# Compound of twelve pentagrammic retroprisms

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Compound of twelve pentagrammic retroprisms
Rank3
TypeUniform
SpaceSpherical
Notation
Bowers style acronymGidasid
Elements
Components12 pentagrammic retroprisms
Faces120 triangles, 24 pentagrams as 12 stellated decagrams
Edges120+120
Vertices120
Vertex figureCrossed isosceles trapezoid, edge length 1, 1, 1, (5–1)/2
Measures (edge length 1)
Circumradius${\displaystyle \sqrt{\frac{5-\sqrt5}{8}} ≈ 0.58779}$
Volume${\displaystyle 2(5-2\sqrt5) ≈ 1.05573}$
Dihedral angles3–3: ${\displaystyle \arccos\left(\frac{\sqrt5}3\right) ≈ 41.81031^\circ}$
5/2–3: ${\displaystyle \arccos\left(\sqrt{\frac{5+2\sqrt5}{15}}\right) ≈ 37.37737^\circ}$
Central density36
Related polytopes
ArmySemi-uniform Grid
RegimentGidasid
DualCompound of twelve pentagrammic concave antitegums
ConjugateCompound of twelve pentagonal antiprisms
Abstract & topological properties
OrientableYes
Properties
SymmetryH3, order 120
ConvexNo
NatureTame

The great inverted disnub dodecahedron, gidasid, or compound of twelve pentagrammic retroprisms is a uniform polyhedron compound. It consists of 120 triangles and 24 pentagrams (whcih fall in pairs in the same plane and combine into 12 stellated decagrams), with one pentagram and three triangles joining at a vertex.

This compound has rotational freedom, represented by an angle θ. we start at θ = 0° with all the pentagrammic retroprisms inscribed in a great icosahedron, and rotate pairs of retroprisms in opposite directions. At θ = 36° the retroprisms coincide by pairs, resulting in a double cover of the great inverted snub dodecahedron.

## Vertex coordinates

The vertices of a great inverted disnub dodecahedron of edge length 1 and rotation angle θ are given by all even permutations of:

• ${\displaystyle \left(±\frac{\sqrt5+(5-\sqrt5)\cos(\theta)}{10},\,±\sqrt{\frac{5-\sqrt5}{10}}\sin(\theta),\,±\frac{5-\sqrt5-4\sqrt5\cos(\theta)}{20}\right),}$
• ${\displaystyle \left(±\frac{2\sqrt5+2\sqrt5\cos(\theta)+(1-\sqrt5)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20},\,±\frac{-10(1-\sqrt5)\cos(\theta)+(1+\sqrt5)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta))}{20},\,±\frac{5-\sqrt5-(5+\sqrt5)\cos(\theta)-2\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20}\right),}$
• ${\displaystyle \left(±\frac{2\sqrt5-(5-3\sqrt5)\cos(\theta)+2\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20},\,±\frac{-10\cos(\theta)+(1-\sqrt5)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20},\,±\frac{5-\sqrt5+(5-\sqrt5)\cos(\theta)+(1+\sqrt5)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20}\right),}$
• ${\displaystyle \left(±\frac{2\sqrt5-(5-3\sqrt5)\cos(\theta)-2\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20},\,±\frac{10\cos(\theta)+(\sqrt5-1)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20},\,±\frac{5-\sqrt5+(5-\sqrt5)\cos(\theta)+(1+\sqrt5)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20}\right),}$
• ${\displaystyle \left(±\frac{2\sqrt5+2\sqrt5\cos(\theta)-(\sqrt5-1)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20},\,±\frac{10(\sqrt5-1)\cos(\theta)+(1+\sqrt5)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20},\,±\frac{5-\sqrt5-(5+\sqrt5)\cos(\theta)+2\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20}\right).}$