\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{USEMO 2020/3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 34}
\maketitle

\section*{Problem}
Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$.
Let $\Gamma$ denote the circumcircle of triangle $ABC$,
and $N$ the midpoint of $\ol{OH}$.
The tangents to $\Gamma$ at $B$ and $C$,
and the line through $H$ perpendicular to line $AN$,
determine a triangle whose circumcircle we denote by $\omega_A$.
Define $\omega_B$ and $\omega_C$ similarly.

Prove that the common chords of $\omega_A$,
$\omega_B$, and $\omega_C$ are concurrent on line $OH$.

\section*{Video}
\href{https://www.youtube.com/watch?v=5a_XCGKiXnI}{\texttt{https://youtu.be/5a\_XCGKiXnI}}

\section*{External Link}
\url{https://aops.com/community/p18471120}

\newpage

\section*{Solution}
We begin by introducing several notations.
The orthic triangle is denoted $DEF$ and the tangential
triangle is denoted $T_a T_b T_c$.
The reflections of $H$ across the sides are denoted $H_a$, $H_b$, $H_c$.
We also define the crucial points $P$ and $Q$ as the poles of $\ol{H_c B}$ and $\ol{H_b C}$
with respect to $\Gamma$.

The solution, based on the independent solutions
found by Anant Mudgal and Nikolai Beluhov, hinges on two central claims:
that $\omega_A$ is the circumcircle of $\triangle T_a PQ$,
and that $\ol{EF}$ is the radical axis of $\Gamma$ and $\omega_A$.
We prove these two claims in turn.
\begin{center}
\begin{asy}
unitsize(2.5cm);
pair A = dir(75);
pair B = dir(200);
pair C = dir(-20);

pair T_a = 2*B*C/(B+C);
pair T_b = 2*C*A/(C+A);
pair T_c = 2*A*B/(A+B);
pair D = foot(A, B, C);
pair E = foot(B, C, A);
pair F = foot(C, A, B);
pair H = A+B+C;
pair H_a = -B*C/A;
pair H_b = -C*A/B;
pair H_c = -A*B/C;

pair Q = 2*C*H_b/(H_b+C);
pair P = 2*B*H_c/(H_c+B);

filldraw(T_a--T_b--T_c--cycle, opacity(0.1)+pink, magenta);
draw(A--B--C--cycle, blue);
draw(A--H_a, lightblue);
draw(B--H_b, lightblue);
draw(C--H_c, lightblue);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);

draw(H_b--Q, red);
draw(H_c--P, red);
draw(Q--P, dashed+red);
pair O = origin;
pair N = midpoint(O--H);
draw(A--N, red);

filldraw(circumcircle(T_a, P, Q), opacity(0.02)+yellow, deepgreen);
pair R1 = IP(unitcircle, circumcircle(T_a, P, Q));
pair R2 = OP(unitcircle, circumcircle(T_a, P, Q));
draw(R1--R2, deepgreen+1);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$T_a$", T_a, dir(T_a));
dot("$T_b$", T_b, dir(T_b));
dot("$T_c$", T_c, dir(T_c));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$H$", H, dir(H));
dot("$H_a$", H_a, dir(H_a));
dot("$H_b$", H_b, dir(H_b));
dot("$H_c$", H_c, dir(H_c));
dot("$Q$", Q, dir(Q));
dot("$P$", P, dir(P));
dot("$O$", O, dir(270));
dot("$N$", N, dir(270));
dot(R1);
dot(R2);

/* TSQ Source:

!unitsize(2.5cm);
A = dir 75
B = dir 200
C = dir -20

T_a = 2*B*C/(B+C)
T_b = 2*C*A/(C+A)
T_c = 2*A*B/(A+B)
D = foot A B C
E = foot B C A
F = foot C A B
H = A+B+C
H_a = -B*C/A
H_b = -C*A/B
H_c = -A*B/C

Q = 2*C*H_b/(H_b+C)
P = 2*B*H_c/(H_c+B)

