# Math on cub 3

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.