Let's call P-ABCDE-A'B'C'D'E'-P' one of the icosahedrons. At least three of the vertices of the octahedron are vertices of some of the icosahedrons. Notice that, choosing 3 vertices of the octahedron, the possible ratios of the sides of the triangle formed by them are [1:1:1] and [1:1:sqrt(2)]. And, choosing 3 vertices of the icosahedron, the only possible ratio in common is [1:1:1]. Then, the vertices of the octahedron were splitted: three are vertices of the "first" icosahedron, and the three others are vertices of the "second" icosahedron, and they are vertices of equilateral triangles in their respective icosahedrons. In P-ABCDE-A'B'C'D'E'-P', calling L the edge of the icosahedron, we have: 1) PA=PB=PC=PD=PE=L 2) PA'=PB'=PC'=PD'=PE'= phi.L [phi, the golden number] 3) PP'=L.sqrt(phi²+1) Choosing P to be one of the three vertices, one can see that we cannot choose P' to be another vertex. If one chooses A, then the remaining one must be E or B. [L,L,L will be the sizes of the triangle]. If one chooses A', then the remaining one must be C' or D'. [phi.L, phi.L, phi.L will be the sizes of the triangle]. Then, we must have one of the following: L=L' or phi.L=L', or L=phi. L', or phi.L=phi.L'. The possible ratios of the edges of the icosahedrons must be [1:1] or [phi:1]. One can give good positioning to them in order to construct the existence of both cases. =P