Another olympiad problem with a complete quadrilateral

Question

Let $ABCD$ be a convex quadrilateral whose diagonals intersect at $P$. Let $X$ and $Y$ be points such that quadrilaterals $ABPX$, $CDXP$, $BCPY$, and $DAYP$ are cyclic. Lines $AB$ and $CD$ intersect at $Q$, $BC$ and $DA$ intersect at $R$, $XR$ and $YQ$ intersect at Z. Prove that $X$, $Y$, $Z$, $P$ are concyclic.

This problem appeared on the Croatian IMO TST in 2021. It's very similar to this problem that appeared on the MEMO TST in the same year. I'm stuck. I'm guessing the solution involves spiral similarities and properties of Miquel points, but I can't seem to make any progress. Again, any insight is appreciated.

Answer 1

Let RXZ cut the circles ABPX and CDXP at S and T respectively.

Join CT. By “properties of cyclic quad.”, $\angle 1 = \angle 2 = \angle 3$. This means SA // CT. Then, $\alpha = \beta$.

Extend SA to cut the circle ADPY at U. After joining DU, we have $\beta = \gamma$.

$\alpha = \gamma$ implies TDU is a straight line.

Through the similar logic, (I think) we should get the result of VCT and VBS are all straight lines.

Required result follows because $\angle red = \angle pink$ and $\angle pink = \angle orange$.

Answer 2

This solution is very similar to the solution to the linked problem.

Let $K$, $L$, $M$, $N$, $E$, $F$ be the midpoints of segments $AB$, $CD$, $AD$, $CB$, $AC$, $BD$ respectively. Let $W$ be the other intersection of the circumcircles of triangles $RAB$ and $QAD$. $W$ also lies on the circumcircles of triangles $RCD$ i $QCB$. Similarly, $X$ lies on the circumcircles of triangles $QAC$ and $QBD$ as well, and $Y$ also lies on the circumcircles of triangles $RAC$ and $RBD$.

$W$ is the center of the spiral similarity taking $A$ to $B$ and $D$ to $C$, so $\triangle WAD\sim\triangle WBC$. We have $$\frac{KE}{KF}=\frac{BC}{AD}=\frac{WC}{WD}$$ and $\sphericalangle(EK, KF)=\sphericalangle(EK, KA) + \sphericalangle(BK, KF)=\sphericalangle(CB, BA) + \sphericalangle(BA, AD)=\sphericalangle(CR, RD)=\sphericalangle(CW, WD)$, so $\triangle KEF\sim\triangle WCD$. $X$ is the center of the spiral similarity taking $A$ to $B$ and $C$ to $D$, so that spiral similarity takes $E$ to $F$. It follows that $X$ is also the center of the spiral similarity taking $E$ to $C$ and $F$ to $D$, therefore $\triangle XEF\sim\triangle XCD$. We have $$\frac{EX}{EK}=\frac{EX}{EF}\cdot \frac{EF}{EK}=\frac{CX}{CD}\cdot \frac{CD}{CW}=\frac{CX}{CW}$$ and $\sphericalangle(XE, EK) = \sphericalangle(XE, EF) + \sphericalangle(FE, EK) = \sphericalangle(XC, CD) + \sphericalangle(DC, CW) = \sphericalangle(XC, CW)$, hence $\triangle XKE\sim\triangle XWC$.

By similar reasoning, $\triangle XKE\sim\triangle XBY$, therefore $\triangle XWC\sim\triangle XBY$. $X$ is the center of the spiral similarity taking $C$ to $Q$ and $P$ to $B$, so $\triangle XCP\sim\triangle XQB$. Now we have $$\frac{XW}{XP} = \frac{XW}{XC}\cdot \frac{XC}{XP} = \frac{XB}{XY}\cdot\frac{XQ}{XB} = \frac{XQ}{XY}$$ and $\sphericalangle(WX, XP) = \sphericalangle(WX, XC) + \sphericalangle(CX, XP) = \sphericalangle(BX, XY) + \sphericalangle(QX, XB) = \sphericalangle(QX, XY)$, therefore $\triangle XWP\sim\triangle XQY$. Similarly, $\triangle YWP\sim\triangle YRX$. Finally, $\sphericalangle(XP, PY) = \sphericalangle(XP, PW) + \sphericalangle(WP, PY) = \sphericalangle(XY, YQ) + \sphericalangle(RX, XY) = \sphericalangle(XZ, ZY)$, Hence $X$, $Y$, $Z$, $P$ are concyclic.

$\blacksquare$

---

A nice result that follows from this is that $E$ and $F$ (as defined above) also lie on $(XYZP)$. Also, as was noticed by Mick in his answer, if we let $S$ and $T$ be the second intersections of $XR$ with $(PAB)$ and $(PCD)$ respectively, and we let $U$ and $V$ be the second intersections of $YQ$ with $(PAD)$ and $(PCB)$ respectively, then $SVTU$ is a parallelogram whose sides contain points $A$, $B$, $C$ and $D$. Furthermore, $SVTU$ is similar to the parallelogram $MLNK$.