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

\begin{document}
\title{Sharygin 2019/23}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 62}
\maketitle

\section*{Problem}
In the plane, let $a$, $b$ be two closed broken lines (possibly
self-intersecting), and $K$, $L$, $M$, $N$ be four points. The vertices
of $a$, $b$ and the points $K$, $L$, $M$, $N$ are in general position
(i.e.\ no three of these points are collinear, and no three segments
between them concur at an interior point). Each of segments $KL$ and
$MN$ meets $a$ at an even number of points, and each of segments $LM$
and $NK$ meets $a$ at an odd number of points. Conversely, each of
segments $KL$ and $MN$ meets $b$ at an odd number of points, and each of
segments $LM$ and $NK$ meets $b$ at an even number of points. Prove that
$a$ and $b$ intersect.

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



\newpage

\section*{Solution}
Assume for contradiction this is not so.

\begin{claim*}
  [Well-known]
  The curve $a$ encloses a region
  (meaning one can discuss being inside or outside $a$),
  and similarly for $b$.
\end{claim*}

Now:
\begin{itemize}
  \ii Since $KN$ intersects $a$ an odd number of times,
  exactly one of the two points is inside $a$.
  WLOG $K$ is inside $a$ and $N$ is outside.
  \ii Following through, $M$ is outside, so $L$ is inside.
  \ii But then $KL$ can't intersect $b$ at all, contradiction.
\end{itemize}

\end{document}
