Problem: Quadrilateral $APBQ$ is inscribed in circle $\omega$ with
$\angle P = \angle Q = 90^{\circ}$ and $AP = AQ < BP$.
Let $X$ be a variable point on segment $\ol{PQ}$.
Line $AX$ meets $\omega$ again at $S$ (other than $A$).
Point $T$ lies on arc $AQB$ of $\omega$ such that $\ol{XT}$
is perpendicular to $\ol{AX}$.
Let $M$ denote the midpoint of chord $\ol{ST}$.

As $X$ varies on segment $\ol{PQ}$, show that $M$ moves along a circle.

Solution: We present three solutions,
one by complex numbers, two more synthetic.
(A fourth solution using median formulas is also possible.)
Most solutions will prove that the center of the fixed circle
is the midpoint of $\ol{AO}$
(with $O$ the center of $\omega$);
this can be recovered empirically by letting
\begin{itemize}
  \ii $X$ approach $P$ (giving the midpoint of $\ol{BP}$)
  \ii $X$ approach $Q$ (giving the point $Q$), and
  \ii $X$ at the midpoint of $\ol{PQ}$ (giving the midpoint of $\ol{BQ}$)
\end{itemize}
which determines the circle;
this circle then passes through $P$ by symmetry
and we can find the center by taking the intersection
of two perpendicular bisectors (which two?).

\paragraph{Complex solution (Evan Chen).}
Toss on the complex unit circle with $a = -1$, $b=1$, $z = -\tfrac12$.
Let $s$ and $t$ be on the unit circle. We claim $Z$ is the center.

It follows from standard formulas that
\[ x = \frac 12 \left( s + t - 1 + s/t \right) \]
thus
\[ 4\operatorname{Re} x + 2
  = 2\left( x + \frac 1x \right) + 2
  = s + t + \frac 1s + \frac 1t + \frac st + \frac ts \]
which depends only on $P$ and $Q$, and not on $X$.
Thus
\[ 4\left\lvert z-\frac{s+t}{2} \right\rvert^2
  = \left\lvert s+t+1 \right\rvert^2 = 3 + (4\operatorname{Re}x+2) \]
does not depend on $X$, done.

\paragraph{Homothety solution (Alex Whatley).}
Let $G$, $N$, $O$ denote the centroid, nine-point center,
and circumcenter of triangle $AST$, respectively.
Let $Y$ denote the midpoint of $\ol{AS}$.
Then the three points $X$, $Y$, $M$
lie on the nine-point circle of triangle $AST$,
which is centered at $N$ and has radius $\frac 12 AO$.

\begin{center}
\begin{asy}
size(9cm);

pair A = dir(90);
pair B = dir(-90);
pair S = dir(-50);
pair T = dir(170);
pair O = midpoint(A--B);
pair X = foot(T, A, S);
pair E = dir(0);

pair P = IP(unitcircle, X--(X-2*E));
pair Q = IP(unitcircle, X--(X+2*E));

filldraw(unitcircle, opacity(0.2)+mediumcyan, mediumblue);

pair M = midpoint(S--T);
pair G = centroid(A, S, T);
pair N = 3/2*G;

filldraw(A--P--B--Q--cycle, opacity(0.1)+lightblue, mediumblue);
draw(A--B, mediumblue+dotted);
draw(P--Q, mediumblue+dotted);

filldraw(A--S--T--cycle, opacity(0.4)+mediumgreen, heavygreen);
filldraw(CP(N, M), opacity(0.6)+lightred, red);

draw(T--X, heavygreen);
draw(A--M, heavygreen);
pair Y = midpoint(A--S);

draw(O--N, red);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$S$", S, dir(S));
dot("$T$", T, dir(T));
dot("$O$", O, dir(-45));
dot("$X$", X, dir(45));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$M$", M, dir(M));
dot("$G$", G, dir(30));
dot("$N$", N, dir(N));
dot("$Y$", Y, dir(Y));

/* Source generated by TSQ

!size(9cm);

A = dir 90
B = dir -90
S = dir -50
T = dir 170
O = midpoint A--B R-45
X = foot T A S R45
E := dir 0

P = IP unitcircle X--(X-2*E)
Q = IP unitcircle X--(X+2*E)

unitcircle 0.2 mediumcyan / mediumblue

M = midpoint S--T
G = centroid A S T R30
N = 3/2*G

A--P--B--Q--cycle 0.1 lightblue / mediumblue
A--B mediumblue dotted
P--Q mediumblue dotted

