# Truncated Archimedeans

In a previous post,we looked at the Platonics as portions were cut off (truncated). There are four objects that are made from further removal of edges/corners.

Unfortunately, these are not “perfect” truncations.

If you take a cuboctahedron and cut off each corner at the midpoint of each edge, you then end up with smaller triangle faces ¼th the original area and inverted and smaller squares ½ the original area and rotated 45°. You also get 12 new faces that are rectangles. This does not fit the definition of a “regular polyhedron” as rectangles are not regular polygons, they have sides that are of different lengths.

In order to get this object to be regular, you have to distort it slightly to make the rectangles into squares. Then you have a small rhombicuboctahedron. It is called this because the faces lie in the same plane as the rhombic dodecahedron, the dual of the cuboctahedron. (No one ever said that mathematicians were good at naming things.)

A better method of construction is a sort of “double truncation.” take a cube and chamfer all the edges at 45° where the cut is $$(edge length\cdot [\sqrt2-1] )$$ from each edge (Fig.1). You then truncate a second time cutting off each end of the elongated hexagons, meeting each square at the very tip of the corners, creating equilateral triangles (Fig.2).

Looking at any square face straight on, the object has the outline of a regular octagon, therefore the dihedral angle between any two squares is 135°.

 Figure 1: Cube, edge chamfered Figure 2: Small Rhombicuboctahedron Figure 3: Great Rhombicuboctahedron

Likewise, you can cut off the corners of a cuboctahedron to turn the triangle faces into hexagons and square faces into octagons, but again the new faces are rectangles, not squares. Distortion will then give you a great rhombicuboctahedron. It is also called by some a truncated cuboctahedron.

The octagon-square dihedral angle is 135°.

The white faces of both are parallel to those of an octahedron, so both can also be made from chamfering an octahedron’s edges and truncating the corners.

Another two objects can be made starting with an icosidodecahedron, instead of a cuboctahedron, the small and great rhombicosidodecahedrons and truncating appropriately.

Regardless of how you construct the objects, all faces should be regular polygons and all edges are the same length.

# Prisms and Antiprisms

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.

 (n=3) Triangular Prism (n=4) Square Prism, aka Cube (n=5) Pentagonal Prism (n=6) Hexagonal Prism

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

 (n=2) Digonal Antiprism, aka Tetrahedron (n=3) Triangular Antiprism, aka Octahedron (n=4) Square Antiprism (n=5) Pentagonal Antiprism (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.

A regular polygon always has the following properties:

\begin{align} radius\ r&=DB=\frac{s}{2\sin\frac{180^\circ}{n}} = \frac{a}{\cos\frac{180^\circ}{n}} \\ apothem\ a &=\overline{DE}=r\cdot\cos\frac{180^\circ}{n}=\frac{s}{2\tan\frac{180^\circ}{n}} \\ side\ s&=r\cdot2\sin\frac{180^\circ}{n}=a\cdot 2\tan\frac{180^\circ}{n}=AB=BC=FG \\ angle\ \theta &=\frac{360^\circ}{n} \\sagitta\ v&=r-a=r(1-\cos\frac{360}{2n})=r(2\sin^2\frac{360}{4n})=\frac{s(1-\cos\frac{360}{2n})}{2\sin\frac{360}{2n}}=\frac{s(\sin^2\frac{360}{4n})}{\sin\frac{360}{2n}}=s\frac12\tan\frac{90}{n}=\overline{BE} \end{align}

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 ($$versin\ \mu=1-\cos\mu=2\sin^2\frac12\mu)$$. 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, $$\frac{\sqrt3}{2}$$. We do not yet know the height of the antiprism ($$\overline{E_1E_2}$$), but can find it.

\begin{align} (\overline{BE_1})^2+(\overline{E_1 E_2})^2 &=(\overline{BE_2})^2 \\ (\overline{E_1 E_2})^2 &=(\overline{BE_2})^2-(\overline{BE_1})^2 \\ &=\left(\frac{\sqrt3}{2}\right)^2-\left(s\frac12\tan\frac{90}{n}\right)^2 \\ &=\frac34-s^2\cdot\frac14\tan^2\frac{90}{n} \\ &=\frac{3-s^2\tan^2\frac{90}{n}}{4} \\ \overline{E_1 E_2} &=\frac12\sqrt{3-s^2\tan^2\frac{90}{n} } \end{align}

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 $$\frac12 s\sqrt3$$, the height of a regular triangle.

