"dy/dx" <email@example.com> wrote in message news:firstname.lastname@example.org... > 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.)
..or it is easy to think of example curves where the curvature stays arbitrarily low (e.g. sort of resembling a large 3-leafed clover).