Take the vertices of the cube to be the eight points (so its edges have length 2). Then the vertices of the octahedron will be the six points , , . Now work out the coordinates of the centres of the faces of the octahedron, and compare them with those of the original cube.

The point at which you will need to try to visualise the problem is that you need to decide which of the possible triples of vertices of the octahedron form the three vertices of one of the eight triangular faces. The key thing here is that opposite vertices of the octahedron cannot belong to the same face. So for example of one of the vertices of a face is (1,0,0), then that face cannot contain the vertex (–1,0,0).