Given $n$ circles of radius $r$, arranged so that line segments connecting the centers from one circle to the next form a convex polygon and none of the circles intersect, the length of a string stretched tightly around the outside of all the circles is:

\begin{equation} length = 2 \pi r + \sum_{i=0}^{n-1} dist(i, (i+1)_n)\end{equation} where $dist(i, j)$ is the distance between the centers of circles $i$ and $j$.

It's easy to see this for $n=1$ and $n=2$, but for $n > 2$ my intuition fails me and it seems like it would be a pretty hairy proof. Is there a fairly straightforward explanation or is it very involved? Thanks for any pointers.


Consider two adjacent circles with centres $C_1$ and $C_2$. Let $AB$ be the straight stretch of string between the first circle and the second. Then $C_1ABC_2$ is a rectangle with sides $r$ and $d$, where $d$ is the distance between $C_1$ and $C_2$. (A sketch is helpful here.) The straight bits of string therefore have a total length equal to the sum of the distances between the centres of adjacent circles.

The curved bits of string, where the string is in contact with the circles, must add up to one full circumference. Imagine that the parts of the circumferences of the circles that are in contact with the string are painted red. Now translate the circles to a common centre without rotating them: the red bits must exactly cover the circumference of the resulting circle, without overlap. Alternatively, imagine walking counterclockwise along the string: by the time you return to your starting point, you must have made one complete counterclockwise revolution around your vertical axis on account of the turning to your left that you’ve done in going around bits of circles.

  • $\begingroup$ Thanks the walking analogy helps explain the curved bits. $\endgroup$ – Bogatyr Sep 16 '12 at 22:44

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