Suppose that $v_i$, for $i \in \{1, 2, \ldots 11, 12\}$, are twelve unit length vectors based at the origin in $R^3$. Suppose that $|v_i - v_j| \geq 1$ for all $i \neq j$. What arrangement of the $v_i$ maximizes the number of pairs $\{i,j\}$ so that $|v_i - v_j| = 1$?

If C is a cube of sidelength $\sqrt{2}$ centered at the origin then we can place the $v_i$ at the midpoints of the twelve edges. Taking the convex hull of the $v_i$ gives a cube-octahedron of edge-length one. See here for a picture. If you cut the cubeoctahedron along a hexagonal equator and rotate the top half by sixty degrees you get another polyhedron. Both of these have 24 edges. Are these the unique maximal solutions to the above problem?

Notice that if you place the $v_i$ at the arguably nicer vertices of a icosahedron then the $v_i$ become too widely separated. It is easy to check this by making a physical model!

I spent some time thinking about areas of spherical polygons and restrictions on the graph of edges (and its dual graph) coming from the Euler characteristic. However, I don't think I got very far - in particular ruling out pentagons seems to be a crucial point that I couldn't deal with. Finally, to explain the problem title: instead of thinking of unit vectors with spacing restrictions, consider the (equivalent) problem of placing twelve identical spherical caps, of radius $\pi/12$, on the unit sphere with disjoint interiors in such a way as to maximize the number of points of tangency.

This question was asked of me by an applied mathematician. It comes from a problem involving packing balls in three-space, minimizing some quantity that is computed by knowing pairwise distances. The solution to the kissing problem thus justifies the "twelve" appearing in the problem statement. The projection of surrounding balls to a central one gives the spherical caps.