<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

Let $P$ be a hyperbolic 11-gon in the Poincar\'e disc ${\bf D}$,
with vertices at the points $\frac{1}{2} \exp\left(\frac{2\pi}{11}
k \right)$ for $0\le k\le 10$.  Calculate the hyperbolic length of
a side of $P$.
\medskip

{\bf Answer}

\begin{center}
\epsfig{file=407-9-6.eps, width=80mm}
\end{center}

\un{$a=\ln\left(\ds\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right)=\ln(3)$}

\un{$\cosh(a)=\ds\frac{1}{2}(e^a+e^{-a})=\ds\frac{1}{2}(3+\frac{1}{3})=\ds\frac{5}{3}$}

$\cosh(b)=\cosh(a)\cosh(a)-\sinh(a)\sinh(a)\cos\left(\ds\frac{2\pi}{11}\right)$
\un{(by lcI)}

$\cosh(b)=\ds\frac{25}{9}-\ds\frac{16}{9}\cos\left(\ds\frac{2\pi}{11}\right)
= 1.2822$

$b=\ln(1.2822+\sqrt{(1.2822)^2-1})$

\un{$b=0.73465$}

Note that the sides of $P$ all have the same length, since
rotation by $\ds\frac{2\pi}{11}$ takes $P$ to $P$, and so the
length of a side of $P$ is \un{$b=0.73465$}.
\end{document}
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