Apollonian Circle Packings

Abstract

Circle packings are a particularly elegant and simple way to construct quite complicated and elaborate sets in the plane. One systematically constructs a countable family of tangent circles whose radii tend to zero. Although there are many problems in understanding all of the individual values of their radii, there is a particularly simple asymptotic formula for the radii of the circles, originally due to Kontorovich and Oh. In this partly expository note we will discuss the history of this problem, explain the asymptotic result and present an alternative approach.

Mathematics Subject Classification (2000). Primary 52C26, 37C30; Secondary 11K55, 37F35, 37D35

I am very grateful to Richard Sharp for many discussions on this approach and the details. I would also like to thank Christoph Bandt and the referee for many useful comments on the presentation.

Appendix: The Case of Reciprocal Integer Circles

The following is an interesting corollary to Descartes’ Theorem.

Corollary 4.8

If $\frac{1} {a_{0}}, \frac{1} {a_{1}}, \frac{1} {a_{2}}, \frac{1} {a_{3}} \in \mathbb{Z}$ then $\frac{1} {a_{4}} \in \mathbb{Z}$ .

Proof

In particular, this is a quadratic polynomial in $\frac{1} {a_{4}} > 0$, so given the radii of the initial circles $a_{1},a_{2},a_{3}$ we have two possible solutions

$$\displaystyle{ \frac{1} {a_{1}} + \frac{1} {a_{2}} + \frac{1} {a_{3}} \pm 2\sqrt{ \frac{1} {a_{1}a_{2}} + \frac{1} {a_{2}a_{3}} + \frac{1} {a_{3}a_{4}}}.}$$

and we denote these $\frac{1} {a_{4}} > 0$ (and $\frac{1} {a_{0}} < 0$). We use the convention that the smaller inner circle has radius a ₄ > 0 and the larger outer circle has a negative “radius” a ₄ (meaning its actually radius is | a ₄ |  > 0 and the negative sign just tells us it is the outer circle). Adding these two solutions gives:

$$\displaystyle{\begin{array}{ll} \frac{1} {a_{0}} + \frac{1} {a_{4}} & = 2\left ( \frac{1} {a_{1}} + \frac{1} {a_{2}} + \frac{1} {a_{3}} \right ) \end{array} }$$

from which we easily deduce the result. □ 

Proceeding inductively, for any subsequent configuration of four circles with radii $a_{n},a_{n+1},a_{n+2},a_{n+3}$, for n ≥ 0, we can similarly write

$$\displaystyle{ \frac{1} {a_{n+4}} = 2\left ( \frac{1} {a_{n+1}} + \frac{1} {a_{n+2}} + \frac{1} {a_{n+3}}\right ) - \frac{1} {a_{n}}.}$$

Proceeding inductively, then one gets infinitely many circles. Moreover, if the reciprocals of the initial four circles are integers then we easily see that this is true for all subsequence circles.

Corollary 4.9

If the four initial Apollonian circles have that their radii $a_{0},a_{1},a_{2},a_{3}$ are reciprocals of integers then all of the circles in $\mathcal{C}$ have that the reciprocals of their radii a _n , n ≥ 4, are integers.

Example 4

Let us consider the example starting with $a_{0} = -\frac{1} {3}$, $a_{1} = \frac{1} {5}$, $a_{2} = \frac{1} {8}$, and $a_{3} = \frac{1} {8}$. In Fig. 11 below we illustrate the iterative process of inscribing circles into each curved triangle formed by three previously constructed tangent circle and write $\frac{1} {a_{n}}$ inside the corresponding circle of radius a _n .

Fig. 11

We iteratively inscribe additional circles starting with circles of radii $a_{0} = -\frac{1} {3}$, $a_{1} = \frac{1} {5}$, $a_{2} = \frac{1} {8}$, $a_{3} = \frac{1} {8}$

Example 5

Let us also consider the example with $a_{0} = -\frac{1} {2}$, $a_{1} = \frac{1} {3}$, $a_{2} = \frac{1} {6}$, $a_{3} = \frac{1} {7}$. In Fig. 12 below we illustrate the iterative process of inscribing circles into each curved triangle formed by three previously constructed tangent circle and write $\frac{1} {a_{n}}$ inside the corresponding circle of radius a _n .

Fig. 12

We iteratively inscribe additional circles starting with circles with radii $a_{0} = -\frac{1} {2}$, $a_{1} = \frac{1} {3}$, $a_{2} = \frac{1} {6}$, $a_{3} = \frac{1} {7}$

Remark 7

An easy consequence of the fact δ > 1 is that then $\frac{1} {a_{n}} \in \mathbb{N}$ some value must necessarily have high multiplicity (since we need to fit approximately C ε ^−δ inverse diameters into the first [ε ⁻¹] natural numbers and the “pigeonhole principle” applies). In subsequent work, Oh-Shah showed that similar results are true for other sorts of circle packing [12]. Oh-Shah also gave an alternative approach to the original proof of Kontorovich-Oh using ideas of Roblin.

Remark 8

Another question we might ask is: It we remove the repetitions in the sequence (a _n ) then how many distinct diameters are greater than ε? The following result was proved by Bourgain and Fuchs [2]: There exists C > 0 such that

$$\displaystyle{\#\{\mbox{ distinct diameters }a_{n}\mbox{: }a_{n} \geq \epsilon \}\geq \frac{C} {\epsilon } }$$

for all sufficiently large ε.s Previously, Sarnak [17] had proved the slightly weaker result that there exists C > 0 such that

$$\displaystyle{\#\{\mbox{ distinct diameters }a_{n}\mbox{: }a_{n} \geq \epsilon \}\geq \frac{C} {\epsilon \sqrt{\log \epsilon }}.}$$