24-CellMake a tower of three triangular antiprisms (i.e., octahedra), each with b3 equilateral bases. The bottom one has r2 slanting edges, the middle one has r1+r2 for its slanting edges (r1 below r2), and the top one has y1+y2 for its slanting edges (y1 below y2). Turn it over and make the same thing superimposed upon itself, reusing just the bottom and top b3 triangles, and adding two new horizontal b3 triangles where needed, partway up. (You will see how its slanting edges intersect those you made earlier, at the x1+x2 junctions.) Then, using b3's, r3's, and y3's, complete the octahedra around the sides so there are three octahedra around every concave b3 edge. If you do this correctly, you will add a b3 hexagon "equator."Notice that each edge is a side of three different triangles, any two of which determine an octahedron. This is an orthogonal projection, so the octahedra are compressed and slanted, but parallel lines remain parallel; so the three square equators of each octahedron show up as parallelograms. Verify that there are 24 vertices and 24 octahedral cells, with eight edges per vertex. (It is self-dual.) Find "rings" of six octahedra in various directions, e.g., the six blue triangles indicate one such ring.