# The question is below?

## Let ${A}_{1} , {A}_{2} , \ldots . {A}_{n}$ be the vertices of n-sided regular polygon such that $\frac{1}{{A}_{1} {A}_{2}} = \frac{1}{{A}_{1} {A}_{3}} + \frac{1}{{A}_{1} {A}_{4}}$. Find the value of n.

##### 1 Answer
Jun 6, 2018

Let each side of the regular n-sided polygon be $a$ and the radius of the circumcircle of the polygon be R. Each side of the polygon will subtend angle $\theta = \frac{2 \pi}{n}$ at the center $O$ of the circumcircle.

The perpendicular dropped from $O$ to ${A}_{1} {A}_{2}$ will bisect the $\angle {A}_{1} O {A}_{2} = \theta$ as well as side ${A}_{1} {A}_{2}$

So $\frac{{A}_{1} {A}_{2}}{2} = R \sin \left(\frac{\theta}{2}\right)$

Similarly the perpendicular dropped from $O$ to ${A}_{1} {A}_{3}$ will bisect the $\angle {A}_{1} O {A}_{3} = 2 \theta$ as well as side ${A}_{1} {A}_{3}$

So $\frac{{A}_{1} {A}_{3}}{2} = R \sin \left(\frac{2 \theta}{2}\right)$

And also the perpendicular dropped from $O$ to ${A}_{1} {A}_{4}$ will bisect the $\angle {A}_{1} O {A}_{4} = 3 \theta$ as well as side ${A}_{1} {A}_{4}$

So $\frac{{A}_{1} {A}_{4}}{2} = R \sin \left(\frac{3 \theta}{2}\right)$

Now given condition is

$\frac{1}{{A}_{1} {A}_{2}} = \frac{1}{{A}_{1} {A}_{3}} + \frac{1}{{A}_{1} {A}_{4}}$

Substituting the values of
${A}_{1} {A}_{2} , {A}_{1} {A}_{3} \mathmr{and} {A}_{1} {A}_{4}$ we have

$\frac{1}{2 R \sin \left(\frac{\theta}{2}\right)} = \frac{1}{2 R \sin \left(\frac{2 \theta}{2}\right)} + \frac{1}{2 R \sin \left(\frac{3 \theta}{2}\right)}$

$> \frac{1}{\sin \left(\frac{2 \theta}{2}\right)} = \frac{1}{\sin \left(\frac{\theta}{2}\right)} - \frac{1}{\sin \left(\frac{3 \theta}{2}\right)}$

$\implies \frac{1}{\sin \left(\frac{2 \theta}{2}\right)} = \frac{\sin \left(\frac{3 \theta}{2}\right) - \sin \left(\frac{\theta}{2}\right)}{\sin \left(\frac{3 \theta}{2}\right) \sin \left(\frac{\theta}{2}\right)}$

$\implies \frac{1}{\sin \left(\frac{2 \theta}{2}\right)} = \frac{2 \cos \left(\frac{2 \theta}{2}\right) \sin \left(\frac{\theta}{2}\right)}{\sin \left(\frac{3 \theta}{2}\right) \sin \left(\frac{\theta}{2}\right)}$

$\implies \sin \left(\frac{3 \theta}{2}\right) = 2 \cos \left(\frac{2 \theta}{2}\right) \sin \left(\frac{2 \theta}{2}\right)$

$\implies \sin \left(\frac{3 \theta}{2}\right) = \sin \left(\frac{4 \theta}{2}\right)$

$\implies \sin \left(\pi - \frac{3 \theta}{2}\right) = \sin \left(\frac{4 \theta}{2}\right)$

$\implies \pi - \frac{3 \theta}{2} = \frac{4 \theta}{2}$

$\implies \frac{7 \theta}{2} = \pi$

$\implies \frac{7}{2} \cdot \frac{2 \pi}{n} = \pi$

$\implies n = 7$