antiprism

In his 1619 book Harmonices Mundi, Johannes Kepler observed the existence of the infinite family of antiprisms.{{cite book|title=Harmonices Mundi|first=Johannes|last=Kepler|author-link=Johannes Kepler|title-link=Harmonices Mundi|year=1619|contribution=Book II, Definition X|language=la|page=49|contribution-url=https://archive.org/details/ioanniskepplerih00kepl/page/n65}} See also [https://archive.org/details/ioanniskepplerih00kepl/page/n75 illustration A], of a heptagonal antiprism. 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 the net of a hexagonal antiprism has been attributed to Hieronymus Andreae, who died in 1556.{{cite journal

| last1 = Schreiber | first1 = Peter

| last2 = Fischer | first2 = Gisela | author2-link = Gisela Fischer

| last3 = Sternath | first3 = Maria Luise

| date = July 2008

| issue = 4

| journal = Archive for History of Exact Sciences

| jstor = 41134285

| pages = 457–467

| title = New light on the rediscovery of the Archimedean solids during the Renaissance

| volume = 62| doi = 10.1007/s00407-008-0024-z

}}

The German form of the word "antiprism" was used for these shapes in the 19th century; Karl Heinze credits its introduction to {{ill|Theodor Wittstein|de}}.{{cite book|title=Genetische Stereometrie|first=Karl|last=Heinze|editor-first=Franz|editor-last=Lucke|publisher=B. G. Teubner|year=1886|language=de|page=14|url=https://books.google.com/books?id=rZALAAAAYAAJ&pg=PA14}} Although the English "anti-prism" had been used earlier for an optical prism used to cancel the effects of a primary optical element,{{cite journal

| last = Smyth | first = Piazzi

| doi = 10.1017/s0080456800029112

| issue = 1

| journal = Transactions of the Royal Society of Edinburgh

| pages = 419–425

| title = XVII. On the Constitution of the Lines forming the Low-Temperature Spectrum of Oxygen

| volume = 30

| year = 1881}} the first use of "antiprism" in English in its geometric sense appears to be in the early 20th century in the works of H. S. M. Coxeter.{{cite journal

| last = Coxeter | first = H. S. M.

| date = January 1928

| doi = 10.1017/s0305004100011786

| issue = 1

| journal = Mathematical Proceedings of the Cambridge Philosophical Society

| pages = 1–9

| title = The pure Archimedean polytopes in six and seven dimensions

| volume = 24| bibcode = 1928PCPS...24....1C

}}

For an antiprism with Regular polygon bases, one usually considers the case where these two copies are twisted by an angle of {{math|{{sfrac|180|n}}}} degrees. The axis of a regular polygon is the line perpendicular to the polygon plane and lying in the polygon centre.

For an antiprism with congruent regular {{mvar|n}}-gon bases, twisted by an angle of {{math|{{sfrac|180|n}}}} degrees, more regularity is obtained if the bases have the same axis: are coaxial; i.e. (for non-coplanar bases): if the line connecting the base centers is perpendicular to the base planes. Then the antiprism is called a right antiprism, and its {{math|2n}} side faces are isosceles triangles.{{cite book

| last1 = Alsina | first1 = Claudi

| last2 = Nelsen | first2 = Roger B.

| year = 2015

| title = A Mathematical Space Odyssey: Solid Geometry in the 21st Century

| publisher = Mathematical Association of America

| url = https://books.google.com/books?id=FEl2CgAAQBAJ&pg=PA87

| page = 87

| isbn = 978-1-61444-216-5

| volume = 50

}}

The symmetry group of a right {{mvar|n}}-antiprism is {{math|1=D{{sub|nd}}}} of order {{math|4n}} known as an antiprismatic symmetry, because it could be obtained by rotation of the bottom half of a prism by $\backslash pi/n$ in relation to the top half. A concave polyhedron created in this way would have this symmetry group, hence prefix "anti" before "prismatic".{{cite book

| last1 = Flusser | first1 = J.

| last2 = Suk | first2 = T.

| last3 = Zitofa | first3 = B.

| year = 2017

| title = 2D and 3D Image Analysis by Moments

| publisher = John Wiley & Sons

| isbn = 978-1-119-03935-8

| url = https://books.google.com/books?id=jwKLDQAAQBAJ&pg=PA126

| page = 126

}} There are two exceptions having groups different than {{math|D_nd}}:

