Consider the Pascal Hexagon as shown cryptically in Figure 1(a).
We are going to construct an equilateral triangle
OO¢O¢¢,
of side k+l, k > l,
as in Section 2 but now with its vertices one
in each of the zero sextants.
We now cut off equal equilateral triangles OAF,
O¢BC,
O¢¢DE, of side l,
so that A, B lie in sextant III, C, D lie
in sextant II, and E, F; lie in sextant I
(see Figure 6). Precisely we
have^{4)}
O 
æ
ç
è 
n
r 
ö
÷
ø 
, O¢ 
æ
ç
è 
nkl
r 
ö
÷
ø 
, O" 
æ
ç
è 
n
r+k+l 
ö
÷
ø 
, 

A 
æ
ç
è 
nl
r 
ö
÷
ø 
, B 
æ
ç
è 
nk
r 
ö
÷
ø 
, C 
æ
ç
è 
nk
r+l 
ö
÷
ø 
, 

D 
æ
ç
è 
nl
r+k 
ö
÷
ø 
, E 
æ
ç
è 
n
r+k 
ö
÷
ø 
, F 
æ
ç
è 
n
r+l 
ö
÷
ø 
, 

where, to ensure that the points be in their assigned sextants, we require
0 £ n <
l, l £
r < 0, s ³
k. (3.1) 

Then O, O', O'' are zero vertices.
This is obvious for O which is in the sextant Z_{1};
as to O¢
we have nkl < 0, r < 0, skl < 0, since it follows from
(3.1)that s < 2l, so that O¢is in sextant Z_{3}; and O¢¢ is in sextant Z_{2} since n < 0, r+k+l > 0, skl < 0. Thus we can get no information by
sliding parallelograms like OADE in the direction of constant s.
However, we do get a Star of David Theorem. For if we compute
we obtain from (1.3),
(1.7) and (1.8),

D
A ×E 
= 
(1)^{r+k}(s+k+l1)!
(r+k)!(n+l1)! 
. 
(1)^{sl}(sl)!(n+l+1)!
(r1)! 
. 
(r+k)!(sk)!
n! 


= 
(1)^{n+k+l}(s+k+l1)!(sk)!(sl)!
(r1)!n! 
(3.2) 

It is plain that
is symmetric in k and l. However, the interchange of k
and l exchanges A and B, D and C, E and F.
Since A, B lie in
the same sextant, and likewise D, C and E, F, it follows
that
or
yielding a Star of David Theorem. This is truly remarkable since, from an
analytic point of view, the vertices A, B come into play precisely
when the vertices C, D are not relevant (i. e., in the expansions of
(a+b)^{nk} and
(a+b)^{nl} with a > b).It seems
that the geometry exercises a very powerful influence, blotting out any
analytic scruples.
Notice also, intriguingly, that we obtain (3.3) by cutting out the zero
factors from zero weights!
Figure 6: Unexpectedly the relation A×C×E = B×D×F survives.
^{4}
Notice that here, and henceforth, we drop the s from the symbol
for the binomial coefficient.
