Editing Antiprism (section)

==Special cases==
=== Right antiprism ===
For an antiprism with [[Regular polygon|regular {{mvar|n}}-gon]] 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 [[Congruence (geometry)|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-[[Coplanarity|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|2''n''}} side faces are [[isosceles triangle]]s.<ref name=oh>{{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
}}</ref>

The [[point groups in three dimensions|symmetry group]] of a right {{mvar|n}}-antiprism is {{math|1=D{{sub|''n''d}}}} of order {{math|4''n''}} known as an [[antiprismatic symmetry]], because it could be obtained by rotation of the bottom half of a prism by <math> \pi/n </math> in relation to the top half. A concave polyhedron created in this way would have this symmetry group, hence prefix "anti" before "prismatic".<ref>{{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
}}</ref> There are two exceptions having groups different than {{math|D<sub>''n''d</sub>}}:
*{{math|1=''n'' = 2}}: the regular [[tetrahedron]], which has the larger symmetry group {{math|T<sub>d</sub>}} of order [[List of spherical symmetry groups#Polyhedral symmetry|{{math|24}}]], which has three versions of {{math|D<sub>2d</sub>}} as subgroups;
*{{math|1=''n'' = 3}}: the regular [[octahedron]], which has the larger symmetry group {{math|O<sub>h</sub>}} of order {{math|48}}, which has four versions of {{math|D<sub>3d</sub>}} as subgroups.<ref>{{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
}}</ref>
If a right 2- or 3-antiprism is not uniform, then its symmetry group is {{math|D<sub>2d</sub>}} or {{math|D<sub>3d</sub>}} as usual.<br />
The symmetry group contains [[Inversion in a point|inversion]] [[if and only if]] {{mvar|n}} is odd.

The [[Rotation group SO(3)|rotation group]] is {{math|D<sub>''n''</sub>}} of order {{math|2''n''}}, 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|D<sub>2</sub>}};
*{{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|D<sub>3</sub>}} as subgroups.
If a right 2- or 3-antiprism is not uniform, then its rotation group is {{math|D<sub>2</sub>}} or {{math|D<sub>3</sub>}} as usual.<br />
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}}.

=== Uniform antiprism ===
A '''[[Prismatic uniform polyhedron|uniform]] {{mvar|n}}-antiprism''' has two [[Congruence (geometry)|congruent]] [[Regular polygon|''regular'']] {{mvar|n}}-gons as base faces, and {{math|2''n''}} [[Equilateral triangle|''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).<ref name=oh/>

{{UniformAntiprisms}}

The [[Schlegel diagram]]s of these semiregular antiprisms are as follows:

{| class=wikitable
|- align=center
|[[File:Triangular antiprismatic graph.png|100px]]<BR>A3
|[[File:Square antiprismatic graph.png|100px]]<BR>A4
|[[File:Pentagonal antiprismatic graph.png|100px]]<BR>A5
|[[File:Hexagonal antiprismatic graph.png|100px]]<BR>A6
|[[File:Heptagonal antiprism graph.png|100px]]<BR>A7
|[[File:Octagonal antiprismatic graph.png|100px]]<BR>A8
|}

===Cartesian coordinates===
[[Cartesian coordinates]] for the vertices of a [[#Right antiprism|right]] {{mvar|n}}-antiprism (i.e. with regular {{mvar|n}}-gon bases and {{math|2''n''}} isosceles triangle side faces, circumradius of the bases equal to 1) are:
:<math>\left( \cos\frac{k\pi}{n}, \sin\frac{k\pi}{n}, (-1)^k h \right)</math>

where {{math|0 ≤ ''k'' ≤ 2''n'' – 1}};

if the {{mvar|n}}-antiprism is uniform (i.e. if the triangles are equilateral), then:
<math display=block>2h^2 = \cos\frac{\pi}{n} - \cos\frac{2\pi}{n}.</math>

===Volume and surface area===
Let {{mvar|a}} be the edge-length of a [[Uniform polyhedron|uniform]] {{mvar|n}}-gonal antiprism; then the volume is:
<math display=block>V = \frac{n\sqrt{4\cos^2\frac{\pi}{2n}-1}\sin \frac{3\pi}{2n} }{12\sin^2\frac{\pi}{n}}~a^3,</math>
and the surface area is:
<math display=block>A = \frac{n}{2} \left( \cot\frac{\pi}{n} + \sqrt{3} \right) a^2.</math>
Furthermore, the volume of a regular [[#Right antiprism|right {{mvar|n}}-gonal 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]}}
<math display=block>V = \frac{nhl^2}{12} \left( \csc\frac{\pi}{n} + 2\cot\frac{\pi}{n}\right).</math>

==== Derivation ====
The circumradius of the horizontal circumcircle of the regular <math>n</math>-gon at the base is
:<math>
R(0) = \frac{l}{2\sin\frac{\pi}{n}}.
</math>
The vertices at the base are at
:<math>\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;</math>
the vertices at the top are at
:<math>\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.</math>
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
:<math>\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\le z\le h, m=0..n-1</math>
and at
:<math>\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\le z\le h, m=0..n-1.</math>
By building the sums of the squares of the <math>x</math> and <math>y</math> coordinates in one of the previous two vectors,
the squared circumradius of this section at altitude <math>z</math> is
:<math>
R(z)^2 = \frac{R(0)^2}{h^2}[h^2-2hz+2z^2+2z(h-z)\cos\frac{\pi}{n}].
</math> 
The horizontal section at altitude <math>0\le z\le h</math> above the base is a <math>2n</math>-gon (truncated <math>n</math>-gon)
with <math>n</math> sides of length <math>l_1(z)=l(1-z/h)</math> alternating with <math>n</math> sides of length <math>l_2(z)=lz/h</math>.
(These are derived from the length of the difference of the previous two vectors.)
It can be dissected into <math>n</math> isoceless triangles of edges <math>R(z),R(z)</math> and <math>l_1</math> (semiperimeter <math>R(z)+l_1(z)/2</math>)
plus <math>n</math>
isoceless triangles of edges <math>R(z),R(z)</math> and <math>l_2(z)</math> (semiperimeter <math>R(z)+l_2(z)/2</math>).
According to Heron's formula the areas of these triangles are
:<math>
Q_1(z) = \frac{R(0)^2}{h^2} (h-z)\left[(h-z)\cos\frac{\pi}{n}+z\right] \sin\frac{\pi}{n}
</math>
and
:<math>
Q_2(z) = 
\frac{R(0)^2}{h^2} z\left[z\cos\frac{\pi}{n}+h-z\right] \sin\frac{\pi}{n}
. 
</math>
The area of the section is <math>n[Q_1(z)+Q_2(z)]</math>, and the volume is
:<math>
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}}
. 
</math>

The volume of a right {{mvar|n}}-gonal [[Prism (geometry)|prism]] with the same {{mvar|l}} and {{mvar|h}} is:
<math display=block>V_{\mathrm{prism}}=\frac{nhl^2}{4} \cot\frac{\pi}{n}</math>
which is smaller than that of an antiprism.