Pentagonal gyrobicupola

{{Infobox polytope The pentagonal gyrobicupola, or pegybcu, is one of the 92 Johnson solids (J31). It consists of 10 triangles, 10 squares, and 2 pentagons. It can be constructed by attaching two pentagonal cupolas at their decagonal bases, such that the two pentagonal bases are rotated 36º with respect to each other.
 * type=CRF
 * img=Pentagonal gyrobicupola 2.png
 * 3d=J31 pentagonal gyrobicupola.stl
 * dim = 3
 * obsa = Pegybcu
 * faces = 10 triangles, 10 squares, 2 pentagons
 * edges = 10+10+20
 * vertices = 10+10
 * verf = 10 trapezoids, edge lengths 1, $\sqrt{2}$, (1+$\sqrt{5}$}0/2, $\sqrt{2}$
 * verf2 = 10 rectangles, edge lengths 1 and $\sqrt{2}$
 * coxeter = xxo5oxx&#xt
 * army=Pegybcu
 * reg=Pegybcu
 * symmetry = I2(10)×A1+, order 20
 * volume = (5+4$\sqrt{5}$)/3 ≈ 4.64809
 * dih = 3–4 cupolaic: acos(–($\sqrt{3}$+$\sqrt{15}$)/6) ≈ 159.09484º
 * dih2 = 4–5: acos(–$\sqrt{(5+√5)/10}$) ≈ 148.28253º
 * dih3 = 3–4 join: acos(($\sqrt{15}$–$\sqrt{3}$)/6) ≈ 69.09484º
 * smm = Yes
 * conjugate = Retrograde pentagrammic gyroobicupola
 * conv=Yes
 * orientable=Yes
 * nat=Tame}}

If the cupolas are joined in the same orientation, the result is the pentagonal orthobicupola.

Vertex coordinates
A pentagonal gyrobicupola of edge length 1 has vertices given by the following coordinates:


 * (±1/2, –$\sqrt{(5+2√5)/20}$, $\sqrt{(5–√5)/10}$),
 * (±(1+$\sqrt{5}$)/4, $\sqrt{(5+√5)/40}$, $\sqrt{(5–√5)/10}$),
 * (0, $\sqrt{(5+√5)/10}$, $\sqrt{(5–√5)/10}$),
 * (±1/2, $\sqrt{(5+2√5)/20}$, –$\sqrt{(5–√5)/10}$),
 * (±(1+$\sqrt{5}$)/4, –$\sqrt{(5+√5)/40}$, –$\sqrt{(5–√5)/10}$),
 * (0, –$\sqrt{(5+√5)/10}$, –$\sqrt{(5–√5)/10}$),
 * (±1/2, ±$\sqrt{(5+2√5)}$/2, 0),
 * (±(3+$\sqrt{5}$)/4, ±$\sqrt{(5+√5)/8}$, 0),
 * (±(1+$\sqrt{5}$)/2, 0, 0).

Related polyhedra
A decagonal prism can be inserted between the two halves of the pentagonal orthobicupola to produce the elongated pentagonal gyrobicupola..