five (5) athletes entered for the relay event. In addition of five, NOCs can nominate a maximum of one alternate athlete for each team.\n", + "The qualifying period was originally from 1 May 2019 to 29 June 2020. Due to the COVID-19 pandemic, the period was suspended from 6 April 2020 to 30 November 2020, with the end date extended to 29 June 2021. The qualifying time standards could be obtained in various meets during the given period that have the approval of the IAAF. Both indoor and outdoor meets are eligible. The most recent Area Championships may be counted in the ranking, even if not during the qualifying period.\n" + ] + } + ], + "source": [ + "from answers_with_ft import create_context, answer_question\n", + "print(create_context(\"Where did women's 4 x 100 metres relay event take place during the 2020 Summer Olympics?\", olympics_search_fileid, max_len=400))" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "' Japan National Stadium'" + ] + }, + "execution_count": 8, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "answer_question(olympics_search_fileid, \"davinci-instruct-beta-v2\", \n", + " \"Where did women's 4 x 100 metres relay event take place during the 2020 Summer Olympics?\")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "After we fine-tune the model for Q&A we'll be able to use it instead of [`davinci-instruct-beta-v2`](https://beta.openai.com/docs/engines/instruct-series-beta), to obtain better answers when the question can't be answered based on the context. We see a downside of [`davinci-instruct-beta-v2`](https://beta.openai.com/docs/engines/instruct-series-beta), which always attempts to answer the question, regardless of the relevant context being present or not. (Note the second question is asking about a future event, set in 2024.)" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "' Japan National Stadium'" + ] + }, + "execution_count": 9, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "answer_question(olympics_search_fileid, \"davinci-instruct-beta-v2\", \n", + " \"Where did women's 4 x 100 metres relay event take place during the 2048 Summer Olympics?\", max_len=1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can see that davinci has a tendency to answer the question, even if the question can't be answered given the context provided. Note the question asked regarding 2048 Summer Olympics, which didn't happen yet, and the retrieved content has only returned results for 2020." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 2.7 (Optional) Investigation into how likely the search endpoint is to return the relevant context" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(0, 58)\n" + ] + } + ], + "source": [ + "def check_context(title, heading, question, max_len=1800, search_model='ada', max_rerank=10):\n", + " \"\"\"\n", + " Evaluate the performance of the search model in retrieving the correct context\n", + "\n", + " Parameters\n", + " ----------\n", + " title: str\n", + " The title of the Wikipedia page\n", + " heading: str\n", + " The heading of the Wikipedia section\n", + " qusetion: str\n", + " The question\n", + " max_len: int\n", + " The maximum length of the context\n", + " search_model: str\n", + " The search model to use - `ada` is most cost effective\n", + " max_rerank: int\n", + " The maximum number of reranking documents to use the search model on\n", + "\n", + " Returns\n", + " -------\n", + " rank: int\n", + " The rank of the correct context\n", + " token_length: int\n", + " The number of tokens needed to obtain the correct context\n", + " \"\"\"\n", + " \n", + " try:\n", + " results = openai.Engine(search_model).search(\n", + " search_model=search_model, \n", + " query=question, \n", + " max_rerank=max_rerank,\n", + " file=olympics_search_fileid,\n", + " return_metadata=True\n", + " )\n", + " index=-1\n", + " returns = []\n", + " cur_len = 0\n", + " for result in results['data']:\n", + " cur_len += int(result['metadata']) + 4 # we add 4 tokens for the separator `\\n\\n###\\n\\n`\n", + " if cur_len > max_len:\n", + " break\n", + " returns.append(result['text'])\n", + " res = result['text'].split('\\n')\n", + " if res[0] == title and res[1] == heading:\n", + " index = len(returns) - 1\n", + " break\n", + " return index, cur_len\n", + " except Exception as e:\n", + " #print (e)\n", + " return []\n", + "print(check_context(\"Athletics at the 2020 Summer Olympics – Women's 4 × 100 metres relay\", \"Summary\", \"Where did women's 4 x 100 metres relay event take place during the 2020 Summer Olympics?\", max_len=10000))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We utilize the generated questions based on context to estimate how often we can retrieve the original context. These questions are noisy, so this is not a perfect estimate.\n", + "\n", + "Our questions and answers are prefixed with numbered bullet points, however due to the way they were generated, they are missing the first number, hence we add \"1.\" to the list of questions (and answers).\n", + "\n", + "We calculate the rank of the section retrieved using ada search, and the number of tokens in the context needed to retrieve the relevant section in full." + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0 [(132, 27104), (-1, 22939), (8, 2151), (2, 121...\n", + "1 [(4, 1737), (0, 130), (8, 744), (96, 17208), (...\n", + "2 [(0, 373), (0, 373), (-1, 40610), (1, 570)]\n", + "3 [(0, 302), (0, 302), (5, 968), (8, 1425)]\n", + "4 [(0, 167), (0, 167), (2, 1442)]\n", + "Name: ada, dtype: object" + ] + }, + "execution_count": 12, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "ada_results = df.apply(lambda x: [\n", + " check_context( x.title, \n", + " x.heading, \n", + " q[3:], # remove the number prefix\n", + " max_len=1000000, # set a large number to get the full context \n", + " search_model='ada', \n", + " max_rerank=200,\n", + " ) \n", + " for q in (x.questions).split('\\n') # split the questions\n", + " if len(q) >10 # remove the empty questions\n", + " ], axis=1)\n", + "ada_results.head()" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": {}, + "outputs": [], + "source": [ + "out = pd.concat([ada_results], axis=1)\n", + "out.columns = ['ada']\n", + "out.to_csv('olympics-data/search_engine_results.csv')" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": {}, + "outputs": [], + "source": [ + "def expand_lists(out):\n", + " \"\"\"\n", + " Expand a pandas series containing lists into a series, where each list element becomes a value on its own\n", + "\n", + " Input is a row per paragraph, which has multiple questions\n", + " Output is a row per question\n", + " \"\"\"\n", + " cols = [pd.DataFrame(out[name].tolist()).stack().reset_index(level=1, drop=True).rename(name) for name in out.columns] \n", + " return pd.concat(cols, axis=1)\n", + "\n", + "out_expanded = expand_lists(out)\n", + "out_expanded['rank'] = out_expanded.ada.apply(lambda x: x[0] if x != [] else -2)\n", + "out_expanded['tokens'] = out_expanded.ada.apply(lambda x: x[1] if x != [] else -2)\n" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "74.3% of relevant paragraphs are retrieved within the first 2k tokens\n" + ] + } + ], + "source": [ + "within_2k = (out_expanded.tokens < 2000).mean()\n", + "print(f\"{within_2k*100:.1f}% of relevant paragraphs are retrieved within the first 2k tokens\")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The relevant context can be obtained 74% of the time on this dataset" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "7.4% of relevant paragraphs are not retrieved within the first 200 results\n" + ] + } + ], + "source": [ + "outside_200 = (out_expanded['rank'] == -1).mean()\n", + "print(f\"{outside_200*100:.1f}% of relevant paragraphs are not retrieved within the first 200 results\")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "7.4% of the time, this is due to the keyword search part of the search algorithm not retrieving the relevant context within the first 200 results.\n", + "18.3% of the time this is due to the semantic search not placing the relevant context within the first 2000 tokens." + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": {}, + "outputs": [ + { + "data": { + "image/png": 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hcDqwCngp1SUi2iRdDTyY6l0VEW1VbreZmXVS1YQREY8CzV1MOqWLugFM72Y5s4BZA9o4MzPrE3/T28zMsjhhmJlZFicMMzPL4oRhZmZZnDDMzCyLE4aZmWVxwjAzsyxOGGZmlsUJw8zMsjhhmJlZFicMMzPL4oRhZmZZnDDMzCyLE4aZmWVxwjAzsyxOGGZmlsUJw8zMsjhhmJlZFicMMzPL4oRhZmZZqpowJK2WtEzSo5IeSmWjJS2StDL9HZXKJemrklZJekzSMaXlTE31V0qaWs02m5lZ12pxhnFSRBwdEc1p/DJgcURMABancYDTgAnpNQ24BYoEA8wAjgOOBWZ0JBkzM6udweiSmgzMTsOzgSml8jlRWAKMlHQgcCqwKCLaImILsAiYVOM2m5nt8hQR1Vu49AywBQjgGxExU9LWiBiZpgvYEhEjJd0NXBcR96dpi4FLgRZgr4i4JpV/Hng5Iq7vtK5pFGcmNDU1vWvevHkVt3tz2zY2vdx7vaPG7FfxOmqpvb2dESNGDHYzBkyjxQONF1OjxQONF1NX8Zx00klLS71BOxlW5TadGBHrJf0RsEjSr8oTIyIkDUjGioiZwEyA5ubmaGlpqXhZN89dwA3Let80q8+pfB211NraSn+2x1DTaPFA48XUaPFA48VUSTxV7ZKKiPXp72bgLoprEJtSVxPp7+ZUfT0wrjT72FTWXbmZmdVQ1RKGpOGS9ukYBiYCjwMLgY47naYCC9LwQuC8dLfU8cC2iNgA3ANMlDQqXeyemMrMzKyGqtkl1QTcVVymYBhwe0T8SNKDwHxJFwJrgDNT/R8CpwOrgJeACwAiok3S1cCDqd5VEdFWxXabmVkXqpYwIuJp4B1dlD8HnNJFeQDTu1nWLGDWQLfRzMzy+ZveZmaWxQnDzMyyOGGYmVkWJwwzM8vihGFmZlmcMMzMLIsThpmZZXHCMDOzLE4YZmaWxQnDzMyyOGGYmVkWJwwzM8vihGFmZlmcMMzMLIsThpmZZXHCMDOzLE4YZmaWxQnDzMyyZCUMSYtzyszMrHH1+JvekvYC3gjsL2kUoDRpX2BMldtmZmZDSG9nGB8HlgJHpL8drwXAP+asQNLukh6RdHcaP1TSA5JWSfqOpD1T+RvS+Ko0fXxpGZen8iclndrnKM3MrN96TBgRcVNEHApcEhFvjohD0+sdEZGVMICLgBWl8S8BN0bEYcAW4MJUfiGwJZXfmOoh6UjgLOBtwCTga5J2z1y3mZkNkKxrGBFxs6Q/k/RXks7rePU2n6SxwPuBb6ZxAScDd6Yqs4EpaXhyGidNPyXVnwzMi4hXI+IZYBVwbFZ0ZmY2YHq8htFB0reBtwCPAq+l4gDm9DLrV4C/A/ZJ428CtkbE9jS+jh3XQsYAawEiYrukban+GGBJaZnlecptnAZMA2hqaqK1tTUntC417Q0XH7W913r9WUcttbe3101bczRaPNB4MTVaPNB4MVUST1bCAJqBIyMichcs6QPA5ohYKqmlT62qQETMBGYCNDc3R0tL5au8ee4CbljW+6ZZfU7l66il1tZW+rM9hppGiwcaL6ZGiwcaL6ZK4slNGI8Dfwxs6MOy3w38paTTgb0o7qy6CRgpaVg6yxgLrE/11wPjgHWShgH7Ac+VyjuU5zEzsxrJ/eLe/sATku6RtLDj1dMMEXF5RIyNiPEUF61/EhHnAPcCZ6RqUynuuAJYmMZJ03+SzmgWAmelu6gOBSYAv8hst5mZDZDcM4wrBnCdlwLzJF0DPALcmspvBb4taRXQRpFkiIjlkuYDTwDbgekR8drOizUzs2rKShgR8dP+rCQiWoHWNPw0XdzlFBGvAB/pZv5rgWv70wYzM+uf3LukXqC4KwpgT2AP4MWI2LdaDTMzs6El9wyj47ZYSt+NOL5ajTIzs6Gnz0+rjcL3AT+iw8xsF5LbJfXh0uhuFN/LeKUqLTIzsyEp9y6pD5aGtwOrKbqlzMxsF5F7DeOCajfEzMyGttwfUBor6S5Jm9Pru+nBgmZmtovIvej9LYpvXB+UXv83lZmZ2S4iN2EcEBHfiojt6XUbcEAV22VmZkNMbsJ4TtK56dfzdpd0LsWDAc3MbBeRmzD+GjgT2EjxxNozgPOr1CYzMxuCcm+rvQqYGhFbACSNBq6nSCRmZrYLyD3DeHtHsgCIiDbgndVpkpmZDUW5CWM3SaM6RtIZRu7ZiZmZNYDcN/0bgJ9L+tc0/hH8uHEzs11K7je950h6CDg5FX04Ip6oXrPMzGyoye5WSgnCScLMbBfV58ebm5nZrskJw8zMsjhhmJlZlqolDEl7SfqFpF9KWi7pylR+qKQHJK2S9B1Je6byN6TxVWn6+NKyLk/lT0ryL/2ZmQ2Cap5hvAqcHBHvAI4GJkk6HvgScGNEHAZsAS5M9S8EtqTyG1M9JB0JnAW8DZgEfE3S7lVst5mZdaFqCSP99nd7Gt0jvYLi1tw7U/lsYEoanpzGSdNPkaRUPi8iXo2IZ4BVwLHVareZmXWtqt/WTmcCS4HDgH8CngK2RsT2VGUdMCYNjwHWAkTEdknbgDel8iWlxZbnKa9rGjANoKmpidbW1orb3bQ3XHzU9l7r9WcdtdTe3l43bc3RaPFA48XUaPFA48VUSTxVTRgR8RpwtKSRwF3AEVVc10xgJkBzc3O0tLRUvKyb5y7ghmW9b5rV51S+jlpqbW2lP9tjqGm0eKDxYmq0eKDxYqoknprcJRURW4F7gROAkZI63o3HAuvT8HpgHECavh/Fb278obyLeczMrEaqeZfUAenMAkl7A+8DVlAkjjNStanAgjS8MI2Tpv8kIiKVn5XuojoUmAD8olrtNjOzrlWzS+pAYHa6jrEbMD8i7pb0BDBP0jXAI8Ctqf6twLclrQLaKO6MIiKWS5pP8ViS7cD01NVlZmY1VLWEERGP0cVvZkTE03Rxl1NEvELxFNyulnUtfjqumdmg8je9zcwsixOGmZllccIwM7MsThhmZpbFCcPMzLI4YZiZWRYnDDMzy+KEYWZmWZwwzMwsixOGmZllccIwM7MsThhmZpbFCcPMzLI4YZiZWRYnDDMzy+KEYWZmWZwwzMwsixOGmZllccIwM7MsVUsYksZJulfSE5KWS7oolY+WtEjSyvR3VCqXpK9KWiXpMUnHlJY1NdVfKWlqtdpsZmbdq+YZxnbg4og4EjgemC7pSOAyYHFETAAWp3GA04AJ6TUNuAWKBAPMAI4DjgVmdCQZMzOrnaoljIjYEBEPp+EXgBXAGGAyMDtVmw1MScOTgTlRWAKMlHQgcCqwKCLaImILsAiYVK12m5lZ1xQR1V+JNB64D/hT4NcRMTKVC9gSESMl3Q1cFxH3p2mLgUuBFmCviLgmlX8eeDkiru+0jmkUZyY0NTW9a968eRW3d3PbNja93Hu9o8bsV/E6aqm9vZ0RI0YMdjMGTKPFA40XU6PFA40XU1fxnHTSSUsjorm7eYZVu1GSRgDfBT4TEc8XOaIQESFpQDJWRMwEZgI0NzdHS0tLxcu6ee4CbljW+6ZZfU7l66il1tZW+rM9hppGiwcaL6ZGiwcaL6ZK4qnqXVKS9qBIFnMj4nupeFPqaiL93ZzK1wPjSrOPTWXdlZuZWQ1V8y4pAbcCKyLiy6VJC4GOO52mAgtK5eelu6WOB7ZFxAbgHmCipFHpYvfEVGZmZjVUzS6pdwMfBZZJejSVfRa4Dpgv6UJgDXBmmvZD4HRgFfAScAFARLRJuhp4MNW7KiLaqthuMzPrQtUSRrp4rW4mn9JF/QCmd7OsWcCsgWudmZn1lb/pbWZmWZwwzMwsixOGmZllccIwM7MsThhmZpbFCcPMzLI4YZiZWRYnDDMzy+KEYWZmWZwwzMwsixOGmZllqfrvYTSy8Zf9ILvu6uveX8WWmJlVn88wzMwsixOGmZllccIwM7MsThhmZpbFCcPMzLI4YZiZWRYnDDMzy+KEYWZmWaqWMCTNkrRZ0uOlstGSFklamf6OSuWS9FVJqyQ9JumY0jxTU/2VkqZWq71mZtazap5h3AZM6lR2GbA4IiYAi9M4wGnAhPSaBtwCRYIBZgDHAccCMzqSjJmZ1VbVEkZE3Ae0dSqeDMxOw7OBKaXyOVFYAoyUdCBwKrAoItoiYguwiJ2TkJmZ1UCtnyXVFBEb0vBGoCkNjwHWluqtS2Xdle9E0jSKsxOamppobW2tvJF7w8VHba94/q70pz391d7ePqjrH2iNFg80XkyNFg80XkyVxDNoDx+MiJAUA7i8mcBMgObm5mhpaal4WTfPXcANywZ206w+p2VAl9cXra2t9Gd7DDWNFg80XkyNFg80XkyVxFPru6Q2pa4m0t/NqXw9MK5Ub2wq667czMxqrNYJYyHQcafTVGBBqfy8dLfU8cC21HV1DzBR0qh0sXtiKjMzsxqrWpeUpDuAFmB/Seso7na6Dpgv6UJgDXBmqv5D4HRgFfAScAFARLRJuhp4MNW7KiI6X0g3M7MaqFrCiIizu5l0Shd1A5jezXJmAbMGsGlmZlYBf9PbzMyy+CdaayT351z9U65mNlT5DMPMzLI4YZiZWRYnDDMzy+KEYWZmWZwwzMwsixOGmZll8W21Q0zu7bfgW3DNrLZ8hmFmZlmcMMzMLIsThpmZZfE1jDqWe73jtknDq9wSM9sV+AzDzMyy+AxjF7Bs/TbO98MPzayffIZhZmZZfIZhr+PHsJtZd5wwrCL+gqHZrscJw6quL8klh+/6MhscThhWd6pxEd9nTGa9q5uEIWkScBOwO/DNiLhukJtkdWCgz26qtcyLj9qelQT7kqx8PapxDJUPNHWRMCTtDvwT8D5gHfCgpIUR8cTgtsysthotAVZLNc4sBzOpDxV1kTCAY4FVEfE0gKR5wGTACcPMdjKYb8T1lgT6QhEx2G3olaQzgEkR8bE0/lHguIj4VKnONGBaGj0ceLIfq9wfeLYf8w81jmfoa7SYGi0eaLyYuornkIg4oLsZ6uUMo1cRMROYORDLkvRQRDQPxLKGAscz9DVaTI0WDzReTJXEUy/f9F4PjCuNj01lZmZWI/WSMB4EJkg6VNKewFnAwkFuk5nZLqUuuqQiYrukTwH3UNxWOysilldxlQPStTWEOJ6hr9FiarR4oPFi6nM8dXHR28zMBl+9dEmZmdkgc8IwM7MsThglkiZJelLSKkmXDXZ7BoKk1ZKWSXpU0kOD3Z6+kjRL0mZJj5fKRktaJGll+jtqMNvYV93EdIWk9Wk/PSrp9MFsY19IGifpXklPSFou6aJUXpf7qYd46nkf7SXpF5J+mWK6MpUfKumB9J73nXRTUffL8TWMQnr8yH9SevwIcHa9P35E0mqgOSLq8gtHkt4DtANzIuJPU9k/AG0RcV1K7KMi4tLBbGdfdBPTFUB7RFw/mG2rhKQDgQMj4mFJ+wBLgSnA+dThfuohnjOp330kYHhEtEvaA7gfuAj4W+B7ETFP0teBX0bELd0tx2cYO/zh8SMR8Tug4/EjNogi4j6grVPxZGB2Gp5N8c9cN7qJqW5FxIaIeDgNvwCsAMZQp/uph3jqVhTa0+ge6RXAycCdqbzXfeSEscMYYG1pfB11fpAkAfxY0tL0+JRG0BQRG9LwRqBpMBszgD4l6bHUZVUX3TedSRoPvBN4gAbYT53igTreR5J2l/QosBlYBDwFbI2I7alKr+95ThiN78SIOAY4DZieukMaRhR9qo3Qr3oL8BbgaGADcMOgtqYCkkYA3wU+ExHPl6fV437qIp663kcR8VpEHE3xpIxjgSP6ugwnjB0a8vEjEbE+/d0M3EVxoNS7TamfuaO/efMgt6ffImJT+of+PfDP1Nl+Sv3i3wXmRsT3UnHd7qeu4qn3fdQhIrYC9wInACMldXyBu9f3PCeMHRru8SOShqeLdkgaDkwEHu95rrqwEJiahqcCCwaxLQOi4401+RB1tJ/SBdVbgRUR8eXSpLrcT93FU+f76ABJI9Pw3hQ396ygSBxnpGq97iPfJVWSbpP7CjseP3Lt4LaofyS9meKsAorHwNxebzFJugNooXgU8yZgBvB9YD5wMLAGODMi6uYicjcxtVB0dQSwGvh4qf9/SJN0IvDvwDLg96n4sxT9/nW3n3qI52zqdx+9neKi9u4UJwrzI+Kq9B4xDxgNPAKcGxGvdrscJwwzM8vhLikzM8vihGFmZlmcMMzMLIsThpmZZXHCMDOzLE4YZkNAehLqJYPdDrOeOGGYDTAV/L9lDccHtdkAkDQ+/ZbKHIpvAN8q6aHybw+keqslXSnpYRW/U7LT83wk/U9J/y99I9dsyBjWexUzyzQBmBoRSySNjoi29DsriyW9PSIeS/WejYhjJH0SuAT4WMcCJH2K4rENU3r6xq3ZYPAZhtnAWRMRS9LwmZIepnjcwtuAI0v1Oh7OtxQYXyo/j+Kpwmc4WdhQ5IRhNnBehOJnLynOHE6JiLcDPwD2KtXrSAav8fqz/GUUCWRs1VtqVgEnDLOBty9F8tgmqYnirCHHI8DHgYWSDqpW48wq5YRhNsAi4pcUb/6/Am4HftaHee+nODv5gaT9q9NCs8r4abVmZpbFZxhmZpbFCcPMzLI4YZiZWRYnDDMzy+KEYWZmWZwwzMwsixOGmZll+f/cJDsrqw+q3wAAAABJRU5ErkJggg==", + "text/plain": [ + "
