We have three circles $w_1$, $w_2$ and $\Gamma$ at the same side of line $l$ such that $w_1$ and $w_2$ are tangent to $l$ at $K$ and $L$ and to $\Gamma$ at $M$ and $N$, respectively. We know that $w_1$ and $w_2$ do not intersect and they are not in the same size. A circle passing through $K$ and $L$ intersect $\Gamma$ at $A$ and $B$. Let $R$ and $S$ be the reflections of $M$ and $N$ with respect to $l$. Prove that $A, B, R, S$ are concyclic.
Problem
Source: Turkey TST 2022 P4 Day 2
Tags: geometry, geometry proposed, circles, tangent circles
electrovector
13.03.2022 14:39
Let $O_1, O, O_2$ be the centers of $w_1, w_2$ and $\Gamma$. $\angle MKL= \frac{\angle KO_1M}{2}=\frac{360-\angle MON-\angle NO_2L}{2}=\angle ONM+\angle O_2NL$ which proves that $KMNL$ is concyclic. Also we have that $ABLK$ and $MNBA$ are concyclic which means $AB$, $KL$ and $MN$ intersects at some point, $T$. It is also easy to see that $MN$, $RS$ and $KL$ intersect at some point which means $T \in RS$. Now we have $TR \cdot TS= TM \cdot TN= TA \cdot TB$ which means $ABSR$ is concyclic.
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badumtsss
20.03.2022 17:28
Because of my amazing geometry skills, I couldn't see the direct angle chasing. Anyway, by Monge Theorem, we know that the center of homotety sending $\omega_1$ to $\omega_2$ is the intersection of $KL$ and $MN$. Take the second intersections $X$, $Y$ of $MN$ and $\omega_1$, $\omega_2$, angle chasing shows that $KLXY$ is cyclic, which also shows that $KLMN$ is cyclic. $MN$ and $RS$ intersect on $KL$ which ends the proof.