## 4 Solutions of the 3H = 6T Puzzle  
Now, let us go back to the 3H = 6T puzzle. Recall that it is clear that one can place eight of the pieces into the box and our question is how can we put nine of them. We consider a coordinate system with the origin the center of the box. For \(x,y,z\in\{+,-\}\), we denote by \(C(x,y,z)\) the unit cube which is the intersection of the box and the corresponding octant. As we have observed, a copy of T can be placed at the center of the box so that its v-vertices are on \((-1,0,0),(0,-1,0)\), and \((0,0,-1)\) and e-vertices are on \((1/2,0,0),(0,1/2,0)\), and \((0,0,1/2)\), as in Figure 7(b). We say that two congruent objects have the same direction if one is mapped to the other one by a translation. We denote by \(T_{3}\) copies of T with the same direction as this central piece. One can see that the eight faces of this copy of \(T_{3}\) exist in all the eight unit cubes. Therefore, the eight cubes are truncated by this object. The cube \(C(+,+,+)\) is truncated by an (e, e, e) face, \(C(-,-,-)\) is truncated by a (v, v, v) face, \(C(+,-,-)\), \(C(-,+,-)\), and \(C(-,-,+)\) are truncated by (v, v, e) faces, and \(C(+,+,-)\), \(C(-,+,+)\), and \(C(+,-,+)\) are truncated by (v, e, e) faces. On the other hand, as we observed in the previous section, T is obtained by truncating a cube with faces of type (v, v, v), (e, e, e), and (v, v, e), and H is obtained by truncating a cube with faces of type (v, e, e). Therefore we can put five copies of T and three copies of H in the corresponding truncated cubes to form a solution (Figure 8(c)). Figure 8: **(a)** The lower four pieces of the solution; **(b)** The locations of \(H\), \(T_{1}\), and \(T_{2}\) pieces with the lower (upper) square showing the lower (upper) four unit cubes, respectively; **(c)** The solution of this puzzle; **(d)** Copies of H and T with directions \(H\), \(T_{1}\), and \(T_{2}\). Copies of \(T\) in \(C(+,+,+)\) and \(C(-,-,-)\) have the same direction which is opposite to \(T_{3}\) (\(T_{1}\) in Figure 8(d)). The three copies of T in \(C(+,-,-)\), \(C(-,+,-)\), and \(C(-,-,+)\) have the same direction which is obtained by rotating \(T_{3}\) by 60 degrees around its 3-fold rotational axis (\(T_{2}\) in Figure 8(d)). The three copies of H have the same direction (\(H\) in Figure 8(d)). The directions of the eight pieces are shown in Figure 8(b). This solution has the 3-fold symmetry as Figure 9 shows. It is immediate from this construction that it satisfies the star-polyhedron condition. One can also see that it has the polytopal complex condition. Note that this is the only solution extending the central \(T_{3}\) piece under both of the conditions. Since we have eight ways of placing the central \(T_{3}\) piece, we have eight rotationally equivalent solutions. We prove the uniqueness of this solution under the two conditions. We start with the following observation. **Lemma 6**. :
1. _If two copies of H are connected through a (v, e, e) face, then it is impossible to fill with puzzle pieces the grooves around the three shared edges._
2. _If two copies of T are connected through a (v, e, e) face, then it is impossible to fill with puzzle pieces the grooves around the three shared edges so that they satisfy the polytopal complex condition._  
**Theorem 7**. : _Under the polytopal complex condition and the star-polyhedron condition, this is the only solution of the puzzle modulo rotations of the cube._  
Proof. : We denote by \(O\) the center of the box. It is immediate that \(O\) is not shared by all the three copies of H in a solution. Therefore, consider a copy \(h\) of H which does not contain \(O\). Since the distance from the center of an H piece to each of its face is 1/2, the center of \(h\) is located in the unit cube with  
Figure 9: The solution looked at from a vertex. the center \(O\), and the distance from \(O\) to the center of \(h\) is no more than \(\sqrt{3}/2\). Let \(P\) be a face of \(h\) which the line segment between \(O\) and the center of \(h\) passes.