NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Consider the point $N$ on the side $AC$ such that $nAN=AC$. The configuration is symmetric with respect to the perpendicular bisector of the segment $BC$, implying $\angle M{P}_{i}A=\angle N{P}_{n-i}A$, $i=1,2,\dots ,n-1$. The claim is equivalent to $\angle M{P}_{1}N+\angle M{P}_{2}N+\cdots +\angle M{P}_{n-1}N=\angle BAC$.

Notice that $B{P}_1=P_1{P}_2=P_2{P}_3=\cdots =P_{n-1}C=MN$. Set $P_0=B$, and notice that in the parallelograms $P_{i}MN{P}_{i+1}$ one has $\angle M{P}_{i+1}N=\angle P_{i}M{P}_{i+1}$, $i=0,1,\dots ,n-2$. Therefore the sum of those angles is equal to $\angle P_{0}M{P}_{n-1}$, in turn equal to $\angle BAC$, since $M{P}_{n-1}\parallel AC$.