Here are a few of the first heights for n:

 n θ Height 2 45 $$\frac1{\sqrt2}$$ 3 60 $$\sqrt{\frac23}$$ 4 90 $$\sqrt[4]{\frac12}$$ 5 36 $$\sqrt{\frac{5+\sqrt5}{10}}$$ 6 30 $$\sqrt{\sqrt3-1}$$ 8 22.5 $$\sqrt{\sqrt{5+\frac{7\sqrt2}{2}} -1-\sqrt2}$$ 10 18 $$\frac12\sqrt{-8-4\sqrt5-2(1+\sqrt5)\sqrt{5+2\sqrt5}}$$

For the dihedral angles, ∠E1BE2 is the same as ∠Ψ, and $$\cos\angle BE_2D_2 =-\cos\angle\Psi$$.

\begin{align} \cos\angle E_1BE_2 &=\frac{(BE_1)^2+(BE_2)^2-(E_1 E_2)^2}{2[(BE_1)(BE_2)]} \\ &=\frac{\left(\frac12\tan\frac{90}{n}\right)^2+\left(\frac{\sqrt3}{2}\right)^2-\left(\frac12\sqrt{3-\tan^2\frac{90}{n} }\right)^2}{2\cdot\frac12\tan\frac{90}{n}\cdot\frac{\sqrt3}{2}} \\ &=\frac{ \frac{\tan^2\frac{90}{n}}{4}+\frac34-\frac{3-\tan^2\frac{90}{n} }{4} }{\frac{\sqrt3}{2}\cdot\tan\frac{90}{n}} \\ &=\frac{ \frac{2\tan^2\frac{90}{n}}{4} }{\frac{\sqrt3}{2}\cdot\tan\frac{90}{n}}=\frac{ \frac12\tan^2\frac{90}{n} }{\frac{\sqrt3}{2}\cdot\tan\frac{90}{n}} \\ &= \frac{\tan^2\frac{90}{n}}{2} \cdot \frac{2}{\sqrt3\cdot\tan\frac{90}{n}} \\ &= \frac{\tan^2\frac{90}{n}}{\sqrt3\cdot\tan\frac{90}{n}} \\ &= \frac{\tan\frac{90}{n}}{\sqrt3} \\ &= \frac1{\sqrt3}\cdot\tan\frac{90}{n} \\ \cos\angle BE_2D_2&=-\frac1{\sqrt3}\cdot\tan\frac{90}{n} \end{align}

Here are the first few dihedrals for n:

 n $$\angle E_1BE_2$$ $$\angle BE_2D_2$$ 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 $$\frac32\theta$$, or $$\frac{540}{n}$$. Connecting F and K, we get a line of $$2r\sin\frac{540}{2n}= s\cdot\frac{sin\frac{270}{n}}{\sin\frac{180}{n}}$$. Note $$\overline{FK}$$ is coplanar with G and E2, but does not go through either point.

We can now find the distance between C and F,

\begin{align} (CF)^2&=(CK)^2+(FK)^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}} \\ (CF) &=\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{align}

In Figure 2, we see that $$CL=LF=BE_2=\frac{\sqrt3}{2}$$. We can now find ∠CLF, the dihedral, with s=1 factored out,

\begin{align} \cos\angle CLF &=\frac{(CL)^2+(LF)^2- (CF)^2}{2\cdot CL\cdot LF} \\ &=\frac{(\frac{\sqrt3}{2})^2+(\frac{\sqrt3}{2})^2-(CF)^2 }{2\cdot (\frac{\sqrt3}{2})(\frac{\sqrt3}{2})} \\ &=\frac{\frac34+\frac34-(CF)^2 }{\frac32} \\ &=\left[\frac64-(CF)^2\right]\div\frac32=\left[\frac32-(CF)^2\right]\cdot\frac23 \\ &=1-\frac23(CF)^2 \\ &=1-\frac23\left(\frac{3-\tan^2\frac{90}{n}}{4} + \frac{\sin^2\frac{270}{n}}{\sin^2\frac{180}{n}} \right) \end{align}

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 $$\cos\angle CLF$$ ∠CLF 2 $$\frac13$$ 70.528779 3 $$-\frac13$$ 109.47122 4 $$\frac{1-2\sqrt2}{3}$$ 127.552 5 $$-\frac{\sqrt5}{3}$$ 138.189685

# Truncated Platonics

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)

Truncated Tetrahedron

Tetratetrahedron
aka Octahedron

Truncated Tetrahedron

Truncated Cube

Cuboctahedron

Truncated Octahedron

Truncated Dodecahedron

Icosidodecahedron

###### 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 $$\frac13$$ the original edge; if square, $$\sqrt\frac12$$; and if a pentagon, $$\sqrt\frac15$$.

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: $$\Large{\cos\theta=\frac{\cos\epsilon}{\cos(\frac{\mu}{2})}}$$, 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.

