A point $P$ in the plane and a $2017$-gon For a regular 2017-gon $A_1A_2...A_{2017}$ in the plane, show that there exists a point P in the plane such that the following is true:
$$
\sum_{i=1}^{2017}i\frac{\mathbf{PA_i}}{|\mathbf{PA_i}|^5}=\mathbf{0}.
$$
Here is my thought: I think that the number 5 has no special meanings and that if I can answer the problem $$
\sum_{i=1}^{2017}i\frac{\mathbf{PA_i}}{|\mathbf{PA_i}|^2}=\mathbf{0},
$$
then the original can be proved by the same method. I did not do it for number 1 because I think that plugging 1 there makes the vector a unit vector and in general it shouldn't be. Then I find out that the number 2017 is not special so I tried 3 for simplicity. But I find that solving the problem by brute force even if it is only a triangle is not that easy because of the i before the term. 
I saw this problem in The Simon Marais Mathematics Competition (held on 7th, October, 2017) but honestly I don't know what is the correct approach to this problem? Should I try to come up with a system of equations and try to show that there has to be a solution? Or should I just use induction (I don't think this is the proper way since I cannot find a direct relation between $P_{n+1}$ and $P_n$).
Please can someone give me some insights?
 A: Let $A = \{ A_1, A_2, \ldots, A_{2017} \}$ be the set of vertices.
Let $D = \bar{B}(0,1)$ be the closed unit disc centered at origin.
WOLOG, we will assume the vertices lie on the unit circle $S^1 = \partial D$.
Consider the function $f : D \to \mathbb{R} \cup \{ +\infty \}$ defined by
$$D \ni p\quad \mapsto\quad f(p) = \begin{cases}
\sum\limits_{k=1}^{2017} \frac{k}{|pA_k|^3}, & p \in D \setminus A\\ \\
+\infty, & p \in A
\end{cases}$$
The function $f$ is non-negative and continuous on $D$ and smooth on $D \setminus A$.
Notice we can cover $S^1 = \partial D$ by a bunch of closed discs centered at $A_k$  with radius $r = \sin\frac{\pi}{2017}$.
There means for any $p \in \partial D$, there is at lease one $A_{k_p}$ with $|pA_{k_p}| \le r$. This leads to
$$f(p) = \sum_{k=1}^{2017} \frac{k}{|pA_k|^3} \ge \frac{k_p}{|pA_{k_p}|^3} \ge \frac{1}{r^3} = \frac{1}{\sin^3\frac{\pi}{2017}}$$
Notice at the origin $O$, all $|OA_k| = 1$, we have
$$f(O) = \sum_{k=1}^{2017} k = \frac{2017(2018)}{2} \approx \frac12 (2017)^2$$
Since $\frac{1}{r^3} \approx \left(\frac{2017}{\pi}\right)^3$ is much larger than $f(O)$, we can conclude $f$ achieve its absolute minimum over $D$ at some point $P \in {\rm int} D \subset D \setminus A$. Since $f(p)$ is smooth at $P$, the gradient vanishes there.
$$\left.\nabla f(p)\right|_{P} = 0
\quad\iff\quad \sum_{k=1}^{2017} k \frac{PA_k}{|PA_k|^5} = 0$$
In the above proof, neither the number of sides $2017$, the exponent $5$ nor the weights $k$ attached to line segment $PA_k$ is that special. What we need is the the number of sides and exponent are large enough to force
the smallest value of $f(p)$ on $\partial D$ larger than the value at some interior point. This will "push" the absolute minimum of $f(p)$ into the interior of $D$ and allow us to have at lease one point where the gradient of $f$ vanishes.
A: Here is an alternative solution (pdf 1.6Mb)
