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Show that there exists a square frame and a set of identical circular coins, such that the coins can be rigidly packed in the frame in more than one arrangement.

Remarks:

  • "Rigidly packed" means no coin can move independently.
  • If two arrangements are the same upon rotation or reflection, then they are the same arrangement.
  • In each arrangement, all of the coins must be used.
  • The coins are placed flat in the frame without overlapping. (This is not a trick question.)

Example 1 of a non-solution:

Here are four coins and a square frame. The coins can be rigidly packed, as shown. But this is the only arrangement in which the coins can be rigidly packed in the frame, so this is not a solution.

enter image description here

Example 2 of a non-solution:

Here are six coins and a square frame. The coins can be rigidly packed in only one arrangement, as shown. The second diagram is just a rotation of the first diagram, so this is not a solution.

enter image description here

Example 3 of a non-solution:

Here are three coins and a parallelogram frame. The coins can be rigidly packed in two different arrangements, as shown. But the frame is not a square, so this is not a solution.

enter image description here

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4 Answers 4

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enter image description here

$\space\space\space\space\space$

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    $\begingroup$ This is awesome! $\endgroup$ Commented Jan 15, 2025 at 17:03
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    $\begingroup$ This is probably the best answer $\endgroup$ Commented Jan 15, 2025 at 17:46
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    $\begingroup$ Nice! Fun fact: In both arrangements, the coins are "rigidly packed" according to the definition in the OP, but only in the left arrangement can the coins simultaneously move (see here for a similar case of simultaneous movement). $\endgroup$ Commented Jan 15, 2025 at 19:48
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    $\begingroup$ Wonderfully simple, great solution $\endgroup$ Commented Jan 15, 2025 at 21:38
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New answer: here’s a solution with

8 coins

coins-square3 coins-square4

Old answer: here’s a solution with

26 coins

coins-square1 coins-square2

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    $\begingroup$ Nice! Classic "Why didn't I think of that?" solution. I will post my solution now. $\endgroup$ Commented Jan 15, 2025 at 7:32
  • $\begingroup$ Do you have a proof of optimality? $\endgroup$ Commented Jan 15, 2025 at 14:45
  • $\begingroup$ @bobble not at the moment. $\endgroup$ Commented Jan 15, 2025 at 15:06
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    $\begingroup$ @Pranay I deleted my comment after reading what exactly meant rigidly packed. You're right $\endgroup$ Commented Jan 15, 2025 at 15:45
  • $\begingroup$ @bobble Not asked for by OP. $\endgroup$ Commented Jun 30, 2025 at 8:25
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enter image description here $\space\space\space\space\space$

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  • $\begingroup$ +1. rot13(Lbh pbhyq erzbir bar bs gur gjb pbvaf va gur zvqqyr bs gur obggbz ebj naq fgvyy trg n evtvq pbasvthengvba.) $\endgroup$ Commented Jan 14, 2025 at 23:53
  • $\begingroup$ @Dan is 11 the minimum? $\endgroup$ Commented Jan 15, 2025 at 0:40
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    $\begingroup$ @Pranay I don't know. $\endgroup$ Commented Jan 15, 2025 at 0:41
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Here is my solution.

It uses 11 coins.

enter image description here enter image description here

The dotted line is there just to help you see the symmetry.

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    $\begingroup$ Nice -- this seemed pretty complicated at first. But the idea is clear when you remove the top 2 and right 2 coins (and shrink the square), and realize that the two orange coins can coexist. $\endgroup$ Commented Jan 15, 2025 at 15:39

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