IMO 1989 SL 24

Origin: POL

Problem

For points A1, . . . , A5 on the sphere of radius 1, what is the maximum value that min1\leqi,j\leq5 AiAj can take? Determine all configura- tions for which this maximum is attained. (Or: determine the diameter of any set {A1, . . . , A5} for which this maximum is attained.)

Solution

Instead of Euclidean distance, we will use the angles \angleAiOAj, O de- noting the center of the sphere. Let {A1, . . . , A5} be any set for which mini̸=j \angleAiOAj \geq\pi/2 (such a set exists: take for example five vertices of an octagon). We claim that two of the Ai’s must be antipodes, thus implying that mini̸=j \angleAiOAj is exactly equal to \pi/2, and consequently that mini̸=j AiAj = \sqrt 2. Suppose no two of the five points are antipodes. Visualize A5 as the south pole. Then A1, . . . , A4 lie in the northern hemisphere, including the equa- tor (but excluding the north pole). No two of A1, . . . , A4 can lie in the interior of a quarter of this hemisphere, which means that any two of them differ in longitude by at least \pi/2. Hence, they are situated on four meridians that partition the sphere into quarters. Finally, if one of them does not lie on the equator, its two neighbors must. Hence, in any case there will exist an antipodal pair, giving us a contradiction.