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Pentagon $ABCDE$ is inscribed in a circle. For any edge of $ABCDE$, we can draw the line perpendicular to that edge that contains the centroid of the remaining three vertices. Show that these 5 lines are concurrent.

We just finished a unit on vectors in my Pre-Calculus class but I have no idea on how to do this final problem. Any help would be greatly appreciated!

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  • $\begingroup$ Have you seen complex numbers with their geometrical interpretation, and solve geometrical problems in that way ? $\endgroup$ – Jean Marie Sep 20 '17 at 6:22
  • $\begingroup$ I have added a picture. $\endgroup$ – Jean Marie Sep 20 '17 at 6:40
  • $\begingroup$ Yes I have. Thank you! $\endgroup$ – Archie Sep 21 '17 at 3:34
  • $\begingroup$ @JeanMarie I don't understand a thing about your picture. What are $G_1$, $G_2$, $I$, what are the lines? $\endgroup$ – JiK Sep 10 '18 at 13:06
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    $\begingroup$ @JiK Take the case of $G_1$, which is the centroid of triangle $ABE$ : the building lines for $G_1$ are thus the median lines in this triangle. This point $G_1$ being built, we can draw through it the perpendicular line to the segment made by the two remaining points $C, D$. Similar construction for $G_2$. $\endgroup$ – Jean Marie Sep 12 '18 at 5:47
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WOLOG, assume the pentagon is inscribed in a circle of radius $R$ centered at origin $O$.
Let $p_1, p_2, p_3, p_4, p_5$ be the vectors correspond to vertices $A,B,C,D,E$ respectively. We have

$$|p_1|^2 = |p_2|^2 = |p_3|^2 = |p_4|^2 = |p_5|^2 = R^2$$

Let $G$ be the vertex centroid of the pentagon and $P$ be a point at the five-third mark on the ray $OG$. If $p$ is the corresponding vector, we have $$p = \frac53 \left(\frac{p_1 + p_2 + p_3 + p_4 + p_5}{5}\right) = \frac{p_1 + p_2 + p_3 + p_4 + p_5}{3}$$ Let $[5] = \{ 1, 2, 3, 4, 5 \}$. For any distinct $i, j \in [5]$, let $\ell, m, n$ be the rest of indices from $[5]$.
i.e. $[5]$ is a disjoint union of $\{ i, j \}$ and $\{ \ell, m, n \}$. We have

$$(p_i - p_j) \cdot \left( p - \frac{p_\ell + p_m + p_n}{3}\right) = \frac{(p_i - p_j) \cdot (p_i + p_j)}{3} = \frac{|p_i|^2 - |p_j|^2}{3} = 0$$

This is the equation for point $p$ lying on a line passing through $\frac{p_\ell + p_m + p_n}{3}$ (the centroid of triangle with vertices $p_\ell$, $p_m$, $p_n$ ) perpendicular to line $p_i p_j$.

Substitute $p_i,p_j$ by $p_1,p_2$, we find $P$ lie on a line perpendicular to $AB$ which passes through the centroid of triangle $CDE$. Same thing happens to other sides of the pentagon. As a result, the $5$ lines mentioned in question are concurrent and meet at this specific point $P$.

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  • $\begingroup$ [+1] Remarkable answer, once more! It was not evident to characterize point P as you have done, and then to use orthogonality in an adequate way.. $\endgroup$ – Jean Marie Sep 21 '17 at 4:24

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