Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.3 Shortlist Prove it Turkey

The circles $\omega_1$ and $\omega_2$ which do not intersect and which have different sizes are tangent to the line $\ell$ at $K$ and $L$ , and are tangent to the circle $\Gamma$ at $M$ and $N$ respectively, such that all three circles lie on the same side of $\ell$ . A circle which passes through $K$ and $L$ intersects $\Gamma$ at $A$ and $B$ . The reflections of $M$ and $N$ over $\ell$ are $R$ and $S$ respectively. Show that the points $A, B, R, S$ are concyclic.

Solution

Denote the tangent lines to $\Gamma$ at $M$ and $N$ by $m$ and $n$ , and let
$\alpha := \angle LKM = \angle (KM, m), \\ \beta := \angle (m, MN) = \angle (MN, n), \quad \gamma := \angle (n, NL) = \angle NLK.$
The internal angles of $KMNL$ add up to $360^\circ$ , thus $\alpha + \beta + \gamma = 180^\circ$ , hence $KMNL$ is cyclic. Now the lines $AB, KL, MN$ are the pairwise radical axes of the circles $KLMN, ABMN, ABKL$ , thus they concur at a point $T$ . The line $RS$ is the reflection of $MN$ in $KL$ , therefore $T \in RS$ and then $TR \cdot TS = TM \cdot TN = TA \cdot TB$ , hence $ABRS$ is cyclic.

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