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

\begin{document}
\title{EGMO 2023/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 124}
\maketitle

\section*{Problem}
We are given an acute triangle $ABC$.
Let $D$ be the point on its circumcircle such that $AD$ is a diameter.
Suppose that points $K$ and $L$ lie on segments $AB$ and $AC$, respectively,
and that $DK$ and $DL$ are tangent to circle $AKL$.
Show that line $KL$ passes through the orthocenter of triangle $ABC$.

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

\section*{External Link}
\url{https://aops.com/community/p27523016}

\newpage

\section*{Solution}
The problem consists of two disjoint ``halves'':
the symmedian picture with $\triangle AKL$ and symmedian $\ol{AD}$,
and the orthocenter picture with $\triangle ABC$ and diameter $\ol{AD}$.
They are colored blue and red below, respectively.

\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 = (7.08,4.8);
pair B = (3.,-5.);
pair C = (16.,-5.);
pair O = (9.5,-1.95681);
pair D = (11.92,-8.71363);
pair M = (7.08,-1.28636);
pair K = (3.60268,-3.55236);
pair L = (10.55731,0.97963);

import graph;
size(8cm);
pen zzttqq = rgb(0.6,0.2,0.);
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(A--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--A, linewidth(0.6) + zzttqq);
draw(circle(O, 7.17711), linewidth(0.6));
draw(K--L, linewidth(0.6));
draw(circle((6.01909,0.34164), 4.58283), linewidth(0.6) + blue);
draw(D--K, linewidth(0.6) + blue);
draw(D--L, linewidth(0.6) + blue);
draw(A--D, linewidth(0.6) + qqwuqq);
draw(A--M, linewidth(0.6) + qqwuqq);

dot("$A$", A, dir((10.540, 24.280)));
dot("$B$", B, dir(225));
dot("$C$", C, dir(315));
dot("$O$", O, dir(200));
dot("$D$", D, dir(320));
dot("$M$", M, dir(270));
dot("$K$", K, dir(100));
dot("$L$", L, dir((8.869, 19.117)));
\end{asy}
\end{center}

Let $M$ be the midpoint of $\ol{KL}$.
\begin{claim*}
  [Blue part of the figure]
  Lines $\ol{AM}$ and $\ol{AD}$ are isogonal to $\angle A$,
  and $AM = AD \cdot \cos A$.
\end{claim*}
\begin{proof}
First part follows from $\ol{AD}$ being the $A$-symmedian.
It's also a standard fact that $AM \cdot AD = AK \cdot AL$
(e.g.\ by $\sqrt{bc}$ inversion);
so it suffices to prove $AM^2 = AK \cdot AL \cos A$.
This follows by Stewart's theorem to get $AM$ together with
the law of cosines $\cos A = \frac{AK^2+AL^2-KL^2}{2\cdot AK \cdot AL}$.
\end{proof}

Let $H$ denote the orthocenter.
\begin{claim*}
  [Red part of the figure]
  Lines $\ol{AH}$ and $\ol{AO}$ are isogonal to $\angle A$,
  and $AH = AD \cdot \cos A$.
\end{claim*}
\begin{proof}
  Both parts are well-known.
\end{proof}

The two claims together imply $H=M$ and we're done.

\begin{remark*}
  The fact that $MBDC$ is a parallelogram is a known result
  and can be proven by more classical means such as angle chasing;
  see \href{https://aops.com/community/c6h535000p3067553}{Russia 2013 / 9.2}.
  This lemma is also the key result needed for
  \url{https://aops.com/community/p15680911}{AIME II 2020 Problem 15}.
\end{remark*}

\end{document}
