Let $\triangle XOY$ be a [right-angled triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle) with $m\angle XOY = 90^{\circ}$. Let $M$ and $N$ be the [midpoints](https://artofproblemsolving.com/wiki/index.php/Midpoint) of legs $OX$ and $OY$, respectively. Given that $XN = 19$ and $YM = 22$, find $XY$. 
$\mathrm{(A)}\ 24 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 28 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 32$
[2002 12B AMC-20.png](https://artofproblemsolving.com/wiki/index.php/File:2002_12B_AMC-20.png)
Let $OM = x$, $ON = y$. By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) on $\triangle XON, MOY$ respectively,
\begin{align*} (2x)^2 + y^2 &= 19^2\\ x^2 + (2y)^2 &= 22^2\end{align*}
Summing these gives $5x^2 + 5y^2 = 845 \Longrightarrow x^2 + y^2 = 169$. 
By the Pythagorean Theorem again, we have 
\[(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = \sqrt{4(169)} = \sqrt{676} = \boxed{\mathrm{(B)}\ 26}\]
Alternatively, we could note that since we found $x^2 + y^2 = 169$, segment $MN=13$. Right triangles $\triangle MON$ and $\triangle XOY$ are similar by Leg-Leg with a ratio of $\frac{1}{2}$, so $XY=2(MN)=\boxed{\mathrm{(B)}\ 26}$