From: robbins@idacrd.UUCP (David Robbins)
Newsgroups: sci.math
Subject: Re: Area of five sided polygon needed
Date: 19 Oct 92 13:13:06 GMT

In article <1992Oct13.040617.16321@husc3.harvard.edu> kubo@birkhoff.harvard.edu (Tal Kubo) writes:
>In article <1992Oct12.200125.826@altair.selu.edu> fcs$1224@altair.selu.edu writes:
>>Given a five sided polygon, is there a formula, in terms
>>of its sides, for the area?
>
>No, the sides don't uniquely determine the area, but...
>
>3. I've heard of Brahmagupta-type polynomial equations for the areas of
>   *cyclic* N-gons, in terms of their sides. For N=3 (Heron) and N=4
>   (Brahmagupta), the equations are well-known and of degree 4 in the
>   sides.  For N=5,6,7 the degrees were said to be 7,7, and 38.
>   Unfortunately I didn't see the formulas.   Can someone reading this
>   provide more information?
>
I think I may have been the ultimate source of this information.  I am
writing a paper on this subject.  In more detail it gives evidence for
the conjecture that if K is the function giving the area of a convex
cyclic n-gon with sides a_1,a_2,...,a_n, then 16K^2 is integral over
the domain of the symmetric polynomials in a_1,...,a_n with integer
coefficients.  The (monic) minimum polynomial is of degree D_m in
16K^2 when n=2m+1 and n=2m+2, and D_1,D_2,... = 1,7,38,... or, more
generally,

            D_m=/sum_{i=0}^m-1 (m-i) {2m+1 \choose i} .

For the case n=5 and n=6 it gives the formula explicity (thus proving
the conjecture in this case).  Though difficult to discover the formulas
for n=5 and n=6 can be written in a remarkable compact form which also
makes them relatively easy to verify:  Here is the formula for n=6.

Let s_1,s_2,s_3,s_4,s_5 be the first five elementary symmetric
functions of the squares of the six sides.  (For example
s_1=a_1^2+...+a_6^2.).  Let S_6 be the product of the six sides and,
for brevity, denote 16K^2 by A.

Next define

        t_2 = A - (4s_2 - s_1^2)
        t_3 = 8s_1s_3 + s_1t_2 + 16S_6
        t_4 = t_2^2 - 64s_4 - 64s_1S_6
        t_5 = 128s_5 - 32t_2S_6

Then the equation satisfied by the area is

     -t_3^2t_4^2 - At_4^3 + 16t_3^3t_5 + 18t_3t_4t_5 + 27 A^2t_5^2 = 0

The "easy" verification (not given here) of the formula derives from a
very remarkable observation of Bradley Brock that the left side of the
preceding formula, which was found through a rather complicated
computational process, is essentially the discriminant of the cubic
(in z)

        z^3 + 2t_3z^2 - At_4z + 2A^2t_5 .

The formula for the pentagon can be obtained by setting S_6=0.

David Robbins    robbins@ccr-p.ida.org.
>-Tal   kubo@math.harvard.edu