USA Tstst

Henceforth assume $\angle A \neq 60^\circ$; we prove the concurrence. Let $L$ denote the center of $\omega$, which is the midpoint of minor arc $BC$.

Claim. Let $K$ be the point on $\omega$ such that $\overline{KL} \parallel \overline{AB}$ and $\overline{KC} \parallel \overline{AL}$. Then $\overline{KA}$ is tangent to $\gamma$, and we may put
$$x = KA = LB = LC = LX = LY = KX = KY.$$
Proof. By construction, $KA = LB = LC$. Also, $\overline{MO}$ is the perpendicular bisector of $\overline{KL}$ (since the chords $\overline{KL}$, $\overline{AB}$ of $\omega$ are parallel) and so $KXLY$ is a rhombus as well.
Moreover, $\overline{KA}$ is tangent to $\gamma$ as well since
$$\angle KAD = \angle KAL = \angle KAC + \angle CAL = \angle KBC + \angle ABK = \angle ABC.$$

Up to now we have not used the existence of $Q$; we henceforth do so.
Note that $Q \neq O$, since $\angle A \neq 60^\circ \implies O \notin \omega$. Moreover, we have $\angle AOM = \angle ACB$ too. Since $O$ and $Q$ both lie inside $\triangle ABC$, this implies that $A, M, O, Q$ are concyclic. As $Q \neq O$ we conclude $\angle CQA = 90^\circ$.
The main claim is now:
Claim. Assuming $Q$ exists, the rhombus $LXKY$ is a square. In particular, $\overline{KX}$ and $\overline{KY}$ are tangent to $\omega$.

First proof of Claim, communicated by Milan Haiman. Observe that $\triangle QLC \sim \triangle LOC$. Hence, $CL^2 = CO \cdot CQ$. Then,
$$x^2 = CL^2 = CO \cdot CQ = CN \cdot CA = \frac{1}{2}CA^2 = \frac{1}{2}LK^2$$
where we have also used the fact $AQON$ is cyclic. Thus $LK = \sqrt{2}x$ and so the rhombus $LXKY$ is actually a square. $\square$

Second proof of Claim, Evan Chen. Observe that $Q$ lies on the circle with diameter $\overline{AC}$, centered at $N$, say. This means that $O$ lies on the radical axis of $\omega$ and $(N)$, hence $\overline{NL} \perp \overline{CO}$ implying
$$\begin{align*} NO^2 + CL^2 &= NC^2 + LO^2 = NC^2 + OC^2 = NC^2 + NO^2 + NC^2 \ \implies x^2 &= 2NC^2 \ \implies x &= \sqrt{2}NC = \frac{1}{\sqrt{2}}AC = \frac{1}{\sqrt{2}}LK. \end{align*}$$
So $LXKY$ is a rhombus with $LK = \sqrt{2}x$. Hence it is a square. $\square$

Third proof of Claim. A solution by trig is also possible. As in the previous claims, it suffices to show that $AC = \sqrt{2}x$.

First, we compute the length $CQ$ in two ways; by angle chasing one can show $\angle CBQ = 180^\circ - (\angle BQC + \angle QCB) = \frac{1}{2}\angle A$, and so
$$\begin{aligned} AC \sin B &= CQ = \frac{BC}{\sin(90^\circ + \frac{1}{2}\angle A)} \cdot \sin \frac{1}{2}\angle A \\ \Leftrightarrow \sin^2 B &= \frac{\sin A \cdot \sin \frac{1}{2}\angle A}{\cos \frac{1}{2}\angle A} \\ \Leftrightarrow \sin^2 B &= 2 \sin^2 \frac{1}{2}\angle A \\ \Leftrightarrow \sin B &= \sqrt{2} \sin \frac{1}{2}\angle A \\ \Leftrightarrow 2R \sin B &= \sqrt{2} \left( 2R \sin \frac{1}{2}\angle A \right) \\ \Leftrightarrow AC &= \sqrt{2}x \end{aligned}$$
as desired (we have here used the fact $\triangle ABC$ is acute to take square roots).

We finish by proving that
$$KD = KA$$
and hence line $\overline{KD}$ is tangent to $\gamma$. Let $E = \overline{BC} \cap \overline{KL}$. Then
$$LE \cdot LK = LC^2 = LX^2 = \frac{1}{2}LK^2$$
and so $E$ is the midpoint of $\overline{LK}$. Thus $\overline{MXOY}$, $\overline{BC}$, $\overline{KL}$ are concurrent at $E$. As $\overline{DL} \parallel \overline{KC}$, we find that $DLCK$ is a parallelogram, so $KD = CL = KA$ as well. Thus $\overline{KD}$ and $\overline{KA}$ are tangent to $\gamma$.