Difference between revisions of "2008 AMC 12A Problems/Problem 13"

Revision as of 17:45, 17 February 2008

Problem

Points $A$ and $B$ lie on a circle centered at $O$ , and $\angle AOB = 60^\circ$ . A second circle is internally tangent to the first and tangent to both $\overline{OA}$ and $\overline{OB}$ . What is the ratio of the area of the smaller circle to that of the larger circle?

$\textbf{(A)}\ \frac {1}{16} \qquad \textbf{(B)}\ \frac {1}{9} \qquad \textbf{(C)}\ \frac {1}{8} \qquad \textbf{(D)}\ \frac {1}{6} \qquad \textbf{(E)}\ \frac {1}{4}$

Solution

$[asy] size(300); defaultpen(0.8); pair O=(0,0), A=(3,0), B=(3/2,3/2*3^.5), C=(3^.5,1), D=(3^.5,0), F=(1.5*3^.5,1.5); picture p = new picture; draw(p,Circle(O,0.2)); clip(p,O--C--A--cycle); add(p); draw(Circle(O,3)); dot(A); dot(B); dot(C); dot(O); draw(A--O--B); draw(O--C--D); draw(C--F); draw(D-(0.2,0)--D-(0.2,-0.2)--D-(0,-0.2)); draw(Circle(C,1)); label("$30^{\circ}$",(.53,.1),O); label("$r$",(C+D)/2,E); label("$2r$",(O+C)/2,SE); label("$O$",O,SW); label("$r$",(C+F)/2,SE); label("$R$",(O+A)/2-(0,0.3),S); label("$P$",C,NW); label("$Q$",D,SE); [/asy]$

Let $P$ be the center of the small circle with radius $r$ , and let $Q$ be the point where the small circle is tangent to $OA$ , and finally, let $C$ be the point where the small circle is tangent to the big circle with radius $R$ . Then $PQO$ is a right triangle, and a 30-60-90 triangle at that. So, $OP = 2PQ$ . Since $OP = OC - PC = OC - r = R - r$ , we have $R - r = 2PQ$ , or $R - r = 2r$ , or $\frac {1}{3} = \frac {r}{R}$ . Then the ratio of areas will be $\frac {1}{3}$ squared, or $\frac {1}{9}$ .

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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Difference between revisions of "2008 AMC 12A Problems/Problem 13"

Revision as of 17:45, 17 February 2008

Problem

Solution

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