T_a--T_b--T_c--cycle 0.1 pink / magenta
A--B--C--cycle blue
A--H_a lightblue
B--H_b lightblue
C--H_c lightblue
unitcircle 0.1 lightcyan / blue

H_b--Q red
H_c--P red
Q--P dashed red
O = origin R270
N = midpoint O--H R270
A--N red

circumcircle T_a P Q 0.02 yellow / deepgreen
R1 .= IP unitcircle circumcircle T_a P Q
R2 .= OP unitcircle circumcircle T_a P Q
R1--R2 deepgreen+1

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

\newpage

\begin{claim*}
  [Characterization of ${\omega_A}$]
  Line $PQ$ passes through $H$ and is perpendicular to $\ol{AN}$.
\end{claim*}
\begin{proof}
  The fact that $H$ lies on line $PQ$ is immediate by Brokard's theorem.

  Showing the perpendicularity is the main part.
  Denote by $B'$ and $C'$ the antipodes of $B$ and $C$ on $\Gamma$.
  Also, define $L = \ol{H_cC'} \cap \ol{H_bB'}$
  and $K = \ol{BH_c} \cap \ol{CH_b}$, as shown.
  \begin{center}
  \begin{asy}
  unitsize(2.5cm);
  pair A = dir(75);
  pair B = dir(200);
  pair C = dir(-20);

  pair T_a = 2*B*C/(B+C);
  pair T_b = 2*C*A/(C+A);
  pair T_c = 2*A*B/(A+B);
  pair D = foot(A, B, C);
  pair E = foot(B, C, A);
  pair F = foot(C, A, B);
  pair H = A+B+C;
  pair H_a = -B*C/A;
  pair H_b = -C*A/B;
  pair H_c = -A*B/C;
  pair Bp = -B;
  pair Cp = -C;

  pair Q = 2*C*H_b/(H_b+C);
  pair P = 2*B*H_c/(H_c+B);
  pair K = extension(B, H_c, C, H_b);
  pair L = extension(Cp, H_c, Bp, H_b);

  filldraw(T_a--T_b--T_c--cycle, opacity(0.1)+pink, magenta);
  draw(A--B--C--cycle, blue);
  draw(A--H_a, lightblue);
  draw(B--H_b, lightblue);
  draw(C--H_c, lightblue);
  filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
  draw(B--Bp, cyan);
  draw(C--Cp, cyan);

  draw(B--K--C, lightgreen);
  draw(Cp--L--Bp, deepgreen);
  draw(H_b--Q, red);
  draw(H_c--P, red);
  draw(Q--P, dashed+red);
  pair O = origin;
  pair N = midpoint(O--H);
  draw(A--N, red);
  draw(O--K, red+dashed);

  dot("$A$", A, dir(A));
  dot("$B$", B, dir(B));
  dot("$C$", C, dir(C));
  dot("$T_a$", T_a, dir(T_a));
  dot("$T_b$", T_b, dir(T_b));
  dot("$T_c$", T_c, dir(T_c));
  dot("$D$", D, dir(D));
  dot("$E$", E, dir(E));
  dot("$F$", F, dir(F));
  dot("$H$", H, dir(H));
  dot("$H_a$", H_a, dir(H_a));
  dot("$H_b$", H_b, dir(H_b));
  dot("$H_c$", H_c, dir(H_c));
  dot("$B'$", Bp, dir(Bp));
  dot("$C'$", Cp, dir(Cp));
  dot("$Q$", Q, dir(Q));
  dot("$P$", P, dir(P));
  dot("$K$", K, dir(K));
  dot("$L$", L, dir(135));
  dot("$O$", O, dir(270));
  dot("$N$", N, dir(270));

  /* TSQ Source:

  !unitsize(3.5cm);
  A = dir 75
  B = dir 200
  C = dir -20

  T_a = 2*B*C/(B+C)
  T_b = 2*C*A/(C+A)
  T_c = 2*A*B/(A+B)
  D = foot A B C
  E = foot B C A
  F = foot C A B
  H = A+B+C
  H_a = -B*C/A
  H_b = -C*A/B
  H_c = -A*B/C
  B' = -B
  C' = -C

