Gruber P. Convex and Discrete Geometry

P
A(P) = 7 1 2 , L(P) = 15, Lb (P) = 13
Fig. 19.1. Pick’s theorem
19 Lattice Polytopes 317
By the latter we mean the following: If P, Q ∈ J Z 2 are such that P ∪ Q ∈ J Z 2 and
P ∩ Q is a polygonal Jordan arc, then
M(P ∪ Q) = M(P) + M(Q).
To see (2), assume that the common arc P ∩ Q of P and Q contains m points of Z 2 .
Then
This, in turn, implies (2):
L(P ∪ Q) = L(P) + L(Q) − m,
L b (P ∪ Q) = L b (P) + L b (Q) − 2m + 2.
M(P ∪ Q) = L(P ∪ Q) − 1
2 Lb (P ∪ Q) − 1
= L(P) − 1
2 Lb (P) + L(Q) − 1
2 Lb (Q) − m + m − 1 − 1
= M(P) + M(Q).
To prove the theorem, it is sufficient to show the following:
(3) Let n = 3, 4,... Then (1) holds for each P ∈ J Z 2 with L(P) = n.
The proof is by induction on n. Ifn = 3, then P is a lattice triangle which contains
no point of Z 2 , except its vertices. We may assume that P = conv{o, b1, b2}. The
triangle −P + b1 + b2 is also a lattice triangle which contains no point of Z 2 , except
its vertices. Hence o is the only point of Z 2 in the parallelogram {α1b1 + α2b2 : 0 ≤
αi < 1} ⊆P ∪ (−P + b1 + b2). Thus {b1, b2} is a basis of Z 2 and therefore,
A(P) = 1
2 | det(b1, b2)| = 1
2
1
1
= 3 − 3 − 1 = L(P) −
2 2 Lb (P) − 1.
Assume now that n > 3 and that (3) holds for 3, 4,...,n − 1. Let q be a vertex
of conv P and let p and r be the points of Z 2 on bd P just before and after q.