Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Ireland

A fixed point $O$ is joined to any point $P$ on a given line which does not contain $O$ . On $OP$ a point $Q$ is taken such that $|OP| \cdot |OQ|$ is constant. Show $Q$ lies on a fixed circle if $P$ varies. Examine all possible cases.

Solution

Let $R$ be the foot of the perpendicular from $O$ on the given line and let $S$ be chosen on $OR$ such that $|OR| \cdot |OS| = |OP| \cdot |OQ|$ , i.e. $S$ is a point on the locus. The value of the constant determines the position of $S$ , either between $O$ and $R$ or not. Then $P, Q, S$ and $R$ lie on a circle in all cases (see formulation of the hint). Hence $\angle PQS = 90^\circ$ since $\angle PRO = 90^\circ$ . This implies that $Q$ lies on the circumference of the circle on $OS$ as diameter which is a fixed circle. In every case $Q$ can be taken on either side of $O$ on the line $OP$ . The analysis in every case (see drawings below) is similar to above.

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