Imagine you are suspending a cube from one vertex (corner) and
allowing it to hang freely. Now imagine you are lowering it into
water until it is exactly half submerged. What shape does the
surface of the water make around the cube?
It is known that the area of the largest equilateral triangular
section of a cube is 140sq cm. What is the side length of the cube?
The distances between the centres of two adjacent faces of another
cube is 8cms. What is the side length of this cube? Another cube
has an edge length of 12cm. At each vertex a tetrahedron with three
mutually perpendicular edges of length 4cm is sliced away. What is
the surface area and volume of the remaining solid?
Imagine you have six different colours of paint. You paint a cube
using a different colour for each of the six faces. How many
different cubes can be painted using the same set of six colours?
There are $12$ edges on a cube so there are $12$ winning lines along edges.
There are $6$ faces on a cube, and $4$ winning lines that pass through the middle of each face, so there are $24$ winning lines through the middle of faces.
Finally we need to consider the winning lines that go through the centre cube:
vertex to opposite vertex: $4$
middle of edge to middle of opposite edge: $6$
middle of face to middle of opposite face: $3$
In total, there are $12 + 24 + 4 + 6 + 3 = 49$ winning lines.
From a vertex there are $7$ other vertices that you can join to in order to make a winning line. $7 \times 8 = 56$ lines, but this counts each line from both ends, so there are $28$ 'vertex' winning lines.
From the middle of an edge there are $3$ other middles-of-edges that you can join to in order to make a winning line. $3 \times 12 = 36$ lines, but this counts each line from both ends so there are $18$ 'middle of edge' winning lines.
From the centre of each face there is one winning line, joining to the opposite face, so there are $3$ 'centre of face' winning lines.
So in total, there are $28 + 18 + 3 = 49$ winning lines.
On a plane there are $8$ winning lines.
In the cube, there are $3$ horizontal planes, so $8 \times 3 = 24$ winning lines.
There are also $3$ vertical planes going from left to right, but now with only $5$ new winning lines per plane, as the $3$ horizontal lines have already been counted. So $5 \times 3 = 15$ winning lines.
On the $3$ vertical planes going from front to back, we now only have $2$ new (diagonal) winning lines per plane. So $2 \times 3 = 6$ winning lines.
Finally, there are also diagonal planes to consider. There are $4$ winning lines going from corner to diagonally opposite corner.
In total, there are $24 + 15 + 6 + 4 = 49$ winning lines.