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
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