On 2/7/11, Bill Gosper <billgosper@gmail.com> wrote:
Ah, from 3.5 yrs ago, ...
If there is a connection with the original thread, I'm afraid it has escaped me. However ...
This is analogous to the maximal area of a quadrilateral with sides a,b,c,d (or a,b,d,c):
sqrt((c + b + a - d) (d - c + b + a) (d + c - b + a) (d + c + b - a)) ---------------------------------------------------------------------, 4 with diagonals
(a d + b c) (b d + a c) (b d + a c) (c d + a b) sqrt(-----------------------) and sqrt(-----------------------). c d + a b a d + b c
Whose formula is this? (Superhero of Alexandria?)
See http://en.wikipedia.org/wiki/Brahmagupta's_formula http://en.wikipedia.org/wiki/Cyclic_quadrilateral
And what about arbitrary pentagons, etc? Are the maximal areas fixed w.r.t. permuting the sides? The formulas must be doozies. --rwg
It doesn't seem generally appreciated that this result leads directly to the isoperimetric property of the circle. Given the side-lengths of any plane polygon, its area may be increased by adjusting any 4 successive angles until their vertices are concyclic. Hence the area is a maximum when the polygon is concyclic; now let the number of sides go to infinity. I seem to have missed the angular analogue below (Billygosper's formula?) last time round. Does it lead to some sort of dual to the isoperimetric theorem, I wonder ... WFL
From gosper@alum.mit.edu Mon Jul 16 08:47:39 2007 Date: Mon, 16 Jul 2007 10:39:48 -0400 (EDT) Message-ID: <23173122.9411184596788302.JavaMail.gbourne@brunch.mit.edu> From: Bill Gosper <gosper@alum.mit.edu> To: math-fun@mailman.xmission.com Lemma: The apical solid angle of an isosceles trapezoidal pyramid with vertex angles a,b,a,c is
2 c b c 2 b - sin (-) + (cos(a) - 1) sin(-) sin(-) - sin (-) + cos(a) + 1 2 2 2 2 (d79) 2 acos(-------------------------------------------------------------) b c (cos(a) + 1) cos(-) cos(-) 2 2
I derived this by splitting the general a,b,a,c hedral vertex into trihedrals a,b,d and d,a,c, then maximizing over d, getting
b c (d38) cos(d) = cos(a) - 2 sin(-) sin(-) 2 2
Clearer might have been the difference between two isosceles trihedrals:
solid_angle(d,d,b)-solid_angle(d-a,d-a,c)
with d chosen to equalize the dihedrals, using
dihedral(a,b,c):=acos(csc(a)*csc(b)*cos(c)-cot(a)*cot(b)),
which gives sin(a) d = atan(----------------------), b c cos(a) - csc(-) sin(-) 2 2
and eventually the same solid angle formula. But the max technique generalizes from trapezoidal pyramids to arbitrary quadrilateral pyramids with apex angles a,b,c,d, whose maximal solid angle is
cos(d) + cos(c) + cos(b) + cos(a) a b c d --------------------------------- - tan(-) tan(-) tan(-) tan(-) a b c d 2 2 2 2 4 cos(-) cos(-) cos(-) cos(-) 2 2 2 2
Note we get the old formula when d=0. Note also the symmetry, giving the same result for a,b,d,c. In this maximal case, the "diagonal angles", i.e. the apex angles exposed by splitting into two triangular pyramids, are a d b c b d a c 2 (sin(-) sin(-) + sin(-) sin(-)) (sin(-) sin(-) + sin(-) sin(-)) 2 2 2 2 2 2 2 2 acos(1 - -----------------------------------------------------------------) c d a b sin(-) sin(-) + sin(-) sin(-) 2 2 2 2
and same(a<->c).