Question

State true or false:

Assume that $A_i(i=1,2,....n)$ are the vertices of a regular n-sided polygon inscribed in a circle of radius unity. Then,

${\left|A_1{A}_2\right|}^2+{\left|A_1{A}_3\right|}^2+\cdots +{\left|A_1{A}_n\right|}^2=-1$. Type 1 for true and 0 for false

Answer 1

With origin as the centre of the circle of redius unity, let $z_1,z_2,.....,z_n$ represent the vertices $A_1,A_2,....,A_n$ of the n-gon.Then we easily get
$z_2=z_1{e}^{2\pi i/n},z_3=z_1{e}^{4\pi i/n},.....,$
$z_n{=}_1{e}^{2(n-1)\pi i/2n}$
Now ${\left|A_1{A}_r\right|}^2={|z_1-z_r|}^2$
$={|z_1-z_1{e}^{2(r-1)\pi i/n}|}^2$
$={\left|z_1\right|}^2{|1-e^{2(r-1)\pi i/n}|}^2.$
$={|1-\cos \frac{2(r-1)\pi }{n}-i\sin \frac{2(r-1)\pi }{n}|}^2$
[$\because \left|z_1\right|=$ radius of the circle=1]
$={(1-\cos \frac{2(r-1)\pi }{n})}^2+{\sin }^2\frac{2(r-1)\pi }{n}$
$=2-2\cos \frac{2(r-1)\pi }{n}$ ...(A)
Hence $\sum _{r=2}^n{\left|A_1{A}_r\right|}^2=2(n-1)-2\sum _{r=2}^n\left[\cos \frac{2(r-1)\pi }{n}\right]$ ...(1)
Let $S=\sum _{r=2}^n\cos \frac{2(r-1)\pi }{n}$ ...(2)
$=\cos \frac{2\pi }{n}+\cos \frac{4\pi }{n}+\cos \frac{6\pi }{n}+\cdots +\cdots +\cos \frac{2(n-1)}{n}\pi$
Formula:
$\cos A+\cos (A+B)+\cos (A+2B)+......+\cos (A+\overline{n-1}B)$
$\sin A+\sin (A+B)+\sin (A+2B)+....+\sin (A+\overline{n-1}B)$
There are n angles in A.P. of common diff. B.
$S=\frac{\sin n.\frac{B}{2}}{\sin \frac{B}{2}}\cos$ or $\sin \left\{\frac{1stang.+lastang.}{2}\right\}$
In (2) there are (n-1) terms and angles are in A.P. of common difference $\frac{2\pi }{n}.$
$S=\frac{\sin [(n-1).\pi /n]}{\sin \left(\pi /n\right)}\cos \left\{\frac{\frac{2\pi }{n}+\frac{2(n-1)\pi }{n}}{2}\right\}$
$=\frac{\sin (\pi -\pi /n)}{\sin \left(\pi /n\right)}\cos \pi$
$=\frac{\sin \left(\pi /n\right)}{\sin \left(\pi /n\right)}(-1)=-1$
Hence $\sum _{r=2}^n{\left|A_1{A}_r\right|}^2=2(n-1)-2S,$ by (1)
$=2n-2+2=2n$ as $S=-1$