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