On Tue, Aug 5, 2014 at 1:34 PM, Bill Gosper <billgosper@gmail.com> wrote:

at the vertices of some octahedron. A fifth and sixth sphere are tangent to all four, a little one inside, a big one engulfing all. Then the sum of the curvatures of the fifth and sixth = the sum of the curvatures of the original four. http://www.tweedledum.com/rwg/Sodddy.htm --rwg

Dear Bill, Surely this is well-known? [Attached: Two notes on the 2D case] Mike Hirschhorn. -------------- Dear Mike, at best it's an immediate consequence of (d2) k5,k6 = (±3 rt3 k1 k2 k3 k4 V + k1 + k2 + k3 + k4)/2 (from http://www.tweedledum.com/rwg/Sodddy.htm), where V is the volume of the tetrahedron formed by the centers of the original four, and k6 *usually* must be taken negative. E.g., if 1 = k1 = k2 = k3 = k4, then k5,k6 = 2 ± rt6. (And the original theorem becomes 4 = 4.) Perhaps mildly interesting is that this 3D case is a little nicer than in 2D, where we have an average instead of a sum: (k4 + k5)/2 = k1 + k2 + k3 vs k5 + k6 = k1 + k2 + k3 +k4 . --rwg