[math-fun] Re: solid conclusion(?)
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? Brahmagupta's (cyclic case). His formula makes it clear that the area is maximized when the vertices lie on a circle. And what about arbitrary pentagons, etc? Are the maximal areas fixed w.r.t. permuting the sides? If the area is maximized when the vertices lie on a circle, we can freely swap adjacent sides, preserving the chord (and circumscribing) arc, and thus area. --rwg
Given a set of sides for a polygon, and suppose the polygon is inscribed in a circle. Bill's remark below, about swapping adjacent sides leaving the area unchanged, means 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? If D = diameter, a side of length A will subtend an angle of 2 arcsin(a/D), and we need the sum arcsin (s/D) = pi as s iterates through the sides. This can be converted to a complicated algebraic equation by requiring the product of (sqrt(D^2-s^2) + i s) to have imaginary part 0. (And taking D as the largest real root.) As the number of sides is increased, this seems unlikely to be solvable in radicals, although it's pretty easy to tackle numerically. Q, repeated: If there are more than 4 sides, is the polygon of maximal area the one inscribed in a circle (and hence independent of the side ordering)? Q2: Given a set of sides, is the polygon of maximal area convex? Suppose you use various plausible definitions of area? Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of Bill Gosper Sent: Tue 7/17/2007 12:05 AM To: math-fun@mailman.xmission.com Cc: rwmgosper@yahoo.com Subject: [math-fun] Re: solid conclusion(?)
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? Brahmagupta's (cyclic case). His formula makes it clear that the area is maximized when the vertices lie on a circle. And what about arbitrary pentagons, etc? Are the maximal areas fixed w.r.t. permuting the sides? If the area is maximized when the vertices lie on a circle, we can freely swap adjacent sides, preserving the chord (and circumscribing) arc, and thus area. --rwg
Rich asked about the area of a polygon inscribed in a circle. The canonical reference is Robbins's 1993 paper cleverly titled "Areas of polygons inscribed in a circle", which takes Heron's formula -- that gives the area of a triangle inscribed in a circle, after all! -- and Brahmagupta's cyclic quadrilateral area formula, generalizes them to 5- and 6-gons, and makes some conjectures about the general case. 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. As Rich guesses, the formalism is to give the polynomial eq'n satisfied by the side lengths and the area; its largest real root is the one you thought of originally, but its other roots give you the area when you inscribe it in smaller circles that give your polygon winding number >1, or where some edges go backwards. (In both cases you define area appropriately -- the integral over the plane of the "winding number around this point" function). And yes, the cyclic polygon is the largest-area one with given side lengths. --Michael Kleber On 7/18/07, Schroeppel, Richard <rschroe@sandia.gov> wrote:
Given a set of sides for a polygon, and suppose the polygon is inscribed in a circle. Bill's remark below, about swapping adjacent sides leaving the area unchanged, means 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? If D = diameter, a side of length A will subtend an angle of 2 arcsin(a/D), and we need the sum arcsin (s/D) = pi as s iterates through the sides. This can be converted to a complicated algebraic equation by requiring the product of (sqrt(D^2-s^2) + i s) to have imaginary part 0. (And taking D as the largest real root.) As the number of sides is increased, this seems unlikely to be solvable in radicals, although it's pretty easy to tackle numerically. Q, repeated: If there are more than 4 sides, is the polygon of maximal area the one inscribed in a circle (and hence independent of the side ordering)? Q2: Given a set of sides, is the polygon of maximal area convex? Suppose you use various plausible definitions of area?
Rich
-----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of Bill Gosper Sent: Tue 7/17/2007 12:05 AM To: math-fun@mailman.xmission.com Cc: rwmgosper@yahoo.com Subject: [math-fun] Re: solid conclusion(?)
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? Brahmagupta's (cyclic case). His formula makes it clear that the area is maximized when the vertices lie on a circle. And what about arbitrary pentagons, etc? Are the maximal areas fixed w.r.t. permuting the sides? If the area is maximized when the vertices lie on a circle, we can freely swap adjacent sides, preserving the chord (and circumscribing) arc, and thus area. --rwg
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Rich asked about the area of a polygon inscribed in a circle...
Incidentally, this touches on a question I posted here a few months ago, about what I then called "pythagorean quadrilaterals" but which are already called "semicyclic polygons", according to that paper of Igor Pak's I mentioned. These are ones in which one of the sides is a diameter of the circumscribed circle; the relation between that diameter and the other sides generalizes Pythagoras, of course. No one bit last time; I'll quote that mail below, just for good measure. I meant to ask Igor about it, but never did... --Michael Kleber On 4/16/07, Michael Kleber <michael.kleber@gmail.com> wrote:
A friend recently mentioned to me the following ARML problem from 1989: "A convex hexagon is inscribed in a circle. If its successive sides are 2, 2, 7, 7, 11, 11, compute the diameter of the circumscribed circle."
(Anyone who wants to solve this on their own may go do so now, and come back later for the rest of my question.)
Since we can reorder sides of this cyclic hexagon, the question is the same as asking for the diameter D of a circle in which you can inscribe a quadrilateral with side lengths 2,7,11,D. So maybe we should generalze from Pythagorean triangle, and say that a convex n-gon is Pythagorean if it has integral "hypotenuse" D and edges a1, a2,...,a_{n-1}, and can be inscribed in a semi-circle with diameter D. The obvious question is, what's the generalization of the Pythagorean theorem?
In the quadrilateral case, the relation you get -- necessraily symmetric on the edges a_i -- turns out to be D * (a1^2 + a2^2 + a3^2) + 2 a1 a2 a3 = D^3 It's maybe more elegant-looking if you say that you're searching for rational solutions to the D=1 version, a^2 + b^2 + c^2 + 2abc = 1 but maybe not, since the LHS isn't homogeneous.
Solutions include all Pythagorean triangles, by setting a3=0, and lots of solutions like 2-2-7-8, whose trigonometric meaning I'll leave you to ponder. Hmm, I suppose the regular hexagon's sol'n 1-1-1-2 also falls into this category.
But the interesting solutions start with the 2-7-11-D from the ARML, then 2-9-12-16, 6-11-14-21, 1-12-22-26, 3-14-25-30, and so on. Do these come up anywhere else?
And what's the relation between the edges and the hypotenuse for the n-gon case?
Likely this is all well-known to those who know it, but not to me...
--Michael
-- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
participants (3)
-
Bill Gosper -
Michael Kleber -
Schroeppel, Richard