Triaugmented triangular prism: Difference between revisions

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m Construction: put middle coordinate of the third type over common denominator to match format of the 4th type
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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}
\left(0,\frac2{\sqrt3},\pm1 \right),\qquad & \left(\pm1,-\frac1{\sqrt3},\pm1 \right),\\
\left(0,-\sqrt2-frac{1+\frac1sqrt6}{\sqrt3},0 \right),\qquad & \left(\pm\frac{1+\sqrt6}{2},\frac{1+\sqrt6}{2\sqrt3},0\right).\\
\end{align}</math>