
> 
> 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](https://en.wikipedia.org/wiki/Cuboctahedron)
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](https://en.wikipedia.org/wiki/Kissing_number_problem) thus justifies the "twelve" appearing in the problem statement. The projection of surrounding balls to a central one gives the spherical caps.



Packing twelve spherical caps to maximize tangencies
Interesting question. I can find answer using my program, which was made for solving Tammes problem for 13 points. But I need some time for answer.


UPD: I wrote program. Result: 24 is a maximal number of edges.
I did in three steps.
First, I enumerated planar graphs with 12 vertices with at least 25 edges, at most 5 edges in a vertex and at most hexagonal faces.
Total number of suc graphs is 67497.


Second, I eliminated by linear programming by considering values of face angles as variables. 
My constrains was:
1. angle in triangle is ~1.2310
2. each angle no less than 1.2310
3. sum of angles around vertex is 2\*pi
4. opposite angles of rectangle are equal
5. sum of non-opposite angles in rectangle between 3.607 and 3.8213


I solve feasibility of this LP problem (with some tolerance)
After this step all graph were eliminated.


