Pentakis dodecahedron

3D view of a pentakis dodecahedron ( animation )

The pentakis dodecahedron is a convex polyhedron , which is composed of 60 isosceles triangles and belongs to the Catalan solids . It is dual to the truncated icosahedron and has 32 vertices and 90 edges. The name is from the Greek words πεντάκις ( pentakis , five times) and δωδεκάεδρον ( dodekaedron , dodecahedron) together.

Emergence

The basic body is the dodecahedron with side length , on the 12 boundary surfaces of which a pyramid with a pentagonal base and the side length is placed. A pentakis dodecahedron arises from this construction if and only if the following condition is met: ${\displaystyle a}$ ${\displaystyle b}$

${\displaystyle {\frac {a}{10}}{\sqrt {50+10{\sqrt {5}}}}<b<{\frac {a}{4}}{\sqrt {10+2{\sqrt {5}}}}}$

• For the aforementioned minimum value of , the pyramids on top have the height 0, so that only the dodecahedron with the edge length remains. ${\displaystyle b}$ ${\displaystyle a}$
• The special pentakis dodecahedron with equal face angles is created when is. ${\displaystyle b={\tfrac {3}{38}}\,a\,(9+{\sqrt {5}})\approx 0{,}887058\cdot a}$
• Takes the above maximum value, the degenerate pentakis dodecahedron to a rhombic triacontahedron with an edge length . ${\displaystyle b}$ ${\displaystyle b}$
• If the maximum value is exceeded , the polyhedron is no longer convex and finally degenerates into a dodecahedron star . ${\displaystyle b}$ ${\displaystyle b={\frac {a}{2}}\left(1+{\sqrt {5}}\right)}$

Formulas

General

Sizes of a pentakis dodecahedron with edge lengths a , b
volume	${\displaystyle V={\frac {a^{2}}{4}}\left(a\left(15+7{\sqrt {5}}\right)+2{\sqrt {\left(10+4{\sqrt {5}}\right)\left(10b^{2}-a^{2}(5+{\sqrt {5}})\right)}}\right)}$
Surface area	${\displaystyle A_{O}=15a\,{\sqrt {4b^{2}-a^{2}}}}$
Pyramid height	${\displaystyle k={\sqrt {b^{2}-{\frac {a^{2}}{10}}\left(5+{\sqrt {5}}\right)}}}$
Inc sphere radius	${\displaystyle \rho \,={\frac {a\,{\sqrt {(14+6{\sqrt {5}})(a+2b)^{2}-4(4b^{2}-a^{2})}}}{4a+8b}}}$
Dihedral angle  (over edge a )	${\displaystyle \cos \,\alpha _{1}={\frac {4b^{2}{\sqrt {5}}-a^{2}(4+3{\sqrt {5}})-4a{\sqrt {4b^{2}(5+2{\sqrt {5}})-2a^{2}(7+3{\sqrt {5}})}}}{5(4b^{2}-a^{2})}}}$
Dihedral angle  (over edge b )	${\displaystyle \cos \,\alpha _{2}={\frac {b^{2}(1-{\sqrt {5}})-a^{2}}{4b^{2}-a^{2}}}}$

Special

Network of Pentakisdodekaeders

Sizes of a pentakis dodecahedron with edge length a
volume	${\displaystyle V={\frac {15}{76}}\,a^{3}(23+11{\sqrt {5}})}$
Surface area	${\displaystyle A_{O}={\frac {15}{19}}\,a^{2}{\sqrt {413+162{\sqrt {5}}}}}$
Pyramid height	${\displaystyle k={\frac {a}{19}}{\sqrt {\frac {65+22{\sqrt {5}}}{5}}}}$
Inc sphere radius	${\displaystyle \rho ={\frac {3}{2}}\,a\,{\sqrt {\frac {81+35{\sqrt {5}}}{218}}}}$
Edge ball radius	${\displaystyle r={\frac {a}{4}}\left(3+{\sqrt {5}}\right)}$
Face angle  ≈ 156 ° 43 ′ 7 ″	${\displaystyle \cos \,\alpha ={\frac {-1}{109}}\,(80+9{\sqrt {5}})}$

Remarks

1. ↑ ${\displaystyle {\tfrac {a}{10}}{\sqrt {50+10{\sqrt {5}}}}<b<{\tfrac {a}{4}}{\sqrt {10+2{\sqrt {5}}}}}$
2. ↑ ; ${\displaystyle b={\tfrac {3}{38}}\,a\,(9+{\sqrt {5}})}$ ${\displaystyle a>b}$

Web links

Commons : Pentakis dodecahedron  - collection of images, videos and audio files

• Eric W. Weisstein : Pentakis dodecahedron . In: MathWorld (English).