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Gruber P. Convex and Discrete Geometry

Gruber P. Convex and Discrete Geometry

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a<br />

f<br />

b<br />

o<br />

e<br />

2D<br />

c<br />

Fig. 30.3. Disc <strong>and</strong> triangular disc<br />

In the second step of the proof, the following will be shown:<br />

d<br />

30 Packing of <strong>Convex</strong> Bodies 453<br />

(9) Let D be an o-symmetric strictly convex disc. Then there is a triangular<br />

disc C such that D = 1 2 (C − C) (Fig. 30.3).<br />

To prove this, we first show that there is an affine regular convex hexagon H inscribed<br />

in 2D. Fora ∈ bd 2D let H be the unique convex hexagon H inscribed in 2D with<br />

opposite vertices a <strong>and</strong> −a such that its edges parallel to the line segment [a, −a]<br />

both have length �a�. Since 2D is strictly convex <strong>and</strong> symmetric in o, these edges<br />

are also symmetric in o. Thus H is affine regular.<br />

Choose an affine regular hexagon with consecutive vertices a, b, c, d, e, f <strong>and</strong><br />

centre at o which is inscribed in 2D. Translate the triangles with vertices a, b, o; o, c,<br />

d; f, o, e along with the adjacent lunae cut off from 2D by the line segments<br />

[a, b], [c, d] <strong>and</strong> [ f, e], such that the triangles coincide. Let C be the union of the<br />

translated triangles <strong>and</strong> lunae. Considering an o-symmetric convex hexagon circumscribed<br />

to 2D the edges of which touch 2D at the points a,..., f, it follows that C<br />

is triangular. Clearly, 2D = C − C, concluding the proof of (9).<br />

In the third step of the proof, we show the following proposition:<br />

(10) Let D be an o-symmetric strictly convex disc. Then the upper density of a<br />

packing of translates of D never exceeds the density of the densest lattice<br />

packing of D.<br />

Let {D + t : t ∈ T } be a packing. By Proposition 30.2, its upper density is A(D)δ,<br />

where δ is the upper density of T . Choose a triangular disc C such that D = 1 2 (C −<br />

C). This is possible by (9). By Proposition 30.4, {C + t : t ∈ T } is also a packing.<br />

Its upper density is A(C)δ. By (8), there is a packing lattice L of C <strong>and</strong> A(C)δ ≤<br />

A(C)/d(L) by Corollary 30.1. Thus δ ≤ 1/d(L). Proposition 30.4 then shows that<br />

L is also a packing lattice of D. Thus {D + l : l ∈ L} is a lattice packing of D<br />

of density A(D)/d(L) ≥ A(D)δ. Now note that A(D)δ is the upper density of the<br />

packing {D + t : t ∈ T }.<br />

C

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