2010 AMC 10B Problem 20

Let the hexagon have side $s$. Extend $CB$ past $B$ and $FA$ past $A$; because the exterior angles of the hexagon are $60^\circ$, they meet at a point $P$ with $\triangle PAB$ equilateral of side $s$. The lines $PB$ and $PA$ form a $60^\circ$ angle at $P$.

Any circle tangent to both lines has its center on the bisector of that angle, and since half the angle is $30^\circ$, the center is at distance $2r$ from $P$ (the radius is opposite a $30^\circ$ angle).

First circle (radius $r_1$): it sits inside $\triangle PAB$, tangent to $AB$. Along the bisector, the distance from $P$ to $AB$ is the triangle's height $\frac{s\sqrt3}{2}$, so $2r_1 + r_1 = \frac{s\sqrt3}{2}$ and $r_1 = \frac{s\sqrt3}{6}$.

Second circle (radius $r_2$): it is tangent to $DE$ and lies outside the hexagon, on the far side of line $DE$ from $P$. The distance from $P$ to line $DE$ is $\frac{s\sqrt3}{2}$ (to $AB$) plus $s\sqrt3$ (the hexagon's width between parallel sides), i.e. $\frac{3s\sqrt3}{2}$. So $2r_2 = \frac{3s\sqrt3}{2} + r_2$, giving $r_2 = \frac{3s\sqrt3}{2}$.

Ratio of radii: $\dfrac{r_2}{r_1} = \dfrac{3s\sqrt3/2}{s\sqrt3/6} = 9$. Ratio of areas: $9^2 = 81$.

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

Two circles tangent to the same pair of lines are homothetic from the lines' intersection, so their radii are proportional to their distances from that point. Finding $P$ and measuring along the angle bisector turns a tangency problem into two one-line equations. Remember that areas scale as the square of lengths.

The first circle is the incircle of an equilateral triangle of side $s$ (inradius $\frac{s\sqrt3}{6}$); the second is the excircle of the equilateral triangle of side $3s$ formed by lines $BC$, $FA$, $DE$ (exradius $\frac{3s\sqrt3}{2}$). Same ratio $9$, area ratio $81$.

• Taking the second circle to be the hexagon's incircle (tangent to DE, BC, FA but inside the hexagon), which gives a ratio of 9 instead of 81.
• Reporting the ratio of radii ($9$) or the ratio of the triangle sides ($3$) instead of the ratio of areas.