$\triangle{ABC}$ is isosceles$(AB=AC)$. Points $P$ and $Q$ exist inside the triangle such that $Q$ lies inside $\widehat{PAC}$ and $\widehat{PAQ} = \frac{\widehat{BAC}}{2}$. We also have $BP=PQ=CQ$.Let $X$ and $Y$ be the intersection points of $(AP,BQ)$ and $(AQ,CP)$ respectively. Prove that quadrilateral $PQYX$ is cyclic. (20 Points)
Problem
Source: Iranian 3rd round Geometry exam P2 - 2014
Tags: geometry, geometric transformation, reflection, circumcircle, angle bisector, geometry proposed
ThirdTimeLucky
28.09.2014 01:31
Let $ R $ be the reflection of $ B $ across $ AP $. Then, $ AR=AB=AC $, $ \angle RAQ = \frac{\hat{A}}{2} - \angle PAR = \frac{\hat{A}}{2} - \angle PAB = \angle CAQ $, i.e $ QA $ is the angle bisector in isosceles $ \triangle CAR $ $ \implies $ $ R $ is the reflection of $ C $ in $ AQ $ too. Now, $ \angle PQX=\angle PBX = \angle PRX $ $ \implies $ $ P,Q,R,X $ are concyclic, $ \angle QPY= \angle QCY = \angle QRY $ $ \implies $ $ P,Q,R,Y $ are concyclic $ \implies $ $ P,Q,R,X,Y $ are concylic.
AHZOLFAGHARI
06.12.2014 20:03
P in near from B and Q is near from C. We choose Q' and P' out of triangle such that $ACQ \cong ABQ'$ and $ABP \cong ACP'$. ($AP = AP'$) and ($AQ = AQ'$ ) $\angle Q'AQ = \angle PAP' = \angle A$ so $\angle AQQ' =\angle B$. We can easily know that $P$ is Circumcenter of $\Delta BQQ'$ and BPQ' is equilateral . so $\angle BQQ' = 30$ so $\angle AQB = 30+B , \angle XQY=150-B$ and Similarly we can see $ \angle XPY=150-B$ so X,P,Q,Y are on a circle .
Mahdi_Mashayekhi
28.05.2022 08:02
Let $B'$ be reflection of $B$ across $AX$. Note that $\angle PAB' = \angle PAB$ so $\angle QAB' = \angle QAC$ and $AB = AB' = AC$ so $B'$ is reflection of $C$ across $AY$. Note that $\angle XQP = \angle XBP = \angle XB'P \implies XQB'P$ is cyclic and $\angle QPY = \angle QCY = \angle QB'Y \implies QPB'Y$ is cyclic so $PXQY$ is cyclic. we're Done.