Soddy's theorem on mutually tangent circles, generalized to n dimensions.

(c5)
(makelist(k[n],n,0,3),%%^^2=(%%.[1,1,1,1])^2/2)

k_i is the curvature (reciprocal radius) of circle_i

The solution for one curvature in terms of the others is a slightly messy radical, but very nice as

where A is the area of the triangle joining the other three centers. Negating A gives the "circumcurvature" for the large, surrounding circle.

(c1)
load("\\macsyma2\\user\\sphod.mac")$

(c2)
hero(a,b,c):=sqrt((a+b-c)*(a-b+c)*(b+c-a)*(a+b+c))/4

(c27)
block([plotnum:5],sphod(1.,.75,2/3.,1/3.))$

The three dimensional case relates the curvatures of five mutually tangent spheres.

(c5)
lasttime

(d5) [3906159, 0]

(c1)
(makelist(k[n],n,0,4),3*%%^^2=(%%.[1,1,1,1,1])^2)

(c2)
k[0]=(sum(k[n],n,1,4)+3*sqrt(3)*prod(k[n],n,1,4)*\v)/2

(c14)
fundef(arcfrac)

(c19)

(c28)

(c3)
block([plotnum:5],sphod(1.,2/7.,.75,2/3.))$

Again, the expression for one curvature in terms of the rest is messy, unless you use

where V is the volume of the tetrahedron formed by the other four centers. Without V,

(c4)

In n dimensions, n+2 mutually tangent hyperspheres must have curvatures satisfying

Solving for k₀

(c3)
k[0]=(funmake_no_simp("*",[sqrt(n),sqrt(sum(k[i],i,1,n+1)^2-(n-1)*sum(k[i]^2,i,1,n+1))])+sum(k[i],i,1,n+1))/(n-1)

Note that the radicand vanishes exactly in the n-1 dimensional case,

i.e., the n+1 other (hyper)spheres are co(hyper)planar.

(c5)
(sum(k[i]^2*prod(n+(k[j]+k[i])/(k[j]-k[i]),j,0,n+1),i,0,n+1)=0)=-(n-1)^n*lhs(d1)

where the product skips j=i. (This lhs is a "non-local derangement" of the rhs, and is thus an alternative form for the Soddy condition.)