AoPS - Problem

Source: 2021 Mexico Center Zone Regional Olympiad, problem 3

Tags: Mexico, geometry, cyclic quadrilateral, tangential quadrilateral, collinear

Justpassingby

17.01.2022 12:58

Let $W,X,Y$ and $Z$ be points on a circumference $\omega$ with center $O$, in that order, such that $WY$ is perpendicular to $XZ$; $T$ is their intersection. $ABCD$ is the convex quadrilateral such that $W,X,Y$ and $Z$ are the tangency points of $\omega$ with segments $AB,BC,CD$ and $DA$ respectively. The perpendicular lines to $OA$ and $OB$ through $A$ and $B$, respectively, intersect at $P$; the perpendicular lines to $OB$ and $OC$ through $B$ and $C$, respectively, intersect at $Q$, and the perpendicular lines to $OC$ and $OD$ through $C$ and $D$, respectively, intersect at $R$. $O_1$ is the circumcenter of triangle $PQR$. Prove that $T,O$ and $O_1$ are collinear. Proposed by CDMX

Let $S$ be the intersection of the lines perpendicular to $OA$ and $OD$ trough $A$ and $D$ respectively. Claim 1: $PQRS\sim WXYZ$. In fact they are homothetic (through $O_2$ say) Proof: All sides are parallel. $\blacksquare$ So we define $\Phi : WXYZ\mapsto PQRS $. We also contend Claim 2: $P-O-R$ are colinear (and so are $Q-O-S$) Proof: Consider $O_3:=BC\cap AD$. Then $P$ is the incenter of $\triangle O_3AB$, $O$ is the incenter of $\triangle O_3CD$ and $R$ is the $O_3$-excenter of $\triangle O_3CD$. They all lie on the internal bisector of $\angle AO_3B$. $\blacksquare$ We can now finish by homothety at $O_2$. In fact $O_2-T-O-O_1$ are colinear: $\Phi(T)=O$ so $O_2-T-O$ and $\Phi(O)=O_1$ so $O_2-O-O_1$. We are done. $\blacksquare$