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 $\Gamma$.