IMO Shortlist, 2019

Step 1. The external bisector of $\angle B A C$ is the line through $A$ perpendicular to $I A$. Let $D I$ meet this line at $L$ and let $D I$ meet $\omega$ at $K$. Let $N$ be the midpoint of $E F$, which lies on $I A$ and is the pole of line $A L$ with respect to $\omega$. Since $A N \cdot A I = A E^{2} = A R \cdot A P$, the points $R$, $N, I$, and $P$ are concyclic. As $I R = I P$, the line $N I$ is the external bisector of $\angle P N R$, so $P N$ meets $\omega$ again at the point symmetric to $R$ with respect to $A N$ - i.e. at $K$.
Let $D N$ cross $\omega$ again at $S$. Opposite sides of any quadrilateral inscribed in the circle $\omega$ meet on the polar line of the intersection of the diagonals with respect to $\omega$. Since $L$ lies on the polar line $A L$ of $N$ with respect to $\omega$, the line $P S$ must pass through $L$. Thus it suffices to prove that the points $S, Q$, and $P$ are collinear.

Step 2. Let $\Gamma$ be the circumcircle of $\triangle B I C$. Notice that
$$\begin{aligned} \angle(B Q, Q C) = \angle & (B Q, Q P) + \angle(P Q, Q C) = \angle(B F, F P) + \angle(P E, E C) \\ & = \angle(E F, E P) + \angle(F P, F E) = \angle(F P, E P) = \angle(D F, D E) = \angle(B I, I C) \end{aligned}$$
so $Q$ lies on $\Gamma$. Let $Q P$ meet $\Gamma$ again at $T$. It will now suffice to prove that $S, P$, and $T$ are collinear. Notice that $\angle(B I, I T) = \angle(B Q, Q T) = \angle(B F, F P) = \angle(F K, K P)$. Note $F D \perp F K$ and $F D \perp B I$ so $F K \parallel B I$ and hence $I T$ is parallel to the line $K N P$. Since $D I = I K$, the line $I T$ crosses $D N$ at its midpoint $M$.

Step 3. Let $F^{\prime}$ and $E^{\prime}$ be the midpoints of $D E$ and $D F$, respectively. Since $D E^{\prime} \cdot E^{\prime} F = D E^{\prime 2} = B E^{\prime} \cdot E^{\prime} I$, the point $E^{\prime}$ lies on the radical axis of $\omega$ and $\Gamma$; the same holds for $F^{\prime}$. Therefore, this radical axis is $E^{\prime} F^{\prime}$, and it passes through $M$. Thus $I M \cdot M T = D M \cdot M S$, so $S, I, D$, and $T$ are concyclic. This shows $\angle(D S, S T) = \angle(D I, I T) = \angle(D K, K P) = \angle(D S, S P)$, whence the points $S, P$, and $T$ are collinear, as desired.

We start as in Solution 1. Namely, we introduce the same points $K, L, N$, and $S$, and show that the triples $(P, N, K)$ and $(P, S, L)$ are collinear. We conclude that $K$ and $R$ are symmetric in $A I$, and reduce the problem statement to showing that $P, Q$, and $S$ are collinear.

Step 1. Let $A R$ meet the circumcircle $\Omega$ of $A B C$ again at $X$. The lines $A R$ and $A K$ are isogonal in the angle $B A C$; it is well known that in this case $X$ is the tangency point of $\Omega$ with the $A$-mixtilinear circle. It is also well known that for this point $X$, the line $X I$ crosses $\Omega$ again at the midpoint $M^{\prime}$ of $\operatorname{arc} B A C$.

Step 2. Denote the circles $B F P$ and $C E P$ by $\Omega_{B}$ and $\Omega_{C}$, respectively. Let $\Omega_{B}$ cross $A R$ and $E F$ again at $U$ and $Y$, respectively. We have
$$\angle(U B, B F) = \angle(U P, P F) = \angle(R P, P F) = \angle(R F, F A),$$
so $U B \parallel R F$.