I’ve already done the math for the tetrahedron, the cosine of the dihedral is $$\frac13$$, or $$70.528779\ldots^\circ$$, so the dihedral cosine for the hexagon-triangle faces of the truncated tetrahedron would be $$-\frac13$$, or $$109.47122\ldots^\circ$$. 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 $$cos^{-1}(\frac{1}{\sqrt3})$$, so the octagon-triangle dihedral would be $$cos^{-1}(\frac{-1}{\sqrt3})$$, 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 $$\cos^{-1}(\frac{1}{\sqrt3})$$, so the trunc. octahedron hex-square dihedral would be $$\cos^{-1}(\frac{-1}{\sqrt3})$$, 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{align}\cos\theta &=\frac{\cos108}{\cos\frac{108}{2}}=\frac{-\cos72}{\cos54} \\&=-\frac{\frac{\sqrt5-1}{4}}{\sqrt{\frac{5-\sqrt5}{8}}} \\&=-\frac{\sqrt5-1}{4}\cdot\sqrt{\frac{8}{5-\sqrt5}} \\&=-\frac{\sqrt8 \cdot (\sqrt5-1)}{4\sqrt{5-\sqrt5}} \\&=-\frac{\sqrt8 \cdot \sqrt{(\sqrt5-1)^2}}{4\sqrt{5-\sqrt5}} \\&=-\frac{\sqrt8 \cdot \sqrt{6-2\sqrt5} }{4\sqrt{5-\sqrt5}} \\&=-\frac{\sqrt8 \cdot \sqrt2\cdot\sqrt{3-\sqrt5} }{4\sqrt{5-\sqrt5}} \\&=-\frac{\sqrt{16} \cdot\sqrt{3-\sqrt5} }{4\sqrt{5-\sqrt5}} \\&=-\frac{4 \cdot\sqrt{3-\sqrt5} }{4\sqrt{5-\sqrt5}} \\&=-\frac{\sqrt{3-\sqrt5} }{\sqrt{5-\sqrt5}} \\&=-\sqrt{\frac{3-\sqrt5 }{5-\sqrt5} } \\&=-\sqrt{\frac{3-\sqrt5 }{5-\sqrt5} \cdot \frac{5+\sqrt5}{5+\sqrt5} } \\&=-\sqrt{\frac{10-2\sqrt5 }{20} } \\&=-\sqrt{\frac{5-\sqrt5 }{10} } \\ \end{align}

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 $$Phi\ (\phi)\ or\ \frac{1+\sqrt5}{2}$$. 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 $$DF=\phi-1 = \frac{\sqrt5-1}{2}$$. ∠DGF is 90°, we can find FG,

\begin{align}DG^2+FG^2 &=DF^2 \\FG^2&=DF^2-DG^2 \\&=\left(\frac{\sqrt5-1}{2}\right)^2-\frac1{2^2} \\&=\frac{6-2\sqrt5}{4} – \frac14 \\&=\frac{5-2\sqrt5}{4} \\FG&=\frac{\sqrt{5-2\sqrt5}} {2} \end{align}

The distance between F and C is

\begin{align}CF^2&=CD^2+DF^2 \\&=1+(\frac{\sqrt5-1}{2})^2 \\&=\frac44 + \frac{6-2\sqrt5}{4} \\&=\frac{10-2\sqrt5}{4} \\CF&=\frac{\sqrt{10-2\sqrt5}} {2} \end{align}

Line CG is the same as the height of the triangle, $$\frac{\sqrt3}{2}$$. Now we can find the dihedral (∠FGC)

\begin{align}\cos\angle FGC&=\frac{FG^2+CG^2-CF^2}{2\cdot FG\cdot CG} \\&=\frac{ \frac{5-2\sqrt5}{4}+\frac34-\frac{10-2\sqrt5}{4}} {2\cdot\frac{\sqrt{5-2\sqrt5}} {2}\cdot\frac{\sqrt3}{2}} \\&=\frac{ \frac{5-2\sqrt5+3-10+2\sqrt5}{4}} {(\sqrt{5-2\sqrt5})\cdot\frac{\sqrt3}{2}} \\&=\frac{-\frac24}{\sqrt{5-2\sqrt5}\cdot\frac{\sqrt3}{2}} = -\frac24\div \left(\frac{\sqrt3\sqrt{5-2\sqrt5}} {2}\right) \\&=-\frac12 \cdot \left(\frac{2}{\sqrt3\sqrt{5-2\sqrt5}}\right) \\&=\frac{-1}{\sqrt3 \sqrt{5-2\sqrt5}} \\&=\frac{-1}{\sqrt3 \sqrt{5-2\sqrt5}}\cdot \frac{\sqrt{5+2\sqrt5}}{\sqrt{5+2\sqrt5}} \\&=\frac{-\sqrt{5+2\sqrt5}}{\sqrt3 \sqrt{25-4\sqrt5-20-4\sqrt5}} \\&=\frac{-\sqrt{5+2\sqrt5}}{\sqrt3 \sqrt5} \\&=-\sqrt{\frac{5+2\sqrt5}{15} } \\\angle FGC&=142.62263\ldots^\circ \end{align}

 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 $$\sqrt2$$. The green line is also $$\sqrt2$$ 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 $$\frac{\sqrt2}{2}\ or \frac1{\sqrt2}$$.

