# How many unit cylinders can touch a unit ball?

What is the maximum number $$k$$ of unit radius, infinitely long cylinders with mutually disjoint interiors that can touch a unit ball?

By a cylinder I mean a set congruent to the Cartesian product of a line and a circular disk. The illustrations, copied from 2, show various configurations of six cylinders, perhaps all possible - up to isometry. The question is about 25 years old; the answer is conjectured to be 6, but the conjecture is still unconfirmed. Heppes and Szabó 1 proved in 1991 that $$k\le 8$$; nine years later Brass and Wenk 2 improved this result to $$k\le 7$$. Also, one would like to know:

What does the configuration space of six unit cylinders touching a unit ball look like?

In particular:

Is the configuration space connected?

Even more specifically, referring to the configurations shown below:

Within the configuration space, is a continuous transition possible from the configuration in Figures 1 and 2 to the configuration in Figure 3?

The last question may be possible to verify by a natural candidate, but the necessary computations seem too tedious to me...  1 Heppes, Aladar and Szabó, Laszló. "On the number of cylinders touching a ball." Geom. Dedicata 40 (1991), no. 1, 111–116; MR1130481.

2 Brass, Peter and Wenk, Carola. "On the number of cylinders touching a ball." Geom. Dedicata 81 (2000), no. 1-3, 281–284; MR1772209.

• (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.) – Joseph O'Rourke Jan 28 '14 at 22:17
• 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. – Will Sawin Jan 29 '14 at 4:13
• 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. – James Cranch Jan 29 '14 at 16:07
• @JamesCranch: Will Sawin says "looks to me", you are not convinced. I have the same two (mixed) feelings, and there is no contradiction. – Wlodek Kuperberg Jan 29 '14 at 16:15
• 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. – James Cranch Jan 29 '14 at 16:22

Here is an idea. Consider the following parameterization, which is supposed to cover the configuration space in question.

$$\mathcal{C}_7:=\left\{\pmatrix{x_k\\y_x\\z_k},\pmatrix{a_k\\b_k\\c_k}_{1\leq k\leq 7}\in{\mathbb{R}^{3\times 2}}^7\,\middle |\, \text{such that conditions 1.-4. are satisfied} \right\}$$

Conditions:

1. $x_k^2+y_k^2+z_k^2=1$
2. $\left\langle\pmatrix{x_k\\y_k\\z_k},\pmatrix{a_k\\b_k\\c_k} \right\rangle=0$
3. $a_k^2+b_k^2+c_k^2=1$
4. $d(l_i,l_j)\geq 2$ for $1\leq i<j\leq 7,$ where we define the line $$l_k:=\left\{2\pmatrix{x_k\\y_k\\z_k}+\alpha\pmatrix{a_k\\b_k\\c_k}\,\middle|\,\alpha\in\mathbb{R} \right\}$$ and denote with $d(\cdot,\cdot)$ the distance between two lines.

Note that condition 4. can be rewritten as polynomial inequalities. Hence $\mathcal{C}_7$ is a semi-algebraic set in $\mathbb{R}^{42}$.

The $(x,z,y)$ are the points, where the unit cylinder is tangent to the unit sphere. The corresponding $(a,b,c)$ gives the direction in the tangent space and the lines $l$ are the cores of the cylinders. (Note that $(-a,-b,-c)$ gives the same cylinder.)

The question "Is $\mathcal{C}_7$ empty?" should be decidable. Maybe an algorithmic approach could help from here.

For the other questions the study of an analogues defined $\mathcal{C}_6$, which we know to be non-empty might be worthwhile.

I wrote a little program that tries to find points in the described semi-algebraic sets. Here's what it found for $\mathcal{C}_6$ (click here for an animation). Let's take a slightly different point of view. Fix the radius of the ball to be $1$, but let the radii of the $k$ cylinders vary while making sure all cylinders have the same radius. We can then ask: What is the largest radius $r_k$, so that we can find $k$ non-overlapping cylinders of radius $r_k$, that touch the unit ball? Hence the question is: $r_7\geq 1?$

An obvious lower bound on $r_k$ comes from the packing that allows a equatorial section which is a circle packing, as for $k=6$ in figure 1 and figure 2 in the question post. We therefore have: $$r_k\geq \frac{\operatorname{sin}(\frac{\pi}{k})}{1-\operatorname{sin}(\frac{\pi}{k})}$$ Here's a list of decimal approximations for small $k$s: $$\begin{array}{c|cccccc}k&3&4&5&6&7&8\\\hline \frac{\operatorname{sin}(\frac{\pi}{k})}{1-\operatorname{sin}(\frac{\pi}{k})} &6.464101& 2.414213& 1.425919& 1& 0.766421& 0.619914\end{array}$$ A perhaps surprising result of my calculations is the fact that $r_6>1$, indeed $$r_6> 1.04965$$ So in other words there is configuration of $6$ cylinders where the cylinders have radius larger than $1.04965$. Here is a picture of the configuration (again click here for an animation): I also drew cylinders of radius $1$ with the same tangent points, so one can see the difference.

The configuration space can be viewed as subset of the the $6$th power of the unit tangent bundle of the sphere $(T^1(S^2))^6$ (see conditions 1.-4. and Henrik Rüping's comment).

The upshot of finding a configuration with larger radius is: the configuration space contains an open subset of $(T^1(S^2))^6$ and hence is $18$-dimensional locally.

