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

**Solution**

Let the points be (x_{i}, y_{i}, z_{i}) for i = 1, 2, ... n. The square of the distance between the pair i, j is (x_{i} - x_{j})^{2} + (y_{i} - y_{j})^{2} + (z_{i} - z_{j})^{2} = (x_{i}^{2} + y_{i}^{2} + z_{i}^{2}) + (x_{j}^{2} + y_{j}^{2} + z_{j}^{2}) - 2(x_{i}x_{j} + y_{i}y_{j} + z_{i}z_{j}) = 2 - 2 (x_{i}x_{j} + y_{i}y_{j} + z_{i}z_{j}). Hence the sum of the squares of the distances is n(n-1) - 2 ∑ (x_{i}x_{j} + y_{i}y_{j} + z_{i}z_{j}).

Now (x_{1} + x_{2} + ... + x_{n})^{2} = ∑ x_{i}^{2} + 2 ∑ x_{i}x_{j}. So 2 ∑ (x_{i}x_{j} + y_{i}y_{j} + z_{i}z_{j}) = X^{2} + Y^{2} + Z^{2} - ∑ (x_{i}^{2} + y_{i}^{2} + z_{i}^{2}) = X^{2} + Y^{2} + Z^{2} - n, where X = ∑ x_{i}, Y = S y_{i}, Z = ∑ z_{i}. Hence the sum of the squares is n^{2} - (X^{2} + Y^{2} + Z^{2}), which is at most n^{2}.

[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.]

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002