• {{math|1=n = 2}}: the regular tetrahedron, which has the larger symmetry group {{math|T_d}} of order List of spherical symmetry groups#Polyhedral symmetry, which has three versions of {{math|D_2d}} as subgroups;
• {{math|1=n = 3}}: the regular octahedron, which has the larger symmetry group {{math|O_h}} of order {{math|48}}, which has four versions of {{math|D_3d}} as subgroups.{{cite book

| last1 = O'Keeffe | first1 = Michael

| last2 = Hyde | first2 = Bruce G.

| title = Crystal Structures: Patterns and Symmetry

| year = 2020

| url = https://books.google.com/books?id=_MjPDwAAQBAJ&pg=PA140

| page = 140

| publisher = Dover Publications

| isbn = 978-0-486-83654-6

}}

If a right 2- or 3-antiprism is not uniform, then its symmetry group is {{math|D_2d}} or {{math|D_3d}} as usual.

The symmetry group contains inversion if and only if {{mvar|n}} is odd.

The rotation group is {{math|D_n}} of order {{math|2n}}, except in the cases of:

• {{math|1=n = 2}}: the regular tetrahedron, which has the larger rotation group {{math|T}} of order {{math|12}}, which has only one subgroup {{math|D₂}};
• {{math|1=n = 3}}: the regular octahedron, which has the larger rotation group {{math|O}} of order {{math|24}}, which has four versions of {{math|D₃}} as subgroups.

If a right 2- or 3-antiprism is not uniform, then its rotation group is {{math|D₂}} or {{math|D₃}} as usual.

The right {{mvar|n}}-antiprisms have congruent regular {{mvar|n}}-gon bases and congruent isosceles triangle side faces, thus have the same (dihedral) symmetry group as the uniform {{mvar|n}}-antiprism, for {{math|n ≥ 4}}.

A uniform {{mvar|n}}-antiprism has two congruent regular {{mvar|n}}-gons as base faces, and {{math|2n}} equilateral triangles as side faces. As do uniform prisms, the uniform antiprisms form an infinite class of vertex-transitive polyhedra. For {{math|n {{=}} 2}}, one has the digonal antiprism (degenerate antiprism), which is visually identical to the regular tetrahedron; for {{math|n {{=}} 3}}, the regular octahedron is a triangular antiprism (non-degenerate antiprism).

{{UniformAntiprisms}}

The Schlegel diagrams of these semiregular antiprisms are as follows:

Cartesian coordinates for the vertices of a right {{mvar|n}}-antiprism (i.e. with regular {{mvar|n}}-gon bases and {{math|2n}} isosceles triangle side faces, circumradius of the bases equal to 1) are:

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

where {{math|0 ≤ k ≤ 2n – 1}};

if the {{mvar|n}}-antiprism is uniform (i.e. if the triangles are equilateral), then:

$$2h^2\; =\; \backslash cos\backslash frac\{\backslash pi\}\{n\}\; -\; \backslash cos\backslash frac\{2\backslash pi\}\{n\}.$$

Let {{mvar|a}} be the edge-length of a uniform {{mvar|n}}-gonal antiprism; then the volume is:

$$V\; =\; \backslash frac\{n\backslash sqrt\{4\backslash cos^2\backslash frac\{\backslash pi\}\{2n\}-1\}\backslash sin\; \backslash frac\{3\backslash pi\}\{2n\}\; \}\{12\backslash sin^2\backslash frac\{\backslash pi\}\{n\}\}~a^3,$$

and the surface area is:

$$A\; =\; \backslash frac\{n\}\{2\}\; \backslash left(\; \backslash cot\backslash frac\{\backslash pi\}\{n\}\; +\; \backslash sqrt\{3\}\; \backslash right)\; a^2.$$

Furthermore, the volume of a regular #Right antiprism with side length of its bases {{mvar|l}} and height {{mvar|h}} is given by:{{sfnp|Alsina|Nelsen|2015|p=[http://books.google.com/books?id=FEl2CgAAQBAJ&pg=PA88 88]}}

$$V\; =\; \backslash frac\{nhl^2\}\{12\}\; \backslash left(\; \backslash csc\backslash frac\{\backslash pi\}\{n\}\; +\; 2\backslash cot\backslash frac\{\backslash pi\}\{n\}\backslash right).$$

The circumradius of the horizontal circumcircle of the regular $n$-gon at the base is

:$$

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

The vertices at the base are at

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

the vertices at the top are at

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

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

are at

:$\backslash left(\backslash begin\{array\}\{c\}$

\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\le z\le h, m=0..n-1

and at

:$\backslash left(\backslash begin\{array\}\{c\}$

\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\le z\le h, m=0..n-1.