The distance between T and X is

\begin{align}TX^2&=WX^2+TW^2 \\&=1+(\frac1{\sqrt2})^2 \\&=\frac22 + \frac12 \\&=\frac32 \\TX&=\sqrt{\frac32} \end{align}

Line SX is the same as the height of the triangle, $$\frac{\sqrt3}{2}$$. Now we can find the dihedral (∠TSX)

\begin{align}\cos\angle TSX&=\frac{ST^2+SX^2-TX^2}{2\cdot ST\cdot SX}=\frac{ \left(\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2-\left(\sqrt\frac32\right)^2} {2\cdot\frac12\cdot\frac{\sqrt3}{2}} \\&=\frac{\frac14+\frac34-\frac64} {\frac{\sqrt3}{2}} \\&=\frac{-\frac24} {\frac{\sqrt3}{2}} = -\frac24 \cdot \frac2{\sqrt3} \\&=\frac{-1}{\sqrt3} \\\angle TSX&=125.2643897\ldots^\circ \end{align}

# Icosahedron

The icosahedron has 12 equilateral triangles as faces.

It can be split into 3 parts, a pentagonal anti-prism and two pentagonal pyramids. We will start by looking at a pentagonal pyramid.

We have already done most of the work earlier.

To find ∠ACB,

\begin{align}\cos \angle ACB &= \frac{1^2 + S^2 – H^2}{2\cdot1\cdot S}\\ &= \frac{1^2 + \frac34 – \frac{5+2\sqrt5}{4}}{2\cdot 1\cdot \frac{\sqrt3}{2}}\\ &= \frac{\frac{2-2\sqrt5}{4} }{\sqrt3} \\ &= \frac{2-2\sqrt5}{4\sqrt3} \\ &= \frac{1-\sqrt5}{2\sqrt3}\\ACB &= 110.905157\ldots^\circ \end{align}

This is the vertex angle from one face to the opposite edge.

Taking two triangle faces that are abutting, the tips would be Phi (Φ) apart, you can draw a triangle of sides S, S, and Φ, going through the center of each equilateral triangle and the tips.

\begin{align}Icosahedron\ Dihedral \angle &= \cos^{-1}\Bigg(\frac{S^2 + S^2 – \Phi^2}{2\cdot S\cdot S}\Bigg)\\ &=\cos^{-1}\Bigg(\frac{ \frac34 + \frac34 – \frac{6+2\sqrt5}{4} }{ 2\cdot \frac{\sqrt3}{2} \cdot \frac{\sqrt3}{2} }\Bigg)\\ &=\cos^{-1}\Bigg(\frac{ – \frac{2\sqrt5}{4} }{ 2\cdot \frac34 }\Bigg)\\ &=\cos^{-1}\Bigg(\frac{ – 2\sqrt5 }{ 2\cdot 3 }\Bigg)\\ &=\cos^{-1}\Bigg(\frac{-\sqrt5}{3}\Bigg)\\ &= 138.1896851\ldots^\circ \end{align}

# Dodecahedron

The dodecahedron is a regular polyhedron made of 12 regular pentagon faces.

A D12 die is often used in certain role playing games.

Since a dodecahedron has 12 sides, you can print out a calendar, one month per face, and have a conversation piece on your desk. (My advice, use heavy paper and score lines before folding.)

The vertex angles are $$cos^{-1} ( \frac{\cos 108}{\cos \frac{108}{2}} )$$ or 121.7174744…°.

 Figure 1 Figure 2 Figure 3: Pentagon lengths

To find the dihedral angles, we will start with a dodecahedron with edge length of 1. Let’s take Figure 1 and slice through the drawn lines going across four of the faces. We get a shape similar to some house roofs, with a square base with sides of length Phi (Φ) or $$\frac{1+\sqrt5}{2}$$, as this is the width of a regular pentagon. There are also two triangles and two trapezoids, each with one edge length of Φ and all others 1. In other words, AE=BE=EF=CF=DF=1 and AB=BD=CD=AC=Φ.

Lengths of the pentagon are in Fig. 3, exact representations are not needed at this time.

If we take triangle ABE and “slide” it up to have it meet triangle CDF at point F, shown in red. We will call the points of the new triangle A’, B’, and F, (those are read as A-prime and B-prime). This gives us a rectangular pyramid A’B’CDF.

We know that line EF is 1, so lines AA’ and BB’ are also 1. Lines AC and BD are Φ, so lines A’C and B’D are length Φ-1 or $$\frac{\sqrt5-1}{2}$$, so B’H and HD are each $$\frac{\sqrt5-1}{4}$$.

