Gruber P. Convex and Discrete Geometry

506 Geometry of Numbers
v1
v∞
G
v3
v2
Dv 1
D∞
Dv 3
Fig. 34.5. Graph and corresponding primal-dual circle packing
(7) Every 4-cycle in G ∧ is country.
If xvyw is such a cycle, assume that x, y correspond to countries of G. These countries
then have the vertices v,w in common. Since G is 3-connected, this is possible
only if v,w are connected in G by an edge and that this edge is the common
edge of the countries. This means that xvyw is a country cycle in G ∧ , concluding
the proof of (7). Next note that by the Corollary 15.1 of the Euler polytope formula
(8) 2#S − #E � G ∧ (S) � ≥ 4
with equality if and only if � G ∧ (S) is 2-connected and � all countries of G ∧ (S) are
quadrangles. G ∧ is connected. Since S�V(G ∧ ), there is an edge of G ∧ which is incident
with a vertex of S but is not an edge of G ∧ (S). This edge meets the interior
of one of the countries of G ∧ (S). Thus, if there is equality in (8), one of the country
4-cycles in G ∧ (S) is not a country 4-cycle of G ∧ , in contradiction to (7). Hence, there
is inequality in (8), concluding the proof of proposition (6).
In the seventh, and last, step of the proof we have to show that there is a list
ρ = � ρv : v ∈ V(G ∧ ) � of radii satisfying (1)(i,ii). More precisely, the following has
to be shown:
(9) Let v1v2 ···vk be the outer cycle of G and let 0 < α1,...,αk < π
be such that α1 + ··· + αk = (k − 2)π. Then there is a list ρ =
� ρv : v ∈ V(G ∧ ) � of positive numbers such that the following statements
hold:
(i)
�
vw∈E(G ∧ )
(ii) 2 �
v i w∈E(G ∧ )
w�=∞
arctan ρw
ρv
arctan ρw
ρvi
Given a list ρ = � ρv : v ∈ V(G ′ ) � , define
Dv 2
= π for v ∈ V(G ∧ ) \{v1,...,vk, ∞}.
= αi for i = 1,...,k.