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

\begin{document}
\title{Shortlist 2020 G7}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 80}
\maketitle

\section*{Problem}
Let $P$ be a point on the circumcircle of acute triangle $ABC$.
Let $D,E,F$ be the reflections of $P$ in the $A$-midline, $B$-midline, and $C$-midline.
Let $\omega$ be the circumcircle of the triangle formed by the perpendicular bisectors of $AD, BE, CF$.

Show that the circumcircles of $\triangle ADP$, $\triangle BEP$, $\triangle CFP$,
and $\omega$ share a common point.

\section*{Video}
\href{https://www.youtube.com/watch?v=0kODbd66TdE&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/0kODbd66TdE}}

\section*{External Link}
\url{https://aops.com/community/p22698237}

\newpage

\section*{Solution}
\emph{The following solution was given by Ankan Bhattacharya.}

Call the formed triangle $XYZ$.
Also denote by $O_a$, $O_b$, $O_c$ the centers of $PAD$, $PBE$, $PCF$.
Let $L_a$, $L_b$, $L_c$ be the altitude feet.
Also, let ray $PH$ meet the nine-point circle again at $Q$.
We contend that $Q$ is the desired point.

\begin{center}
\begin{asy}
import graph; size(12cm);
pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215,0); pen zzttff = rgb(0.6,0.2,1); pen ccqqqq = rgb(0.8,0,0); pen fuqqzz = rgb(0.95686,0,0.6); pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw((-1.23265,1.71708)--(-2.19748,-0.20728)--(0.20251,-0.20728)--cycle, linewidth(1) + zzttqq);

draw((-1.23265,1.71708)--(-2.19748,-0.20728), linewidth(1) + zzttqq);
draw((-2.19748,-0.20728)--(0.20251,-0.20728), linewidth(1) + zzttqq);
draw((0.20251,-0.20728)--(-1.23265,1.71708), linewidth(1) + zzttqq);
draw(circle((-0.99748,0.39512), 1.34271), linewidth(1));
draw((-0.43383,-0.82356)--(-0.43383,2.33336), linewidth(1) + qqwuqq);
draw((-0.43383,-0.82356)--(-0.56753,-0.92327), linewidth(1) + qqwuqq);
draw((-0.43383,-0.82356)--(-1.82851,-0.12430), linewidth(1) + qqwuqq);
draw((-1.23265,1.71708)--(-0.43383,2.33336), linewidth(1) + zzttqq);
draw((-2.19748,-0.20728)--(-0.56753,-0.92327), linewidth(1) + zzttqq);
draw(circle((0.14678,0.75490), 1.68186), linewidth(1) + linetype("2 2") + zzttff);
draw(circle((-1.14082,-0.01508), 1.07399), linewidth(1) + linetype("2 2") + zzttff);
draw(circle((-1.11506,0.45369), 0.67135), linewidth(1) + ccqqqq);
draw((0.14678,0.75490)--(-1.14082,-0.01508), linewidth(1) + fuqqzz);
draw((-0.43383,-0.82356)--(-1.51851,0.99031), linewidth(1));
draw((-0.45814,1.53901)--(-1.14082,-0.01508), linewidth(1) + fuqqzz);
draw((-0.72900,1.89010)--(-0.79847,0.18963), linewidth(1) + fuqqzz);
draw((-0.72900,1.89010)--(0.14678,0.75490), linewidth(1) + fuqqzz);
dot("$A$", (-1.23265,1.71708), dir((1.108, 2.769)));
dot("$B$", (-2.19748,-0.20728), dir((1.224, 2.749)));
dot("$C$", (0.20251,-0.20728), dir((1.033, 2.749)));
dot("$P$", (-0.43383,-0.82356), dir((0.978, 2.902)));
dot("$D$", (-0.43383,2.33336), dir((0.978, 2.894)));
dot("$E$", (-0.56753,-0.92327), dir((1.056, 2.904)));
dot("$F$", (-1.82851,-0.12430), dir((1.157, 2.759)));
dot("$Z$", (-0.45814,1.53901), dir((1.193, 2.300)));
dot("$Y$", (-0.72900,1.89010), dir((1.141, 2.082)));
dot("$X$", (-0.77259,0.82317), dir((1.070, 2.162)));
dot("$H$", (-1.23265,0.51227), dir((1.108, 2.238)));
dot("$Q$", (-1.51851,0.99031), dir((1.171, 2.894)));
dot("$L_a$", (-1.23265,-0.20728), dir((1.108, 2.195)));
dot("$P_a$", (-0.43383,0.40899), dir((0.978, 2.320)));
dot("$O_a$", (0.14678,0.75490), dir((1.068, 2.343)));
dot("$O_b$", (-1.14082,-0.01508), dir((1.063, 2.082)));
dot("$O_c$", (-0.79847,0.18963), dir((1.166, 2.102)));
dot("$L_b$", (-0.65526,0.94288), dir((1.244, 2.099)));
\end{asy}
\end{center}

\begin{claim*}
  $AQL_aPD$ is cyclic, and so on.
\end{claim*}
\begin{proof}
  $ADPL_a$ is cyclic because it is an isosceles trapezoid,
  while $AQL_aP$ is cyclic by power of a point from $H$.
\end{proof}

\begin{claim*}
  $Q$ is the Miquel point of complete quadrilateral $XYZO_aO_bO_c$.
\end{claim*}
\begin{proof}
  It is sufficient to show that $Q$ lies on $(O_a O_b Z)$.
  This is an angle chase:
  \begin{align*}
    \dang O_b Q O_a &= \dang O_b Q P + \dang P Q O_a
      = (90\dg - \dang P L_b Q) + (90 \dg - \dang Q L_a P) \\
    &= \dang Q L_b P + \dang P L_a Q
      = \dang L_a P L_b + \dang L_b Q L_a \\
    &= \dang L_a P L_b + 2 \dang ACB \\
    % --------
    \dang O_b Z O_a &=  \dang(BE, AD) = \arg (AD) - \arg (BE) \\
    &= (2\arg(BC) - \arg(PL_a)) - (2\arg(AC) - \arg(PL_b)) \\
    &= 2\dang ACB - \dang L_b P L_a
  \end{align*}
  as needed.
\end{proof}

\end{document}
