Gruber P. Convex and Discrete Geometry

33 Optimum Quantization 495
In (4) equality holds precisely in case where Hn contains B 2 . Since inequality is
excluded by Propositions (2)–(4), the hexagon Hn contains the circular disc B 2 . Then
A(Hn) ≥ 2√3 π A(B2 ) and thus nπ π
≤ √ ,
4τ 2
12
concluding the proof of statement (1).
Second, the following stronger statement will be shown:
(5) Let {B 2 + s : s ∈ S} be a packing. Then
1
4τ 2
�
A � (B 2 + s) ∩ τ K � ≤ π
√ + O
12 � 1 �
.
τ
s∈S
To see this, rewrite the left side of this inequality in the form
1
4τ 2
�
A(B 2 + s) + 1
4τ 2
�
A � (B 2 + s) ∩ τ K ).
s∈S
B 2 +s⊆τ K
s∈S
B 2 +s�⊆τ K
(B 2 +s)∩τ K �=∅
Apply (1) to the first expression and note that in the second expression only discs are
considered which intersect bd τ K . Since we consider a packing, these discs do not
overlap. The total area of these discs is thus O(τ). This gives the second term on the
right side of the inequality in (5).
Proposition (5) implies that δT (B 2 ) ≤ π/ √ 12. Since there is a lattice packing of
B 2 of density π/ √ 12, the equality δL(B 2 ) = δL(B 2 ) = π/ √ 12 follows. ⊓⊔
Similar arguments lead to the following counterpart of the result of Thue and
Fejes Tóth due to Kershner [578]. It says that the minimum density of a covering of
E 2 with circular discs equals the minimum lattice covering density.
Corollary 33.2. ϑT (B 2 ) = ϑL(B 2 ) = 2π
√ 27 = 1.209 199 ...
Proof. First, we show the following:
(6) Let τ>0 and consider a covering of τ K with n translates of B 2 . Then,
nπ 2π
≥ √ .
4τ 2
27
We may assume that all centres of these translates are in τ K .LetSnbe the set of
centres. Fejes Tóth’s inequality then yields
�
�
min{
f (�x − s�)} dx ≥ n f (�x�) dx
s∈Sn
τ K
with f and Hn as before. Since the translates of B 2 cover τ K , the integrand of the
first integral is 0. Being non-negative, the integrand of the second integral thus is 0
too. Taking into account the definition of f , this implies that Hn ⊆ B 2 which, in
turn, yields (6).
Second, the following statement holds:
Hn