Finding $n$ such that in a regular $n$-gon $A_1A_2\ldots A_n$ we have $\frac1{A_1A_2}=\frac1{A_1A_3}+\frac1{A_1A_4}$

Question

INMO '92 Question 9:

Find $n$ such that in a regular $n$-gon $A_1A_2 ...A_n$ we have $$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$$

I tried the following

Assume it is inscribed in a circle.

Then length of chord is $2\sin(\theta)$ where $\theta$ is half the angle subtended at the center between consecutive points. So, $\theta=\frac{180^\circ}{n}$.

Then we get $$\csc(\theta)=\csc(2\theta)+\csc(3\theta)$$ Not sure quite how to proceed from there- using double and triple angle formulae doesn't seem to work

Answer 1

Since $2\cos\theta\sin3\theta=\sin2\theta+\sin3\theta$, $\sin2\theta+\sin4\theta=\sin2\theta+\sin3\theta$, so $4\theta+3\theta=\pi$. So $n=\pi/\theta=7$.