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November 7

120 Cell Question

The Maximum number of pentagons which can be selected from a dodecahedron with no elements (points, lines) shared is 3. 4 identical copies of such a set will make the dodecahedron. What is the maximum number of dodecahedra sharing no elements can be selected from the 120 cell and can X copies of that maximum set make the entire 120 cell?Naraht (talk) 02:09, 7 November 2021 (UTC)

A maximal such set of dodecahedra in a 120-cell would immediately give a maximal vertex set of the dual 600-cell. However, it is already known that the maximal such set consists of 24 vertices. So one obvious candidate comes from the compound of five 24-cells in a 120-cell 600-cell (sorry!) Double sharp (talk) 22:16, 9 November 2021 (UTC). From the illustrations here, it seems that if you take a single dodecahedron of a 120-cell, then it is entirely closed off by the 12 dodecahedra surrounding it (connected at each face). So I'd expect that the 24-cell arrangement works and leads to completely nonadjacent dodecahedra. Sorry that my 4D visualisation is a bit rusty right now. :( Double sharp (talk) 11:38, 7 November 2021 (UTC)

Saul Schleimer's answer. —Tamfang (talk) 19:09, 7 November 2021 (UTC)

So I was right! :D Double sharp (talk) 21:02, 7 November 2021 (UTC)

Template:UI'm sorry, I can't read twitter on this computer. Can it be copied here?Naraht (talk) 16:49, 9 November 2021 (UTC)

Template:Ping It says: Template:Tq Double sharp (talk) 20:06, 9 November 2021 (UTC) —Template:Small

Template:U, The 120 cell itself is by layers 1, 12, 20,12,30,12,20, 12,1. I'm guessing that the non-touching are the pole end, 8 of the 20, 6 of the 30, 8 of the 20 and the other pole. the 20 dodecahedra on that layer line up with the vertices in the pole dodec, so the 8 in the non-touching are presumably at the points corresponding to one of the nested cubes of the pole dodec. This and then the 6 in the middle would nicely correspond to an embedded 24-cell. Should their be an article for the compound of five 24-cells in a 120-cell, the way we do for the compound of five cubes?Naraht (talk) 20:19, 9 November 2021 (UTC)

I'm not Tamfang, but yes, I think that's correct. The compound of five 24-cells is for now on WP just one of the entries at Polytope compound#4-polytope compounds and List of regular polytopes and compounds#Four dimensional compounds. I'm not sure how much there would be to hang an article on, though. The regular polychoron compounds were actually until recently not even fully enumerated: Coxeter in his Regular Polytopes got almost all of them by systematically faceting a 120-cell, but in 2018 McMullen found that Coxeter had missed a few while doing so. He did prove that the list is now complete: his proof heavily uses quaternions. Double sharp (talk) 22:10, 9 November 2021 (UTC)

P.S. BTW, I misspoke: it's not a compound of five 24-cells in a 120-cell, but a compound of five 24-cells in a 600-cell. (Of course it is, since five 24-cells have in total 5×24=120 vertices, same as a 600-cell.) What I meant to refer to is that it is a (chiral) stellation of the 120-cell, instead, which is why you can naturally partition the original cells into five equal sets by considering which 24-cell they stellate into. So, kind of like how you can do this trick for an icosahedron by considering the compound of five tetrahedra, which is a stellation of the icosahedron, but whose vertices are those of the dual dodecahedron. Sorry about that. Double sharp (talk) 22:12, 9 November 2021 (UTC)

Inverse function

If x, y, and z are positive integers and z is a function of x and y, z(x,y) = (y^2-3y)/2+x+1, given z, what are x and y? (Actually, if you have y, x is trivial to get.) Bubba73 ^{You talkin' to me?} 02:32, 7 November 2021 (UTC)

In general, given Template:Math, there are many pairs Template:Math satisfying the equation

Template:Math.

The number of such pairs grows roughly like Template:Math.

There are three variables. If you treat two of the variables as knowns, you can solve for the third.

The equation has the particularity that whenever it is satisfied by some triple Template:Math, then each triple Template:Math also satisfies the equation. The study of the equation can be simplified by the substitution Template:Math, introducing a new variable Template:Math, which need not be positive. Applying this and simplifying the result gives us this:

Template:Math.

