Difference between revisions of "1990 AIME Problems/Problem 12"
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*The length of each of the 12 diagonals that span across 3 edges is <math>2 \cdot 12\sin 45 = 12\sqrt{2}</math> (or notice that the triangle formed is a <math>45 - 45 - 90</math> [[right triangle]]). | *The length of each of the 12 diagonals that span across 3 edges is <math>2 \cdot 12\sin 45 = 12\sqrt{2}</math> (or notice that the triangle formed is a <math>45 - 45 - 90</math> [[right triangle]]). | ||
*The length of each of the 12 diagonals that span across 4 edges is <math>2 \cdot 12\sin 60 = 12\sqrt{3}</math>. | *The length of each of the 12 diagonals that span across 4 edges is <math>2 \cdot 12\sin 60 = 12\sqrt{3}</math>. | ||
− | *The | + | *The length of each of the 12 diagonals that span across 5 edges is <math>2 \cdot 12\sin 75 = 24\sin (45 + 30) = 24\frac{\sqrt{6}+\sqrt{2}}{4} = 6(\sqrt{6}+\sqrt{2})</math>. |
− | *The length of each of the 6 [[diameter]]s | + | *The length of each of the 6 [[diameter]]s is <math>2 \cdot 12 = 24</math>. |
Adding all of these up, we get <math>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}</math>. Thus, the answer is <math>144 \cdot 5 = 720</math>. | Adding all of these up, we get <math>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}</math>. Thus, the answer is <math>144 \cdot 5 = 720</math>. |
Revision as of 12:49, 27 October 2007
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 where , , , and are positive integers. Find .
Solution
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 .
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |