Yesterday Zorn Leung conjectured that if a plane curve has curvature , then it is contained in a circle (of some radius). I will give an example to show that this is actually false.

The example is the following curve:

This curve is called the prolate cycloid (I’ve just learned this name today). Geometrically it is the path traced out by a fixed point at a radius (here ) , where is the radius of a rolling circle (the animation is not of the same scale as my example because I am too lazy to make the animation myself, I just download it from the net):

It is easy to calculate that

noting that

It is not necessary to calculate exactly the minimum value of , just observe that is periodic and that for all , this implies for some constant .

But it is easy to see that is unbounded, and in particular it cannot be contained in any circle, no matter how large we choose the radius.

Remark 1The conjecture should be true if the curve is simple closed and the radius of the circle can be chosen to be . I haven’t written down the full details of the proof, though. I think it would be a nice exercise for the geometry students ☺

If the curve is closed, then it is compact. Therefore, we do not need the curvature condition. Perhaps, the interesting case is that the curve is simple (no self-intersection) and has a positive lower bound of curvature. It should be a consequence of Fundamental Theorem of DG.

Of course a closed curve must be contained in a large circle, since it’s compact as you described (and in the remark I’ve actually forgotten to state that it is simple), the emphasis is that the radius can be . It would be nice if someone (perhaps… you?) can post a proof of it here.

And by the way, forgive my ignorance, what is the fundamental theorem of differential geometry?

The curvature function determines a plane curve up to rigid motion.