Problem: Let $ABC$ be an acute triangle
and let $I_B$, $I_C$, and $O$ denote its
$B$-excenter, $C$-excenter, and circumcenter, respectively.
Points $E$ and $Y$ are selected on $\ol{AC}$ such that
$\angle ABY = \angle CBY$ and $\ol{BE} \perp \ol{AC}$.
Similarly, points $F$ and $Z$ are selected on $\ol{AB}$ such that
$\angle ACZ = \angle BCZ$ and $\ol{CF} \perp \ol{AB}$.

Lines $I_B F$ and $I_C E$ meet at $P$.
Prove that $\ol{PO}$ and $\ol{YZ}$ are perpendicular.

Solution: We present two solutions.

\paragraph{First solution.}
Let $I_A$ denote the $A$-excenter and $I$ the incenter.
Then let $D$ denote the foot of the altitude from $A$.
Suppose the $A$-excircle is tangent to $\ol{BC}$, $\ol{AB}$, $\ol{AC}$
at $A_1$, $B_1$, $C_1$ and let
$A_2$, $B_2$, $C_2$ denote the reflections of $I_A$ across these points.
Let $S$ denote the circumcenter of $\triangle I I_B I_C$.

\begin{center}
\begin{asy}
size(14cm);

pair A = dir(110);
pair B = dir(208);
pair C = dir(332);
draw(A--B--C--cycle, blue+1);
pair I = incenter(A, B, C);
pair O = circumcenter(A, B, C);
pair E = foot(B, C, A);
pair F = foot(C, A, B);
pair M_A = extension(A, I, O, B+C);
pair I_A = 2*M_A-I;
pair M_B = extension(B, I, O, C+A);
pair I_B = 2*M_B-I;
pair M_C = extension(C, I, O, A+B);
pair I_C = 2*M_C-I;

pair P = extension(E, I_C, F, I_B);
draw(A--I_A, blue+dashed);

pair D = foot(A, B, C);
filldraw(I--I_B--I_C--cycle, opacity(0.2)+lightred, red);
filldraw(unitcircle, opacity(0.1)+lightblue, blue);

pair A_1 = foot(I_A, B, C);
pair B_1 = foot(I_A, A, B);
pair C_1 = foot(I_A, C, A);
draw(arc(I_A, abs(I_A-A_1), 15, 165), blue+dotted);
draw(C--C_1, blue+dashed);
draw(B--B_1, blue+dashed);
pair A_2 = 2*A_1-I_A;
pair B_2 = 2*B_1-I_A;
pair C_2 = 2*C_1-I_A;
draw(D--A_2, heavygreen);
draw(I_B--B_2, heavygreen);
draw(I_C--C_2, heavygreen);
draw(I_A--A_2, heavygreen+dashed);
draw(I_A--B_2, heavygreen+dashed);
draw(I_A--C_2, heavygreen+dashed);

draw(I_B--I_A--I_C, red+dashed);
draw(B--I--C, red+dashed);
filldraw(A_2--B_2--C_2--cycle, opacity(0.2)+lightgreen, deepgreen+1);
pair S = circumcenter(I, I_B, I_C);
draw(S--I_A, orange);
draw(arc(S, abs(S-I), 185, 355), red+dotted);

pair Y = extension(B, I, A, C);
pair Z = extension(C, I, A, B);
draw(Y--Z, red);

dot("$A$", A, dir(A));
dot("$B$", B, dir(180));
dot("$C$", C, dir(0));
dot("$I$", I, dir(270));
dot("$O$", O, dir(-45));
dot("$E$", E, dir(30));
dot("$F$", F, dir(F));
dot("$I_A$", I_A, dir(I_A));
dot("$I_B$", I_B, dir(I_B));
dot("$I_C$", I_C, dir(I_C));
dot("$P$", P, dir(135));
dot("$D$", D, dir(-90));
dot("$A_1$", A_1, dir(A_1));
dot("$B_1$", B_1, dir(200));
dot("$C_1$", C_1, dir(20));
dot("$A_2$", A_2, dir(A_2));
dot("$B_2$", B_2, dir(B_2));
dot("$C_2$", C_2, dir(C_2));
dot("$S$", S, dir(S));
dot("$Y$", Y, dir(350));
dot("$Z$", Z, dir(110));

/* Source generated by TSQ

!size(14cm);

A = dir 110
B = dir 208 R180
C = dir 332 R0
A--B--C--cycle blue+1
I = incenter A B C R270
O = circumcenter A B C R-45
E = foot B C A R30
F = foot C A B
M_A := extension A I O B+C
I_A = 2*M_A-I
M_B := extension B I O C+A
I_B = 2*M_B-I
M_C := extension C I O A+B
I_C = 2*M_C-I