A--S--T--cycle 0.4 mediumgreen / heavygreen
CP N M 0.6 lightred / red

T--X heavygreen
A--M heavygreen
Y = midpoint A--S

O--N red

*/
\end{asy}
\end{center}

Let $R$ denote the radius of $\omega$.
Note that the nine-point circle of $\triangle AST$
has radius equal to $\half R$, and hence is independent of $S$ and $T$.
Then the power of $A$ with respect to the nine-point circle equals
\[ AN^2 - \left( \half R \right)^2 = AX \cdot AY
  = \frac 12 AX \cdot AS = \frac 12 AQ^2 \]
and hence
\[ AN^2 = \left( \half R \right)^2 + \frac 12 AQ^2 \]
which does not depend on the choice of $X$.
So $N$ moves along a circle centered at $A$.

Since the points $O$, $G$, $N$ are collinear on the Euler line of $\triangle AST$
with \[ GO = \frac 23 NO \] it follows by homothety
that $G$ moves along a circle as well,
whose center is situated one-third of the way from $A$ to $O$.
Finally, since $A$, $G$, $M$ are collinear with \[ AM = \frac 32 AG \]
it follows that $M$ moves along a circle
centered at the midpoint of $\ol{AO}$.

\paragraph{Power of a point solution (Zuming Feng, official solution).}
Let $Y$ be the foot of the altitude from $S$ to $\ol{AT}$.
Then $\ol{XY} \perp \ol{AO}$, so $Y$ lies on line $PQ$ too.
We then complete the picture by letting $K$ be the foot of $A$ to $\ol{ST}$.

\begin{center}
\begin{asy}
size(9cm);

pair A = dir(125);
pair B = -A;
pair S = dir(210);
pair T = dir(330);
pair O = midpoint(A--B);
pair X = foot(T, A, S);
pair E = dir(0);

filldraw(unitcircle, opacity(0.2)+mediumcyan, mediumblue);
pair M = midpoint(S--T);

filldraw(A--S--T--cycle, opacity(0.4)+mediumgreen, heavygreen);

draw(T--X, heavygreen);
draw(A--M, heavygreen);

pair Y = foot(S, A, T);
pair K = foot(A, S, T);
filldraw(circumcircle(X, Y, M), opacity(0.1)+yellow, red);
draw(S--Y, heavygreen);
draw(A--K, heavygreen);

pair P = IP(unitcircle, X--(3*Y-2*X));
pair Q = IP(unitcircle, Y--(3*X-2*Y));

pair V = extension(P, Q, S, T);
draw(P--Q, blue);
draw(A--B, blue);
draw(Q--V, blue);
draw(V--S, heavygreen);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$S$", S, dir(S));
dot("$T$", T, dir(T));
dot("$O$", O, dir(45));
dot("$X$", X, dir(135));
dot("$M$", M, dir(M));
dot("$Y$", Y, dir(70));
dot("$K$", K, dir(K));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(150));
dot("$V$", V, dir(V));

/* TSQ Source:

!size(9cm);

A = dir 125
B = -A
S = dir 210
T = dir 330
O = midpoint A--B R45
X = foot T A S R135
E := dir 0

unitcircle 0.2 mediumcyan / mediumblue
M = midpoint S--T

A--S--T--cycle 0.4 mediumgreen / heavygreen

T--X heavygreen
A--M heavygreen

Y = foot S A T R70
K = foot A S T
circumcircle X Y M 0.1 yellow / red
S--Y heavygreen
A--K heavygreen

P = IP unitcircle X--(3*Y-2*X)
Q = IP unitcircle Y--(3*X-2*Y) R150

V = extension P Q S T
P--Q blue
A--B blue
Q--V blue
V--S heavygreen

*/
\end{asy}
\end{center}

The main claim is:
\begin{claim*}
  Quadrilateral $PQKM$ is cyclic.
\end{claim*}
\begin{proof}
  To see this, we use power of a point: let $V = \ol{QXYP} \cap \ol{SKMT}$.
  One approach is that since $(VK;ST) = -1$
  we have $VQ \cdot VP = VS \cdot VT = VK \cdot VM$.
  A longer approach is more elementary:
  \[ VQ \cdot VP = VS \cdot VT = VX \cdot VY = VK \cdot VM \]
  using the nine-point circle, and the circle with diameter $\ol{ST}$.
\end{proof}
But the circumcenter of $PQKM$,
is the midpoint of $\ol{AO}$,
since it lies on the perpendicular bisectors of $\ol{KM}$ and $\ol{PQ}$.
So it is fixed, the end.