If $ (A_1A_2)^2 + (A_1A_3)^2.......... + (A_1A_n)^2= 14r^2$, then prove that the number of sides is 7. Let $A_1, A_2,\ldots,A_n$ be the vertices of a regular $n$ sided polygon inscribed in a circle of radius r.
If
 $ (A_1A_2)^2 + (A_1A_3)^2+\ldots + (A_1A_n)^2= 14r^2$,
then prove that the number of sides is 7.
I used sine law for each $A_1A_2 , A_1A_3$ till $A_1A_n$
And then wrote a relation with r and theta using Sine law.
Here theta is $\dfrac{360}{n}.$
 A: The angle $A_1OA_{k+1}$ is $2k\pi/n$.  The distance between two vertices is given by
$$A_1A_{k+1}=2r\sin\Bigl(\frac{k\pi}{n}\Bigr)\ .$$
So the given sum is
$$\sum_{k=0}^{n-1} 4r^2\sin^2\Bigl(\frac{k\pi}{n}\Bigr)
  =2r^2\sum_{k=0}^{n-1}\left(1-\cos\Bigl(\frac{2k\pi}{n}\Bigr)\right)\ .$$
The sum of the cosine terms is zero: this can be proved by using complex numbers or by converting it into a telescoping sum.  Therefore the entire sum is $2nr^2$.  The result follows easily.
Edit.  Here is the telescoping sum.  Writing $\theta=2\pi/n$ we have
$$\eqalign{\sum_{k=0}^{n-1}\cos k\theta
  &=\frac{1}{2\sin\theta}\sum_{k=0}^{n-1}2\cos k\theta\sin\theta\cr
  &=\frac{1}{2\sin\theta}\sum_{k=0}^{n-1}\bigl(\sin(k+1)\theta-\sin(k-1)\theta\bigr)\cr
  &=\frac{1}{2\sin\theta}\bigl(\sin\theta+\sin2\theta+\cdots+\sin n\theta
    -\sin(-\theta)-\sin0-\cdots-\sin(n-2)\theta\bigr)\cr
  &=\frac{1}{2\sin\theta}\bigl(\sin n\theta+\sin(n-1)\theta+\sin\theta\bigr)\cr
  &=\frac{1}{2\sin\theta}\bigl(\sin2\pi+\sin(2\pi-\theta)+\sin\theta\bigr)\cr
  &=0\ .\cr}$$
