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Difference between revisions of "1990 AIME Problems/Problem 12"
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== Solution ==
 
== Solution ==
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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 [[radius|radii]] that meet the endpoints of the sides/diagonals; this will form [[isosceles triangle]]s. Drawing the [[altitude]] of those [[triangle]]s and then solving will yield the respective lengths.
 
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 [[radius|radii]] that meet the endpoints of the sides/diagonals; this will form [[isosceles triangle]]s. Drawing the [[altitude]] of those [[triangle]]s and then solving will yield the respective lengths.

Revision as of 20:25, 26 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 $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},$ where $a^{}_{}$, $b^{}_{}$, $c^{}_{}$, and $d^{}_{}$ are positive integers. Find $a + b + c + d^{}_{}$.

Solution

1990 AIME-12.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 $144 \cdot 5 = 720$.

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
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