    " + ] + }, + "metadata": { + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "import matplotlib.pyplot as plt\n", + "\n", + "# plot a histogram, and add axis descriptions and title\n", + "out_expanded[(out_expanded['rank'] >=0)&(out_expanded['rank'] <30)]['rank'].hist(bins=29)\n", + "plt.xlabel('rank')\n", + "plt.ylabel('count')\n", + "plt.title('Histogram of ranks of retrieved paragraphs')\n", + "plt.show()" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": {}, + "outputs": [ + { + "data": { + "image/png": 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    " + ] + }, + "metadata": { + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "out_expanded[(out_expanded.tokens>=0)&(out_expanded.tokens < 2000)]['tokens'].hist(bins=29)\n", + "plt.xlabel('tokens')\n", + "plt.ylabel('count')\n", + "plt.title('Histogram of the number of minimum tokens needed')\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can observe that the context is most likely to be returned as one of the first results, and most likely to be returned within the first 200-500 tokens." + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "-2 0.000063\n", + "-1 0.074428\n", + " 0 0.453420\n", + " 1 0.089515\n", + " 2 0.047146\n", + " 3 0.032437\n", + " 4 0.024139\n", + " 5 0.019676\n", + " 6 0.015967\n", + " 7 0.013452\n", + " 8 0.011189\n", + " 9 0.009869\n", + " 10 0.009178\n", + "Name: rank, dtype: float64" + ] + }, + "execution_count": 20, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# normalized value_counts\n", + "out_expanded['rank'].value_counts(normalize=True).sort_index()[:13]" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "probabilities of the relevant context being returned at each rank. (-2 means a processing error, -1 means the rank is >200)" + ] + } + ], + "metadata": { + "interpreter": { + "hash": "be4b5d5b73a21c599de40d6deb1129796d12dc1cc33a738f7bac13269cfcafe8" + }, + "kernelspec": { + "display_name": "Python 3.7.3 64-bit ('base': conda)", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.7.3" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/examples/fine-tuned_qa/olympics-3-train-qa.ipynb b/examples/fine-tuned_qa/olympics-3-train-qa.ipynb new file mode 100644 index 0000000..76fb3f3 --- /dev/null +++ b/examples/fine-tuned_qa/olympics-3-train-qa.ipynb @@ -0,0 +1,637 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# 3. Train a fine-tuning model specialized for Q&A\n", + "This notebook will utilize the dataset of context, question and answer pairs to additionally create adversarial questions and context pairs, where the question was not generated on that context. In those cases the model will be prompted to answer \"No sufficient context for answering the question\". We will also train a discriminator model, which predicts whether the question can be answered based on the context or not.\n", + "\n", + "We will add hard adversarial examples as well, which will be based either on semantically similar sections, or neighbouring sections, originating from the same article." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": {}, + "outputs": [ + { + "data": { + "text/html": [ + "
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    titleheadingcontenttokenscontextquestionsanswers
    02020 Summer OlympicsSummaryThe 2020 Summer Olympics (Japanese: 2020年夏季オリン...7132020 Summer Olympics\\nSummary\\n\\nThe 2020 Summ...1. What is the 2020 Summer Olympics?\\n2. When ...1. The 2020 Summer Olympics is an internationa...
    12020 Summer OlympicsHost city selectionThe International Olympic Committee (IOC) vote...1262020 Summer Olympics\\nHost city selection\\n\\nT...1. \\n2. \\n3. \\n4.1. What is the International Olympic Committee...
    22020 Summer OlympicsImpact of the COVID-19 pandemicIn January 2020, concerns were raised about th...3692020 Summer Olympics\\nImpact of the COVID-19 p...1. What was the COVID-19 pandemic?\\n2. How did...1. The COVID-19 pandemic was a pandemic that o...
    32020 Summer OlympicsQualifying event cancellation and postponementConcerns about the pandemic began to affect qu...2982020 Summer Olympics\\nQualifying event cancell...1. What was the original location of the Asia ...1. The original location of the Asia & Oceania...
    42020 Summer OlympicsEffect on doping testsMandatory doping tests were being severely res...1632020 Summer Olympics\\nEffect on doping tests\\n...1. What was the COVID-19 pandemic?\\n2. What di...1. The COVID-19 pandemic was a pandemic that o...
    \n", + "
    " + ], + "text/plain": [ + " title heading \\\n", + "0 2020 Summer Olympics Summary \n", + "1 2020 Summer Olympics Host city selection \n", + "2 2020 Summer Olympics Impact of the COVID-19 pandemic \n", + "3 2020 Summer Olympics Qualifying event cancellation and postponement \n", + "4 2020 Summer Olympics Effect on doping tests \n", + "\n", + " content tokens \\\n", + "0 The 2020 Summer Olympics (Japanese: 2020年夏季オリン... 713 \n", + "1 The International Olympic Committee (IOC) vote... 126 \n", + "2 In January 2020, concerns were raised about th... 369 \n", + "3 Concerns about the pandemic began to affect qu... 298 \n", + "4 Mandatory doping tests were being severely res... 163 \n", + "\n", + " context \\\n", + "0 2020 Summer Olympics\\nSummary\\n\\nThe 2020 Summ... \n", + "1 2020 Summer Olympics\\nHost city selection\\n\\nT... \n", + "2 2020 Summer Olympics\\nImpact of the COVID-19 p... \n", + "3 2020 Summer Olympics\\nQualifying event cancell... \n", + "4 2020 Summer Olympics\\nEffect on doping tests\\n... \n", + "\n", + " questions \\\n", + "0 1. What is the 2020 Summer Olympics?\\n2. When ... \n", + "1 1. \\n2. \\n3. \\n4. \n", + "2 1. What was the COVID-19 pandemic?\\n2. How did... \n", + "3 1. What was the original location of the Asia ... \n", + "4 1. What was the COVID-19 pandemic?