# 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.

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 :)

• Beautiful solution! I too attempted but unsuccessfully. I should have tried harder. Thank you for your answer! – cosmo5 Sep 22 '20 at 14:54
• 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. – Math Lover Sep 23 '20 at 11:50
• @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 :) – Anand Sep 23 '20 at 14:05
• Yes ok I see now! – 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)

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).