February 2015 – PolyhedraMath

In geometry, a prism is a polyhedron with an n-sided polygonal base, another congruent parallel base (with the same rotational orientation), and n other faces (necessarily all parallelograms) joining corresponding sides of the two bases. Prisms are named for their base, so a prism with a pentagonal base is called a pentagonal prism.

In regular prisms, all side faces (all squares) are at right angles to the bases. The number of prisms is infinite, approaching a circular base.

The interior angles of the base faces is $\frac{180 (n - 2)}{n}$ , where n is the number of sides. This is really the only “difficult” math property of prisms.

Some prisms are not normally considered or called prisms, such as a square prism is just a cube.

If instead of connecting the bases with squares, you connect them with equilateral triangles, you then get antiprisms.

Antiprism 2 — (n=2) Digonal Antiprism, aka Tetrahedron

Antiprism 3 — (n=3) Triangular Antiprism, aka Octahedron

Antiprism 5 — (n=5) Pentagonal Antiprism

Antiprism 12 — (n=12) Dodecagonal Antiprism

For n=2, the digonal antiprism is a degenerate case, as there is no base polygon, only a line segment. Otherwise the top face is a duplicate of the bottom, only rotated 180°/n.

Figure 1: Polygon Dimensions

For dihedral angles for antiprisms, things are a lot more complicated than prisms.

Lets start with looking at the base polygon, Figure 1.

A regular polygon always has the following properties:

$\begin{aligned} r a d i u s r & = D B = \frac{s}{2 \sin \frac{180^{\circ}}{n}} = \frac{a}{\cos \frac{180^{\circ}}{n}} \\ a p o t h e m a & = \overset{―}{D E} = r \cdot \cos \frac{180^{\circ}}{n} = \frac{s}{2 \tan \frac{180^{\circ}}{n}} \\ s i d e s & = r \cdot 2 \sin \frac{180^{\circ}}{n} = a \cdot 2 \tan \frac{180^{\circ}}{n} = A B = B C = F G \\ a n g l e θ & = \frac{360^{\circ}}{n} \\ s a g i t t a v & = r - a = r (1 - \cos \frac{360}{2 n}) = r (2 \sin^{2} \frac{360}{4 n}) = \frac{s (1 - \cos \frac{360}{2 n})}{2 \sin \frac{360}{2 n}} = \frac{s (\sin^{2} \frac{360}{4 n})}{\sin \frac{360}{2 n}} = s \frac{1}{2} \tan \frac{90}{n} = \overset{―}{B E} \end{aligned}$

The apothem is the distance from the center to the midpoint of any side. The difference between the radius and apothem is called the sagitta, also sometimes called the versine function ( $v e r s i n μ = 1 - \cos μ = 2 \sin^{2} \frac{1}{2} μ)$ . The antiprism’s sides will be deflected away from 90° by the length of BE (or v). The dotted line is the location of the second base, always rotated θ/2 to that of the first base.

Normally it would be odd trying to calculate the radius of a circle, but we are defining our polygon based on its side length being one unit.

Figure 2

Figure 3

If we draw a line at the midpoint of each triangle all the way around, we get a regular polygon with 2n sides, parallel to each base.

The height of the triangle is line BE₂, $\frac{\sqrt{3}}{2}$ . We do not yet know the height of the antiprism ( $\overset{―}{E_{1} E_{2}}$ ), but can find it.

$\begin{aligned} (\overset{―}{B E_{1}})^{2} + (\overset{―}{E_{1} E_{2}})^{2} & = (\overset{―}{B E_{2}})^{2} \\ (\overset{―}{E_{1} E_{2}})^{2} & = (\overset{―}{B E_{2}})^{2} - (\overset{―}{B E_{1}})^{2} \\ = {(\frac{\sqrt{3}}{2})}^{2} - {(s \frac{1}{2} \tan \frac{90}{n})}^{2} \\ = \frac{3}{4} - s^{2} \cdot \frac{1}{4} \tan^{2} \frac{90}{n} \\ = \frac{3 - s^{2} \tan^{2} \frac{90}{n}}{4} \\ \overset{―}{E_{1} E_{2}} & = \frac{1}{2} \sqrt{3 - s^{2} \tan^{2} \frac{90}{n}} \end{aligned}$

This is as simplified as we can get without deciding on how many sides the base will have. The height of an antiprism must always be less than $\frac{1}{2} s \sqrt{3}$ , the height of a regular triangle.

