## 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+l 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+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 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 O¢is 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.

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.