(New page: Choose 13 real numbers <math>x_1,x_2,\ldots,x_{13}\in\mathbb{R}</math> with <math>x_i\neq x_j</math> if <math>i\neq j</math>. For these 13 numbers there exist at least two numbers amongst ...) |
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Choose 13 real numbers <math>x_1,x_2,\ldots,x_{13}\in\mathbb{R}</math> with <math>x_i\neq x_j</math> if <math>i\neq j</math>. For these 13 numbers there exist at least two numbers amongst them such that | Choose 13 real numbers <math>x_1,x_2,\ldots,x_{13}\in\mathbb{R}</math> with <math>x_i\neq x_j</math> if <math>i\neq j</math>. For these 13 numbers there exist at least two numbers amongst them such that | ||
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It's a tricky one to prove, unless you are one endowed with a certain amount of intuition (exempli gratia, Uli, et alii). Simple trigonometry and the ever overlooked Pigeonhole Principle are key tools to solving it. | It's a tricky one to prove, unless you are one endowed with a certain amount of intuition (exempli gratia, Uli, et alii). Simple trigonometry and the ever overlooked Pigeonhole Principle are key tools to solving it. | ||
| − | :--[[User:Narupley| | + | :--[[User:Narupley|Nick Rupley]] 10:13, 18 January 2009 (UTC) |
Latest revision as of 05:15, 18 January 2009
Choose 13 real numbers $ x_1,x_2,\ldots,x_{13}\in\mathbb{R} $ with $ x_i\neq x_j $ if $ i\neq j $. For these 13 numbers there exist at least two numbers amongst them such that
$ 0 < \frac{x_i-x_j}{1+x_ix_j} \leq 2-\sqrt{3} $
It's a tricky one to prove, unless you are one endowed with a certain amount of intuition (exempli gratia, Uli, et alii). Simple trigonometry and the ever overlooked Pigeonhole Principle are key tools to solving it.
- --Nick Rupley 10:13, 18 January 2009 (UTC)