  Q = 2*C*H_b/(H_b+C)
  P = 2*B*H_c/(H_c+B)
  K = extension B H_c C H_b
  L = extension Cp H_c Bp H_b R135

  T_a--T_b--T_c--cycle 0.1 pink / magenta
  A--B--C--cycle blue
  A--H_a lightblue
  B--H_b lightblue
  C--H_c lightblue
  unitcircle 0.1 lightcyan / blue
  B--Bp cyan
  C--Cp cyan

  B--K--C lightgreen
  Cp--L--Bp deepgreen
  H_b--Q red
  H_c--P red
  Q--P dashed red
  O = origin R270
  N = midpoint O--H R270
  A--N red
  O--K red dashed

  */
  \end{asy}
  \end{center}
  We observe that:
  \begin{itemize}
    \ii We have $\ol{OK} \perp \ol{PQ}$
    since $K$ is the pole of line $\ol{PQ}$ (again by Brokard).
    \ii The points $O$, $K$, $L$ are collinear by Pascal's theorem on $BH_cC'CH_bB'$.
    \ii The point $L$ is seen to be the
    reflection of $H$ across $A$, so it follows $\ol{AN} \parallel \ol{OL}$
    by a $\half$-factor homothety at $H$.
  \end{itemize}
  Putting these three observations together completes the first claim.
\end{proof}

\newpage

\begin{remark*}
  [First claim is faster with complex numbers]
  It is also straightforward to prove the first claim by using complex numbers.
  Indeed, in the usual setup,
  we have that the intersection of the tangents at $B$ and $H_c$
  is given explicitly by
  \[ p = \frac{2b \cdot \left( -\frac{ab}{c} \right)}{b - \frac{ab}{c}}
    = \frac{2ab}{a-c} \]
  and one explicitly checks $p-(a+b+c) \perp (b+c-a)$, as needed.
\end{remark*}

\begin{claim*}
  [Radical axis of $\omega_A$ and $\Gamma$]
  Line $EF$ coincides with the radical axis of $\omega_A$ and $\Gamma$.
\end{claim*}
\begin{proof}
  Let lines $EF$ and $T_a T_c$ meet at $Z$.
  It suffices to show $Z$ lies on the radical axis,
  and then repeat the argument on the other side.

  \begin{center}
  \begin{asy}
    unitsize(2.5cm);
    pair A = dir(75);
    pair B = dir(200);
    pair C = dir(-20);

    pair T_a = 2*B*C/(B+C);
    pair T_b = 2*C*A/(C+A);
    pair T_c = 2*A*B/(A+B);
    pair D = foot(A, B, C);
    pair E = foot(B, C, A);
    pair F = foot(C, A, B);
    pair H = A+B+C;
    pair H_a = -B*C/A;
    pair H_b = -C*A/B;
    pair H_c = -A*B/C;

    pair Q = 2*C*H_b/(H_b+C);
    pair P = 2*B*H_c/(H_c+B);

    filldraw(T_a--T_b--T_c--cycle, opacity(0.1)+pink, magenta);
    draw(A--B--C--cycle, blue);
    draw(A--H_a, lightblue);
    draw(B--H_b, lightblue);
    draw(C--H_c, lightblue);
    filldraw(unitcircle, opacity(0.1)+lightcyan, blue);

    draw(H_b--Q, deepgreen);
    draw(H_c--P, deepgreen);
    draw(Q--P, deepgreen);
    pair O = origin;

    filldraw(circumcircle(T_a, P, Q), opacity(0.02)+yellow, deepgreen);
    pair X = midpoint(B--C);
    pair Y = midpoint(B--H_c);
    pair Z = extension(E, F, B, P);
    pair Z2 = extension(E, F, C, Q);

    draw(B--H_c, blue);
    draw(Z--Z2, red);
    draw(X--Z, orange);
    draw(P--O--T_a, orange);
    filldraw(circumcircle(B, X, Y), opacity(0.1)+lightcyan, lightblue);

