49th Austrian Mathematical Olympiad, National Competition (Final Round, part 1)

For $AB=AC$, we get $D=M=P$, so the points $M$, $N$ and $P$ are trivially collinear. We will now only prove the case $AB>AC$ as $AB<AC$ is completely analogous. Let $\alpha$, $\beta$ and $\gamma$ denote the interior angles of the triangle in $A$, $B$ and $C$ respectively, as usual; see Figure 1. We note that $$\angle BDP=\angle CDE=90^{\circ }-\frac{\gamma }{2}\text{and}\angle BIP=\frac{\alpha +\beta }{2}=90^{\circ }-\frac{\gamma }{2}$$ certainly hold, which implies that the quadrilateral $BPDI$ is inscribed. We therefore have $90^{\circ }=\angle IDB=\angle IPB$, which implies that the triangle $APB$ is right. Since $N$ is the mid-point of its hypotenuse, it must be its circumcenter, which yields $\angle BNP=2\cdot \angle BAP=\alpha$. We see that $PN$ is parallel to $AC$, and since $MN$ is also parallel to $AC$ by virtue of the fact that $M$ and $N$ are the mid-points of their respective sides of $ABC$, we see that $M$, $N$ and $P$ must be collinear, as claimed.