Warren, nice analysis for the triangle case. But we'd like something that also worked for polygons with more than 3 vertices. My first email stated that using csc(alpha/2) for 'curvature' at a vertex would keep polygons self-similar as they shrunk, but that is false. Consider a rectangle--the shorter sides would collapse to a point before the longer sides. We could get self-similarity (at least for the rectangle case) if we defined the vector curvature at a vertex as the sum of the two vector sides from that vertex. Clearly a nonconvex polygon would not remain self-similar. Is there some definition of [vector] curvature that would cause arbitrary polygons to circularize? Maybe a sum of the two vectors weighted by their relative lengths, or some convex function of their relative lengths? I hacked up a little demo at https://ms0.github.io/polycirc.html --ms On 17-Feb-17 12:19, Warren D Smith wrote:

The distance D from a triangle vertex to the triangle's incenter is D = csc(alpha/2) if the inradius=1.

So I presume you meant that alpha is the vertex angle, not pi-alpha, in your claim about csc(alpha/2) being the velocity of the vertex's infall?

Then the most natural candidate functions would clearly seem to be csc(alpha/2)^2 and exp(csc(alpha/2)).

Any F(csc(alpha/2)) where F(x) is a continuous concave-U monotonically increasing function will cause the triangle to become continually "rounder" during the flow, where "roundness" is the variance of the vertex-incenter distances if triangle scaling normalized so inradius=1 always, and is 0 (roundest possible) for an equilateral triangle, else positive.

With this normalization the process continues forever, and roundness continues to improve monotonically forever. Can the process ever get "hung up" approaching from above, but never reaching, some positive roundness value?

No. Because by continuity there would have to be a limit-triangle in that case, which would be a fixed point of the flow. It is easy to see that to be such a limit-triangle you need to have all three incenter-vertex distances equal. The only triangle for which that is true is equilateral.

So ANY such flow, for any such F, always equilateralizes triangles. QED