In triangle $ABC$ , $w$ is a circle which passes through $B,C$ and intersects $AB,AC$ at $E,F$ respectively. $BF,CE$ intersect the circumcircle of $ABC$ at $B',C'$ respectively. Let $A'$ be a point on $BC$ such that $\angle C'A'B=\angle B'A'C$ . Prove that if we change $w$, then all the circumcircles of triangles $A'B'C'$ passes through a common point.
Problem
Source: Iranian 3rd round 2016 final geometry exam problem 1
Tags: geometry
andria
31.08.2016 16:03
Note that $\angle B'BA=\angle C'CA\Longrightarrow AB'=AC'$. Let $\ell$ be the perpendicular bisector of $B'C'$ and $\ell\cap BC=X$; since $AB'=AC'\Longrightarrow A\in \ell$. easy too see $X\in \odot(B'A'C')$. since $\angle EFB=\angle C'CB=\angle C'B'B\Longrightarrow B'C'\parallel EF\Longrightarrow AX\perp EF\Longrightarrow \angle XAC=90^{\circ}-B\Longrightarrow X$ is fixed. Q.E.D
Garfield
31.08.2016 16:11
Could someone add Iran MO 2016 (2nd and 3rd round) in contest colection?
madmathlover
26.08.2018 08:18
Let $O$ be the circumcenter of $\triangle ABC$. $\angle B'C'A =\angle B'BA = \angle FBE = \angle ECF = \angle C'CA = \angle C'B'A$ $\Rightarrow$ $AB' = AC'$. So, $AO$ is the perpendicular bisector of segment $B'C'$. Let $(A'B'C') \cap BC = D$. Then $\angle DC'B' = \angle DA'B' = \angle BA'C' = \angle C'B'D$ $\Rightarrow$ $DC' = DB'$ $\Rightarrow$ $D = AO \cap BC$ which is fixed.