Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Many years ago I picked up a little book by J. E. Littlewood and was baffled by part of a question he posed:

"Is it possible in 3-space for seven infinite circular cylinders of unit radius each to touch all the others? Seven is the number suggested by counting constants."

It is the bit in italics which baffled me then (and still does). Can anyone explain how he gets 7 by "counting constants"?

P.S. For completeness, the book is "Some problems in Real and Complex Analysis" (1968)

share|cite|improve this question
I do not know about this problem but there is a related problem for spheres posed by Donald Coxeter: see and the linked explanation. The answer for spheres is 5. I do not know if he ever considered cylinders. – Ronnie Brown Oct 31 '12 at 17:13
Many thanks. I definitely like that sculpture! – Old John Oct 31 '12 at 17:19
Should we think of the cylinders as being solid (like the spheres in the sculpture), or are they allowed to intersect? The terms "touch" implies tangency to me, but I thought I'd check. – Fly by Night Oct 31 '12 at 17:28
He doesn't make it clear in the book, but I would assume he means solid and that he is talking about tangency. – Old John Oct 31 '12 at 17:35
up vote 18 down vote accepted

Here is my take: There are $4$ degrees of freedom in selecting the center line of each cylinder, for a total of $4n$ degrees of freedom. Subtract from this the $6$ degrees of freedom given by the Euclidean motions (rotations and translations in space), as applied to the total configuration – for a total of $4n-6$ degrees of freedom.

For two cylinders to touch, the minimal distance between points on their respective center lines must be $2$. This results in $\binom{n}{2}$ equations. To be able to satisfy all these equations, we must probably have $4n-6\ge\binom{n}{2}$, which holds for $n\le7$.

share|cite|improve this answer
Interesting! I think this might be a tough one to decide which to accept :) – Old John Oct 31 '12 at 18:06
@OldJohn So wait for the dust to settle before deciding. – Harald Hanche-Olsen Oct 31 '12 at 18:21
+1 for brevity :) – alex.jordan Oct 31 '12 at 21:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.