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Antiprism

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Polyhedron with parallel bases connected by triangles

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Ingeometry, ann-gonal antiprism orn-antiprism is apolyhedron composed of twoparallel direct copies (not mirror images) of ann-sidedpolygon, connected by an alternating band of2ntriangles. They are represented by theConway notationAn.

Antiprisms are a subclass ofprismatoids, and are a (degenerate) type ofsnub polyhedron.

Antiprisms are similar toprisms, except that the bases are twisted relatively to each other, and that the side faces (connecting the bases) are2n triangles, rather thannquadrilaterals.

Thedual polyhedron of ann-gonal antiprism is ann-gonaltrapezohedron.

History

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In his 1619 bookHarmonices Mundi,Johannes Kepler observed the existence of the infinite family of antiprisms.^[1] This has conventionally been thought of as the first discovery of these shapes, but they may have been known earlier: an unsigned printing block for thenet of ahexagonal antiprism has been attributed toHieronymus Andreae, who died in 1556.^[2]

The German form of the word "antiprism" was used for these shapes in the 19th century; Karl Heinze credits its introduction toTheodor Wittstein [de].^[3] Although the English "anti-prism" had been used earlier for anoptical prism used to cancel the effects of a primary optical element,^[4] the first use of "antiprism" in English in its geometric sense appears to be in the early 20th century in the works ofH. S. M. Coxeter.^[5]

Special cases

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Right antiprism

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For an antiprism withregularn-gon bases, one usually considers the case where these two copies are twisted by an angle of⁠180/n⁠ degrees. The axis of a regular polygon is the lineperpendicular to the polygon plane and lying in the polygon centre.

For an antiprism withcongruent regularn-gon bases, twisted by an angle of⁠180/n⁠ degrees, more regularity is obtained if the bases have the same axis: arecoaxial; i.e. (for non-coplanar bases): if the line connecting the base centers is perpendicular to the base planes. Then the antiprism is called aright antiprism, and its2n side faces areisosceles triangles.^[6]

Thesymmetry group of a rightn-antiprism isD_nd of order4n known as anantiprismatic symmetry, because it could be obtained by rotation of the bottom half of a prism by $\pi /n$ in relation to the top half. A concave polyhedron obtained in this way would have this symmetry group, hence prefix "anti" before "prismatic".^[7] There are two exceptions having groups different thanD_nd:

n = 2: the regulartetrahedron, which has the larger symmetry groupT_d of order24 = 3 × (4 × 2), which has three versions ofD_2d as subgroups;
n = 3: the regularoctahedron, which has the larger symmetry groupO_h of order48 = 4 × (4 × 3), which has four versions ofD_3d as subgroups.^[8]

The symmetry group containsinversion if and only ifn is odd.

Therotation group isD_n of order2n, except in the cases of:

n = 2: the regular tetrahedron, which has the larger rotation groupT of order12 = 3 × (2 × 2), which has three versions ofD₂ as subgroups;
n = 3: the regular octahedron, which has the larger rotation groupO of order24 = 4 × (2 × 3), which has four versions ofD₃ as subgroups.

The rightn-antiprisms have congruent regularn-gon bases and congruent isosceles triangle side faces, thus have the same (dihedral) symmetry group as the uniformn-antiprism, forn ≥ 4.

Uniform antiprism

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Auniformn-antiprism has twocongruent regularn-gons as base faces, and2nequilateral triangles as side faces. As do uniform prisms, the uniform antiprisms form an infinite class of vertex-transitive polyhedra. Forn = 2, one has the digonal antiprism (degenerate antiprism), which is visually identical to the regulartetrahedron; forn = 3, the regularoctahedron is atriangular antiprism (non-degenerate antiprism).^[6]

Family ofuniformn-gonalantiprisms
v
t
e
Antiprism name	Digonal antiprism	(Trigonal) Triangular antiprism	(Tetragonal) Square antiprism	Pentagonal antiprism	Hexagonal antiprism	Heptagonal antiprism	...	Apeirogonal antiprism
Polyhedron image							...
Spherical tiling image							Plane tiling image
Vertex config.	2.3.3.3	3.3.3.3	4.3.3.3	5.3.3.3	6.3.3.3	7.3.3.3	...	∞.3.3.3

TheSchlegel diagrams of these semiregular antiprisms are as follows:

Cartesian coordinates

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Cartesian coordinates for the vertices of arightn-antiprism (i.e. with regularn-gon bases and2n isosceles triangle side faces, circumradius of the bases equal to 1) are:

