IMO 1983 SL 3

Origin: BEL

Problem

We say that a set E of points of the Euclidian plane is “Pythagorean” if for any partition of E into two sets A and B, at least one of the sets contains the vertices of a right-angled triangle. Decide whether the following sets are Pythagorean: (a) a circle; (b) an equilateral triangle (that is, the set of three vertices and the points of the three edges).

Solution

(a) A circle is not Pythagorean. Indeed, consider the partition into two semicircles each closed at one and open at the other end. (b) An equilateral triangle, call it PQR, is Pythagorean. Let P ′, Q′, and R′ be the points on QR, RP, and PQ such that PR′ : R′Q = QP ′ : P ′R = RQ′ : Q′P = 1 : 2. Then Q′R′ \perpPQ, etc. Suppose that PQR is not Pythagorean, and consider a partition into A, B, neither of which contains the vertices of a right-angled triangle. At least two of P ′, Q′, and R′ belong to the same class, say P ′, Q′ \inA. Then [PR] \ {Q′} \subsetB and hence R′ \inA (otherwise, if R′′ is the foot of the perpendicular from R′ to PR, \triangleRR′R′′ is right-angled with all vertices in B). But this implies again that [PQ] \ {R′} \subsetB, and thus B contains vertices of a rectangular triangle. This is a contradiction.