\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{USAMO 1996/3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 81}
\maketitle
\section*{Problem}
Let $ABC$ be a triangle. Prove that there is a line $\ell$ (in the plane
of triangle $ABC$) such that the intersection of the interior of
triangle $ABC$ and the interior of its reflection $A'B'C'$ in $\ell$ has
area more than $\frac23$ the area of triangle $ABC$.
\section*{Video}
\href{https://www.youtube.com/watch?v=30XpnQ__wzs&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/30XpnQ\_\_wzs}}
\section*{External Link}
\url{https://aops.com/community/p353052}
\newpage
\section*{Solution}
All that's needed is:
\begin{claim*}
If $ABC$ is a triangle where $\half < \frac{AB}{AC} < 1$, then
the $\angle A$ bisector works.
\end{claim*}
\begin{proof}
Let the $\angle A$-bisector meet $BC$ at $D$.
The overlapped area is $2[ABD]$ and
\[ \frac{[ABD]}{[ABC]} = \frac{BD}{BC} = \frac{AB}{AB+AC} \]
by angle bisector theorem.
\end{proof}
In general, suppose $x < y < z$ are sides of a triangle.
Then $\half < \frac yz < 1$ by triangle inequality as needed.
\end{document}