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 n^2.

 

unit_sphere_segments 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 n^2.

 

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 vec{v_1} be the vector connecting O and a point 1 (one of the n points). Then the vector vec{d_12} = vec{v_1}-vec{v_2} (see the image below). A given segment’s length is just the difference in length of the two vectors radiating from the origin:  <vec{v_i} - vec{v_j},vec{v_i} - vec{v_j}> = delim {|} { d_ij } {|} ^ 2″ title=”<vec{v_i} - vec{v_j},vec{v_i} - vec{v_j}> = delim {|} { d_ij } {|} ^ 2″/></li>
</ol>
<h3></h3>
<h3>Adolescent Steps</h3>
<p><a href=unit_sphere_vectors_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:

    S = sum {ij} {} {delim{|}{d_i * d_j}{|} ^ 2} 

    1. Now, the trick is to realize that 2 times the Sum : 2*S = sum {i=1}{n}{<v_i-v_1, v_i-v_1>} +  sum {i=1}{n}{<v_i-v_2, v_i-v_2>} + cdots + sum {i=1}{n}{<v_i-v_n, v_i-v_n>}” title=”2*S = sum {i=1}{n}{<v_i-v_1, v_i-v_1>} +  sum {i=1}{n}{<v_i-v_2, v_i-v_2>} + cdots + sum {i=1}{n}{<v_i-v_n, v_i-v_n>}”/>
<li>Once we have the above expression for 2S, the rest is just algebra: </li>
</ol>
<p><img src=

      2S = 2 * n * (delim{|} {v_1}{|} ^ 2 + cdots + delim{|} {v_n}{|} ^ 2  - 2*sum{i,j} {} {< v_j, v_i>})” title=”2S = 2 * n * (delim{|} {v_1}{|} ^ 2 + cdots + delim{|} {v_n}{|} ^ 2  – 2*sum{i,j} {} {< v_j, v_i>})”/></p>
<p><img src=

      2S = 2 * n * sum{i=1} {n} {delim{|}(v_i){|}^2}-2*<v,v>” title=”2S = 2 * n * sum{i=1} {n} {delim{|}(v_i){|}^2}-2*<v,v>“/>  </p>
<p><img src=

      2 S = (2*n ^ 2)- (2*delim{|} v {|} ^ 2)

      Final Step

      The above expression shows that 2 * S <= 2 n ^ 2. Hence,

      S <= n ^ 2   i.e. the sum of the squares of all segments is less than    n ^ 2.

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