Augmented sphenocorona

To calculate Cartesian coordinates for the augmented sphenocorona, one may start by calculating the coordinates of the sphenocorona. Let k ≈ 0.85273 be the smallest positive root of the quartic polynomial

${\displaystyle 60x^{4}-48x^{3}-100x^{2}+56x+23.}$

Then, Cartesian coordinates of a sphenocorona with edge length 2 are given by the union of the orbits of the points

${\displaystyle \left(0,1,2{\sqrt {1-k^{2}}}\right),\,(2k,1,0),\left(0,1+{\frac {\sqrt {3-4k^{2}}}{\sqrt {1-k^{2}}}},{\frac {1-2k^{2}}{\sqrt {1-k^{2}}}}\right),\,\left(1,0,-{\sqrt {2+4k-4k^{2}}}\right)}$

under the action of the group generated by reflections about the xz-plane and the yz-plane.[2] Calculating the centroid and the normal unit vector of one of the square faces gives the location of its last vertex as

${\displaystyle \left(k+{\sqrt {2-2k^{2}}},0,k+{\sqrt {2-2k^{2}}}\right).}$

One may then calculate the surface area of a snub square of edge length a as

${\displaystyle A=\left(1+4{\sqrt {3}}\right)a^{2}\approx 7.92820a^{2},}$[3]

and its volume as

${\displaystyle V=\left({\frac {1}{2}}{\sqrt {1+3{\sqrt {\frac {3}{2}}}+{\sqrt {13+3{\sqrt {6}}}}}}+{\frac {1}{3{\sqrt {2}}}}\right)a^{3}\approx 1.75105a^{3}.}$[4]

• Weisstein, Eric W., "Augmented sphenocorona" ("Johnson solid") at MathWorld.