Show equality of polygonal commodity

Question

$ A_1, A_2, ...., A_n $ is a regular polygon inscribed on a radius circumference $ r $ and center $ O $, $ P $ is a point over $ OA_1 $. Show that:$$\prod_{k=1}^n PA_k=OP^n-r^n$$

I really am without effective ideas. Can anyone give me a hint?

Answer 1

I can't duplicate the post, so I will expect a different solution from mine now:

Let $ z $ be the complex representative of $ P $ in the complex plane, and $ A_1, A_2, ... A_n $ the nth roots of $ r $, representing the geometric situation of the problem. With this: $ A_1 = r \in \mathbb {R} , OP = z \in \mathbb {R} $

Calculating the product: $(PA_1) \cdot(PA_2) \cdot (PA_3) \cdot....\cdot(PA_n)$ $$(PA_1) \cdot(PA_2) \cdot (PA_3) \cdot....\cdot(PA_n)=$$ $$=|r-z| \cdot \left|r \cdot cis\left(\frac{2\pi}{n}\right)-z\right| \cdot \left|r \cdot cis \left(\frac{4 \pi}{n}\right)-z\right|\cdot....\cdot\left|r \cdot cis \left(\frac{2(n-1)\pi}{n}\right)-z\right|$$ $$=|r-z| \cdot |w-z|\cdot |w^2-z|\cdot ....\cdot |w^n-z|$$ $$|(z-r)\cdot (z-r \cdot w)\cdot (z-r \cdot w^2)\cdot ....\cdot (z-r \cdot w^n)|$$ Consider the polynomial: $ P (Z) = Z ^ n-r ^ n $ that has roots: $$P(Z)=z^n-r^n=0 \Leftrightarrow z^n=r^n \Leftrightarrow z= r \cdot cis\left(\frac{2k \pi}{n}\right), k=0,1,...,n-1$$ Factoring: $$z^n-r^n=(z-r) \cdot (z-r \cdot w) \cdot (z- r \cdot w^2) \cdot... \cdot (z- r \cdot w^n)$$

And with that, remembering that $ z ^ n, r ^ n \in \mathbb {R} $ since $ z, r \in \mathbb {R} $: $$(PA_1) \cdot(PA_2) \cdot (PA_3) \cdot....\cdot(PA_n)=|z^n-r^n|=OP^n-r^n$$