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 1

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 is $2 \cdot 12 = 24$ .

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 $144 \cdot 5 = \boxed{720}$ .

Solution 2

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 $12$ , the Law of Cosines can be applied to this isosceles triangle, to give:

$a^2 = 12^2 + 12^2 - 2\cdot 12\cdot 12\cdot \cos \theta$

$a^2 = 2\cdot 12^2 - 2\cdot 12^2 \cos \theta$

$a^2 = 2\cdot 12^2 (1 - \cos \theta)$

$a = 12\sqrt{2} \cdot \sqrt{1 - \cos \theta}$

There are six lengths of sides/diagonals, corresponding to $\theta = {30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 150^{\circ}, 180^{\circ}}$

Call these lengths $a_1, a_2, a_3, a_4, a_5, a_6$ from shortest to longest. The total length $l$ that is asked for is

$l = 12(a_1 + a_2 + a_3 + a_4 + a_5) + 6a_6$ , noting that $a_6$ as written gives the diameter of the circle, which is the longest diagonal.

$l = 12[12\sqrt{2} (\sqrt{1 - \cos 30^{\circ}}+\sqrt{1-\cos 60^{\circ}}+\sqrt{1-\cos 90^{\circ}}+\sqrt{1-\cos 120^{\circ}}+\sqrt{1 - \cos 150^{\circ}})] + 6 \cdot 24$

$l = 144\sqrt{2} \left(\sqrt{1 - \frac{\sqrt{3}}{2}} + \sqrt{1 - \frac{1}{2}} + \sqrt{1-0} + \sqrt{1 + \frac{1}{2}} + \sqrt{1 + \frac{\sqrt{3}}{2}}\right) + 144$

$l = 144\sqrt{2} \left(\sqrt{1 - \frac{\sqrt{3}}{2}} + \frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{6}}{2} + \sqrt{1 + \frac{\sqrt{3}}{2}}\right) + 144$

To simplify the two nested radicals, add them, and call the sum $x$ :

$x = \sqrt{1 - \frac{\sqrt{3}}{2}} + \sqrt{1 + \frac{\sqrt{3}}{2}}$

Squaring both sides, the F and L part of FOIL causes the radicals to cancel, leaving $2$ :

$x^2 = 2 + 2\sqrt{\left(1 - \frac{\sqrt{3}}{2}\right)\left(1 + \frac{\sqrt{3}}{2}\right)}$

$x^2 = 2 + 2\sqrt{1 - \frac{3}{4}}$

$x^2 = 2 + 2\sqrt{\frac{1}{4}}$

$x^2 = 2 + 2\cdot\frac{1}{2}$

$x = \sqrt{3}.$

Plugging that sum $x$ back into the equation for $l$ , we find

$l = 144\sqrt{2} \left(\frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{6}}{2} + \sqrt{3}\right) + 144$

$l = 144 + 144\sqrt{2} + 144\sqrt{3} + 144\sqrt{6} + 144$ .

Thus, the desired quantity is $144\cdot 5 = \boxed{720}$ .

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

Contents

Problem

Solution 1

Solution 2

See also