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2002 AMC 10B Problems/Problem 22

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Problem

Let \triangle{XOY} be a right-triangle with m\angle{XOY}=90^\circ. Let M and N be the midpoints of the legs OX and OY, respectively. Given XN=19 and YM=22, find XY.

Solution

Let OM=MX=x and ON=NY=y. By the Pythagorean Theorem, x^2+4y^2=484 and 4x^2+y^2=361 We wish to find \sqrt{4x^2+4y^2}. So, we add the two equations, multiply by \frac{4}{5}, and take the squareroot to get XY=26

Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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