, 2021

Let $AI, BI$, and $CI$ meet $\Gamma$ again at $D, E$, and $F$, respectively. Let $\ell$ be the common tangent to $\omega_B$ and $\omega_C$ at $I$. We always denote by $\angle(p, q)$ the directed angle from a line $p$ to a line $q$, taken modulo $180^\circ$.

*Step 1: We show that $Y$ lies on $\Gamma$.*
Recall that any chord of a circle makes complementary directed angles with the tangents to the circle at its endpoints. Hence
$$\angle(BY, BI) + \angle(CI, CY) = \angle(IB, \ell) + \angle(\ell, IC) = \angle(IB, IC).$$
Therefore,
$$\begin{aligned} \angle(BY, BA) + \angle(CA, CY) &= \angle(BI, BA) + \angle(BY, BI) + \angle(CI, CY) + \angle(CA, CI) \\ &= \angle(BC, BI) + \angle(IB, IC) + \angle(CI, CB) = 0, \end{aligned}$$
which yields $Y \in \Gamma$.

Let $X_* = \ell \cap EF$. To prove our claim, it suffices to show that $X_*$ lies on both $PM$ and $QN$; this will yield $X_* = X$. Due to symmetry, it suffices to show $X_* \in QN$.
Notice that
$$\angle(IX_*, IQ) = \angle(CI, CQ) = \angle(CF, CQ) = \angle(EF, EQ) = \angle(EX_*, EQ);$$
therefore, the points $X_*, I, Q$, and $E$ are concyclic (if $Q = E$, then the direction of $EQ$ is supposed to be the direction of a tangent to $\Gamma$ at $Q$; in this case, the equality means that the circle ($X_*IQ$) is tangent to $\Gamma$ at $Q$). Then we have
$$\angle(QX_*, QI) = \angle(EX_*, EI) = \angle(EF, EB) = \angle(CA, CF) = \angle(CN, CI) = \angle(QN, QI),$$

Step 3: We finally show that A, X, and Y are collinear.
Recall that $I$ is the orthocenter of triangle $DEF$, and $A$ is symmetric to $I$ with respect to $EF$. Therefore,
$$\angle(AX, AE) = \angle(IE, IX) = \angle(BI, \ell) = \angle(BY, BI) = \angle(BY, BE) = \angle(AY, AE),$$
which yields the desired collinearity.