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

\begin{document}
\title{EGMO 2024/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 146}
\maketitle

\section*{Problem}
Let $ABC$ be a triangle with $AC>AB$,
and denote its circumcircle by $\Omega$ and incenter by $I$.
Let its incircle meet sides $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively.
Let $X$ and $Y$ be two points on minor arcs $\widehat{DF}$ and $\widehat{DE}$ of the incircle,
respectively, such that $\angle BXD = \angle DYC$.
Let line $XY$ meet line $BC$ at $K$.
Let $T$ be the point on $\Omega$ such that $KT$ is tangent to $\Omega$
and $T$ is on the same side of line $BC$ as $A$.
Prove that lines $TD$ and $AI$ meet on $\Omega$.

\section*{Video}
\href{https://www.youtube.com/watch?v=Q1CLri5l28I&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/Q1CLri5l28I}}

\section*{External Link}
\url{https://aops.com/community/p30433677}

\newpage

\section*{Solution}
We redefine the points in the reverse order:
let $M$ be the arc midpoint of minor arc $BC$,
let $T$ be the second intersection of ray $TD$ with $\Omega$,
and let $K$ be the point on line $BC$ such that $\ol{KT}$ is tangent to $\Omega$.
Then we will show that if a line through $K$ meets the incircle again at $X$ and $Y$,
then $\angle BXD = \angle DYC$.
A phantom point argument will then show the originas result.

\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 = (-3.15888,1.43930);
pair B = (-4.5,-2.5);
pair C = (2.5,-2.5);
pair I = (-2.36683,-0.97282);
pair D = (-2.36683,-2.5);
pair E = (-1.49432,0.28055);
pair F = (-3.81252,-0.48065);
pair M = (-1.,-5.13543);
pair T = (-3.97684,0.60428);
pair K = (-6.16456,-2.5);
pair X = (-3.60416,-1.86796);
pair Y = (-0.85509,-1.18934);

import graph;
size(12cm);
pen zzttqq = rgb(0.6,0.2,0.);
pen yqqqqq = rgb(0.50196,0.,0.);
pen qqwuqq = rgb(0.,0.39215,0.);
pen aqaqaq = rgb(0.62745,0.62745,0.62745);
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((-1.,-1.49361), 3.64181), linewidth(0.6) + yqqqqq);
draw(circle(I, 1.52717), linewidth(0.6) + yqqqqq);
draw(A--M, linewidth(0.6) + yqqqqq);
draw(K--B, linewidth(0.6));
draw(A--D, linewidth(0.6));
draw(T--M, linewidth(0.6) + yqqqqq);
draw(T--K, linewidth(0.6) + qqwuqq);
draw(B--X, linewidth(0.6) + gray);
draw(X--D, linewidth(0.6) + gray);
draw(D--Y, linewidth(0.6) + gray);
draw(C--Y, linewidth(0.6) + gray);
draw(K--Y, linewidth(0.6) + aqaqaq);

dot("$A$", A, dir(130));
dot("$B$", B, dir(225));
dot("$C$", C, dir(315));
dot("$I$", I, dir((3.539, 8.205)));
dot("$D$", D, dir(225));
dot("$E$", E, dir((4.017, 7.482)));
dot("$F$", F, dir((-23.362, 0.859)));
dot("$M$", M, dir((-6.535, -33.120)));
dot("$T$", T, dir((-20.887, 9.005)));
dot("$K$", K, dir((-34.397, -11.544)));
dot("$X$", X, dir((-42.204, 3.012)));
dot("$Y$", Y, dir((8.881, -1.047)));
\end{asy}
\end{center}

\begin{claim*}
  $KT = KD$.
\end{claim*}
\begin{proof}
  Note $\angle KTD = \half \arc{TM} = \half(\arc{TB} + \arc{MC}) = \angle TDK$.
\end{proof}
Hence, it follows that
\[ KX \cdot KY = KD^2 = KT^2 = KB \cdot KC \]
so quadrilateral $BXYC$ is cyclic.
From this we deduce the inverse similarities
\begin{align*}
  \triangle KXD &\overset{-}{\sim} \triangle KDY \\
  \triangle KXB &\overset{-}{\sim} \triangle KCY.
\end{align*}
Hence,
\[ \angle BXD = \angle XKD - \angle KXB = \angle YDK - \angle YCK = \angle DYC \]
as needed.

\end{document}