 Figure 4 Figure 5 Figure 6

Taking the midpoints of A’C and B’D (points G and H, Fig. 5) we can use the Pythagorean theorem to find lengths of these lines,

\begin{align}\left(\frac{\sqrt5-1}{4}\right)^2 + FH^2 &= 1^2 \\ \frac{6-2\sqrt5}{16} + FH^2 &= 1\\ FH^2 &= 1- \frac{6-2\sqrt5}{16} \\ &= \frac{16}{16} – \frac{6-2\sqrt5}{16} \\ &=\frac{10+2\sqrt5}{16} \\ &= \frac{5+\sqrt5}{8} \\ FH &= \sqrt{\frac{5+\sqrt5}{8}} = Cos 18^\circ \end{align}

FG = FH and GH=CD=Φ. This gives us all three sides of triangle GFH, so we use the law of cosines to find ∠GFH,

\begin{align} \cos\angle GFH &= \frac{FG^2 + FH^2 – \Phi^2}{2\cdot FG\cdot FH} \\ &=\frac{\frac{5+\sqrt5}{8} + \frac{5+\sqrt5}{8} – \frac{6+2\sqrt5}{4} }{ 2\cdot \sqrt{\frac{5+\sqrt5}{8}} \cdot \sqrt{\frac{5+\sqrt5}{8}} } \\ &=\frac{\frac{10+2\sqrt5}{8} – \frac{12+4\sqrt5}{8} }{ 2\cdot \frac{5+\sqrt5}{8} } \\ &=\frac{\frac{-2-2\sqrt5}{8} }{ 2\cdot \frac{5+\sqrt5}{8} } \\ &=\frac{-2-2\sqrt5 }{ 2\cdot (5+\sqrt5) } \\ &=\frac{-1-\sqrt5 }{ (5+\sqrt5) } \\ &=\frac{-1-\sqrt5 }{ 5+\sqrt5 }\cdot \frac{-1+\sqrt5}{-1+\sqrt5} \\ &=\frac{1-5+\sqrt5-\sqrt5 }{-5+5-\sqrt5+5\sqrt5} \\ &=\frac{-4}{4\sqrt5}\\ &=\frac{-1}{\sqrt5} \\ \angle GFH &=116.56505\ldots^\circ \end{align}

This is the dodecahedron’s dihedral angle.

# SketchUp Platonics

I’ve been playing with Google’s Sketchup for a week or so and I had the thought of doing some polyhedra. It is often hard to mentally visualize a 3D object when looking at a 2D image, so I figured these would help.

Here is the first SketchUp file, the five Platonic solids.

You will need to install the program before you can view the file, but don’t worry, SketchUp is free.

Each of the edge lengths is the same for all five objects, just to give good reference to size.

In SketchUp, cubes are super-simple, so I purposely made it more difficult for myself (just for fun) by building the cube on-point.

You can use the Orbit tool (just hit “o”) to pan and rotate.

I will add more files as I make them.

# Octahedron

The octahedron is a Platonic solid, which means it is made of all regular polygons for each face, being eight equilateral triangles arranged four at each vertex.

With edge length of S, the surface area would be that of 8 equilateral triangles, $$8\cdot S^2\cdot \frac{\sqrt3}{4}=2\cdot S^2\sqrt3$$.

If we slice one in half, through 4 co-planar vertices, we get two 4-sided pyramids. Since we know that all the edges are the same length and all the angles are the same, the only possible shape for the base would have to be a square.

If there were only three triangles meeting at the apex, instead of four, we could use the formula found earlier to get the vertex edge angle, but we don’t need to recalculate a new formula. Slicing through any co-planar vertices will result in a square-based pyramid. Therefore, at each vertex, each edge is 60° to the ones on either side of it and 90° to the fourth.

Given an octahedron with edge length of 1 and vertices A-F, we can draw diagonals (AF, BD, CE) across opposite corners, each being the same length $$\sqrt2$$, and meeting at point G, the center of the octahedron. Half of these lengths is $$\frac{\sqrt2}{2} \text{, or } \frac{1}{\sqrt2}$$, which is the radius of the circumscribed sphere.

The midpoint of line BC is H. Line AH and FH are length $$\frac{\sqrt3}{2}$$. We now have a triangle AHF. We then can use the law of cosines to find ∠AHF,

\begin{align}\cos\angle AHF &=\frac{(AH)^2+(FH)^2-(AF)^2}{2(AH)(AF)}\\ &=\frac{\left(\frac{\sqrt3}{2}\right)^2+\left(\frac{\sqrt3}{2}\right)^2 – \sqrt2^2}{2\cdot\frac{\sqrt3}{2}\cdot\frac{\sqrt3}{2}} \\ &=\frac{\frac34+\frac34-2}{\frac32}\\ &=\left(\frac64-\frac84\right)\cdot\frac23 \\ &= -\frac24\cdot\frac23 \\ &= -\frac{4}{12}\\ &=-\frac13\end{align}

So $$\cos\angle AHF =-\frac13$$, ∠AHF= 109.4712206…°, the dihedral angle.

