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 $\sqrt{1156} = \boxed{34}$ .

2008 iTest (Problems)
Preceded by: Problem 54	Followed by: Problem 56
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100

Page

Toolbox

Search

2008 iTest Problems/Problem 55

Problem

Solution

See Also