Points $A$ and $B$ lie on a [circle](https://artofproblemsolving.com/wiki/index.php/Circle) centered at $O$, and $\angle AOB = 60^\circ$. A second circle is internally [tangent](https://artofproblemsolving.com/wiki/index.php/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?
$\mathrm{(A)}\ \frac{1}{16}\qquad\mathrm{(B)}\ \frac{1}{9}\qquad\mathrm{(C)}\ \frac{1}{8}\qquad\mathrm{(D)}\ \frac{1}{6}\qquad\mathrm{(E)}\ \frac{1}{4}$
Let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is [tangent](https://artofproblemsolving.com/wiki/index.php/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 $\frac{1}{3}=\frac{r}{R}$.
Ratio of areas of circles is ratio of radii squared, so the answer is $\left(\frac{1}{3}\right)^2 = \frac{1}{9} \Rightarrow \boxed{B}$