Let \(A_1,A_2,...A_n\) be a regular polygon of \(n\) sides whose centre is origin \(O\).Llet the complex numbers representing the vertices \(A_1,A_2.....A_n\)be\(z_1,z_2....z_n\) respectively.The radius is of length unity.Then, \[\prod_{n=2}^{n}|A_1A_n|=\mathbb{?}\]

## Comments

Sort by:

TopNewestLet the circumradius be \(R\). Now we consider all \(z\) that satisfy the circle and write a certain expansion:

\[z^n-R^n=(z-R)(z^{n-1} + Rz^{n-2}+Rz^{n-3} + \dots + zR^{n-2}+R^{n-1})\]

Then we have \(z_1=R\) and \(,z_2,z_3\dots z_n\) as roots of \((z^{n-1} + Rz^{n-2}+Rz^{n-3} + \dots + zR^{n-2}+R^{n-1})\) , So we can write :

\[(z-z_2)(z-z_3) \dots (z-z_n)=z^{n-1} + Rz^{n-2}+Rz^{n-3} + \dots + zR^{n-2}+R^{n-1})\]

Since \(z_1=R\) is one of the \(z\)'s , we substitute \(z=R\) in RHS . Thus the equation becomes:

\[|(z-z_1)||(z-z_2)||(z-z_3)| \dots |(z-z_n)|=nR^{n-1} \dots (1)\]

To calculate \(R\) we use cosine rule and get \(R=\dfrac{1}{2\sin\left(\dfrac{\pi}{n}\right)}\). So finally

\[\displaystyle\prod_{i=2}^n |A_1A_n|=\dfrac{n}{2\sin^{n-1}\left(\dfrac{\pi}{n}\right)}\]

I may be wrong... Adarsh please check

Edit:The problem is edited and the circumradius is unity and not the side. Hence substituting \(R=1\) in \((1)\) we get:\[ \displaystyle\prod_{i=2}^n |A_1A_n|=n\times 1^{n-1} = \boxed{n}\] – Nihar Mahajan · 10 months, 3 weeks ago

Log in to reply

– Adarsh Kumar · 10 months, 3 weeks ago

Answer is \(n\).Log in to reply

– Nihar Mahajan · 10 months, 3 weeks ago

Are you sure that the side equals unity? Or the circumradius?Log in to reply

– Adarsh Kumar · 10 months, 3 weeks ago

Oh!Sorry!R=1!Log in to reply

– Nihar Mahajan · 10 months, 3 weeks ago

Yes , now the answer is \(n\).Log in to reply

I asked this question a while ago and Alan Yan wrote a nice solution – Otto Bretscher · 10 months, 3 weeks ago

Log in to reply

– Adarsh Kumar · 10 months, 3 weeks ago

Thanks a lot sir!Log in to reply

– Nihar Mahajan · 10 months, 3 weeks ago

Thank you for the link!Log in to reply