We revert here to the notation

for the trinomial coefficient, so that we can compare the analogue more easily with the Pascal Hexagon in Figure 1. To see how we might find the appropriate polyhedron to accommodate the trinomial coefficients, when n, r, s, t are any integers subject to r+s+t=n, we make the following observations about the Pascal Hexagon in 2 dimensions:

(1) There are precisely 2³-2=6 regions in the plane, since it is possible to have all combinations of assignments of signs to n, r, s, except n³0, r<0, s<0 and n<0, r³0, s³0.

(2) The lines separating the 6 sextants may be thought of as the extensions of the sides of an equilateral triangle having 1's at each of its vertices, as shown in Figure 1(a).

(3) The three non-zero sextants are as symmetrically arranged as possible and each one contains two boundaries emanating from the center.

(4) Precisely one of the three inequalities changes whenever a line between sextants of the hexagon is crossed (see Figure 6(a)). So we begin our search for the 3-dimensional analogue by observing that for the trinomial coefficients

(1') There are precisely 2⁴-2=14 regions in space, since it is possible to have all combinations of assignments of signs to n, r, s, t except n ³ 0, r < 0, s < 0, t < 0 and n < 0, r ³ 0, s ³ 0, t ³ 0.

Now, what we need is a symmetric polyhedron whose extended face planes produce 14 distinct regions in space satisfying the analogue of (2) in a satisfactory way. So we ask first: What is the 3-dimensional analogue of an equilateral triangle? The natural answer is the regular tetrahedron (and we know that, when n,r,s,t ³ 0, the trinomial coefficients can be arranged in a regular tetrahedral region as illustrated in Figure 5). Now if we extend the face planes of a regular tetrahedron we obtain

Figure 5: The Pascal Tetrahedron.

Voila! It turns out that, when a centrally symmetric boundary is placed around the configuration with flat faces covering each region, the extended face planes of the regular tetahedron produce a polyhedron with 8 equilateral triangular faces and 6 square faces. This is, in fact, the cuboctahedron. Can the cuboctahedron satisfy the rest of our conditions? Perhaps the 4 non-zero regions are trihedral regions off the vertices of a regular tetrahedron with 1's at each vertex and the other regions are the zero regions. We check to see if this is so and to see if we get an arrangement of faces (representing regions) consistent with the analogous statements of (3) and (4). The net diagram of the cuboctahedron in Figure 6(b) shows that it is, indeed, possible to arrange the 14 sets of inequalities so that precisely one inequality changes whenever an edge of the polyhedron is crossed.

Figure 6b: Net diagram for the Pascal Cuboctahedron. Notice that precisely one inequality changes whenever an edge of the polyhedron is crossed.

By copying the diagram in Figure 6(b) and actually constructing the model (it is advisable to draw a tab on every other exterior boundary before cutting the pattern out), one sees that the non-zero regions are located so that they may be thought of as occupying the 4 trihedral regions off the vertices of the central tetrahedron. But there is even more confirmation that this is the right polyhedron. By examining the four hexagonal cross-sections1) of this model one sees that when t=0 the cross-section actually is the Pascal Hexagon! Moreover, the two other hexagons, obtained when s=0 or when r=0, have the identical numerical values. A fourth non-zero hexagonal cross-section occurs when n=-1, and some of its values are shown in Figure 7.

This is certainly the polyhedron we seek. We call it the Pascal Cuboctahedron.