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In a circular tray of radius $1$, arrange coins of radius $\frac12,\frac13,\frac14,\frac15$ - at least one of each, and no other kind of coin - so that none of them can move independently, i.e. if any coin moves, then at least one other coin must simultaneously move.

  • By "move", I am referring to movement of a coin's centre relative to an arbitrary fixed point on the tray.
  • The coins are circular, rigid and smooth.
  • Coins must be placed flat in the tray. This is not a trick question.

For a hint, see my comments to @Weather Vane's answer.

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    $\begingroup$ @Dan Welcome to PSE (Puzzling Stack Exchange)! $\endgroup$ Commented Jun 27 at 18:21
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    $\begingroup$ This does not help, but I just realized today we can pack exactly 2 x 1/2 coins and 2 x 1/3 coins in the 1-radius circle! So nice... $\endgroup$ Commented Jun 28 at 13:53
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    $\begingroup$ @Prim3numbah I found one solution. I don't know whether there are others. $\endgroup$
    – Dan
    Commented Jun 29 at 13:24
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    $\begingroup$ @Prim3numbah I've added a comment to the accepted answer, explaining why. $\endgroup$
    – Dan
    Commented Jul 1 at 12:33
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    $\begingroup$ @bobble I don't see why you deleted "Edit 2". It's clearly for those interested who may not otherwise be aware of the new, closely related, problem. $\endgroup$ Commented Jul 1 at 14:15

2 Answers 2

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OP's hint quickly leads to this:

enter image description here

red = 1/2, green = 1/3, blue = 1/4, purple = 1/5

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  • $\begingroup$ Yes, this is the solution I found. If you try to arrange the coins like this (with actual coins, or with a computer), you will initially find that there are small gaps between coins. Then push the 1/2 (red) coin down as far as possible, and all the gaps will have disappeared. (In the final arrangement, the distance between the tray's centre and the 1/2 coin's centre is somewhere between $0.34716$ and $0.34717$.) $\endgroup$
    – Dan
    Commented Jun 30 at 23:17
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    $\begingroup$ Some people may have read my MSE question linked to in the OP, and may be wondering why this arrangement does not satisfy $n=5$ in the MSE question. The reason is that, in this arrangement, the 1/2 coin can move up while the 1/5 and the 1/4 coins also move; this kind of movement is not allowed in the MSE question. $\endgroup$
    – Dan
    Commented Jul 1 at 12:30
  • $\begingroup$ I had assumed you were asking for a stable arrangement, where the only possible movement is rotation of the whole group within the dish. $\endgroup$ Commented Jul 1 at 13:22
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Back to 6 coins, the simplest and tightest yet.

But there is still a $0.005576$ gap between coins A and F.

enter image description here

Edit: the best fit I can find so far is still imperfect.

There are 8 coins, but the gaps between F/H and G/H are $0.008277$
I placed coins A to G first, so everything touches.
Then coin H was placed touching coins B and C.

enter image description here

Previous: I have a close but not exact solution with 6 coins.
There didn't seem to be a solution using 4 coins (by pencil and paper trials) and I skipped 5 coins.

The best fit I can find is this:

The tray has a radius $1.032859$

enter image description here

I found this by placing the three central coins, all touching, and three more coins around them, each touching two of the first three. Then make sure that none of the middle three can 'escape', by checking its centre lies on the correct side of a line joining the other centres. Finally find the circumcircle.

Repeating for all valid permutations, this one's circumcircle is closest to 1.0 radius.

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  • $\begingroup$ Nice try, but not quite. The radius of the tray should be exactly $1$. (Also, it looks like your $1/3$ circle can roll along the tray wall anti-clockwise without disturbing any other coin.) $\endgroup$
    – Dan
    Commented Jun 28 at 15:51
  • $\begingroup$ Is there an exact solution? $\endgroup$ Commented Jun 28 at 16:08
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    $\begingroup$ If by "exact solution" you mean an arrangement in which the coins perfectly fit together, with no coin being able to move independently, then yes. $\endgroup$
    – Dan
    Commented Jun 28 at 16:28
  • $\begingroup$ Hm, yes. Rolling the 1/3 coin clockwise around the rightmost 1/5 coin from the positive x-axis, it hits the rim at the same place (approx -52°), and when rolling it anticlockwise around the same 1/5 coin from the negative x-axis it first hits the rim at about -77°. So it isn't trapped. $\endgroup$ Commented Jun 28 at 17:44
  • $\begingroup$ Would you like a hint? $\endgroup$
    – Dan
    Commented Jun 30 at 21:41

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