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Geometry Difficulty 5.2 AIME, harder Prove it Taiwan

In the plane there is $\triangle ABC$ and a point $O$ , and $\Gamma$ is the circumcircle of $\triangle ABC$ . Let line $CO$ meet line $AB$ at point $D$ , and let line $BO$ meet line $CA$ at point $E$ . Let line $AO$ meet $\Gamma$ again at point $F$ . Let point $I$ be the second intersection point of $\Gamma$ and the circumcircle of $\triangle ADE$ , let point $Y$ be the second intersection point of line $BE$ and the circumcircle of $\triangle CEI$ , and let point $Z$ be the second intersection point of line $CD$ and the circumcircle of $\triangle BDI$ . On $\Gamma$ , draw the two tangent lines with tangent points $B$ , $C$ respectively, and let them meet at point $T$ . Let line $TF$ meet $\Gamma$ again at point $U$ , and let $G$ be the reflection of $U$ with respect to line $BC$ .

Prove that: the six points $F, I, G, O, Y, Z$ are concyclic.

Solution

Solution: Let $BC$ and $ED$ meet at point $X$ . Since $I$ is the Miquel point of $BCED$ , $X$ lies on $\odot(BDI)$ , $\odot(CEI)$ . Since $BFCU$ is a harmonic quadrilateral, we have
$A(B, C; F, U) = -1 = A(B, C; O, X),$
hence $X \in AU$ .

Since
$\angle BYX = \angle ACB = \angle XUB = \angle BGX,$
we get $X \in \odot(BGY)$ .

Similarly we obtain that $X$ lies on $\odot(CGZ)$ , hence
$\angle YGZ = \angle YGX + \angle XGZ = \angle YBX + \angle XCZ = \angle YOZ,$
that is, $O, G, Y, Z$ are concyclic.

On the other hand, from
$\angle IYO = \angle ICA = \angle IFO,$
we get $Y \in \odot(FIO)$ .

Similarly we obtain that $Z \in \odot(FIO)$ , hence $F, I, O, G, Y, Z$ are concyclic, and the proof is complete.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.