Gruber P. Convex and Discrete Geometry

32 Tiling with Convex Polytopes 473
projection “ ′ ”ofE d onto the 2-dimensional subspace orthogonal to G. Then P ′
is a centrally symmetric convex 2k-gon. The edges of P ′ are the projections of the
facets of the belt corresponding to G and the vertices of P ′ are the projections of
the sub-facets parallel to G, more precisely, which are translates of G or −G. The
dihedral angle of P at two adjacent facets of the belt corresponding to G equals
the internal angle of the corresponding edges of the centrally symmetric convex
2k-gon P ′ . Since k ≥ 4, the following statements hold:
(1) The dihedral angle of P at a sub-facet parallel to G is less than π.
(2) The sum of the dihedral angles of P at two non-opposite sub-facets parallel
to G is greater than π and less than 2π.
(3) The sum of the dihedral angles of P at three pairwise non-opposite subfacets
parallel to G is greater than 2π.
Choose a point g ∈ relint G which is not contained in a face of any tile P + t,
t ∈ T of dimension less than d − 2orofdimensiond − 2 but not parallel to G.
If g ∈ relint F where F is a facet of a tile P +s, s ∈ T \{o}, Proposition (1) applied
to G, shows that there is a tile P +t, t ∈ T \{o, s}, which contains g and is contained
in the wedge determined by the facet F of P + s and a facet E of P where G is a
sub-facet of E. Considering P ′ , P ′ + s ′ , P ′ + t ′ , we see that P has dihedral angles
at two non-opposite sub-facets parallel to G with sum at most π, in contradiction
to (2). By our choice of g and the case just settled, we see, if g is contained in a
tile P + s, s ∈ T \{o}, then g is contained in the relative interior of a sub-facet of
P + s which is parallel to G. The sum of the dihedral angles at g of such tiles is 2π.
By (1)–(3) we see that g belongs to at least 3 sub-facets parallel to G. We distinguish
the following two cases: first, g belongs to 2 opposite sub-facets. Then g belongs to
2 pairs of non-opposite sub-facets which is impossible by (2). Second, g belongs to
3 pairwise non-opposite sub-facets which is impossible by (3). This shows that our
assumption that k ≥ 4 is wrong and thus concludes the proof of (iv). ⊓⊔
P ′
E ′
P ′ + t ′
F ′
G ′ P ′ + s ′
Fig. 32.5. On the proof of the Venkov–McMullen theorem