2010 AMC 10B Problem 20

The ratio does not depend on the size of the hexagon, so give it side length $2$. A regular hexagon of side $2$ is made of six equilateral triangles of side $2$, each with height $\sqrt3$. So opposite sides, like $AB$ and $DE$, are $2\sqrt3$ apart.

1. Build a corner. Extend $CB$ past $B$ and $FA$ past $A$ until they meet at a point $P$. Each angle of the hexagon is $120^\circ$, so the angles of triangle $PAB$ at $A$ and $B$ are $180^\circ - 120^\circ = 60^\circ$. Triangle $PAB$ is equilateral with side $2$, and its height from $P$ down to $AB$ is $\sqrt3$.
2. Both circles sit in the $60^\circ$ corner at $P$. Each circle touches both lines $PB$ and $PA$, so its center is on the line that cuts the corner in half. Along that line, the center, $P$ and the point where the circle touches line $PA$ form a right triangle with a $30^\circ$ angle at $P$. The radius $r$ is opposite the $30^\circ$ angle, so the distance from $P$ to the center is $2r$ (in a $30$-$60$-$90$ triangle the longest side is twice the shortest).
3. The line that halves the corner at $P$ goes straight down through the middle of $AB$, through the center of the hexagon, and through the middle of $DE$, meeting both sides at right angles. Measure everything along this line.
4. First circle (radius $r_1$): it touches $AB$ and lies inside triangle $PAB$. From $P$ to the center is $2r_1$, and from the center down to $AB$ is another $r_1$. That covers the height of the triangle:
$$3r_1 = \sqrt3 \quad\Longrightarrow\quad r_1 = \frac{\sqrt3}{3}.$$
5. Second circle (radius $r_2$): it touches $DE$ from outside the hexagon, so it is on the far side of $DE$. The distance from $P$ to $DE$ is $\sqrt3$ (to $AB$) plus $2\sqrt3$ (across the hexagon), which is $3\sqrt3$. The center is another $r_2$ beyond $DE$, and it is also $2r_2$ from $P$:
$$2r_2 = 3\sqrt3 + r_2 \quad\Longrightarrow\quad r_2 = 3\sqrt3.$$
6. Compare. The radii are in the ratio $\frac{r_2}{r_1} = \frac{3\sqrt3}{\sqrt3/3} = 9$. Area grows with the square of the radius, so the area ratio is $9^2 = 81$.

The answer is $\boxed{\textbf{(D)}\ 81}$.

Any circle that touches both sides of an angle has its center on the line that halves the angle, and for a $60^\circ$ angle the center is exactly twice the radius away from the corner. Finding that corner turns a hard-looking tangent-circle problem into measuring distances along one straight line. Remember to square the ratio of radii to get the ratio of areas.

Lines $BC$, $FA$ and $DE$ form a big equilateral triangle with side $3 \cdot 2 = 6$ (three hexagon sides fit along each of its sides). Its height is $3\sqrt3$. The first circle is the circle inside the small triangle $PAB$, whose radius is one third of that triangle's height $\sqrt3$. The second circle sits outside the big triangle beyond side $DE$, touching that side and the other two sides extended; in an equilateral triangle this outside circle has radius equal to the full height, $3\sqrt3$. The ratio of radii is again $9$, so the areas are in the ratio $81$.