New observations on an old puzzle:

A plane configuration of 16 points with 15 lines of four

An old puzzle (due to Sam Loyd?) asks for 16 points in the plane for which there are 15 lines each containing (at least) four of the points. The solution (GIF, PDF) generalizes: for each even n there is a similar configuration of n2 points with 3n+3 lines each containing n of the points. The configuration has the dihedral symmetry group of a regular (n+1)-gon. Still, the case of n=4 is special, in that it has extra symmetries when viewed projectively: the 10-element group of Euclidean symmetries extends to a 60-element group of projective symmetries, isomorphic with the alternating group A5!

This is the group of rotational symmetries of the regular icosahedron and dodecahedron in Euclidean 3-space; the associated projective 2-space inherits this symmetry group. The vertices of these regular solids then map to A5 orbits of six and ten points respectively, which together constitute our configuration of 16 points. In particular, the central point is projectively equivalent to the other five points circled in the picture. There is just one more orbit of points with a nontrivial stabilizer: the 15 points coming from the edge centers of either the icosahedron or the dodecahedron.

Dually, the action of A5 on the lines of the projective plane has three orbits with nontrivial stabilizer, also of size 6, 10, and 15; the 15-line orbit is the set of four-point lines in our configuration. Here (GIF, PDF) is a picture that shows those lines, and also the 10-line orbit (in green) and the 6-lines orbit (the five red lines plus the line at infinity). The 6-point orbit is again circled, and the 10-point orbit is readily located from the previous picture. Where are the points in the 15-point orbit?

Further remarks:

  1. For n=2 we also get more symmetries than we bargained for: the full group S4 of permutations of our 22=4 points, rather than the 6-element dihedral group. But this is automatic. All ordered quadruples of points in the projective plane, no three of of which are collinear, are projectively equivalent, and any two such ordered quadruples are equivalent under a unique projective transformation. In particular, each permutation of the four points is realized by a unique projective transformation. More generally, in projective space of dimension n-1 over any field F, the group PGLn(F) of projective linear transformations acts simply transitively on ordered (n+1)-tuples of points no n of which are on the same hyperplane; so in particular any unordered such (n+1)-tuple yields a copy of the symmetric group Sn+1 in PGLn(F).
  2. What about odd n? Ignoring the silly n=1, we have n=3, for which nine points can span 12 lines of three -- but not in the real projective plane! Complex coordinates are needed to realize this configuration, which consists of the nine inflection points of any smooth cubic curve (a.k.a. the 3-torsion points of an elliptic curve). In the real plane the maximal number of three-point lines is 10. For n=5, I found many years ago a configuration of 25 points with 18 five-point lines. But this configuration (PS, PDF) does not obviously generalize to any other n, and as far as I know the problem is still open for n=7,9,11,... Nor do I know whether it has been proven that no more than 3n+3 lines can be attained for any n>3.

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