Combinatorics question about alternately-coloured diagonal halves of sides of a cube

Question

Diagonal halves of each side of a cube are painted in alternate colours. Let the vertex at which such a half forms a right angle be its base vertex. What is the minimum number and the maximum number possible of common base vertices of three halves of the same colour?

Answer 1

There are $4^6$=4,096 possible different cubes, ignoring any duplicates that are due to orientation or reflection. I was going to simply have a program loop through each of them and count the common base vertices of 3 halves of the same color, but before I could do that, I realized that the minimum number of base vertices is 0, as in the example below,

and, when you force the constraint that each side of the cube MUST have 2 different colors, then the maximum number of common base vertices of 3 halves of the same color is 2, as in the example below

The red circles mark the common base vertices of 3 halves of the same color. Why can't we have more than 2? Because of the alternate coloring constraint. The 2 base vertices circled in orange and green can never be the same color. (Note, the bottom green circle is referring only to the upper box for this point.)

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EDIT: Another way to think about it:
If we remove the coloring constraint, and each triangle could be the same color, then all 8 points would be a common base vertex with 3 identical colors.

However, let's say we then introduce the rule that each side must contain both colors. As you create a cube with one of the vertices all the same color, you must realize that you are also forcing the other color to be part of the adjacent group of three vertices, and only one of those remaining 3 groups can contain triangles all of the same color.

For example, take the top right red circle in the picture. Let's say that that you set those three halves to blue to create your first group of 3. In doing this, you are forcing one member of each of the 3 adjacent groups to be white (top orange circle, bottom red circle, and top green circle). Since there are 12 triangles total, you might first think (as I did) that there are at most 4 groups of three with each group having the same color, but achieving this is not possible because the two colors circled in orange will always be different, because of the coloring constraint. Same thing for green.

In other words, you can choose a corner to give 3 identical colors to, and then you may choose a 2nd corner to give three identical colors to, but in choosing the second corner, you guarantee that the remaining 2 corners cannot ever have 3 triangles of the same color.