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{\bf 
On An Intriguing Property of the Center of Mass of Points on a  Sphere in $R^d$ 
}
\bigskip
\centerline
{\it Doron ZEILBERGER}

{\bf 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 {\it 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.


{\bf Theorem 1:} Let $X_1, \dots, X_n$ be $n \geq 2$ points on a  sphere in $R^d$, and $C$ be their geometric center of mass. Denote by $Y_1, \dots, Y_n$ the second points of intersection
of the lines $X_1 C, X_2 C, \dots, X_n C$ with the sphere, respectively, then
$$
\sum_{i=1}^{n} \frac{X_iC}{CY_i} \, = \, n \quad .
$$


{\bf Theorem 2:} Let $X_1, \dots, X_n$ be $n \geq 2$  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  $Y_1, \dots, Y_n$ be the second points of intersection of $PX_i$ with the sphere.
The set of points $P$ for which
$$
\sum_{i=1}^{n} \frac{X_iP}{PY_i} \, = \, n \quad ,
$$
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.


{\bf 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
$$
\frac{XP}{PY} = \frac{2(X,P)-(P,P)-1}{(P,P)-1}
\quad .
$$

{\bf 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
$$
((1-s)X+sP,(1-s)X+sP)-1=0 \quad .
$$
Expanding, we get
$$
(1-s)^2 \cdot (X,X) + 2(1-s)s \cdot (X,P)+ s^2 \cdot (P,P)-1=0 \quad .
$$
Using $(X,X)=1$ we get
$$
s(-2+s+ 2(1-s) \cdot(X,P)+ s \cdot (P,P))=0 \quad.
$$
This equation has two solutions: $s=0$ corresponds to $X$,  and the one corresponding to $Y$ is:
$$
s= \frac{2(1-(X,P))}{1-2(X,P)+(P,P)} \quad .
$$
Hence
$$
\frac{XP}{PY} = \frac{1}{s-1}=\frac{2(X,P)-(P,P)-1}{(P,P)-1} \quad \quad. \quad  \halmos 
$$

{\bf 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
$$
C:=\frac{1}{n} \left ( \sum_{i=1}^{n} X_i \right )\quad .
$$

The condition
$$
\sum_{i=1}^{n} \frac{X_iP}{PY_i} \, = \, n \quad ,
$$
thanks to the lemma, is
$$
\sum_{i=1}^{n} \, \frac{2(X_i, P)-(P,P)-1}{(P,P)-1}  = \, n \quad ,
$$
which is easily seen to be equivalent to
$$
(P-C/2,P-C/2)=(C/2,C/2) \quad . \quad \halmos
$$


{\bf Reference}

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


\bigskip
\hrule
\bigskip
Doron Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen
Rd., Piscataway, NJ 08854-8019, USA. \hfill\break
Email:{\tt DoronZeil] at gmail dot com}   \quad .


Written: {\bf May 2, 2025}.

{\bf Exclusively published in the Personal Journal of Shalosh B. Ekhad and Doron Zeilberger}

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