Figure 3: A configuration in the (positive) Pascal Triangle.
Now consider Figure 3. We have W1(OADE)
= W1(OBCF) (invariance under flipping). Thus
or
The geometric meaning of this is displayed in Figure 4. Although
we arrived at (2.2) by starting with a parallelogram, we may
view this result as follows: We start with an equilateral triangle
OO¢O¢¢,
of side k+l, with k
> l, and sides parallel to the directions for
which r, s and n are constant.
We cut off equal equilateral triangles, of side l, from
each corner. The result is a semi-regular hexagon ABCDEF,
where AB = CD = EF( = k-l), BC = DE = FA( = l),
all interior angles are 120 degrees and therefore opposite edges
are parallel (see Figure 4). Then (2.2) holds, which
means that the product of the vertices of the two triangles shown in Figure
4 are equal. We call this the Generalized Star of David Theorem,
featured with the invariance of weights in [5]. We should add
that our original equilateral triangle could just as well have pointed
downward as upward (see Figure 5
which shows the two equilateral triangles
that could have produced the same case of our Generalized Star of David
Theorem).
We obtain a genuine Star of David if and only if k = 2l.
Then ABCDEF is a regular hexagon, so that Figure
4 shows a genuine Star of David; moreover, the two equilateral triangles
of Figure 5 themselves form a Star of David.
Actually the original
Star of David Theorem (see [1, 7])
even required k = 2, l = 1, and did not use the
term3)
'Star of David'.
Figure 4: Highlighting the star of
Figure 3, where A×C×E = B×D×F.
Figure 5: Showing the star of Figure 4 could have come from either of
two different equilateral triangles, one pointing up and the other pointing down.
We come now to the question motivating this note. What happens
if we enlarge the positive Pascal Triangle to the whole hexagon?
As you might expect, our invariance of weight theorems remain valid if
we stay within a (non-zero) sextant of the hexagon. Moreover, and
a little more remarkably, we may slide our parallelograms, in the allowed
direction, from a non-zero sextant, through a zero 'no mans land', into
another non-zero sextant and the weight remains invariant. These
claims can be verified by straightforward calculation. However, what
is even more remarkable is that we may consider parallelograms - or, equivalently,
equilateral triangles - which straddle the three non-zero sextants of the
Pascal Hexagon, and again get invariance theorems and Star of David Theorems.
This we will now show.
2
Recall again - and for the rest of this paper - that s = n-r.
3
We discuss the history of the Star of David Theorem in a brief appendix
to this paper.