    dot("$A$", A, dir(A));
    dot("$B$", B, dir(B));
    dot("$C$", C, dir(C));
    dot("$T_a$", T_a, dir(T_a));
    dot("$T_b$", T_b, dir(T_b));
    dot("$T_c$", T_c, dir(T_c));
    dot("$D$", D, dir(D));
    dot("$E$", E, dir(E));
    dot("$F$", F, dir(F));
    dot("$H$", H, dir(H));
    dot(H_a);
    dot(H_b);
    dot("$H_c$", H_c, dir(H_c));
    dot("$Q$", Q, dir(Q));
    dot("$P$", P, dir(P));
    dot("$O$", O, dir(20));
    dot("$X$", X, dir(315));
    dot("$Y$", Y, dir(200));
    dot("$Z$", Z, dir(170));
    dot(Z2);

    /* TSQ Source:

    !unitsize(2.5cm);
    A = dir 75
    B = dir 200
    C = dir -20

    T_a = 2*B*C/(B+C)
    T_b = 2*C*A/(C+A)
    T_c = 2*A*B/(A+B)
    D = foot A B C
    E = foot B C A
    F = foot C A B
    H = A+B+C
    H_a .= -B*C/A
    H_b .= -C*A/B
    H_c = -A*B/C

    Q = 2*C*H_b/(H_b+C)
    P = 2*B*H_c/(H_c+B)

    T_a--T_b--T_c--cycle 0.1 pink / magenta
    A--B--C--cycle blue
    A--H_a lightblue
    B--H_b lightblue
    C--H_c lightblue
    unitcircle 0.1 lightcyan / blue

    H_b--Q deepgreen
    H_c--P deepgreen
    Q--P deepgreen
    O = origin R20

    circumcircle T_a P Q 0.02 yellow / deepgreen
    X = midpoint B--C R315
    Y = midpoint B--H_c R200
    Z = extension E F B P R170
    Z2 .= extension E F C Q

    B--H_c blue
    Z--Z2 red
    X--Z orange
    P--O--T_a orange
    circumcircle B X Y 0.1 lightcyan / lightblue

    */
  \end{asy}
  \end{center}
  Since $\dang FBZ = \dang ABZ = \dang BCA = \dang EFA = \dang ZFB$,
  it follows $ZB = ZF$.
  We introduce two other points $X$ and $Y$ on the perpendicular
  bisector of $\ol{BF}$: they are the midpoints of $\ol{BC}$ and $\ol{BH_c}$.

  Since $OX \cdot OT_a = OB^2 = OY \cdot OP$, it follows that $XYPT_a$ is cyclic.
  Then \[ ZP \cdot Z T_a = ZX \cdot ZY = ZB^2 \]
  with the last equality since the circumcircle of $\triangle BXY$ is tangent
  to $\Gamma$ (by a $\half$-homothety at $B$).
  So the proof of the claim is complete.
\end{proof}

\newpage

Finally, we are ready to finish the problem.
\begin{claim*}
  Line $DT_a$ coincides with the radical axis of $\omega_B$ and $\omega_C$.
\end{claim*}
\begin{proof}
  The point $D$ already coincides with the radical axis
  because it is the radical center of $\Gamma$, $\omega_B$ and $\omega_C$.
  As for the point $T_a$, we let the tangent to $\Gamma$ at $H_a$
  meet $\ol{T_a T_c}$ at $U$ and $V$;
  by the first claim, these lie on $\omega_C$ and  $\omega_B$ respectively.
  \begin{center}
  \begin{asy}
    unitsize(2.5cm);
    pair A = dir(75);
    pair B = dir(200);
    pair C = dir(-20);

    pair T_a = 2*B*C/(B+C);
    pair T_b = 2*C*A/(C+A);
    pair T_c = 2*A*B/(A+B);
    pair D = foot(A, B, C);
    pair E = foot(B, C, A);
    pair F = foot(C, A, B);
    pair H = A+B+C;
    pair H_a = -B*C/A;
    pair H_b = -C*A/B;
    pair H_c = -A*B/C;

