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

The following problem is from both the 2008 AMC 12A #13 and 2008 AMC 10A #16, so both problems redirect to this page.

## 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?

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

## Solution 1

$[asy]size(200); defaultpen(fontsize(10)); 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("$$30^{\circ}$$",(0.65,0.15),O); label("$$r$$",(C+D)/2,E); label("$$2r$$",(O+C)/2,NNE); 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 $OA$. Also, let $C$ be the point where the small circle is tangent to the big circle with radius $R$.

Then $PQO$ is a right triangle. Angle $POQ$ is $30$ degrees because line $OP$ bisects angle $AOB$ (this can be proved by dropping a perpendicular line from $P$ to line $OB$, letting their intersection be point $S$, and proving triangles $PQO$ and $PSO$ congruent), meaning that $PQO$ is a $30-60-90$ triangle. Therefore, $OP=2PQ$.

Since $OP=OC-PC=OC-r=R-r$, we have $R-r=2PQ$, or $R-r=2r$, or $\frac{1}{3}=\frac{r}{R}$.

## Solution 2

Like in Solution 1, 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 $OA$.

Let $N$ be the tangency point of the two circles. As shown in Solution 1, $POQ = 30$ degrees, so angle $NOA$ is also $30$ degrees. Let the line tangent to the two circles at $N$ intersect $OA$ and $OB$ at points $C$ and $D$, respectively. Since line $CD$ is tangent to circles $O$ and $P$, it must be perpendicular to $ON$, meaning that angle $ONC$ must be $90$ degrees. Because angle $NOA$ is $30$ degrees, angle $DCO$ is $180-30-90 = 60$ degrees. Angle $DOC$ is also $60$ degrees, so triangle $DOC$ is equilateral.

Note that an equilateral triangle's incenter is also its centroid. This means the center of the inscribed circle is also the centroid. From properties of median lengths, the radius of the large circle is 3 times the radius of the small circle.

Then the ratio of areas will be $\frac{1}{3}$ squared, or $\frac{1}{9}\Rightarrow \boxed{\text{B}}$.