I don't think this has been asked before...

Suppose you have an open circular disk [imath]D[/imath] in the Euclidean plane.

Suppose I give you a finite collection [imath]\{D_1, D_2, \ldots, D_n\}[/imath] of open circular disks whose total area is at most half the area of [imath]D[/imath]. Can you always arrange these disks to fit inside of [imath]D[/imath], without any overlaps between the [imath]D_i[/imath]?

## Fitting circles into a circle

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- BlackSails
**Posts:**5315**Joined:**Thu Dec 20, 2007 5:48 am UTC

### Re: Fitting circles into a circle

Lets look at the limit of big and small subcircles. In the limit of small circles, you just have points with total area equal to half the circle, and they easily fit. In the limit of big circles, you have one circle with half the area, which easily fits. In the middle, you have a bunch of circles of medium size, which I guess is that hard case to do with rigor, but the small and big circles both fit, so I dont see why the intermediate case wouldnt.

### Re: Fitting circles into a circle

Either this is one of those things where it seems obvious but is almost impossible to prove, there's some quirk of geometry I'm not familiar with that makes the problem with an open disc more complicated than just a circle, or it's just a trivial result of circular packing density being higher than 50%. As these are the XKCD forums, it is almost certainly one of the first two.

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- NathanielJ
**Posts:**882**Joined:**Sun Jan 13, 2008 9:04 pm UTC

### Re: Fitting circles into a circle

BlackSails wrote:Lets look at the limit of big and small subcircles. In the limit of small circles, you just have points with total area equal to half the circle, and they easily fit. In the limit of big circles, you have one circle with half the area, which easily fits. In the middle, you have a bunch of circles of medium size, which I guess is that hard case to do with rigor, but the small and big circles both fit, so I dont see why the intermediate case wouldnt.

But you could use that exact same logic to say that in the two "extreme" cases, as long as the combined area of the D_i is no greater than the large circle, they fit. This clearly is not the case for the in-between cases though.

Edit: If it matters, I believe that the conjecture is true for density 50%. However, it should be noted that there are simple counterexamples for any density greater than 50% (and thus circle packing likely doesn't come into play).

### Re: Fitting circles into a circle

The first example to think about is two circles of half radius of the original. Their total area is exactly half of the original circle, and they only just fit, so 50% is an upper bound on what could be possible.

- BlackSails
**Posts:**5315**Joined:**Thu Dec 20, 2007 5:48 am UTC

### Re: Fitting circles into a circle

Yeah, I was thinking only about density <= 50%.

- eta oin shrdlu
**Posts:**451**Joined:**Sat Jan 19, 2008 4:25 am UTC

### Re: Fitting circles into a circle

As a simple special case, if all the [imath]D_i[/imath] have the same radius [imath]\frac1{\sqrt{2n}}[/imath], then you can always fit the n disks in concentric annuli (starting from the outside of the large disk, and working inward as necessary--for 12 or fewer you just need one ring, for 13-48 you need two rings, and so on). For large n this gives a packing density of [imath]\frac\pi4>\frac12[/imath], not as good as the hexagonal close-packing arrangement, but it's easier to handle the circular boundary this way; use inequalities to show it works for all n>N, then just check explicitly that it works up to N.

I don't have any good ideas about handling the general case, though.

I don't have any good ideas about handling the general case, though.

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