1991 AIME Problems/Problem 11

Problem

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 from $\pi(a-b\sqrt{c})$ , where $a,b,c^{}_{}$ are positive integers and $c^{}_{}$ is not divisible by the square of any prime. Find $a+b+c^{}_{}$ .

Solution

Let $R_{}^{}$ and $r_{}^{}$ denote the radii of the large and small circles ( $R_{}^{}>r_{}^{}$ ), respectively. Suppose that there are $n_{}^{}$ circles of radius $r_{}^{}$ centered on the circumference of the circle having radius $R_{}^{}$ . Let $O_{}^{}$ , $P_{}^{}$ , and $Q_{}^{}$ label the vertices of the triangle with $O_{}^{}$ being at the center of the large circle, whereas $P_{}^{}$ and $Q_{}^{}$ are the tangential points of one of the $n_{}^{}$ small circles with its two other neighbours, and $S_{}^{}$ its center. The angle subtended by $P_{}^{}OQ$ is $2\pi/n_{}^{}$ . The segments $O_{}^{}P$ and $S_{}^{}P$ are perpendicular. Therefore, triangle $P_{}^{}OS$ is rectangular and the angle subtended by $P_{}^{}OS$ equals $\pi/n_{}^{}$ . Hence, the radius $r_{}^{}=R\sin(\pi/n)$ . The total area $A_{n}^{}$ of the $n_{}^{}$ circles is thus given by

$A_{n}^{}=n\pi r^{2}=n\pi R^{2}\sin^{2}\left(\frac{\pi}{n}\right)=\frac{1}{2}n\pi R^{2}\left[1-\cos\left(\frac{2\pi}{n}\right)\right]\, .$

In the present problem, $n_{}^{}=12$ and $R_{}^{}=1$ . It follows that $A_{12}^{}=6\pi\left[1-\cos(\pi/6)\right]=\pi\left(6-3\sqrt{3}\right)\equiv \pi\left(a-b\sqrt{c}\right)$ .

In summary, $a_{}^{}+b+c=12$ .

1991 AIME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1991 AIME Problems/Problem 11

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