1991 AIME Problems/Problem 11
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Problem
Twelve congruent disks are placed on a circle 
of radius 1 in such a way that the twelve disks cover , 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 
, where 
are positive integers and 
is not divisible by the square of any prime. Find 
.
Solution
Note that the above image is not completely accurate with regard to the problem description.
We wish to find the radius of one circle, so that we can find the total area.
Notice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle. Now consider two adjacent smaller circles. This means that the line connecting the radii is a segment of length 
that is tangent to the larger circle at the midpoint of the two centers. Thus, we have essentially have a regular dodecagon whose vertices are the centers of the smaller triangles circumscribed about a circle of radius 
.
We thus know that the apothem of the dodecagon is equal to . To find the side length, we make a triangle consisting of a vertex, the midpoint of a side, and the center of the dodecagon, which we denote 
and 
respectively. Notice that 
, and that 
is a right triangle with hypotenuse 
and 
. Thus 
, which is the radius of one of the circles. The area of one circle is thus 
, so the area of all 
circles is 
, giving an answer of 
.
See also
| 1991 AIME (Problems • Resources) | ||
| Preceded by Problem 10 | Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
