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

\begin{document}
\title{H3167860}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 133}
\maketitle

\section*{Problem}
Let triangle $\triangle ABC$ have orthocenter $H$.
Let $K=X(69)$ be its isotomic conjugate.
Let $U$ and $V$ be the intersection of $HA$ with the $B$-median
and $HK$ with the $C$-median respectively.
Suppose that $UV$ intersects $AC$ at $X$, and that $XK$ intersects $BC$ at $Y$.
Show that $\angle ABU = \angle CYX$.

\section*{Video}
\href{https://www.youtube.com/watch?v=E3bzg5HKpCY&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/E3bzg5HKpCY}}

\section*{External Link}
\url{https://aops.com/community/p28831427}

\newpage

\section*{Solution}
Let $\infty$ be the point at infinity along the $B$-symmedian.
We will show that $X$, $K$, $\infty$ are collinear, solving the problem.

\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 = (-2.30297,4.61391);
pair B = (-5.,-1.5);
pair C = (3.,-1.5);
pair H = (-2.30297,0.83929);
pair K = (-0.38766,0.12033);
pair M = (0.34851,1.55695);
pair U = (-2.30297,0.04148);
pair N = (-3.65148,1.55695);
pair V = (-1.14048,0.40292);
pair X = (0.82061,1.01265);
pair Y = (-2.58172,-1.5);
pair R = (-6.16847,-4.14883);

import graph;
size(10cm);
pen zzttqq = rgb(0.6,0.2,0.);
pen yqqqqq = rgb(0.50196,0.,0.);
pen qqwuqq = rgb(0.,0.39215,0.);
pen qqffff = rgb(0.,1.,1.);
pen xfqqff = rgb(0.49803,0.,1.);
pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw(A--B--C--cycle, linewidth(0.6) + zzttqq);
draw(arc(Y,0.32788,0.,36.44611)--Y--cycle, linewidth(0.6) + qqwuqq);
draw(arc(B,0.32788,29.75024,66.19635)--B--cycle, linewidth(0.6) + qqwuqq);

draw(A--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--A, linewidth(0.6) + zzttqq);
draw(C--N, linewidth(0.6) + yqqqqq);
draw(B--M, linewidth(0.6) + yqqqqq);
draw(A--U, linewidth(0.6) + zzttqq);
draw(U--X, linewidth(0.6) + qqwuqq);
draw(H--K, linewidth(0.6) + qqffff);
draw(X--Y, linewidth(0.6) + qqwuqq);

dot("$A$", A, dir((4.371, 10.929)));
dot("$B$", B, dir(160));
dot("$C$", C, dir((0.871, -32.351)));
dot("$H$", H, dir((-39.346, -14.905)));
dot("$K$", K, dir(10));
dot("$M$", M, dir((4.807, 8.415)));
dot("$U$", U, dir((-36.068, 4.765)));
dot("$N$", N, dir((-44.392, -1.421)));
dot("$V$", V, dir((-6.955, -37.937)));
dot("$X$", X, dir((4.594, 8.198)));
dot("$Y$", Y, dir(90));
// dot("$R$", R, dir((-25.490, -23.217)));
\end{asy}
\end{center}

We use barycentric coordinates.
The coordinates of all relevant points are:
\begin{align*}
H &= \left( \frac{1}{S_a} : \frac{1}{S_b} : \frac{1}{S_c} \right) \\
K &= \left( S_a : S_b : S_c \right) \\
U &= \left( \frac{1}{S_c} : \frac{1}{S_b} : \frac{1}{S_c} \right)
  = \left( S_b : S_c : S_b \right) \\
V &= (1:1:v) \implies \left( \frac{S_b}{S_a}-\frac{S_a}{S_b} \right)v
  + \frac{S_c}{S_b}-\frac{S_b}{S_c}+\frac{S_a}{S_c}-\frac{S_c}{S_a} = 0 \\
  \implies (S_b^2-S_a^2)v &= \frac{S_a S_b (S_b-S_a)}{S_c} + S_c(S_b-S_a)
  \implies v = \frac{S_aS_b+S_c^2}{c^2S_c} \\
X &= (S_c-S_b : 0 : vS_c-S_b)
  = \left( b^2-c^2 : 0 : \frac{S_aS_b-S_bc^2+S_c^2}{c^2} \right) \\
  &= \left( b^2-c^2 : 0 : \frac{S_c^2-S_b^2}{c^2} \right) \\
  &= \left( c^2 : 0 : a^2  \right) \\
\infty &= (a^2 : -(a^2+c^2) : c^2).
\end{align*}
Now,
\begin{align*}
\det(\infty,X,K) =
\det \begin{bmatrix}
  a^2 & -(a^2+c^2) & c^2 \\
  c^2 & 0 & a^2 \\
  S_a & S_b & S_c \\
\end{bmatrix}
&= \det \begin{bmatrix}
  a^2 & -(a^2+c^2) & c^2 \\
  c^2 & 0 & a^2 \\
  \frac{b^2+c^2}{2} & -\frac{b^2}{2} & \frac{a^2+b^2}{2} \\
\end{bmatrix} \\
&= \det \begin{bmatrix}
  a^2 & -(a^2+c^2) & c^2 \\
  c^2 & 0 & a^2 \\
  \frac{b^2}{2} & -\frac{b^2}{2} & \frac{b^2}{2} \\
\end{bmatrix} = 0.
\end{align*}

\end{document}
