IMO Shortlist, 2016

Let the perpendicular from $S$ to $X Y$ meet line $Q M$ at $S'$. Let $E$ be the foot of altitude from $B$ to side $A C$. Since $Q$ and $S$ lie on the perpendicular bisector of $A D$, using directed angles, we have
$$\begin{aligned} \measuredangle S D Q & =\measuredangle Q A S=\measuredangle X A S-\measuredangle X A Q=\left(\frac{\pi}{2}-\measuredangle A Y X\right)-\measuredangle B A P=\measuredangle C B A-\measuredangle A Y X \\ & =(\measuredangle C B A-\measuredangle A C B)-\measuredangle B C A-\measuredangle A Y X=\measuredangle P E M-(\measuredangle B C A+\measuredangle A Y X) \\ & =\measuredangle P Q M-\measuredangle(B C, X Y)=\frac{\pi}{2}-\measuredangle\left(S' Q, B C\right)-\measuredangle(B C, X Y)=\measuredangle S S' Q \end{aligned}$$
This shows $D, S', S, Q$ are concyclic.

Let the perpendicular from $N$ to $B C$ intersect line $S S'$ at $O_1$. (Note that the two lines coincide when $S$ is the midpoint of $A O$, in which case the result is true since the circumcentre of triangle $X S Y$ must lie on this line.) It suffices to show that $O_1$ is the circumcentre of triangle $X S Y$ since $N$ lies on the perpendicular bisector of $P M$. From
$$\measuredangle D S' O_1=\measuredangle D Q S=\measuredangle S Q A=\angle(S Q, Q A)=\angle\left(O D, O_1 N\right)=\measuredangle D N O_1$$
since $S Q \parallel O D$ and $Q A \parallel O_1 N$, we know that $D, O_1, S', N$ are concyclic. Therefore, we get
$$\measuredangle S D S'=\measuredangle S Q S'=\angle\left(S Q, Q S'\right)=\angle\left(N D, N S'\right)=\measuredangle D N S'$$
so that $S D$ is a tangent to the circle through $D, O_1, S', N$. Then we have
$$\begin{equation*} S S' \cdot S O_1=S D^2=S X^2 \tag{1} \end{equation*}$$
Next, we show that $S$ and $S'$ are symmetric with respect to $X Y$. By the Sine Law, we have
$$\frac{S S'}{\sin \angle S Q S'}=\frac{S Q}{\sin \angle S S' Q}=\frac{S Q}{\sin \angle S D Q}=\frac{S Q}{\sin \angle S A Q}=\frac{S A}{\sin \angle S Q A} .$$
It follows that
$$S S'=S A \cdot \frac{\sin \angle S Q S'}{\sin \angle S Q A}=S A \cdot \frac{\sin \angle H O A}{\sin \angle O H A}=S A \cdot \frac{A H}{A O}=S A \cdot 2 \cos A,$$
which is twice the distance from $S$ to $X Y$. Note that $S$ and $C$ lie on the same side of the perpendicular bisector of $P M$ if and only if $\angle S A C<\angle O A C$ if and only if $\angle Y X A>\angle C B A$. This shows $S$ and $O_1$ lie on different sides of $X Y$. As $S'$ lies on ray $S O_1$, it follows that $S$ and $S'$ cannot lie on the same side of $X Y$. Therefore, $S$ and $S'$ are symmetric with respect to $X Y$.
Let $d$ be the diameter of the circumcircle of triangle $X S Y$. As $S S'$ is twice the distance from $S$ to $X Y$ and $S X=S Y$, we have $S S'=2 \frac{S X^2}{d}$. It follows from (1) that $d=2 S O_1$. As $S O_1$ is the perpendicular bisector of $X Y$, point $O_1$ is the circumcentre of triangle $X S Y$.

Denote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $O_1$ be the circumcentre of triangle $X S Y$. Consider two other possible positions of $S$. We name them $S'$ and $S''$ and define the analogous points $X', Y', O_1', X'', Y'' O_1''$ accordingly. Note that $S, S', S''$ lie on the perpendicular bisector of $A D$.
As $X X'$ and $Y Y'$ meet at $A$ and the circumcircles of triangles $A X Y$ and $A X' Y'$ meet at $D$, there is a spiral similarity with centre $D$ mapping $X Y$ to $X' Y'$. We find that
$$\measuredangle S X Y=\frac{\pi}{2}-\measuredangle Y A X=\frac{\pi}{2}-\measuredangle Y' A X'=\measuredangle S' X' Y'$$
and similarly $\measuredangle S Y X=\measuredangle S' Y' X'$. This shows triangles $S X Y$ and $S' X' Y'$ are directly similar. Then the spiral similarity with centre $D$ takes points $S, X, Y, O_1$ to $S', X', Y', O_1'$. Similarly, there is a spiral similarity with centre $D$ mapping $S, X, Y, O_1$ to $S'', X'', Y'', O_1''$. From these, we see that there is a spiral similarity taking the corresponding points $S, S', S''$ to points $O_1, O_1', O_1''$. In particular, $O_1, O_1', O_1''$ are collinear.

It now suffices to show that $O_1$ lies on the perpendicular bisector of $P M$ for two special cases.
Firstly, we take $S$ to be the midpoint of $A H$. Then $X$ and $Y$ are the feet of altitudes from $C$ and $B$ respectively. It is well-known that the circumcircle of triangle $X S Y$ is the nine-point circle of triangle $A B C$. Then $O_1$ is the nine-point centre and $O_1 P=O_1 M$. Indeed, $P$ and $M$ also lies on the nine-point circle.
Secondly, we take $S'$ to be the midpoint of $A O$. Then $X'$ and $Y'$ are the midpoints of $A B$ and $A C$ respectively. Then $X' Y' \parallel B C$. Clearly, $S'$ lies on the perpendicular bisector of $P M$. This shows the perpendicular bisectors of $X' Y'$ and $P M$ coincide. Hence, we must have $O_1' P=O_1' M$.