### Engaging math activities for the summer break - Day 11

**The setup** A cube whose faces have been partitioned by the midlines into four squares.

**The activity** Color the squares in as many colors as you wish, with a single caveat: no two adjacent squares, i.e., squares that share an edge, may be of the same color.

**The task** Determine the maximum possible number of squares of the same color.

It seems rather natural to focus on coloring a face at a time. However, *shifting a view point* leads to a faster track. Focus on the squares that form a cube's vertex. There are 8 vertices and at each 3 pairwise adjacent squares. This means that at a every vertex there could be only one square of a particular color, meaning that there could be no more than 8 squares of the same color.

Here is a cube's net with vertices numbered 1 through 8. It is convenient in case you do not plan to color (or create for that purpose) a physical cube.

Having 24 squares and 8 as the maximum number of squares of the same color suggests that 3 colors may suffice to color the cube. This is indeed so but needs to be verified - it does not automatically follows from what has been said so far.

Here is a coloring my 12 years old boy came up with:

**Related problems for a follow-up** The square faces of a cube could be cut by the diagonals into four triangles. The cube with so partitioned faces could be colored with just 2 colors each covering 12 triangles. This is the maximum possible because the center of each of the six faces is shared by 4 triangles so that at most two may be of the same color.

Further, a similar activity could be carried on a tetrahedron whose face have been partitioned by the midlines into four triangles:

For convenience the vertices and and additional nodes could be numbered:

Here is a coloring with 6 red triangles, 5 blue and 5 yellow triangles. Is 6 the possible maximum?