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Here are a few of the first heights for n:

n	θ	Height
2	45	$\frac{1}{\sqrt{2}}$
3	60	$\sqrt{\frac{2}{3}}$
4	90	$\sqrt[4]{\frac{1}{2}}$
5	36	$\sqrt{\frac{5 + \sqrt{5}}{10}}$
6	30	$\sqrt{\sqrt{3} - 1}$
8	22.5	$\sqrt{\sqrt{5 + \frac{7 \sqrt{2}}{2}} - 1 - \sqrt{2}}$
10	18	$\frac{1}{2} \sqrt{- 8 - 4 \sqrt{5} - 2 (1 + \sqrt{5}) \sqrt{5 + 2 \sqrt{5}}}$

For the dihedral angles, ∠E₁BE₂ is the same as ∠Ψ, and $\cos ∠ B E_{2} D_{2} = - \cos ∠ Ψ$ .

$\begin{aligned} \cos ∠ E_{1} B E_{2} & = \frac{(B E_{1})^{2} + (B E_{2})^{2} - (E_{1} E_{2})^{2}}{2 [(B E_{1}) (B E_{2})]} \\ = \frac{{(\frac{1}{2} \tan \frac{90}{n})}^{2} + {(\frac{\sqrt{3}}{2})}^{2} - {(\frac{1}{2} \sqrt{3 - \tan^{2} \frac{90}{n}})}^{2}}{2 \cdot \frac{1}{2} \tan \frac{90}{n} \cdot \frac{\sqrt{3}}{2}} \\ = \frac{\frac{\tan^{2} \frac{90}{n}}{4} + \frac{3}{4} - \frac{3 - \tan^{2} \frac{90}{n}}{4}}{\frac{\sqrt{3}}{2} \cdot \tan \frac{90}{n}} \\ = \frac{\frac{2 \tan^{2} \frac{90}{n}}{4}}{\frac{\sqrt{3}}{2} \cdot \tan \frac{90}{n}} = \frac{\frac{1}{2} \tan^{2} \frac{90}{n}}{\frac{\sqrt{3}}{2} \cdot \tan \frac{90}{n}} \\ = \frac{\tan^{2} \frac{90}{n}}{2} \cdot \frac{2}{\sqrt{3} \cdot \tan \frac{90}{n}} \\ = \frac{\tan^{2} \frac{90}{n}}{\sqrt{3} \cdot \tan \frac{90}{n}} \\ = \frac{\tan \frac{90}{n}}{\sqrt{3}} \\ = \frac{1}{\sqrt{3}} \cdot \tan \frac{90}{n} \\ \cos ∠ B E_{2} D_{2} & = - \frac{1}{\sqrt{3}} \cdot \tan \frac{90}{n} \end{aligned}$

Here are the first few dihedrals for n:

n	$∠ E_{1} B E_{2}$	$∠ B E_{2} D_{2}$
2	54.735°	125.264°
3	70.528°	109.471°
4	76.163°	103.836°
5	79.187°	100.812°
6	81.101°	98.8994°
8	83.405°	96.5945°

For the side-to-side dihedrals, let’s draw a line from point C, straight down, equal to the height, and call this point K. K is in the same plane as the bottom base. Looking at Figure 1, we can see that ∠KDF is $\frac{3}{2} θ$ , or $\frac{540}{n}$ . Connecting F and K, we get a line of $2 r \sin \frac{540}{2 n} = s \cdot \frac{s i n \frac{270}{n}}{\sin \frac{180}{n}}$ . Note $\overset{―}{F K}$ is coplanar with G and E₂, but does not go through either point.

We can now find the distance between C and F,

$\begin{aligned} (C F)^{2} & = (C K)^{2} + (F K)^{2} \\ = \frac{3 - s^{2} \tan^{2} \frac{90}{n}}{4} + s^{2} \cdot \frac{\sin^{2} \frac{270}{n}}{\sin^{2} \frac{180}{n}} \\ (C F) & = \sqrt{\frac{3 - s^{2} \tan^{2} \frac{90}{n}}{4} + s^{2} \cdot \frac{\sin^{2} \frac{270}{n}}{\sin^{2} \frac{180}{n}}} \end{aligned}$

In Figure 2, we see that $C L = L F = B E_{2} = \frac{\sqrt{3}}{2}$ . We can now find ∠CLF, the dihedral, with s=1 factored out,

