The original puzzle
Two diagonals on the sides of a cube meet in a corner. What is the size of the angle between them ?
This is a real beauty !
The answer to the original question is not too difficult to find,
but it leads to all sorts of new questions concerning volume.
Each of the 2 given diagonals, on adjacent faces of a cube, has 2 ends -
one end at which they meet, and another open, "loose" end.
If a line is drawn between these 2 "loose" ends,
it will be another (third) diagonal on a face adjacent to the other 2 faces.
These 3 diagonals form an equilateral triangle,
giving us the answer to our puzzle.
This equilateral triangle of diagonals, acts as the base of a tetrahedron,
whose sides are 3 right-angle isosceles triangles,
each occupying half of a cube face, 1.5 cube faces in all.
There are 4 such tetrahedrons occupying the total surface of the 6 faces of the cube.
There also seems to be a fifth tetrahedron, this one perfectly regular,
on the inside, enclosed in the 4 others.
How is the volume of the cube distributed among these 5 tetrahedrons ?
It might be fun for the children to cut up a soft-plastic (or styrofoam) cube,
along the six diagonals of the cube (one on each face),
to remove 4 pyramids from the corners
(all built on an equilateral base with 3 isosceles right-angle sides),
and discover another pyramid "hidden" inside
(perfectly equilateral, base and sides),
larger than the other 4 exterior pyramids.