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

\begin{document}
\title{EGMO 2023/1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 164}
\maketitle

\section*{Problem}
There are $n \ge 3$ positive real numbers $a_1$, $a_2$, \dots, $a_n$.
For each $1 \le i \le n$ we let $b_i = \frac{a_{i-1} + a_{i+1}}{a_i}$
with indices modulo $n$.
Assume that for all $i$ and $j$ in the range $1$ to $n$,
we have $a_i \le a_j$ if and only if $b_i \le b_j$.
Prove that $a_1 = a_2 = \dots = a_n$.



\section*{External Link}
\url{https://aops.com/community/p27522955}

\newpage

\section*{Solution}
\begin{claim*}
  We have $\max b_i \le 2$.
\end{claim*}
\begin{proof}
  Suppose $M = \opname{argmax}_i a_i$ (i.e.\ that $a_M = \max a_i$).
  Since the $a_i$'s and $b_i$'s have the same ordering, then we should also have
  \[ b_M = \max b_i. \]
  But \[ b_M = \frac{a_{M-1}+a_{M+1}}{a_M} \le 2. \]
\end{proof}
For the same reason we conclude
\[ \min b_i \ge 2. \]
So all the $b_i$'s must be equal to $2$, hence all $a_i$'s are equal.

\end{document}