Edit:

Here is list of lower bounds on $r_k$ for small $k$:

• For $k=3$ and $k=4$ I conjecture the trivial bound for $r_k$ given above to be sharp.
• For $k=5$ one can find a configuration that shows: $r_5>1.45289>1.425919$
• For $k=6$ we have $r_6>1.04965 >1$ as mentioned above.
• For $k=7$ I found a configuration that shows $r_7>0.846934>0.766421$. Here is a picture of this configuration (again click here for an animation): • I do not understand, why you need condition (3). I would just normalize the tangent vector $(a_k,b_k,c_k)$ instead. – HenrikRüping Apr 1 '14 at 8:26
• Furthermore if we drop condition (4), we would just get a manifold (the 6-th power of the unit tangent bundle of the sphere $T^1(S^2)^6$). So in order to see that Figure (3) is an isolated point, one could compute the differentials of the polynomial functions appearing in (4) at that point. – HenrikRüping Apr 1 '14 at 8:31
• Nice idea, fantastic graphics! – Wlodek Kuperberg Apr 2 '14 at 0:58
• What a marvelous surprise! The configuration of $6$ cylinders with radius exceeding $1.04965$ is astounding! – Wlodek Kuperberg Apr 3 '14 at 0:54
• I doubt that I can get a better answer any time soon, therefore I am awarding the bounty to Moritz Firsching. Thank you, Moritz! – Wlodek Kuperberg Apr 4 '14 at 1:51

For a javascript rendered 3D model of the path of configurations reported by Ogievetsky and Shlosman in arXiv:1805.09833 see this page on my website.

The configurations in this path are composed of two "three-pole teepees" pointed in antipodal directions. This is also true of the Figure 3 configuration in the question. But a crucial difference is that in Ogievetsky and Shlosman's configurations, the two teepees are both right-handed. In Figure 3, the two teepees have different handedness.

You can set up the calculation of OS in the case of opposite-handed teepees. Just switch $\delta$ to $-\delta$ for $D$, $E$, and $F$ in their Equation (5). Now solving for $d_{AB}=d_{AD}=d_{BD}$ is even easier. From $d_{AD}=d_{BD}$ I get $\varkappa=0$, and from $d_{AB}=d_{AD}$, I get $$s+\frac{3(s-1)}{t+4-3s}+\frac{s-s^2}{s+t}=0\text,$$ where $s=(\sin\phi)^2$ and $t=(\tan\delta)^2$.

There are two real solutions for $t$ when $0\le s\le \tfrac13$, and then the solutions become complex for $\tfrac13<s\le1$. All the real solutions give $d_{AB}=d_{AD}=d_{BD}\le1$, with equality only for the double solution $s=\tfrac13$, $t=1$, which is Figure 3.

• Nice demo! It would be nice if there were a live display showing the gap size. Or if this is too hard to calculate, at least some numerical way of tracking where you are in the path so that it's easier to find and jump back to the point of largest separation. – j.c. Jun 3 '18 at 6:52
• @j.c. The gap size is not hard to calculate. I will add a field displaying it this evening. – Yoav Kallus Jun 4 '18 at 11:42
• @j.c. now updated – Yoav Kallus Jun 4 '18 at 23:44

This is not a complete answer, but a suggestion of approach to prove that the configuration shown in figure 3 is locked. I hope it is not to naive.

We consider a continuous perturbation of the configuration and want to prove it is obtained by applying a continuous deformation of identity in $\mathrm{SO}(3)$.

Step 1: look at two opposite cylinders, and show that their contact points with the sphere must stay antipodal. This should hold because otherwise one of the other four cylinders would probably not have enough room left.

Step 2: consider again two opposite cylinders, and show that they stay parallel. This should hold because of step 1: we can assume that their contact points with the sphere are fixed, and then if they rotate in opposite direction then another pair of opposite cylinders must also rotate in opposite directions, which in turns constrains the movement of the last pair; if I am right, this last pair in turn constrains the first pair to twist in the opposite direction it does twist. In this argument, I guess that the parity of the dimension matters, so the $4$-dimensional version of the problem could have a different solution.

Step 3: conclude the proof. We can assume that a given pair of opposite cylinders is fixed; then the second pair is constrained to rotate in the plane containing its axes (at least at first order), which rotation is prevented by the third pair.

• Step 1 is crucial, and not obvious at all. I have mixed feelings about it: sometimes I feel it must be that way; sometimes I don't. – Wlodek Kuperberg Mar 31 '14 at 15:56
• There is an "invisible" cube circumscribed about the ball in Figure 3; look at the three cylinders "adjacent" to the cube's front-upper-right vertex. Sometimes I think that the configuration of the three cylinders can be "twisted" continuously, so that their points of tangency move closer to the said vertex, allowing for the other three cylinders to twist symmetrically to the ball's center, while all six cylinders remain non-overlapping, but I have no patience for the necessary computations... Maybe someone would like to try a computer experiment...? – Wlodek Kuperberg Mar 31 '14 at 16:10
• Consider making six copies of a spherical "2-tile". The bottom layer of the tile is that section of a sphere between the center and 'the closest' cylinder, and the second layer represents the section of the sphere between the center and any point of the same cylinder. Note the first layers remain disjoint, and the second layers can slide over the first layers of a different tile. Now try moving these tiles around on a basketball. Gerhard "Solving Problems Using Analog Computers" Paseman, 2014.03.31 – Gerhard Paseman Mar 31 '14 at 18:16
• @WlodekKuperberg: if the two triples of cylinders move symmetrically, then this movement is satisfying step 1. This does not imply that it cannot happen, but that the statement "step 1 => lockedness" would rule out such a movement. – Benoît Kloeckner Mar 31 '14 at 18:17
• @BenoîtKloeckner You're right: the points of contact would remain antipodal. But none of the cylinders would remain parallel. So, even if Step 1 were correct, Step 2 could still fail. – Wlodek Kuperberg Mar 31 '14 at 18:48