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March 28

A (probably) simple problem in plane differential geometry

Take a plane, smooth, non self-intersecting curve C and a circle of fixed radius whose center moves along C. Suppose that C always cuts the circle in two simply connected regions of equal area. Is it true that the curve must be a straight line?

93.150.177.220 (talk) 18:34, 28 March 2019 (UTC)[reply]

I would think so. The slightly weaker statement where we require this to hold for circles of arbitrarily small radius is definitely true (since an arbitrarily small circle can be fit to only enclose a part of C that is strictly concave or convex). It is also true for when C has everywhere nonzero curvature. The only difficult case is when C has zero curvature at some point within every possible placement of the circle. I'm not sure how to prove it rigorously but I think using some calculus of variations would go a long way. We could pick a parametrization of C, when the difference between the two regions' areas is then a functional of the parametrization of C. It would then be sufficient to show that when the parameter is arc length, the second derivative vanishes (equivalent to zero curvature).--Jasper Deng (talk) 20:08, 28 March 2019 (UTC)[reply]

I doubt the fact that the result is intuitive makes it a simple problem. Especially when it comes to differential geometry, where the obvious statement "a closed continuous non self-intersecting curve divides the plane into an interior and an exterior" is devilishly difficult to prove. Tigraan^{Click here to contact me} 12:59, 29 March 2019 (UTC)[reply]
- I think the Jordan curve theorem is not so difficult if limited to smooth curves (like in the OP's question), polygonal paths, or whatever. The difficulty comes from pathological nowhere-differentiable curves and stuff like that. 67.164.113.165 (talk) 15:58, 2 April 2019 (UTC)[reply]

~~The answer is "no". Consider for example radius $2\pi$ and curve $y=\sin(x)$ .~~ Sorry, that doesn't work. However, I see no reason to believe that it couldn't be tweaked to give something that does work. --JBL (talk) 00:51, 3 April 2019 (UTC)[reply]