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Let me first give some of a background as to where I got this problem. I had a math teacher ask me a few months ago: "How many 1 unit by 1 unit squares could one fit on a sphere with a radius of 32 units?" The only problem was that it was a high school math course, and that I, as a high school student, had no way of answering the question in its true sense, as the teacher simply wanted a division of surface area.

But the problem remains unanswered, and it has been bothering me for some time now. How many mirrors can you fit on a disco ball? This isn't a question for the construction of a disco ball simply the construction of a theoretical "perfect" disco ball given the parts. My teacher didn't realize that the problem answer she was expecting was for an entirely different problem relating to how many areas of 1 unit squared you could create from the surface. Given my severely limited math knowledge, I am simply unable to answer a problem as complex as this.

Here is the problem in its entirety:

If you have a sphere of radius $r$ and squares of length $l$, then how many squares can you fit on it in tangent with the sphere with only the centers of each square touching the sphere, and with no square overlap?

This is a disco ball so the parameters get a bit tight:

  1. The squares can only be attached by their centers

  2. The squares can't overlap

  3. The squares can't go into the sphere

  4. The squares can't be broken or bent (don't want bad luck)

  5. The squares go on the outside of the sphere

I don't know what math this would take to solve this, but I know that it has a solution lurking in the voids of some imagination if so gifted to me.

What this question might mean to a mathematician:

In this given situation the square mirrors could be interchanged with spheres without any adverse effects to the problem. So where this problem might find a solution could be in the already well-defined field of sphere packing, the only caveat being the fact that it's packing with set radii ratios and where non-touching spheres are discarded.

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  • $\begingroup$ About twelve thousand, eight hundred and sixty,. If you want a more accurate answer, get thirteen thousand such squares and an appropriate sized ball, and some glue. This is the wrong forum for your question. Gerhard "Let Us Know The Result" Paseman, 2018.09.19. $\endgroup$ – Gerhard Paseman Sep 19 '18 at 23:16
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    $\begingroup$ @Gerhard, questions about packing things in other things generally involve a lot more than just calculating some areas and doing a division, and can certainly be subject to serious mathematical research. $\endgroup$ – Gerry Myerson Sep 19 '18 at 23:20
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    $\begingroup$ As a starting point, one could use a greedy algorithm: attach the first square at the north pole, and then keep attaching others at points as far north as possible, preferring the westernmost location when there's a choice. I hope someone will code it and graph it! $\endgroup$ – Matt F. Sep 20 '18 at 1:09
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    $\begingroup$ I disagree with @GerhardPaseman's opinion, and I think it is an excellent research-level question. On the other hand, I don't think the mathematical community has the theoretical or computational tools to give a precise answer with proof at this point in history. $\endgroup$ – S. Carnahan Sep 20 '18 at 3:22
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    $\begingroup$ This is a variant of the Thomson problem (en.wikipedia.org/wiki/Thomson_problem) with a potential that is not radial. One can presumably get good solutions along similar lines. $\endgroup$ – Steve Huntsman Sep 20 '18 at 16:23
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This is not a direct answer, but it's close. The following paper partitions a sphere into equal-area, near-squares:

"The system has several degrees of freedom that can be fixed, for instance, by enforcing the cells aspect ratios. Therefore, the cells can have very comparable forms, i.e. close to the square."

Beckers, Benoit, and Pierre Beckers. "A general rule for disk and hemisphere partition into equal-area cells." Computational Geometry 45, no. 7 (2012): 275-283. (Elsevier link.)


          Fig11
          Figure 11. Dome, equal areas.


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  • $\begingroup$ This answer seems to side with my initial comment and against Gerry's comment. For then, it becomes a matter of dividing the surface area of a sphere by 1 unit squared, and ignoring the error. Gerhard "Probably What Disco-Ball Makers Do" Paseman, 2018.09.19. $\endgroup$ – Gerhard Paseman Sep 20 '18 at 4:00
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There are two interpretations of "no overlap". One is suggested by the original problem, in that two mirror tiles do not physically intersect, while still having their centers attached to the sphere. For a sphere of radius 1/2 times 1/$\sqrt{3}$, there are at least two ways to glue three unit tiles for this definition of overlap: one where the tiles form a triangular prism, and one where one tile is given a 45 degree twist. Both of these do not overlap in the first sense, but the second one fails to not overlap in a second sense, where spherical projections of each tile are supposed to be disjoint. While both problems are of interest, it would be good to resolve this ambiguity and focus on just one version. This is one reason why I think the original version was not good for this forum: the problem is not clear enough. I encourage the poster to think more and revise the problem further toward greater clarity.

I took the trouble to make some computations for the first sense of overlap and small numbers of tiles. I suspect that the radii computed are the smallest that can accommodate the given number of unit square tiles, but the amount of effort to prove it for any number greater than two tiles is substantial. I invite readers to contribute a brief proof even for $t=3$, and remember to use overlap in the first sense.

$$t = (1,2,3, 4, 5, 6, 12, 18)$$

$$r= 1/2 (\epsilon,\epsilon,1/\sqrt{3}, 1/\sqrt{\phi},1,1,\sqrt{3},1+ \sqrt{2})$$

For larger values of t the two senses of overlap seem to converge, and that the smallest radius having a disco ball of $t$ tiles in either sense approach each other fairly rapidly. However, as there is still work to do on packing unit squares into large squares of nonintegral side length, I hesitate to declare any results on this problem as other than temporary.

Gerhard "Do Not Overlap With Ambiguity" Paseman, 2018.09.19.

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  • $\begingroup$ Indeed, for the first sense of overlap, I expect rapid phase shifts where the minimal radius stays constant for t=18 through t=26, while there is less constancy in the radius as t grows for the second sense of overlap. However this is all conjecture. Further my objection is not to the idea of the problem being posted on MathOverflow; I objected to the initial version where it appeared that not enough thought was put into the question. Gerhard "Just So it Is Clear" Paseman, 2018.09.19. $\endgroup$ – Gerhard Paseman Sep 20 '18 at 3:40
  • $\begingroup$ Here is a brief proof for t=3: take a plane running through the centers of the three square tiles. The (portion of the) triangle formed with the intersection of the plane with the squares has sides of length at least 1, so the incircle has radius at least 1/2sqrt(3). Gerhard "Good Luck With Four Tiles" Paseman, 2018.09.19. $\endgroup$ – Gerhard Paseman Sep 20 '18 at 3:51
  • $\begingroup$ I think there's a third sense of "no overlap" that can be considered (especially applicable in the context of a disco ball), that is the squares (mirrors) are required to lie on the boundary of the convex hull of the set-up. This is a slightly more stringent condition than your second sense. $\endgroup$ – Willie Wong Sep 20 '18 at 13:32
  • $\begingroup$ @Willie, hmm, I am seeing it as equivalent, not stronger. It might be stronger if the tiles were of different sizes or shapes. Of course, if you have an example with square tiles (all the same size) to clarify, I would like to see it. Gerhard "Sometimes Finds One Mirror Troubling" Paseman, 2018.09.20. $\endgroup$ – Gerhard Paseman Sep 20 '18 at 14:13
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    $\begingroup$ Happy ending. I even get to return the snake I bought. Gerhard "Back To Understanding No Overlap" Paseman, 2018.09.20. $\endgroup$ – Gerhard Paseman Sep 20 '18 at 21:49

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