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?