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

\begin{document}
\title{Twitch 137.2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 137}
\maketitle

\section*{Problem}
In triangle $ABC$, let $H$ and $O$ denote the orthocenter and circumcenter, respectively.
Define $D = \ol{CH} \cap \ol{AB}$, $E = \ol{AH} \cap \ol{CO}$,
$F = \ol{AO} \cap \ol{BC}$, $P = \ol{FH} \cap \ol{CO}$, $Q = \ol{DE} \cap \ol{BO}$.
Prove that $\angle BAP=\angle CAQ$.

\section*{Video}
\href{https://www.youtube.com/watch?v=1VW8EToKfWs&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/1VW8EToKfWs}}

\section*{External Link}
\url{https://aops.com/community/p29310655}

\newpage

\section*{Solution}
The problem is true for any pair of isogonal conjugates in place of $(H,O)$.
We use trilinear coordinates, with $H = (u:v:w)$ and $O=(1/u:1/v:1/w)$.
\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 C = (5.,-2.);
pair B = (-4.,-2.);
pair A = (-2.43184,3.32955);
pair H = (-2.48776,0.25406);
pair O = (0.68947,-0.52487);
pair D = (-3.26768,0.48885);
pair E = (-2.48219,0.56051);
pair F = (1.88402,-2.);
pair P = (-4.26623,1.17103);
pair Q = (6.84509,1.41143);

import graph;
size(13.93225cm);
pen zzttqq = rgb(0.6,0.2,0.);
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(C--D, linewidth(0.6));
draw(A--H, linewidth(0.6));
draw(A--F, linewidth(0.6));
draw(C--P, linewidth(0.6));
draw(B--Q, linewidth(0.6));
draw(D--Q, linewidth(0.6));
draw(F--P, linewidth(0.6));

dot("$C$", C, dir((-8.384, -58.269)));
dot("$B$", B, dir((-16.948, -54.127)));
dot("$A$", A, dir((5.739, 13.564)));
dot("$H$", H, dir((-14.904, -40.655)));
dot("$O$", O, dir(70));
dot("$D$", D, dir((-48.756, -14.426)));
dot("$E$", E, dir((5.251, 11.547)));
dot("$F$", F, dir((-10.227, -61.031)));
dot("$P$", P, dir((-44.176, -9.462)));
dot("$Q$", Q, dir((-14.770, 24.492)));
\end{asy}
\end{center}

Compute the points:
\begin{align*}
  D &= (u:v:0) \\
  E &= (v^2:uv:uw) \\
  F &= (0:w:v) \\
  P &= (v:u:k) \implies \det(PHF) = 0 \\
  Q &= (w:\ell:u) \implies \det(DEQ) = 0
\end{align*}
We solve for $k$ and $\ell$ in turn: for $k$ we have
\[
  0 = \det \begin{bmatrix}
    v & u & k \\
    u & v & w \\
    0 & w & v
  \end{bmatrix}
  = v(v^2-w^2)+u(wk-uv)
\]
hence
\[ k = v \cdot \frac{w^2-v^2+u^2}{uw}. \]
Meanwhile for $\ell$ we have:
\[
  0 = \det \begin{bmatrix}
    u & v & 0 \\
    v^2 & uv & uw \\
    w & \ell & u \\
  \end{bmatrix}
  = u(uuv-v^3) + uw(vw-\ell u)
\]
hence
\[ \ell = v \cdot \frac{u^2-v^2+w^2}{uw} \]
We see that $k = \ell$. The isogonality $\angle BAP = \angle CAQ$ follows.

\end{document}
