All-Soviet-Union Mathematical Olympiad

Solution:

If the tangent at $B$ is parallel to $AC$, then $\angle NBC = \angle BCA$ (parallel lines) and $\angle NBC = \angle BAC$ ($NB$ tangent), so $BCA$ is isosceles and $BC = BA$. Hence the figure is symmetrical about the line $PB$ and so $BP$ bisects $MPN$.

So assume $AC$ is not parallel to the tangent at $B$. Assume it meets it at $L$ on the same side of $B$ as $N$. Take $A'$ on the line $AC$ so that $MA = MA'$. We show that $LMA'$ and $LNC$ are similar. Obviously the angles at $L$ are the same. $\angle MA'L = \angle MAL$ ($MA = MA'$) $= \angle ABC$ ($MA$ tangent) $= \angle LCN$ ($NC$ tangent). So the triangles are similar. Hence $LN / NC = LM / MA'$. But $NC = NB$ and $MA' = MA = MB$, so $LN / NB = LM / MB$ and hence $LM / LN = MB / NB$. So the circle on $LB$ as diameter has all points $Q$ on it satisfying $QM / QN = BM / BN$. But $\angle LPB = 90^{\circ}$, so $P$ must lie on the circle and hence $PM / PN = BM / BN$. Hence $PB$ is the angle bisector of $MPN$.