2002 AMC 10B Problem 22

1. Name the pieces. $M$ is the middle of $OX$, so call $OM = MX = x$; then $OX = 2x$. Likewise call $ON = NY = y$; then $OY = 2y$.
2. Use the Pythagorean theorem on the two given segments.
- $XN$ is the long side of right triangle $XON$, whose legs are $OX = 2x$ and $ON = y$:
$$(2x)^2 + y^2 = 19^2, \qquad\text{so}\qquad 4x^2 + y^2 = 361.$$
- $YM$ is the long side of right triangle $YOM$, whose legs are $OM = x$ and $OY = 2y$:
$$x^2 + (2y)^2 = 22^2, \qquad\text{so}\qquad x^2 + 4y^2 = 484.$$
3. Decide what we need. $XY$ is the long side of the big triangle with legs $2x$ and $2y$, so
$$XY^2 = 4x^2 + 4y^2 = 4(x^2 + y^2).$$
We only need $x^2 + y^2$, not $x$ and $y$ on their own.
4. Add the two equations. The $4$s swap places, so adding treats $x$ and $y$ equally:
$$5x^2 + 5y^2 = 361 + 484 = 845, \qquad x^2 + y^2 = 169.$$
5. So $XY^2 = 4 \cdot 169 = 676$ and $XY = 26$.

The answer is $\boxed{\textbf{(B)}\ 26}$.

The two equations were mirror images of each other, and the thing we wanted, $x^2 + y^2$, treats $x$ and $y$ the same way. Adding mirror-image equations is the natural move in that situation. Before solving for each unknown, always check whether the question only needs a combination of them.

After step 4, notice that $x^2 + y^2$ is the square of $MN$, since triangle $MON$ has legs $x$ and $y$. So $MN = 13$. Triangle $MON$ is the big triangle $XOY$ shrunk by half (both legs are halved), so $XY = 2 \cdot 13 = 26$.