Placing point J at the circumcenter of ΔABC, gives a line AJ of length $$\sqrt{\frac13}$$, and ∠AJG is 90°. Solving for GJ gives:

\begin{align}(GJ)^2+(AJ)^2&=(AG)^2\GJ)^2 + \left(\frac{1}{\sqrt3}\right)^2 &= \left(\frac{1}{\sqrt2}\right)^2\\(GJ)^2 &= \left(\frac{1}{\sqrt2}\right)^2-\left(\frac{1}{\sqrt3}\right)^2\\(GJ)^2 &= \frac12-\frac13\\ &=\frac16 \end{align} \(GJ=\sqrt\frac16, the radius of an inscribed sphere. Joining the centers of each triangle, gives the edges a cube, the dual of the octahedron. # The Cube The cube is probably the most recognized and best known of and 3D shape. Kids young enough not to even know how to talk will still know about cubes, they play with multicolored wooden blocks. Older kids may tackle the Rubik’s Cube puzzle or play Yahtzee with dice. Then they may graduate to working in a cubicle as adults. We use the cube as a representation of the space around us. Pepsi was briefly sold in 24-can “cubes.” Apple’s G4 computer was cubic. Al Kaaba (literal English, the cube) is a cuboid shaped building in Mecca and the most sacred site to Islam. Salt grows into cubic crystals. Stones chipped and shaped into cubes have been found that date back 100’s of thousands of years. Star Trek’s fictional, I hope, Borg traveled through space in a cube shaped ship. There are several prisons with the nickname “the cube,” despite not having walls as tall as they are wide. They are literally, all around us. We have have such a familiarity with the cube that we don’t even have to think about its qualities or structure. We just know it. Therefore, having a page devoted to the cube may seem like a waste of space, but I’m certain that there is still knowledge we can gain from re-examining the world around us. The cube is a 3 dimensional solid object made of 6 squares (variously called faces, facets, or sides), each meeting at right (90°) angles with two others. It is one of the 5 Platonic solids and can also be referred to as a regular hexahedron or a square prism. Like a sphere is the set of all points equidistant from the origin, the cube is the set of points that make up the square faces, it has no volume, although we usually refer to the volume it contains. If we call the edge length S, then each face will have area $$S^2$$, so the cube will have surface area $$6*S^2$$. The internal volume is $$S^3$$. Many different shapes can tile 2D space, squares, hexagons, various triangles. The cube is the only object that can regularly tile Euclidean 3D space. Using formulas already mentioned, vertex edge angles are: $$\Large\cos\mu=\frac{cos90}{cos45}=\frac{0}{\frac{1}{\sqrt2}}=0, \mu=90^\circ$$. Using a cube with vertexes named A-H, as in the image, with edge length of 1, we can then calculate other lengths, easily done with right triangles and the Pythagorean theorem. Actually, it is even easier as almost all will be 45° right triangles. Drawing a diagonal across a face from B to D, we get a right triangle with BD length $$\sqrt2$$. Drawing diagonal AC crosses at the midpoint J. Drawing a spacial diagonal through the cube from A to G, triangle AEG lengths would be $$1,\sqrt2, and \sqrt3$$. Midpoint of AG is K which is also the spacial midpoint of the cube. AK and GK are $$\frac{\sqrt3}{2}$$, the radius of a sphere surrounding the cube. To find JK:\begin{align}
(AJ)^2+(JK)^2&=(AK)^2\\
(JK)^2&=(AK)^2-(AJ)^2\\
&=\left(\frac{\sqrt3}{2}\right)^2-\left(\frac{\sqrt2}{2}\right)^2\\
