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For what values of $n$ can circular coins of radius $\frac12,\frac13,\frac14,...,\frac1n$ (at least one of each, and no other kind of coin) be held rigidly in a circular tray of radius $1$?

By "held rigidly", I mean that no coin can move, except that each coin can rotate about its own centre, and all the coins can rotate collectively about the tray's centre.

My attempt

I know that $n$ can equal $2,3,4$:

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For $n>4$, I suspect there are no solutions, but I cannot prove it.

For $n=5$, the "closest" I have gotten is the following (here is the desmos graph).

enter image description here

Below is a perfect fit with $\frac12,\frac13,\frac14,\frac16,\frac17$. (Missing $\frac15$, so no good.)

enter image description here

Context:

This is the latest question of mine related to rigidly held coins.

The original question was: "Five circles in a rectangle: Can the circles move?"

The previous question was on Puzzling Stack Exchange: "In a circular tray of radius 1, arrange coins of radius $\frac12,\frac13,\frac14,\frac15$ so that none of them can move independently" (Note: The Puzzling question uses a different kind of rigidity than my question here.)

Dan
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  • Somewhat related, but not quite what you want. – Jyrki Lahtonen Jul 28 '25 at 03:31
  • Why is this considered a good question if I may ask ? I mean it already breaks down at $n=5$. And then we are suppose to ignore that and go on with $n>5$ ? Why would not all $n$ (that exist !) be rigid ? I mean just add enough small ones to fill in the empty space and/or cherry pick the configuration. Just like a box of pebbles with different sizes, but many sizes similar to each other, rigid is expected. Especially if we try to make it rigid by selecting the number of each size of pebbles. And especially if we ignore some sizes of pebbles ( analogue to the $n=5$ has no solution). – mick Jul 30 '25 at 23:46
  • Proving packing problems without a defined number of fixed variables is notoriously hard. Probably as hard as a random diophantine equation. – mick Jul 30 '25 at 23:49
  • @mick Perhaps we have a different understanding of the requirements in the question. It is not clear to me whether or not there exists is a solution with $n=5$ (I haven't found one yet). It is conceivable that the answer to my question turns out to be $2,3,4,15$ or something like that, and that this is provable. You cannot just "add enough small ones to fill in the empty space", because the result might not be rigid. We cannot "ignore some sizes": all sizes from $\frac12$ to $\frac{1}{n}$ must be present (at least one of each, with no other size). – Dan Jul 31 '25 at 00:00
  • @mick I realize that a complete answer to my question might be very difficult, but I would also welcome solutions for individual values of $n$. For example, can anyone find a solution for $n=5$? Or $n=6$? – Dan Jul 31 '25 at 00:21
  • @Dan here are some conjectures : conjecture A) All rigid solutions are symmetric. Conjecture B) All rigid solution have that the $1/2$ circles must always touch exactly $n$ circles. – mick Jul 31 '25 at 00:47
  • And conjecture C : all rigid solutions satisfy : $(1/k)$ must touch the same number of other circles. – mick Jul 31 '25 at 00:52
  • This would probably be a really nice result if it were possible for all $n$... (I'm hoping it is) – Kevin Chen Sep 14 '25 at 16:42

0 Answers0