Given n > 1, find the maximum value of sin2x1 + sin2x2 + ... + sin2xn, where xi are non-negative and have sum π.
Solution
Solution by Demetres Christofides
Answer: 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 2x1, 2x2, ... , 2xn at the center. If n > 3, then this n-gon must have an angle ≥ 90o. Suppose it is the angle at B between the sides AB and AC. Then by the cosine rule AC2 ≥ AB2 + BC2, 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 xi = 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 PA2 + PB2 + PC2 = 3 GP2 + GA2 + GB2 + GC2. Putting P = O gives 3R2 + 3OG2 = GA2 + GB2 + GC2. Putting P = A gives AB2 + AC2 = 4GA2 + GB2 + GC2. Putting P = B and P = C and adding the three equations gives: AB2 + BC2 + CA2 = 3(GA2 + GB2 + GC2). Hence AB2 + BC2 + CA2 = 9R2 - 9 OG2 ≤ 9R2 = 9/4 for R = 1/2. [Note that equality is certainly achieved by x1 = x2 = x3 = 60o and other xi = 0.]
© John Scholes
jscholes@kalva.demon.co.uk
11 Sep 2002