$I,\Omega$ are the incenter and the circumcircle of triangle $ABC$, respectively, and the tangents of $B,C$ to $\Omega$ intersect at $L$. Assume that $P\neq C$ is a point on $\Omega$ such that $CI,AP$, and the circle with center $L$ and radius $LC$ are concurrent. Let the foot from $I$ to $AB$ be $F$, the midpoint of $BC$ be $M$, $X$ is a point on $\Omega$ s.t. $AI,BC,PX$ are concurrent. Prove that the lines $AI,AX,MF$ form an isosceles triangle. Proposed by ckliao914
Problem
Source: IRN-SGP-TWN 2023 friendly contest p5
Tags: geometry
Euler365
16.07.2023 12:35
We'll prove that angle betn $AI$ and $AX$ is equal to angle betn $AI$ and $MX$. Let $D$ be 2nd intersection pt. of $AI$ with $\Omega$ and let $DP\cap BC=Q$. Also let $AI\cap BC= E$ and $AP\cap CI= K$. Then the angle between $AI$ and $AX$ is $\angle DAX=\angle DPX=\angle DPE$. Also $\angle AEB=\angle APD$. So all we need to show is that $\angle FMB=\angle APE$. $\textbf{Claim 1: }$ $\angle APE=\angle AQE$ $\textbf{Proof: }$ $\angle AEQ=\angle APQ$ and hence $AEPQ$ is cyclic and the claim follows. $\textbf{Claim 2: }$ $\angle FMB=\angle AQE$ $\textbf{Proof: }$ By ratio lemma, $\frac{BQ}{QC}=\frac{BP}{PC}\frac{BD}{DC}=\frac{BP}{PC}=\frac{sin(\angle PCB)}{sin(\angle PBC)}=\frac{sin(\angle PAB)}{sin(\angle PAC)}$. But now $\sin(\angle PAB)=\frac{BKsin(\angle KBA)}{AK}$ and $\sin(\angle PAC)=\frac{CKsin(\angle KCA)}{AK}$ and hence $\frac{sin(\angle PAB)}{sin(\angle PAC)}=\frac{BKsin(\angle KBA)}{CKsin(\angle KCA)}=\frac{sin(\angle KCB)sin(\angle KBA)}{sin(\angle KBC)sin(\angle KCA)}=\frac{sin(\angle KBA)}{sin(\angle KBC)}=\frac{sin(\frac{B+A}{2})}{sin(\frac{B-A}{2})}$. But the last ratio is nothing but $\frac{c}{b-a}$. It follows easily that $\frac{BM}{MQ}=\frac{a+c-b}{b+c-a}=\frac{BF}{FA}$, which means that $FM\parallel AQ$ and the claim follows. Claims 1 and 2 imply that $\angle FMB=\angle APE$ which finishes the proof.