Timeline for How many unit cylinders can touch a unit ball?
Current License: CC BY-SA 4.0
34 events
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| Jul 22, 2019 at 1:31 | history | edited | Wlodek Kuperberg | CC BY-SA 4.0 |
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| Jun 5, 2018 at 20:11 | history | edited | Wlodek Kuperberg | CC BY-SA 4.0 |
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| Jun 4, 2018 at 14:29 | history | edited | Wlodek Kuperberg | CC BY-SA 4.0 |
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| Jun 3, 2018 at 11:44 | comment | added | Andreas Blass | @smci The cylinders are supposed to be infinitely long --- the product of a disk and a line, not a line segment. Of course, the pictures show only a finite part of each cylinder. | |
| Jun 3, 2018 at 1:40 | comment | added | smci | I don't understand the Brass&Wenk disproof you can't achieve 8: in the top right picture with six parallel cylinders in a hexagonal prism, can't you slide 2 extra cylinders inside, one from the top, one from the bottom? | |
| Jun 3, 2018 at 0:33 | answer | added | Yoav Kallus | timeline score: 7 | |
| May 29, 2018 at 22:40 | comment | added | Wlodek Kuperberg | Yes, @Yoav, the article improves on Moritz's lower bound $r_6>1.04965$, namely $r_6>{1\over8}(3+3\sqrt3)\approx{1.093070331}$. The main question is still unanswered. | |
| May 29, 2018 at 13:38 | comment | added | Yoav Kallus | A recent preprint improves on Moritz's result arxiv.org/abs/1805.09833 | |
| Apr 4, 2014 at 13:40 | history | edited | Ricardo Andrade |
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| S Apr 4, 2014 at 1:55 | history | bounty ended | Wlodek Kuperberg | ||
| S Apr 4, 2014 at 1:55 | history | notice removed | Wlodek Kuperberg | ||
| Apr 4, 2014 at 1:49 | vote | accept | Wlodek Kuperberg | ||
| Mar 31, 2014 at 23:20 | answer | added | Moritz Firsching | timeline score: 47 | |
| Mar 31, 2014 at 20:07 | comment | added | Pietro Majer | A possible approach is perhaps looking at plane sections. A first point to understand seems, what are the possible configurations obtained cutting two cylinders and a sphere, mutually tangent, by a plane passing by the three tangency points (and necessarily the center of the sphere). | |
| Mar 31, 2014 at 18:06 | comment | added | Pietro Majer | (yes, I was following similar thoughts) | |
| Mar 31, 2014 at 18:03 | comment | added | Pietro Majer | Maybe in fig 2.2 the front cylinder can move its lower part inside the hollow formed by the rotating group of three? Or maybe a torsion from fig 1.3, like in a mikado game? | |
| Mar 31, 2014 at 18:03 | comment | added | Wlodek Kuperberg | @PietroMajer: Perhaps by squeezing and twisting the top part of three non-adjacent cylinders, making the points of tangency move upwards, at the same time doing it symmetrically (with respect to the ball's center) with the other three...? Again, as in my comment to Benoit Kloeckner's answer, this would require some tedious computations... Comupter-aided experiments? Anyone willing and able...? | |
| Mar 31, 2014 at 17:54 | comment | added | Pietro Majer | I can't imagine how to deform the first configuration either, apart than rotating a group of three cylinders as shown in figure 1. | |
| Mar 31, 2014 at 15:17 | answer | added | Benoît Kloeckner | timeline score: 5 | |
| S Mar 31, 2014 at 13:58 | history | bounty started | Wlodek Kuperberg | ||
| S Mar 31, 2014 at 13:58 | history | notice added | Wlodek Kuperberg | Draw attention | |
| Jan 29, 2014 at 16:22 | comment | added | James Cranch | I agree! I can't back up my intuition either: I've just tried doing it with pens, and learned only that I cannot manipulate six pens simultaneously. | |
| Jan 29, 2014 at 16:15 | comment | added | Wlodek Kuperberg | @JamesCranch: Will Sawin says "looks to me", you are not convinced. I have the same two (mixed) feelings, and there is no contradiction. | |
| Jan 29, 2014 at 16:07 | comment | added | James Cranch | I'm not convinced, Will. What happens if I divide the cylinders into two sets of three, where each set contains one cylinder in each of the three directions? It's not at all obvious to me that I can't twist the two sets so as to deform the configuration to Figure 1. | |
| Jan 29, 2014 at 15:56 | history | edited | Wlodek Kuperberg | CC BY-SA 3.0 |
Title made more informative.
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| Jan 29, 2014 at 9:44 | comment | added | Campello | If you project the first configuration along the vector that determines the cylinder line, you get a cell of the Hexagonal lattice. Probably the belief that $k = 6$ has to do with the solution to the kissing number problem. | |
| Jan 29, 2014 at 4:13 | comment | added | Will Sawin | It looks to me like the final configuration is locked - there is no motion of the six cylinders where they remain attached to the sphere other than rotations. If one could prove this, it would obviously negatively answer the second question. | |
| Jan 29, 2014 at 3:04 | history | edited | Wlodek Kuperberg | CC BY-SA 3.0 |
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| Jan 28, 2014 at 23:09 | comment | added | Wlodek Kuperberg | @HughThomas: Thanks for pointing this out. | |
| Jan 28, 2014 at 23:07 | history | edited | Wlodek Kuperberg | CC BY-SA 3.0 |
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| Jan 28, 2014 at 23:01 | comment | added | Hugh Thomas | The sentence just above the three figures is unfinished. | |
| Jan 28, 2014 at 22:17 | comment | added | Joseph O'Rourke | (This is a useless tangent, but the rightmost image in Fig.3 occurred in another MO question asking a different question: Blocking visibility with cylinders.) | |
| Jan 28, 2014 at 22:14 | history | edited | Wlodek Kuperberg | CC BY-SA 3.0 |
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| Jan 28, 2014 at 22:09 | history | asked | Wlodek Kuperberg | CC BY-SA 3.0 |