    filldraw(T_a--T_b--T_c--cycle, opacity(0.1)+pink, magenta);
    draw(A--B--C--cycle, blue);
    draw(A--H_a, lightblue);
    draw(B--H_b, lightblue);
    draw(C--H_c, lightblue);
    filldraw(unitcircle, opacity(0.1)+lightcyan, blue);

    pair U = 2*B*H_a/(B+H_a);
    pair V = 2*C*H_a/(C+H_a);
    pair U1 = extension(U, H, T_b, T_c);
    pair V1 = extension(V, H, T_b, T_c);
    draw(U--U1, dotted+deepgreen);
    draw(V--V1, dotted+deepgreen);
    filldraw(circumcircle(U, U1, T_c), opacity(0.02)+yellow, deepgreen);
    filldraw(circumcircle(V, V1, T_b), opacity(0.02)+yellow, deepgreen);
    draw(U--V, orange);

    pair Z1 = IP(circumcircle(T_c, U, U1), circumcircle(T_b, V, V1));
    pair Z2 = OP(circumcircle(T_c, U, U1), circumcircle(T_b, V, V1));
    draw(Z1--T_a, red+1);

    dot("$A$", A, dir(A));
    dot("$B$", B, dir(B));
    dot("$C$", C, dir(C));
    dot("$T_a$", T_a, dir(T_a));
    dot("$T_b$", T_b, dir(T_b));
    dot("$T_c$", T_c, dir(T_c));
    dot("$D$", D, dir(D));
    dot("$E$", E, dir(E));
    dot("$F$", F, dir(F));
    dot("$H$", H, dir(180));
    dot("$H_a$", H_a, dir(H_a));
    dot(H_b);
    dot(H_c);
    dot("$U$", U, dir(U));
    dot("$V$", V, dir(V));
    dot(Z1);
    dot(Z2);

    /* TSQ Source:

    !unitsize(2.5cm);
    A = dir 75
    B = dir 200
    C = dir -20

    T_a = 2*B*C/(B+C)
    T_b = 2*C*A/(C+A)
    T_c = 2*A*B/(A+B)
    D = foot A B C
    E = foot B C A
    F = foot C A B
    H = A+B+C R180
    H_a = -B*C/A
    H_b .= -C*A/B
    H_c .= -A*B/C

    T_a--T_b--T_c--cycle 0.1 pink / magenta
    A--B--C--cycle blue
    A--H_a lightblue
    B--H_b lightblue
    C--H_c lightblue
    unitcircle 0.1 lightcyan / blue

    U = 2*B*H_a/(B+H_a)
    V = 2*C*H_a/(C+H_a)
    U1 := extension U H T_b T_c
    V1 := extension V H T_b T_c
    U--U1 dotted deepgreen
    V--V1 dotted deepgreen
    circumcircle U U1 T_c 0.02 yellow / deepgreen
    circumcircle V V1 T_b 0.02 yellow / deepgreen
    U--V orange

    Z1 .= IP circumcircle T_c U U1 circumcircle T_b V V1
    Z2 .= OP circumcircle T_c U U1 circumcircle T_b V V1
    Z1--T_a red+1

    */
  \end{asy}
  \end{center}
  We need to show $T_aU \cdot T_a T_c = T_aV \cdot T_aT_b$.

  But $UVT_bT_c$ is apparently cyclic:
  the sides $\ol{T_b T_c}$ and $\ol{UV}$ are reflections
  across a line perpendicular to $\ol{AH_a}$,
  while the sides $\ol{UT_c}$ and $\ol{VT_b}$ are reflections
  across a line perpendicular to $\ol{BC}$.
  So this is true.
\end{proof}

Now since $\triangle DEF$ and $\triangle T_a T_b T_c$ are homothetic
(their opposite sides are parallel),
and their incenters are respectively $H$ and $O$,
the problem is solved.

