$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$ If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).
I know that sides of a regular polygon are equal but i could not relate$A_1A_3$ and $A_1A_4$ with the side length.Can someone assist me in solving this problem?

$Outline:$
I'm sorry I cannot draw a diagram and explain, though that would be very neat. Let $r$ be the radius of the circumcircle, and let $2\theta$ be the angle subtended at the center by the adjacent vertices. If $a$ is the length of any side of the regular polygon, then $a = 2r\sin\theta$.  Now $\theta = \frac{\pi}{n}$, then the given equation becomes $\frac{1}{\sin\theta} = \frac{1}{\sin2\theta} + \frac{1}{\sin3\theta}$. Simplify this, you will get $\sin4\theta = \sin3\theta$, gives $\theta = \frac{\pi}{7}$. So $n = 7$ is the final answer. You may fill in the gaps in the solution.
Since the answer got a few upvotes, I looked up a little bit more and found that this problem was asked in IIT JEE 1994 (entrance exam for Indian Institute of Technology) and again a variant of it in 2011. Prof K. D. Joshi, in his book Educative JEE, has outlined 3 solutions, one of them is like mine (only better), then the Ptolemy one given by user260674 as well as the one using complex numbers given by Seven. You can refer to page 18 and 19 of the original script here
