3. The Big Zero Triangle


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 have4)

O æ
ç
è
n
r
ö
÷
ø
,     O¢ æ
ç
è
n-k-l
r
ö
÷
ø
,     O" æ
ç
è
n
r+k+
ö
÷
ø
,
A æ
ç
è
n-l
r
ö
÷
ø
,     B æ
ç
è
n-k
r
ö
÷
ø
,     C æ
ç
è
n-k
r+l
ö
÷
ø
,
D æ
ç
è
n-l
r+k
ö
÷
ø
,     E æ
ç
è
n
r+k
ö
÷
ø
,     F æ
ç
è
n
r+
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 Z1; as to O we have n-k-l < 0, r < 0, s-k-l < 0, since it follows from (3.1)that s < 2l, so that Ois in sextant Z3; and O is in sextant Z2 since n < 0, r+k+l > 0, s-k-l < 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

D
A
×E

we obtain from (1.3), (1.7) and (1.8),

D
A
×E
(-1)r+k(-s+k+l-1)!
(r+k)!(-n+l-1)!
(-1)s-l(s-l)!(-n+l+1)!
(-r-1)!
(r+k)!(s-k)!
n!
(-1)n+k+l(-s+k+l-1)!(s-k)!(s-l)!
(-r-1)!n!
       (3.2)

It is plain that

D
A
×E

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

D
A
×E
C
B
×F

or

A×C ×E = B ×D ×F        (3.3)

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)n-k and (a+b)n-l 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!


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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.

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