This is an exercise from the Kürschák Mathematics competition from 1947:

The radius of each small disc is half of the large disc. How many small disks are needed to cover the large disc completely?

I assume they ask for the minimal number of small disks.

So, to cover the plane with disks with "minimal overlapping" we have to do something like:

The question is how to place the large disk on this picture so that it overlaps with the least amount of small disks.

If we place the large disk so that its middlepoint is in the middle as so:

then it overlaps with all the $12$ small circles on the picture but I recon we can do better. If we shift the large disk so its centerpoint coincides with one of the small disks' then it will overlap with $9$ small disks.

this looks minimal, but I have a hard time finding a rigorous way showing this. I tried looking at the lattice the small disks' middlepoints define and reason just using the middlepoints but without any good conclusion.

I have solutions on the back of the book (where I found the exercise) but it would be better if I myself (with maybe some help from MSE) could cook up something reasonable.

Is my approach good at all? All the above pictures look really symmetrical, maybe it is better to approach the problem in a "less symmentric way".

Could anyone give me some HINTS how to think here?

Thank you!

  • $\begingroup$ Hint: You are on the right track but you can do better than 9 small circles. Note that in the two coverings you present here the centres of the circles lie on a square grid. Is that necessary or can you think of something else? $\endgroup$ – Ronald Blaak Nov 13 '17 at 13:46
  • $\begingroup$ @RonaldBlaak you mean like for example what if they lie on the large disks boundary? $\endgroup$ – Vinyl_cape_jawa Nov 13 '17 at 14:08

The minimal number is $7$.

Let $D \subset \mathbb{R}^2$ be the larger disk. Let $r$, $O$ and $C$ be its radius, center and boundary respectively.

Any disk of radius $\frac12 r$ has diameter $r$ and it can cover at most $\frac16$ of $C$. One need at least $6$ disks just to cover $C$. Let's say one use $6$ disks $D_1, D_2, \ldots D_6$ to cover $C$. For each small disk $D_i$, $D_i \cap C$ will be an arc exactly $\frac16$ of $C$. The end points of that arc will form a diameter of $D_i$. Such a disk will be too far away from $O$. This means one need at least one more disk to cover $O$. As a result, it takes at least $7$ disks to cover $D$.

As illustrated by picture at end, $7$ is indeed possible.

Just inscribe a regular hexagon of side $r$ into $C$. $D$ can be covered by one small disk of radius $\frac12 r$ at $O$ and six more centered at the edge centers of the hexagon.

$\hspace0.75in$ Cover a disk by 7 smaller disks


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