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 $\dfrac{360}{n}.$


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 $$\eqalign{\sum_{k=0}^{n-1}\cos k\theta &=\frac{1}{2\sin\theta}\sum_{k=0}^{n-1}2\cos k\theta\sin\theta\cr &=\frac{1}{2\sin\theta}\sum_{k=0}^{n-1}\bigl(\sin(k+1)\theta-\sin(k-1)\theta\bigr)\cr &=\frac{1}{2\sin\theta}\bigl(\sin\theta+\sin2\theta+\cdots+\sin n\theta -\sin(-\theta)-\sin0-\cdots-\sin(n-2)\theta\bigr)\cr &=\frac{1}{2\sin\theta}\bigl(\sin n\theta+\sin(n-1)\theta+\sin\theta\bigr)\cr &=\frac{1}{2\sin\theta}\bigl(\sin2\pi+\sin(2\pi-\theta)+\sin\theta\bigr)\cr &=0\ .\cr}$$

  • $\begingroup$ i had reached here..i want the last step plz.!!! $\endgroup$ – maths lover Jul 2 '14 at 8:42
  • $\begingroup$ @mathslover Use $\cos\Bigl(\frac{(n-k)\pi}{n}\Bigr)+\cos\Bigl(\frac{k\pi}{n}\Bigr)=0$ $\endgroup$ – N. S. Jul 2 '14 at 8:56
  • $\begingroup$ thanx.!!!!!!!!! $\endgroup$ – maths lover Jul 2 '14 at 9:01
  • 1
    $\begingroup$ @mathslover, sorry, there was a slip in the last line of my answer: $\cos(k\pi/n)$ should have been $\cos(2k\pi/n)$. Unfortunately this also means NS's comment does not help any more. I have added a bit more to my answer. $\endgroup$ – David Jul 2 '14 at 13:14

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