A simple problem on regular polygon inscribed in a unit circle Let $A_1,A_2,...,A_n$ be the vertice of an regular polygon inscribed in a unit circle, $P$ is a point on the circle.
What is the maximum and the minimum values of $\sum_{k=1}^n|PA_k|$?
My idea : Maybe we can define $\omega=e^{\frac{2\pi i}n}$, $A_k=\omega^k$ and $P=e^{i\theta}$, then
$$\sum_{k=1}^n|PA_k|=\sum_{k=1}^n|e^{i\theta}-e^{\frac{2k\pi i}n}|=\sum_{k=1}^ne^{i(-\arg (P-A_k))}(P-A_k)=...$$
In this way, perhaps we can write the final result in the form of trigonometric functions, so that it is convenient to find its maximum and minimum values. But I failed to expand and simplify it, can anyone help me?
 A: Using simple trigonometry, we have
$$S(\theta)=\sum_{k=1}^n|PA_k|=2\sum_{k=0}^{n-1}\sin{\frac{\theta+\frac{2k\pi}{n}}{2}}, \text{where } 0\leq\theta\leq\frac{\pi}{n}.$$
Then
$$\frac{\text{d}S}{\text{d}\theta}=\sum_{k=0}^{n-1}\cos{\frac{\theta+\frac{2k\pi}{n}}{2}}$$
Based on symmetry in the unit circle,
$$\theta=\dfrac{\pi}{n}\implies \frac{\text{d}S}{\text{d}\theta}=\sum_{k=0}^{n-1}\cos{\left((1+2k)\frac{\pi}{2n}\right)}=0$$
As $\theta$ decreases from $\dfrac{\pi}{n}$ to $0$, every term in the series $\dfrac{\text{d}S}{\text{d}\theta}$ increases. From this it follows that $S_\text{min}=S(0)$ and $S_\text{max}=S\left(\dfrac{\pi}{n}\right).$
We know that the mean distance between two points on the unit circle is $\dfrac{4}{\pi}$. From this it follows that as $n\to\infty$, both $S_\text{min}$ and $S_\text{max}$ approach $\dfrac{4n}{\pi}$.
A: EDIT1:

If centered at origin, the vector sum of all polar symmetric vectors or forces $AP$ ( inclined at same $\pi/k$ to each next/consecutive vector) vanishes. So,
$$=\sum_{k=1}^n|OA_k|+\sum_{k=1}^n| A_k P|== n|OA| +0 $$
where $\vec {OA}$ is the constant eccentric vector. Since polygon vectors sum up to zero (no matter radius length), the summation to $A$ or $P$ is same. The former constant eccentric vector is summed $n$ times.
