HMMT Spring Team Round, 2021

Solution:

Let $H$ be the orthocenter of $\triangle D E F$. We first claim that $O, I, H$ are collinear. We present two proofs.

Proof 1. Invert about $\omega$. Circle $(A B C)$ inverts to a circle with center on $O I$, but $A, B, C$ invert to the midpoints of $E F, F D, D E$, respectively, so the nine-point center of $\triangle D E F$ is on $O I$. As this center is the midpoint of $I H$, we get that $H, I, O$ are collinear.

Proof 2. Let $Q_{A}, Q_{B}, Q_{C}$ be the second intersections of the $D$-, $E$-, and $F$-altitudes, respectively, in $\triangle D E F$ with $\omega$. We claim $\triangle Q_{A} Q_{B} Q_{C}$ is homothetic with $\triangle A B C$. Indeed, as $Q_{B}$ is the reflection of $H$ over $D F$ and $Q_{C}$ is the reflection of $H$ over $D E$, $D Q_{B} = D Q_{C}$, so the perpendicular bisector of $Q_{B} Q_{C}$ is line $I D$. As $I D \perp B C$, $Q_{B} Q_{C} \parallel B C$, whence the homothety follows. This homothety takes the incircle to the circumcircle, so it is centered on line $O I$. However, it also takes the incenter $H$ of $Q_{A} Q_{B} Q_{C}$ to the incenter $I$ of $A B C$, so it is centered on line $I H$. So, $O, I, H$ are collinear.

As $P$ is the midpoint of $Q H$, it suffices to show that $P$ is on the circle with diameter $Q H$, or that $\angle Q T H = 90^{\circ}$. As $A T \perp \overline{T H} = \overline{I O}$, it suffices to show that $Q$ is on line $A T$. We also present two proofs of this.

Proof 1. Let $D'$ be the antipode of $D$, and let $A D'$ intersect $B C$ at $X$. As $X$ is the $A$-extouch point, the midpoint $M$ of $D X$ is also the midpoint of $B C$. We have
$$\frac{O M}{M X} = \frac{I D}{\frac{D X}{2}} = \frac{D D'}{D X}$$
and $\angle O M X = \angle D' D X = 90^{\circ}$, so $D', O, X$ are collinear, so $D'$ is on line $A O$. As $Q D' \parallel E F$, $A Q$ and $A D'$ are isogonal in $\angle B A C$, so $A Q$ and $A O$ are isogonal, which means $Q$ is on the $A$-altitude, finishing the proof.

Proof 2. Let $\Gamma$ denote the circumcircle of $\triangle A B C$, and let $M$ be the midpoint of arc $B C$ on $\Gamma$ not containing $A$.

Lemma. The intersection $T'$ of $M D$ and the $A$-altitude to $B C$ is on the line through $I$ parallel to $B C$.

Proof. Let $D' = M A \cap B C$. As $\angle D' B M = \angle C B M = \angle C A M = \angle M A B$, $\triangle D' B M \sim \triangle B A M$, and
$$M I^{2} = M B^{2} = M D' \cdot M A$$
Since $A T' \parallel I D$, we have
$$\frac{M T'}{M D} = \frac{M A}{M I} = \frac{M I}{M D'}$$
so $I T' \parallel D D'$, finishing the proof.

By the above lemma, $T$ is on $M D$. Consider a homothety centered at $T$ that takes $D$ to $M$. It takes $\omega$ to a circle centered on line $I T$ that is tangent to $\Gamma$ at $M$; since $O$ is on line $I T$ this circle must be $\Gamma$ itself. So, $T$ is the exsimilicenter of $\Gamma$ and $\omega$. By Proof 2 above, $T$ is the center of the homothety which sends $Q_{A} Q_{B} Q_{C}$ to $A B C$, so $T, Q = Q_{A}$, and $A$ are collinear, finishing the proof.