Show that 10 lines pass through the same point.

Let $$A$$, $$B$$, $$C$$, $$D$$, and $$E$$ be five 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 $$AC$$.) In this way, we draw a total of $$\binom{5}{3}= 10$$ lines. Show that all $$10$$ lines pass through the same point. (Here is the image for three lines. I took it from another post for the same question, that has answer only using complex numbers: Prove the lines are concurrent (using vectors)) I do think that solution should be done using vectors. However I do not feel very comfortable juggling with vectors. I tried to do it by placing the center of the circle at the origin and putting on of the points at $$(1, 0)$$ and tried to work with Cartesian coordinates. The equations didn't work out well, so I gave up on this way. Some ideas how to use vectors? I know position vector for the centroid, but how to proceed from there onwards?

By the way how does one proof that for centroid $$G$$ in triangle $$ABC$$: $$G=\dfrac {\vec A+\vec B+\vec C}3$$I know it is a famous result, but I do not know the proof.

• There is not substantial difference between vectors and complex numbers (apart from the latter being easier to rotate, at most), so abel's proof in the linked question can be transposed into the language of vectors. But why to do so, since the proof with complex numbers is crystal clear? – Jack D'Aurizio Jun 21 '15 at 22:20
• To answer myself: because the answer without complex numbers is crystal clear as well. – Jack D'Aurizio Jun 21 '15 at 23:45

Assume $A,B,C,D,E\in S^1$, with $O$ (the origin) being the circumcenter of $ABCDE$.

We want to prove that $\color{red}{\frac{A+B+C+D+E}{3}}$ in the point of concurrency of our ten lines, or: $$\frac{A+B+C+D+E}{3}-\frac{A+B+C}{3}\perp (D-E)$$ that is equivalent to $\frac{D+E}{2}\perp (D-E)$, that is trivial since the line connecting the circumcenter $O$ of $ABCDE$ with the midpoint of $DE$ is the perpendicular bisector of $DE$, since $OD=OE$.

About the centroid being the arithmetic mean of the vertices, you simply have to check that: $$A,\quad G=\frac{A+B+C}{3},\quad \frac{B+C}{2}$$ are collinear to prove that $G$ lies on the median from $A$. That is also trivial since: $$G = \frac{1}{3}\cdot A+\frac{2}{3}\left(\frac{B+C}{2}\right)$$ hence $G$ is a convex combination of the vertex $A$ and the midpoint of the opposite side.

In both cases, you finish by considering every possible permutation of the letters.

• If the location of the concurrency wasn't known in advance. How can one go about finding it? Thanks! – Ekushkebi Jun 22 '15 at 0:07
• @Ekushkebi: sometimes we have got to guess. Given the high symmetry of the configuration, a point of the form $\lambda(A+B+C+D+E)$ is the best candidate. We just have to notice that with the choice $\lambda=\frac{1}{3}$ the orthogonality condition boils down to something really trivial, et voilà. – Jack D'Aurizio Jun 22 '15 at 1:18
• @g.kov: I forgot to mention that the origin is took in the circumcenter of $ABCDE$, but that is also the most trivial choice. – Jack D'Aurizio Jun 22 '15 at 11:03 The centroid of any three points, for example, $B,C,E$ is obviously a point $U=\frac{1}{3}(B+C+E)$, which is a convex combination of points $B,C,E$. The vector that is perpendicular to the line, passing through the other two points ($A$ and $D$) is simply $\vec{OQ}=\vec{OA}+\vec{AQ}=\vec{OD}+\vec{DQ}=\vec{OA}+\vec{OD}$. Consider the point \begin{align} X&=U+\frac{1}{3}\vec{OQ} \\ &=\frac{1}{3}(B+C+E)+\frac{1}{3}(\vec{OA}+\vec{AQ}) \\ &=\frac{1}{3}(B+C+E)+\frac{1}{3}(\vec{OA}+\vec{OD}) \\ &=\frac{1}{3}(B+C+E)+\frac{1}{3}(A-O)+\frac{1}{3}(D-O) \\ &=\frac{1}{3}(A+B+C+D+E-2O), \end{align} For any combination of three points, point $X$ can be considered as a centroid of the three selected points plus one third of the sum of two vectors from the center of the circle $O$ to the other two points, and the line through the centroid and $X$ is perpendicular to the line through of the other two points. Thus all 10 lines pass through the same point $X$.

Another correct form of expression for the point of intersection $X$ would be

\begin{align} X&=O+\frac{5}{3}\vec{OV}, \\ \vec{OV}&=\frac{1}{5}(\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}+\vec{OE}) =M-O, \\ M&=\frac{1}{5}(A+B+C+D+E). \end{align} where $\vec{OV}$ is an average vector from the center of the circle to the points on the circle, $M$ is the centroid point of all five points.

This form suggests a straightforward generalization to $n$ points $P_1,\dots,P_n$ on a circle, where instead of the centroid of three points, centroid of $n-2$ points is used:

\begin{align} X&=O+\frac{n}{n-2}\vec{OV}, \\ \vec{OV}&=\frac{1}{n} \sum_{k=1}^{n} \vec{OP_k} =M-O, \\ M&=\frac{1}{n}\sum_{k=1}^{n}P_k. \end{align}