Berkeley Math Circle Monthly Contest 3

Solution:

Without loss of generality, $AB \leq AC$. Let $B_1$ and $C_1$ be the projections of $C$ and $B$ onto lines $BI$ and $CI$, respectively. Then $\angle IEC = \angle IB_1C = 90^{\circ}$, so quadrilateral $IEB_1C$ is cyclic. But then
$$\angle B_1EC = \angle B_1IC = 180^{\circ} - \angle BIC = 180^{\circ} - \angle FEC$$
and so we find that $B_1$ lies on line $EF$. Similarly, $C_1$ lies on line $EF$.

Observe that $I$ is the orthocenter of triangle $HBC$. Then it is well-known that $I$ is the incenter of triangle $DB_1C_1$. Moreover, $H$ is the intersection of the external angle bisectors of $\angle DC_1B_1$ and $\angle DB_1C_1$. So it is the center of the circle $\Gamma$ tangent to $B_1C_1$ and the extensions of rays $DB_1$ and $DC_1$ past $B_1$ and $C_1$.

Let $N$ be the foot of the altitude from $I$ to $\overline{B_1C_1}$, noting that $N$ is in fact the midpoint of $\overline{EF}$. Let $K$ be the tangency point of the $\Omega$ to $\overline{B_1C_1}$. Consider the homothety which takes the incircle of $\triangle DB_1C_1$ to the circle $\Gamma$; it sends $N$ to the point diametrically opposite $K$ on $\Omega$. Call this point $L$; we see $D, N, L$ are collinear by the homothety mentioned. Moreover, $M, N, K$ are collinear. As $H$ is the midpoint of $\overline{KL}$ and $P$ is the midpoint of $\overline{DM}$, it follows that the points $H, N, P$ are collinear, which is what we wanted to prove.