On An Intriguing Property of the Center of Mass of Points on a Sphere in R^{d}

Doron ZEILBERGER

Abstract: We present a short proof of an intriguing result proved in 2007 by Leonid Hanin, Robert Fisher, and Boris Hanin.

In a delightful article [HFH], the authors first stated, and gave a beautiful synthetic proof, of the case d = 2 of the following two theorems regarding points on a circle. They later considered general quadratic surfaces, focusing on two and three dimensions, and remarked that their reasoning is true for any dimension.

Theorem 1: Let X_{1}, …, X_{n} be n ≥ 2 points on a sphere in R^{d}, and C be their geometric center of mass. Denote by Y_{1}, …, Y_{n} the second points of intersection of the lines X_{1}C, X_{2}C, …, X_{n}C with the sphere, respectively, then

∑_{i=1}^{n} \frac{X_{i}C}{CY_{i}} = n .

Theorem 2: Let X_{1}, …, X_{n} be n ≥ 2 points on a sphere in R^{d}, with center O, and let C be their geometric center of mass. For any point P inside the sphere, let Y_{1}, …, Y_{n} be the second points of intersection of P X_{i} with the sphere. The set of points P for which

∑_{i=1}^{n} \frac{X_{i}P}{P Y_{i}} = n ,

is a sphere with diameter OC.

In this note I will give a short, self-contained, proof of their result for general d, but for the sake of simplicity will stick to spheres. Of course, by a change of variables every quadratic surface can be transformed to a sphere, if you don’t mind “virtual points”.

Since Theorem 2 implies Theorem 1 we will only prove the former. We need the following simple lemma.

Lemma: For any point X on the unit d-dimensional sphere, and any point P inside the sphere , let Y be the second point of intersection of the line XP with the sphere, then

\frac{XP}{P Y} = \frac{2(X, P) − (P, P) − 1}{(P, P) − 1} .

Proof of the Lemma: Every point on the line joining X and P has the form X + s(P − X) = (1 − s)X + sP where s = 0, corresponds to X, and s = 1 corresponds to P. It meets the sphere when s satisfies

((1 − s)X + sP, (1 − s)X + sP) − 1 = 0 .

Expanding, we get

(1 − s)^{2} · (X, X) + 2(1 − s)s · (X, P) + s^{2} · (P, P) − 1 = 0 .

Using (X, X) = 1 we get

s(−2 + s + 2(1 − s) · (X, P) + s · (P, P)) = 0 .

This equation has two solutions: s = 0 corresponds to X, and the one corresponding to Y is:

s = \frac{2(1 − (X, P))}{1 − 2(X, P) + (P, P)} .

Hence

\frac{XP}{P Y} = \frac{1}{s − 1} = \frac{2(X, P) − (P, P) − 1}{(P, P) − 1} .

Proof of Theorem 2: Without loss of generality, the sphere is centered at the origin, and has radius 1. The center of mass of the points X_{i} is

C := \frac{1}{n} (∑_{i=1}^{n} X_{i}) .

The condition

∑_{i=1}^{n} \frac{X_{i}P}{P Y_{i}} = n ,

thanks to the lemma, is

∑_{i=1}^{n} \frac{2(X_{i}, P) − (P, P) − 1}{(P, P) − 1} = n ,

which is easily seen to be equivalent to

(P − C/2, P − C/2) = (C/2, C/2) .

Reference

[HFH] Leonid G. Hanin, Robert J. Fisher, and Boris L. Hanin, An intriguing property of the center of mass for points on quadratic curves and surfaces, Mathematics Magazine 80, No. 5, (December 2007), 353-362.

Doron Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen Rd., Piscataway, NJ 08854-8019, USA.

Email:DoronZeil] at gmail dot com .

Written: May 2, 2025.

Exclusively published in the Personal Journal of Shalosh B. Ekhad and Doron Zeilberger