Prove that 5 lines are concurrent, and find the expression for the position vector of the point they all go through. Pentagon $ABCDE$ is inscribed in a circle centered at the origin. Define the lines \begin{align*}
\ell_{ABC} &= \text{Line through the centroid of $\triangle ABC$ perpendicular to $\overline{DE}$},\\
\ell_{BCD} &= \text{Line through the centroid of $\triangle BCD$ perpendicular to $\overline{AE}$}, \\
\ell_{CDE} &= \text{Line through the centroid of $\triangle CDE$ perpendicular to $\overline{AB}$}, \\
\ell_{DEA} &= \text{Line through the centroid of $\triangle DEA$ perpendicular to $\overline{BC}$}, \\
\ell_{EAB} &= \text{Line through the centroid of $\triangle EAB$ perpendicular to $\overline{CD}$}. \\
\end{align*}
These are lines going through the centroid of a triangle formed by three consecutive vertices, perpendicular to the line segment formed by the other two vertices. Here's $\ell_{ABC}$ in the picture:

Prove that $\ell_{ABC}, \ell_{BCD}, \ell_{CDE},\ell_{DEA}$ and $\ell_{EAB}$ are concurrent, and find the expression for the position vector of the point they all go through.
I truly have no idea how to approach this problem. Please help!
 A: WLOG, say the center of the circle ($O$) is at the origin. Vertices of the pentagon $ABCDE$ are represented by position vectors $\overline{a}, \overline{b}, \overline{c}, \overline{d}$ and $\overline{e}$.
Centroid of $\triangle ABC, \, \overline {g} = \frac{\overline{a} + \overline{b} + \overline{c}}{3}$
Line $DE = \overline{d} - \overline{e}$
As points $A, B, C, D, E$ are concyclic with center at $O$
$|\overline{a}|^2 = |\overline{b}|^2 = |\overline{c}|^2 = |\overline{d}|^2 = |\overline{e}|^2$ ...(i)
If a point $P$ with position vector $\overline{p} \,$ is on the perpendicular line from the centroid of $\triangle ABC$ to the line $DE$,
$(\overline{p}-\overline{g}) \cdot (\overline{d} - \overline{e}) = 0$
Based on (i) one of the ways for the dot product to be zero is
$(\overline{p}-\overline{g}) = n_1 (\overline{d}+\overline{e}) \,$  (you can easily show why $\overline{p} = \overline{g}$ will not give you the concurrent point by symmetry)
$\overline{p}-\overline{g} = \overline{p}-\frac{\overline{a} + \overline{b} + \overline{c}}{3} = n_1 (\overline{d}+\overline{e})$ ...(ii)
Similarly,
$\overline{p}-\frac{\overline{b} + \overline{c} + \overline{d}}{3} = n_2 (\overline{e}+\overline{a})$ ...(iii)
From (ii)-(iii), you get one solution when $n_1 = n_2 = \frac{1}{3}$ and
$\overline {p} = \frac{\overline{a} + \overline{b} + \overline{c} + \overline{d} + \overline{e}}{3}$
Now we need to prove this point is the point of concurrency for other $3$ lines too. So we take the lines from centroids of $\triangle CDE, \triangle DEA, \triangle EAB$ through point $\overline {p}$ and show each of them is perpendicular to the line segment made by other two vertices.
$(\overline{p}- \frac{\overline{c} + \overline{d} + \overline{e}}{3}) \cdot   (\overline{a} - \overline{b}) = 0$
$(\overline{p}- \frac{\overline{d} + \overline{e} + \overline{a}}{3}) \cdot   (\overline{b} - \overline{c}) = 0$
$(\overline{p}- \frac{\overline{e} + \overline{a} + \overline{b}}{3}) \cdot   (\overline{c} - \overline{d}) = 0$
which is easy to show given (i).
A: We denote orthocenter of any triangle $XYZ$ by $H_{XYZ}$ and the centeroid by $G_{XYZ}$. We also denote midpoint of any two points $XY$ by $M_{XY}$.

Proof. Without loss of generality, let $\odot(ABCDE)$ be the unit circle centered at origin of complex plane. We claim that the point $G_{DEH_{ABC}}$ is a symmetric point with respect to points $A,B,C,D,E$. To see this, let complex number at points $\{A,B,C,D,E\}$ be $\{a,b,c,d,e\}$. Thus, $H_{ABC}:a+b+c$ and so, $$G_{DEH_{ABC}}: \frac{a+b+c+d+e}{3}$$which is symmetric with respect to points $\{A,B,C,D,E\}$. As this point is symmetric, we call it $P$. Thus, $H_{DEA}-P-M_{BC}$ are collinear. Also, as $A-G_{ABC}-M_{BC}$ are collinear as well, thus, using the fact that centroid divides medial line in $2:1$ ratio, Thales' theorem gives us $G_{ABC}P\|AH_{ADE}\implies G_{ABC}P\perp DE$ and thus, by symmetry, we get all the lines $\ell_{whatever}$ will concur at $P$.$\tag*{$\blacksquare$}$
PS: Really cute problem :)
