\documentclass{amsart}

\usepackage{graphics}

\title{Construction of the Pentagon}

\begin{document}

\maketitle

\begin{enumerate}

\item Construct a circle and a diameter $AB$.

\item Construct a another diameter perpendicular to $AB$ which meets
the circle at $D$. (Find the point $C$ by marking an arc centered at $A$
with radius $AB$ and then doing the same with $B$ as the center.)

\item  Bisect the segment $DO$ at the point~$G$. (Find the points $E$
and $F$ on the circle by marking arcs centered at $D$ with radius $DO$.)

\item Construct an arc centered at $G$ with radius $GB$ so that it meets the
perpendicular diameter at a point~$H$.

\item Construct an arc centered at $A$ with radius $AH$ so that it meets the
circle at the point~$I$. Mark off the point $J$ with an arc centered at $A$
with radius~$AI$. Mark off the points $K$ and $L$ with arcs centered at
$I$ and $J$, respectively, and radius~$AI$.  The points $A$,$I$, $J$, $K$,
and $L$ are the vertices of a regular pentagon.

\end{enumerate}

\begin{center}
\resizebox{3.5in}{!}{\includegraphics{pentagon.epsf}}
\end{center}

\begin{proof}
Let us assume for simplicity that the radius of the circle is~1. (For a
circle of radius $R$, all lengths below would be scaled by a factor
of $R$.)  We first calculate some lengths in the diagram. From the
right triangle $GOB$, we compute that $|GB| =\sqrt{(1/2)^2 + 1} =
\sqrt{5}/2$. Thus, $|DH| = 1/2 + \sqrt{5}/2 = \phi$, the golden mean,
and $|OH| = |DH| - 1 = -1/2 + \sqrt{5}/2$. From the right triangle $AOH$,
$|AH| = \sqrt{1 + (-1/2+\sqrt{5}/2)^2} =\sqrt{(5-\sqrt{5})/2}$.
Let $AIJKL$ be the pentagon inscribed in the circle as indicated in
the diagram.  The proof will be complete if we show that $|AH| = |AI|$.

Let $s = |AI|$.  Consider the isosceles triangle $AOI$. Since the pentagon
in comprised of 5 of these triangles all sharing the vertex~$O$, the
angle $AOI$ is clearly~$2\pi/5$.  Bisecting this angle produces two
equal right triangles in $AOI$ and we conclude $s/2 = \sin(\pi/5)$.
Applying the formula $\sin(\alpha+\beta) =
\sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)$  repeatedly, we arrive
at the equation
\[
\sin(5\theta) = 16\sin^5(\theta) - 20\sin^3(\theta) + 5\sin(\theta)
\]
Letting $\theta = \pi/5$ we obtain
\[
0 = 16(s/2)^5 - 20(s/2)^3 + 5(s/2) = (s/2)(s^4 - 5s^2 + 5)
\]
Since $0 < \sin(\pi/5) < \sin(\pi/4) = \sqrt{2}/2$, we are forced to
conclude using the quadratic formula that $s^2 = (5 - \sqrt{5})/2$,
and $s = \sqrt{(5 - \sqrt{5})/2}$.  Therefore, $|AH| = s$ as claimed.

\end{proof}


\end{document}