Next, we show that the points $B, I, U$, and $X$ are concyclic. Since
$$\angle(U B, U X) = \angle(R F, R X) = \angle(A F, A R) + \angle(F R, F A) = \angle\left(M^{\prime} B, M^{\prime} X\right) + \angle(D R, D F),$$
it suffices to prove $\angle(I B, I X) = \angle\left(M^{\prime} B, M^{\prime} X\right) + \angle(D R, D F)$, or $\angle\left(I B, M^{\prime} B\right) = \angle(D R, D F)$. But both angles equal $\angle(C I, C B)$, as desired. (This is where we used the fact that $M^{\prime}$ is the midpoint of arc $B A C$ of $\Omega$.)

It follows now from circles BUIX and BPUFY that
$$\begin{aligned} \angle(I U, U B) = \angle(I X, B X) = \angle\left(M^{\prime} X, B X\right) = & \frac{\pi-\angle A}{2} \\ & = \angle(E F, A F) = \angle(Y F, B F) = \angle(Y U, B U), \end{aligned}$$
so the points $Y, U$, and $I$ are collinear.

Let $E F$ meet $B C$ at $W$. We have
$$\angle(I Y, Y W) = \angle(U Y, F Y) = \angle(U B, F B) = \angle(R F, A F) = \angle(C I, C W),$$
so the points $W, Y, I$, and $C$ are concyclic.

Similarly, if $V$ and $Z$ are the second meeting points of $\Omega_{C}$ with $A R$ and $E F$, we get that the 4-tuples ( $C, V, I, X$ ) and ( $B, I, Z, W$ ) are both concyclic.

Step 3. Let $Q^{\prime} = C Y \cap B Z$. We will show that $Q^{\prime} = Q$.

First of all, we have
$$\begin{aligned} \angle\left(Q^{\prime} Y, Q^{\prime} B\right) = \angle(C Y, Z B) & = \angle(C Y, Z Y) + \angle(Z Y, B Z) \\ & = \angle(C I, I W) + \angle(I W, I B) = \angle(C I, I B) = \frac{\pi-\angle A}{2} = \angle(F Y, F B) \end{aligned}$$
so $Q^{\prime} \in \Omega_{B}$. Similarly, $Q^{\prime} \in \Omega_{C}$. Thus $Q^{\prime} \in \Omega_{B} \cap \Omega_{C} = \{P, Q\}$ and it remains to prove that $Q^{\prime} \neq P$. If we had $Q^{\prime} = P$, we would have $\angle(P Y, P Z) = \angle\left(Q^{\prime} Y, Q^{\prime} Z\right) = \angle(I C, I B)$. This would imply
$$\angle(P Y, Y F) + \angle(E Z, Z P) = \angle(P Y, P Z) = \angle(I C, I B) = \angle(P E, P F),$$
so circles $\Omega_{B}$ and $\Omega_{C}$ would be tangent at $P$. That is excluded in the problem conditions, so $Q^{\prime} = Q$.

Step 4. Now we are ready to show that $P, Q$, and $S$ are collinear.

Notice that $A$ and $D$ are the poles of $E W$ and $D W$ with respect to $\omega$, so $W$ is the pole of $A D$. Hence, $W I \perp A D$. Since $C I \perp D E$, this yields $\angle(I C, W I) = \angle(D E, D A)$. On the other hand, $D A$ is a symmedian in $\triangle D E F$, so $\angle(D E, D A) = \angle(D N, D F) = \angle(D S, D F)$. Therefore,
$$\begin{aligned} \angle(P S, P F) = \angle(D S, D F) = \angle(D E, D A) = & \angle(I C, I W) \\ & = \angle(Y C, Y W) = \angle(Y Q, Y F) = \angle(P Q, P F) \end{aligned}$$
which yields the desired collinearity.