Pentagonal antiprismatic pyramid

{{Infobox polytope The pentagonal antiprismatic pyramid, or pappy, is a CRF segmentochoron (designated K-4.80 on Richard Klitzing's list). It consists of 10 regular tetrahedra, 2 pentagonal pyramids, and 1 pentagonal antiprism. As the name suggests, it is a pyramid based on the pentagonal antiprism.
 * type=Segmentotope
 * dim = 4
 * obsa = Pappy
 * cells = 10 tetrahedra, 2 pentagonal pyramids, 1 pentagonal antiprism.
 * faces = 10+10+10 triangles, 2 pentagons
 * edges = 10+10+10
 * vertices = 1+10
 * verf = 1 pentagonal antiprism, edge length 1
 * verf2 =10 isosceles trapezoidal pyramids, base edge lengths 1, 1, 1, (1+$\sqrt{5}$)/2, side edge length 1
 * coxeter=oxo5oox&#x
 * army=Pappy
 * reg =Pappy
 * symmetry = (I{{sub|2}}(10)×A{{sub|1}})/2×I, order 20
 * circum = $$\frac{1+\sqrt5}{2} ≈ 1.61083$$
 * height = Peg atop gyro peppy: $$\frac12 = 0.5$$
 * height2 = point atop pap: $$\frac{\sqrt5-1}{4} ≈ 0.30902$$
 * hypervolume = $$\fracP5+3\sqrt5}{96} ≈ 0.12196$$
 * dich = Tet–3–tet: $$\arccos\left(-\frac{1+3\sqrt5}{8}\right) ≈ 164.47751º$$
 * dich2 = Pap–5–peppy: 36º
 * dich3 = Tet–3–peppy: $$\arccos\left(\frac{\sqrt{7+3\sqrt5}}{4}\right) ≈ 22.23876º$$
 * dich4 = Pap–3–tet: $$\arccos\left(\frac{\sqrt{7+3\sqrt5}}{4}\right) ≈ 22.23876º$$
 * dual=Pentagonal antitegmatic pyramid
 * conjugate=Pentagrammic retroprismatic pyramid
 * conv = Yes
 * orientable=Yes
 * nat=Tame}}

It can be obtained as the middle piece of an icosahedral pyramid, with the remainder formed by augmenting two pentagonal scalenes onto the pentagonal pyramids. It also occurs as a part of icosahedron atop dodecahedron, as the vertex pyramid of the icosahedral vertices.

Vertex coordinates
The vertices of a pentagonal antiprismatic pyramid of edge length 1 are given by:
 * $$±\left(±\frac12,\,-\sqrt{\frac{5+2\sqrt5}{20}},\,\sqrt{\frac{5+\sqrt5}{40}},\,0\right),$$
 * $$±\left(±\frac{1+\sqrt5}{4},\,\sqrt{f\frac{5-\sqrt5}{40}},\,\sqrt{\frac{5+\sqrt5}{40}},\,0\right),$$
 * $$±\left(0,\,\sqrt{\frac{5+\sqrt5}{10}},\,\sqrt{\frac{5+\sqrt5}{40}},\,0\right),$$
 * $$\left(0,\,0,\,0,\,\frac{\sqrt5-1}{4}\right).$$