# Prove the lines are concurrent (using vectors)

Problem: Let $A$, $B$, $C$, $D$, and $E$ be points on a circle. For any three points, we draw the line going through the centroid of the triangle formed by these three points that is perpendicular to the line passing through the other two points. (For example, we draw the line going through the centroid of $\triangle BDE$ that is perpendicular to $\overline{AC}$.) In this way, we draw a total of $\binom{5}{3} = 10$ lines. Show that all 10 lines pass through the same point.

My work so far: I know the centroid of $\triangle ABC$ can be expressed as $\overrightarrow{G_1}=\frac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3}$. I believe the equation for the line that through from this point perpendicular to $\overrightarrow{DE}$ can be expressed as $\overrightarrow{l}=(\text{proj}_{\overrightarrow{DE}}\overrightarrow{G_1}-\overrightarrow{G_1})t+G_1$. Is this correct? If so, how can I use this to prove all 10 lines are congruent? Thank you in advance.

Note: By "$\text{proj}_{\overrightarrow{DE}}\overrightarrow{G_1}$" I mean the projection of $\overrightarrow{G_1}$ onto $\overrightarrow{DE}$.

• Some geometrical problems, like this one, without a diagram can be pretty tough to follow. – Timbuc Nov 29 '14 at 2:07
• Sorry, a diagram was not provided with the problem and I do not know how to make one. – Andrew Wang Nov 29 '14 at 2:21
• Perhaps this can help meta.matheducators.stackexchange.com/questions/93/… I just am afraid that without a diagram not many will have the patience to read, understand and work out the details of your question. – Timbuc Nov 29 '14 at 2:26
• I withdrew my comment that I had a counterexample. – LouisB Nov 29 '14 at 5:21

claim: all ten lines go through the point $\frac{1}{3}(a+b+c+d+e).$
proof: i will use complex numbers instead of vectors. chooses a coordinate system so that the circle has unit radius and the point $D$ is at $1.$ let the complex numbers $a, b, c=e^{2i\gamma}, d = 1, e = e^{2i\epsilon}.$ the line through the center of the triangle $ABC$ and orthogonal to line $DE$ has the parametric form $$\mbox{ line1:} \frac{1}{3}(a+b+c) +\frac{s}{3}e^{i\epsilon} , s \mbox{ real}$$ in the same way the line through the center of the triangle $ABE$ orthogonal to $DC$ has the parametric form $$\mbox{ line2:} \frac{1}{3}(a+b+e) + \frac{t}{3}e^{i\gamma}, t \mbox{ real}$$
solving the two equation and their complex conjugates, we find that $$s =e^{i\epsilon} + e^{-i\epsilon} \mbox{ and } t = e^{i\gamma} + e^{-i\gamma}$$ and the common points of intersection as claimed.
The above is a diagram of 3 of the 10 triangles. For example $$C_1$$ is the centroid of the blue triangle ABE with the blue line perpendicular to line CD. As shown, the 3 centroid lines all pass through a single point. I think your approach is promising. You just need to intersect 2 of the centroid lines and show this intersection is independent of the particular triangles.
• off by a factor $5/3.$ see my answer. – abel Nov 30 '14 at 2:54