\left(\cos {\frac {k\pi }{n}},\sin {\frac {k\pi }{n}},(-1)^{k}h\right)

where0 ≤k ≤ 2n – 1;

if then-antiprism is uniform (i.e. if the triangles are equilateral), then: $2h^{2}=\cos {\frac {\pi }{n}}-\cos {\frac {2\pi }{n}}.$

Volume and surface area

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Leta be the edge-length of auniformn-gonal antiprism; then the volume is: $V={\frac {n{\sqrt {4\cos ^{2}{\frac {\pi }{2n}}-1}}\sin {\frac {3\pi }{2n}}}{12\sin ^{2}{\frac {\pi }{n}}}}~a^{3},$ and the surface area is: $A={\frac {n}{2}}\left(\cot {\frac {\pi }{n}}+{\sqrt {3}}\right)a^{2}.$ Furthermore, the volume of a regularrightn-gonal antiprism with side length of its basesl and heighth is given by:^[9] $V={\frac {nhl^{2}}{12}}\left(\csc {\frac {\pi }{n}}+2\cot {\frac {\pi }{n}}\right).$

Derivation

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The circumradius of the horizontal circumcircle of the regular $n {\displaystyle n}$ -gon at the base is

R(0)={\frac {l}{2\sin {\frac {\pi }{n}}}}.

The vertices at the base are at

\left({\begin{array}{c}R(0)\cos {\frac {2\pi m}{n}}\\R(0)\sin {\frac {2\pi m}{n}}\\0\end{array}}\right),\quad m=0..n-1;

the vertices at the top are at

\left({\begin{array}{c}R(0)\cos {\frac {2\pi (m+1/2)}{n}}\\R(0)\sin {\frac {2\pi (m+1/2)}{n}}\\h\end{array}}\right),\quad m=0..n-1.

Via linear interpolation, points on the outer triangular edges of the antiprism that connect vertices at the bottom with vertices at the topare at

\left({\begin{array}{c}{\frac {R(0)}{h}}[(h-z)\cos {\frac {2\pi m}{n}}+z\cos {\frac {\pi (2m+1)}{n}}]\\{\frac {R(0)}{h}}[(h-z)\sin {\frac {2\pi m}{n}}+z\sin {\frac {\pi (2m+1)}{n}}]\\\\z\end{array}}\right),\quad 0\leq z\leq h,m=0..n-1

and at

\left({\begin{array}{c}{\frac {R(0)}{h}}[(h-z)\cos {\frac {2\pi (m+1)}{n}}+z\cos {\frac {\pi (2m+1)}{n}}]\\{\frac {R(0)}{h}}[(h-z)\sin {\frac {2\pi (m+1)}{n}}+z\sin {\frac {\pi (2m+1)}{n}}]\\\\z\end{array}}\right),\quad 0\leq z\leq h,m=0..n-1.

By building the sums of the squares of the $x {\displaystyle x}$ and $y {\displaystyle y}$ coordinates in one of the previous two vectors,the squared circumradius of this section at altitude $z {\displaystyle z}$ is

R(z)^{2}={\frac {R(0)^{2}}{h^{2}}}[h^{2}-2hz+2z^{2}+2z(h-z)\cos {\frac {\pi }{n}}].

The horizontal section at altitude $0\leq z\leq h$ above the base is a $2n$ -gon (truncated $n {\displaystyle n}$ -gon)with $n {\displaystyle n}$ sides of length $l_{1}(z)=l(1-z/h)$ alternating with $n {\displaystyle n}$ sides of length $l_{2}(z)=lz/h$ .(These are derived from the length of the difference of the previous two vectors.)It can be dissected into $n {\displaystyle n}$ isoceless triangles of edges $R(z),R(z)$ and $l_{1}$ (semiperimeter $R(z)+l_{1}(z)/2$ )plus $n {\displaystyle n}$ isoceless triangles of edges $R(z),R(z)$ and $l_{2}(z)$ (semiperimeter $R(z)+l_{2}(z)/2$ ).According to Heron's formula the areas of these triangles are

Q_{1}(z)={\frac {R(0)^{2}}{h^{2}}}(h-z)\left[(h-z)\cos {\frac {\pi }{n}}+z\right]\sin {\frac {\pi }{n}}

and

Q_{2}(z)={\frac {R(0)^{2}}{h^{2}}}z\left[z\cos {\frac {\pi }{n}}+h-z\right]\sin {\frac {\pi }{n}}.

The area of the section is $n[Q_{1}(z)+Q_{2}(z)]$ , and the volume is

V=n\int _{0}^{h}[Q_{1}(z)+Q_{2}(z)]dz={\frac {nh}{3}}R(0)^{2}\sin {\frac {\pi }{n}}(1+2\cos {\frac {\pi }{n}})={\frac {nh}{12}}l^{2}{\frac {1+2\cos {\frac {\pi }{n}}}{\sin {\frac {\pi }{n}}}}.

