Pentagonal-hendecagonal duoprism Rank 4 Type Uniform Notation Bowers style acronym P ahen dip Coxeter diagram x5o x11o ( Elements Cells 11 pentagonal prisms , 5 hendecagonal prisms Faces 55 squares , 11 pentagons , 5 hendecagons Edges 55+55 Vertices 55 Vertex figure Digonal disphenoid , edge lengths (1+√5 )/2 (base 1), 2cos(π/11) (base 2), and √2 (sides)Measures (edge length 1) Circumradius
5
+
5
10
+
1
4
sin
2
π
11
≈
1.96807
{\displaystyle {\sqrt {{\frac {5+{\sqrt {5}}}{10}}+{\frac {1}{4\sin ^{2}{\frac {\pi }{11}}}}}}\approx 1.96807}
Hypervolume
11
25
+
10
5
16
tan
π
11
≈
16.11337
{\displaystyle {\frac {11{\sqrt {25+10{\sqrt {5}}}}}{16\tan {\frac {\pi }{11}}}}\approx 16.11337}
Dichoral angles Pip–5–pip:
9
π
11
≈
147.27273
∘
{\displaystyle {\frac {9\pi }{11}}\approx 147.27273^{\circ }}
Henp–11–henp: 108° Pip–4–henp: 90° Central density 1 Number of external pieces 16 Level of complexity 6 Related polytopes Army Pahendip Regiment Pahendip Dual Pentagonal-hendecagonal duotegum Conjugates Pentagonal-small hendecagrammic duoprism , Pentagonal-hendecagrammic duoprism , Pentagonal-great hendecagrammic duoprism , Pentagonal-grand hendecagrammic duoprism , Pentagrammic-hendecagonal duoprism , Pentagrammic-small hendecagrammic duoprism , Pentagrammic-hendecagrammic duoprism , Pentagrammic-great hendecagrammic duoprism , Pentagrammic-grand hendecagrammic duoprism Abstract & topological properties Flag count1320 Euler characteristic 0 Orientable Yes Properties Symmetry H2 ×I2 (11) , order 220Flag orbits 6 Convex Yes Nature Tame
The pentagonal-hendecagonal duoprism or pahendip , also known as the 5-11 duoprism , is a uniform duoprism that consists of 5 hendecagonal prisms and 11 pentagonal prisms , with two of each joining at each vertex.
The coordinates of a pentagonal-hendecagonal duoprism, centered at the origin and with edge length 2sin(π/11), are given by:
(
±
sin
π
11
,
−
5
+
2
5
5
sin
π
11
,
1
,
0
)
{\displaystyle \left(\pm \sin {\frac {\pi }{11}},-{\sqrt {\frac {5+2{\sqrt {5}}}{5}}}\sin {\frac {\pi }{11}},1,0\right)}
(
±
sin
π
11
,
−
5
+
2
5
5
sin
π
11
,
cos
(
j
π
11
)
,
±
sin
(
j
π
11
)
)
{\displaystyle \left(\pm \sin {\frac {\pi }{11}},-{\sqrt {\frac {5+2{\sqrt {5}}}{5}}}\sin {\frac {\pi }{11}},\cos \left({\frac {j\pi }{11}}\right),\pm \sin \left({\frac {j\pi }{11}}\right)\right)}
(
±
1
+
5
2
sin
π
11
,
5
−
5
10
sin
π
11
,
1
,
0
)
{\displaystyle \left(\pm {\frac {1+{\sqrt {5}}}{2}}\sin {\frac {\pi }{11}},{\sqrt {\frac {5-{\sqrt {5}}}{10}}}\sin {\frac {\pi }{11}},1,0\right)}
(
±
1
+
5
2
sin
π
11
,
5
−
5
10
sin
π
11
,
cos
(
j
π
11
)
,
±
sin
(
j
π
11
)
)
{\displaystyle \left(\pm {\frac {1+{\sqrt {5}}}{2}}\sin {\frac {\pi }{11}},{\sqrt {\frac {5-{\sqrt {5}}}{10}}}\sin {\frac {\pi }{11}},\cos \left({\frac {j\pi }{11}}\right),\pm \sin \left({\frac {j\pi }{11}}\right)\right)}
(
0
,
2
5
+
5
10
sin
π
11
,
1
,
0
)
{\displaystyle \left(0,2{\sqrt {\frac {5+{\sqrt {5}}}{10}}}\sin {\frac {\pi }{11}},1,0\right)}
(
0
,
2
5
+
5
10
sin
π
11
,
cos
(
j
π
11
)
,
±
sin
(
j
π
11
)
)
{\displaystyle \left(0,2{\sqrt {\frac {5+{\sqrt {5}}}{10}}}\sin {\frac {\pi }{11}},\cos \left({\frac {j\pi }{11}}\right),\pm \sin \left({\frac {j\pi }{11}}\right)\right)}
where j = 2, 4, 6, 8, 10.
