Problem: The isosceles triangle $\triangle ABC$, with $AB=AC$,
is inscribed in the circle $\omega$.
Let $P$ be a variable point on the arc $BC$
that does not contain $A$,
and let $I_B$ and $I_C$ denote the incenters of
triangles $\triangle ABP$ and $\triangle ACP$, respectively.
Prove that as $P$ varies, the circumcircle
of triangle $\triangle PI_{B}I_{C}$ passes through a fixed point.

Solution: Let $M$ be the midpoint of arc $BC$ not containing $A$.
We claim $M$ is the desired fixed point.

\begin{center}
\begin{asy}
pair A = dir(90);
pair M_B = dir(152);
pair B = M_B*M_B/A;
pair M_C = A*A/M_B;
pair C = A*A/B;
pair P = dir(250);

draw(unitcircle, blue);
filldraw(A--B--P--cycle, opacity(0.1)+lightblue, blue);
filldraw(A--C--P--cycle, opacity(0.1)+lightblue, blue);
draw(B--C, blue);

pair M = -A;
pair I_B = incenter(P, A, B);
pair I_C = incenter(P, A, C);

draw(I_B--P--I_C, blue);
draw(A--P, blue);

filldraw(incircle(A, B, P), opacity(0.05)+heavycyan, heavycyan+dotted);
filldraw(incircle(A, C, P), opacity(0.05)+heavycyan, heavycyan+dotted);


filldraw(M_B--I_B--M--cycle, opacity(0.2)+yellow, orange+1);
filldraw(M_C--I_C--M--cycle, opacity(0.2)+yellow, orange+1);

draw(circumcircle(P, I_B, I_C), red+dashed);

dot("$A$", A, dir(A));
dot("$M_B$", M_B, dir(M_B));
dot("$B$", B, dir(B));
dot("$M_C$", M_C, dir(M_C));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$M$", M, dir(M));
dot("$I_B$", I_B, dir(160));
dot("$I_C$", I_C, dir(170));

/* TSQ Source:

A = dir 90
M_B = dir 152
B = M_B*M_B/A
M_C = A*A/M_B
C = A*A/B
P = dir 250

unitcircle blue
A--B--P--cycle 0.1 lightblue / blue
A--C--P--cycle 0.1 lightblue / blue
B--C blue

M = -A
I_B = incenter P A B R160
I_C = incenter P A C R170

I_B--P--I_C blue
A--P blue

incircle A B P 0.05 heavycyan / heavycyan dotted
incircle A C P 0.05 heavycyan / heavycyan dotted


M_B--I_B--M--cycle 0.2 yellow / orange+1
M_C--I_C--M--cycle 0.2 yellow / orange+1

circumcircle P I_B I_C red dashed

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

Let $M_B$, $M_C$ be the second intersections of $PI_B$ and $PI_C$ with circumcircle.
\begin{claim*}
  We have $\triangle I_{B}M_{B}M \cong \triangle I_{C}M_{C}M$.
\end{claim*}
\begin{proof}
  Note that
  \begin{align*}
    M_{B}I_{B} &= M_{B}B = M_{C}C = M_{C}I_{C} \\
    MM_B &= MM_C \\
    \angle I_{B}M_{B}M &= \frac{1}{2} \widehat{PM} = \angle I_{C}M_{C}M.
  \end{align*}
  This implies the desired congruence.
\end{proof}

Since $\angle MPA = 90^\circ$ and ray $PA$ bisects $\angle I_{B}PI_{C}$,
the conclusion $MI_B = MI_C$ finishes the problem.

\begin{remark}
  Complex in the obvious way DOES NOT WORK,
  because the usual claim (``the fixed point is arc midpoint'') is FALSE
  if the hypothesis that $P$ lies in the interior of the arc is dropped.
  See figure below.
  \begin{center}
  \begin{asy}
  pair A = dir(90);
  pair M_B = dir(152);
  pair B = M_B*M_B/A;
  pair M_C = A*A/M_B;
  pair C = A*A/B;
  pair P = dir(140);

  draw(unitcircle, blue);
  filldraw(A--B--P--cycle, opacity(0.1)+lightblue, blue);
  filldraw(A--C--P--cycle, opacity(0.1)+lightblue, blue);
  draw(B--C, blue);

  pair M = -A;
  pair I_B = incenter(P, A, B);
  pair I_C = incenter(P, A, C);

  draw(A--P, blue);

  filldraw(incircle(A, B, P), opacity(0.05)+heavycyan, heavycyan+dotted);
  filldraw(incircle(A, C, P), opacity(0.05)+heavycyan, heavycyan+dotted);

  draw(circumcircle(P, I_B, I_C), red+dashed);

  dot("$A$", A, dir(A));
  dot("$B$", B, dir(B));
  dot("$C$", C, dir(C));
  dot("$P$", P, dir(P));
  dot("$M$", M, dir(M));
  dot("$I_B$", I_B, dir(100));
  dot("$I_C$", I_C, dir(340));

  /* TSQ Source:

  A = dir 90
  M_B := dir 152
  B = M_B*M_B/A
  M_C := A*A/M_B
  C = A*A/B
  P = dir 140

  unitcircle blue
  A--B--P--cycle 0.1 lightblue / blue
  A--C--P--cycle 0.1 lightblue / blue
  B--C blue

  M = -A
  I_B = incenter P A B R100
  I_C = incenter P A C R340

  A--P blue

  incircle A B P 0.05 heavycyan / heavycyan dotted
  incircle A C P 0.05 heavycyan / heavycyan dotted

  circumcircle P I_B I_C red dashed

  */
  \end{asy}
  \end{center}
  Fun story, I pointed this out to Zuming during grading;
  I was the only one that realized the subtlety.
\end{remark}