What is the minimum area of a rectangle containing all circles of radius $1/n$?

Question

What is the minimum area of a rectangle containing all (non-overlapping) circles of radius $1/n$, $n\in\mathbb{N}$ ?

The total area of the circles is finite: $\sum\limits_{n=1}^\infty \frac{\pi}{n^2}=\frac{\pi^3}{6}\approx5.168$.

Below I show the rectangle of smallest area that can contain circles with radii $1$ and $\frac{1}{2}$. This rectangle has height $2$ and width $\frac{3}{2}+\sqrt2$, so the area is $3+2\sqrt2\approx5.828$. I placed circles with radii $1, \frac{1}{2}, \frac{1}{3}, ..., \frac{1}{20}$ in no particular pattern.

Here is the desmos graph.

My intuition tells me that the other circles can also fit. The others would occupy only $\dfrac{\pi^3/6-\sum_{n=1}^{20}\pi/n^2}{3+2\sqrt2-\sum_{n=1}^{20}\pi/n^{2}}\approx 0.188$ of the remaining space. But how could we prove that all the circles can fit in this rectangle?

My question was inspired by a question about fitting harmonic circles on a unit disk.

Answer 1

One can tackle this question by forming an offset grid of squares of radii $1/k$ in a roughly triangular fashion that fits inside the top left sector of the rectangle, whereby the squares can be filled with circles and we are done.

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To formalize this idea, let us begin by giving ourselves enough leeway to only need the top left sector. The calculation in the post notes we don't need much of the remaining area if we fill in the first $n = 20$ balls, but to give ourselves way more room (and since we need it later anyway), let's instead fill in the first $65$ balls by hand so we can start with $n=66$:

(Desmos graph here, modified from the OP's graph)

Now that we have this, our goal will be to construct a triangle in the top left sector out of boxes, with radii as such:

There is, of course, the question of how to make the squares tangent when the radii are decreasing, so we stipulate that at each stage we first put a box as far left and then as far up as possible, and then put the remaining boxes flush against the top right of the associated box from the prior step. Graphically, where a box is labeled $m$ if it's added in the $m$th step, this looks like:

If we draw a line with angle $45 \deg$ at the bottom right of the bottom-most box, we get a bounding box for a triangle:

The short side length of this isosceles triangle is given by first summing the diameters of the column of leftmost boxes and then adding the diameter of the bottom-left box. Specifically, if we realize the triangular numbers $T_n = \frac{n(n+1)}{2}$ are involved, we can bound the side length $S_m$ at the $m$th stage of the construction as: $$S_m = \frac{2}{n + \frac{m(m-1)}{2}} + \sum_{k=0}^{m-1} \frac{2}{n + \frac{k(k+1)}{2}} \le \frac{2}{n} + \sum_{k=0}^{\infty} \frac{2}{n + \frac{k(k+1)}{2}}$$

Indeed, here we use the fact $n = 66$ to determine:

$$S_m \le \frac{2}{66} + \sum_{k=0}^{\infty} \frac{2}{66 + \frac{k(k+1)}{2}} = \frac{2}{66} + \frac{4\pi \tanh \left(\frac{\sqrt{527} \pi}{2} \right)}{\sqrt{527}} = 0.577703\ldots$$

Now, this is important, because it turns out to be a good enough bound. Indeed, if one draws the tangent line to the unit circle in the OP's diagram at the point $(\cos(3\pi/4), \sin(3\pi/4))$ and sees where it intersects the vertical boundary on the left, a quick calculation (using the fact the line has equation $y = x + \sqrt{2}$) shows the largest possible isosceles triangle which can fit has short side length $2 - \sqrt{2} = 0.585786\ldots$, so our triangle constructed above fits!

As a sanity check, we can now put the remaining circles of radius $1/k$ into the boxes of radius $1/k$ and determine the resulting area. Indeed, the area of the boxes can be calculated as $\sum_{k = 66}^\infty \left(\frac{2}{k}\right)^2 \approx 0.061$ while the area of the top-left sector is given by $1 - \frac{\pi}{4} \approx 0.21$, so we end up with plenty of area to spare.

Answer 2

For suitable $a$ and $N$, we can place disjoint disks of radii $\frac 1n$, $n\ge N$ into an $a\times a$ square, column by column, where column $k$ ($k=0,1,2,\ldots$), consists only of disks of radius $\le\frac 1{n_k}$.

We clearly have $n_0=N$, and as $\frac 2{n_k}$ fits at least $n_k\cdot\frac a2-1$ times into $a$, we have $$n_{k+1}\ge \left(1+\frac a2\right) n_k-1.$$ This makes $$ n_{k+1}-\frac2a\ge \left(1+\frac a2\right)\left(n_k-\frac 2a\right)$$ so that $$ n_k\ge \left(1+\frac a2\right)^k\left(N-\frac2a\right)+\frac2a>\left(1+\frac a2\right)^k\left(N-\frac2a\right).$$ Then the combined width of all columns is $$ \sum_{k=0}^\infty\frac2{n_k}\le \frac2{N-\frac2a}\sum_{k=0}^\infty\left(1+\frac a2\right)^{-k}=\frac{2}{N-\frac2a}\cdot\frac1{1-\frac1{1+\frac a2}}=\frac{2+\frac4a}{N-\frac2a}.$$ To fit into our $a\times a$ square, it is sufficient that $\frac{2+\frac4a}{N-\frac2a}\le a$, i.e., that $$\tag1 N\ge \frac{4(a+1)}{a^2}=\left(1+\frac2a\right)^2-1.$$ We can also solve this for $a$ and pick the minimal value for $a$ compatible with $(1)$: $$\tag2a=\frac{2}{\sqrt{N+1}-1}.$$

Now cut this square along a diagonal into two halves and assign any disk that overlaps the cut line to a part that contains at least half of the disk. Thereby, if we rearrange the two halves into an isosceles right triangle of height $a\sqrt 2$, all disk are contained in a slightly bigger isosceles right triangle of height $$ h=a\sqrt 2+\frac{\sqrt2}N=\frac{2}{\sqrt{N+1}-1}+\frac{\sqrt2}N.$$ By picking $N=39$, we arrive at $$ h = \frac2{\sqrt{40}-1}+\frac{\sqrt2}{39}=0.41188\ldots<\sqrt2-1.$$

We conclude that all disks of radius $\frac1n$, $n\ge39$, fit into the gap on the top left of the your design. (With only slightly more effort, e.g., using estimates for harmonic numbers to better fill the columns, one might lower the $39$ down to $33$, and that is still a very crude bound). The image below shows the original design plus a yellow triangle to contain all disks $\le\frac1{39}$ disks, plus possible of nine red disks of radius $\frac1{20}$ (so big enough to place $\frac1{21},\ldots,\frac1{29}$) and nine green disks of radius $\frac1{30}$ (so big enough to place $\frac1{30},\ldots,\frac1{38}$). And there is still quite some leeway.