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Difference between revisions of "1991 AHSME Problems/Problem 22"

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== Solution ==
 
== Solution ==
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<math>\fbox{B}</math>
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Using the tangent-tangent theorem, <math>PA=AB=PA'=A'B'=4</math>. 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 <math>S</math> and the center of the larger circle be point <math>L</math>. If we let the radius of the larger circle be <math>x</math> and the radius of the smaller circle be <math>y</math>, we can see that, using similar triangle, <math>x=2y</math>. In addition, the total hypotenuse of the larger right triangles equals <math>2(x+y)</math> since half of it is <math>x+y</math>, so <math>y^2+4^2=(3y)^2</math>. If we simplify, we get <math>y^2+16=9y^2</math>, so <math>8y^2=16</math>, so <math>y=\sqrt2</math>. This means that the smaller circle has area <math>\fbox{2\pi}</math>, which is answer choice <math>\fbox{B}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:07, 11 October 2016

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 $\fbox{2\pi}$ (Error compiling LaTeX. Unknown error_msg), which is answer choice $\fbox{B}$.

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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