1990 AIME Problems/Problem 12
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 . .
- The length of each of the 12 diagonals that span across 2 edges is (or notice that the triangle formed is equilateral).
- The length of each of the 12 diagonals that span across 3 edges is (or notice that the triangle formed is a right triangle).
- The length of each of the 12 diagonals that span across 4 edges is .
- The length of each of the 12 diagonals that span across 5 edges is .
- The length of each of the 6 diameters is .
Adding all of these up, we get
. Thus, the answer is .
A second method involves drawing a triangle connecting the center of the 12-gon to two vertices of the 12-gon. Since the distance from the center to a vertex of the 12-gon is , the Law of Cosines can be applied to this isosceles triangle, to give:
There are six lengths of sides/diagonals, corresponding to
Call these lengths from shortest to longest. The total length that is asked for is
, noting that as written gives the diameter of the circle, which is the longest diagonal.
To simplify the two nested radicals, add them, and call the sum :
Squaring both sides, the F and L part of FOIL causes the radicals to cancel, leaving:
Plugging that sum back into the equation for , we find
Thus, the desired quantity is .
Begin as in solution 2, drawing a triangle connecting the center of the 12-gon to two vertices of the 12-gon. Apply law of cosines on to get where is the diagonal or sidelength distance between two points on the 12-gon. Now, . Instead of factoring out as in solution 2, factor out instead-the motivation for this is to make the expression look like the half angle identity, and the fact that is an integer doesn't hurt. Now, we have that , which simplifies things quite nicely. Continue as in solution 2, computing half-angle sines instead of nested radicals, to obtain .
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