Dan Asimov writes:
This "puzzle" was evidently harder and/or of less interest than I thought.
Speaking for myself, I thought it was quite interesting; I just didn't see how to get started with it!
Theorem: Given two convex curves A,B in the plane with B lying inside A, then length(B) < length(A).
Several posters have remarked that this theorem was known to Archimedes, and some have asked the question of how he proved it. The answer is, he didn't prove it; he assumed it as an axiomatic property of arc-length. I mean, what else could he have done? Lacking the notion of a limit, I don't think you can really give a satisfactory definition of length, let alone prove theorems about it. I once heard my teacher Andrew Gleason speak very approvingly of Archimedes' approach to arc-length. The monotonicity property of arc-length for convex curves that Dan cites is simple enough to be intuitive, and powerful enough to enable one to prove things about arc-length of various curves of interest (e.g., it enables one to bound the lengths of curves above and below by lengths of polygonal paths).
... 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. ...
A nice argument. Wish I'd thought of that.
It's easy to see there exist such H's if the C^2 condition is dropped.
Details, please?
(This observation was made in collaboration with Joseph Gerver.)
Please say hi to Joe for me. (He and Jamie Simpson and I collaborated on my very first published paper, way back when.) Jim