Let $H$ be the orthocenter and $I$ be the incenter of $\triangle ABC$. Key Claim: $\angle MON = 90^\circ$ iff $IO=IH$. Proof: It is well known that the homothety at $M$ with ratio $-2$ takes $O \to H$ and $I \to N$. Let$X,Y$ be the midpoints of segments $HN,HO$, respectively. Clearly $XY \parallel ON$. Now $IONX$ is a parallelogram, thus $IX \parallel ON$. Hence points $I,X,Y$ lie on a line parallel to $ON$. Finally, $$ \angle MON = 90^\circ \iff IY \perp OH \iff \overline{IY} \text{ is perpendicular bisector of segment } OH \iff IO = IH $$$\square$ Now it isn't hard to show $IO = IH$ iff some angle of the triangle is $60^\circ$.