8 Aug
2014
8 Aug
'14
4:36 p.m.
On Fri, Aug 8, 2014 at 7:05 AM, Bill Gosper <billgosper@gmail.com> wrote:
If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg
Found nothing for n=5, but for n=6, {k1,...,k6} bounded by k0 and k7, k1+k4=k2+k5=k3+k6 (opposite pairs)=3(k0-k7), *and* k1+k3+k5=k2+k4+k6. (Four relations.) --rwg