On 2/20/07, Daniel Asimov <dasimov@earthlink.net> wrote:
(*) 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 (*).
This seems unlikely: a plane meeting a curve in (say) 4 points has freedom 3; but those points forming a square imposes 4 constraints. Let's consider a concrete example. Consider the closed curve comprising the twisted cubic (x,y,z) = (t, t^2, t^3) for 0 <= t <= 1, together with the segment of line x = y = z joining its ends. As was pointed out quite recently in another thread, no 4 points of the cubic are coplanar: so a general plane meets the curve in 3 points of the cubic and 1 point of the line. [If the entire line lies on the plane, there is only 1 other common point on the cubic; so no square is possible.] Let the plane meet the cubic at points U,V,W, where t = u,v,w, resp: then its equation is ax+by+cz+d = 0 with a = u*w+v*u+v*w -b = u+w+v c = 1 -d = u*w*v; and it meets the line at point S = (s,s,s), where (a+b+c)s + d = 0, s = -d/(a+b+c). To form a square, these points must satisfy the 4 equations in 3 variables u,v,w, |U - V|^2 = |U - W|^2, |U - V|^2 = |S - W|^2, |V - W|^2 = 2|U - V|^2, |U - V|^2 = |S - V|^2. Maple's approximate solver fsolve() reports that even the first three of these equations have no solution in the interval 0 <= u,v,w <= 1. In other words, there does not even exist a quadrilateral with 3 equal sides and one right-angle. [One might take this with a grain of salt, since there is obviously the trivial solution u = v = w = 0: but further experimentation strongly suggests that no solution with distinct u,v,w exists; and with sufficient industry the computation could presumably be converted into a pukka proof of a counterexample.] Shame about that, Dan! Fred Lunnon