\\n2. What di... \n", + "\n", + " answers \n", + "0 1. The 2020 Summer Olympics is an internationa... \n", + "1 1. What is the International Olympic Committee... \n", + "2 1. The COVID-19 pandemic was a pandemic that o... \n", + "3 1. The original location of the Asia & Oceania... \n", + "4 1. The COVID-19 pandemic was a pandemic that o... " + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "import openai\n", + "import pandas as pd\n", + "df = pd.read_csv('olympics-data/olympics_qa.csv')\n", + "olympics_search_fileid = \"file-c3shd8wqF3vSCKaukW4Jr1TT\"\n", + "df.head()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Split the sections into a training and testing set" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "(3014, 754)" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "from sklearn.model_selection import train_test_split\n", + "train_df, test_df = train_test_split(df, test_size=0.2, random_state=42)\n", + "len(train_df), len(test_df)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "we check that he separator we intend to use isn't present within the contexts" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "df.context.str.contains('->').sum()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 3.1 Create the fine-tuning datasets for Q&A and discriminator models\n", + "The fine-tuning dataset is created in the following way. For every corresponding question, answer and context pair we create:\n", + "- Positive example: correct question, answer, context pair\n", + "- Negative examples:\n", + " - random negative example, where the random context is paired with the question \n", + " - two hard negative examples\n", + " - one originating from the same wikipedia article\n", + " - another, which is most similar to the correct context\n", + "\n", + "This process is noisy, as sometimes the question might be answerable given a different context, but on average we hope this won't affect the peformance too much.\n", + "\n", + "We apply the same process of dataset creation for both the discriminator, and the Q&A answering model. We apply the process separately for the training and testing set, to ensure that the examples from the traing set don't feature within the test set." + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": {}, + "outputs": [], + "source": [ + "import random\n", + "\n", + "def get_random_similar_contexts(question, context, file_id=olympics_search_fileid, search_model='ada', max_rerank=10):\n", + " \"\"\"\n", + " Find similar contexts to the given context using the search file\n", + " \"\"\"\n", + " try:\n", + " results = openai.Engine(search_model).search(\n", + " search_model=search_model, \n", + " query=question, \n", + " max_rerank=max_rerank,\n", + " file=file_id\n", + " )\n", + " candidates = []\n", + " for result in results['data'][:3]:\n", + " if result['text'] == context:\n", + " continue\n", + " candidates.append(result['text'])\n", + " random_candidate = random.choice(candidates)\n", + " return random_candidate\n", + " except Exception as e:\n", + " print(e)\n", + " return \"\"\n", + "\n", + "def create_fine_tuning_dataset(df, discriminator=False, n_negative=1, add_related=False):\n", + " \"\"\"\n", + " Create a dataset for fine tuning the OpenAI model; either for a discriminator model, \n", + " or a model specializing in Q&A, where it says if no relevant context is found.\n", + "\n", + " Parameters\n", + " ----------\n", + " df: pd.DataFrame\n", + " The dataframe containing the question, answer and context pairs\n", + " discriminator: bool\n", + " Whether to create a dataset for the discriminator\n", + " n_negative: int\n", + " The number of random negative samples to add (using a random context)\n", + " add_related: bool\n", + " Whether to add the related contexts to the correct context. These are hard negative examples\n", + "\n", + " Returns\n", + " -------\n", + " pd.DataFrame\n", + " The dataframe containing the prompts and completions, ready for fine-tuning\n", + " \"\"\"\n", + " rows = []\n", + " for i, row in df.iterrows():\n", + " for q, a in zip((\"1.\" + row.questions).split('\\n'), (\"1.\" + row.answers).split('\\n')):\n", + " if len(q) >10 and len(a) >10:\n", + " if discriminator:\n", + " rows.append({\"prompt\":f\"{row.context}\\nQuestion: {q[2:].strip()}\\n Related:\", \"completion\":f\" yes\"})\n", + " else:\n", + " rows.append({\"prompt\":f\"{row.context}\\nQuestion: {q[2:].strip()}\\nAnswer:\", \"completion\":f\" {a[2:].strip()}\"})\n", + "\n", + " for i, row in df.iterrows():\n", + " for q in (\"1.\" + row.questions).split('\\n'):\n", + " if len(q) >10:\n", + " for j in range(n_negative + (2 if add_related else 0)):\n", + " random_context = \"\"\n", + " if j == 0 and add_related:\n", + " # add the related contexts based on originating from the same wikipedia page\n", + " subset = df[(df.title == row.title) & (df.context != row.context)]\n", + " \n", + " if len(subset) < 1:\n", + " continue\n", + " random_context = subset.sample(1).iloc[0].context\n", + " if j == 1 and add_related:\n", + " # add the related contexts based on the most similar contexts according to the search\n", + " random_context = get_random_similar_contexts(q[2:].strip(), row.context, search_model='ada', max_rerank=10)\n", + " else:\n", + " while True:\n", + " # add random context, which isn't the correct context\n", + " random_context = df.sample(1).iloc[0].context\n", + " if random_context != row.context:\n", + " break\n", + " if discriminator:\n", + " rows.append({\"prompt\":f\"{random_context}\\nQuestion: {q[2:].strip()}\\n Related:\", \"completion\":f\" no\"})\n", + " else:\n", + " rows.append({\"prompt\":f\"{random_context}\\nQuestion: {q[2:].strip()}\\nAnswer:\", \"completion\":f\" No appropriate context found to answer the question.\"})\n", + "\n", + " return pd.DataFrame(rows) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We apply the same process of dataset creation for both the discriminator, and the Q&A answering model. We apply the process separately for the training and testing set, to ensure that the examples from the traing set don't feature within the test set." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [] + }, + "execution_count": 5, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "for name, is_disc in [('discriminator', True), ('qa', False)]:\n", + " for train_test, dt in [('train', train_df), ('test', test_df)]:\n", + " ft = create_fine_tuning_dataset(dt, discriminator=is_disc, n_negative=1, add_related=True)\n", + " ft.to_json(f'{name}_{train_test}.jsonl', orient='records', lines=True)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We formatted the data according to the recommendations from the fine-tuning tool, which is available using\n", + "> openai tools fine_tunes.prepare_data -f qa_train.jsonl\n", + "\n", + "We highly recommend that you use this tool, which suggests improvements in your data formatting for fine-tuning.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 3.2 Submit the datasets for fine-tuning" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "!openai api fine_tunes.create -t \"olympics-data/discriminator_train.jsonl\" -v \"olympics-data/discriminator_test.jsonl\" --batch_size 16 --compute_classification_metrics --classification_positive_class \" yes\" --model ada" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [] + }, + "execution_count": 7, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "!openai api fine_tunes.create -t \"olympics-data/qa_train.jsonl\" -v \"olympics-data/qa_test.jsonl\" --batch_size 16" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 3.3 Using the fine-tuned models\n", + "\n", + "We will now use the fine-tuned discriminator and the fine-tuned Q&A model. By requesting logprobs, we can see how certain the discriminator is in a `yes` vs `no` answer." + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[ JSON: {\n", + " \" no\": -10.819577,\n", + " \" yes\": -2.045765e-05\n", + " }]" + ] + }, + "execution_count": 8, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "ft_discriminator = \"curie:ft-openai-internal-2021-08-23-23-58-57\"\n", + "ft_qa = \"curie:ft-openai-internal-2021-08-23-17-54-10\"\n", + "\n", + "def apply_ft_discriminator(context, question, discriminator_model):\n", + " \"\"\"\n", + " Apply the fine tuned discriminator to a question, to assess whether it can be answered from the context.