Taiwan IMO Selection Camp

(∠ denotes directed angles.)

Let $M$, $N$ be the midpoints of $\overline{BC}$, $\overline{IJ}$ respectively, let $N'$ be the reflection of $N$ over $M$, and let $D$ be the foot of the perpendicular from $J$ to $BC$. It is well known that $IM \parallel AD$, and also $MN \perp BC \perp DJ$, so $\triangle ADJ$ and $\triangle IMN$ are homothetic, hence since $T$ is the reflection of $J$ over $D$, we know $AT \parallel IN'$. Let $O$ be the circumcenter of $\triangle ABC$; from $\angle YON = 2\angle YAN = \angle YIN$ we obtain that $Y,I,N,O$ are concyclic. From $\angle CN'N = \angle N'NC = \angle ONC = \angle NCO$ and the Shooting Lemma (雞爪定理), we know $NO \cdot NN' = \overline{NC}^2 = \overline{NI}^2$, that is, $AI$ is tangent to $\odot(ION')$. Let $Q$ be the intersection of $NY$ and $BC$; from $\angle QCN = \angle BAN = \angle NAC = \angle NYC$ and the Shooting Lemma, we know $NQ \cdot NY = \overline{NC}^2 = \overline{NI}^2$, so $\odot(IQY)$ is tangent to $AI$. From $\angle QIN = \angle NYI = \angle NOI = \angle N'IN$ we obtain that $Q,I,N'$ are collinear. Also $\angle IXQ = \angle IAP = \angle NIN' = \angle ION = \angle IYQ$, so $X \in \odot(IQY)$, hence $\odot(IXY)$ is tangent to $AI$.