\begin{remark*}
  [Barycentric approaches with respect to $\triangle T_a T_b T_c$]
  Because the first claim is so explicit,
  it is possible to calculate the length of the segment $PB$.
  This opens the possibility of using barycentric coordinates
  with respect to the reference triangle $T_a T_b T_c$,
  and in fact some contestants were able to complete this approach.
  Writing $a = T_b T_c$, $b = T_c T_a$, $c = T_a T_b$
  one can show that the radical center is the point
  \[ \left( \frac{a}{s-a} : \frac{b}{s-b} : \frac{c}{s-c} \right) \]
  which is checked to be collinear with
  the circumcenter and incenter of $\triangle T_a T_b T_c$.
\end{remark*}

\paragraph{Alternate inversion approach replacing the last two claims, by Serena An.}
After finding $P$ and $Q$,
it's also possible to solve the problem by using inversion.
This eliminates the need to identify line $EF$ as the radical axis
of $\omega_A$ and $\Gamma$.

Inverted points are denoted with $\bullet^\ast$ as usual,
but we will only need two points:
$P^\ast$, the midpoint of $\ol{BH_c}$,
and $Q^\ast$, the midpoint of $\ol{CH_b}$.
Now, let $M_a = T_a^\ast$ denote the midpoint $\ol{BC}$ and
let $K$ be a point on ray $HA$ with \[ KH = \frac 32 AH. \]

\begin{center}
\begin{asy}
size(9cm);
pair A = dir(75);
pair B = dir(200);
pair C = dir(-20);

pair D = foot(A, B, C);
pair E = foot(B, C, A);
pair F = foot(C, A, B);
pair H = A+B+C;
pair H_a = -B*C/A;
pair H_b = -C*A/B;
pair H_c = -A*B/C;

pair Q = 2*C*H_b/(H_b+C);
pair P = 2*B*H_c/(H_c+B);

draw(A--B--C--cycle, blue);
draw(A--H_a, paleblue);
draw(B--H_b, paleblue);
draw(C--H_c, paleblue);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);

draw(H_b--Q--C, deepgreen);
draw(H_c--P--B, deepgreen);
draw(Q--P, deepgreen);
pair O = origin;
draw(B--H_c, red);
draw(C--H_b, red);
pair L = 2*A-H;
pair K = midpoint(L--A);
pair M_a = midpoint(B--C);
pair Ps = midpoint(B--H_c);
pair Qs = midpoint(C--H_b);
// H_c--L--H_b deepgreen
// A--L deepgreen
draw(A--K, lightblue);
draw(M_a--Ps--K--Qs--cycle, red);

filldraw(circumcircle(Ps, Qs, D), opacity(0.1)+yellow, red);

dot("$A$", A, dir(45));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(80));
dot("$F$", F, dir(F));
dot("$H$", H, dir(H));
dot("$H_a$", H_a, dir(H_a));
dot("$H_b$", H_b, dir(H_b));
dot("$H_c$", H_c, dir(H_c));
dot("$Q$", Q, dir(Q));
dot("$P$", P, dir(P));
dot("$O$", O, dir(270));
dot("$K$", K, dir(45));
dot("$M_a$", M_a, dir(M_a));
dot("$P^\ast$", Ps, dir(Ps));
dot("$Q^\ast$", Qs, dir(Qs));

/* TSQ Source:

!size(9cm);
A = dir 75 R45
B = dir 200
C = dir -20

D = foot A B C
E = foot B C A R80
F = foot C A B
H = A+B+C
H_a = -B*C/A
H_b = -C*A/B
H_c = -A*B/C

Q = 2*C*H_b/(H_b+C)
P = 2*B*H_c/(H_c+B)

A--B--C--cycle blue
A--H_a paleblue
B--H_b paleblue
C--H_c paleblue
unitcircle 0.1 lightcyan / blue

