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

\begin{document}
\title{IMO 1990/6}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 27}
\maketitle

\section*{Problem}
Prove that there exists a convex equiangular $1990$-gon
whose side lengths are $1^2$, $2^2$, $3^2$, \dots, $1990^2$ in some order.

\section*{Video}
\href{https://www.youtube.com/watch?v=rTb16n8eAU8&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/rTb16n8eAU8}}

\section*{External Link}
\url{https://aops.com/community/p366472}

\newpage

\section*{Solution}
Throughout this solution,
$\omega$ denotes a primitive $995$th root of unity.

We first commit to placing $1^2$ and $2^2$ on opposite sides,
$3^2$ and $4^2$ on opposite sides, etc.
Since $2^2-1^2=3$, $4^2-3^2=7$, $6^2-5^2=11$, etc.,
this means the desired conclusion is equivalent to
\[ 0 = \sum_{n=0}^{994} c_n \omega^n \]
being true for some permutation
$(c_0, \dots, c_{994})$ of $(3, 7, 11, \dots, 3979)$.

Define $z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15
\omega^{597} + 19 \omega^{796}$.
Then notice that
\begin{align*}
  z &= 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796} \\
  \omega^5 z &= 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801} \\
  \omega^{10} z &= 43 \omega^{10} + 47 \omega^{209} + 51\omega^{408} + 55 \omega^{607} + 59 \omega^{806} \\
  &\vdotswithin{=}
\end{align*}
and so summing yields the desired conclusion,
as the left-hand side becomes
\[ (1+\omega^5+\omega^{10}+\dotsb+\omega^{990})z = 0 \]
and the right-hand side is the desired expression.

\end{document}
