proofs

Hilbert’s third problem: decomposing polyhedra 55
˜P = P ∪ P ′ 1 ∪ · · · ∪ P ′ m
˜Q = Q ∪ Q ′ 1 ∪ · · · ∪ Q ′ m
P ′ 1
Q ′ 1
P
Q
P ′ 2 Q ′ 2
˜P = P 1 ′′ ∪ · · · ∪ P n
′′
˜Q = Q ′′
1 ∪ · · · ∪ Q′′ n
P ′′
1
P ′′
2
Q ′′
2 Q ′′
4
P 3 ′′ P 4
′′
Q ′′
1 Q ′′
3
For a parallelogram P and a non-convex
hexagon Q that are equicomplementary,
this figure illustrates the four decompositions
we refer to.
The Pearl Lemma. If P and Q are equidecomposable, then one can place
a positive numbers of pearls (that is, assign positive integers) to all the
segments of the decompositions P = P 1 ∪ · · · ∪P n and Q = Q 1 ∪ · · · ∪Q n
in such a way that each edge of a piece P k receives the same number of
pearls as the corresponding edge of Q k .
Proof. Assign a variable x i to each segment in the decomposition of P
and a variable y j to each segment in the decomposition of Q. Now we have
to find positive integer values for the variables x i and y j in such a way
that the x i -variables corresponding to the segments of any edge of some
P k yield the same sum as the y j -variables assigned to the segments of the
corresponding edge of Q k . This yields conditions that require that “some
x i -variables have the same sum as some y j -values”, namely
∑
x i − ∑
y j = 0
i:s i⊆e j:s ′ j ⊆e′
where the edge e ⊂ P k decomposes into the segments s i , while the corresponding
edge e ′ ⊂ Q k decomposes into the segments s ′ j . This is a linear
equation with integer coefficients.
We note, however, that positive real values satisfying all these requirements
exist, namely the (real) lengths of the segments! Thus we are done, in view
of the following lemma.
□
The polygons P and Q considered in the figure above are, indeed, equidecomposable.
The figure to the right illustrates this, and shows a possible
placement of pearls.
P 1
P 2
Q 1
P 3
P 4
Q 3
Q 4
Q 2