1) and 2) Let us consider the 24-cell given by the 16 points of the first kind and 8 points of the second kind that you described. Call the the set of 8 points of the second kind . Now partition the 16 points of the first kind into two sets: the points with an odd number of minus signs (call this set ), and the ones with an even number of minus signs (call this set ). Now we have partitioned the points of the 24-cell into three sets of size 8. It is straightforward to check that the inner products within these sets are all or , and that the inner products between two points in different sets is . This shows that a 24-cell contains exactly 3 orthoplexes and thus has only one partition into three orthoplexes. Actually, this also shows that a 24-cell is exactly a set three mutually unbiased bases in (plus the antipodes).

3) This one is not too bad, you can basically just count them. First, note that an orthoplex is just an orthonormal basis plus the antipodes of all the vectors in the basis. To make things a bit simpler, we will consider a vector/point of the 600-cell and its antipode the same. So the 120 vertices of the 600-cell become 60 lines through the origin, and an orthoplex is a set of four mutually orthogonal lines. We will just pick the lines for our orthoplex one at a time. First, we have 60 choices for the first one, and they are all equivalent since the symmetry group of the 600-cell can map any one to any other one. So now we must pick the second element of our orthoplex. To determine how many choices we have here, we can assume without loss of generality that our first choice was the point (and its antipode). There are then 3 choices for the second point/line coming from the second kind of points (6 but divided by 2 because antipodes are the same), and there are choices of points/lines of the third kind (all the ones with 0 in the first coordinate). So in total we have choices for the second element of our orthoplex. Now, note that again the symmetry of the 600-cell says that all choices for these first two elements are the same, since any two orthogonal vectors from the 600-cell can be mapped to any other two (in any order). So we may assume that the first two chosen are and . It is easy to see that there are only 2 choices for the third element of our orthoplex, and then the final choice is determined for us. Thus we had choices for our orthoplex. But we could have picked the same four elements in any order! So we actually have orthoplexes contained in the 600-cell. You mentioned that you can obtain a partition of the 600-cell into 15 orthoplexes by considering the cosets of the subgroup . Is it possible to find five distinct isomorphic copies of whose cosets would then give all of the 75 orthoplexes? This might be of help with puzzle (4).

4) I don’t have a human proof here. I’m still thinking about it, but I think it might be difficult. The only idea I really have is to consider the graph that I mentioned last time: the vertices are the points of the 600-cell, two points are adjacent if their inner product is *not* or . Then the maximum independent sets in are exactly the orthoplexes of the 600-cell, and the 15-colorings of are exactly the partitions of the 600-cell into 15 orthoplexes. We can maybe make things easier for ourselves by merging every point with its antipode. This doesn’t really change anything because every point always receives the same color as its antipode in a 15-coloring of (since such a coloring is a partition into orthoplexes). All this does is give us a smaller graph, call it , to work with, and the 15-colorings of are in one-to-one correspondence to the 15-colorings of . In any case, the value of the Lovasz theta number of the complement of (or ) is equal to 15, the chromatic number of these graphs. The Lovasz theta number of the complement of a graph is always a lower bound on the chromatic number, but in this case they are equal, and this means that a 15-coloring of or gives an optimal solution for the Lovasz theta number of their complements. I won’t go into much detail here, because it would go on forever, but the set of optimal solutions for (a certain formulation of) the Lovasz theta number forms a convex set, and I would be willing to bet that the 15-colorings of or are extreme points of this set (but I do not have a proof of this). So investigation of this convex set may be of some help, but again, this seems difficult. Chris Godsil, Brendan Rooney, Robert Samal, Antonios Varvitsiotis, and myself have done some work on describing the set of all optimal solutions for Lovasz theta in some cases: https://arxiv.org/abs/1512.04972 and https://arxiv.org/abs/1610.10002. But honestly I think there is probably a better approach using group theory.

5) We can use our previous solutions to help us here. An understanding of the proof of (3) lets us know that the symmetry group of the 600-cell acts transitively on the set of orthoplexes inscribed in the 600-cell (it acts transitively on pairs of orthogonal points, and once two orthogonal points are fixed the remaining elements of the orthoplex are determined). So all the orthoplexes are “the same”. Let us consider the one given by the eight points of the second kind. Since there are some 24-cells in the 600-cell, it must be possible to complete this orthoplex to a 24-cell. How many ways are there to do this? The proof of (1) and (2) tells us that each of the additional 16 points we add must have inner product with all of the 8 points we started with. Clearly, the only possible choices are the 16 points of the first kind, and we must take all of them. So every orthoplex that is inscribed in the 600-cell can be completed in a unique way to an inscribed 24-cell. So we get a 24-cell for each orthoplex, which gives us 75. But wait! We over-counted again. Of course, some orthoplexes will be completed to the same 24-cell. Here we use our solution to (1) and (2) again. Every 24-cell is the disjoint union of three orthoplexes, so these three orthoplexes will all be completed to the same 24-cell. Thus we must divide by 3, and we get that there are 24-cells inscribed in the 600-cell. These must also be exactly the 25 that you get by considering the right cosets of the 5 different copies of the binary tetrahedral group that you described. This also means that the left cosets give the same 25 (maybe this is obvious but I am not as familiar as you with the group theory here). Actually, question for you: how do we know that the left cosets give different partitions of the 600-cell into 24-cells from the partitions given by the right cosets? I guess this is equivalent to the subgroup not being normal.

6) I think you have a typo in the post, because the last puzzle is actually labeled (4). Again, I do not have a human proof here, but it seems more tractable than (4). Since the symmetry group acts transitively on the set of orthoplexes inscribed in the 600-cell, the argument from (5) shows that the same is true of the 24-cells, which may be useful. We know how to construct 10 partitions of the 600-cell into 24-cells by taking the left and right cosets of the five copies of the binary tetrahedral group sitting inside the binary icosahedral group , but we want to show that these are all the ways to do this. We also know from (5) that the only 24-cells are the cosets of the five copies of contained in . I checked in Sage, and it turns out that given any two *disjoint* 24-cells inscribed in the 600-cell, there are exactly three 24-cells disjoint from both of these, and together these five 24-cells partition the 600-cell (so the three additional ones are also mutually disjoint). Given that we know that the only 10 partitions into 24-cells come from the coset construction, this implies that given any two disjoint 24-cells, they are either both left or both right cosets of the same copy of . Maybe proving this by hand would not be too hard? Suppose we can prove this, and consider a partition of the 600-cell into five 24-cells. By what we have assumed, any two of these five 24-cells must both be left or both be right cosets of some fixed copy of . But then we are done, since we can fix one 24-cell and then apply this statement to it and the other four 24-cells in turn. So we just need to show that given two disjoint cosets and of two, possibly different, copies of , there must be some copy of for which and $latex are either both left or both right cosets. By symmetry, we may even assume that is one of the copies of (and thus is a proper coset of a different copy of in the nontrivial case).