$\begin{aligned} \cos ∠ C L F & = \frac{(C L)^{2} + (L F)^{2} - (C F)^{2}}{2 \cdot C L \cdot L F} \\ = \frac{(\frac{\sqrt{3}}{2})^{2} + (\frac{\sqrt{3}}{2})^{2} - (C F)^{2}}{2 \cdot (\frac{\sqrt{3}}{2}) (\frac{\sqrt{3}}{2})} \\ = \frac{\frac{3}{4} + \frac{3}{4} - (C F)^{2}}{\frac{3}{2}} \\ = [\frac{6}{4} - (C F)^{2}] \div \frac{3}{2} = [\frac{3}{2} - (C F)^{2}] \cdot \frac{2}{3} \\ = 1 - \frac{2}{3} (C F)^{2} \\ = 1 - \frac{2}{3} (\frac{3 - \tan^{2} \frac{90}{n}}{4} + \frac{\sin^{2} \frac{270}{n}}{\sin^{2} \frac{180}{n}}) \end{aligned}$

Multiplying out the CF² is really messy, best to reduce the other parts as much as possible, then square.

Here are the first few dihedral angles for n:

n	$\cos ∠ C L F$	∠CLF
2	$\frac{1}{3}$	70.528779
3	$- \frac{1}{3}$	109.47122
4	$\frac{1 - 2 \sqrt{2}}{3}$	127.552
5	$- \frac{\sqrt{5}}{3}$	138.189685

There are several Archimedean solids that are formed by the truncation (cutting off) of each corner of a Platonic solid.

These can be shown in successive truncations from one shape to its dual.

Original	Truncation	Rectification	Bitruncation	Birectification (dual)
Tetrahedron	Truncated Tetrahedron	Tetratetrahedron aka Octahedron	Truncated Tetrahedron	Tetrahedron
Cube	Truncated Cube	Cuboctahedron	Truncated Octahedron	Octahedron
Dodecahedron	Truncated Dodecahedron	Icosidodecahedron	Truncated Icosahedron	Icosahedron
Above images from Wikipedia.com

Initial truncations remove a pyramid from each corner the original object, the base of which is a regular polygon. If the original face is a triangle, the resultant edges will be $\frac{1}{3}$ the original edge; if square, $\sqrt{\frac{1}{2}}$ ; and if a pentagon, $\sqrt{\frac{1}{5}}$ .

Note: Each truncation makes the object smaller, but for the math below, the presumption is an object with edge length of 1.

Complete truncation (also called “rectified”) completely removes original edges, new faces meet a the midpoint of original edges.

Note that the left objects are “sitting” on a face, and the right objects are “standing” on a vertex. This is the result of matching the vertices of one to the midpoints of the dual.

In quasi-regular polyhedra, truncation is not necessarily resultant in regular faces. Adjustments are made to make the faces become regular, these are called rhombi-truncations.

The dihedral angles of the “main” faces (red/red or yellow/yellow) of the truncated object are the same as the original object’s dihedral angles. If there are only three faces at a vertex, the vertex angles are easily found with the formula: $\cos θ = \frac{\cos ϵ}{\cos (\frac{μ}{2})}$ , where θ is the vertex angle of the newly created face to the opposing edge, ε is the interior angle of the new main face, and μ is the interior angle of the newly created face.

The dihedral angles of the new faces are a bit harder. We have to do each separately.

Since each truncation removes a pyramid with a regular base, we can calculate the dihedral angle of one side to the base, giving us the complimentary angle to that of the dihedral angle of the new/main faces of the polyhedra. Remember that if A+B=180, then cos A = —cos B, so we can find the cosine of the pyramid’s dihedral angle, then negate it.

Let’s start with the simplest, the truncation of a tetrahedron.

I’ve already done the math for the tetrahedron, the cosine of the dihedral is $\frac{1}{3}$ , or $70.528779 \dots^{\circ}$ , so the dihedral cosine for the hexagon-triangle faces of the truncated tetrahedron would be $- \frac{1}{3}$ , or $109.47122 \dots^{\circ}$ . The two do add up to 180° and the latter is the same as the rectified tetrahedron (aka octahedron), therefore we are on the right track.

For the trunc. cube, a triangular based pyramid with right angles at the apex, would have a base dihedral of $c o s^{- 1} (\frac{1}{\sqrt{3}})$ , so the octagon-triangle dihedral would be $c o s^{- 1} (\frac{- 1}{\sqrt{3}})$ , or 125.2643897…°.

