, 2008

Solution:

Consider the points $M'$ on the ray $B A$ (after $A$), $N'$ on the ray $C B$ (after $B$), and $P'$ on the ray $A C$ (after $C$), so that $A M = A M'$, $B N = B N'$, $C P = C P'$. Since $A M - B C = B N - A C = B N' - A C$, we get $C M = A C + A M = B C + B N' = C N'$. Thus triangle $M C N'$ is isosceles, so the perpendicular bisector of $[M N']$ bisects angle $A C B$ and hence passes through the incenter $I$ of triangle $A B C$.

Arguing similarly, we may conclude that $I$ lies also on the perpendicular bisectors of $[N P']$ and $[P M']$. On the other side, $I$ clearly lies on the perpendicular bisectors of $[M M']$, $[N N']$, and $[P P']$. Thus the hexagon $M' M N' N P' P$ is cyclic.

Then angle $P M N$ equals angle $P N' N$, which measures $90^{\circ} - \frac{\beta}{2}$ (the angles of triangle $A B C$ are $\alpha, \beta, \gamma$). In the same way, angle $M N P$ measures $90^{\circ} - \frac{\gamma}{2}$ and angle $M P N$ measures $90^{\circ} - \frac{\alpha}{2}$.