Question

Twelve congruent disks are placed on a circle $C$ of radius $1$ in such a way that the twelve disks cover $C,$ no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below. The sum of the areas of the twelve disks can be written in the form $\pi(a-b\sqrt c),$ where $a,$ $b,$ $c$ are positive integers and $c$ is not divisible by the square of any prime. Find $a+b+c.$

Answer

Answer: 135

Concepts:circle tangent circles special right triangle

Difficulty rating: 2200

Small Hint:

Join the center of $C$ to the centers and tangency point of two neighboring disks

Big Hint:

The resulting right triangle has angle $15^\circ$ , adjacent leg $1$ , and opposite leg equal to a disk radius

Solution:

Let $O$ be the center of $C,$ let $U$ and $V$ be the centers of two neighboring disks, and let $T$ be their tangency point. By the $12$ -fold symmetry, $\angle UOV=30^\circ,$ and $OT$ bisects that angle. Also $T$ is the midpoint of $\overline{UV},$ so triangle $OUT$ is right at $T.$ Since $T$ lies on $C,$ $OT=1,$ while $UT$ is the disk radius $r.$ Thus $r=\tan15^\circ=2-\sqrt3.$ The total area is $\begin{aligned}12\pi r^2&=12\pi(7-4\sqrt3)\\&=\pi(84-48\sqrt3).\end{aligned}$ Therefore $a+b+c=84+48+3=135.$

1991 AIME Problem 11

Answer: 135

Difficulty rating: 2200

Small Hint:

Big Hint:

Solution:

Problem 11 in Other Years