26th IMO 1985 shortlist Problem 3 Given n > 1, find the maximum value of sin 2 x 1 + sin 2 x 2 + ... + sin 2 x n , where x i are non-negative and have sum π. Solution Solution by Demetres Christofides
: 2 for n = 2; 9/4 for n > 2. The result for n = 2 is obvious. So assume n > 2. The expression is the sum of the squares of the sides of an n-gon inscribed in a circle radius 1/2 whose sides subtend angles 2x 1 , 2x 2 , ... , 2x n at the center. If n > 3, then this n-gon must have an angle ≥ 90 o . Suppose it is the angle at B between the sides AB and AC. Then by the cosine rule AC 2 ≥ AB 2 + BC 2 , so the expression is not reduced if we drop one of the vertices to get an (n-1)-gon. Hence the maximum is achieved by taking at least n-3 angles x i = 0. So we need only consider the case n = 3. So let ABC be a triangle with circumcenter O, centroid G and circumradius R. Take G as the origin and P as any point. Using vectors it is immediate that PA 2 + PB 2 + PC 2 = 3 GP 2 + GA 2 + GB 2 + GC 2 . Putting P = O gives 3R 2 + 3OG 2 = GA 2 + GB 2 + GC 2 . Putting P = A gives AB 2 + AC 2 = 4GA 2 + GB 2 + GC 2 . Putting P = B and P = C and adding the three equations gives: AB 2 + BC 2 + CA 2 = 3(GA 2 + GB 2 + GC 2 ). Hence AB 2 + BC 2 + CA 2 = 9R 2 - 9 OG 2 ≤ 9R 2 = 9/4 for R = 1/2. [Note that equality is certainly achieved by x 1 = x 2 = x 3 = 60 o and other x i = 0.] 26th IMO shortlist 1985 © John Scholes jscholes@kalva.demon.co.uk 11 Sep 2002