\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{IMO 1999/1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 98}
\maketitle
\section*{Problem}
A set $S$ of points from the space will be called
completely symmetric if it has at least three elements
and fulfills the condition that for every two distinct points
$A$ and $B$ from $S$,
the perpendicular bisector plane of the segment $AB$
is a plane of symmetry for $S$.
Prove that if a completely symmetric set is finite,
then it consists of the vertices of either a regular polygon,
or a regular tetrahedron or a regular octahedron.
\section*{Video}
\href{https://www.youtube.com/watch?v=QIkoSVCHrR8&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/QIkoSVCHrR8}}
\section*{External Link}
\url{https://aops.com/community/p131833}
\newpage
\section*{Solution}
Let $G$ be the centroid of $S$.
\begin{claim*}
All points of $S$ lie on a sphere $\Gamma$ centered at $G$.
\end{claim*}
\begin{proof}
Each perpendicular bisector plane passes through $G$.
So if $A,B \in S$ it follows $GA = GB$.
\end{proof}
\begin{claim*}
Consider any plane passing through three or more points of $S$.
The points of $S$ in the plane form a regular polygon.
\end{claim*}
\begin{proof}
The cross section is a circle because we are intersecting
a plane with sphere $\Gamma$.
Now if $A$, $B$, $C$ are three adjacent points on this circle,
by taking the perpendicular bisector we have $AB=BC$.
\end{proof}
If the points of $S$ all lie in a plane, we are done.
Otherwise, the points of $S$ determine a polyhedron
$\Pi$ inscribed in $\Gamma$.
All of the faces of $\Pi$ are evidently regular polygons,
of the same side length $s$.
\begin{claim*}
Every face of $\Pi$ is an equilateral triangle.
\end{claim*}
\begin{proof}
Suppose on the contrary some face $A_1 A_2 \dots A_n$
has $n > 3$.
Let $B$ be any vertex adjacent to $A_1$ in $\Pi$
other than $A_2$ or $A_n$.
Consider the plane determined by $\triangle A_1 A_3 B$.
This is supposed to be a regular polygon,
but arc $A_1 A_3$ is longer than arc $A_1 B$,
and by construction there are no points inside these arcs.
This is a contradiction.
\end{proof}
Hence, $\Pi$ has faces all congruent equilateral triangles.
This implies it is a regular polyhedron --- either
a regular tetrahedron, regular octahedron,
or regular icosahedron.
We can check the regular icosahedron fails by
taking two antipodal points as our counterexample.
This finishes the problem.
\end{document}