Question

Assume that $A_i(i=1,2,.....,n)$ are the vertices of a regular polygon inscribed in a circle of radius unity. Find the value of $|A_1{A}_2{|}^2+|A_1{A}_3{|}^2+.....+|A_1{A}_n{|}^2$.

• A. $n$
• B. $n-1$
• C. $4n$
• D. $2n$

Answer 1

The correct option is D $2n$

Given that $A_i(i=1,2,......,n)$ are regular vertices of regular polygon ${\left|{A_{1}A}_2\right|}^2+{\left|{A_{1}A}_3\right|}^2+.....+{\left|{A_{1}A}_n\right|}^2$

For a polygon, ${A_{1}A}_2=2R\sin \alpha .$

$\Rightarrow {A_{2}A}_3=2R\sin \left(2\alpha \right)$ and ${A_{1}A}_4=2R\sin \left(3\alpha \right)$

$\Rightarrow {A_{1}A}_n=2R\sin \left(n\alpha \right)$

Where $\alpha =\frac{\pi }{n}$

$\Rightarrow \frac{1}{2R\sin \pi }=\frac{1}{2R\sin 2\pi }+\frac{1}{2R\sin 3\pi }$

$\Rightarrow \frac{1}{\sin \alpha }=\frac{1}{\sin 2\alpha }+\frac{1}{\sin 3\alpha }$

$\Rightarrow \frac{1}{\sin 2\alpha }=\frac{1}{\sin \alpha }-\frac{1}{\sin 3\alpha }$

$\Rightarrow \frac{1}{\sin 2\alpha }=\frac{\sin 3\alpha -\sin \alpha }{\sin \alpha .\sin 3\alpha }$

$\Rightarrow \frac{1}{\sin 2\alpha }=\frac{2\sin \alpha \times \cos 2\alpha }{\sin \alpha .\sin 3\alpha }$

$\Rightarrow \sin 3\alpha =2\cos \left(2\alpha \right)\sin \left(2\alpha \right)$

$\Rightarrow \sin 3\alpha =\sin \left(4\alpha \right)$

$\Rightarrow 3\alpha =4\alpha$

$\Rightarrow 3\alpha -4\alpha =\pi$

$\therefore -\alpha =\pi$

$\therefore \pi +\alpha =0$

$\therefore {\left|{A_{1}A}_2\right|}^2+{\left|{A_{1}A}_3\right|}^2+.....+{\left|{A_{1}A}_n\right|}^2=2n$

Hence, the ansswer is $2n.$