# 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}

# 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.