Question

Let \(P_1, P_2,\) \(\dots\), \(P_n\) be points on a circle of radius 1, centered at \(O\). Let \(G\) be the point such that
\(\overrightarrow{OG} = \frac{\overrightarrow{OP_1} + \overrightarrow{OP_2} + \dots + \overrightarrow{OP_n}}{n},\)
and let \(d = OG\). Express
\(\sum_{1 \le i < j \le n} (P_i P_j)^2\)
in terms of \(n\) and \(d\).

Answer 1

The length of a vector \(\vec v\) using the dot product: \(|\vec v| = \left< \vec v, \vec v\right>\). Note also that if the points \(P_i\) are on the unit circle, then \(OP_i = 1\).

Using the dot product:
\((P_1P_2)^2
\\\quad= \left< \vec{OP_1}-\vec{OP_2} , \vec{OP_1}-\vec{OP_2} \right>
\\\quad= \left< \vec{OP_1},\vec{OP_1}\right> + \left< \vec{OP_1},\vec{OP_1}\right> - 2\left< \vec{OP_1},\vec{OP_2}\right>
\\\quad= (OP_1)^2 + (OP_2)^2 - 2\left< \vec{OP_1},\vec{OP_2}\right>
\\\quad= 1 + 1 - 2\left< \vec{OP_1},\vec{OP_2}\right>
\\\quad = 2 - 2\left< \vec{OP_1},\vec{OP_2}\right>
\)
Summing over all couples \((i,j)\) with \(i<j\), one obtains
\[
\sum_{1\le i < j \le n} (P_iP_j)^2 = \sum_{1\le i < j \le n} \left(2-2\left<\vec{OP_1},\vec{OP_2}\right>\right) \qquad\qquad(*_1)
\]
On the other hand, \(d^2 = (OG)^2\). So, using the same method, we can also write \(d^2\) as
\(
\left< \vec{OG},\vec{OG}\right>
\\\quad= \left< \frac{\vec{OP_1}+\dotsb+\vec{OP_n}}{n}, \frac{\vec{OP_1}+\dotsb+\vec{OP_n}}{n}\right>
%\\\quad= \left< \vec{OP_1},\vec{OP_1}\right> + \left< \vec{OP_2},\vec{OP_2}\right> + \dotsb + \left< \vec{OP_n},\vec{OP_n}\right> + \sum_{1\le i < j \le n} 2\left< \vec{OP_i},\vec{OP_j}\right>
\\\quad= a + b\sum_{1\le i < j \le n} \left< \vec{OP_i},\vec{OP_j}\right> \qquad\qquad(*_2)
\)
I let out a few things. You have to find the value of \(a\) and \(b\)

Then you will be able to make a link between \((*_1)\) and \((*_2)\).

Answer 2

--> on line \((*_1)\), \(P_1\) should be \(P_i\), and \(P_2\) should be \(P_j\).

Answer 3

--> on first line, I meant \(|\vec v|^2 = \left< \vec v, \vec v\right>\).