\n", + " \"\"\"\n", + " prompt = f\"{context}\\nQuestion: {question}\\n Related:\"\n", + " result = openai.Completion.create(model=discriminator_model, prompt=prompt, max_tokens=1, temperature=0, top_p=1, n=1, logprobs=2)\n", + " return result['choices'][0]['logprobs']['top_logprobs']\n", + "\n", + "apply_ft_discriminator('The first human-made object in space was the Soviet Union satellite Sputnik 1 on 4 October 1957.', \n", + " 'What was the first human-made object in space?', ft_discriminator)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can see that the model can generalize well to different contexts and questions. " + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "' The first human-made object in space was the Soviet Union satellite Sputnik 1 on 4 October 1957'" + ] + }, + "execution_count": 9, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "def apply_ft_qa_answer(context, question, answering_model):\n", + " \"\"\"\n", + " Apply the fine tuned discriminator to a question\n", + " \"\"\"\n", + " prompt = f\"{context}\\nQuestion: {question}\\nAnswer:\"\n", + " result = openai.Completion.create(model=answering_model, prompt=prompt, max_tokens=30, temperature=0, top_p=1, n=1, stop=['.','\\n'])\n", + " return result['choices'][0]['text']\n", + "\n", + "apply_ft_qa_answer('The first human-made object in space was the Soviet Union satellite Sputnik 1 on 4 October 1957.', \n", + " 'What was the first human-made object in space?', ft_qa)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can see that the model can answer the question, when the context is appropriate." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "' The Soviet Union was the first country to successfully launch a satellite into space'" + ] + }, + "execution_count": 10, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "apply_ft_qa_answer('The first human-made object in space was the Soviet Union satellite Sputnik 1 on 4 October 1957.',\n", + " 'What is impressive about the Soviet Union?', ft_qa)" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "' No appropriate context found to answer the question'" + ] + }, + "execution_count": 11, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "apply_ft_qa_answer('The first human-made object in space was the Soviet Union satellite Sputnik 1 on 4 October 1957.',\n", + " 'How many cars were produced in the Soviet Union in 1970?', ft_qa)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can see that the model knows when to answer the question, and when to say that insufficient context is present to answer the question." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We can also combine a discriminator and a base model, or a fine-tuned Q&A model. Discriminator can essentially serve as a decision whether the question can be answered given the context or not." + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "' Weather could cause a sport event to have no crowd'" + ] + }, + "execution_count": 12, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "def answer_question_conditionally(answering_model, discriminator_model, context, question, discriminator_logprob_yes_modifier=0):\n", + " logprobs = apply_ft_discriminator(context, question, discriminator_model)\n", + " yes_logprob = logprobs[' yes'] if ' yes' in logprobs else -100\n", + " no_logprob = logprobs[' no'] if ' no' in logprobs else -100\n", + " if yes_logprob + discriminator_logprob_yes_modifier < no_logprob:\n", + " return \" No appropriate context found to answer the question based on the discriminator.\"\n", + " return apply_ft_qa_answer(context, question, answering_model)\n", + "answer_question_conditionally(ft_qa, ft_discriminator, \n", + " \"Crowdless games are a rare although not unheard-of occurrence in sports. \\\n", + " When they do occur, it is usually the result of events beyond the control \\\n", + " of the teams or fans, such as weather-related concerns, public health concerns, \\\n", + " or wider civil disturbances unrelated to the game. For instance, \\\n", + " the COVID-19 pandemic caused many sports leagues around the world \\\n", + " to be played behind closed doors.\",\n", + " \"Could weather cause a sport event to have no crowd?\")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The above function illustrates how to potentially combine a discriminator and a fine-tuned Q&A model. This gives a more fine-grained control over how certain we want the model to be before it answers the question.\n", + "\n", + "We'll now take a look on how answers endpoint works - combining search to retrieve the relevant context from a knowledge base, and then using the fine-tuned Q&A model to answer the question." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 3.4 Answering the question based on a knowledge base\n", + "Finally we can use a logic similar to the [/answers](https://beta.openai.com/docs/api-reference/answers) endpoint, where we first search for the relevant context, and then ask a Q&A model to answer the question given that context. If you'd like to see the implementation details, check out the [`answers_with_ft.py`](answers_with_ft.py) file." + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "\" Canada won the Women's football tournament at the 2020 Olympic games\"" + ] + }, + "execution_count": 13, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "from answers_with_ft import answer_question\n", + "answer_question(olympics_search_fileid, ft_qa, \"Which country won the Women's football tournament at the 2020 Olympic games?\")" + ] + } + ], + "metadata": { + "interpreter": { + "hash": "be4b5d5b73a21c599de40d6deb1129796d12dc1cc33a738f7bac13269cfcafe8" + }, + "kernelspec": { + "display_name": "Python 3.7.3 64-bit ('base': conda)", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.7.3" + }, + "orig_nbformat": 4 + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/transition_guides_for_deprecated_API_endpoints/README.md b/transition_guides_for_deprecated_API_endpoints/README.md new file mode 100644 index 0000000..60d296a --- /dev/null +++ b/transition_guides_for_deprecated_API_endpoints/README.md @@ -0,0 +1,50 @@ +# Deprecation of Answers, Classification, and Search + +In 2021, OpenAI released specialized endpoints in beta for Answers, Classification, and Search. + +While these specialized endpoints were convenient, they had two drawbacks: + +1. These specialized endpoints were eclipsed by techniques that achieved better results. +2. These specialized endpoints were more difficult to customize and optimize for individual use cases. + +As a result, **the Answers, Classifications, and Search endpoints are being deprecated.** + +## Timeline of deprecation + +For those who have not used these endpoints, nothing will change except that access will no longer be available. + +**For existing users of these endpoints, access will continue until December 3, 2022.