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
Problem
Source: 2019 RMM Shortlist G4, version 2 , generalized
Tags: geometry, equal angles, circles
WizardMath
19.06.2020 01:18
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.
Aryan-23
19.06.2020 09:26
Let $CK,CL$ hit $\odot (CAB)$ at $L',K'$ . Note that if $E'$ is the $C$-mixtillinear incircle touch point of $\Delta CK'L' $ ,then under $\sqrt{ab}$ inversion at $C$ followed by a flip over $CD$ , we have that the pairs $\{K,K'\}$,$\{L,L'\}$ and $\{E,E'\}$ are swapped . Also since lines isogonal lines with respect to $\measuredangle{ACB}$ are swapped under this map , hence we just need that $C,E', P$ are collinear. Now we use the same trick as in the easier version , we allow $\gamma$ to be the incircle of $\Delta CKL$ .Also let $\omega'$ be the $C$- mixtilinear incircle of $\Delta CK'L'$ .Conclude by Monge on the triples of circles $\{ \Omega ,\Omega_1 ,\gamma \}$ and $\{ \Omega ,\Omega_2 ,\gamma \}$ and $\{ \Omega ,\omega' ,\gamma \}$
Can you please tell me what the "Three Homothey Lemma" is . Kindly leave a reference
CyclicISLscelesTrapezoid
18.12.2022 02:56
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 $FNM$.