\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{IMO 1973/5}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 132}
\maketitle
\section*{Problem}
Let $G$ be a group, each of whose elements is a nonconstant linear function
from $\RR$ to itself, whose operation is function composition.
Suppose that all elements of $G$ have a fixed point.
Show that all elements of $G$ share a fixed point.
\section*{Video}
\href{https://www.youtube.com/watch?v=mqkpYly4eg8&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/mqkpYly4eg8}}
\section*{External Link}
\url{https://aops.com/community/p357910}
\newpage
\section*{Solution}
Note that the only element of $G$ of the form $x \mapsto x+\lambda$ is the identity.
Now consider any two non-identity functions of $G$, say
\begin{align*}
f(x) &= ax+b \\
g(x) &= cx+d
\end{align*}
so that $a \neq 1$ and $c \neq 1$.
The (unique) fixed point of $ax+b$ is $\frac{b}{a-1}$
and the (unique) fixed point of $cx+d$ is $\frac{d}{c-1}$, so we just need to show
\[ \frac{b}{a-1} = \frac{d}{c-1}. \]
Note the inverses $f^{-1}(x) = \frac{x-b}{a}$ and $g^{-1}(x) = \frac{x-d}{c}$
should also be in $G$.
Then consider the composition
\begin{align*}
g(f(g^{-1}(f^{-1}(x))))
&= c \cdot \left[ a \cdot \frac{\frac{x-b}{a}-d}{c} + b \right] + d \\
&= x - b - ad + bc + d.
\end{align*}
Since this is an element of $G$, as we commented before we must have
\[ b - ad + bc + d = 0 \]
which was what we wanted.
\end{document}