Puzzles of the {0, 1, 2}^3 grid:                   December 1992
--------------------------------------------------------------------
  Select as much as possible vertices from the grid (3*3*3)
  such that there are no three in a line.

  Solution: maximal number is 16.  (Achim Flammenkamp)

        . o o        o . o        o o .       12 edges &
        o . o        . . .        o . o        4 vertives (tetrahedron)
        o o .        o . o        . o o

--------------------------------------------------------------------
  Select as much as possible vertices from the grid (3*3*3)
  such that there are no four in a plane.

  Solution: maximal number is 8.  (Achim Flammenkamp, Torsten Sillke)
    there is only one solution. It has threefold symmetry.

        o . .        . o o        . . .
        . o .        o . .        o . .
        . o .        . . .        . . o

  Proof of the upper bound 8:   (Achim Flammenkamp)
    As in each layer you can place maximal 3 vertices, you get an upper
    bound of 9. But 9 is impossible. If you place in each layer 3 points,
    you have in each layer 3 lines through the 3 pairs of points. So there
    are 9 lines in the 3 layers. Now count the different slopes a line can
    have in a 3*3 layer. As there are only 8 different slopes, you have two
    parallel lines by the piginhole pinciple. But this means you have four
    on a plane. I assumed, that there are no 3 points in a line in a layer,
    but if there were 3 in a line in the cube then each additional point
    gives a 4 in the plane configuration.