Triaugmented triangular prism: Difference between revisions

Revision as of 04:58, 11 December 2023

Revision as of 07:17, 3 January 2024

Line 25:

One possible system of [[Cartesian coordinates]] for the vertices of a triaugmented triangular prism, giving it edge length 2, is:{{r|shdc}}

<math display="block"> \displaystyle \begin{align}

{{bi|left=1.6|<!-- avoid unnecessary interior scrollbar --><math>\displaystyle

\left(0,\frac2{\sqrt3},\pm1 \right),\qquad & \left(\pm1,-\frac1{\sqrt3},\pm1 \right),\\

\begin{align}

\left(0,-\frac2sqrt2-\frac1{\sqrt3},0 \pm1right),\qquad & \left(\pm\frac{1+\sqrt6}{2},\frac{1+\sqrt6}{2\sqrt3},0\right).\\

\end{align}</math>}}

(\pm1,-\frac1{\sqrt3},\pm1),\\

(0,-\sqrt2-\frac1{\sqrt3},0),\qquad &

(\pm\frac{1+\sqrt6}{2},\frac{1+\sqrt6}{2\sqrt3},0).\\

\end{align}</math>}}

==Properties==