Packing twelve spherical caps to maximize tangencies - MathOverflow most recent 30 from http://mathoverflow.net2013-05-21T08:51:33Zhttp://mathoverflow.net/feeds/question/25042http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/25042/packing-twelve-spherical-caps-to-maximize-tangenciesPacking twelve spherical caps to maximize tangenciesSam Nead2010-05-17T17:59:15Z2010-05-18T15:37:38Z
<blockquote>
<p>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 $|v_i - v_j| = 1$?</p>
</blockquote>
<p>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 <a href="http://en.wikipedia.org/wiki/Cuboctahedron%20" rel="nofollow">here</a>
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?</p>
<p>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!</p>
<p>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. </p>
<p>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 <a href="http://en.wikipedia.org/wiki/Kissing_number_problem" rel="nofollow">kissing problem</a> thus justifies the "twelve" appearing in the problem statement. The projection of surrounding balls to a central one gives the spherical caps. </p>
http://mathoverflow.net/questions/25042/packing-twelve-spherical-caps-to-maximize-tangencies/25124#25124Answer by Alexey Tarasov for Packing twelve spherical caps to maximize tangenciesAlexey Tarasov2010-05-18T13:48:32Z2010-05-18T15:37:38Z<p>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.</p>
<p>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.</p>
<p>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</p>
<p>I solve feasibility of this LP problem (with some tolerance)
After this step all graph were eliminated.</p>