1991 AIME Problem 11

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.