IMO 3J, Mock Exam 1

Solution. Method one: Let $M$ be the intersection of $AT$ with the circumcircle $\odot(ABC)$. Then $TM \cdot TA = TB \cdot TC = TI^2$, so $\angle AMI = \angle AIT = 90^\circ$, that is, $M \in \odot(AEF) \cap \odot(ABC)$. By Miquel's theorem (or since $\angle MFG = \angle MAC = \angle MBG$), $M$ lies on $\odot(BFG)$, hence, if we let $W$ be the intersection of $XY$ and $BC$, then
$$\angle WGM = \angle AFM = \angle AIM = \angle ITM = \angle WZM,$$
that is, $W, M, G, Z$ are concyclic.
Let $D$ be the point of tangency of $\omega$ with $BC$, and let $M_E, M_F$ be the midpoints of $\overline{FD}, \overline{DE}$ respectively. Then $M_E D \cdot M_E F = M_E B \cdot M_E I$, that is, $M_E$ lies on the radical axis $XY$ of $\omega$ and $\odot(BIC)$. Similarly, $M_F$ also lies on $XY$. Therefore $W = M_E M_F \cap BC$ is the midpoint of $\overline{DG}$. It is well known that $MD$ bisects $\angle BMC$:
Since $\angle FMB = \angle EMC, \angle MEB = \angle MFC, \triangle MBF \sim \triangle MCE$, hence $\overline{MB} = \overline{BF} = \overline{BD}$, so $MD$ bisects $\angle BMC$.
And since $G, D$ harmonically divide $B, C$, $\angle DMG$ is a right angle. So $W$ is the circumcenter of $\triangle MGD$.
Finally, we prove that $\angle IAZ = \angle (GZ, AI)$, which is equivalent to
$$\angle WGM = \angle WZM = \angle IAZ + 90^\circ = \angle (GZ, AI) + 90^\circ = \angle GZW = \angle GMW,$$
that is, $\triangle WMG$ is isosceles with apex $W$, and this follows because $W$ is the circumcenter of $\triangle MGD$.

Method two: Let $M$ be the second intersection of $\odot(AEF)$ and $\odot(ABC)$. Likewise, we have that $A, M, T$ are collinear. Let $L$ be the intersection of $AZ$ and $EF$. Then we only need to prove that $\triangle ZLG$ is isosceles with apex $L$. Let $H$ be the foot of the perpendicular from $I$ to $EF$. It is well known that $I, H, M$ are collinear:
Since $\angle DMG = 90^\circ = \angle DHG$, $D, H, M, G$ are concyclic. Combining this with $B, F, M, G$ concyclic, $C, E, M, G$ concyclic, and the properties of spiral similarity, we get $\triangle MFE \cup H \sim \triangle MBC \cup D$. Hence $\angle FMH = \angle BMD = \angle BAI = \angle FMI$.
So we obtain $GT \perp DI, TL \perp IH, LG \perp HD$, and this tells us $\triangle LGT \sim \triangle HID$.
Let $S$ be the intersection of $HI$ and $XY$. Then from $\frac{HI}{IS} = \frac{LT}{TZ}$, we get $\triangle LGT \cup Z \sim \triangle HID \cup S$.
Since $XY$ is the perpendicular bisector of $\overline{DH}$, $\triangle ZLG \sim \triangle SHD$ is isosceles with apex $Z$, which completes the proof.