P = extension E I_C F I_B R140
A--I_A blue dashed

D = foot A B C R-90
I--I_B--I_C--cycle 0.2 lightred / red
unitcircle 0.1 lightblue / blue

A_1 = foot I_A B C
B_1 = foot I_A A B R200
C_1 = foot I_A C A R20
!draw(arc(I_A, abs(I_A-A_1), 15, 165), blue+dotted);
C--C_1 blue dashed
B--B_1 blue dashed
A_2 = 2*A_1-I_A
B_2 = 2*B_1-I_A
C_2 = 2*C_1-I_A
D--A_2 heavygreen
I_B--B_2 heavygreen
I_C--C_2 heavygreen
I_A--A_2 heavygreen dashed
I_A--B_2 heavygreen dashed
I_A--C_2 heavygreen dashed

I_B--I_A--I_C red dashed
B--I--C red dashed
A_2--B_2--C_2--cycle 0.2 lightgreen / deepgreen+1
S = circumcenter I I_B I_C
S--I_A orange
!draw(arc(S, abs(S-I), 185, 355), red+dotted);

Y = extension B I A C R350
Z = extension C I A B R110
Y--Z red



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

We begin with the following observation:
\begin{claim*}
  Points $D$, $I$, $A_2$ are collinear.
  Similarly, points $E$, $I_C$, $C_2$ are collinear
  and points $F$, $I_B$, $B_2$ are collinear.
\end{claim*}

\begin{proof}
  This basically follows from the ``midpoints of altitudes'' lemma.
  To see $D$, $I$, $A_2$ are collinear,
  recall first that $\ol{IA_1}$ passes
  through the midpoint $M$ of $\ol{AD}$.
  \begin{center}
  \begin{asy}
    pair A = dir(125);
    pair B = dir(220);
    pair C = dir(320);
    draw(A--B--C--cycle, blue);
    pair I = incenter(A, B, C);
    pair O = circumcenter(A, B, C);
    pair M_A = extension(A, I, O, B+C);
    pair I_A = 2*M_A-I;
    filldraw(incircle(A, B, C), opacity(0.2)+lightblue, blue);

    pair D = foot(A, B, C);
    filldraw(unitcircle, opacity(0.1)+palecyan, blue);
    draw(A--D, heavygreen);

    pair A_1 = foot(I_A, B, C);
    draw(arc(I_A, abs(I_A-A_1), 0, 180), blue+dashed);
    pair B_1 = foot(I_A, A, B);
    pair C_1 = foot(I_A, C, A);
    draw(C--C_1, blue+dotted);
    draw(B--B_1, blue+dotted);

    pair A_2 = 2*A_1-I_A;
    draw(A--I_A, blue);
    draw(D--A_2, red);
    draw(I_A--A_2, blue+dotted);

    pair P = foot(I, B, C);
    pair Q = 2*I-P;
    draw(A--A_1, heavygreen);
    draw(P--Q, heavygreen);

    pair U = extension(Q, Q+B-C, A, B);
    pair V = extension(Q, Q+C-B, A, C);
    draw(U--V, heavygreen);

    pair M = midpoint(A--D);
    draw(M--A_1, dotted);

    dot(P); dot(Q);


    dot("$A$", A, dir(A));
    dot("$B$", B, dir(180));
    dot("$C$", C, dir(0));
    dot("$I$", I, dir(10));
    dot("$I_A$", I_A, dir(I_A));
    dot("$D$", D, dir(-90));
    dot("$A_1$", A_1, dir(-45));
    dot("$B_1$", B_1, dir(200));
    dot("$C_1$", C_1, dir(20));
    dot("$A_2$", A_2, dir(A_2));
    dot("$M$", M, dir(0));

    /* TSQ Source:

    A = dir 125
    B = dir 220 R180
    C = dir 320 R0
    A--B--C--cycle blue
    I = incenter A B C R10
    O := circumcenter A B C R45
    M_A := extension A I O B+C
    I_A = 2*M_A-I
    incircle A B C 0.2 lightblue / blue

    D = foot A B C R-90
    unitcircle 0.1 palecyan / blue
    A--D heavygreen

    A_1 = foot I_A B C R-45
    !draw(arc(I_A, abs(I_A-A_1), 0, 180), blue+dashed);
    B_1 = foot I_A A B R200
    C_1 = foot I_A C A R20
    C--C_1 blue dotted
    B--B_1 blue dotted

    A_2 = 2*A_1-I_A
    A--I_A blue
    D--A_2 red
    I_A--A_2 blue dotted

    P := foot I B C
    Q := 2*I-P
    A--A_1 heavygreen
    P--Q heavygreen

    U := extension Q Q+B-C A B
    V := extension Q Q+C-B A C
    U--V heavygreen

    M = midpoint A--D R0
    M--A_1 dotted

    !dot(P); dot(Q);

    */

  \end{asy}
  \end{center}
  Now since $\ol{AD} \parallel \ol{I_A A_2}$,
  and $M$ and $A_1$ are the midpoints of $\ol{AD}$ and $\ol{I_A A_2}$,
  it follows from the collinearity of $A$, $I$, $I_A$
  that $D$, $I$, $A_2$ are collinear as well.

