Let $ABC$ be a triangle and let $M$ be the midpoint of the segment $BC$. Let $X$ be a point on the ray $AB$ such that $2 \angle CXA=\angle CMA$. Let $Y$ be a point on the ray $AC$ such that $2 \angle AYB=\angle AMB$. The line $BC$ intersects the circumcircle of the triangle $AXY$ at $P$ and $Q$, such that the points $P, B, C$, and $Q$ lie in this order on the line $BC$. Prove that $PB=QC$. Proposed by Dominik Burek, Poland
Problem
Source: 2021 MEMO T-6
Tags: geometry, power of a point, circumcircle, MEMO 2021, memo
timon92
05.09.2021 21:22
This problem was proposed by Burii.
khina
05.09.2021 22:37
whoa this is cool
It suffices to prove that $AB \cdot BX = AC \cdot CY$.
Construct the point $K$ on $AM$ extended beyond $M$ such that $MK = MB = MC$. Note that $ACKX$, $ABKY$ cyclic. Say $(ACK)$ and $(ABK)$ hit $BC$ again at $F$ and $E$. Now, note that $MC = MK$, so $MA = MF = ME$. Thus, $BF = CE$, and so $AB \cdot BX = FB \cdot BE = EC \cdot CF = YC \cdot CA$, as desired.
VicKmath7
05.09.2021 22:48
Indeed, pretty cool. Here's a sketch. Let the internal angle bisector of angle $AMC$ intersect $AB$ at $R$ and $S$ on $AC$ is defined analogously. It suffices to show by PoP that $CA/BA=BX/CY=CS/BR$, where the last follows because of PoP for the cyclic quads $RXMC$ and $BMYS$. Now finish using the angle bisectors in ratios.
Kimchiks926
06.09.2021 18:20
Let $E,F$ be points on the sides $AB, AC$ such that $ME$, $MF$ are angle bisectors of $\angle BMA$ and $\angle CMA$ respectively. From Angle bisector theorem we have: $$ \frac{AE}{EB} = \frac{AF}{FC} = \frac{AM}{MB} \implies EF \parallel BC$$We also introduce points $S=FM \cap AB$ and $T=EM \cap AC$. Since $\angle EMF =90^{\circ}$, we have: $$ -1=(A,B;E,S) = (A,C;F,T) $$Therefore by Prism's lemma $EF,BC,ST$ are concurrent (possibly at infinity), meaning that $ST \parallel EF \parallel BC$. Claim 1: $BX \cdot BS = CY \cdot CT$. Proof: Note that given angle relations reveals that quadrilaterals $SXMC$ and $BMYT$ are cyclic. Thus by PoP we have: $$ BM \cdot BC = BX \cdot BS = CY \cdot CT $$as desired. Claim 2: $AB \cdot BX = AC \cdot CY$ Proof: Just note that: $$ \frac{CY}{BX}=\frac{BS}{CT}=\frac{AB}{AC} $$where the last equality follows from the fact $ST \parallel BC$. Now from PoP once more we have: $$ AB \cdot BX = AC \cdot CY = BP \cdot BQ = QC \cdot CP $$This is enough to conclude that $BP=CQ$ as desired.
X.Allaberdiyev
16.08.2023 23:49
Nice problem! Let $XC \cap BY=T$ and $L$ be the point on $AM$ such that $\angle BLC =90^{\circ}$, then from $MB$=$ML$=$MC$ we have $2\angle AXC=\angle AMC=2\angle ALC$, which means that $(A X L C)$ is cyclic, similarly we have $(A Y L B)$ is cyclic, which gives us that $L$ is Miquel point of $ABXTYC$ $(1)$. Then from ratio lemma in $ABC$ respect to $AM$ gives us $AB/AC$=$sin(\angle BAL)/sin(\angle LAC)= sin(\angle BYL)/sin(\angle LBY)=BL/LY$ $(2)$. Using $(1)$ we know that $L$ is intersection of circles $(BXT)$ and $(CTY)$, then it is well known that $L$ is the center of spiral similarity which sends $XB$ to $CY$, which means that $BL/LY=BX/CY$, then from $(2)$ $BX/CY=AB/AC$, then $PB*BQ=AB*BX=AC*CY=CQ*PC$, then $PB*BQ=CQ*PC$ which gives that $PB=CQ$, so we are done.