60th Belarusian Mathematical Olympiad

Let $P$ and $Q$ be the midpoints of $BC$ and $AC$ respectively. Since $CM=MB$ and $CN=NA$, we see that $M$ and $N$ lie on the perpendicular bisectors of the sides $BC$ and $AC$ respectively. Since $AB$ is the smallest side, the distance between $A$ and $B$ is less than the distance between $A$ and $C$, so $A$ and $C$ lie in the different half-planes with respect to the perpendicular bisector of the side $BC$. Thus $M$ lies on the side $AC$. Similarly, $N$ lies on the side $BC$. By condition, $CM=MB$, it follows that the triangle $BMC$ is isosceles and $\angle BCM=\angle MBC$. Similarly, $\angle ACN=\angle NAC$. Since $\angle BCM=\angle ACN$, we have



$$\angle NAM=\angle NAC=\angle MBC=\angle MBN.$$

Therefore, $A$, $B$, $M$, $N$ lie on the same circle $\Gamma$ (since the angles $NAM$ and $MBN$ are subtended by the same segment, $MN$). It remains to note that the angles $NOM$ and $ACB$ are the angles with mutually perpendicular sides, so either $\angle NOM+\angle ACB=\angle NOM+\angle MBN=180^{\circ }$ (see Fig. 1), or $\angle NOM=\angle ACB=\angle MBN$ (see Fig. 2). It follows that $O$ lie on $\Gamma$.