This "puzzle" was evidently harder and/or of less interest than I thought. This may be in part because it depends on a theorem of Cauchy: ----- Theorem: Given two convex curves A,B in the plane with B lying inside A, then length(B) < length(A). ----- If such a mapping as H exists, then since it is C^2, the curvature of the closed curves H(S^1, t) for 1 <= t <= 2 must vary continuously, and in particular for parameter values t close enough to 2, the curvature of H(S^1, t) must remain everywhere close to the boundary component of A -- a circle of radius 2. Thus for t close to 2, the curvature of H(S^1,t) must be close to 1/r = 1/2, and hence positive. So this curve is convex. Then by Cauchy's theorem, the length of H(S^1,t) for t close enough to 2 must be strictly less than L(2) = 4pi, contradicting the condition on lengths. Thus there is no such H, QED. ----- It's easy to see there exist such H's if the C^2 condition is dropped. I find it surprising that such an innocuous- sounding condition as constant lengths of H(S^1,t) implies H must have less than C^2 differentiability. (This observation was made in collaboration with Joseph Gerver.) --Dan << So let me ask a simpler and straightforward version of the question: ----------------------------------------- Let A = { p in R^2 : 1 < ||p|| <= 2 }. PUZZLE: Prove there is no C^2 homeomorphism H : S^1 x (1,2] -> A such that if L(r) denotes the length of the curve H(S^1, r) in A, then L(r) is independent of r. -----------------------------------------
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele