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]](http://latex.artofproblemsolving.com/c/9/6/c96955003b9b7250ca252b97c1660969fc7f5725.png)
Two circles are externally tangent. Lines 
and 
are common tangents with 
and 
on the smaller circle 
and 
on the larger circle. If 
, then the area of the smaller circle is
Solution
Using the tangent-tangent theorem, 
. 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 
and the center of the larger circle be point 
. If we let the radius of the larger circle be 
and the radius of the smaller circle be 
, we can see that, using similar triangle, 
. In addition, the total hypotenuse of the larger right triangles equals 
since half of it is 
, so 
. If we simplify, we get 
, so 
, so 
. This means that the smaller circle has area $\fbox{2\pi}$ (Error compiling LaTeX. Unknown error_msg), which is answer choice 
.
See also
| 1991 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
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