1990 AIME Problems/Problem 12

Problem

A regular 12-gon is inscribed in a circle of radius 12. The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},$ where $a^{}_{}$ , $b^{}_{}$ , $c^{}_{}$ , and $d^{}_{}$ are positive integers. Find $a + b + c + d^{}_{}$ .

Solution

http://img294.imageshack.us/img294/907/dodecahedronjg8.png

The easiest way to do this seems to be to find the length of each of the sides and diagonals. To do such, draw the radii that meet the endpoints of the sides/diagonals; this will form isosceles triangles. Drawing the altitude of those triangles and then solving will yield the respective lengths.

The length of each of the 12 sides is $2 \cdot 12\sin 15$ . $24\sin 15 = 24\sin (45 - 30) = 24\frac{\sqrt{6} - \sqrt{2}}{4} = 6(\sqrt{6} - \sqrt{2})$ .
The length of each of the 12 diagonals that span across 2 edges is $2 \cdot 12\sin 30 = 12$ (or notice that the triangle formed is equilateral).
The length of each of the 12 diagonals that span across 3 edges is $2 \cdot 12\sin 45 = 12\sqrt{2}$ (or notice that the triangle formed is a $45 - 45 - 90$ right triangle).
The length of each of the 12 diagonals that span across 4 edges is $2 \cdot 12\sin 60 = 12\sqrt{3}$ .
The length of each of the 12 diagonals that span across 5 edges is $2 \cdot 12\sin 75 = 24\sin (45 + 30) = 24\frac{\sqrt{6}+\sqrt{2}}{4} = 6(\sqrt{6}+\sqrt{2})$ .
The length of each of the 6 diameters are $2 \cdot 12 = 24$ .

Adding all of these up, we get $12[6(\sqrt{6} - \sqrt{2}) + 12 + 12\sqrt{2} + 12\sqrt{3} + 6(\sqrt{6}+\sqrt{2})] + 6 \cdot 24 = 12(12 + 12\sqrt{2} + 12\sqrt{3} + 12\sqrt{6}) + 144 = 288 + 144\sqrt{2} + 144\sqrt{3} + 144\sqrt{6}$ . Thus, the answer is $144 \cdot 5 = 720$ .

1990 AIME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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1990 AIME Problems/Problem 12

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