The trunc. octahedron has a square pyramid (Johnson solid J₁) removed that equates to a half octahedron. The dihedral for J₁ is $\cos^{- 1} (\frac{1}{\sqrt{3}})$ , so the trunc. octahedron hex-square dihedral would be $\cos^{- 1} (\frac{- 1}{\sqrt{3}})$ , or 125.2643897…°.

Yes, these are the same two angles, which can be seen in the cuboctahedron, all edges are square-triangles, so must be the same. It is also evident when you realize that corresponding faces are parallel from one shape to the next during truncation.

For a trunc. dodecahedron, the decagon-decagon faces have the same dihedral as the dodecahedron. The decagon-triangle edges have the same dihedral as the pent-hex edges of the trunc. icosahedron, 142.622632…°, but we will check this later below.

The vertex angle θ is found by

$\begin{aligned} \cos θ & = \frac{\cos 108}{\cos \frac{108}{2}} = \frac{- \cos 72}{\cos 54} \\ = - \frac{\frac{\sqrt{5} - 1}{4}}{\sqrt{\frac{5 - \sqrt{5}}{8}}} \\ = - \frac{\sqrt{5} - 1}{4} \cdot \sqrt{\frac{8}{5 - \sqrt{5}}} \\ = - \frac{\sqrt{8} \cdot (\sqrt{5} - 1)}{4 \sqrt{5 - \sqrt{5}}} \\ = - \frac{\sqrt{8} \cdot \sqrt{(\sqrt{5} - 1)^{2}}}{4 \sqrt{5 - \sqrt{5}}} \\ = - \frac{\sqrt{8} \cdot \sqrt{6 - 2 \sqrt{5}}}{4 \sqrt{5 - \sqrt{5}}} \\ = - \frac{\sqrt{8} \cdot \sqrt{2} \cdot \sqrt{3 - \sqrt{5}}}{4 \sqrt{5 - \sqrt{5}}} \\ = - \frac{\sqrt{16} \cdot \sqrt{3 - \sqrt{5}}}{4 \sqrt{5 - \sqrt{5}}} \\ = - \frac{4 \cdot \sqrt{3 - \sqrt{5}}}{4 \sqrt{5 - \sqrt{5}}} \\ = - \frac{\sqrt{3 - \sqrt{5}}}{\sqrt{5 - \sqrt{5}}} \\ = - \sqrt{\frac{3 - \sqrt{5}}{5 - \sqrt{5}}} \\ = - \sqrt{\frac{3 - \sqrt{5}}{5 - \sqrt{5}} \cdot \frac{5 + \sqrt{5}}{5 + \sqrt{5}}} \\ = - \sqrt{\frac{10 - 2 \sqrt{5}}{20}} \\ = - \sqrt{\frac{5 - \sqrt{5}}{10}} \end{aligned}$

So the vertex angle is 121.7174744…°

The trunc. icosahedron we already did the work for, the vertex angles are 148.2825255…°.

We already know the dihedrals of the cuboctahedron and icosidodecahedron are the same as their truncated counterparts, but what if we wanted to verify that?

Figure 1: Icosidodecahedron vertex

As Fig. 1 shows, we can draw a rectangle onto the surface of a icosidodecahedron, with points ABCD, which creates a pyramid with apex E. As lines AD and BC cross the width of the pentagons, their length is then $P h i (ϕ) o r \frac{1 + \sqrt{5}}{2}$ . The green line is also Φ and crosses AD at point F. Point G is the midpoint of line DE, making DG=EG=1/2.

Notice that the blue and green lines are parallel to the upper and left sides of the pentagon, making a parallelogram, making AF=1, so $D F = ϕ - 1 = \frac{\sqrt{5} - 1}{2}$ . ∠DGF is 90°, we can find FG,

$\begin{aligned} D G^{2} + F G^{2} & = D F^{2} \\ F G^{2} & = D F^{2} - D G^{2} \\ = {(\frac{\sqrt{5} - 1}{2})}^{2} - \frac{1}{2^{2}} \\ = \frac{6 - 2 \sqrt{5}}{4} - \frac{1}{4} \\ = \frac{5 - 2 \sqrt{5}}{4} \\ F G & = \frac{\sqrt{5 - 2 \sqrt{5}}}{2} \end{aligned}$

