IMO 1989 SL 28

Origin: ROM

Problem

Consider in a plane \Pi the points O, A1, A2, A3, A4 such that \sigma(OAiAj) \geq1 for all i, j = 1, 2, 3, 4, i ̸= j. Prove that there is at least one pair i0, j0 \in{1, 2, 3, 4} such that \sigma(OAi0Aj0) \geq \sqrt 2. (We have denoted by \sigma(OAiAj) the area of triangle OAiAj.)

Solution

Assume w.l.o.g. that the rays OA1, OA2, OA3, OA4 are arranged clock- wise. Setting OA1 = a, OA2 = b, OA3 = c, OA4 = d, and \angleA1OA2 = x, \angleA2OA3 = y, \angleA3OA4 = z, we have S1 = \sigma(OA1A2) = 1 2ab| sin x|, S2 = \sigma(OA1A3) = 1 2ac| sin(x + y)|, S3 = \sigma(OA1A4) = 1 2ad| sin(x + y + z)|, S4 = \sigma(OA2A3) = 1 2bc| sin y|, S5 = \sigma(OA2A4) = 1 2bd| sin(y + z)|, S6 = \sigma(OA3A4) = 1 2cd| sin z|. Since sin(x+y +z) siny + sin x sin z = sin(x+y) sin(y +z), it follows that there exists a choice of k, l \in{0, 1} such that S1S6 + (−1)kS2S5 + (−1)lS3S4 = 0. For example (w.l.o.g.), if S3S4 = S1S6 + S2S5, we have  max 1\leqi\leq6 Si 2 \geqS3S4 = S1S6 + S2S5 \geq1 + 1 = 2, i.e., max1\leqi\leq6 Si \geq \sqrt 2 as claimed.