** Before that date, we strongly encourage developers to switch over to newer techniques which produce better results. + +## How to transition + +We've written guides and code examples for transitioning from the deprecated API endpoints to better methods. + +### Answers + +[Guide: How to transition off the Answers endpoint](https://help.openai.com/en/articles/6233728-answers-transition-guide) + +* Option 1: transition to embeddings-based search **(recommended)** + * Example code: [Semantic_text_search_using_embeddings.ipynb](../examples/Semantic_text_search_using_embeddings.ipynb) + +* Option 2: reimplement Answers endpoint functionality + * Example code: [answers_functionality_example.py](answers_functionality_example.py) + +### Classification + +[Guide: How to transition off the Classifications endpoint](https://help.openai.com/en/articles/6272941-classifications-transition-guide) + +* Option 1: transition to fine-tuning **(recommended)** + * Example code: [Classification.ipynb](../examples/Classification.ipynb) +* Option 2: transition to embeddings + * Example code: [Semantic_text_search_using_embeddings.ipynb](../examples/Semantic_text_search_using_embeddings.ipynb) +* Option 3: reimplement Classifications endpoint functionality + * Example code: [classification_functionality_example.py](classification_functionality_example.py) + +### Search + +[Guide: How to transition off the Search endpoint](https://help.openai.com/en/articles/6272952-search-transition-guide) + +* Option 1: transition to embeddings-based search **(recommended)** + * Example code: [Semantic_text_search_using_embeddings.ipynb](../examples/Semantic_text_search_using_embeddings.ipynb) +* Option 2: reimplement Search endpoint functionality + * Example code: [search_functionality_example.py](search_functionality_example.py) diff --git a/transition_guides_for_deprecated_API_endpoints/answers_functionality_example.py b/transition_guides_for_deprecated_API_endpoints/answers_functionality_example.py new file mode 100644 index 0000000..7c69a44 --- /dev/null +++ b/transition_guides_for_deprecated_API_endpoints/answers_functionality_example.py @@ -0,0 +1,304 @@ +from transformers import GPT2TokenizerFast + +import openai + +tokenizer = GPT2TokenizerFast.from_pretrained("gpt2") + +MAX_TOKENS_LIMIT = 2048 +ANSWERS_INSTRUCTION = "Please answer the question according to the above context.\n" +CONTEXT_TEMPLATE = "===\nContext: {context}\n===\n" + + +def extract_instruction(instruction): + """ + Extract `instruction` parameter and format it properly. + If not exist, return empty string. + """ + if instruction is None: + return "" + + return f"{instruction.strip()}\n\n" + + +def semantic_search( + search_model, query_for_search, file_id=None, max_documents=None, examples=None +): + """ + :param examples: A list of {"text":...} or {"text": ..., "label": ...}. + :return: + a list of semantic search result dict of documents sorted by "score": + [ + { + "document": ..., + "object": "search_result", + "score": ..., + "text": ..., + }, + ... + ] + """ + assert (examples is None) ^ (file_id is None) # xor + + if file_id is not None: + # This is where you'd do an elastic search call. Since there isn't an example of this + # we can query, we'll raise an error. + # The return value from this would be a list of examples + raise NotImplementedError() + + # This isn't quite accurate since Search is also being deprecated. See our search guide for more + # information. + + search_result = openai.Search.create( + model=search_model, + documents=[x["text"] for x in examples], + query=query_for_search, + ) + + info_dict = {d["document"]: d for d in search_result["data"]} + sorted_doc_ids = sorted( + info_dict.keys(), key=lambda x: info_dict[x]["score"], reverse=True + ) + if max_documents: + sorted_doc_ids = sorted_doc_ids[:max_documents] + return [info_dict[i] for i in sorted_doc_ids] + + +def select_by_length( + sorted_doc_infos, + max_token_len, + lambda_fn=None, +): + """ + Give a list of (document ID, document content in string), we will select as many + documents as possible as long as the total length does not go above `max_token_len`. + + :param sorted_doc_infos: A list of semantic search result dict of documents sorted by "score". + :param max_token_len: The maximum token length for selected documents. + :param lambda_fn: A function that takes in search results dict and output a formatted + example for context stuffing. + :return: A tuple of ( + A concatenation of selected documents used as context, + A list of selected document IDs + ) + """ + if not sorted_doc_infos: + return "", [] + + selected_indices = [] + total_doc_tokens = 0 + doc_dict = {} + for i, doc_info in enumerate(sorted_doc_infos): + doc = lambda_fn(doc_info) if lambda_fn else doc_info["text"] + n_doc_tokens = len(tokenizer.encode(doc)) + if total_doc_tokens + n_doc_tokens < max_token_len: + total_doc_tokens += n_doc_tokens + selected_indices.append(i) + doc_dict[i] = doc + + # The top ranked documents should go at the end. + selected_indices = selected_indices[::-1] + + context = "".join([doc_dict[i] for i in selected_indices]) + selected_doc_infos = [sorted_doc_infos[i] for i in selected_indices] + return context, selected_doc_infos + + +def answers( + examples, + question, + model, + examples_context, + file_id=None, + documents=None, + logit_bias=None, + max_rerank=200, + max_tokens=16, + alternative_question=None, + search_model="ada", + temperature=0.0, + logprobs=0, + stop=None, + n=1, +): + """ + Given a prompt, a question, a list of (question, answer) pairs as examples, and + a list of documents for context, it tries to include all the QA examples and top + relevant context documents. + + The constructed prompt for the final completion call: + ``` + Please answer the question according to the above context. + + === + Context: {{ the context for example QA pairs. }} + === + Q: example 1 question + A: example 1 answer + --- + Q: example 2 question + A: example 2 answer + === + Context: {{ a list of relevant documents sorted via search(question, documents) }} + === + Q: question + A: + ``` + + The returned object has a structure like: + { + "answers": [ + "Beijing", + "Beijing, China" + ], + "completion_id": "xxx-xxx", + "object": "answer", + "selected_documents": [ + { + "document": ..., # document index, same as in search/ results. + "object": "search_result", + "text": ..., + }, + ... + ], + } + """ + + examples = examples if examples else [] + + example_prompts = [f"Q: {x}\nA: {y}" for x, y in examples] + prompt = f"Q: {question}\nA:" + + # Append all the QA examples into the prompt. + if examples_context: + examples_context = CONTEXT_TEMPLATE.format(context=examples_context) + instruction = ( + ANSWERS_INSTRUCTION + examples_context + "\n---\n".join(example_prompts) + "\n" + ) + + logit_bias = logit_bias if logit_bias is not None else {} + + if file_id is None and documents is None: + raise Exception("Please submit at least one of `documents` or `file`.") + if file_id is not None and documents is not None: + raise Exception("Please submit only one of `documents` or `file`.") + + instruction = extract_instruction(instruction) + + n_instruction_tokens = len(tokenizer.encode(instruction)) + n_prompt_tokens = len(tokenizer.encode(prompt)) + n_query_tokens = len(tokenizer.encode(question)) + n_context_tokens = len(tokenizer.encode(CONTEXT_TEMPLATE.format(context=""))) + + if documents is not None: + documents = [doc.strip() + " " for doc in documents] + n_docs_tokens = [len(tokenizer.encode(doc)) for doc in documents] + + # Except all the required content, how many tokens left for context stuffing. + leftover_token_len = MAX_TOKENS_LIMIT - ( + n_instruction_tokens + n_context_tokens + n_prompt_tokens + max_tokens + ) + sorted_doc_infos = [] + + question_for_search = ( + alternative_question if alternative_question is not None else question + ) + if file_id is not None: + search_model_, sorted_doc_infos = semantic_search( + search_model, + question_for_search, + file_id=file_id, + max_documents=max_rerank, + ) + + elif len(documents) == 0: + # If no context document is provided, do nothing. + pass + + elif min(n_docs_tokens) >= leftover_token_len: + # If there is no room for adding any context doc. + pass + + elif (max_rerank is None or max_rerank >= len(documents)) and sum( + n_docs_tokens + ) < leftover_token_len: + # If the total length of docs is short enough to be added all. + selected_indices = list(range(len(documents))) + + sorted_doc_infos = [ + {"document": i, "text": documents[i]} for i in selected_indices + ] + + elif n_query_tokens + max(n_docs_tokens) >= MAX_TOKENS_LIMIT: + # If the prompt and the longest document together go above the limit. + total_tokens = n_query_tokens + max(n_docs_tokens) + raise Exception( + f"The longest document and prompt pair together contains {total_tokens} " + f"tokens, above the limit {MAX_TOKENS_LIMIT} for semantic search. Please consider " + f"shortening the prompt or the longest document." + ) + + else: + # If we can add some context documents but not all of them, we should + # query search endpoint to rank docs by score. + sorted_doc_infos = semantic_search( + search_model, + question_for_search, + examples=[{"text": doc} for doc in documents], + max_documents=max_rerank, + ) + + # Select documents w.r.t. the context length limitation. + context, sorted_doc_infos = select_by_length( + sorted_doc_infos, + leftover_token_len, + lambda_fn=lambda x: x["text"].strip() + " ", + ) + + # Add instruction before the context and the prompt after the context. + if context: + context = CONTEXT_TEMPLATE.format(context=context.strip()) + full_prompt = instruction + context + prompt + + completion_result = openai.Completion.create( + engine=model, + prompt=full_prompt, + logit_bias=logit_bias, + temperature=temperature, + n=n, + max_tokens=max_tokens, + stop=stop, + logprobs=logprobs, + ) + + completion_result["selected_documents"] = sorted_doc_infos + + result = dict( + object="answer", + selected_documents=completion_result.pop("selected_documents"), + completion=completion_result["id"], + ) + + result["answers"] = [ + item["text"].replace("A:", "").split("Q:")[0].strip() + for item in completion_result["choices"] + ] + + return result + + +print( + answers( + examples=[ + ["What is the capital of Washington", "Olympia"], + ["What is the capital of Oregon", "Salem"], + ], + question="What is the capital of China?", + examples_context="I am a bot that names country capitals", + documents=["I am a bot that names country capitals"], + model="davinci", + search_model="ada", + alternative_question="different test", + max_tokens=16, + stop=["\n\n"], + ) +) diff --git a/transition_guides_for_deprecated_API_endpoints/classification_functionality_example.py b/transition_guides_for_deprecated_API_endpoints/classification_functionality_example.py new file mode 100644 index 0000000..e847fb2 --- /dev/null +++ b/transition_guides_for_deprecated_API_endpoints/classification_functionality_example.py @@ -0,0 +1,302 @@ +import itertools +from collections import defaultdict + +from transformers import GPT2TokenizerFast + +import openai + +tokenizer = GPT2TokenizerFast.from_pretrained("gpt2") + +MAX_TOKENS_LIMIT = 2048 + + +def create_instruction(labels) -> str: + """ + Construct an instruction for a classification task. + """ + instruction = f"Please classify a piece of text into the following categories: {', '.join(labels)}." + + return f"{instruction.strip()}\n\n" + + +def semantic_search( + search_model, query_for_search, file_id=None, max_documents=None, examples=None +): + """ + :param examples: A list of {"text":...} or {"text": ..., "label": ...}. + :return: + a list of semantic search result dict of documents sorted by "score": + [ + { + "document": ..., + "object": "search_result", + "score": ..., + "text": ..., + }, + ... + ] + + """ + assert (examples is None) ^ (file_id is None) # xor + + if file_id is not None: + # This is where you'd do an elastic search call. Since there isn't an example of this + # we can query, we'll raise an error. + # The return value from this would be a list of examples + raise NotImplementedError() + + # This isn't quite accurate since Search is also being deprecated. See our search guide for more + # information. + + search_result = openai.Search.create( + model=search_model, + documents=[x["text"] for x in examples], + query=query_for_search, + ) + + info_dict = {d["document"]: d for d in search_result["data"]} + sorted_doc_ids = sorted( + info_dict.keys(), key=lambda x: info_dict[x]["score"], reverse=True + ) + if max_documents: + sorted_doc_ids = sorted_doc_ids[:max_documents] + return [info_dict[i] for i in sorted_doc_ids] + + +def select_by_length( + sorted_doc_infos, + max_token_len, + lambda_fn=None, +): + """ + Give a list of (document ID, document content in string), we will select as many + documents as possible as long as the total length does not go above `max_token_len`. + + :param sorted_doc_infos: A list of semantic search result dict of documents sorted by "score". + :param max_token_len: The maximum token length for selected documents. + :param lambda_fn: A function that takes in search results dict and output a formatted + example for context stuffing. + :return: A tuple of ( + A concatenation of selected documents used as context, + A list of selected document IDs + ) + """ + if not sorted_doc_infos: + return "", [] + + selected_indices = [] + total_doc_tokens = 0 + doc_dict = {} + for i, doc_info in enumerate(sorted_doc_infos): + doc = lambda_fn(doc_info) if lambda_fn else doc_info["text"] + n_doc_tokens = len(tokenizer.encode(doc)) + if total_doc_tokens + n_doc_tokens < max_token_len: + total_doc_tokens += n_doc_tokens + selected_indices.append(i) + doc_dict[i] = doc + + # The top ranked documents should go at the end. + selected_indices = selected_indices[::-1] + + context = "".join([doc_dict[i] for i in selected_indices]) + selected_doc_infos = [sorted_doc_infos[i] for i in selected_indices] + return context, selected_doc_infos + + +def format_example_fn(x: dict) -> str: + return "Text: {text}\nCategory: {label}\n---\n".format( + text=x["text"].replace("\n", " ").strip(), + label=x["label"].replace("\n", " ").strip(), + ) + + +def