The distance between F and C is

$\begin{aligned} C F^{2} & = C D^{2} + D F^{2} \\ = 1 + (\frac{\sqrt{5} - 1}{2})^{2} \\ = \frac{4}{4} + \frac{6 - 2 \sqrt{5}}{4} \\ = \frac{10 - 2 \sqrt{5}}{4} \\ C F & = \frac{\sqrt{10 - 2 \sqrt{5}}}{2} \end{aligned}$

Line CG is the same as the height of the triangle, $\frac{\sqrt{3}}{2}$ . Now we can find the dihedral (∠FGC)

$\begin{aligned} \cos ∠ F G C & = \frac{F G^{2} + C G^{2} - C F^{2}}{2 \cdot F G \cdot C G} \\ = \frac{\frac{5 - 2 \sqrt{5}}{4} + \frac{3}{4} - \frac{10 - 2 \sqrt{5}}{4}}{2 \cdot \frac{\sqrt{5 - 2 \sqrt{5}}}{2} \cdot \frac{\sqrt{3}}{2}} \\ = \frac{\frac{5 - 2 \sqrt{5} + 3 - 10 + 2 \sqrt{5}}{4}}{(\sqrt{5 - 2 \sqrt{5}}) \cdot \frac{\sqrt{3}}{2}} \\ = \frac{- \frac{2}{4}}{\sqrt{5 - 2 \sqrt{5}} \cdot \frac{\sqrt{3}}{2}} = - \frac{2}{4} \div (\frac{\sqrt{3} \sqrt{5 - 2 \sqrt{5}}}{2}) \\ = - \frac{1}{2} \cdot (\frac{2}{\sqrt{3} \sqrt{5 - 2 \sqrt{5}}}) \\ = \frac{- 1}{\sqrt{3} \sqrt{5 - 2 \sqrt{5}}} \\ = \frac{- 1}{\sqrt{3} \sqrt{5 - 2 \sqrt{5}}} \cdot \frac{\sqrt{5 + 2 \sqrt{5}}}{\sqrt{5 + 2 \sqrt{5}}} \\ = \frac{- \sqrt{5 + 2 \sqrt{5}}}{\sqrt{3} \sqrt{25 - 4 \sqrt{5} - 20 - 4 \sqrt{5}}} \\ = \frac{- \sqrt{5 + 2 \sqrt{5}}}{\sqrt{3} \sqrt{5}} \\ = - \sqrt{\frac{5 + 2 \sqrt{5}}{15}} \\ ∠ F G C & = 142.62263 \dots^{\circ} \end{aligned}$

Figure 2: Cuboctahedron vertex

Just like the icosidodecahedron, we can draw a rectangle onto the surface of a cuboctahedron (Fig. 2), with points WXYZ, which creates a pyramid with apex V. As lines WZ and XY cross the diagonals of the squares, their length is then $\sqrt{2}$ . The green line is also $\sqrt{2}$ and crosses WZ at point T. Point S is the midpoint of line WV, making WS=VS=1/2.

Line ST is half the width of the square, so its length is ½. Lines TV and TW are $\frac{\sqrt{2}}{2} o r \frac{1}{\sqrt{2}}$ .

The distance between T and X is

$\begin{aligned} T X^{2} & = W X^{2} + T W^{2} \\ = 1 + (\frac{1}{\sqrt{2}})^{2} \\ = \frac{2}{2} + \frac{1}{2} \\ = \frac{3}{2} \\ T X & = \sqrt{\frac{3}{2}} \end{aligned}$

Line SX is the same as the height of the triangle, $\frac{\sqrt{3}}{2}$ . Now we can find the dihedral (∠TSX)

$\begin{aligned} \cos ∠ T S X & = \frac{S T^{2} + S X^{2} - T X^{2}}{2 \cdot S T \cdot S X} = \frac{{(\frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2} - {(\sqrt{\frac{3}{2}})}^{2}}{2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2}} \\ = \frac{\frac{1}{4} + \frac{3}{4} - \frac{6}{4}}{\frac{\sqrt{3}}{2}} \\ = \frac{- \frac{2}{4}}{\frac{\sqrt{3}}{2}} = - \frac{2}{4} \cdot \frac{2}{\sqrt{3}} \\ = \frac{- 1}{\sqrt{3}} \\ ∠ T S X & = 125.2643897 \dots^{\circ} \end{aligned}$

PolyhedraMath

A guide to the world of polyhedrons.

Month: February 2015

Prisms and Antiprisms

Truncated Platonics

Above images from Wikipedia.com