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Say that a region $R$ is covered with multiplicity by some pieces $P_1,\ldots,P_n$ if $\sum_{i=1}^n\text{Area}(P_i\cap R)\ge \text{Area}(R)$ - ie, there's enough total overlap of $R$, it just isn't necessarily spread out evenly. This is obviously a weaker notion than covering in the normal sense.

In this answer to a question about covering disks, I proved that it was not possible to cover a unit disk with a specified set of small disks by showing that there were central rings of radii $x_1, x_2, x_3, x_4\le 1$ for which it was not possible for the smaller disks to simultaneously cover those rings with multiplicity.

I'm curious for a packing problem of this form - a finite set of small disks for which the task is to cover a unit disk - which is impossible, but cannot be disproven via this method because there is an arrangement of the small disks $D_1,\ldots,D_n$ such that for every $r\in[0,1]$ we have

$$\sum_{i=1}^n\text{Length}(D_i\cap \{x,y: x^2+y^2=r^2\}) \ge 2\pi r$$

My strong intuition is that such arrangements ought to exist - surely this type of argument isn't powerful enough to solve all circle packing problems! - but I've had some trouble finding a good construction.

RavenclawPrefect
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  • The main issue is that it's really hard or bothersome in most cases to prove that a particular collection of discs can't pack the circle, isn't it? – Chris Sanders Jan 25 '24 at 12:28
  • I hope you find a paper with a readymade example that you can tweak a bit to get what you want. – Chris Sanders Jan 25 '24 at 12:43
  • At the beginning you define covering with multiplicity in terms of areas, but at the end you use it in terms of covering the length of any circle. Also, you seem to be asking about covering any arbitrary single circle, but surely this has trivial counterexample, for example 100 discs with diameter $2\pi/100$, can cover any circle with $r<1$ but don't have enough area to cover the unit disc. Your proof used a set of 4 circles to be covered simultaneously. I would not at all be surprised that having to be able to cover any finite set of circles is sufficient to allow covering the disc. – Jaap Scherphuis Jan 25 '24 at 13:47
  • @JaapScherphuis: the length/area equivocation is a little sloppy, but I think in any particular case one can rescue it by using a very thin annulus instead. I'm not asking about covering a single circle, but about covering all circles with $r<1$ simultaneously (but only in this weaker sense of having enough total overlap with those rings). – RavenclawPrefect Jan 25 '24 at 19:02
  • @ChrisSanders One avenue that might be promising is taking a look at the records for circles covering circles, which includes some proofs of optimality (implicitly a proof of impossibility for any larger covering target) and some conjectured best possible solutions. I'd be reasonably happy with an answer that exhibited something no one was able to find a solution for, even if they didn't have a rigorous proof of impossibility. – RavenclawPrefect Jan 25 '24 at 19:05

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