2002 AMC 10B Problem 22

Let $OM = MX = x$ and $ON = NY = y$, so the legs are $OX = 2x$ and $OY = 2y$.

Segment $XN$ is the hypotenuse of right triangle $XON$ with legs $2x$ and $y$; segment $YM$ is the hypotenuse of right triangle $YOM$ with legs $x$ and $2y$. The Pythagorean theorem gives
$$4x^2 + y^2 = 19^2 = 361, \qquad x^2 + 4y^2 = 22^2 = 484.$$

The two equations are mirror images, so add them rather than solving:
$$5x^2 + 5y^2 = 845 \quad\Longrightarrow\quad x^2 + y^2 = 169.$$

The hypotenuse we want satisfies $XY^2 = (2x)^2 + (2y)^2 = 4(x^2 + y^2) = 4 \cdot 169 = 676$, so $XY = 26$.

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

Midpoints on the legs create two auxiliary right triangles whose hypotenuses are given, and the target hypotenuse is a third combination of the same two unknowns. Because the target $x^2 + y^2$ is symmetric while the data are swapped versions of each other, adding is the natural move; the individual values of $x$ and $y$ are never needed.

Note $x^2 + y^2 = MN^2$, so $MN = 13$. Triangle $MON$ is a half-scale copy of triangle $XOY$ (both legs halved), so $XY = 2 \cdot MN = 26$.

• Solving for x^2 and y^2 separately (they come out as 64 and 105, easy to botch) instead of adding the equations, or forgetting the factor 4 in XY^2 = (2x)^2 + (2y)^2.
• Assigning $XN$ to the triangle with legs $x$ and $2y$ (mixing up which median goes with which leg); the sum trick still rescues the answer, but any other method breaks.