The volume of a rightn-gonalprism with the samel andh is: $V_{\mathrm {prism} }={\frac {nhl^{2}}{4}}\cot {\frac {\pi }{n}}$ which is smaller than that of an antiprism.

Generalizations

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In higher dimensions

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Four-dimensional antiprisms can be defined as having twodual polyhedra as parallel opposite faces, so that eachthree-dimensional face between them comes from two dual parts of the polyhedra: a vertex and a dual polygon, or two dual edges. Every three-dimensional convex polyhedron is combinatorially equivalent to one of the two opposite faces of a four-dimensional antiprism, constructed from itscanonical polyhedron and its polar dual.^[10] However, there exist four-dimensional polychora that cannot be combined with their duals to form five-dimensional antiprisms.^[11]

Self-crossing polyhedra

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Further information:Prismatic uniform polyhedron

3/2-antiprism nonuniform	5/4-antiprism nonuniform	5/2-antiprism	5/3-antiprism
9/2-antiprism	9/4-antiprism	9/5-antiprism

All the non-star and star uniform antiprisms up to 15 sides, together with those of a 29-gon (oricosaenneagon). For example, the icosaenneagrammic crossed antiprism (29/q) with the greatestq, such that it can be uniform, hasq = 19 and is depicted at the bottom right corner of the image. Forq ≥ 20 up to28 the crossed antiprism cannot be uniform.
Note: Octagrammic crossed antiprism (8/5) is missing.

Uniform star antiprisms are named by theirstar polygon bases,{p/q}, and exist in prograde and in retrograde (crossed) solutions. Crossed forms have intersectingvertex figures, and are denoted by "inverted" fractions:p/(p –q) instead ofp/q; example: (5/3) instead of (5/2).

Aright starn-antiprism has twocongruent coaxial regular convex orstar polygon base faces, and2nisosceles triangle side faces.

Any star antiprism withregular convex or star polygon bases can be made aright star antiprism (by translating and/or twisting one of its bases, if necessary).

In the retrograde forms, but not in the prograde forms, the triangles joining the convex or star bases intersect the axis of rotational symmetry. Thus:

Retrograde star antiprisms with regular convex polygon bases cannot have all equal edge lengths, and so cannot be uniform. "Exception": a retrograde star antiprism with equilateral triangle bases (vertex configuration: 3.3/2.3.3) can be uniform; but then, it has the appearance of an equilateral triangle: it is a degenerate star polyhedron.

Similarly, some retrograde star antiprisms with regular star polygon bases cannot have all equal edge lengths, and so cannot be uniform. Example: a retrograde star antiprism with regular star {7/5}-gon bases (vertex configuration: 3.3.3.7/5) cannot be uniform.

Also, star antiprism compounds with regular star{p/q}-gon bases can be constructed ifp andq have common factors. Example: a star (10/4)-antiprism is the compound of two star (5/2)-antiprisms.

Number of uniform crossed antiprisms

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If the notation(p/q) is used for an antiprism, then forq >p/2 the antiprism is crossed (by definition) and forq <p/2 is not. In this section all antiprisms are assumed to be non-degenerate, i.e.p ≥ 3,q ≠p/2. Also, the condition(p,q) = 1 (p andq are relatively prime) holds, as compounds are exluded from counting. The number of uniform crossed antiprisms for fixedp can be determined using simple inequalities. The condition on possibleq is

⁠p/2⁠ <q <⁠2/3⁠p and(p,q) = 1.

Examples:

p = 3:⁠p/2⁠ = 1.5 and⁠2/3⁠p = 2, so 2 ≤q ≤ 1 – a uniform triangular crossed antiprism does not exist.
p = 5:⁠p/2⁠ = 2.5 and⁠2/3⁠p =⁠10/3⁠ =⁠3+1/3⁠, so 3 ≤q ≤ 3 – one antiprism of the type (5/3) can be uniform.
p = 29:⁠p/2⁠ = 14.5 and⁠2/3⁠p =⁠58/3⁠ =⁠19+1/3⁠, 15 ≤q ≤ 19 – there are five possibilities shown in the rightmost column, below the (29/1) convex antiprism, on the image above.
p = 15:⁠p/2⁠ = 7.5 and⁠2/3⁠p = 10, 8 ≤q ≤ 9 – antiprism withq = 8 is a solution, butq = 9 must be rejected, as (15,9) = 3 and⁠15/9⁠ =⁠5/3⁠. The antiprism (15/9) is a compound of three antiprisms (5/3). Since 9 satisfies the inequalities, the compound can be uniform, and if it is, then its parts must be. Indeed, the antiprism (5/3) can be uniform by example 2.

