For a circle-Apollonian-packing problem, finding the smallest circle such that all natural numbers may be packed.

Question

In section $1$ of the paper "The sequence of radii of the apollonian packing" of David W. Boyd, the author said that the question for the smallest circle into which disks of radius $1/n,n=1,2,...$ can all be packed is posted by H.S.M. Coxeter in 1979, the paper named "Problem P.276", and Boyd found the answer is $3/2=1.5$ and published it in the paper named "Solution to problem P.276" in 1980. I tried to find the paper, but I cannot. So I wonder whether anyone may suggest proof, a reference, or a source for the question. Thanks.

Something here we can know straight forward is that the radius of the smallest circle is $\ge \frac{\pi }{{\sqrt 6 }} \approx 1.28254983016$ (quite closed to the answer) because the Riemann zeta function $\zeta (2) = \frac{{{\pi ^2}}}{6}$.

Integer circle packing in higher dimensions should be very relevant to the values of the zeta function. But I cannot find many papers towards to the relation.

Answer 1

This is a very interesting problem and just for the fun, let's prove it ourselves.

First, it is impossible to pack all those circles inside a circle of radius $< \frac32$. If a circle of radius $R$ is large enough to hold the circle of radius $1$ and $\frac12$, then $2R$ will be larger than the length of the line segment passing through the centers of these two circles. Since the length of that line segment is at least $3 = 2 +1$. This forces $2R \ge 3 \implies R \ge \frac32$.

Next, start from an outer circle $\mathcal{C}_V$ of radius $\frac32$, we can place a circle $\mathcal{C}_U$ of radius $1$ and a circle $\mathcal{C}_0$ of radius $\frac12$ along a diameter of $\mathcal{C}_V$. If one work out the radii of circles in the Pappus chain associated with the pair $\mathcal{C}_U, \mathcal{C}_V$. The radii are given by the formula

$$\frac{3}{6+ n^2}\quad\text{ for }\quad n \in \mathbb{Z}$$ The $n = 0$ corresponds to our circle $\mathcal{C}_0$. Aside from this, one can find

• two circles $\mathcal{C}_{\pm 1}$ with radius $\frac37$,
• two circles $\mathcal{C}_{\pm 2}$ with radius $\frac3{10}$,
• two circles $\mathcal{C}_{\pm 3}$ with radius $\frac15$,
• two circles $\mathcal{C}_{\pm 4}$ with radius $\frac3{22}$.

Using the lemma at end of this answer, we can place all circles of radii $\frac19, \frac1{10}, \frac1{11}, \ldots$ inside $\mathcal{C}_1$. For the remaining $6$ circles, we can place

• the circle with radius $\frac13$ into $\mathcal{C}_{-1}$.
• the circles with radii $\frac14$ and $\frac15$ into $\mathcal{C}_{\pm 2}$.
• the circles with radius $\frac16$ and $\frac17$ into $\mathcal{C}_{\pm 3}$.
• the circle with radius $\frac18$ into $\mathcal{C}_{4}$.

In this manner, we can place all the circles inside a single circle of radius $\frac32$.

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Lemma

It is possible to pack mutually disjoint open disks of radii $\frac19$, $\frac1{10},\ldots$ inside a circle of radius $\frac37$.

Proof of Lemma

Let $A = \frac37$, $\kappa = 3$ and $B = \frac{\kappa+1}{\kappa-1} = 2$.

For $n \ge 9$, let $r_n = A - \frac{B}{n}$, $\rho_n = nr_n = An - B$ and $\delta_n = \sin^{-1}\frac{1}{\rho_n}$.
Define angles $\theta_n$ by $\theta_9 = 0$ and $$\theta_{n} - \theta_{n-1} = \delta_n + \delta_{n-1}\quad\text{ for }n > 9\tag{*1}$$ Let $D_n$ be the open disc with radius $\frac1n$ centered at $(r_n\cos\theta_n,r_n\sin\theta_n)$. Notice for these $n$, $r_n > \frac1n$ and disc $D_n$ subtend a cone $C_n$ of angle $2\delta_n$ at origin $O$. The relation $(*1)$ forces $$C_n \cap C_{n+1} \subset \{O\} \implies D_n \cap D_{n+1} = \emptyset$$ ie. $D_n$ will be disjoint from its neighbours $D_{n-1}$ and $D_{n+1}$.

