## Problem of the Day #58: Pythagorean TriplesMay 16, 2011

Posted by Seungln in : potd , trackback

Let $ALBERT$ be a hexagon inscribed in the circle $O$ with radius $r$. $ALBERT$ has several interesting properties.

1. $AL = ER, LB = RT, BE = TA$

2. $\cos ALE = \cos LBR = \cos BET$

$= \cos ERA = \cos RTL = \cos TAB= 0$

3. $\cos AEL = \cos EAR = \frac{12}{13}$

4. $\cos LRB = \cos RLT = \frac{15}{17}$

5. $\cos BTE = \cos TBA = \frac{84}{85}$

6. $3 | r$

If $r$ is the hypotenuse of a Pythagorean triple, then there are four possible values for the length of the shorter leg, only one of which is not the length of one of the sides of $ALBERT$. What is the value?

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