Taiwan IMO Selection Camp

First we prove a lemma.

Lemma 1. Let $I, O$ be the incenter and circumcenter of $\triangle ABC$ respectively. Let points $E, F$ lie on rays $CA, BA$ respectively, satisfying $BF = BC = CE$. Then $EF \perp IO$.

Proof of Lemma. Let $M$ be the midpoint of arc $BC$ containing $A$ on the circumcircle of $\triangle ABC$, and let $X, Y, Z$ be the circumcenters of triangles $AEF, CIA, AIB$ respectively. Since $MB = MC$, $BF = CE$, and $\angle MCE = \angle MBF$, triangles $\triangle MBF$ and $\triangle MCE$ are directly congruent. From this we know
$$\angle MEA = \angle MFA,$$
so $M$ lies on the circumcircle of $\triangle AEF$, hence $OX \perp YZ$. On the other hand, from $\angle CFA = \angle CIA$, we know that $F$ lies on the circumcircle of $\triangle CIA$. Similarly, $E$ lies on the circumcircle of $\triangle AIB$. Therefore
$$XY \perp AB, \quad XZ \perp CA \implies OY \parallel XZ, \quad OZ \parallel XY,$$
so $OYXZ$ is a rhombus. Since $YZ$ is parallel to the angle bisector of $\angle EAF$, and $AX, IO$ are symmetric with respect to $YZ$, we get $EF \perp IO$, and the lemma is proved.

Returning to the original problem, let $M, N, P$ be the midpoints of $AB, CZ, XY$ respectively. Let $U$ be the reflection of $Z$ about $M$, $V$ be the reflection of $C$ about $P$, and let $AB$ and $VX$ meet at $W$. Since $G$ lies on $CM$ with $\frac{GM}{CG} = \frac{1}{2}$, $G$ is the centroid of $\triangle CUZ$, so $G, N, U$ are collinear, with $\frac{GN}{UG} = \frac{1}{2}$. Since $F$ lies on $ZP$ with $\frac{FP}{ZF} = \frac{1}{2}$, $F$ is the centroid of $\triangle CVZ$, so $F, N, V$ are collinear, with $\frac{FN}{VF} = \frac{1}{2}$. On the other hand, note that $\triangle ABC$ and $\triangle WBX$ are homothetic, and $BU = BX = XV$, so by the lemma, $UV \perp IO$. This completes the proof.