# 2008 iTest Problems/Problem 89

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## Problem

Two perpendicular planes intersect a sphere in two circles. These circles intersect in two points, $A$ and $B$, such that $AB=42$. If the radii of the two circles are $54$ and $66$, find $R^2$, where $R$ is the radius of the sphere.

## Solution

Let $O$ be the center of the sphere, $X$ be the center of the circle radius $54$, $Y$ be the center of the circle radius $66$, and $M$ be the midpoint of $AB$. Since $XM \perp AB$ and $YM \perp AB$, by the Pythagorean Theorem, $XM = \sqrt{54^2 - 21^2} = 15\sqrt{11}$ and $YM = \sqrt{66^2 - 21^2} = 3\sqrt{435}$.

Additionally, by symmetry, the plane containing $X, M, Y$ must also contain $O$. Since the two planes are perpendicular, $XM \perp YM$. Because $OX \perp XM$ and $OY \perp YM$, $OXMY$ is a rectangle, so $OX = 3\sqrt{435}$. $[asy] draw(circle((0,0),100)); draw((-100,0)--(10,99.499)); draw((95,-31.225)--(-20,97.980)); dot((0,0)); label("O",(0,0),S); pair X=(-45,49.750),Y=(37.5,33.376); draw(X--(0,0)--Y); dot(X); label("X",X,NW); dot(Y); label("Y",Y,NE); draw((0,0)--(-100,0),dotted); [/asy]$

Thus, by the Pythagorean Theorem the radius of the circle is $\sqrt{54^2 + 3915} = \sqrt{6831}$, so $R^2 = \boxed{6831}$.

## See Also

 2008 iTest (Problems) Preceded by:Problem 88 Followed by:Problem 90 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
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