IMO 1983 SL 25

Origin: USS

Problem

Prove that every partition of 3-dimensional space into three disjoint subsets has the following property: One of these subsets contains all possible distances; i.e., for every a \inR+, there are points M and N inside that subset such that distance between M and N is exactly a.

Solution

Suppose the contrary, that R3 = P1 \cupP2 \cupP3 is a partition such that a1 \inR+ is not realized by P1, a2 \inR+ is not realized by P2 and a3 \inR+ not realized by P3, where w.l.o.g. a1 \geqa2 \geqa3. If P1 = \emptyset= P2, then P3 = R3, which is impossible. If P1 = \emptyset, and X \inP2, the sphere centered at X with radius a2 is included in P3 and a3 \leqa2 is realized, which is impossible. If P1 ̸= \emptyset, let X1 \inP1. The sphere S centered in X1, of radius a1 is included in P2 \capP3. Since a1 \geqa3, S ̸\subsetP3. Let X2 \inP2 \capS. The circle {Y \inS | d(X2, Y ) = a2} is included in P3, but a2 \leqa1; hence it has radius r = a2

1 −a2 2/(4a2

1. \geqa2 \sqrt 3/2 and a3 \leqa2 \leqa2 \sqrt 3 < 2r; hence a3 is realized by P3.