Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 5.5 AIME, harder Prove it Ireland

Circles $S$ and $T$ intersect at $P$ and $Q$ , with $S$ passing through the centre of $T$ . Distinct points $A$ and $B$ lie on $S$ , inside $T$ , and are equidistant from the centre of $T$ . The line $PA$ meets $T$ again at $D$ . Prove that $|AD| = |PB|$ .

Solution

Let $O$ be the centre of the circle $T$ . Extend $PB$ to meet the circle at $C$ . Let $\angle QPA = \alpha$ and let $\angle BPO = \beta$ . Since $|AO| = |OB|$ the join of the two centres is perpendicular to $AB$ . The join of the two centres is also perpendicular to $PQ$ . Thus $PQ \parallel AB$ . Hence $\angle BAP = \alpha$ . Hence $|PB| = |QA|$ .

Since $|AO| = |OB|$ , $\angle APO = \angle BPO = \beta$ . Hence $|PC| = |PD|$ . $\angle PCD = \pi/2 - \beta$ , $\angle DQP = \pi/2 + \beta$ . Also $\angle PDC = \angle PCD = \pi/2 - \beta$ .

As $QDCP$ is a cyclic quadrilateral
$\angle QDP = \pi = (\alpha + 2\beta + \pi/2 - \beta) = \pi/2 - (\alpha + \beta)$
Also $\angle AQP = \angle BPQ = \alpha + 2\beta$ . Then since $\angle DQP + \angle DCP = \pi$ we get
$\angle DQA = \pi - (\pi/2 - \beta + \alpha + 2\beta = \pi/2 - (\alpha + \beta) = \angle QDP)$
Hence $|AD| = |QA| = |PB|$ .

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