  The other two claims follow in a dual fashion.
  For example, using the homothety taking the $A$ to $C$-excircle,
  we find that $\ol{C_1I_C}$ bisects the altitude $\ol{BE}$.
  (The homothety at $B$ will send $C_1$ to the point on the $C$-excircle which is
  diametrically opposite the foot from $I_C$ to $AC$. This diameter of the
  $C$-excircle is then mapped through homothety through $C_1$ onto the $B$-altitude.)
  Then since $I_C$, $B$, $I_A$ are collinear the same argument
  now gives $I_C$, $E$, $C_2$ are collinear.
  The fact that $I_B$, $F$, $B_2$ are collinear is symmetric.
\end{proof}

Observe that $\ol{B_2C_2} \parallel \ol{B_1C_1} \parallel \ol{I_B I_C}$.
Proceeding similarly on the other sides,
we discover $\triangle I I_B I_C$ and $\triangle A_2 B_2 C_2$ are homothetic.
Hence $P$ is the center of this homothety (in particular, $D$, $I$, $P$, $A_2$ are collinear).
Moreover, $P$ lies on the line joining $I_A$ to $S$,
which is the Euler line of $\triangle I I_B I_C$,
so it passes through the nine-point center of $\triangle I I_B I_C$,
which is $O$. Consequently, $P$, $O$, $I_A$ are collinear as well.

To finish, we need only prove that $\ol{OS} \perp \ol{YZ}$.
In fact, we claim that $\ol{YZ}$ is the radical axis
of the circumcircles of $\triangle ABC$ and $\triangle I I_B I_C$.
Actually, $Y$ is the radical center of these two circumcircles
and the circle with diameter $\ol{II_B}$
(which passes through $A$ and $C$).
Analogously $Z$ is the radical center of the circumcircles
and the circle with diameter $\ol{II_C}$, and the proof is complete.

\paragraph{Second solution (barycentric, outline, Colin Tang).}
we are going to use barycentric coordinates
to show that the line through $O$ perpendicular to $\ol{YZ}$
is concurrent with $\ol{I_B F}$ and $\ol{I_C E}$.

The displacement vector $\overrightarrow{YZ}$
is proportional to $(a(b-c):-b(a+c):c(a+b))$,
and so by strong perpendicularity criterion
and doing a calculation gives the line
\[ x(b-c)bc(a+b+c) + y(a+c)ac(a+b-c) + z(a+b)ab(-a+b-c) = 0. \]
On the other hand, line $I_C E$ has equation
\[
  0 =
  \det \begin{bmatrix}
    a &  b & -c \\
    S_C & 0 & S_A \\
    x & y & z
  \end{bmatrix}
  = bS_a \cdot x + (-cS_C-aS_A) \cdot y + (-bS_C) \cdot z
\]
and similarly for $I_B F$.
Consequently, concurrence of these lines is equivalent to
\[
  \det \begin{bmatrix}
    bS_A & -cS_C - aS_A & -bS_C \\
    cS_A & -cS_B & -aS_A - bS_B \\
    (b-c)bc(a+b+c) & (a+c)ac(a+b-c) & (a+b)ab(-a+b-c)
  \end{bmatrix}
  = 0
\]
which is a computation.

\paragraph{Authorship comments.}
I was intrigued by a Taiwan TST problem which implied that,
in the configuration above, $\angle I_B D I_C$ was bisected by $\ol{DA}$.
This motivated me to draw all three properties above where $I_A$ and $P$
were isogonal conjugates with respect to $DEF$.
After playing around with this picture for a long time,
I finally noticed that $O$ was on line $PI_A$.
(So the original was to show that $I_B F$, $I_C E$, $DA_2$ concurrent).
Eventually I finally noticed in the picture that $PI_A$
actually passed through the circumcenter of $ABC$ as well.
This took me many hours to prove.

The final restatement (which follows quickly from $P$, $O$, $I_A$ collinear)
was discovered by Telv Cohl when I showed him the problem.