2008 iTest Problems/Problem 55

Problem

Let $\triangle XOY$ be a right-angled triangle with $\angle XOY=90^\circ$. Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$, respectively. Find the length $XY$ given that $XN=22$ and $YM=31$.

Solution

[asy]  pair O=(0,0),X=(0,60),Y=(80,0),M=(0,30),N=(40,0); draw((0,5)--(5,5)--(5,0)); draw(O--X--Y--O); draw(X--N); draw(Y--M); dot(O); label("O",O,SW); dot(M); label("M",M,W); dot(X); label("X",X,NW); dot(N); label("N",N,S); dot(Y); label("Y",Y,SE);  label("a",(0,15),W); label("a",(0,45),W); label("b",(20,0),S); label("b",(60,0),S);  [/asy]

Let $a = MO = MX$ and $b = NO = NY$. By the Pythagorean Theorem, \[4a^2 + b^2 = 22^2\] \[a^2 + 4b^2 = 31^2\] Add the two equations to get \[5a^2 + 5b^2 = 22^2 + 31^2\] \[5a^2 + 5b^2 = 1445\] Multiply both sides by $\tfrac45$ to get \[4a^2 + 4b^2 = 1156\] \[(2a)^2 + (2b)^2 = 1156\] Thus, the length of $XY$ is $\sqrt{1156} = \boxed{34}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 54
Followed by:
Problem 56
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