Circle $\omega$ is tangent to sides $AB, AC$ of triangle $ABC$. A circle $\Omega$ touches the side $AC$ and line $AB$ (produced beyond $B$), and touches $\omega$ at a point $L$ on side $BC$. Line $AL$ meets $\omega, \Omega$ again at $K, M$. It turned out that $KB \parallel CM$. Prove that $\triangle LCM$ is isosceles.
Problem
Source: All-Russia 2018 Grade 9 P2
Tags: geometry, inscribed circles, homothety
WolfusA
28.04.2018 23:12
Shouldn't it be "touches $\omega$ at a point $L$ on side $AC$? I don't get the configuration at all.
falantrng
28.04.2018 23:14
WolfusA wrote: Shouldn't it be "touches $\omega$ at a point $L$ on side $AC$? But $AL$ cannot intersects $\omega,\Omega.$ Problem condition is wrong.Please be careful.
anantmudgal09
28.04.2018 23:16
Note that dilation at $L$ maps $\omega \mapsto \Omega$ hence $K \mapsto M$ and $B \mapsto C$. Thus the line $CA'$ through $C$ parallel to $AB$ is a tangent to $\Omega$. Let $O$ be the center of $\Omega$ then line $CO$ bisects angle $A'CA$. Meanwhile line $AO$ bisects angle $BAC$ hence $\angle AOC=90^{\circ}$. Since $ML$ is a diameter of $\Omega$ and $OC \perp ML$ we conclude $CL=CM$.
Mprog.
15.08.2018 18:27
anantmudgal09 wrote: Note that dilation at $L$ maps $\omega \mapsto \Omega$ hence $K \mapsto M$ and $B \mapsto C$. Thus the line $CA'$ through $C$ parallel to $AB$ is a tangent to $\Omega$. Let $O$ be the center of $\Omega$ then line $CO$ bisects angle $A'CA$. Meanwhile line $AO$ bisects angle $BAC$ hence $\angle AOC=90^{\circ}$. Since $ML$ is a diameter of $\Omega$ and $OC \perp ML$ we conclude $CL=CM$. Can someone explain this solution in detail, please, or show another solution?
MeineMeinung
18.12.2018 11:27
Let $O_1,O_2$ and $r_1,r_2$ be the center and radius of $\omega,\Omega$ respectively. Let $\omega$ and $\Omega$ touch side $AC$ at $X$ dan $Y$ respectively.
It is easy to see that $O_1X \parallel O_2Y$ and since $O_1XL$ and $O_2YM$ are both isosceles, we have $\triangle XO_1L \sim \triangle YO_2M$. Hence $\frac{r_2}{r_1} = \frac{YM}{XL}$. But we also have $\triangle AXL \sim \triangle AYM$ which implies that $\frac{r_2}{r_1} = \frac{YM}{XL} = \frac{AM}{AL}$.
Next is to notice that $AL$ is the angle bisector of $\angle BAC$. Since $KB \parallel CM$ and $AL$ bisects $\angle BAC$, we have $\frac{AC}{AB}=\frac{CL}{LB} = \frac{LM}{LK} = \frac{2r_2}{2r_1} = \frac{r_2}{r_1} = \frac{AM}{AL}$. Since $\angle MAC = \angle LAB$, we get $\triangle ACM \sim \triangle ABL$. Thus, $\angle CLM = \angle ALB = \angle LMC$ and $\triangle LCM$ is isosceles.