H_b--Q--C deepgreen
H_c--P--B deepgreen
Q--P deepgreen
O = origin R270
B--H_c red
C--H_b red
L := 2*A-H
K = midpoint L--A R45
M_a = midpoint B--C
P* = midpoint B--H_c
Q* = midpoint C--H_b
// H_c--L--H_b deepgreen
// A--L deepgreen
A--K lightblue
M_a--Ps--K--Qs--cycle red

circumcircle Ps Qs D 0.1 yellow / red

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

\begin{claim*}
  The points $D$, $P^\ast$, $Q^\ast$ lie on the circle
  with diameter $\ol{K M_a}$.
\end{claim*}
\begin{proof}
  Consider the homothety at $H$ with scale factor $\frac 32$.
  It maps $F$ to the midpoint of $\ol{FH_c}$ and $A$ to $K$,
  so we find $\ol{KP^\ast}$ is the perpendicular bisector of $\ol{FH_c}$.
  As $\ol{M_a P^\ast} \parallel \ol{CH}$, we conclude
  $\angle KP^\ast M_a = 90\dg$.

  Similarly $\angle K Q^\ast M_a = 90\dg$. And $\angle KDM_a = 90\dg$ is given.
\end{proof}

\begin{claim*}
  Line $\ol{CH}$ coincides with the radical axis
  of $\omega_A^\ast$ and $\omega_B^\ast$.
  In particular, the circles $\omega_A$, $\omega_B$, $(COH_c)$ are coaxial.
\end{claim*}
\begin{proof}
  Letting $\Gamma$ and $\Gamma_9$ denote circumcircle and nine-point circle,
  \begin{align*}
  \opname{Pow}(C,\omega_A^\ast) &= CD \cdot CM_a = \opname{Pow}(C, \Gamma_9) \\
  \opname{Pow}(H,\omega_A^\ast) &= HM \cdot HD = \tfrac32 HA \cdot \tfrac12 HH_a
    = \tfrac34 \opname{Pow}(H, \Gamma).
  \end{align*}
  The same calculation holds with $\omega_B^\ast$.
  Now line $CH$ inverts to $(COH_c)$, as needed.
\end{proof}
Since the circles $(AOH_a)$, $(BOH_b)$, $(COH_c)$
have common radical axis equal to line $OH$, the problem is solved.

\begin{remark*}
  [Nikolai Beluhov --- generalization with variable $ABC$ and fixed $H$]
  Take a fixed circle $\Gamma$ and a fixed point $H$ in its interior. Then there exist infinitely many triangles $ABC$ with orthocenter $H$ and circumcircle $\Gamma$. In fact, for every point $A$ on $\Gamma$ we get a unique pair of $B$ and $C$, determined as follows: Let line $AH$ meet $\Gamma$ again at $S_A$, and take $B$ and $C$ to be the intersection points of the perpendicular bisector of segment $HS_A$ with $\Gamma$.

  With this framework, the following generalization is true:
  The radical center $W$ of $\omega_A$, $\omega_B$, and $\omega_C$ is the same point for all such triangles. Indeed, the Euler circle $e$ of triangle $ABC$ is constant because it depends only on $H$ and $\Gamma$. Let inversion relative to $\Gamma$ map $e$ onto $\Omega$. Then all three of $T_a$, $T_b$, and $T_c$ lie on $\Omega$, and so, by the solution, $W$ is the homothety center of $e$ and $\Omega$. (We take the homothety center with positive ratio when triangle $ABC$ is acute, and with negative ratio when it is obtuse. When triangle $ABC$ is right-angled, $\Omega$ degenerates into a straight line.)

  Explicitly, let $\Gamma$ be the unit circle and put $H$ on the real axis at $h$. Then $W$ is also on the real axis, at $4h/(h^2 + 3)$.

  Furthermore, it turns out the power of $W$ with respect to $\omega_A$, $\omega_B$, and $\omega_C$ is constant as well. This, however, is much tougher to prove; we are not aware of a purely geometric proof at this time.
  Explicitly, in the setting above where $\Gamma$ is the unit circle
  and $H$ is on the real axis at $h$, the power of $W$ with respect to
  $\omega_A$, $\omega_B$, and $\omega_C$ equals $12(h^2 - 1)/(h^2 + 3)^2$.
\end{remark*}

\end{document}
