The rectified 24-cell r{3,4,3}
The vertices of the rectified 24-cell (hereafter shortly called the r{3,4,3}) can be found, starting from the vertices of a 24-cell, by running along the 96 edges until you reach the midpoint of the edge. The coloring of the vertices is the same as the edges they lie on. This results in 32 red ones, 32 green ones and 32 blue ones. The cells of the r{3,4,3} are 24 cubes and 24 cuboctahedra. Firstly I show the complete object having 96 vertices, 24x12=288 edges (the 24 cubes have no edges in common), 24x6=144 square faces and 24x8/2=96 triangles (the cuboctahedra are linked by a triangular face). Then I show in three different figures 3x8 cubes. Each set of 8 cubes is made with 64 vertices of two colors, thus leaving out one color. The 8 centres of each set of 8 cubes are the vertices of a 16-cell (hyperoctahedron).
In the next two figures you can see two sets of 6 cuboctahedra. Each set of 6 lies around the vertices of a octahedral cell of a 24-cell that is the dual figure of the original 24-cell. The 6 cuboctahedra are grouped around one cube with which they share a square face. In the second (right) picture the 6 cuboctahedra are completely flat on the faces of a cube because of the stereographic projection (of the 4-dimensional sphere to the 3-dimensional space) that is used.
In the next picture one can see the other 12 of the 24 cuboctahedra lying around the vertices of a cuboctahedron that is part of a 24-cell that is the dual figure of the original 24-cell. Four of them are colored yellow to discern them more clearly. There are two similar groups of 4. As you see these cuboctahedra make a connection between the 2 groups of 6 cuboctahedra mentioned before. As the cuboctahedra are connected by means of the triangular faces, the yellow ones (and alike) share 2 triangles with the outer cuboctahedra and 2 with the inner.
A set of 4x6 =24 diagonals of cubes can be selected as can be seen in the two pictures below:
The black line is a diagonal of the biggest and smallest cube. It has 6 vertices lying on it. This same line is also the diagonal of 4 cubes selected from the other sets (with red-blue vertices and red-green vertices). It forms the thread of a "necklace" of 6 cubes connected by vertices. The same properties can be seen for the other 3 diagonals. This amounts to 4x6=24 vertices lying on this set of 4 diagonals. There are 4 choices for the black diagonal of which 2 instances are given in the 2 pictures. On the 4 sets of 4 diagonals we can find 4x24=96 vertices forming the complete set of vertices of the r{3,4,3}. The 4 diagonals of one set lie on 4 circles that are part of one Hopf fibration. Actually, in the spherical case, the diagonals ARE the (equatorial) circles. The 3 colored circles lie on one torus of which the black "circle" is the symmetry axis. The 24 vertices of the 24-cell of which the r{3,4,3} is the vertex figure, lie on one set of diagonals. If we start running from these 24 vertices along these 4 (circular) diagonals and stop half way the edge, we have 24 of the 96 vertices of the r{3,4,3}.
The fun has only just begun...
Let's make another selection: only the blue vertices connected by diagonals of square faces of the cubes. We see a net of 16 tetrahedra of which 8 are colored and the other 8 colourless (so that no two colored ones connect).
In the next 2 pictures only the green (red) vertices connected by diagonals of square faces of the cubes.
Summing up, there are 2x8=16 tetrahedra with red vertices, 16 tetrahedra with green vertices and 16 tetrahedra with blue vertices. A total of 3x16=48 tetrahedra forming the vertices of 24 cubes in pairs of different color. Now select one vertex of each tetrahedron by picking out those that lie on one Hopf flow. This gives us 4x2=8 red ones, 8 green and 8 blue ones. We have selected 24 vertices in this way.