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

\begin{document}
\title{EGMO 2023/6}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 123}
\maketitle

\section*{Problem}
Let $ABC$ be a triangle with circumcircle $\Omega$.
Let $S_b$ and $S_c$ respectively denote the midpoints of the arcs
$AC$ and $AB$ that do not contain the third vertex.
Let $N_a$ denote the midpoint of arc $BAC$ (the arc $BC$ including $A$).
Let $I$ be the incenter of $ABC$.
Let $\omega_b$ be the circle that is tangent to $AB$
and internally tangent to $\Omega$ at $S_b$,
and let $\omega_c$ be the circle that is tangent to $AC$
and internally tangent to $\Omega$ at $S_c$.
Show that the line $IN_a$, and the lines through the intersections of
$\omega_b$ and $\omega_c$, meet on $\Omega$.

\section*{Video}
\href{https://www.youtube.com/watch?v=ZmG11r8tQNI&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/ZmG11r8tQNI}}

\section*{External Link}
\url{https://aops.com/community/p27522960}

\newpage

\section*{Solution}
Let $Z = \ol{S_b S_b} \cap \ol{S_c S_c}$,
$X = \ol{AB} \cap \ol{S_B S_c}$ (which lies on $\omega_b$ by shooting lemma),
and $Y = \ol{AC} \cap \ol{S_B S_c}$ (which lies on $\omega_c$ similarly).

\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 = (5.88727,2.48727);
pair B = (3.,-4.);
pair C = (14.,-4.);
pair I = (6.85661,-1.49511);
pair S_b = (12.05038,1.87825);
pair S_c = (3.30622,-0.25014);
pair M = (8.5,-8.24666);
pair N_a = (8.5,3.12323);
pair X = (4.83450,0.12185);
pair Y = (7.90938,0.87030);
pair T = (4.90886,-6.96880);
pair Z = (6.30002,6.47648);

import graph;
size(10cm);
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(A--B--C--cycle, linewidth(0.6) + zzttqq);

draw(A--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--A, linewidth(0.6) + zzttqq);
draw(circle((8.5,-2.56171), 5.68494), linewidth(0.6));
draw(S_b--S_c, linewidth(0.6));
draw(circle((9.12052,-1.78571), 4.69135), linewidth(0.6) + qqwuqq);
draw(circle((6.04036,-1.46701), 2.99270), linewidth(0.6) + qqwuqq);
draw(T--N_a, linewidth(0.6) + gray);
draw(A--T, linewidth(0.6) + qqwuqq);
draw(A--M, linewidth(0.6) + gray);
draw(Z--S_c, linewidth(0.6) + qqffff);
draw(S_b--Z, linewidth(0.6) + qqffff);
draw(Z--A, linewidth(0.6) + qqwuqq);

dot("$A$", A, dir((8.312, 20.785)));
dot("$B$", B, dir((-89.492, -30.422)));
dot("$C$", C, dir((43.229, -57.583)));
dot("$I$", I, dir((24.204, -61.528)));
dot("$S_b$", S_b, dir((8.361, 16.917)));
dot("$S_c$", S_c, dir((-95.041, 29.177)));
dot("$M$", M, dir(-90));
dot("$N_a$", N_a, dir((-21.042, 44.942)));
dot("$X$", X, dir((-38.934, 19.140)));
dot("$Y$", Y, dir((8.769, 17.422)));
dot("$T$", T, dir((-48.459, -63.660)));
dot("$Z$", Z, dir((8.824, 16.752)));
\end{asy}
\end{center}

\begin{claim*}
  Point $A$ lies on the radical axis of $\omega_b$ and $\omega_c$.
\end{claim*}
\begin{proof}
  It's well known $\ol{S_b S_c}$ is the perpendicular bisector of $\ol{AI}$,
  so we get $AX = AY$ and hence $AX^2 = AY^2$.
\end{proof}

\begin{claim*}
  Point $Z$ lies on the radical axis of $\omega_b$ and $\omega_c$.
\end{claim*}
\begin{proof}
  It's the radical center of $\omega_b$, $\omega_c$, $\Omega$.
\end{proof}

\begin{claim*}
  Quadrilateral $AS_bTS_c$ is harmonic.
\end{claim*}
\begin{proof}
  $(AT;S_bS_c) \overset{I}{=} (MN_a;BC) = -1$.
\end{proof}

The third claim implies $A$, $T$, $Z$ are collinear, solving the problem.

\end{document}
