Double duals of polyhedra

The previous post mentioned the dual of a tetrahedron is another tetrahedron. The dual of a cube is an octahedron and the dual of an octahedron is a cube. And the dual of a dodecahedron is an icosahedron, and the dual of an icosahedron is a dodecahedron.

So if you take the dual of a regular solid twice, you get back another regular solid of the same type. But you do not get the same regular solid back. To see this, let's look at how you form the dual of a polyhedron.

To find the dual of a polyhedron, you make a new polyhedron whose vertices are the centroids of the faces of the original polyhedron. The vertices of a regular polyhedron all lie on the surface of a sphere. A face of the polygon lies inside the sphere, so the centroids of the solid are all inside the sphere. Since the centroids form the verticies of a regular solid, they also all lie on a sphere, but a smaller sphere inside the first sphere.

Size of duals

If you take the double dual of a regular solid, the dual of the dual, then you get a smaller solid of the same type. How much smaller? We can get Mathematica to calculate this for us.

The function DualPolyhedron does what the name implies. Let's apply this to the default tetrahedron. When we enter

 DualPolyhedron[ Tetrahedron[] ]

we get back

 Tetrahedron[{2 Pi/3}, Pi}, 1/3]

The default tetrahedron, returned by calling Tetrahedron[], has vertices

• (1, 0, 0)
• (0, 0, 1)
• (1, 0, 1)
• (1, 1, 1)

Each edge has length 1. Its dual has been rotated by 2/3 about the z axis and by about the y axis, and has edges of length 1/3. Taking the dual of a tetrahedron rotates the tetrahedron and shrinks it by a factor of 3 (in each dimension).

Size of double duals

If we take the double dual of a tetrahedron, we get a tetrahedron nine times smaller. Let's see what happens to the rest of the regular solids.

 DualPolyhedron[DualPolyhedron[#]] & /@ {Tetrahedron[], Octahedron[], Cube[], Dodecahedron[], Icosahedron[]}

This tells us again what we found above for the double dual of a tetrahedron. It also tells us that taking the double dual of an octahedron or cube results in an octahedron or cube 3 times smaller. And it tells us that taking the double dual of a dodecahedron or icosahedron reduces its size by a factor of

(5 + 2 5)/15 = 0.6314...

Taking the double dual shrinks a regular solid, and the amount of shrinkage goes down as the number of faces goes up.

Orientation of double duals

The process of taking the double dual may also rotate the solid. The second dual of the standard tetrahedron, for example, is not just 9 times smaller. It's also been rotated. We can see this by calling PolyhedronCoordinates. The code

 PolyhedronCoordinates[ DualPolyhedron[ DualPolyhedron[ Tetrahedron[] ] ] ]

shows that the new vertices are

• (7/9, 2/9, 6/9)
• (7/9, 2/9, 7/9)
• (7/9, 3/9, 7/9)
• (6/9, 2/9, 7/9)

So the second dual of the tetrahedron is not just 9 times smaller.

However, taking the double dual of the rest of the regular solids does not change the orientation. You could verify this for the cube and dodecahedron using Mathematica. It then follows that taking the double dual of octahedra and icosahedra does not change the orientation.

Sixth dual

If you take the dual of a tetrahedron 6 times you do return to the original orientation, but not if you take the dual fewer times. You can verify this by running the following:

 NestList[DualPolyhedron, Tetrahedron[], 6]

Related posts

• Dual polyhedra with Zometool
• Duals and double duals of Banach spaces
• Computing the minimal bounding sphere

The post Double duals of polyhedra first appeared on John D. Cook.