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Geometry Difficulty 5.4 AIME, harder Find the answer United States

Problem:
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

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 $PAO_{2}$ , we can calculate $AP = 4\sqrt{6}$ , so $\sin \angle AO_{2}P = \frac{2\sqrt{6}}{5}$ and $\cos \angle AO_{2}P = \frac{1}{5}$ . Because $\triangle AO_{2}P$ and $\triangle BO_{2}P$ are congruent, we have
$\sin \angle APB = \sin 2\angle AO_{2}P = 2 \sin \angle AO_{2}P \cos \angle AO_{2}P = \frac{4\sqrt{6}}{25}$
implying that
$[APB] = \frac{PA \cdot PB}{2} \sin \angle APB = \frac{192\sqrt{6}}{25}.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.