TITLE: How many unit cylinders can touch a unit ball?
QUESTION [64 upvotes]: 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.

REPLY [6 votes]: 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.