< KCE = < LCP , 4 circles related, hard version

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Source: 2019 RMM Shortlist G4, version 2 , generalized

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#1 h Jun 18, 2020, 9:40 PM

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Let $\Omega$ be the circumcircle of an acute-angled triangle $ABC$ . A point $D$ is chosen on the internal bisector of $\angle ACB$ so that the points $D$ and $C$ are separated by $AB$ . A circle $\omega$ centered at $D$ is tangent to the segment $AB$ 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

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#2 h Jun 18, 2020, 10:18 PM • 1 Y

Y by AlastorMoody

Clearly, $E$ is the $C-$ excircle touchpoint of $CKL$ . By the three homotheties theorem, $LM$ contains the exsimilicenter of the incircle of $CKL$ and the circumcircle of $ABC$ , so by a symmetric argument, $P$ is this exsimilicenter. Now note that $(CKL)$ and $(CAB)$ are homothetic at $C$ due to $CK, CL$ being isogonal in $\angle ACB$ . Thus by the three homotheties theorem again, $P$ lies on the line joining $C$ and the exsimilicenter of the incircle and circumcircle of $CKL$ . However since this line contains the $C-$ mixtilinear incircle touchpoint of $CKL$ (by the three homotheties theorem again), $CP$ passes through the $C-$ mixtilinear incircle touchpoint of $CKL$ , which is the image of $E$ under the $\sqrt{CK \cdot CL}$ inversion at $C$ and a reflection in the angle bisector of $\angle KCL$ , whence we are done.

@below: the three homotheties theorem is a more general theorem about three homotheties than the usual Monge or Monge-d'Alembert's theorem. It says that if we have two homotheties mapping an object $\mathcal{O}_1$ to $\mathcal{O}_2$ and $\mathcal{O}_2$ to $\mathcal{O}_3$ , then the center of the homothety that is the composition of these two homotheties lies on the line joining the centers of these homotheties.

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#3 h Jun 19, 2020, 6:26 AM • 2 Y

Y by AlastorMoody, Mango247

Brief Sketch

@above

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CyclicISLscelesTrapezoid

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CyclicISLscelesTrapezoid

#4 h Dec 17, 2022, 11:56 PM

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https://cdn.discordapp.com/attachments/761371910704857098/1053819698610974821/Screen_Shot_2022-12-17_at_3.41.33_PM.png

By symmetry, $\overline{CK}$ and $\overline{CL}$ are isogonal in $\triangle CAB$ , so it suffices to prove that $\overline{CE}$ and $\overline{CP}$ are isogonal. Let $\overline{CK}$ and $\overline{CL}$ intersect $\Omega$ again at $K'$ and $L'$ , and let the $C$ -mixtilinear incircle $\gamma$ of $CK'L'$ touch $\Omega$ at $F$ . It's well known that $\overline{CE}$ and $\overline{CF}$ are isogonal, so it suffices to prove that $C$ , $F$ , and $P$ are collinear.

Let $X$ , $Y$ , and $Z$ be the exsimilicenters of $\{\Omega_1,\Omega_2\},\{\Omega_1,\gamma\},\{\Omega_2,\gamma\}$ , respectively. Clearly, $X$ is on $\overline{AB}$ , $Y$ is on $\overline{CL}$ , and $Z$ is on $\overline{CK}$ . We have the following:

$M$ , $N$ , and $X$ are collinear by Monge on $\{\Omega,\Omega_1,\Omega_2\}$
$M$ , $F$ , and $Y$ are collinear by Monge on $\{\Omega,\Omega_1,\gamma\}$
$N$ , $F$ , and $Z$ are collinear by Monge on $\{\Omega,\Omega_2,\gamma\}$
$X$ , $Y$ , and $Z$ are collinear by Monge on $\{\Omega_1,\Omega_2,\gamma\}$ .

We are done by Desargues on triangles $CKL$ and $FNM$ .

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