: √2. All maximal configurations have two points at opposite ends of a diameter. The other three points lie on the great circle midway between these two points (such that the distance between any two is at least √2). Suppose the points are such that the distance between any two is at least √2. Then any two must subtend an angle of at least 90 o at the center of the sphere. We claim that two of the points must be at opposite ends of a diameter. Let S be one of the points. Take it at the south pole. So suppose that none of the points is at the north pole N. If the other 4 points all lie on the equator, then they must be equally spaced and so a pair lie at opposite ends of a diameter as claimed. If there is a point X not on the equator, then take a line of longitude L 1 through it. Now take three other lines of longitude L 2 , L 3 , L 4 , so that the 4 lines are equally spaced. They divide the northern hemisphere into 4 equal quadrants (which overlap along the lines of longitude). Each quadrant is a curvilinear triangle with three vertices. Clearly, if no two points in the same quadrant subtend an angle of more than 90 o at the center, and if they subtend an angle of 90 o , then one must be a vertex and the other must be on the opposite (curvilinear) side. Thus (1) the minimum angle is at least 90 o , and (2) each of the two quadrants containing X must contain another of the 4 points (apart from X) and they must lie at P and P'. But now P and P' are opposite ends of a diameter. Now if 2 of the 5 points are opposite ends of a diameter, and the minimum distance is at least √2, then the other three points must lie on the great circle midway between. 30th IMO shortlist 1989 © John Scholes jscholes@kalva.demon.co.uk 28 Dec 2002 Last corrected/updated 28 Dec 02
30th IMO 1989 shortlist Problem 18 Five points are placed on a sphere of radius 1. That is the largest possible value for the shortest distance between two of the points? Find all configurations for which the maximum is attained. Solution