\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

Calculate the perimeter of the regular hyperbolic pentagon
\textbf{R}, all of whose interior angles are $\frac{\pi}{2}$.


\medskip

{\bf Answer}

\begin{center}
\epsfig{file=407-1999-3.eps, width=70mm}
\end{center}

If we are to draw $R$ in the Poinc\'{a}re Disc, we could do so so
that the vertices are at $re^{i\frac{2\pi k}{5}}$ ($0 \leq k \leq
4$) for some $0<r<1$. Joining the angles to the vertices then
breaks $R$ into 5 triangles with interior angles
$\ds\frac{\pi}{4},\ \ds\frac{\pi}{4},\ \ds\frac{2\pi}{5}$.

Let $T$ be one of these triangles; and let $c$ be the side
opposite the vertex with angle $\ds\frac{2\pi}{5}$. Then, by the
law of cosines II:

\begin{eqnarray*}
\cosh(c) & = &
\ds\frac{\cos(\frac{\pi}{4})\cos(\frac{\pi}{4})+\cos(\frac{2\pi}{5})}
{\sin(\frac{\pi}{4})\sin(\frac{\pi}{4})}\\ & = &
\ds\frac{\frac{1}{2}+\cos(\frac{2\pi}{5})}{\frac{1}{2}}=x
\end{eqnarray*}

Then, $\ds\frac{1}{2}(e^c+e^{-c})=x$, so $\begin{array} {l}
e^c+e^{-c}-2x=0\\ e^{2c}-2xe^{c}+1=0 \end{array}$

\begin{eqnarray*} e^c & = & \ds\frac{1}{2}(2x \pm \sqrt{4x^2-4})\\ & =
& x + \sqrt{x^2-1} \end{eqnarray*}

So perimeter of $R$ is \un{$5c=5\ln(x + \sqrt{x^2-1})$}

where \un{$x=1+2\cos(\frac{2\pi}{5}$)}.
\end{document}