Number Theory and the Circle Packings of Apollonius

24 Pages·2011·4.39 MB·English

Number Theory and the Circle Packings of
Apollonius
Peter Sarnak
Mahler Lectures 2011
PeterSarnak MahlerLectures2011 NumberTheoryandtheCirclePackingsofApollonius
Apollonius of Perga
lived from about 262 BC to about 190 BC
Apollonius was known as ‘The Great Geometer’. His famous book
Conics introduced the terms parabola, ellipse, and hyperbola.
PeterSarnak MahlerLectures2011 NumberTheoryandtheCirclePackingsofApollonius
PeterSarnak MahlerLectures2011 NumberTheoryandtheCirclePackingsofApollonius
Scale the picture by a factor of 252 and let
a(c) = curvature of the circle c = 1/radius(c).
The curvatures are displayed. Note the outer one by convention
has a negative sign.
By a theorem of Apollonius, place unique circles in the lines.
PeterSarnak MahlerLectures2011 NumberTheoryandtheCirclePackingsofApollonius
The Diophantine miracle is the curvatures are integers!
PeterSarnak MahlerLectures2011 NumberTheoryandtheCirclePackingsofApollonius
Repeat ad infnitum to get an integral Apollonian packing:
There are infinitely many such P’s.
Basic questions (Diophantine)
Which integers appear as curvatures?
Are there infinitely many prime curvatures, twin primes i.e. pairs of
tangent circles with prime curvature?
PeterSarnak MahlerLectures2011 NumberTheoryandtheCirclePackingsofApollonius
The integral structure — F. Soddy (1936)
Diophantine setup and questions — R. Graham–J.
Lagarias–C. Mellows–L. Wilks–C. Yan (2000)
Many of the problems are now solved
Recent advances in modular forms, ergodic theory, hyperbolic
geometry, and additive combinatorics.
Apollonius’ Theorem
Given three mutually tangent circles c ,c ,c , there are exactly
1 2 3
two circles c and c(cid:48) tangent to all three.
Inversion in a circle takes circles to circles and preserves tangencies
and angles.
PeterSarnak MahlerLectures2011 NumberTheoryandtheCirclePackingsofApollonius
c ,c ,c given invert in ξ (ξ → ∞) yields
1 2 3
Now the required unique circles c(cid:48) and c are clear
(cid:101) (cid:101)
−→ invert back.
PeterSarnak MahlerLectures2011 NumberTheoryandtheCirclePackingsofApollonius
Descartes’ Theorem
Given four mutually tangent circles whose curvatures are
a ,a ,a ,a (with the sign convention), then
1 2 3 4
F(a ,a ,a ,a ) = 0,
1 2 3 4
where F is the quadratic form
F(a) = 2(a2+a2+a2+a2)−(a +a +a +a )2.
1 2 3 4 1 2 3 4
I don’t know of the proof “from the book”. (If time permits, proof
at end.)
Diophantine Property:
Given c ,c ,c ,c mutually tangent circles, a ,a ,a ,a
1 2 3 4 1 2 3 4
curvatures. If c and c(cid:48) are tangent to c ,c ,c , then
1 2 3
F(a ,a ,a ,a ) = 0
1 2 3 4
F(a ,a ,a ,a(cid:48)) = 0
1 2 3 4
So a and a(cid:48) are roots of the same quadratic equation =⇒
4 4
PeterSarnak MahlerLectures2011 NumberTheoryandtheCirclePackingsofApollonius
a +a(cid:48) = 2a +2a +2a (1)
4 4 1 2 3
√
a ,a(cid:48) = a +a +a ±2 ∆
4 4 1 2 3
∆ = a a +a a +a a
1 2 1 3 2 3
(our example (a ,a ,a ) = (21,24,28), ∆ = 1764 = 422)
1 2 3
If c ,c ,c ,c have integral curvatures, then c(cid:48) also does from (1)!
1 2 3 4 4
In this way, every curvature built is integral.
Apollonian Group:
(1) above =⇒ that in forming a new curvature when inserting a
new circle
a(cid:48) = −a +2a +2a +2a
4 4 1 2 3
PeterSarnak MahlerLectures2011 NumberTheoryandtheCirclePackingsofApollonius