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Find the maximum $n$ such that circles of radius $1, \frac12, \frac13, ..., \frac1n$ can be held immobile by a convex frame, or show that there is no maximum.

Here is an example with $n=7$.

enter image description here

By "immobile", I mean no circle can move without overlapping other circles or the frame, either individually or simultaneously. The frame is rigid.

It seems to get increasingly difficult to keep adding circles (and expanding the frame), while maintaining the conditions that the circles are immobile and the frame is convex. Or maybe there is a clever way to arrange the circles so that you can include all of them.

(This question was inspired by What is the minimum area of a rectangle containing all circles of radius $1/n$?)

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    $\begingroup$ You are one of the best askers on the site. $\endgroup$
    – mathlander
    Commented Jan 15, 2023 at 20:36
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    $\begingroup$ @mathlander Thanks. I am awed by the answers that I receive to my questions, and I have learned a lot here. $\endgroup$
    – Dan
    Commented Jan 15, 2023 at 21:09
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    $\begingroup$ I'm glad that people here appreciate my questions. My wife saw my scratch paper for this question and thought I was just drawing paw prints. $\endgroup$
    – Dan
    Commented Jan 16, 2023 at 22:55
  • $\begingroup$ Yep, that diagram up there looks pretty much like a paw print. $\endgroup$
    – mathlander
    Commented Jan 16, 2023 at 22:55
  • $\begingroup$ If you go to 3D, starts of {-1, 2, 3, 5, 6} and {-2, 3, 7, 9, 10} generate all bends except {12, 16, 24}. So, a unit sphere filled with Soddy's hexlet, then fill one of the 1/2 squares, then add those 3 extra spheres somewhere, and you've got a frame for everything in 3D. $\endgroup$
    – Ed Pegg
    Commented Jan 27, 2023 at 23:19

1 Answer 1

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You can start by using the Descartes Circle Theorem. For four mutually touching circles $2 (a^2+b^2+c^2+d^2) = (a+b+c+d)^2$, where the values are 1/radius (the bend). For example, circles with radii {1/2, 1/3, 1/6, 1/23} can all touch each other, but the bends are easier to work with: {2, 3, 6, 23}. Once you have 4 circles working together, you can get infinite circles, but only a third of the possible bends. Mod 12, a set of Descartes circles will only have 4 values.

So, we just need to find three Descartes circle starters that cover the 12 possible mod 12 values. Unfortunately, that may not exist. But here are some of the starter sets you can play with: {{2,3,6,23}, {1,4,12,33},{3,6,7,34}, {4,6,12,46}, {2,8,24,66}, {8,9,17,72}, {14,21,42,161},{7,28,84,231}, {21,42,49,238}, {11,44,132,363}}. Here's a set of four starters that cover all the mod 12 values, but there will be overlap: {{2, 3, 6, 23}, {1, 4, 12, 33}, {7, 10, 27, 90}, {8, 9, 17, 72}}. In the image below, circles are colored by their modulus. Descartes

Here are some starters with negative bends to work with.
{{-1,2,3,6}, {-1,2,6,11}, {-1,3,14,26}, {-1,6,11,30}, {-2,3,6,7}, {-2,3,7,10}, {-2,3,10,15}, {-2,6,7,19}, {-3,4,12,13}, {-3,4,13,16}, {-3,4,16,21}, {-3,5,8,12}, {-3,8,21,44}, {-4,8,9,17}, {-5,6,30,31}, {-7,12,17,20}, {-7,12,17,24}, {-9,14,26,27}, {-11,16,36,37}}

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    $\begingroup$ But if the frame is a circle, the circles can always move by just rotating simultaneously, or is this not what OP meant? $\endgroup$
    – fweth
    Commented Jan 15, 2023 at 18:23
  • $\begingroup$ The negative bends are for setting up predictable series. The entire frame would be irregular, but might include numerous circular arcs. $\endgroup$
    – Ed Pegg
    Commented Jan 15, 2023 at 18:51
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    $\begingroup$ @fweth replace the containing circle with a rubber band and tighten it. (Straightening one arc may be enough) $\endgroup$
    – Hagen von Eitzen
    Commented Jan 15, 2023 at 20:25
  • $\begingroup$ @HagenvonEitzen ah I see, thanks! $\endgroup$
    – fweth
    Commented Jan 16, 2023 at 13:00

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