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

\begin{document}
\title{ELMO SL 2023 G7}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 132}
\maketitle

\section*{Problem}
Let $\mathcal E$ be an ellipse with foci $F_1$ and $F_2$,
and let $P$ be a point on $\mathcal{E}$.
Suppose line $PF_1$ and $PF_2$ intersect $\mathcal E$ again at distinct points $A$ and $B$,
and the tangents to $\mathcal E$ at $A$ and $B$ intersect at point $Q$.
Show that the midpoint of $\ol{PQ}$ lies on the circumcircle of $\triangle PF_1F_2$.

\section*{Video}
\href{https://www.youtube.com/watch?v=S3z2LSt7zzY&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/S3z2LSt7zzY}}

\section*{External Link}
\url{https://aops.com/community/p28033735}

\newpage

\section*{Solution}
Let $\rho = PF_1 + PF_2 = AF_1 + AF_2 = BF_1 + BF_2$.
We start with the following claim:
\begin{claim*}
  The $P$-excircles of triangle $PAF_2$ and $PF_1B$ coincide.
\end{claim*}
\begin{proof}
  This follows from the fact that $\triangle PAF_2$ and $\triangle PB_1$
  have the same perimeter, namely $2\rho$.
\end{proof}

\begin{claim*}
  That common excircle has center $Q$.
\end{claim*}
\begin{proof}
  The tangency of $\ol{AQ}$ to the ellipse at the vertex of $\angle F_1AF_2$
  implies that $\ol{AQ}$ is the external bisector of $\angle F_1AF_2$
  (by the famous ``river problem'').
  Similarly, $\ol{BQ}$ is the external bisector of $\angle F_2BF_1$.
\end{proof}

Let $A_1$ and $A_2$ denote the reflections of $P$ over $F_1$ and $F_2$.

\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 F_1 = (-1.,0.);
pair F_2 = (1.,0.);
pair P = (-0.5,1.5);
pair A = (-1.35410,-1.06230);
pair B = (1.67082,-0.67082);
pair Q = (0.5,-2.73606);
pair A_1 = (-1.5,-1.5);
pair A_2 = (2.5,-1.5);
pair T_1 = (-1.67082,-2.01246);
pair T_2 = (2.11803,-1.11803);

import graph;
size(10cm);
pen ubqqys = rgb(0.29411,0.,0.50980);
pen zzttqq = rgb(0.6,0.2,0.);
pen qqwuqq = rgb(0.,0.39215,0.);
pen qqffff = rgb(0.,1.,1.);
pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw(P--A_1--A_2--cycle, linewidth(0.6) + zzttqq);

draw(Q--A, linewidth(0.6) + ubqqys);
draw(Q--B, linewidth(0.6) + ubqqys);
draw(P--A, linewidth(0.6));
draw(P--B, linewidth(0.6));
draw(P--Q, linewidth(0.6));
draw(F_1--F_2, linewidth(0.6));
draw(A--F_2, linewidth(0.6));
draw(B--F_1, linewidth(0.6));
draw(P--A_1, linewidth(0.6) + zzttqq);
draw(A_1--A_2, linewidth(0.6) + zzttqq);
draw(A_2--P, linewidth(0.6) + zzttqq);
draw(circle((0.5,-0.5), 2.23606), linewidth(0.6) + linetype("2 2") + qqwuqq);
draw(circle(Q, 2.28824), linewidth(0.6) + qqffff);

dot("$F_1$", F_1, dir((-19.952, 12.485)));
dot("$F_2$", F_2, dir((2.622, 6.059)));
dot("$P$", P, dir((2.487, 5.610)));
dot("$A$", A, dir((-10.830, 9.473)));
dot("$B$", B, dir((2.137, 4.792)));
dot("$Q$", Q, dir(-90));
dot("$A_1$", A_1, dir((-28.370, 4.172)));
dot("$A_2$", A_2, dir((6.263, -9.265)));
dot("$T_1$", T_1, dir((-20.635, 12.188)));
dot("$T_2$", T_2, dir((2.398, 4.531)));
\end{asy}
\end{center}

\begin{claim*}
  The points $P$, $A_1$, $Q$, $A_2$ are concyclic.
\end{claim*}
\begin{proof}
  Let $T_1$ and $T_2$ denote the tangency points of $(Q)$ to $\ol{PA}$ and $\ol{PB}$.
  Note that
  \[ PT_1 + PT_2 = 2PT_1 = 2\rho = 2(PF_1+PF_2) = PA_1 + PA_2
    \implies T_1A_1 = T_2A_2 \]
  so it follows that $\triangle Q T_1 A_1 \cong \triangle Q T_2 A_2$
  as right triangles, with the same orientation. In particular,
  \[ \dang PA_1Q = \dang T_1 A_1 Q = \dang T_2 A_2 Q
    = \dang T_1 A_1 Q = \dang P_2 A_2 Q. \qedhere \]
\end{proof}

Finally, apply a homothety at $P$ with ratio $\half$
to deduce the problem statement.

\end{document}
