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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.6 AIME, harder Prove it United States

Problem:

Let $AB$ be the diameter of a semicircle $\Gamma$ . Two circles, $\omega_1$ and $\omega_2$ , externally tangent to each other and internally tangent to $\Gamma$ , are tangent to the line $AB$ at $P$ and $Q$ , respectively, and to semicircular arc $AB$ at $C$ and $D$ , respectively, with $AP < AQ$ . Suppose $F$ lies on $\Gamma$ such that $\angle FQB = \angle CQA$ and that $\angle ABF = 80^\circ$ . Find $\angle PDQ$ in degrees.

Solution

Solution:

Extend the semicircle centered at $O$ to an entire circle $\omega$ , and let the reflection of $F$ over $AB$ be $F'$ . Then $CQF'$ is a straight line. Also, the homothety centered at $C$ taking $\omega_1$ into $\omega$ takes $P$ to a point $X$ on $\omega$ and $AB$ to the parallel line tangent to $\omega$ at $X$ . Therefore, $X$ is the midpoint of semicircle $AXB$ , and $C$ , $P$ , and $X$ lie on a line. Similarly, $D$ , $Q$ , and $X$ lie on a line. So,
$45^\circ = \angle XCB = \angle PCB = \angle PCQ + \angle QCB = \angle PCQ + 10^\circ,$
since $\angle QCB = \angle F'CB = \angle F'AB = \angle FAB = 90^\circ - \angle ABF = 10^\circ$ . Thus $\angle PCQ = 35^\circ$ . We will show that $\angle PCQ = \angle PDQ$ to get that $\angle PDQ = 35^\circ$ .

Note that $\angle XPQ$ subtends the sum of arcs $AC$ and $BX$ , which is equal to arc $XC$ . Therefore $\angle XPQ = \angle CDX$ , so $CDQP$ is cyclic and $\angle PCQ = \angle PDQ$ . The conclusion follows.

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