Eighteenth STARS OF MATHEMATICS Competition

The argument hinges on the four facts below:
(1) $X'$ lies on circle $YMC$ and $Y'$ lies on circle $XMB$.
(2) $S$, $A$ and $M$ are collinear.
(3) $U$ and $V$ both lie on circle $SXY$.
(4) $UV$ is parallel to $BC$.
Assume these facts for the moment, to complete the solution as follows: By (3), the conclusion is equivalent to circles $SUV$ and $SBC$ being tangent; and by (4), these circles are similar from $S$, whence the conclusion.

To prove (1), write $\angle Y'XM = \angle YXM = 180^\circ - \angle PAX - \angle APX = 180^\circ - \angle ABC - \angle BAC = \angle ACB$.
As $YN$ and $NM$ are midlines in triangles $ABY'$ and $ABC$, respectively, $\angle Y'BM = \angle Y'BA + \angle ABC = \angle BNM + \angle ABC = 180^\circ - \angle ACB$.
Consequently, $\angle Y'XM + \angle Y'BM = 180^\circ$, so $Y'$ lies on circle $XMB$. Similarly, $X'$ lies on circle $YMC$. This establishes (1).

To prove (2), note that $SM$ is the radical axis of the circles $YMC$ and $XMB$, so it is sufficient to show that $A$ has equal powers with respect to these circles. By (1), the two circles are $X'MC$ and $Y'MB$, respectively, so $AX' \cdot AY = 2 \cdot AX \cdot AY = AX \cdot 2 \cdot AY = AX \cdot AY'$, as desired. This establishes (2).
To prove (3), note that $YU \parallel Y'B$, as $YN$ is a midline in triangle $ABY'$, so $\angle SUY = \angle SBY'$. By (1), $\angle SBY' = \angle SXY$, so $\angle SUY = \angle SXY$, implying that $U$ lies on circle $SXY$. Similarly, $V$ lies on this circle. This establishes (3).

Finally, to prove (4), it is sufficient to show that $SU/UB = SV/VC$ and then apply Thales. As triangles $MSU$ and $MUB$ have the same $M$-altitude and share the side $MU$,
$$\frac{SU}{UB} = \frac{\text{area } MSU}{\text{area } MUB} = \frac{SM}{MB} \cdot \frac{\sin \angle SMU}{\sin \angle UMB} = \frac{SM}{MB} \cdot \frac{\sin \angle AMN}{\sin \angle NMB},$$
as $A$ lies on $SM$ by (2).
Triangles $MAN$ and $MNB$ have equal areas, as they both have the same $M$-altitude, and $N$ is the midpoint of $AB$. These triangles also share the side $MN$, so
$$MA \cdot MN \cdot \sin \angle AMN = 2 \cdot \text{area } MAN = 2 \cdot \text{area } MNB = MN \cdot MB \cdot \sin \angle NMB.$$
Hence $\sin \angle AMN / \sin \angle NMB = MB/MA$, so, by the preceding,
$$\frac{SU}{UB} = \frac{SM}{MB} \cdot \frac{MB}{MA} = \frac{SM}{MA}.$$
Similarly, $SV/VC = SM/MA$, so $SU/UB = SV/VC$, as stated. This establishes (4) and completes the solution.