Eighteenth STARS OF MATHEMATICS Competition

The argument hinges on the four facts below: (1) $X^{\prime }$ lies on circle $YMC$ and $Y^{\prime }$ 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 $\angle Y^{\prime }XM=\angle YXM=180^{\circ }-\angle PAX-\angle APX=180^{\circ }-\angle ABC-\angle BAC=\angle ACB$. As $YN$ and $NM$ are midlines in triangles $AB{Y}^{\prime }$ and $ABC$, respectively, $\angle Y^{\prime }BM=\angle Y^{\prime }BA+\angle ABC=\angle BNM+\angle ABC=180^{\circ }-\angle ACB$. Consequently, $\angle Y^{\prime }XM+\angle Y^{\prime }BM=180^{\circ }$, so $Y^{\prime }$ lies on circle $XMB$. Similarly, $X^{\prime }$ 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 $X^{\prime }MC$ and $Y^{\prime }MB$, respectively, so $A{X}^{\prime }\cdot AY=2\cdot AX\cdot AY=AX\cdot 2\cdot AY=AX\cdot A{Y}^{\prime }$, as desired. This establishes (2). To prove (3), note that $YU\parallel Y^{\prime }B$, as $YN$ is a midline in triangle $AB{Y}^{\prime }$, so $\angle SUY=\angle SB{Y}^{\prime }$. By (1), $\angle SB{Y}^{\prime }=\angle SXY$, 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.