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Antiprism


A general n-gonal antiprism is a polyhedron consisting of identical top and bottom n-gonal faces whose periphery is bounded by a band of 2n triangles with alternating up-down orientations.

If the top and bottom faces are regular n-gons displaced relative to one another in the direction perpendicular to the plane of the polygons and rotated relative to one another by an angle of 180 degrees/n degrees, then the antiprism is known as a right antiprism and its faces are equilateral triangles.

Antiprism03Antiprism04Antiprism05Antiprism06
Antiprism07Antiprism08Antiprism09Antiprism10
AntiprismNets

A uniform or equilateral antiprism, sometimes simply called an "antiprism" (e.g., Cromwell 1997, p. 85) is a semiregular polyhedron constructed from two regular n-gons and 2n equilateral triangles, where the n-gons are rotated by an angle 180 degrees/n with respect to each other and vertically separated by such a height that the triangular edges connecting top and bottom n-gons have the same length as the n-gon sides. Such antiprisms have the highest degree of symmetry and their nets are particularly simple, consisting of two n-gons on top and bottom, separated by a ribbon of 2n equilateral triangles, with the two n-gons being offset by one ribbon segment.

The duals of the antiprisms are the trapezohedra.

The graph corresponding to the skeleton of an antiprism is known, not surprisingly, as an antiprism graph.

The sagitta of a regular n-gon of side length a has length

 s=1/2atan(pi/(2n)).
(1)

Let d be the length of a lateral edge when the top and bottom bases of a right antiprism separated by a distance h, then

 s^2+(1/2a)^2+h^2=d^2,
(2)

so

 d=1/2sqrt(4h^2+a^2sec^2(pi/(2n))).
(3)

For an equilateral antiprism d=a, so solving for h gives

 h=sqrt(1-1/4sec^2(pi/(2n)))a.
(4)

Consider a unit equilateral antiprism of height h and base circumradius R. The circumradius R_(circ) is then given by

R_(circ)=sqrt((1/2h)^2+R^2)
(5)
=1/4sqrt(4+csc^2(pi/(2n)))a,
(6)

where

 R=1/2csc(pi/n)a
(7)

is the circumradius of one of the bases.

The regular tetrahedron can be considered a degenerate 2-equilateral antiprism, and the 3-equilateral antiprism of height sqrt(6)a/3 (for side length a) is simply the regular octahedron. The first few heights h_n producing equilateral antiprisms for n=3, 4, ... are then

h_3=1/3sqrt(6)a
(8)
h_4=2^(-1/4)a
(9)
h_5=sqrt(1/(10)(5+sqrt(5)))a
(10)
h_6=sqrt(sqrt(3)-1)a
(11)
h_8=sqrt(sqrt(5+7/2sqrt(2))-1-sqrt(2))a.
(12)

The surface area of a right n-gonal antiprism is

S=2A_(n-gon)+2nA_Delta
(13)
=2[1/4na^2cot(pi/n)]+2n(1/2a)sqrt(s^2+h^2)
(14)
=1/2na[acot(pi/n)+2sqrt(h^2+1/4a^2tan^2(pi/(2n)))].
(15)

For an equilateral antiprism, this simplifies to

 S=1/2n[cot(pi/n)+sqrt(3)]a^2.
(16)

The surface area for equilateral antiprisms with n=3, 4, ... are then

S_3=2sqrt(3)a^2
(17)
S_4=2(1+sqrt(3))a^2
(18)
S_5=1/2(5sqrt(3)+sqrt(25+10sqrt(5)))a^2
(19)
S_6=6sqrt(3)a^2
(20)
S_8=4(1+sqrt(2)+sqrt(3))a^2.
(21)
AntiprismLengths

To find the volume of a right antiprism, label vertices as in the above figure. Then the vectors v_1 and v_2 are given by

v_1=(-s,1/2a,h)
(22)
v_2=(-s,-1/2a,h),
(23)

so the normal to one of the lateral facial planes is

 n=v_1xv_2=(ah,0,as),
(24)

and the unit normal is

 n^^=(v_1xv_2)/(|v_1xv_2|)=((ah)/(sqrt(a^2(h^2+s^2))),0,(as)/(sqrt(a^2(h^2+s^2)))).
(25)

The height of a pyramid with apex at the center and having the triangle determined by x_1 and x_2 as the base is then given by the projection of a vector from the origin to a point on the plane onto the normal,

h_(pyr)=u^^·(R-s,-1/2a,1/2h)
(26)
=u^^·(R-s,1/2a,1/2h)
(27)
=u^^·(R,0,1/2h)
(28)
=(a^2hcot(pi/(2n)))/(4sqrt(a^2[h^2+1/4a^2tan^2(pi/(2n))])).
(29)
AntiprismVolume

The total volume of the 2n pyramids having the lateral faces as bases is therefore

V_(pyr)=(2n)[1/3h_(pyr)(1/2asqrt(s^2+h^2))]
(30)
=1/(12)a^2hcot(pi/(2n)).
(31)

For the case of an equilateral antiprism, plug in h from above to obtain

 V_(pyr)=1/(12)ncot(pi/(2n))sqrt(1-1/4sec^2(pi/(2n)))a^3.
(32)

The two pyramids having the upper and lower surfaces as bases contribute a volume

V_(base)=2(1/3)(1/2h)[1/4na^2cot(pi/n)]
(33)
=1/(12)na^2hcot(pi/n).
(34)

Again using h from above and combining the two volumes gives the total volume of an equilateral antiprism as

 V=1/(12)n[cot(pi/(2n))+cot(pi/n)]sqrt(1-1/4sec^2(pi/(2n)))a^3.
(35)

The volumes of the first few equilateral antiprisms are therefore given by

V_3=1/3sqrt(2)a^3
(36)
V_4=1/3sqrt(4+3sqrt(2))a^3
(37)
V_5=1/6(5+2sqrt(5))a^3
(38)
V_6=sqrt(2(1+sqrt(3)))a^3.
(39)

See also

Antiprism Graph, Gyroelongated Pyramid, Octahedron, Prism, Prismoid, Square Antiprism, Trapezohedron

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References

Ball, W. W. R. and Coxeter, H. S. M. "Polyhedra." Ch. 5 in Mathematical Recreations and Essays, 13th ed. New York: Dover, p. 130, 1987.Coxeter, H. S. M. Introduction to Geometry, 2nd ed. New York: Wiley, p. 149, 1969.Cromwell, P. R. Polyhedra. New York: Cambridge University Press, pp. 85-86, 1997.Pedagoguery Software. Poly. http://www.peda.com/poly/.Pegg, E. Jr. "Perfect and Almost Perfect Rings (Chains) of 4-Antiprisms." https://community.wolfram.com/groups/-/m/t/2451238.Webb, R. "Prisms, Antiprisms, and their Duals." http://www.software3d.com/Prisms.html.

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Antiprism

Cite this as:

Weisstein, Eric W. "Antiprism." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/Antiprism.html

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