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Bulgarian Mathematical Competitions

Geometry Difficulty 5.7 AIME, harder Prove it Bulgaria

Problem:

Let $ABC$ be an isosceles triangle with $AC = BC$ and let $k$ be a circle with center $C$ and radius less than the altitude $CH$ , $H \in AB$ . Lines through $A$ and $B$ are tangent to $k$ at points $P$ and $Q$ lying on the same side of the line $CH$ . Prove that the points $P$ , $Q$ and $H$ are collinear.

Solution

Solution:

First solution. Since $CP = CQ$ , $CA = CB$ and $\angle APC = \angle BQC = 90^\circ$ , then $\triangle APC \cong BQC$ . Hence $\angle CAP = \angle CBQ$ . Setting $AP \cap BQ = T$ , it follows that the quadrilateral $ABTC$ is cyclic. Then $\angle BAC = \angle QTC$ and now $\angle TQC = \angle AHC = 90^\circ$ implies that $\angle QCT = \angle ACH$ . The equalities $\angle AHC = \angle APC = \angle CPT = \angle CQT = 90^\circ$ show that $AHPC$ and $CPTQ$ are cyclic quadrilaterals. Thus $\angle APH = \angle ACH$ and $\angle QPT = \angle QCT$ which means that $\angle APH = \angle QPT$ . Hence the points $H$ , $P$ and $Q$ are collinear.

Second solution. Set $S = HQ \cap k$ . Since the quadrilateral $BHCQ$ is cyclic, and the triangles $ABC$ and $CQS$ are isosceles, it follows that $\angle BAC = \angle ABC = \angle HQC = \angle CSQ$ . Then $AHSC$ is a cyclic quadrilateral and therefore $\angle ASC = \angle AHC = 90^\circ$ . Hence $S = P$ , i.e., the points $H$ , $P$ and $Q$ are collinear.

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