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Source: 2019 RMM Shortlist G4, version 1

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#1 h Jun 18, 2020, 9:35 PM • 2 Y

Y by Rounak_iitr, ehuseyinyigit

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

This post has been edited 4 times. Last edited by parmenides51, Oct 9, 2020, 12:32 PM
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#2 h Jun 19, 2020, 2:08 AM • 3 Y

Y by AlastorMoody, A-Thought-Of-God, SerdarBozdag

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 . $\blacksquare$

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#3 h Jun 21, 2020, 7:03 PM

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Also a resubmitting: 2019 Polish Final P6

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ohiorizzler1434

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ohiorizzler1434

#4 h Nov 5, 2024, 1:35 AM

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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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#5 h Nov 23, 2024, 9:18 AM

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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 $CQ$ .

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