Question

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 $YM = 22,$ what is $XY?$

$24$

$26$

$28$

$30$

$32$

Answer: B

Concepts:Pythagorean Theorem median (geometry)system of equations

Difficulty rating: 1690

Solution:

Let $OM = a$ and $ON = b,$ so $OX = 2a$ and $OY = 2b.$ The right angle at $O$ gives $19^2 = (2a)^2 + b^2$ and $22^2 = a^2 + (2b)^2.$

Adding these, $5(a^2 + b^2) = 19^2 + 22^2 = 845,$ so $a^2 + b^2 = 169$ and $MN = \sqrt{a^2 + b^2} = 13.$

Since $\triangle XOY \sim \triangle MON$ with ratio $2,$ we have $XY = 2\cdot MN = 26.$

Thus, the correct answer is B.

2002 AMC 10B Problem 22

Answer: B

Difficulty rating: 1690

Solution:

Problem 22 in Other Years