8 Aug
2014
8 Aug
'14
9:40 p.m.
On Fri, Aug 8, 2014 at 3:35 PM, Bill Gosper <billgosper@gmail.com> wrote:
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,
k1^2 + k2 k3 + k2 k4 + k3 k5 == k1 k2 + k3^2 + k1 k4 + k2 k5 plus nine more, closing over D_5. Huh? And there seems to be a very weird one involving k0 and k6, which may well fail under scrutiny. --rwg 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