Circles $\omega_1$ and $\omega_2$ meet at $P$ and $Q$. Segments $AC$ and $BD$ are chords of $\omega_1$ and $\omega_2$ respectively, such that segment $AB$ and ray $CD$ meet at $P$. Ray $BD$ and segment $AC$ meet at $X$. Point $Y$ lies on $\omega_1$ such that $P Y \parallel BD$. Point $Z$ lies on $\omega_2$ such that $P Z \parallel AC$. Prove that points $Q,~ X,~ Y,~ Z$ are collinear.
Problem
Source: Saudi Arabia BMO TST Day III Problem 2
Tags: Asymptote, geometry unsolved, geometry
Sardor
03.08.2014 22:50
It's old problem. see here http://www.artofproblemsolving.com/Forum/viewtopic.php?p=982011&sid=2690f77818d5d08b5b7c606716f698e6#p982011
jlammy
03.08.2014 23:06
$Q$ is the Miquel point of $\triangle PAB$ w.r.t $\triangle XDC$, so $ABQX$ is cyclic. Hence $\angle QXC = \angle PBQ = \angle CDQ$, which implies $XDQC$ is cyclic. So $\angle DQX = \angle DCX = \angle PCA = \angle DPZ$, and it follows $\angle DQX + \angle DQZ = 180^{\circ}$. Hence $X, Q, Z$ are collinear. Similarly, $Y, X, Q$ are collinear, which gives the desired result.
$Q$ is the Miquel point of $\triangle PAB$ w.r.t $\triangle XDC$, so $ABQX$ is cyclic. By Reim's theorem on $\omega_1$ and $(ABQX)$, $Q,X,Y$ are collinear; by Reim's theorem on $\omega_2$ and $(ABQX)$, $Z,Q,X$ are collinear, and the result follows.
Hmm how did this question appear in USA TST 2007 and in Saudi Arabia BMO TST Day III Problem 2?
Com10atorics
02.10.2020 15:24
$\angle AXD=180-\angle BAC-\angle PBF=\angle PQC-\angle PQD$. Hence $QDXC$ is cyclic. $\angle AXQ=\angle PQD=180-\angle PBQ=180-\angle ABQ$. Which means $ABXQ$ is also cyclic and $\angle QYP=\angle QCP=\angle QCD=\angle QXD$ So $QXY$ is colinear and $\angle QZP=180-\angle QBP=180-\angle QBA=\angle QXA$. So $QZX$ is also colinear and we get the desired concludion