Interesting problem. Let $AM-BC=x$ Let $O$ -incenter of $ABC$ and incircle touch $AC$ at $H$ Then $MH=AM+AH=BC+x+AH=x+BC+\frac{AC+AB-BC}{2}=x+\frac{AB+AC+BC}{2}$ So $OM^2=r^2+(x+s)^2$ where $r,s$ - inradius and semiperimeter of $ABC$ So $O$ - circumcircle of $MON$ Let incircle touch $AB$ at $E$ and $OM$ intersect $AN$ at $T$ Then $\angle OMA=\angle ONB$ and so $\angle OTB=\angle A-\angle OMA \to \angle MON=180-\angle OTB-\angle ONB=180-\angle A$ $\angle MPN= \frac{\angle MON}{2}=90-\frac{\angle A}{2}$ So $MPN$ has angles $90-\frac{\angle A}{2},90-\frac{\angle B}{2},90-\frac{\angle C}{2}$