On 2/20/07, Daniel Asimov <dasimov@earthlink.net> wrote:
I guess the Claim that Robert refers to can be stated rigorously as follows:
Any continuous space curve of the form z = f(exp(i*theta)) (i.e., {cos(theta), sin(theta), z(theta), 0 <= theta <= 2pi}, where z has period theta) contains 4 points in space that form the corners of a square . . . as long as z = f(x+iy) is differentiable and satisfies (dz/dy)/(dz/dy) <= sqrt(1/2). --------------------------------------------------------------------------------------
The misprint fairy's been visiting again here ...
A[n apparently as yet unproved] theorem that would imply both of these is this:
(*) Claim: Any simple closed curve in R^3 contains the 4 corners of a (planar) square.
--Dan
P.S. I will say, smugly, that I believe I know how to prove (*).
Intriguing --- but if (*) were true, why would any restriction on the gradient of the floor (presumably assumed horizontal at some point) be necessary? WFL