RE: [math-fun] Re: solid conclusion(?)
Rich writes>we can freely reorder the sides, so the area
is independent of the order for an inscribed polygon. What's the radius of the circle?
For the quadrilateral, sqrt((a d + b c) (b d + a c) (c d + a b)) r = -----------------------------------------, 4 A (which I overlooked in Mathworld and stupidly rederived) where A is the area sqrt((c + b + a - d) (d - c + b + a) (d + c - b + a) (d + c + b - a)) --------------------------------------------------------------------- 4 (Brahmagupta). (Did he really have the general case with the cos^2(opposite angles)? I.e., he knew from trig?) These conjectures map back to my(?) "Brahmagupta" solid angle formula: is it maximal when the pyramid inscribes in a circular cone? Etc for more sides. The trig for the n=4 case is not, so far, repeatable in polite company. MiKle> Igor Pak's recent "The area of cyclic polygons: Recent progress on Robbins' conjectures", available off his web page ( http://math.mit.edu/~pak ), says just about everything there is to know now. rwg>The formulas must be doozies. Wow, the degrees (in z=16 Area^2) go up exponentially: 1, 2, 7, 14, 38, 76,..., n/2 binom(n-1,floor((n-1)/2)) - 2^(n-2), for n = 3,4,5,... . (Sic 'em, Neil.) But how can n=3 (Heron) and n=4 (Brahmagupta) differ, when their formulas are nearly identical? Because the theorem also assumes that the coeffs are polynomials in the *squares* of the sides, producing, for n=4, the product of Brahmagupta and a conjugate describing the nonconvex case. So possibly the n=6 case also factors? Anyway, the n=5 case is already unsolvable in radicals. The area of a general quadrilateral is very simply half the product of the diagonals times the sine of the angle between them, so there are probably simpler hexagon (e.g.) area formulas involving the chord lengths. --rwg
RWG wrote:
MiKle> Igor Pak's recent "The area of cyclic polygons: Recent progress on Robbins' conjectures", available off his web page ( http://math.mit.edu/~pak ), says just about everything there is to know now.
rwg>The formulas must be doozies.
Wow, the degrees (in z=16 Area^2) go up exponentially:
1, 2, 7, 14, 38, 76,..., n/2 binom(n-1,floor((n-1)/2)) - 2^(n-2),
They must: the complex-numbers-with-product-1 formulation of the problem can't tell whether edges go forwards or backwards or how many times you wrap around the circle, and if you have a bunch of edges with slightly different lengths (plus possibly one much shorter edge, if needed for parity reasons), then every choice for winding number and for which subset face backwards will give you a different area. Robbins's original conjecture of the degree was based on calculating how many areas this required, then conjecturing (with good reason) that the minimal polynomial had no other roots.
... Because the theorem also assumes that the coeffs are polynomials in the *squares* of the sides, producing, for n=4, the product of Brahmagupta and a conjugate describing the nonconvex case. So possibly the n=6 case also factors? Anyway, the n=5 case is already unsolvable in radicals.
Yes, the even-n case always factors into a conjugate pair. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
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Michael Kleber -
Ralph Gosper