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

\begin{document}
\title{GaussMO 2022/3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 113}
\maketitle

\section*{Problem}
On a triangle $\triangle ABC$ let $\omega$ be its circumcircle and let $H_A$
be a point on $\omega$ such that $AH_A \perp BC$. Let the projections from
$B$, $C$ to $AC$, $AB$ be $E$, $F$ respectively.
Let $H_AE$, $H_AF$ meet $\omega$ again at $P$, $Q$ respectively
and let $PQ$ hit $BC$ at $T$.
Show that one of the tangents from $T$ to $\omega$ bisects
the common chord of the circumcircles of triangles $BFP$ and $CEQ$.

\section*{Video}
\href{https://www.youtube.com/watch?v=RsTDs5YNlIM&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/RsTDs5YNlIM}}

\section*{External Link}
\url{https://aops.com/community/p26134873}

\newpage

\section*{Solution}
Let $K$ be a point such that $ABKC$ is a cyclic harmonic quadrilateral.
Let $M$ be the midpoint of $\ol{EF}$.

\begin{claim*}
  [ELMO SL 2019 G1]
  Lines $AK$, $BP$, and $CQ$ meet at $M$.
\end{claim*}
\begin{proof}
  The fact that $\ol{AKM}$ collinear is well-known,
  as is the lemma that $(BFMK)$ and $(CEMK)$ are cyclic.

  We also contend that $EFH_AK$ is cyclic.
  Indeed, $\ol{EF}$ and $\ol{H_AK}$
  concur at the harmonic conjugate of $\ol{AH} \cap \ol{BC}$
  with respect to $\ol{BC}$, so this follows by radical axis.

  Then, we do the angle chase
  \[ \dang KBM = \dang KFM = \dang KFE = \dang KH_AE = \dang KH_AP =
  \dang KBP \]
  to get that $B$, $M$, $P$ are collinear.
\end{proof}

\begin{center}
\begin{asy}
/*
    Converted from GeoGebra by User:Azjps using Evan's magic cleaner
    https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
pair A = (4.21405,12.26079);
pair B = (2.,2.);
pair C = (18.,2.);
pair H = (4.21405,4.97471);
pair H_A = (4.21405,-0.97471);
pair E = (7.70381,9.66338);
pair F = (2.71182,5.29886);
pair O = (10.,5.64304);
pair P = (9.25832,14.40213);
pair Q = (1.84415,8.92236);
pair T = (-7.52186,2.);
pair K = (7.33138,-2.73253);
pair M = (5.20781,7.48112);
pair D = (15.78594,-0.97471);
pair O_1 = (19.86018,-0.12761);
pair O_2 = (6.32129,-2.94254);
pair N = (10.,2.);

size(12cm);
pen zzttqq = rgb(0.6,0.2,0.);
pen qqffff = rgb(0.,1.,1.);
pen yqqqqq = rgb(0.50196,0.,0.);
pen qqwuqq = rgb(0.,0.39215,0.);
pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw(A--B--C--cycle, linewidth(0.6) + zzttqq);
draw(D--N--O--cycle, linewidth(0.6) + qqwuqq);

draw(A--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--A, linewidth(0.6) + zzttqq);
draw(circle(O, 8.79043), linewidth(0.6));
draw(P--H_A, linewidth(0.6) + blue);
draw(Q--H_A, linewidth(0.6) + blue);
draw(P--T, linewidth(0.6) + blue);
draw(T--B, linewidth(0.6));
draw(C--Q, linewidth(0.6) + qqffff);
draw(B--P, linewidth(0.6) + qqffff);
draw(E--F, linewidth(0.6) + qqffff);
draw(B--E, linewidth(0.6) + yqqqqq);
draw(C--F, linewidth(0.6) + yqqqqq);
draw(A--K, linewidth(0.6) + zzttqq);
draw(D--N, linewidth(0.6) + qqwuqq);
draw(N--O, linewidth(0.6) + qqwuqq);
draw(O--D, linewidth(0.6) + qqwuqq);

dot("$A$", A, dir((-63.907, 55.318)));
dot("$B$", B, dir((-73.122, -99.493)));
dot("$C$", C, dir((57.924, -18.915)));
dot("$H$", H, dir((-13.892, -105.215)));
dot("$H_A$", H_A, dir((-100.028, -68.763)));
dot("$E$", E, dir((48.359, 12.196)));
dot("$F$", F, dir((-85.955, -4.258)));
dot("$O$", O, dir(90));
dot("$P$", P, dir((12.387, 21.791)));
dot("$Q$", Q, dir((-96.437, 58.512)));
dot("$T$", T, dir((-21.191, -146.729)));
dot("$K$", K, dir(K-O));
dot("$M$", M, dir((11.767, 22.029)));
dot("$D$", D, dir((7.030, -115.999)));
dot("$O_1$", O_1, dir((10.834, 21.575)));
dot("$O_2$", O_2, dir((-88.467, -38.694)));
dot("$N$", N, dir((-61.781, -107.829)));
\end{asy}
\end{center}

We tie in the radical axis of the two given circles now.
\begin{claim*}
Line $\ol{AMK}$ is both the radical axis of the given circles
and exactly the polar of $T$.
\end{claim*}
\begin{proof}
  Continuing on, we have
  \[ -1 = (AK;BC)_\omega \overset{M}{=} (KA;PQ) \]
  and that's enough to imply that
  $T = \ol{PQ} \cap \ol{BC} \cap \ol{AA} \cap \ol{KK}$
  with respect to $\omega$.
  As $A$ and $M$ have equal power to $(BFP)$ and $(CEQ)$,
\end{proof}

We will now prove $K$ is the midpoint of the common chord;
this will solve the problem (since $\ol{TK}$ is tangent).
Define $N$ as the midpoint of $\ol{BC}$ and $D$ as the antipode.
\begin{claim*}
  Triangles $\triangle AMC$ and $\triangle OND$ are orthologic.
\end{claim*}
\begin{proof}
  The altitude from $A$ to $\ol{ND}$ is well-known to
  pass through $\ol{EF} \cap \ol{BC}$.
  The altitude from $C$ to $\ol{ON}$ is $\ol{BC}$ itself,
  and the altitude from $M$ to $\ol{OD}$ is exactly line $\ol{EF}$
  (since $\ol{AOD} \perp \ol{EF}$).
  So $\ol{EF} \cap \ol{BC}$ is the orthology center.
\end{proof}
But running the orthology the other way,
the perpendicular from $O$ to $\ol{MC}$
and the perpendicular from $N$ to $\ol{AC}$
are exactly the perpendicular bisectors of $\ol{QC}$ and $\ol{CE}$,
and meet at the circumcenter $O_2$ of $\triangle CEQ$, say.
Meanwhile, $\ol{DK} \perp \ol{AM}$.
In other words, $\ol{DO_2K} \perp \ol{AMK}$, and we're done.

\end{document}
