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So I want to find the green area which denotes the "waste" when using circle tiling. Circle radius may be assumed to be 1.

enter image description here

Sure, there are multiple ways of doing it, but I want to try solving this using limits. We will select two axis at an angle of 60 degrees and tile the the domain $M$ steps in one direction and $N$ steps in another direction. It should look something like this

enter image description here

The total area taken by circles is $A^{tile}_{circle} = (M+1)(N+1)A_c$, where $A_c = \pi r^2$.

I can approximate the total area that would have been taken by no-waste packing by a parallelogram, namely $A^{tile}_{par} = MNA_{par}$, where $A_{par} = (2r)^2 \sin 60 = r^2 2\sqrt 2$.

Now, for small number of circles, the area taken by circles looks larger than the area taken by the parallelogram, but that is due to edge effects, which grow linearly with $M$ and $N$. So, if both are very large, I would expect that the edge effects would become negligible, and the ratio of the two areas would allow me to find the desired area of the green segment.

So I have attempted to compute

$$\lim_{M,N\rightarrow \infty} = \frac{A^{tile}_{circle}}{A^{tile}_{par}} = \frac{\pi}{2\sqrt 2}$$

To my surprise, this ratio is greater than 1. According to my logic, it should be less than 1, accounting for the fact that circles have gaps in between them. What went wrong?

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    $\begingroup$ $\sin 60^\circ = \dfrac {\sqrt 3}2$. The rest are fine. $\endgroup$ – player3236 Nov 3 '20 at 18:33
  • $\begingroup$ @player3236 Damn I'm getting old :D $\endgroup$ – Aleksejs Fomins Nov 3 '20 at 18:34
  • $\begingroup$ Using limits does not help, it suffices to consider a single triangular tile. $\endgroup$ – Yves Daoust Nov 3 '20 at 19:06
  • $\begingroup$ @YvesDaoust It does help. You can use a single tile if you wish, in this case it works. But the limits approach should work as well. $\endgroup$ – Aleksejs Fomins Nov 3 '20 at 19:14
  • $\begingroup$ @AleksejsFomins: realize that you are using a limit like $\lim_{M\to\infty}\frac{M+c}{M}$ when you know upfront that it is $1$, and this simplifies no geometric computation. $\endgroup$ – Yves Daoust Nov 3 '20 at 19:16

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