1991 AHSME Problems/Problem 22

Problem

$[asy] draw(circle((0,6sqrt(2)),2sqrt(2)),black+linewidth(.75)); draw(circle((0,3sqrt(2)),sqrt(2)),black+linewidth(.75)); draw((-8/3,16sqrt(2)/3)--(-4/3,8sqrt(2)/3)--(0,0)--(4/3,8sqrt(2)/3)--(8/3,16sqrt(2)/3),dot); MP("B",(-8/3,16*sqrt(2)/3),W);MP("B'",(8/3,16*sqrt(2)/3),E); MP("A",(-4/3,8*sqrt(2)/3),W);MP("A'",(4/3,8*sqrt(2)/3),E); MP("P",(0,0),S); [/asy]$

Two circles are externally tangent. Lines $\overline{PAB}$ and $\overline{PA'B'}$ are common tangents with $A$ and $A'$ on the smaller circle $B$ and $B'$ on the larger circle. If $PA=AB=4$ , then the area of the smaller circle is

$\text{(A) } 1.44\pi\quad \text{(B) } 2\pi\quad \text{(C) } 2.56\pi\quad \text{(D) } \sqrt{8}\pi\quad \text{(E) } 4\pi$

Solution

Using the tangent-tangent theorem, $PA=AB=PA'=A'B'=4$ . We can then drop perpendiculars from the centers of the circles to the points of tangency and use similar triangles. Let us let the center of the smaller circle be point $S$ and the center of the larger circle be point $L$ . If we let the radius of the larger circle be $x$ and the radius of the smaller circle be $y$ , we can see that, using similar triangle, $x=2y$ . In addition, the total hypotenuse of the larger right triangles equals $2(x+y)$ since half of it is $x+y$ , so $y^2+4^2=(3y)^2$ . If we simplify, we get $y^2+16=9y^2$ , so $8y^2=16$ , so $y=\sqrt2$ . This means that the smaller circle has area $2\pi$ , which is answer choice $\fbox{B}$ .

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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1991 AHSME Problems/Problem 22

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