# 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}

# 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.