Mutually tangent ellipsoids in 3 space I recently heard a claim that for any n, it is possible to arrange n ellipsoids in 3 space such that each pair of ellipsoids is kissing.  Is this true, and if so, how?
Edit: By kissing, I mean that I would like the interiors of the ellipsoids to be disjoint, but each pair of ellipsoids should intersect at a point.
 A: Here is the paper that Gerhard remembered, which doesn't answer the question as posed,
but does answer a related question:

Jeff Erickson and Scott Kim,
  "Arbitrarily large neighborly families of congruent symmetric convex 3-polytopes," 2003.
  (link)


      



                  


That their paper does not mention ellipsoids might be taken as indirect evidence that the posed question may not yet have been answered a decade ago.
A: I've never heard of this before, and it sounds quite counterintuitive. If one is granted that this is true, then one can try to work backwards and apply a bit of dimensional analysis to understand how this can be. 
First, note that ellipsoids are quadrics, and there are a $9$ dimensional space of quadrics. Also, note that ellipsoids are invariant under affine (or more generally projective) transformations, so if we have one such configuration, we will expect to get at least a $15$-dimensional family of kissing ellipsoids. 
On the other hand, there are degenerate quadrics for which this is true. Consider $n$ mutually non-parallel lines in the plane. We can consider these as degenerate ellipsoids, with two axes $0$ radius and one axis $\infty$ radius. They the are each mutually tangent, since they each meet in a point. There is an $2n+3$-parameter family of these. One could hope to "regenerate" families of $n$ tangent ellipsoids from these, but there might be restrictions on when this is possible. 
For $n=1$, there is a $9$ dimensional family.
For $n=2$, there is a $17=9+8$ dimensional family. We have $9$ dimensions for the first ellipsoid. Choosing any other ellipsoid with center disjoint from the first, we may rescale it uniquely to be tangent to the first ellipsoid. So this gives us $8$ dimensions for the second ellipsoid. 
For $n=3$, let's assume that one of the ellipsoids is a sphere. Then its center is equidistant from the other two ellipsoids. The space of points equidistant from two kissing ellipsoids is $2$-dimensional. So we get a $17+2=19$-dimensional space of 3 tangent ellipsoids, with one round. We may change the round sphere by an affine transformation based at its center to get any ellipsoid with the same center, and we have a $6$-parameter family of such affine maps (we have $3$ dimensions for the major axis, $2$ for the minor axis, and $1$ for the third axis). However, this over-counts, since a similarity (all $3$ axes equal) will take the sphere to a sphere, so this gives us $19+6-1=24$ dimensions (or we may take a $5$-parameter family of volume-preserving affine transformations to eliminate repetitions). 
For $n=4$, let's again assume that one of the ellipsoids is a sphere. The space of points equidistant from $3$ ellipsoids is $1$-dimensional. As above, we may modify the sphere by a $5$-parameter family of volume-preserving affine transformations to get $24+1+5=30$ dimensions of 4 mutually tangent ellipsoids. 
For $n=5$, we expect finitely many points which are equidistant from $4$ mutually tangent ellipsoids, so the computation gives $35$ dimensions. 
At this point, one expects adding the next sphere will cut down on the dimension of the space of 5 tangent ellipsoids which have an equidistant point. If we continue the sequence $9,17,24,30,35$, then the next term ought to be $39$ dimensions ($42, 44, 45, ?$). Of course, this trend couldn't possibly continue if one expects to have $n$ tangent ellipsoids for all $n$, so there must be some non-generic phenomenon creating the incidence.  
A: I'm assuming "kissing" means osculating, i.e. the ellipsoids intersect at a point where they have second-order contact, i.e. in a coordinate system where the point of contact is the origin and the tangent plane is the $xy$ plane, near the origin we have $z = a x^2 + b x y + c y^2 + O((|x|+|y|)^3)$ for the same $a,b,c$.  Well, e.g. this is true for the ellipsoids
$$ x^2 + y^2 + b (z - 1/b)^2 = 1/b,\ b > 0$$
which are all mutually osculating at the origin.
Or did you want each pair to be osculating at a different point?
