Dlanod is building huts, yet again

This website gives a distressingly complete listing of circle-packing solutions. In particular, it shows how to pack $30$ circles of radius $0.091671057985988438718806599233$ in a square of side $1$ . This means that $30$ huts of radius $3$ metres can be packed inside a square of side $32.725704992501976152306415628$ metres, which means we have $\frac{32.725704992501976152306415628^2}{30} = 35.699058908542292233701667218$ square metres per circle. This makes Dlanod right. Considering the other optimal packings given on that website, $30$ huts is the smallest number for which less than $36$ square metres per hut is possible.

Of course, Dlanod will need the assistance of his colleague Jerry Mandau to get around the building regs for this planned design.

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Here is a theoretical justification, as well.

Start off by packing six columns of five $3$ metre radius huts in a hexagonal array in a $33 \times 33$ metre square. I will show how to adjust this pattern to fit the huts in a $32.8 \times 32.8$ metre square, which gives $35.8613$ square meters per hut (not optimal, but enough to answer the question).

The initial array of huts fills the vertical height of the $33\times33$ square, and there is a strip of width $33 - (6 + 15\sqrt{3}) = 1.01924$ metres free on the right. If I slide the third column of huts $2\sqrt{6^2 - 2.8^2} - 3\sqrt{3}= 0.11045$ metres to the right, then the second column of huts can shift a little to the right and drop down $0.2$ metres (the triangle $ABC$ has been stretched horizontally so that $C$ is now $2.8$ metres above the line $AB$ , instead of its initial $3$ metres). Readjust the fifth and fourth columns similarly, and shift the sixth column to match the second and fourth. The $30$ huts are now placed in a rectangle that is $6+8\sqrt{11} = 32.533$ meters wide and $32.8$ metres high. Thus we can certainly place $30$ huts in a $32.8 \times 32.8$ metre square, and we are done.

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It seems possible that this stretching approach will give us the optimal packing. If we want to fit the huts into an $x \times x$ metre square, we will need to drop $C$ down $3 - (x-30)$ metres, and so we need to shift $B$ a distance of $2\sqrt{36 - (x-30)^2} - 6\sqrt{3}$ metres to the right. If doing this (and performing similar shifts to the other columns) results in the huts fitting exactly inside the $x\times x$ metre square, then we must have $5\sqrt{36 - (x-30)^2} +6 \; = \; x$ which solves to give $x = 32.72570499250195$ .

The plan is to place the huts in a "honeycomb pattern," in $p$ rows with $q$ cottages each, as shown in the attached figure, where $p=6$ and $q=7$ . We can place the huts onto a rectangular plot with length $L=6q+3$ and width $W=6+3\sqrt{3}(p-1)$ . For example, if we choose $p=9$ and $q=7$ , then $L=45$ , and $W<47.6$ , so that we can place these 63 huts on a square plot with side length 47.6 meters. The amount of land we need per cottage will be $\frac{47.6^2}{63}<36$ square meters.

Once again, $\boxed{\text{Dlanod is right}}$ .