Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.3 AMC 10/12 Find the answer

Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY = 90^{\circ}$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Given that $XN = 19$ and $YM = 22$ , find $XY$ :

Pick one

Solution

2002 12B AMC-20.png
Let $OM = x$ , $ON = y$ . By the 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 $XY=2(MN)=\boxed{\mathrm{(B)}\ 26}$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.