Well done to Alex from Stoke-on-Trent Sixth
Form College, and Anurag from Queen Elizabeth's Grammar School,
Horncastle, for their solutions to this problem.
Here is Alex's explanation for the first
part of the problem:
Consider one of the sides painted blue, and rotate the cube so that
this side is on top. Out of the four sides that touch it (and are
rotationally symmetrical), at most three of those are red because
there are three red and three blue, therefore at least one of these
is blue. Rotate the cube so that the top and a selected blue side
is facing. All combinations must be rotatable to this position, so
considering the uncoloured sides is sufficient to deduce the number
of combinations. Out of the four remaining sides, one must be
coloured blue and the other three coloured red. This reduces the
problem to four possible combinations, each found by colouring a
different remaining face blue. Two of these result in three
coloured faces that share a vertex, and the other two form a lines
of colour that bends around the cube with another line of the other
colour in the same shape. Both of these are the same if the colours
are swapped over, so there are exactly two combinations.
This diagram shows the two possibilities.
The left cube has three blue faces meeting at a vertex (and three
red faces on the other side of the cube). The right cube has blue
faces on the front, top, and back of the cube.
Anurag explained very clearly how Cath can
ensure the two cubes are the same:
James has only one possible move first. This is because the cube
could be rotated so the bottom cube is always blue. If Cath now
wants the shared vertex cube, she colours the opposite side to
whichever James paints. This will always produce the desired result
as, in this cube, no opposite sides are the same colour. By
painting the opposite side red, Cath removes James's ability to
colour it blue, thus forcing him to paint a cube where similar
sides meet at a vertex.
However, Cath cannot force a cube in which similar sides are in a
strip. In a strip there must be two similar sides opposite each
other (they share no vertices, and so whatever the third side is, a
strip is formed). Cath must avoid painting any side opposite to a
blue. This is so, on James's last move, his only two options are
the only sides opposite his other, blue coloured side. So, on her
first move, Cath must paint a side adjacent to the side James
painted blue side. But if James now paints the sides opposite to
Cath's, Cath cannot make two of her reds opposite each other, which
she too must need. Which means Cath cannot force the second type of
cube, so to ensure both were the same she would have to make sure
the first cube was the shared vertex type.