Comments on Exam 2

Problem 1: The standard tessellation of space by tetrahedra and octahedra has the arrangement tetra-octa-tetra-octa around each edge. The problem was marked correct if you described this standard tessellation, which is related to "cubic close packing" of spheres. But there is another tessellation of space by tetrahedra and octahedra that has the arrangement tetra-tetra-octa-octa around each edge. Sharon and Dana found this non-standard tessellation, and they earned extra credit for this observation. This second tessellation is related to "hexagonal close packing" of spheres.

You can mix the two types of edge configurations in a single tessellation, so that in fact there are an infinite number of different tessellations of space using tetrahedra and octahedra!

Problem 2: The symmetry group of the box is contained in the symmetry group of the cube. The multiplication of symmetries of the box is commutative: For every pair of symmetries S and T, we have ST = TS. The square of every symmetry of the box is the identity: For every symmetry S, we have S² = I.

The number of symmetries of the box is 8. If you pick one of the 8 vertices of the box, then there is a unique symmetry of the box that takes the chosen vertex to each one of the 8 vertices of the box.

Every symmetry of the box can be written uniquely as a product of some of the three reflection symmetries of the box. If the three reflection symmetries are called F, G, H, then the 8 symmetries of the box correspond to the 8 subsets of the set {F, G, H}: I, F, G, H, FG, FH, GH, FGH. (The identity I corresponds to the empty subset.)