Integral Apollonian circle packings

I have got rather behind with various things. In particular, I had several interesting topics which I wanted to mention here, and it hasn’t got done. So this is not hot news; but still very interesting, I think.

It concerns the notes of an MAA lecture by Peter Sarnak in 2009. One of our Masters students discovered this, and drew his supervisor’s attention to it, from where I learned about it. This post is just a taster for the article, which can be found here, and is highly recommended. It starts from elementary Euclidean geometry and proceeds to deep matters of arithmetic groups, dynamical systems, and the other things that Sarnak juggles so well. I urge you to take a look.

Begin with three two circles in the plane, any two touching externally. There are always exactly two circles touching all three: a small one in the gap between the circles, and a large one enclosing all three. This is a theorem of Apollonius, which Sarnak proves by inversion. He also uses inversion to prove the theorem of Descartes, describing the sizes of the circles. If a1,…,a4 are the curvatures of four mutually touching circles, then

F(a1,a2,a3,a4) = 0,

where F is the quadratic form


Here the curvature of a circle is simply the reciprocal of its radius.

Note that, given three of the curvatures, we get a quadratic equation for the fourth, with two solutions, as Apollonius requires. The discriminant of the quadratic is a1a2+a2a3+a3a1. If a1, a2, a3 are integers and the discriminant is a perfect square, then the two solutions are also integers.

Now we can iterate the construction: given any three circles, there is a circle touching the three and situated in the gap between them. If C is the circle containing the original three circles in its interior, then we end up with a packing of C with circles, a so-called Apollonian circle packing. Moreover, if the discriminant is a square, then all the curvatures we obtain are integers, and the packing is integral.

By coincidence, the curvatures of the United States quarter, nickel and dime are 21, 24 and 28 (taking the unit of length to be 252mm), and the discriminant is 422. So we obtain a circle packing by circles with integral curvature. Sarnak gives pictures of the result after several stages.

He then goes deeper by introducing the Apollonian group. The two solutions a4 and a4 of the quadratic have sum equal to 2(a1+a2+a3); so the 4-tuple (a1,a2,a3,a4) is obtained from (a1,a2,a3,a4) by multiplying by the matrix

1 0 0  2
0 1 0  2
0 0 1  2
0 0 0 −1

This and the three similar matrices generate the Apollonian group.

Thus, the curvatures of the circles in the packing are all the components of the vectors obtained by applying elements of this group to this initial vector.

Now Sarnak and others have used properties of this group, together with Diophantine and Laplacian properties of such arithmetic groups, to get information about the distribution of the curvatures. Among the results are the following:

  • The number of circles of curvature ≤ x grows exponentially, with constant 1.30568….
  • The logarithms of the curvatures of the circles obtained at the nth stage are approximately normally distributed.
  • The set of curvatures has positive density in the natural numbers. In fact, they satisfy certain congruence conditions, and it is conjectured that almost all positive integers satisfying these congruences actually occur.
  • There is a “twin primes” theorem: if the initial curvatures have no common factor > 1, then it happenss infinitely often that tangent circles both have prime curvature.
  • The paper is a fine mixture of exposition, theory and experiment. Go read it!

About Peter Cameron

I count all the things that need to be counted.
This entry was posted in exposition, history. Bookmark the permalink.

4 Responses to Integral Apollonian circle packings

  1. Emil says:

    I made some pictures of integral Apollonian packings for a Theorem of the Day:

    Click to access TotDApollonian.pdf

  2. Yiftach says:

    You have a typo in the second formula should be $2a_4^2$ rather than $2a_1^2$.

  3. Pingback: Weekly Picks « — the Blog

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