2006 AIME I Problems/Problem 12

Problem

Equilateral $\triangle ABC$ is inscribed in a circle of radius 2. Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime, and $q$ is not divisible by the square of any prime, find $p+q+r.$

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Solution

Observe that $2\cos 4x\cos x = \cos 5x + \cos 3x$ by the sum-to-product formulas. Defining $a = \cos 3x$ and $b = \cos 5x$ , we have $a^3 + b^3 = (a+b)^3 \Leftrightarrow ab(a+b) = 0$ . But $a+b = 2\cos 4x\cos x$ , so we require $\cos x = 0$ , $\cos 3x = 0$ , $\cos 4x = 0$ , or $\cos 5x = 0$ .

Hence the solution set is $A = \{150, 112, 144, 176, 112.5, 157.5\}$ and thus $\sum_{x \in A} x = 852$ .

2006 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2006 AIME I Problems/Problem 12

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