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
(1) There are precisely 2 (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 2
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
(b) 4 truncated trihedral regions off the faces, and (c) 6 wedges off the edges (with square cross sections). Figure 5: The Pascal Tetrahedron.
Figure 6a: Notice that precisely one inequality
changes whenever a line between sextants of the hexagon 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-sections
This is certainly the polyhedron we seek. We call it the Figure 7: Hexagonal cross-section of the Pascal Cuboctahedron when n=-1
and -4 £r, s, t
£ 3.^{1}
Here, as in the original case of the hexagon, one approximates to
regularity as the values of
n, r, s, t
tend to ±¥. |