# N Points on a unit sphere

### Problem Statement

Consider n points lying on the Unit Sphere (the image shows a unit circle instead of a sphere). Show that the sum of the squares of the lengths of all segments determined by the n points is less than .

 Consider n points lying on the Unit Sphere (the image shows a unit circle instead of a sphere). Show that the sum of the squares of the lengths of all segments determined by the n points is less than .

### Potential Stumbling Block

n is the NUMBER of points. It seems a bit counter-intuitive to think of the lengths of the segments related to the number of points on the unit sphere. Yet – that is exactly what this problem is asking us to prove.

### Baby Steps

Perhaps the first realization should be that the sum of the squares of lengths is sought. So – we should probably think in terms of the vectors representing the segments – so that we can take an inner product (which will be a squared length).

1. Let O be the origin
2. Let be the vector connecting O and a point 1 (one of the n points). Then the vector (see the image below). A given segment’s length is just the difference in length of the two vectors radiating from the origin:

Now that we have a way to describe the length of an individual segment, let us denote by S the sum of all such segments:

1. Now, the trick is to realize that 2 times the Sum :

### Final Step

The above expression shows that . Hence,

i.e. the sum of the squares of all segments is less than  .

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