From: Jim Ferry <"jferry"@[delete_this]uiuc.edu>
Subject: Re: Kiss Precise
Date: Wed, 15 Dec 1999 10:10:09 0600
Newsgroups: sci.math
Jason B wrote:
>
> Does anyone have any information about this in 3d?
>
> Are there any catches, or does the
>
> 2(a^2 + b^2 + c^2 + d^2) = (a + b + c + d)^2 formula hold,
> where a,b,c,d are the curvature of the spheres?
This is the extension to n dimensions:
Let s_j be signed reciprocals of the radii of n+2 mutually
tangent (n1)dimensional spheres in R^n. Choose the signs
of the s_j such that externally tangent spheres have the
same sign, and internally tangent spheres have opposite signs.
Then: n SUM (s_j)^2 = (SUM s_j)^2
Example: You have a unit sphere and three spheres of equal
size, with centers on the midplane of the unit sphere, all
mutually tangent. What is the radius of a sphere tangent
to all of these?
Solution: Let the reciprocal radii be: 1 for the unit
sphere, s for the 3 midplane spheres, and t for the other
one. s and t are positive.
We solve for s by slicing in the midplane and doing the
2d problem: 2 (3s^2 + (1)^2) = (3s1)^2. The positive
solution is s = 1 + 2/sqrt(3).
Now we solve for t: 3(t^2 + 3s^2 + (1)^2) = (t+3s1)^2.
This reduces to (t  (1+sqrt(3)))^2 = 0. (A double root?
Of course! Why?) So the radius of the final sphere is
1/t = (sqrt(3)  1)/2.
 Jim Ferry  Center for Simulation 
++ of Advanced Rockets 
 http://www.uiuc.edu/ph/www/jferry/ ++
 jferry@[delete_this]uiuc.edu  University of Illinois 
==============================================================================
From: dmoews@xraysgi.ims.uconn.edu (David Moews)
Subject: Re: Kiss Precise
Date: 15 Dec 1999 13:23:24 0800
Newsgroups: sci.math
In article <3856B8F4.A070A6D2@uunet.com>, Jason B wrote:
Does anyone have any information about this in 3d?

Are there any catches, or does the

2(a^2 + b^2 + c^2 + d^2) = (a + b + c + d)^2 formula hold, where a,b,c,d
are the curvature of the spheres?
In general, in n dimensions, you need n+2 pairwise tangent hyperspheres, and
the coefficient is n instead of 2:
n(a_1^2 + ... + a_{n+2}^2) = (a_1 + ... + a_{n+2})^2,
where the a_is are appropriately signed reciprocal radii.

David Moews dmoews@xraysgi.ims.uconn.edu