South-Afrika, 2011

Geometry Difficulty 4.5 AIME Prove it South Africa

Let $PQ$ be the diameter of semicircle $H$ . Circle $\omega$ is internally tangent to $H$ and tangent to $PQ$ at $C$ . Let $A$ be a point on $H$ and $B$ a point on $PQ$ such that $AB$ is perpendicular to $PQ$ and tangent to $\omega$ . Prove that $AC$ bisects $\angle PAB$ .

Solution

In version centred at $C$ with any radius gives $\triangle CAP \sim \triangle CP'A'$ and $\triangle CAB \sim \triangle CB'A'$ . Clearly the line $PQ$ inverts to itself. $\omega$ inverts to $\omega'$ , a line. $H$ inverts to semicircle $H'$ tangent to $\omega'$ with diameter $P'Q'$ . $AB$ inverts to arc $A'B'$ of a circle with diameter $CB'$ tangent to $\omega'$ . Now observe that arc $A'Q'$ and arc $CA'$ are symmetrical w.r.t the perpendicular bisector of $CQ'$ . Which implies $\angle CP'A' = \angle CB'A'$ . Notice that $AC$ bisects $\angle PAB$ iff $\angle CAP = \angle CAB$ iff $\angle CP'A' = \angle CB'A'$ and so we are done.

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