### 29th Putnam 1968

Problem A4

Let S2 be the 2-sphere { (x, y, z) : x2 + y2 + z2 = 1}. Show that for any n points on S2, the sum of the squares of the n(n - 1)/2 distances between them (measured in space, not in S2) is at most n2.

Solution

Let the points be (xi, yi, zi) for i = 1, 2, ... n. The square of the distance between the pair i, j is (xi - xj)2 + (yi - yj)2 + (zi - zj)2 = (xi2 + yi2 + zi2) + (xj2 + yj2 + zj2) - 2(xixj + yiyj + zizj) = 2 - 2 (xixj + yiyj + zizj). Hence the sum of the squares of the distances is n(n-1) - 2 ∑ (xixj + yiyj + zizj).

Now (x1 + x2 + ... + xn)2 = ∑ xi2 + 2 ∑ xixj. So 2 ∑ (xixj + yiyj + zizj) = X2 + Y2 + Z2 - ∑ (xi2 + yi2 + zi2) = X2 + Y2 + Z2 - n, where X = ∑ xi, Y = S yi, Z = ∑ zi. Hence the sum of the squares is n2 - (X2 + Y2 + Z2), which is at most n2.

[Playing with small examples it is tempting to think that the maximum is only achieved when the points are symmetrically distributed, eg at the vertices of a triangle or tetrahedron, but the result shows that this is not true. For n even, for example, we can take the points in diametrically opposite pairs, but otherwise random.]