Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.6 National Olympiad Prove it Mongolia

Let $\gamma_1$ and $\gamma_2$ be externally tangent circles and $S$ be the point of tangency. Let $\omega$ be a circle that touches internally $\gamma_1$ and $\gamma_2$ at $P$ and $Q$ respectively. Denote by $R$ one of the intersection points of $\omega$ and the common tangent line of $\gamma_1$ and $\gamma_2$ that passes through $S$ . Furthermore, the lines $RP$ and $RQ$ intersect $\gamma_1$ and $\gamma_2$ at $A$ and $B$ respectively and the line $PQ$ intersects $\gamma_1$ and $\gamma_2$ at $C$ and $D$ respectively. Prove that the lines $RS$ , $AC$ and $BD$ have a common point.

Solution

Let $(AC) \cap (BD) = M$ . It suffices to prove that $M$ is on the radical axis of $\gamma_1$ and $\gamma_2$ . It's equivalent to $MC \cdot MA = MD \cdot MB$ . This is equivalent to $ACDB$ is a cyclic quadrilateral. Since $PT$ and $TQ$ are tangents to $\omega$ , $\angle TPQ = \angle TQP = \alpha$ . From this $\angle PAC = \alpha = \angle QBD$ implies. Since $R$ is on the radical axis of $\gamma_1$ and $\gamma_2$ , $RA \cdot RP = RB \cdot RQ$ . This means $ABQP$ is a cyclic quadrilateral and $\angle BAD + \angle BQP = 180^\circ$ . $\alpha + \angle BAC + \angle BQP = 180^\circ$ . $x + \angle BQP = \angle QBD + \angle BQP = 180^\circ - \angle QDB$ . Hence $\angle BAC = \angle QDB$ . This means $ABDC$ is cyclic, as desired.

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