VES 176 Final Project Description John M. Sullivan

Home , Stellated octahedron, Vertex (geometry)

VES 176 Final Project Description

John M. Sullivan

We want to consider the family of (3,4) solids related to the cube and the octahedron. If we take a cube and an octahedron with the same center and orientation and consider their intersection, we get a truncated cube, cuboctahedron or truncated octahedron. If instead we consider their convex hull, we get the corresponding dual polyhedra, stellations of the cube or octahedron, with the intermediate form being the rhombic dodecahedron. Let (u, 0, 0) be one vertex of the octahedron and (v, v, v) be one vertex of the cube. Then when u = v we get just a cube as the convex hull, and an octahedron as the intersection. With v < u < 2v the intersection is a truncated octahedron with a typical vertex at (v, u − v, 0). When u = 2v we have a cuboctahedron (vertex (v, v, 0)), and 2v < u < 3v gives a truncated cube with vertex at (v, v, u − 2v). Finally, setting u = 3v gives us a cube. At each stage the convex hull is the dual of the intersection, and its vertices are just the vertices of both polyhedra. Note that in the most general truncation we have eight hexagonal faces (positioned like the faces of an octahedron); if one vertex of the face in the positive octant is (x, y, z) with x > y > z > 0, then the vertices in counterclockwise order (looking from the outside) are (x, y, z), (y, x, z), (z, x, y), (z, y, x), (y, z, x), (x, z, y). In other words, adjacent vertices are obtained by swapping two coordinates (except that interchanging the largest and smallest gives us the opposite corner). Similarly there are six octagonal faces positioned like the faces of a cube. The one on the positive x-axis will have vertices (x, y, z), (x, z, y), (x, z, −y), (x, −y, z), (x, −y, −z), (x, −z, −y), (x, −z, y), (x, y, −z). Finally, a typical rectangular face would be (x, y, z), (x, y, −z), (y, x, −z), (y, x, −z); these are positioned like the edges of a cube or octahedron. We are here interested in cases where z = 0 or x = y, so the rectangular faces disappear, and one of the other sets degenerates (pairs of adjacent vertices in the lists above collapse into one). But the other set is still present, and we will use these coordinates to ﬁnd out when the faces are regular. We can relate these to the jitterbug transformations. If we look at only alternate vertices around each hexagonal face of the truncated octahedron, we get eight equilateral triangles. If we choose the orientations consistently, these form the jitterbug transformation of the octahedron into the cuboctahedron. Similarly taking alternate vertices around each octagonal face of the truncated cube gives us a jitterbug of the cube (although this is not quite as nice since the cube’s vertices have odd valence, so the twisting squares can’t be oriented consistenly). These jitterbug transformations give us a nice way to parametrize the sequence of stages from octahedron to cube described above. To vary u/v from 1 to 3, we change both u and v in such a way that the rotating triangles or squares stay the same size. For the truncated octahedra, let φ be the angle through which the triangles have been twisted. The vertices in the jitterbug must each lie in some co¨ordinate plane, and some

1 simple trigonometry shows that they must have the form (sin(φ + π ), sin(φ), 0). Thus √ 3 π π v = sin(φ + 3 ), and u = 3 sin(φ + 6 ). π The octahedron is obtained when φ = 0 and the cuboctahedron when φ = 3 . The π 1 midpoint φ = 6 is the point where v reaches a maximum, and the vertex is at (1, 2 , 0). This is recognizable as the vertex of the six strut tensegrity structure. In the truncated 2 2 octahedron, the hexagon is regular if√ 2y = 2(x − y) . This happens at this same stage: x/y = 2, or equivalently, tan(φ) = 1/ 3. Another interesting stage occurs when the vertex (x, y, 0) is equidistant from (0, x, y) and (x, −y, 0), i.e., when the coordinates are in the golden ratio and we get an icosahedron from the jitterbug. Since the ratio x/y here is √ sin(φ)/2 + 3 cos(φ)/2 sin(φ) we get the golden ratio when tan2(φ) = 3/5. To get the truncated cubes, we let θ be the rotation angle of the squares in the cube jitterbug. Then one vertex√ of the front one will be at (cos(θ), cos(θ), sin(θ)). Then we find v = cos(θ), and u = 5 sin(θ + arctan(2)). At θ = 0 we get a cuboctahedron with vertex (1, 1, 0). (Note that we have to scale by √ 3 π π a factor of 2 to match up with the cuboctahedron φ = 3 we described before.) At θ = 4 we get a cube. The most interesting intermediate point is the one where we get a (semi-)regular truncated cube. This happens when (x, y, z) is equidistant√ from (x, z, y) and (x, y, −z). 2 2 π Thus we see that z = y −2yz, so z/y = tan(φ) = 2−1, and thus φ = 8 . It is interesting to note that in the parameterizations we have chosen (by rotation angle in the jitterbug), both semiregular forms both occur at the exact half-way points. For the transformation from octahedron to cuboctahedron, images have been produced of the following structures: 1. Solid stellated cube (convex hull). 2. Stick figure stellation with solid cube and octahedron inside. 3. Stick figure cube and octahedron with solid intersection (truncated octahedron) inside. 4. Jitterbug with solid triangular faces. 5. Tensegrity structure (thin stick jitterbug, with thick struts). 6. Truncated octahedron with this stick jitterbug drawn on it. 7. Jitterbug with extra triangles added to make icosahedron. For the transformation from cuboctahedron to cube, we have images corresponding to the first four sets above: 1. Solid stellated octahedron (convex hull). 2. Stick figure stellation with solid cube and octahedron inside. 3. Stick figure cube and octahedron with solid intersection (truncated cube) inside. 4. Cube jitterbug with solid square faces.