## 3 Properties of H and T  
We show some geometrical properties of H and T.  
As Figure 2(a) shows, the dodecahedron H has 6-fold symmetry around the line connecting the two v-vertices, whereas the surrounding cube has 3-fold symmetry around the same line. Consequently, this object can be put into the cube in two different ways modulo rotations of the cube and it has square projections not in three but in six ways. We call an imaginary cube a _double imaginary cube_ if it is an imaginary cube of two differently oriented cubes. This dodecahedron is a double imaginary cube and the two cubes of which it is an imaginary cube share a pair of opposite vertices and one cube is obtained by rotating the other one by 60 degrees. Note that this double imaginary cube is the intersection of the two cubes. It means that it is a maximal double imaginary cube as well as a minimal one and therefore it is the only convex double imaginary cube of the two cubes. The rotation group of this imaginary cube is the dihedral group \(D_{6}\) of order 12, which is not a subgroup of the rotation group of a cube; half of the rotations of this object map one cube to the other cube. Note that the same two vertices are cube-vertices in both of the ways of putting it into a cube and therefore the notion of a v-vertex is well-defined. A sculpture based on a pre-fractal of a fractal object based on H is presented in [4,5]. More generally, consider the intersection of two different cubes which share a pair of opposite vertices. We have a dodecahedron which is a convex double imaginary cube of the two cubes as Figure 6 shows. It belongs to the equivalence class No. 5 as minimal convex imaginary cubes of both of the cubes. In the next proposition, we show that all the convex double imaginary cubes have this form. **Proposition 5**. : _Every convex double imaginary cube is the intersection of two different cubes which share a pair of opposite vertices._  
Proof. : Consider a maximal double imaginary cube \(A\) of two different cubes \(C\) and \(D\). \(A\) is the intersection of \(C\) and \(D\), and every edge of \(C\) intersects with \(D\) and vice versa. Suppose that no vertex of \(C\) is on one side of a plane \(F\) containing a face of \(D\). Then, in order that the four edges of \(D\) on \(F\) intersect with \(C\), a face of \(C\) is contained in \(F\) or an edge of \(C\) contains two opposite vertices of \(D\) on \(F\), from both of which one can easily derive a contradiction. Therefore, each face of \(D\) truncates a vertex of \(C\) to form a triangular face of \(A\). Since \(A\) is also obtained by truncating \(D\) with six faces of \(C\), one  
Figure 6: Double imaginary cubes obtained as the intersection of two cubes. One cube is obtained by rotating the other one by (**a**): 60 degrees, (**b**): 42 degrees. can see that \(A\) is a dodecahedron with twelve triangular faces. Since a square face \(P\) of \(D\) is cut into a triangle \(Q\) so that all the four edges of \(P\) contain a vertex of \(Q\), one vertex \(p\) of \(P\) is a vertex of \(Q\). Since it holds on the other faces of \(D\) which share the vertex \(p\), one can see that \(p\) is a common vertex of \(C\) and \(D\). Through the same argument on other faces of \(D\), we have another shared vertex. Therefore, \(C\) and \(D\) share a pair of opposite vertices. Since this double imaginary cube is minimal as well as maximal, all the double imaginary cubes have this form. Next, we study the octahedron T. One can see through an easy calculation that the three diagonals connecting two opposite vertices intersect at one point and are orthogonal to each other. In addition, the intersection point divides each of the diagonals by the ratio of 1:2. Therefore, four vertices of T share one plane and we have the plane section in Figure 7(a).