Let $ABC$ be a triangle. Let $\omega_1$ be circle tangent to $BC$ at $B$ and passes through $A$. Let $\omega_2$ be circle tangent to $BC$ at $C$ and passes through $A$. Let $\omega_1$ and $\omega_2$ intersect again at $P \neq A$. Let $\omega_1$ intersect $AC$ again at $E\neq A$, and let $\omega_2$ intersect $AB$ again at $F\neq A$. Let $R$ be the reflection of $A$ about $BC$, Prove that lines $BE, CF, PR$ are concurrent.
Problem
Source: Junior Olympiad of Malaysia Shortlist 2015 G6
Tags: geometry
TelvCohl
17.07.2015 18:06
My solution : Let $ M $ be the midpoint of $ BC $ and $ T \equiv AP \cap \odot (ABC) $ . Since $ BC $ is the common tangent of $ \omega_1$ and $ \omega_2 $ , so $ M $ lie on the radical axis $ AP $ of $ \omega_1, \omega_2 $ . i.e. $ M \in AP $ From $ \angle CBP=\angle BAP=\angle BAT=\angle BCT \Longrightarrow BP \parallel CT $ . Similarly, $ CP \parallel BT \Longrightarrow P $ is the reflection of $ T $ in $ M \Longrightarrow P\in \odot (RBC) $ . From $ \angle EBC=\angle BAC=\angle FCB \Longrightarrow EB, FC $ are the tangents of $ \odot (RBC) $ . $(\star) $ From $ TB:TC=AC:AB $ (well-known) $ \Longrightarrow CP:BP=AC:AB=RC:RB $ , so $ PBRC $ is a harmonic quadrilateral $ \Longrightarrow $ combine $ (\star) $ we get $ BE, CF, PR $ are concurrent . Q.E.D
lian_the_noob12
21.04.2023 11:56
$\color{blue} \boxed{\textbf{Angle Chase}}$ $ Q \equiv BE \cap CF $ At first, $B,P,Q,F$ are concyclic. Because, $\angle PFQ=\angle PFC=\angle PAC=\angle PAE=\angle PBE=\angle PBQ$ $\angle BPQ= \angle QFA=\angle AFC$ $B,P,C,R$ concyclic because, $\angle BPC=180°-\angle BCP -\angle CBP=180°-\angle CAP -\angle BAP=180°-\angle BAC=180°-\angle BRC$ So, $\angle BPR= \angle BCR=\angle BCA=\angle BCF+\angle ACF=\angle FAC+\angle ACF=180°-\angle AFC$ And, $\angle BPQ+\angle BPR=\angle AFC +180°-\angle AFC= 180°$ means $R,P,Q$ are collinear that is $RP$ goes through $Q$.