Antiprism

In his 1619 book Harmonices Mundi, Johannes Kepler observed the existence of the infinite family of antiprisms. 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. The German form of the word "antiprism" was used for these shapes in the 19th century; Karl Heinze credits its introduction to . Although the English "anti-prism" had been used earlier for an optical prism used to cancel the effects of a primary optical element, 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. ==Special cases==

Right antiprism For an antiprism with regular -gon bases, one usually considers the case where these two copies are twisted by an angle of 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 -gon bases, twisted by an angle of 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 side faces are isosceles triangles. The symmetry group of a right -antiprism is of order known as an antiprismatic 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 created in this way would have this symmetry group, hence prefix "anti" before "prismatic". There are two exceptions having groups different than : • : the regular tetrahedron, which has the larger symmetry group of order List of spherical symmetry groups#Polyhedral symmetry|, which has three versions of as subgroups; • : the regular octahedron, which has the larger symmetry group of order , which has four versions of as subgroups. If a right 2- or 3-antiprism is not uniform, then its symmetry group is or as usual. The symmetry group contains inversion if and only if is odd. The rotation group is of order , except in the cases of: • : the regular tetrahedron, which has the larger rotation group of order , which has only one subgroup ; • : the regular octahedron, which has the larger rotation group of order , which has four versions of as subgroups. If a right 2- or 3-antiprism is not uniform, then its rotation group is or as usual. The right -antiprisms have congruent regular -gon bases and congruent isosceles triangle side faces, thus have the same (dihedral) symmetry group as the uniform -antiprism, for . Cartesian coordinates for the vertices of a right -antiprism (i.e. with regular -gon bases and 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) where ; if the -antiprism is uniform (i.e. if the triangles are equilateral), then: 2h^2 = \cos\frac{\pi}{n} - \cos\frac{2\pi}{n}. Uniform antiprism A uniform -antiprism has two congruent regular -gons as base faces, and equilateral triangles as side faces. As do uniform prisms, the uniform antiprisms form an infinite class of vertex-transitive polyhedra. For , one has the digonal antiprism (degenerate antiprism), which is visually identical to the regular tetrahedron; for , the regular octahedron is a triangular antiprism (non-degenerate antiprism). The Schlegel diagrams of these semiregular antiprisms are as follows: == Properties ==

Volume and surface area Let be the edge-length of a uniform -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 regular right -gonal antiprism with side length of its bases and height is given by: V = \frac{nhl^2}{12} \left( \csc\frac{\pi}{n} + 2\cot\frac{\pi}{n}\right). Derivation 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 :\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 top are 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\le z\le 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\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\le z\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 -gonal prism with the same and is: V_{\mathrm{prism}}=\frac{nhl^2}{4} \cot\frac{\pi}{n} which is smaller than that of an antiprism. == Generalizations ==

In higher dimensions 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. However, there exist four-dimensional polychora that cannot be combined with their duals to form five-dimensional antiprisms. Self-crossing polyhedra 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: instead of ; example: (5/3) instead of (5/2). A right star -antiprism has two congruent coaxial regular convex or star polygon base faces, and 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). ). For example, the icosaenneagrammic crossed antiprism () with the greatest , such that it can be uniform, has and is depicted at the bottom right corner of the image. For up to the crossed antiprism cannot be uniform.Note: Octagrammic crossed antiprism (8/5) is missing.|500x500pxIn 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 -gon bases (vertex configuration: 3.3.3.7/5) cannot be uniform. Also, star antiprism compounds with regular star -gon bases can be constructed if and have common factors. Example: a star (10/4)-antiprism is the compound of two star (5/2)-antiprisms. The dual of a () antiprism with is a -trapezohedron, and the dual of a () antiprism with , i.e. crossed antiprism, is a () concave trapezohedron, where "concave" refers to 2-dimensional faces of the 3D solid. Number of uniform crossed antiprisms If the notation is used for an antiprism, then for the antiprism is crossed (by definition) and for is not. In this section all antiprisms are assumed to be non-degenerate, i.e. , . Also, the condition ( and are relatively prime) holds, as compounds are excluded from counting. The number of uniform crossed antiprisms for fixed can be determined using simple inequalities. The condition on possible is : and Examples: • = 3: 2 ≤ ≤ 1 – a uniform triangular crossed antiprism does not exist. • = 5: 3 ≤ ≤ 3 – one antiprism of the type (5/3) can be uniform. • = 29: 15 ≤ ≤ 19 – there are five possibilities (15 thru 19) shown in the rightmost column, below the (29/1) convex antiprism, on the image above. • = 15: 8 ≤ ≤ 9 – antiprism with = 8 is a solution, but = 9 must be rejected, as (15,9) = 3 and = . 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. == See also ==