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Suppose you have $n$ identical circular coins and you would like to arrange them on the table so that their edges touch and their centers lie on a circle.

Mathematically, there is no trouble. "Just" put the center of each coin at $re^{2ik\pi/n}$ for $k$ in $\{0, 1\ldots, n-1\}$ and some suitable $r$. But in practice, one can't easily calculate $e^{2ik\pi/n}$ and one wouldn't be able to position the coins even if the coordinates of their centers were given.

What I want are heuristics that allow one to position the coins approximately correctly, which can be executed by someone with an ordinarily good eye and ordinarily good hands, without any special measuring devices.

Good solutions for $n\le 3$ are trivial. There is also a good solution for the special case of $n=6$, which is to arrange the six coins around a seventh. In practice it does not seem too hard to arrange four coins into a square, by first estimating the right angles and then looking to see if the resulting quadrilateral is visibly rhombic. But I would be glad to see a more methodical approach.

This is a soft question. I expect the solution to be informed by mathematics, but not purely mathematical.

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Place the coins in a centrifuge whose circumference you can control... – Qiaochu Yuan Aug 22 '12 at 0:27
Just a comment: this subject is considered in configuration space – Sigur Aug 22 '12 at 0:40
Are the shapes we are looking for not the regular polygons? – Matt Calhoun Aug 22 '12 at 3:29
Yes, the centers of the coins will lie at the vertices of a regular $n$-gon. – MJD Aug 22 '12 at 3:30
How about this for (large) even $n$? Ordinarily-good eyes and ordinarily-good hands can create a reasonably-straight line of $\lceil n/\pi \rceil$ coins, approximating the circle's diameter with, say, end-coins $A$ and $B$. Let coin $C$ touch $A$ such that $\angle ACB$ appears (to ordinarily-good eyes) to be a right angle, so $C$ lies on (or near) a semicircle. Let $D$ touch $C$ such that $\angle ADB$ appears right; and so on: an arc of coins approaches $B$. Do the same on the other side of $AB$. Accumulated error almost-certainly keeps the arcs from meeting properly at $B$, but it's a start. – Blue Aug 22 '12 at 7:16

If you are allowed to use surplus coins, pack them together to form a large hexagonal grid in the (approximate) shape of a circle with centre $C$ and radius $r$. It is then algebraically straightforward to find which of these coins are closest to $re^{2 i k \pi / n}$.

Now remove all the other coins and use them to buy some string. Place a string from $C$ to the centres of each of the remaining coins; contract along the strings.

Here's an example with a few hundred coins and $n = 11$:

enter image description here

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+1 for "use them to buy some string", but -1 because your construction fails: the remaining coins will be touching, so a protruding coin can't be pulled in without disturbing its neighbours. – TonyK Aug 22 '12 at 5:57
Ha! I think I should have made it clear that my radius $r$ is not the same as the $r$ in the original description. Here $r$ is just the size of the circle that you happen to build. I'll add a picture. – Théophile Aug 22 '12 at 19:14
I see! -- you put a napkin ring in the centre and pull all eleven strings up through it together. – TonyK Aug 23 '12 at 7:13
Yes, that should do the trick! – Théophile Aug 23 '12 at 15:51

For a large number of coins, I would apply an iterative scheme inspired by mean curvature flow. In other words, just keep nudging in coins that are jutting out, until the curvature of the whole arrangement is uniform.

We can consider the arrangement of coins as a polygon whose vertices are the centers of the coins, and whose edges connect adjacent coins and are all of equal length. The polygon is regular if and only if all its internal angles are equal, which holds if and only if each angle is equal to (i) the average of all the angles, and/or (ii) the average of the angles at adjacent vertices. If you can compare adjacent angles by eye, or tell which vertices have the largest and smallest angles of all, then you can nudge them in the right direction.

In practice, I found that this has a couple of limitations. First, it's easy to push coins inwards because the rest of them will automatically move out of the way, but if you push a coin outward, you have to carefully reconnect the arrangement. Second, the procedure converges quite slowly. You quickly arrive at a nice smooth oval, after which moving coins one at a time is a frustratingly constrained way to round out the arrangement.

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I think it turns out to be rather difficult to estimate whether the local curvature at a coin is too much or too little. What did you find when you tried? – MJD Aug 22 '12 at 1:41
I just updated my answer. I agree that it can be a little difficult to compare angles when the curvature becomes close to zero, but I found that I could make a decent circle nevertheless using this process. Although it's hard to say whether I didn't start subconsciously pattern-matching against the image of an ideal circle instead of applying these explicit criteria at some point. – Rahul Aug 22 '12 at 6:50
Anyway, I meant this to be a mathematically sound procedure that can be implemented (albeit imperfectly) by hand, rather a foolproof practical procedure. – Rahul Aug 22 '12 at 6:56

As our eyes easily detect (absence of) mirror symmetry I propose the following: Start with a more or less circular layout; then symmetrize iteratively with respect to randomly selected axes.

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The main problem at hand is to compute the radius of the big circle of coins, since if we know this we can just measure out 2 adjacent sides from a central point and go from there.

The plan is to arrange the coins "like a regular n-gon", by which I mean associate the diameter, $D$, of a coin to the unit length of the n-gon.

Assuming that for N coins we can "easily calculate" the roots of unity, we can compute the distance in the complex plane between two adjacent roots of unity, $L$, and scale the whole n-gon formed by the roots of unity so that it each side has length $D$. In other words, the radius we are looking for is $\frac{D}{L}$.

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I think you are missing the point of this question. The question specifies that there are no measuring devices. How do you plan to "measure out 2 adjacent sides"? – MJD Aug 22 '12 at 3:59
You actually said no "special" measuring devices ;) I was trying to answer the question that was asked, given N coins, what is a good heuristic to lay them out in a circle. Well, just measure out 2 sides, then another 2 sides, and so on for a little while; then it will be obvious how to connect everything. – Matt Calhoun Aug 22 '12 at 4:05
Even if you are allowed to measure out two sides, I think by the time you got all the way around the circle you would find that the last coin would not match up with the first. – MJD Aug 22 '12 at 4:51

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