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The problem goes as

Let $P_1$, $P_2$, $P_3$, $\ldots P_n$ be $n$ points on a circle with radius unity. Prove that sum of squares of their mutual distances is not greater than $n^2$.

Using the properties of vectors, I assumed centre of circle as origin and the radius $r_1$, $r_2$, $r_3 \ldots$ as vectors but I'm not able to comprehend after that. My second thought was using complex numbers. Can they be used here?


$|r_1-r_2| +|r_1-r_3| +|r_1-r_4|+\dots|r_1-r_n|+ \\|r_2-r_3| +|r_2-r_4| +|r_2-r_5|+\dots|r_2-r_n|\ +\\.\\..\\...\\|r_{n-1}-r_n|$

I see that on squaring them and adding I get each term $n-1$ times and a huge no. of pairs are subtracted which I don't know how to write in an organized form.


$$D^2= (n-1)(r_1^2+r_2^2 ......,+r_n^2) -x$$ So although it's clear from the expression that distance is certainly less than $n^2$ but I want to know how to find $x$.

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  • $\begingroup$ I’m voting to close this question because it is a copy paste of the same question with the same author asked a week ago here. $\endgroup$
    – Jean Marie
    Commented Oct 17, 2020 at 9:14
  • $\begingroup$ @JeanMarie my previous question wanted my attempt which I've given and there is no appropriate justification to close this question $\endgroup$
    – imposter
    Commented Oct 17, 2020 at 10:12
  • $\begingroup$ @JeanMarie aditya deleted his answer and I'm no more able to see his answer now so for me this question remains unanswered right?. Talking about my previous question it has been deleted $\endgroup$
    – imposter
    Commented Oct 17, 2020 at 10:35
  • $\begingroup$ I am still able to see the "answer" of Aditya which is a copy paste of a text without reference. it is delicate to give it back. $\endgroup$
    – Jean Marie
    Commented Oct 17, 2020 at 10:52
  • $\begingroup$ @JeanMarie I'm not as it is saying account temporarily suspended $\endgroup$
    – imposter
    Commented Oct 17, 2020 at 11:03

2 Answers 2

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Let the center of the circle be an origin and $P_i(x_i,y_i).$

Thus, $x_i^2+y_i^2=1$ and $$\sum_{1\leq i<j\leq n}P_iP_j^2=\sum_{1\leq i<j\leq n}\left((x_i-x_j)^2+(y_i-y_j)^2\right)=$$ $$=n(n-1)-2\sum_{1\leq i<j\leq n}(x_ix_j+y_iy_j)=$$ $$= n(n-1)-\left(\sum_{i=1}^nx_i\right)^2-\left(\sum_{i=1}^ny_i\right)^2+\sum_{i=1}^n(x_i^2+y_i^2)\leq$$ $$\leq n(n-1)+n=n^2.$$

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  • $\begingroup$ [+1] ... In fact "parallel" to the complex numbers solution given by Aditya Dwivedi in his non-referenced text (see the previous answers to this question). $\endgroup$
    – Jean Marie
    Commented Oct 17, 2020 at 9:31
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    $\begingroup$ Equality to $n^2$ looks achieved, for example for regular polygons ? $\endgroup$
    – Jean Marie
    Commented Oct 17, 2020 at 10:28
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    $\begingroup$ @Jean Marie If the center of mass it's the origin: $\frac{\sum\limits_{i=1}^nx_i}{n}=\frac{\sum\limits_{i=1}^ny_i}{n}=0.$ $\endgroup$ Commented Oct 17, 2020 at 11:10
  • $\begingroup$ Please see the reference in the comment I just posted. $\endgroup$
    – Jean Marie
    Commented Oct 17, 2020 at 15:06
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Please neither upvote or downvote this text: it was the only way for me to post an image.

I have found valuable to copy paste this deleted solution given a week ago by Aditya Dwivedi (https://math.stackexchange.com/q/3859542) to a first version of this question. This answer itself was copy-paste of a solution he had found in a un-referenced document. As it has been deleted, it is not visible by all users of this site. I am trying to find back the origin of this excerpt.

Remark: one of the interests of this solution is the fact that if $\sum z_i=0$ (which is the case for regular polygons) the upper bound $n^2$ is reached.

enter image description here

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