2008 AMC 10A Problem 16

Call the big radius $R$ and the small radius $r$. Let $P$ be the center of the small circle.

1. The small circle touches both $OA$ and $OB$, so $P$ is the same distance, $r$, from both lines. That puts $P$ on the line that cuts $\angle AOB$ in half. So $\angle POA = 60^\circ \div 2 = 30^\circ$.
2. Drop a line from $P$ straight down to $OA$ at a right angle. It meets $OA$ where the small circle touches it, so this line has length $r$.
3. Now look at the triangle made by $O$, $P$ and that foot. It has angles $30^\circ$, $60^\circ$, $90^\circ$. In such a triangle, the longest side is twice the side across from the $30^\circ$ angle. The side across from the $30^\circ$ angle is $r$, so
$$OP = 2r.$$
4. Next use the fact that the small circle touches the big circle from inside. Then $O$, $P$ and the touching point $T$ lie on one straight line. The distance from $O$ to $T$ is $R$, and the last piece, from $P$ to $T$, is $r$. So $OP = R - r$.
5. Put the two facts together:
$$2r = R - r, \qquad 3r = R, \qquad r = \frac{R}{3}.$$
6. Areas scale with the square of the radius, so the ratio of areas is
$$\left(\frac{r}{R}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{1}{9}}$.

Two facts do the work. A circle touching both sides of an angle has its center on the line that halves the angle, which gives a right triangle with a known angle. And when one circle sits inside another and touches it, the distance between centers is the big radius minus the small one. Draw the centers and the touching points, and a $30$-$60$-$90$ triangle appears.