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2002 AMC 12B Problems/Problem 20

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Problem

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 $XY$ .

$\mathrm{(A)}\ 24\qquad\mathrm{(B)}\ 26\qquad\mathrm{(C)}\ 28\qquad\mathrm{(D)}\ 30\qquad\mathrm{(E)}\ 32$

Solution

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

$(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = 26 \Rightarrow \mathrm{(B)}$

See also

2002 AMC 12B (Problems • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2002_AMC_12B_Problems/Problem_20"

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