2002 AMC 10B Problems/Problem 22

Revision as of 00:01, 30 December 2008 by Lifeisacircle (talk | contribs) (created page, added problem, solution, need box)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $XY=26$