&=\frac34-\frac24\\
&=\frac14\\
JK&=\frac12\end{align}$$so JK=$$\frac12$$. This is the radius of an inscribed sphere. The midpoint of CD is L. Line JL would 1/2 the edge length, equal to JK, so $$KL=\frac{\sqrt2}{2}=\frac1{\sqrt2}$$. Angle JLK is 45°, one half the dihedral (face to face) angle. If you join the mid-face points (all the J’s), you would then get the cube’s dual shape called the octahedron. Similarly, joining the midpoints of the octahedron’s face would give you a cube. An octahedron inside a cube of edge length S would have edges of length $$S\cdot\frac1{\sqrt2}$$ and an octahedron outside the cube would have edge length of $$S\cdot\sqrt2$$. The octahedra would have volumes of $$S^3\cdot\frac13$$ and $$S^3\cdot\frac23$$, respective. # An unequal pyramid Last post, I worked on the vertex edge angles of a triangular pyramid that had all equal angles originating from the apex. Using the same camera tripod analogy, take one of the legs (line DA) of the tripod and slide it out further from the other two, but keep it equally distant from each. You now have a camera that’s about to crash to the floor, but you also have a pyramid that has two equal apex angles and that leans. A right pyramid is a pyramid for which the apex lies directly above the centroid of the base. I don’t know if there is an actual name for a leaning pyramid, but I will just call it a leaning pyramid. Again, we will work with a pyramid with vertexes A, B, C, & D, with D as the apex (Fig. 1). Again, the legs DA, DB, & DC are all equal length of 1 unit.  Figure 1 Figure 2 We can “unfold” the pyramid (Fig. 2) to show the bisection of the apex angles where they join the base vertexes, the midpoints E, F, & G. We can see BG=CG, and AE=BE=CF=AF, and DE=DF. Since ΔDAB = ΔDAC, we will ignore showing the the latter in the formulas. Both are isosceles, as well as ΔDBC. $$\large{ BG=CG= sin(\frac{\mu}{2})\\ BC=2\cdot sin(\frac{\mu}{2})\\ DG=cos(\frac{\mu}{2})\\ AE=BE= sin(\frac{\epsilon}{2})\\ AB=2\cdot sin(\frac{\epsilon}{2})\\ DE=cos(\frac{\epsilon}{2})}$$ Items that must be true for this to be a real 3D object: $$2\cdot\epsilon>\mu, 2\cdot\epsilon+\mu<360^\circ$$. Figure 3 shows how to find the line AG on the base triangle. Using the Pythagorean theorem, $$AG=\sqrt{4\cdot sin^{2}(\epsilon/2)-sin^{2}(\mu/2)}$$. We split the pyramid in half, giving us cross section ΔADG (Fig. 4).  Figure 3 Figure 4 Using law of cosines, we get:$$\large{\begin{align}\cos\theta
&=\frac{1^2+\cos^{2}(\frac{\mu}{2})-4\cdot \sin^{2}(\frac{\epsilon}{2})+\sin^{2}(\frac{\mu}{2})}{2\cdot 1\cdot \cos(\frac{\mu}{2})}\\
&=\frac{1 +[\cos^{2}(\frac{\mu}{2})+\sin^{2}(\frac{\mu}{2})]-4\cdot \sin^{2}(\frac{\epsilon}{2})}{2\cdot \cos(\frac{\mu}{2})} &\text{Rearrange figures}\\
&=\frac{1+[1]-4\cdot \sin^{2}(\frac{\epsilon}{2})}{2\cdot \cos(\frac{\mu}{2})} &\text{cos²α + sin²α = 1}\\
&=\frac{2-4\cdot \sin^{2}(\frac{\epsilon}{2})}{2\cdot \cos(\frac{\mu}{2})}\\
&=\frac{1-2\cdot \sin^{2}(\frac{\epsilon}{2})}{\cos(\frac{\mu}{2})}\\
&=\frac{\cos(\epsilon)}{\cos(\frac{\mu}{2})} & \text{1-2sin²α = cos 2α}
\end{align}}$$We can now use this new formula to double check our work on the truncated icosahedron. Hexagons have exterior angles of 120° (ε) and pentagons are 108° (μ). So we get:$$cos\theta=\frac{\cos(120)}{\cos(\frac{108}{2})}=\frac{cos(120)}{cos(54)}\\ \theta=148.2825256\ldots^\circ

