1)I have lots of black and white squares that can be joined together to make cubes. How many distinguishable cubes can I make?

2)Now I try it with triangles and regular octahedrons?

3)Now pentagons and regular dodecahedrons?

4)Triangles again but making regular icosahedrons!?

Note: Distinguishable means rotations are the same, but reflections are not.

The octahedron may be all white (1 octahedron).

The octahedron may have one black face (1 octahedron).

The octahedron with two black faces might have them opposite one another, adjacent at an edge or adjacent at a vertex (3 octahedra).

The octahedron with three black faces might have:
A. two of them opposite one another with the third being adjacent to one of them on an edge and the other on a vertex.
B. all three arranged around one vertex with the fourth triangle at that vertex remaining white.
C. the three faces adjacent to one given face by edge.
     (3 octahedra)
    
The octahedron with four black faces and four white faces could have:
A. all four black faces arranged around one vertex.
B. a black face surrounded on each of its 3 edges by the other three black faces.
C. two pairs of opposing black faces.
D. one pair of opposing black faces and two black faces that are adjacent to one another on an edge (one of the latter two will be edge-adjacent to one of the first two, and vertex-adjacent to the other, and the other will be similarly oriented to the other of the first two.).
E. one pair of opposing black faces and two black faces that are adjacent by edge to one of the first two.
    (5 octahedra)
   
The situations for 3, 2, 1 or 0 white faces will have the same numbers (3, 3, 1 and 1) as the corresponding number of black faces.

The total is 1+1+3+3+ 5 +3+3+1+1 = 21 octahedra.

Did I miss any?

Edited on March 4, 2006, 1:24 pm

Posted by Charlie on 2006-03-04 13:23:38