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Pentagonal Dipyramid

\begin{figure}\begin{center}\BoxedEPSF{J13_net.epsf scaled 700}\end{center}\end{figure}

The pentagonal dipyramid is one of the convex Deltahedra, and Johnson Solid $J_{13}$. It is also the Dual Polyhedron of the Pentagonal Prism. The distance between two adjacent Vertices on the base of the Pentagon is

$\displaystyle {d_{12}}^2$ $\textstyle =$ $\displaystyle [1-\cos({\textstyle{2\over 5}}\pi)]^2+\sin^2({\textstyle{2\over 5}}\pi)$  
  $\textstyle =$ $\displaystyle [1-{\textstyle{1\over 4}}(\sqrt{5}-1)]^2+\left[{(1+\sqrt{5})\sqrt{5-\sqrt{5}}\over 4\sqrt{2}\,}\right]^2$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}(5-\sqrt{5}\,),$ (1)

and the distance between the apex and one of the base points is
\begin{displaymath}
{d_{1h}}^2 = (0-1)^2+(0-0)^2+(h-0)^2=1+h^2.
\end{displaymath} (2)

But
\begin{displaymath}
{d_{12}}^2={d_{12}}^2
\end{displaymath} (3)


\begin{displaymath}
{\textstyle{1\over 2}}(5-\sqrt{5})=1+h^2
\end{displaymath} (4)


\begin{displaymath}
h^2={\textstyle{1\over 2}}(3-\sqrt{5}),
\end{displaymath} (5)

and
\begin{displaymath}
h=\sqrt{3-\sqrt{5}\over 2}.
\end{displaymath} (6)

This root is of the form $\sqrt{a+b\sqrt{c}}$, so applying Square Root simplification gives
\begin{displaymath}
h = {\textstyle{1\over 2}}(\sqrt{5}-1)\equiv \phi-1,
\end{displaymath} (7)

where $\phi$ is the Golden Mean.

See also Deltahedron, Dipyramid, Golden Mean, Icosahedron, Johnson Solid, Triangular Dipyramid




© 1996-9 Eric W. Weisstein
1999-05-26