Let $\bar{C}_n$ be the closure of $C_n$. Consider finite union of these closed cones start at some $p \ge 11$. When enough cones are added to the union, it will cover the plane. Let $q$ be the smallest integer for this to happen: $$\bigcup_{\ell=p}^q \bar{C}_\ell \supset \mathbb{R}^2 \quad\iff\quad \pi \le \sum_{\ell=p}^q \delta_\ell\tag{*2}$$ Notice $\delta_\ell = \sin^{-1}\frac{1}{\rho_\ell} = \sin^{-1}\frac{1}{A\ell - b}$ is a decreasing function in $\ell$. By bounding $\delta_\ell$ with corresponding integral over $[\ell,\ell+1]$, one find:

$$\pi \le \sum_{\ell=p}^q \delta_\ell \le \frac1A\int_{x_0}^{kx_0}\sin^{-1}\frac{1}{x} dx\tag{*3} $$ where $x_0 = A(p-1)-B$ and $kx_0= Aq-B$.

Since $p \ge 11$, $x_0 \ge \frac{16}{7} \implies \frac1x \le \frac12$ for $x \in [x_0,kx_0]$. For $y \in (0,\frac12)$, we have following bound for $\sin^{-1}y$,

$$\sin^{-1}y \le y + 4y^3\left(\frac{\pi}{3}-1\right)$$ Substitute $y$ by $\frac1x$ and plug these into RHS of $(*3)$, we obtain:

$$\pi A \le \log k + \frac{2}{x_0^2}\left(\frac{\pi}{3}-1\right)\left(1 - \frac1{k^2}\right) $$ Treat RHS as a function of $k, p$ and call it $E(k,p)$. For fixed $k$, it is a decreasing function in $p$. If $\pi A > E(k,p )$ for a $p$, then $\pi A > E(k,p')$ for all $p' \ge p$.

In particular, we have $\pi A > E(k_0,11)$ for $k_0 = \frac{15}{4}$. This means for any $p \ge 11$ and corresponding $q$ in $(*2)$, we have

$$\frac{Aq-B}{A(p-1)-B} = k > k_0 \implies q - 3p > \frac{9p-199}{12}$$ For $p \ge 21$, RHS of last expression $> -1$. Since $q - 3p$ is an integer, we find $q \ge \kappa p = 3p$ whenever $p \ge 21$. This leads to

$$r_q - r_p = \frac{B}{p} - \frac{B}{q} \ge \frac{B}{p}\left(1 - \frac{1}{\kappa}\right) = \frac{1}{p}\left(1 + \frac{1}{\kappa}\right) \ge \frac1{p} + \frac1q\\ $$ This implies $D_p \cap D_q = \emptyset$. Since $r_n$ is increasing while $\frac1n$ is decreasing, we have a similar inequality for all $n > q$. As a result, $D_p$ and $D_n$ will be disjoint for all $n \ge q$.

Based on this, we find

It is possible to pack mutually disjoint open discs of radii $\frac1{21}, \frac{1}{22}, \ldots$ inside a circle of radius $\frac37$.

If we modify definition of $r_n$ to

$$r_n = \begin{cases} \frac17, & n = 9\\ A - \frac{B}{n} - \frac23(\frac1n - \frac1{21}), & 9 < n < 21\\ A - \frac{B}{n}& n \ge 21 \end{cases}$$ and redefine $\rho_n, \delta_n, \theta_n, D_n$ using same formula as above. It is not hard to see for $n \ge 21$, the new location of $D_n$ can be obtained from the old location by an rotation independent of $n$. This means the new $D_n$ are also mutually disjoint for $n \ge 21$.

By brute force, one can verify $D_n \cap D_m = D_m \cap D_m'$ for any $n \le 21$ and $21 < m \ne m' \le 9$.

Combine all these, the lemma follows...

Sometimes a picture worth more than a thousand words. The figure below illustrate what the configuration looks like.

• the outer circle in red has radius $A = \frac37$.
• the red disc is $D_9$.
• the yellow discs are $D_{10}$ to $D_{20}$
• the green discs are $D_{21}$ to $D_{200}$.

As one can see, aside from those that are adjacent to each other, the rest are clearly mutually disjoint.