For any pair Template:Math satisfying the transformed equation, each triple of the form Template:Math satisfies the original equation, provided that Template:Math. And conversely, each triple Template:Math satisfying the original equation gives rise to the pair Template:Math satisfying the transformed equation.  --Lambiam 04:52, 7 November 2021 (UTC)

Thanks, I forgot one condition that will make the solution unique: x<y. Bubba73 ^{You talkin' to me?} 05:24, 7 November 2021 (UTC)

The following involves some hand-waving. Given Template:Math and Template:Math, we can solve for Template:Math using the quadratic formula, discarding the negative solution. From the formula we see that for fixed Template:Math, as Template:Math increases, Template:Math (not necessarily an integer) decreases. Since Template:Math, a boundary case is obtained when Template:Math, which means that Template:Math. Solving for Template:Math shows now that Template:Math, with equality for Template:Math. Since Template:Math is an integer, we can put

Template:Math.

 --Lambiam 11:24, 7 November 2021 (UTC)

Thanks, I thought that y would involve the integer part of a square root, but I wasn't able to get it. This is likely to be in a new sequence in OEIS soon. Bubba73 ^{You talkin' to me?} 01:18, 8 November 2021 (UTC)

• This would have been easier if I'd thought about it in the right way - y is the index of the largest triangular number that is < z. Bubba73 ^{You talkin' to me?} 06:51, 8 November 2021 (UTC)

Plotting 3-D but 2-df data in 2-D

If you have three variables that sum to some constant, $a+b+c=1$, then you only have two degrees of freedom. So while you could plot them in a cube, you can also plot all three on a ternary diagram. Since the value of each variable can be read off a ternary diagram as proportional to the distance from each corner, and humans are pretty good at judging relative and absolute linear distances, and the internet is still mostly 2-D, this works well.

I need to present some data with three variables and two degrees of freedom in the form $abc=1$, with variables that multiply to a constant. Does anyone have any good suggestions for how to do this? I mean, something more like a ternary diagram than a contour plot. Ideally, I'd like to be able to join such diagrams along the axes, as ternary diagrams are joined in a piper plot (yes, a contour plot would do that, for two axes, but it's harder to read).

As a simple example for illustrative purposes, suppose I had the masses and maximum accellerations of a Saturn V rocket, the QEII, an ice yacht, the latest Tesla, the latest Hummer, and Usain Bolt. I probably do, this is Wikipedia. And knowing that [[Newton's first law|Template:Math]], I wanted to plot all three variables for all of these on a ternary-like plot.

I actually want to plot some data over some equations, comparing the datasets and the fit of the data. I have several equations of form $abc=1$, sharing some variables. HLHJ (talk) 15:56, 7 November 2021 (UTC)

This is tough since the triangular shape of ternary diagrams only works because each variable is bounded (from 0 to 1). I suppose if you can bound your data you could try a logarithmic ternary diagram since the logs would then add to 0. eviolite (talk) 16:01, 7 November 2021 (UTC)

In the typical applications of a plot with $a+b+c=1$ using a triangle-based barycentric coordinate system, the three variables are each bounded by the unit interval, so that the plot fits neatly in a triangle. This is, however, not essential. Using complex coordinates, defining

$u=e^{\frac{3}{6}i\pi },v=e^{\frac{7}{6}i\pi },w=e^{\frac{11}{6}i\pi },$

we obtain a bijective mapping ${ℝ}_{|\mathrm{\Sigma }=1}^3\leftrightarrow {ℝ}^2,$ with

$(a,b,c)\mapsto (x,y)$ where $x+iy=au+bv+cw.$

For example, $(2,-1,0)\mapsto (0.8660,2.5).$ The formulas for the inverse mapping are a bit messier.  --Lambiam 16:51, 7 November 2021 (UTC)

Thank you to you both. I will have a play around in R, and see if I can get these visualizations to work. THis will probably be archived by the time I produce anything, but your advice is much appreciated! HLHJ (talk) 21:13, 7 November 2021 (UTC)