USA IMO TST

First solution using symmedians (Merlijn Staps) Let $\overline{XY}$ be the other tangent from $X$ to $(AMN)$.
Claim. Line $\overline{XM}$ is tangent to $(BMY)$; hence $Y$ lies on $\omega_B$.

Proof. Let $Z$ be the midpoint of $\overline{AY}$. Then $\overline{MX}$ is the $M$-symmedian in triangle $AMY$. Since $\overline{MZ} \parallel \overline{BY}$, it follows that $\angle AMX = \angle ZMY = \angle BYM$. We conclude that $\overline{XM}$ is tangent to the circumcircle of triangle $BMY$. $\square$

Similarly, $\omega_C$ is the circumcircle of triangle $CNY$. As $AMYN$ is cyclic too, it follows that $\omega_B$ and $\omega_C$ intersect on $\overline{BC}$, by Miquel's theorem.

Second solution (Jetze Zoethout) Let $\omega_B$ intersect $\overline{BC}$ again at $S$ and let $\overline{MS}$ intersect $\overline{AC}$ again at $Y$. Angle chasing gives $\angle XMY = \angle XMS = \angle MBS = \angle ABC = \angle XAC = \angle XAY$, so $Y$ is on the circumcircle of triangle $AMX$. Furthermore, from $\angle XMY = \angle ABC$ and $\angle ACB = \angle XAB = \angle XYM$ it follows that $\triangle ABC \sim \triangle XMY$ and from $\angle XAY = \angle MBS$ and $\angle YXA = \angle YMA = \angle BMS$ it follows that $\triangle AXY \sim \triangle BMS$.
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We now find
$$\frac{AN}{AX} = \frac{AN/BM}{AX/BM} = \frac{AC/AB}{MS/XY} = \frac{AB/AB}{MS/XM} = \frac{XM}{MS},$$
which together with $\angle XMS = \angle XAN$ yields $\triangle XMS \sim \triangle XAN$. From $\angle XSY = \angle XSM = \angle XNA = \angle XNY$ we now have that $S$ is on the circumcircle of triangle $XNY$. Finally, we have $\angle XNS = \angle XYS = \angle XYM = \angle ACB = \angle NCS$ so $\overline{XN}$ is tangent to the circle through $C, N$, and $S$, as desired.

Third solution by moving points method Fix triangle $ABC$ and animate $X$ along the tangent at $A$. We let $D$ denote the second intersection point of $\omega_C$ with line $\overline{BC}$.
Claim. The composed map $X \mapsto D$ is a fractional linear transformation (i.e. a projective map) in terms of a real coordinate on line $\overline{AA}, \overline{BC}$.
Proof. Let $\ell$ denote the perpendicular bisector of $\overline{CN}$, also equipped with a real coordinate. We let $P$ denote the intersection of $\overline{XM}$ with $\ell$, $S$ the circumcenter of $\triangle CMD$. Let $T$ denote the midpoint of $\overline{BD}$.
We claim that the composed map
$$\overline{AA} \rightarrow \ell \rightarrow \ell \rightarrow \overline{BC} \rightarrow \overline{BC}$$
by $X \mapsto P \mapsto S \mapsto T \mapsto D$
is projective, by showing each individual map is projective.

* The map $X \mapsto P$ is projective since it is a perspectivity through $N$ from $\overline{AA}$ to $\ell$.
* The map $P \mapsto S$ is projective since it is equivalent to a negative inversion on $\ell$ at the midpoint of $\overline{NC}$ with radius $\frac{1}{2}NC$. (Note $\angle PNS = 90^\circ$ is fixed.)
* The map $S \mapsto T$ is projective since it is a perspectivity $\ell \to \overline{BC}$ through the point at infinity perpendicular to $\overline{BC}$ (in fact, it is linear).
* The map $T \mapsto D$ is projective (in fact, linear) since it is a homothety through $C$ with fixed ratio 2.
Thus the composed map is projective as well. $\square$

Similarly, if we define $D'$ so that $\overline{XM}$ is tangent to $(BMD')$, the map $X \mapsto D'$ is projective as well. We aim to show $D = D'$, and since the maps correspond to fractional linear transformations in projective coordinates, it suffices to verify it for three distinct choices of $X$. We do so:
* If $X = \overline{AA} \cap \overline{MN}$, then $D$ and $D'$ satisfy $MB = MD'$, $NC = ND$. This means they are the feet of the $A$-altitude on $\overline{BC}$.
* As $X$ approaches $A$ the points $D$ and $D'$ approach the infinity point along $\overline{BC}$.
* If $X$ is a point at infinity along $\overline{AA}$, then $D$ and $D'$ coincide with the midpoint of $\overline{BC}$.
This completes the solution.

Fourth solution by isogonal conjugates (Anant Mudgal) Let $Y$ be the isogonal conjugate of $X$ in $\triangle AMN$ and $Z$ be the reflection of $Y$ in $\overline{MN}$. As $\overline{AX}$ is tangent to the circumcircle of $\triangle AMN$, it follows that $\overline{AY} \parallel \overline{MN}$. Thus $Z$ lies on $\overline{BC}$ since $\overline{MN}$ bisects the strip made by $\overline{AY}$ and $\overline{BC}$.

Finally,
$$\angle ZMX = \angle ZMN + \angle NMX = \angle NMY + \angle YMA = \angle NMA = \angle ZBM$$
so $\overline{XM}$ is tangent to the circumcircle of $\triangle ZMB$, hence $Z$ lies on $\omega_B$. Similarly, $Z$ lies on $\omega_C$ and we're done.