BE, CF, PR concurrent

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

BE, CF, PR concurrent

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Junior Olympiad of Malaysia Shortlist 2015 G6

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

92 posts

#1 h Jul 17, 2015, 1:56 PM • 1 Y

Y by Adventure10

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2312 posts

#2 h Jul 17, 2015, 3:06 PM • 2 Y

Y by enhanced, Adventure10

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

lian_the_noob12

173 posts

lian_the_noob12

#3 h Apr 21, 2023, 8:56 AM

Y by

$\color{blue} \boxed{\textbf{Angle Chase}}$

$Q \equiv BE \cap CF$

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 $Q$ .

This post has been edited 12 times. Last edited by lian_the_noob12, Apr 21, 2023, 9:55 AM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a