In the first column of the following table, the symbols are Schoenflies, Coxeter, and orbifold notation, in this order.

Star (p/q)-antiprisms by symmetry, forp ≤ 12
Symmetry group	Uniform stars			Right stars
D_3h [2,3] (2*3)				3.3/2.3.3 Crossed triangular antiprism
D_4d [2⁺,8] (2*4)				3.3/2.3.4 Crossed square antiprism
D_5h [2,5] (*225)	3.3.3.5/2 Pentagrammic antiprism			3.3/2.3.5 Crossed pentagonal antiprism
D_5d [2⁺,10] (2*5)	3.3.3.5/3 Pentagrammic crossed-antiprism
D_6d [2⁺,12] (2*6)				3.3/2.3.6 Crossed hexagonal antiprism
D_7h [2,7] (*227)	3.3.3.7/2 Heptagrammic antiprism (7/2)	3.3.3.7/4 Heptagrammic crossed antiprism (7/4)
D_7d [2⁺,14] (2*7)	3.3.3.7/3 Heptagrammic antiprism (7/3)
D_8d [2⁺,16] (2*8)	3.3.3.8/3 Octagrammic antiprism	3.3.3.8/5 Octagrammic crossed-antiprism
D_9h [2,9] (*229)	3.3.3.9/2 Enneagrammic antiprism (9/2)	3.3.3.9/4 Enneagrammic antiprism (9/4)
D_9d [2⁺,18] (2*9)	3.3.3.9/5 Enneagrammic crossed-antiprism
D_10d [2⁺,20] (2*10)	3.3.3.10/3 Decagrammic antiprism
D_11h [2,11] (*2.2.11)	3.3.3.11/2 Undecagrammic (11/2)	3.3.3.11/4 Undecagrammic (11/4)	3.3.3.11/6 Undecagrammic crossed (11/6)
D_11d [2⁺,22] (2*11)	3.3.3.11/3 Undecagrammic (11/3)	3.3.3.11/5 Undecagrammic (11/5)	3.3.3.11/7 Undecagrammic crossed (11/7)
D_12d [2⁺,24] (2*12)	3.3.3.12/5 Dodecagrammic	3.3.3.12/7 Dodecagrammic crossed
...	...

References

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^Kepler, Johannes (1619)."Book II, Definition X".Harmonices Mundi (in Latin). p. 49. See alsoillustration A, of a heptagonal antiprism.
^Schreiber, Peter;Fischer, Gisela; Sternath, Maria Luise (July 2008). "New light on the rediscovery of the Archimedean solids during the Renaissance".Archive for History of Exact Sciences.62 (4):457–467.doi:10.1007/s00407-008-0024-z.JSTOR 41134285.
^Heinze, Karl (1886). Lucke, Franz (ed.).Genetische Stereometrie (in German). B. G. Teubner. p. 14.
^Smyth, Piazzi (1881). "XVII. On the Constitution of the Lines forming the Low-Temperature Spectrum of Oxygen".Transactions of the Royal Society of Edinburgh.30 (1):419–425.doi:10.1017/s0080456800029112.
^Coxeter, H. S. M. (January 1928). "The pure Archimedean polytopes in six and seven dimensions".Mathematical Proceedings of the Cambridge Philosophical Society.24 (1): 1–9.Bibcode:1928PCPS...24....1C.doi:10.1017/s0305004100011786.
^^a ^bAlsina, Claudi; Nelsen, Roger B. (2015).A Mathematical Space Odyssey: Solid Geometry in the 21st Century. Vol. 50.Mathematical Association of America. p. 87.ISBN 978-1-61444-216-5.
^Flusser, J.; Suk, T.; Zitofa, B. (2017).2D and 3D Image Analysis by Moments.John Wiley & Sons. p. 126.ISBN 978-1-119-03935-8.
^O'Keeffe, Michael; Hyde, Bruce G. (2020).Crystal Structures: Patterns and Symmetry.Dover Publications. p. 140.ISBN 978-0-486-83654-6.
^Alsina & Nelsen (2015), p. 88.
^Grünbaum, Branko (2005)."Are prisms and antiprisms really boring? (Part 3)"(PDF).Geombinatorics.15 (2):69–78.MR 2298896.
^Dobbins, Michael Gene (2017). "Antiprismlessness, or: reducing combinatorial equivalence to projective equivalence in realizability problems for polytopes".Discrete & Computational Geometry.57 (4):966–984.doi:10.1007/s00454-017-9874-y.MR 3639611.