classifications( + query, + model, + search_model="ada", + examples=None, + file=None, + labels=None, + temperature=0.0, + logprobs=None, + max_examples=200, + logit_bias=None, + alternative_query=None, + max_tokens=16, +) -> dict: + """ + Given a prompt, a question and a list of examples, containing (text, label) pairs, + it selects top relevant examples to construct a prompt for few-shot classification. + + The constructed prompt for the final completion call: + ``` + {{ an optional instruction }} + + Text: example 1 text + Category: example 2 label + --- + Text: example 1 text + Category: example 2 label + --- + Text: question + Category: + ``` + + The returned object has a structure like: + { + "label": "Happy", + "model": "ada", + "object": "classification", + "selected_examples": [ + { + "document": ..., # document index, same as in search/ results. + "text": ..., + "label": ..., + }, + ... + ], + } + """ + + query = query.replace("\n", " ").strip() + logit_bias = logit_bias if logit_bias else {} + labels = labels if labels else [] + + if file is None and examples is None: + raise Exception("Please submit at least one of `examples` or `file`.") + if file is not None and examples is not None: + raise Exception("Please submit only one of `examples` or `file`.") + + instruction = create_instruction(labels) + + query_for_search = alternative_query if alternative_query is not None else query + + # Extract examples and example labels first. + if file is not None: + sorted_doc_infos = semantic_search( + search_model, + query_for_search, + file_id=file, + max_documents=max_examples, + ) + + else: + example_prompts = [ + format_example_fn(dict(text=x, label=y)) for x, y in examples + ] + n_examples_tokens = [len(tokenizer.encode(x)) for x in example_prompts] + + query_prompt = f"Text: {query}\nCategory:" + n_instruction_tokens = len(tokenizer.encode(instruction)) + n_query_tokens = len(tokenizer.encode(query_prompt)) + + # Except all the required content, how many tokens left for context stuffing. + leftover_token_len = MAX_TOKENS_LIMIT - ( + n_instruction_tokens + n_query_tokens + max_tokens + ) + + # Process when `examples` are provided but no `file` is provided. + if examples: + if (max_examples is None or max_examples >= len(examples)) and sum( + n_examples_tokens + ) < leftover_token_len: + # If the total length of docs is short enough that we can add all examples, no search call. + selected_indices = list(range(len(examples))) + + sorted_doc_infos = [ + {"document": i, "text": examples[i][0], "label": examples[i][1]} + for i in selected_indices + ] + + elif max(n_examples_tokens) + n_query_tokens >= MAX_TOKENS_LIMIT: + # If the prompt and the longest example together go above the limit: + total_tokens = max(n_examples_tokens) + n_query_tokens + raise Exception( + user_message=f"The longest classification example, query and prompt together contain " + f"{total_tokens} tokens, above the limit {MAX_TOKENS_LIMIT} for semantic search. " + f"Please consider shortening your instruction, query or the longest example." + ) + + else: + # If we can add some context documents but not all of them, we should + # query search endpoint to rank docs by score. + sorted_doc_infos = semantic_search( + search_model, + query_for_search, + examples=[{"text": x, "label": y} for x, y in examples], + max_documents=max_examples, + ) + + # Per label, we have a list of doc id sorted by its relevancy to the query. + label_to_indices = defaultdict(list) + for idx, d in enumerate(sorted_doc_infos): + label_to_indices[d["label"]].append(idx) + + # Do a round robin for each of the different labels, taking the best match for each label. + label_indices = [label_to_indices[label] for label in labels] + mixed_indices = [ + i for x in itertools.zip_longest(*label_indices) for i in x if i is not None + ] + sorted_doc_infos = [sorted_doc_infos[i] for i in mixed_indices] + + # Try to select as many examples as needed to fit into the context + context, sorted_doc_infos = select_by_length( + sorted_doc_infos, + leftover_token_len, + lambda_fn=format_example_fn, + ) + + prompt = instruction + context + query_prompt + + completion_params = { + "engine": model, + "prompt": prompt, + "temperature": temperature, + "logprobs": logprobs, + "logit_bias": logit_bias, + "max_tokens": max_tokens, + "stop": "\n", + "n": 1, + } + + completion_resp = openai.Completion.create( + **completion_params, + ) + + label = completion_resp["choices"][0]["text"] + label = label.split("\n")[0].strip().lower().capitalize() + if label not in labels: + label = "Unknown" + + result = dict( + # TODO: Add id for object persistence. + object="classification", + model=completion_resp["model"], + label=label, + completion=completion_resp["id"], + ) + + result["selected_examples"] = sorted_doc_infos + + return result + + +print( + classifications( + query="this is my test", + model="davinci", + search_model="ada", + examples=[ + ["this is my test", "davinci"], + ["this is other test", "blahblah"], + ], + file=None, + labels=["davinci", "blahblah"], + temperature=0.1, + logprobs=0, + max_examples=200, + logit_bias=None, + alternative_query="different test", + max_tokens=16, + ) +) diff --git a/transition_guides_for_deprecated_API_endpoints/search_functionality_example.py b/transition_guides_for_deprecated_API_endpoints/search_functionality_example.py new file mode 100644 index 0000000..1589435 --- /dev/null +++ b/transition_guides_for_deprecated_API_endpoints/search_functionality_example.py @@ -0,0 +1,76 @@ +from transformers import GPT2TokenizerFast + +import openai + +tokenizer = GPT2TokenizerFast.from_pretrained("gpt2") + +docs = ["test1", "asdklgjnasdv", "banana", "lord lollipop"] +query = "apple orang asdansbdausd" + +print(openai.Search.create(model="davinci", query=query, documents=docs)) + + +def construct_context(query, document): + return "<|endoftext|>{document}\n\n---\n\nThe above passage is related to: {query}".format( + document=document, query=query + ) + + +def get_score(context, query, log_probs, text_offsets) -> float: + SCORE_MULTIPLIER = 100.0 + + log_prob = 0 + count = 0 + cutoff = len(context) - len(query) + + for i in range(len(text_offsets) - 1, 0, -1): + log_prob += log_probs[i] + count += 1 + + if text_offsets[i] <= cutoff and text_offsets[i] != text_offsets[i - 1]: + break + + return log_prob / float(count) * SCORE_MULTIPLIER + + +def search(query, documents, engine): + + prompts = [construct_context(query, doc) for doc in [""] + docs] + + resps = openai.Completion.create( + model=engine, + prompt=prompts, + temperature=1.0, + top_p=1.0, + max_tokens=0, + logprobs=0, + n=1, + echo=True, + ) + + resps_by_index = {choice["index"]: choice for choice in resps["choices"]} + + scores = [ + get_score( + prompts[i], + query, + resps_by_index[i]["logprobs"]["token_logprobs"], + resps_by_index[i]["logprobs"]["text_offset"], + ) + for i in range(len(prompts)) + ] + + # Process results + scores = [score - scores[0] for score in scores][1:] + + return [ + { + "object": "search_result", + "document": document_idx, + "score": round(score, 3), + } + for document_idx, score in enumerate(scores) + ] + + +print(search(query=query, documents=docs, engine="davinci"))