Let $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. Let $D$ be the midpoint of the minor arc $AB$ of $\Omega$. A circle $\omega$ centered at $D$ is tangent to $AB$ at $E$. The tangents to $\omega$ through $C$ meet the segment $AB$ at $K$ and $L$, where $K$ lies on the segment $AL$. A circle $\Omega_1$ is tangent to the segments $AL, CL$, and also to $ \Omega$ at point $M$. Similarly, a circle $\Omega_2$ is tangent to the segments $BK, CK$, and also to $\Omega$ at point $N$. The lines $LM$ and $KN$ meet at $P$. Prove that $\angle KCE = \angle LCP$. Poland
Problem
Source: 2019 RMM Shortlist G4, version 1
Tags: geometry, equal angles, circles
Aryan-23
19.06.2020 05:08
Let $\gamma$ denote the incircle of $\Delta CKL$ Let $T$ be the intersection point of tangents to $\Omega$ at $B$ and $C$. Claim : $\omega$ is the incircle of $\Delta TAB$. Proof: This is just straightforward angle chasing . Just note that $\measuredangle{DAB}=\frac{1}{2}\measuredangle{TAB}$. Now we seek to prove that $CE$ and $CP$ are isogonal in $\measuredangle{ACB}$ . That would clearly finish as we already have that the pairs $\{CA,CB\}$ and $\{CK ,CL\}$ are isogonal in $\measuredangle{ACB}$. Hence we need to prove that $C,P,T$ are collinear . Claim : $P$ is the exsimilicenter of $\gamma$ and $\Omega$ Proof: We use phantom points , and let $P'$ denote the desired exsimilicenter . By Monge on the circles $\{ \Omega ,\Omega_1 ,\gamma \}$ and $\{ \Omega ,\Omega_2 ,\gamma \}$ , we have $$P' \equiv LM \cap KN \implies P' \equiv P .$$ To finish , note that by Monge on $\{ \Omega ,\omega ,\gamma \}$ , we get that $C,P,T$ are collinear, as desired . $\blacksquare$
Rickyminer
21.06.2020 22:03
Also a resubmitting: 2019 Polish Final P6
ohiorizzler1434
05.11.2024 04:35
You want that $CP$ is the symmedian because $E$ is the midpoint. Also notice that $AA \cap BB$ at $F$ gives $ABF$ has incircle $\omega$ by angle chase. Let's prove $F$, $C$, $P$ collinear. Draw the incircle of $CKL$, call it $\omega_2$. Monge on $\omega_2$, $\Omega$ and $\Omega_1$ has $M$, $L$, and wanted exsimilicenter collinear. Similarly with $K$ and $N$. So by reverse reconstruction that wanted point is $P$, and now monge on $\Omega$, $\omega$, and $\omega_2$ finishes.
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TestX01
23.11.2024 12:18
unc skibidi rizz Let $Q$ be such $\omega$ is the incircle of $\triangle ABQ$. By DDIT on $AQBE$ at $C$, since trivially by symmetry, $CL,CK$ are isogonal, $CQ$ is a $C$ symmedian. By isogonals again, it suffices to show $C,P,Q$ collinear. Suppose that $\triangle CKL$ has incircle $\Gamma$. Now, we claim that $P$ is the exsimilicentre of $\Gamma$ and $\Omega$. Yet $P'$, the exsimilicentre must also lie on $KN$, $LM$ by monge on $\Gamma$, $\Omega$, and $\Omega_1$,$\Omega_2$ respectively. By Monge on $\omega$, $\Gamma$, $P$ lies on $CQ$.