Triangular hebesphenorotunda

Cartesian coordinates for the triangular hebesphenorotunda with edge length √5 – 1 are given by the union of the orbits of the points

${\displaystyle \left(0,-{\frac {2}{\tau {\sqrt {3}}}},{\frac {2\tau }{\sqrt {3}}}\right),\left(\tau ,{\frac {1}{{\sqrt {3}}\tau ^{2}}},{\frac {2}{\sqrt {3}}}\right),}$

${\displaystyle \left(\tau ,-{\frac {\tau }{\sqrt {3}}},{\frac {2}{{\sqrt {3}}\tau }}\right),\left({\frac {2}{\tau }},0,0\right),}$

under the action of the group generated by rotation by 120° around the z-axis and the reflection about the yz-plane.[2] Here, 𝜏 = √5 + 1/2 (sometimes written φ) is the golden ratio. The first point generates the triangle opposite the hexagon, the second point generates the bases of the triangles surrounding the previous triangle, the third point generates the tips of the pentagons opposite the first triangle, and the last point generates the hexagon.

One may then calculate the surface area of a triangular hebesphenorotunda of edge length a as

${\displaystyle A=\left(3+{\frac {1}{4}}{\sqrt {1308+90{\sqrt {5}}+114{\sqrt {75+30{\sqrt {5}}}}}}\right)a^{2}\approx 16.38867a^{2},}$[3]

and its volume as

${\displaystyle V={\frac {1}{6}}\left(15+7{\sqrt {5}}\right)a^{3}\approx 5.10875a^{3}.}$[4]

A second, inverted, triangular hebesphenorotunda can be obtained by negating the second and third coordinates of each point. This second polyhedron will be joined to the first at their common hexagonal face, and the pair will inscribe an icosidodecahedron. If the hexagonal face is scaled by the golden ratio, then the convex hull of the result will be the entire icosidodecahedron.

• Weisstein, Eric W., "Triangular hebesphenorotunda" ("Johnson solid") at MathWorld.