On Wed, 14 Nov 2012 09:23:06 +1000, Brad Cooper wrote:
> I expect that this is true... > > We have three points on a Cartesian x-y plane, and the circle that passes through these three points has a constant curvature of k. > > If we have a doubly differentiable curve in the x-y plane that passes through these points, is there always some point on the curve which has curvature k? > > I am finding it tough to prove this. Any help appreciated. > > Cheers, > Brad
If you're having difficulty proving something, it may be worth considering the possibility that it's false.
In this case, if I imagine a V-shaped pair of line segments joining the three points, then rounding the corner of the V so it's twice-differentiable (but widening the V slightly so the curve still goes through the middle point), it's clear that the curvature goes from 0 up through k to a higher value at the middle point, then down through k to 0 again.
However, if I then imagine superimposing a high-frequency "coiling" on this curve, like a telephone cord projected down to 2D, arranging that it still pass through all three points, it seems it should be possible to keep the curvature everywhere higher than some lower bound B > k. (The curve will now self-intersect.)