By building the sums of the squares of the $x$ and $y$ coordinates in one of the previous two vectors,

the squared circumradius of this section at altitude $z$ 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 $0\backslash le\; z\backslash le\; h$ above the base is a $2n$-gon (truncated $n$-gon)

with $n$ sides of length $l\_1(z)=l(1-z/h)$ alternating with $n$ 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 $n$ isoceless triangles of edges $R(z),R(z)$ and $l\_1$ (semiperimeter $R(z)+l\_1(z)/2$)

plus $n$

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 right {{mvar|n}}-gonal prism with the same {{mvar|l}} and {{mvar|h}} is:

$$V\_\{\backslash mathrm\{prism\}\}=\backslash frac\{nhl^2\}\{4\}\; \backslash cot\backslash frac\{\backslash pi\}\{n\}$$

which is smaller than that of an antiprism.

Four-dimensional antiprisms can be defined as having two dual polyhedra as parallel opposite faces, so that each three-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 its canonical polyhedron and its polar dual.{{cite journal | last = Grünbaum | first = Branko | author-link = Branko Grünbaum | issue = 2 | journal = Geombinatorics | mr = 2298896 | pages = 69–78 | title = Are prisms and antiprisms really boring? (Part 3) | url = https://faculty.washington.edu/moishe/branko/BG256.Prisms%20and%20antiprisms.%20Part%203.pdf | volume = 15 | year = 2005}} However, there exist four-dimensional polychora that cannot be combined with their duals to form five-dimensional antiprisms.{{cite journal | last = Dobbins | first = Michael Gene | doi = 10.1007/s00454-017-9874-y | issue = 4 | journal = Discrete & Computational Geometry | mr = 3639611 | pages = 966–984 | title = Antiprismlessness, or: reducing combinatorial equivalence to projective equivalence in realizability problems for polytopes | volume = 57 | year = 2017}}

{{unreferenced section|date=April 2025}}

{{Further|Prismatic uniform polyhedron}}

File:Antiprisms.pdf). For example, the icosaenneagrammic crossed antiprism ({{math|29/q}}) with the greatest {{math|q}}, such that it can be uniform, has {{math|q {{=}} 19}} and is depicted at the bottom right corner of the image. For {{math|q ≥ 20}} up to {{math|28}} the crossed antiprism cannot be uniform.
Note: Octagrammic crossed antiprism (8/5) is missing.]]

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

A right star {{math|n}}-antiprism has two congruent coaxial regular convex or star polygon base faces, and {{math|2n}} isosceles triangle side faces.

Any star antiprism with regular convex or star polygon bases can be made a right 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 {{mset|7/5}}-gon bases (vertex configuration: 3.3.3.7/5) cannot be uniform.

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

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

: {{math|{{sfrac|p|2}} < q < {{sfrac|2|3}} p}} and {{math|1=(p,q) = 1.}}

Examples:

• {{mvar|p}} = 3: 2 ≤ {{mvar|q}} ≤ 1 – a uniform triangular crossed antiprism does not exist.
• {{mvar|p}} = 5: 3 ≤ {{mvar|q}} ≤ 3 – one antiprism of the type (5/3) can be uniform.
• {{mvar|p}} = 29: 15 ≤ {{mvar|q}} ≤ 19 – there are five possibilities (15 thru 19) shown in the rightmost column, below the (29/1) convex antiprism, on the image above.
• {{mvar|p}} = 15: 8 ≤ {{mvar|q}} ≤ 9 – antiprism with {{mvar|q}} = 8 is a solution, but {{mvar|q}} = 9 must be rejected, as (15,9) = 3 and {{sfrac|15|9}} = {{sfrac|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.

• Antiprism graph, graph of an antiprism
• Grand antiprism, a four-dimensional polytope
• Skew polygon, a three-dimensional polygon whose convex hull is an antiprism

{{reflist}}

• {{cite book | author= Anthony Pugh | year= 1976 | title= Polyhedra: A visual approach | publisher= University of California Press Berkeley | location= California | isbn= 0-520-03056-7 }} Chapter 2: Archimedean polyhedra, prisms and antiprisms

• {{Commons category-inline}}
• {{MathWorld|urlname=Antiprism|title=Antiprism}}

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