\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{USEMO 2019/5}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 11}
\maketitle
\section*{Problem}
Let $\mathcal{P}$ be a regular polygon,
and let $\mathcal{V}$ be the set of its vertices.
Each point in $\mathcal{V}$ is colored red, white, or blue.
A subset of $\mathcal{V}$ is \emph{patriotic}
if it contains an equal number of points of each color,
and a side of $\mathcal{P}$ is \emph{dazzling}
if its endpoints are of different colors.
Suppose that $\mathcal{V}$ is patriotic
and the number of dazzling edges of $\mathcal{P}$ is even.
Prove that there exists a line,
not passing through any point of $\mathcal{V}$,
dividing $\mathcal{V}$ into two nonempty patriotic subsets.
\section*{Video}
\href{https://www.youtube.com/watch?v=0rdn01T_q4Y&list=PLi6h8GM1FA6yfhMMNqJi1mht4KmZnXpj6}{\texttt{https://youtu.be/0rdn01T\_q4Y}}
\section*{External Link}
\url{https://aops.com/community/p15425728}
\newpage
\section*{Solution}
We prove the contrapositive:
if there is no way to split $\mathcal V$
into two patriotic sets,
then the number of dazzling edges is odd.
Let $\zeta = -\half + \frac{\sqrt3}{2} i$ be a root of unity.
Read the $n$ vertices of the polygon in order starting from any point.
In the complex plane, start from the origin and,
corresponding to red, white, or blue,
move by $1$, $\zeta$, or $\zeta^2$, respectively, to get a path.
The diagram below shows an example
(where black stands in for white, for legibility reasons).
\begin{center}
\begin{asy}
size(6cm);
real[] S = {0, 0, 2, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 2, 2, 0, 2, 2, 1, 2, 2};
pair P = (0,0);
pair Q;
path r = P;
pen tp = linewidth(1.0);
for (int i=0; i