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: enter image description here

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!


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 :)

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    $\begingroup$ Beautiful solution! I too attempted but unsuccessfully. I should have tried harder. Thank you for your answer! $\endgroup$ – cosmo5 Sep 22 '20 at 14:54
  • $\begingroup$ How do you make use of the fact that it is inscribed in a circle? I see that you start your proof by writing that but not able to see where you use that. $\endgroup$ – Math Lover Sep 23 '20 at 11:50
  • $\begingroup$ @MathLover I used it when I said assume $A,B,C,D,E$ lie on the unit circle. Using this, I was able to comment that $H_{ABC}=a+b+c$. Hope that helps :) $\endgroup$ – Anand Sep 23 '20 at 14:05
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    $\begingroup$ Yes ok I see now! $\endgroup$ – Math Lover Sep 23 '20 at 14:08

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)


$\overline{p}-\frac{\overline{b} + \overline{c} + \overline{d}}{3} = n_2 (\overline{e}+\overline{a})$ ...(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).


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