Start with the Cube and Octahedron, holding them next to each other. There are 8 vertices (corners) on a cube, and 8 faces on an octahedron. Try to picture a cube with one vertex at the center of each face of the octahedron. Now picture an octahedron with one vertex at the center of each face of the cube. Make sure to note that the cube and octahedron each have 12 edges.

Try the same experiments using the icosahedron and dodecahedron. When each face in one solid corresponds to the vertex of another, and each vertex corresponds to the face of tthe other, we call it a "Dual" relationship. What is the Dual of a Tetrahedron? With 4 faces and 4 vertices, you may have already guessed that the dual is another tetrahedron!

Break up the Archimedean solids into three groups, related to 1) the Tetrahedron, 2) the Cube and Octahedron, and 3) the Dodecahedron and Icosahedron. Notice how the corresponding faces on each figure are in the same orientation throughout the entire group.

Let's start with the Dodecahedron, which has 2-fold, 3-fold and 5-fold rotational symmetries. To see one of the 5-fold axes, place the model flat on a desk, resting on one face, and look straight down from above at the top pentagon. If you rotate the model 1/5 of a complete turn, it will look just like it does now. You can do 5 rotations like this before you get back to the original position.

The 3-fold axes run straight through opposite vertices, so we can repeat our experiment by standing the model up on one vertex and looking down from above. The 2-fold axes, run straight through the mid-points of opposite edges.

How many axes of each type does the dodecahedron have? With 12 faces, and each 5-fold axis running through two opposite faces, there must be 6 of those. How many 3-fold and 2-fold?

You will notice that each face of the tetrahedron is in the same plane as its next door neighbor on the octahedron. This is an important relationship that is often used in engineering an architecture.

Continue adding octahedra and tetrahedra together, always making sure to alternate between them. What kinds of shapes and figures can you make with these? Try making a beam by attaching them in a long line. Attach two beams at an angle. Try to attach a third beam and make a triangle.

Try to follow how the list is arranged, and understand why some figures are not included. For example, some figures that you might expect to be there are already listed as Platonic, Archimedean, Prism or Antiprism, or the model is not convex. A Square Dipyramid is the same as an Octahedron, and a Triangular Gyrobicupola is a Cuboctahedron. If you are not careful making the Augmented Truncated Tetrahedron, you could end up with coplanar faces.

By my count there are only 18 solids that cannot be split like this. Which shows that the vast majority can be derived from other models by augmentation, and/or rotation.

Deltahedra - models made only of triangles. There are only 8 convex deltahedra, but any number of non-convex.

Models with tetrahedral, octahedral or icosahedral symmetries. Highly Symmetrical Models - Start by symmetrically augmenting and excavating some existing models, and then try coming up with your own.

Toroidal Models - Models shaped like a donut, or with one or more "handles" (like a coffee cup). Look on the math page for a few samples of these.

To count the number of edges for a model, you can count the number of faces of each kind and multiply by the number of edges that face has. Divide the total by two since each edge is shared by two faces. So for a rhombicuboctahedron, we have E = (3T + 4S)/2, and we have 8 triangles, and 18 squares. So E = (24 + 72)/2 = 48.

Euler's formula indicates that for simple polyhedra (without intersecting faces or toroidal shapes), the following relationship holds: E + 2 = F + V. So for the rhombicosidodecahedron, we have 120 edges, 62 faces and 60 vertices, so 120 + 2 = 62 + 60. Try this out for other figures as well. They don't even need to be convex. Try it out for toroidal models (models with handles), and see that the formula doesn't work. Can you figure out a formula that does, using the number of holes the model has?