International Mathematical Olympiad

We first show that the points $A,M,R,N$ are concyclic. Since $ABC$ is an acute-angled triangle, $M$ and $N$ are on the line segments $AB$ and $AC$ respectively. Let $R_1$ be the point such that the points $A,M,R_1,N$ are concyclic, where $R_1$ is on the ray $AR$. Since $A{R}_1$ bisects $\angle BAC$, we have $R_{1}M=R_{1}N$. Since $M$ and $N$ lie on the circle with centre $O$, we have $OM=ON$. It follows from $OM=ON$ and $R_{1}M=R_{1}N$ that $R_1$ is on the bisector of $\angle MON$. Since $AB\ne AC$, the bisectors of the angles $BAC$ and $MON$ intersect at the unique point $R$, and so $R_1=R$, or $A,M,R,N$ are concyclic.

Let the bisector of $\angle BAC$ meet $BC$ at $K$. Since the points $B,C,N,M$ are concyclic, $\angle MBC=\angle ANM$. Moreover, since $A,M,R,N$ are concyclic, $\angle ANM=\angle MRA$. This implies $\angle MBK=\angle MRA$. Therefore, the points $B,M,R,K$ are concyclic. Using the same argument as above, we obtain that $C,N,R,K$ are concyclic. This completes the solution.