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     2. On Sliding parallelograms and the Star of David Theorem


Let us first confine attention to the positive Pascal Triangle. Consider the parallelogram OADE (this is part of Figure 3) with vertices

O æ
ç
è
n
    s
ö
÷
ø
,     A æ
ç
è
n-l
    s-l
ö
÷
ø
,    D æ
ç
è
n-l
r+k     s-k-l
ö
÷
ø
,     E æ
ç
è
n
r+k    s-l
ö
÷
ø

We define the weight of this parallelogram to be

W1(OADE) =  O×D
A
×E
(s-k)!(s-l)!
s!(s-k-l)!

by straightforward calculation2) Notice that our parallelogram had one pair of sides running in the direction for which r is constant, and the other pair of sides running in the direction for which n is constant. Our result

W1(OADE) =  O×D
A
×E
(s-k)!(s-l)!
s!(s-k-l)!
      (2.1) 

shows that W1 is invariant under sliding in the direction for which s is constant, and under the interchange of k and l, which we call flipping. Exactly similar results apply to parallelograms with their sides running in the directions for which s and n are constant (we then call the weight W2)and parallelograms with their sides running in the directions for which r and s are constant (we then call the weight W3).

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Figure 3: A configuration in the (positive) Pascal Triangle.

Now consider Figure 3. We have W1(OADE) = W1(OBCF) (invariance under flipping). Thus

O×D
A
×E
O×C
B
×F

or

A×C ×E = B ×D ×F                           (2.2)

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'.

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Figure 4: Highlighting the star of Figure 3, where A×C×E = B×D×F.



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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.


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