Show that a disk cannot be tiled completely and without overlap using only finitely many smaller disks. (A disk here contains its boundary)

I came up with this problem while reading about the Apollonian gasket. My little brother's original solution (cleaned up somewhat by myself) went as follows: Assume that we have a valid finite tiling. Note that every nondegenerate disk has infinitely points on its circumference. Consider any disk $O$ in the tiling. Each disk that disk $O$ touches, touches $O$'s circumference at exactly one point. But the space around $O$ must be filled, so every point along its circumference must touch another disk (if it did not, there would necessarily be some $\epsilon>0$ such that for a distance $<\epsilon$ from the circumference of $O$, there exists a point not covered by a disk, which would make it a non-complete tiling, which would be a contradiction). Thus there must be infinitely many disks touching $O$, which contradicts our assumption that there were only finitely many disks in the tiling. Therefore no such tiling exists.

Are there any alternative proofs that there must be infinitely many disks in such a tiling? I was so struck by my brother's elegant proof (slightly nonrigorous as it may be) that I haven't come up with any alternatives, but I'd really like to know if any exist. (Making the above proof more rigorous would be helpful too.)

  • $\begingroup$ are you talking about tiling a disk? $\endgroup$ Oct 16, 2020 at 20:57
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    $\begingroup$ Yes. I thought that was pretty clear. From, well, the whole thing. $\endgroup$ Oct 16, 2020 at 21:03
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    $\begingroup$ More generally, you can't tile a strictly convex set with finitely many smaller strictly convex sets. $\endgroup$ Oct 16, 2020 at 21:09

1 Answer 1


Here's a different proof, though with a similar flavor (we're effectively using the "only one point on the circumference" argument on the big disk, though with more formality).

Without loss of generality, suppose the large disk is the unit disk centered at the origin. Then the maximum radius of any of the smaller disks must be some $x<1$ (since there are finitely many). Then for every $\epsilon$, there will be some $\delta$ such that no disk of radius at most $x$ can cover more than $\epsilon$ fraction of the annulus $\{(x,y)\ |\ 1-\delta\le x^2+y^2\le1\}$. (Proving this rigorously would involve some garbage bashing of trig functions and coordinates, but it shouldn't pose any real difficulty.) Then you just choose $\epsilon$ to be less than the reciprocal of the number of disks in the tiling, and you conclude that not all points on the annulus can be covered.

This proof actually shows the stronger statement that finitely many smaller disks cannot cover the large disk even if they are allowed to overlap each other (but not to cross the boundary of the large disk), since we don't use disjointness of the small tiles at any point. It also gives us concrete lower bounds on how much area is left uncovered, given the list of radii in the tiling.

Another approach: Pick any point $P$ on the boundary of a tile which does not lie on the boundary of the large disk. All 360 degrees of angle around $P$ must be covered by some tile, but it can only be covered by tiles it lies on the boundary of. But since every circle $P$ touches uses up 180 degrees of angle, it must be the meeting point of two circles. But then since every other disk is some positive distance away from $P$, there is some $\epsilon>0$ such that the ball of radius $\epsilon$ around $P$ is only intersected by the two tangent circles mentioned above. But for any such ball, there will be two "slivers" of area left uncovered.

(This proof actually applies to any collection of smooth curves - we just need for there to be no "corners" in the tiles. If our curves have flat sides, we'll need to choose $P$ on a part of the tile with positive curvature.)


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