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

\begin{document}
\title{Twitch 106.4}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 106}
\maketitle

\section*{Problem}
Given an equilateral triangle $ABC$ and a point $P$ within ABC,
construct the perpendicular $\ell_A$ to $AP$ through $P$.
Let $A_1$ and $A_2$ be the intersections of $\ell_A$
through $AB$ and $AC$, respectively,
and let the intersection of $BA_2$ and $CA_1$ be called $A'$.
Construct $B'$ and $C'$ in a similar manner.
Prove that $AA'$, $BB'$, and $CC'$ are concurrent.

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



\newpage

\section*{Solution}
Let line $AA'$ meet line $BC$ at $X$; also let $\ell_A \cap BC = X'$.
\begin{claim}
  $(BC;XX') = -1$.
\end{claim}
\begin{proof}
  Harmonic bundle picture.
\end{proof}
So by Ceva's theorem, to prove the concurrence, it's equivalent
(by Menelaus) to prove the collinearity of $X'Y'Z'$.

But this line has a name: it's the \emph{orthotransversal} of $P$
with respect to $\triangle ABC$.

\end{document}