The formula can be used to quickly find the vertex edge angle of any of the Platonic or Archimedean solids that have only 3 shapes meeting at each vertex, with 2 (or all 3) of them the same. (Although it isn’t as much fun as actually working it out ourselves.) It can also be used on prisms, but that is trivial, as the edge will always be at 90° to the bases.

We can find the angle made by slicing the pyramid through ΔCDE, but the result really doesn’t mean much as the plane made by ΔCDE is not perpendicular to the plane made by the base. ΔADG is perpendicular to the base plane. There would be 2 angles needed to describe the result. Example, looking at the truncated icosahedron, you can see the edges splaying out from the pentagons even with the center lines, sort of like the spokes of a wheel, but looking at the hexagons, the edges don’t line up with the center lines, they are slightly off.

Likewise, in a pyramid with all 3 apex angles different, there would be no cross section that was perpendicular to the base plane, so no real usage here.

# Tetrahedron

I started off posting about the truncated icosahedron, despite being fairly complex compared to the platonic solids, like the tetrahedron.

The regular tetrahedron is probably the simplest 3D shape, except for maybe the sphere. It is made of 4 equilateral triangles, forming a triangular pyramid.

Starting with a tetrahedron with vertexes named A, B, C, & D, with D as the apex (fig. 1), we can bisect ΔABC from point A to line BC and call that point G (fig. 2).

 Figure 1 Figure 2 Figure 3

The bisection of ΔDBC would follow the same, a line DG would also equal AG. Figure 3 shows the cross section of the tetrahedron, ΔADG. We can then use the law of cosines to find ∠θ and ∠η.

$$\Large\cos \eta = \frac{AG^{2} + DG^{2} – AD^{2}}{2\cdot AG \cdot DG} = \frac{\frac34 + \frac34 – 1}{2\cdot\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2}} = \frac{\frac24}{\frac64} = \frac13$$.

The angle (η) between each face, the dihedral angle, is $$cos^{-1} (\frac{1}{3})=70.528779\ldots^\circ$$.

$$\Large\cos \theta = \frac{AG^{2} + AD^{2} – DG^{2}}{2\cdot AG \cdot AD} = \frac{\frac34 + 1 – \frac34}{2\cdot 1\cdot\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt3}$$.

The angle that the edge to face (θ) makes would be $$cos^{-1}\frac{1}{\sqrt{3}}= 54.73561\ldots^\circ$$. Also since ΔADG is isosceles, ∠ADG = ∠DAG = θ.

$$2\cdot\theta+\eta=180^\circ$$

From the above equations and figures, you can then calculate other aspects, such as the height.

But what about non-regular tetrahedra? Let’s think about other pyramids, both fat and short and tall and thin. All the triangles in the regular tetrahedron have three 60° corners.

Let’s suppose that the angles originating at point D are all equal (we will call them λ), and the edges are also equal (we will call them length 1). We would have something like that of a camera tripod.

If we spread the legs, the camera will get lower to the ground and if we bring the legs together, the camera will get higher. The same is true of our pyramid.

 Figure 4 Figure 5 Figure 6

Let us again bisect the angles λ, as shown in fig. 4.

$$BG = CG= \sin(\frac{\lambda}{2})$$, $$BC= 2\cdot \sin(\frac{\lambda}{2})$$, $$DG= \cos(\frac{\lambda}{2})$$.

Looking at the pyramid’s base ΔABC (fig. 5), $$BC = AC= AB = 2\cdot \sin(\frac{\lambda}{2})$$, then $$AG = BG\cdot\sqrt{3} = \sqrt{3}\cdot sin(\frac{\lambda}{2})$$.

We again us the law of cosines to find μ (fig. 6).

$$\Large\cos\mu=\frac{1+cos^{2}\frac{\lambda}{2} – 3\cdot sin^{2}\frac{\lambda}{2}}{2\cdot cos\frac{\lambda}{2}}$$

If we add $$(1\cdot sin^{2}\frac{\lambda}{2} -1\cdot sin^{2}\frac{\lambda}{2})$$, which equals zero, to the top of the equation, we get $$\large\cos\mu=\frac{1+(cos^{2}\frac{\lambda}{2}) – 4\cdot sin^{2}\frac{\lambda}{2}+(sin^{2}\frac{\lambda}{2})}{2\cdot cos\frac{\lambda}{2}}$$. Note the change of the three to four.

The items in parentheses can be combined using the identity cos² α + sin² α = 1:

$$\Large\cos\mu=\frac{1+1-4\cdot sin^{2}\frac{\lambda}{2}}{2\cdot cos\frac{\lambda}{2}}=\frac{2-4\cdot sin^{2}\frac{\lambda}{2}}{2\cdot cos\frac{\lambda}{2}}=\frac{1-2\cdot sin^{2}\frac{\lambda}{2}}{cos\frac{\lambda}{2}}$$.

Using the identity cos 2α = 1 – 2·sin² α, gives: $$\large\cos\mu=\frac{cos\lambda}{cos\frac{\lambda}{2}}$$.

We can then use this formula to verify the regular tetrahedron’s vertex edge angle that we found earlier. If $$\lambda=60^\circ$$, $$\large\cos\mu=\frac{cos60}{cos30}=\frac{\frac12}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}$$, which is the same as θ above.

This formula can also be used on other polyhedra where there are three identical angles joining at a vertex, such as the cube,$$\large\cos\mu=\frac{cos90}{cos45}=\frac{0}{\frac{1}{\sqrt2}}=0, \mu=90^\circ$$ or dodecahedron, $$\large\cos\mu=\frac{cos108}{cos54}=\frac{\frac{1-\sqrt5}{4}}{\frac{\sqrt{10-2\sqrt5}}{4}}=\frac{1-\sqrt5}{\sqrt{10-2\sqrt5}}, \mu=121.71747^\circ$$.

I will get to the dodecahedron soon. The cube is so easy to figure out mathematically, rather boring.

Next posting, I will show you what happens if the pyramid leans a bit, that is if one of the apex angles is different that the other two.