Two circles $\Gamma_1$ and $\Gamma_2$ intersect at points $A$ and $Z$ (with $A\neq Z$). Let $B$ be the centre of $\Gamma_1$ and let $C$ be the centre of $\Gamma_2$. The exterior angle bisector of $\angle{BAC}$ intersects $\Gamma_1$ again at $X$ and $\Gamma_2$ again at $Y$. Prove that the interior angle bisector of $\angle{BZC}$ passes through the circumcenter of $\triangle{XYZ}$. For points $P,Q,R$ that lie on a line $\ell$ in that order, and a point $S$ not on $\ell$, the interior angle bisector of $\angle{PQS}$ is the line that divides $\angle{PQS}$ into two equal angles, while the exterior angle bisector of $\angle{PQS}$ is the line that divides $\angle{RQS}$ into two equal angles.
Problem
Source: Benelux MO 2019 P3
Tags: geometry, circumcircle
MarkBcc168
28.04.2019 16:07
Isn't it just an easy angle chasing?
YangYu
28.04.2019 16:09
∠BZO=90°-∠XZB-∠ZOB=∠XAZ-∠XYZ=∠AZY also we have ∠CZO=∠AZX ∠XAB=∠CAY so∠XZA=∠YZA so∠CZO=∠BZO
Tafi_ak
17.03.2023 09:01
Let the internal angle bisector of $\angle BAC$ intersects $\Gamma_1$ and $\Gamma_2$ at $E$, $F$ respectively. Obviously $X, B, E$ and $Y, C, F$ are collinear. Let $D=XE\cap YF$. Let $O$ be the center of $(XYZ)$. Claim: Points $B, Z, D, C, O$ is cyclic. Proof. Now notice that $Z$ is the miquel point of the complete quadrilateral $AEDY$. So we have $X, Z, D, Y$ is cyclic. We known from the miquel configuration that the miquel point $Z$ and the centers of the four circle are concyclic. Hence $ZBCO$ is cyclic. Notice that $ABDC$ is a parallelogram. So $\angle BZC=\angle BAC=\angle BDC$, means $BZDC$ is cyclic. Which means $BZDCO$ is cyclic. $\square$ Notice that $\angle DXY=\angle CAY=\angle XYD$, so $\triangle DXY$ is isosceles. And $O$ is its center therefore $OD$ bisects $\angle BDC$, means $O$ is the midpoint of the arc $BOC$. As desired.