HMMT November

Geometry Difficulty 5.1 AIME, harder Find the answer

Let $\omega_{1}$ be a circle of radius 5, and let $\omega_{2}$ be a circle of radius 2 whose center lies on $\omega_{1}$ . Let the two circles intersect at $A$ and $B$ , and let the tangents to $\omega_{2}$ at $A$ and $B$ intersect at $P$ . If the area of $\triangle ABP$ can be expressed as $\frac{a \sqrt{b}}{c}$ , where $b$ is square-free and $a, c$ are relatively prime positive integers, compute $100a+10b+c$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let $O_{1}$ and $O_{2}$ be the centers of $\omega_{1}$ and $\omega_{2}$ , respectively. Because $\angle O_{2}AP+\angle O_{2}BP=90^{\circ}+90^{\circ}=180^{\circ}$ quadrilateral $O_{2}APB$ is cyclic. But $O_{2}, A$ , and $B$ lie on $\omega_{1}$ , so $P$ lies on $\omega_{1}$ and $O_{2}P$ is a diameter of $\omega_{1}$ . From the Pythagorean theorem on triangle $PAP_{2}$ , we can calculate $AP=4\sqrt{6}$ , so $\sin \angle AOP=\frac{2\sqrt{6}}{5}$ and $\cos \angle AOP=\frac{1}{5}$ . Because $\triangle AOP$ and $\triangle BOP$ are congruent, we have $\sin \angle APB=\sin 2 \angle AOP=2 \sin \angle AOP \cos \angle AOP=\frac{4\sqrt{6}}{25}$ implying that $[APB]=\frac{PA \cdot PB}{2} \sin \angle APB=\frac{192\sqrt{6}}{25}$

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