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Gyroelongated Square Dipyramid

\begin{figure}\BoxedEPSF{Deltahedron16_net.epsf scaled 600}\end{figure}

One of the eight convex Deltahedra. It consists of two oppositely faced Square Pyramids rotated 45° to each other and separated by a ribbon of eight side-to-side Triangles. It is Johnson Solid $J_{17}$.


Call the coordinates of the upper Pyramid bases (± 1, ± 1, $h_1$) and of the lower ($\pm\sqrt{2}$, 0, $-h_1$) and (0, $\pm\sqrt{2}$, $-h_1$). Call the Pyramid apexes (0, 0, $\pm (h_1+h_2)$). Consider the points (1, 1, 0) and (0, 0, $h_1+h_2$). The height of the Pyramid is then given by

\begin{displaymath}
\sqrt{1^2+1^2+{h_2}^2} = \sqrt{2+{h_2}^2}=2
\end{displaymath} (1)


\begin{displaymath}
h_2=\sqrt{2}\,.
\end{displaymath} (2)

Now consider the points (1, 1, $h_1$) and ($\sqrt{2}$, 0, $-h_1$). The height of the base is given by


\begin{displaymath}
(1-\sqrt{2}\,)^2+1^2+(2h_1)^2 = 1-2\sqrt{2}+2+1+4{h_1}^2 = 4-2\sqrt{2}+4{h_1}^2 = 2^2=4\equm
\end{displaymath}


\begin{displaymath}
4{h_1}^2=2\sqrt{2}
\end{displaymath} (3)


\begin{displaymath}
{h_1}^2 = {\sqrt{2}\over 2} = {1\over\sqrt{2}} = 2^{-{1/2}},
\end{displaymath} (4)

so
$\displaystyle h_1$ $\textstyle =$ $\displaystyle 2^{-1/4}$ (5)
$\displaystyle h_2$ $\textstyle =$ $\displaystyle 2^{1/2}.$ (6)




© 1996-9 Eric W. Weisstein
1999-05-25