2002 AMC 10B Problems/Problem 22

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 $XY$ .

$\mathrm{(A) \ } 24\qquad \mathrm{(B) \ } 26\qquad \mathrm{(C) \ } 28\qquad \mathrm{(D) \ } 30\qquad \mathrm{(E) \ } 32$

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 square root. $\begin{align*} 5x^2+5y^2&=845\\ 4x^2+4y^2&=676\\ \sqrt{4x^2+4y^2}&=\boxed{\mathrm{(B) \ } 26} \end{align*}$

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search

2002 AMC 10B Problems/Problem 22

Problem

Solution

See Also