Truncated Octahedron

Theorem A truncated octahedron may be constructed with a minimum of two tile types.

Proof. The truncated octahedron is an Archimedean solid composed of twenty-four vertices, thirty-six edges, and six squares and eight regular hexagon faces. It requires twenty-four tiles with three arms each. We now present a construction that requires precisely two tile types.

Tile Type One: consists of one cohesive end of type b, one of type a, and one of type â such that the a and â cohesive ends are π/2 apart.

Tile Type Two: consists of one cohesive end of type b̂, one of type a, and one of type â such that the a and â cohesive ends are π/2 apart.

Remark. Notice that the tiles utilize more than one cohesive end type. We do this for the same reason that we used two cohesive end types for the truncated tetrahedron. The truncated octahedron also has two different sorts of edges. Edges of the first sort form the square faces, while edges of the second sort do not border a square face. So, it makes sense to use two separate cohesive end types because arms that make up the two sorts of edges never intermix. Using two cohesive end types decreases the chance that undesirable bonds will form.

Twelve tiles of type one and twelve tiles of type two assemble as shown in Figure 1. There are twenty-four similar bonds composed of a and â arms, and these bonds are equivalent because they orient their members in the same way. Furthermore, there are twelve similar bonds composed of b and b̂ arms which are equivalent because they orient their member tiles in an identical way (see Figure 1).

Notice that it is impossible to form larger or smaller complete structures from these tiles.

We have shown a construction which demonstrates that the absolute minimum we began with is achievable under our present constraints (see Figure 1). We therefore have proven that the minimum number of tiles required to construct the truncated octahedron is two.