Excellent proof! By the way, for the 'obvious' lemma that the girth is an increasing function of the two semi-axes, it follows from: Theorem: If L is a compact subset of R^n and K is a convex compact subset of L, then the surface area of K is bounded above by the surface area of L, with equality if and only if K = L. Proof: The map f : boundary(K) --> boundary(L) given by outward normal projection cannot decrease area. QED. Incidentally, does anyone have a reference for this beautiful theorem? It's well known, but I don't know what it's called, and I'd like to cite it in my thesis. Thank you! Best wishes, Adam P. Goucher
Sent: Monday, June 24, 2019 at 12:19 AM From: "Brad Klee" <bradklee@gmail.com> To: "Dan Asimov" <dasimov@earthlink.net>, math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Is it me, or is math.stackexchange.com controlled by morons?
If we lengthen axis a, we must shorten axis b and c to preserve the ab and ac girths. Consequently the bc girth decreases, a contradiction. The inverse holds for shortening axis a.
A root-solving algorithm should converge to arbitrary precision given enough time and a good enough initial guess.
—Brad
On Jun 23, 2019, at 5:24 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Come to think of it, how can we prove that the mapping (0,oo)^3 —> (0,oo)^3 of positive octants via
(a,b,c) |—> (girth_ab, girth_bc, girth_ca)
is one-to-one? Or is it?
—Dan
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