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Gruber P. Convex and Discrete Geometry

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494 <strong>Geometry</strong> of Numbers<br />

33.4 Packing <strong>and</strong> Covering of Circles, Data Transmission <strong>and</strong> Numerical<br />

Integration<br />

Fejes Tóth’s theorem on sums of moments <strong>and</strong> Zador’s theorem on minimum distortion,<br />

the refinements of the latter by the author [443] <strong>and</strong> the results on the structure<br />

of the minimizing configurations are tools for a series of applications, see the references<br />

cited in the introduction of Sect. 33.<br />

In the following, applications to packing <strong>and</strong> covering with circular discs, to<br />

data transmission <strong>and</strong> to numerical integration are presented. For an application of<br />

Zador’s theorem to the approximation of convex bodies by circumscribed convex<br />

polytopes see the approximation theorem 11.4.<br />

Packing <strong>and</strong> Covering of B 2<br />

Using Fejes Tóth’s theorem on sums of moments, we prove a result of Thue [996,<br />

997] <strong>and</strong> Fejes Tóth [327] which says that the maximum density of a packing in E 2<br />

with circular discs equals the maximum lattice packing density, compare Sects. 29.1<br />

<strong>and</strong> 31.4.<br />

Corollary 33.1. δT (B 2 ) = δL(B 2 ) = π<br />

√ 12 = 0.906 899 ...<br />

Proof. We first show a weaker statement:<br />

(1) Let τ>0 <strong>and</strong> consider a packing of n translates of B 2 in the square τ K ,<br />

where K ={x :|xi| ≤1}. Then<br />

τ K<br />

nπ π<br />

≤ √ .<br />

4τ 2<br />

12<br />

Let Sn be the set of centres of this packing. Fejes Tóth’s inequality on sums of<br />

moments then shows that<br />

�<br />

�<br />

�<br />

0for0≤ t ≤ 1,<br />

(2) min{<br />

f (�x−s�)} dx ≥ n f (�x�) dx where f (t) =<br />

s∈Sn<br />

1fort > 1,<br />

Hn<br />

<strong>and</strong> Hn is a regular hexagon of area A(Hn) = A(H)/n = 4τ 2 /n with centre o.<br />

Note that<br />

�<br />

(3) min{<br />

f (�x − s�)} dx = 4τ<br />

s∈Sn<br />

2 − nπ<br />

<strong>and</strong><br />

τ K<br />

�<br />

(4) n<br />

Hn<br />

f (�x�) dx = nA(Hn\B 2 ) ≥ nA(Hn) − nA(B 2 ) = 4τ 2 − nπ.

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