MHB Do Complex Plane Medians Intersect at Triangle Centroid?

[Euge]

Here is this week's POTW, from Chris L T521:

-----
Prove that if $z_1,z_2,z_3$ are noncollinear points in the complex plane, then the medians of the triangle with vertices $z_1,z_2,z_3$ intersect at the point $\frac{1}{3}(z_1+z_2+z_3)$.
-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

No one answered this week's problem. You can read Chris L T521's solution below.

Spoiler

Suppose that $z_1,z_2,z_3$ are the vertices of the following triangle in the complex plane.

\begin{tikzpicture}
\draw[->] (-.5,0) -- (5.5,0) node

{$\mathrm{Re}\,z$};
\draw[->] (0,-.5) -- (0,5.5) node[above]{$\mathrm{Im}\,z$};
\draw (1,1) -- (5,2.5) -- (2,5) -- cycle;
\fill (1,1) circle (1.25pt) node[below]{$z_1$};
\fill (5,2.5) circle (1.25pt) node

{$z_2$};
\fill (2,5) circle (1.25pt) node[above]{$z_3$};
\draw (1,1) -- (3.5,3.75);
\draw (3,1.75) -- (2,5);
\draw (1.5,3) -- (5,2.5);
\fill (3,1.75) circle (1.25pt) node[below]{$m_1$};
\fill (3.5,3.75) circle (1.25pt) node

{$m_2$};
\fill (1.5,3) circle (1.25pt) node

{$m_3$};
\fill (8/3,17/6) circle (1.25pt) node[left=.1cm,below=.15cm]{$C$};
\end{tikzpicture}
Figure 1: A triangle with vertices $z_1,z_2,z_3$ in the complex plane. Note that $m_1,m_2,m_3$ are the midpoints of each side and $C$ is the centroid (intersection of the medians).

Using Figure 1, we find that $m_1=\frac{1}{2}(z_1+z_2)$, $m_2=\frac{1}{2}(z_2+z_3)$ and $m_3=\frac{1}{2}(z_1+z_3)$. The line segment passing through $z_1$ and $m_2$ is given by

\[\ell_1(t) = (1-t)z_1+tm_1=(1-t)z_1+t\frac{z_2+z_3}{2};\quad t\in[0,1].\]

Similarly, the line segment passing through $z_2$ and $m_3$ is given by

\[\ell_2(t) = (1-t)z_2+tm_3 = (1-t)z_2+t\frac{z_1+z_3}{2};\quad t\in[0,1],\]

and the line segment passing through $z_3$ and $m_1$ is given by

\[\ell_3(t) = (1-t)z_3+tm_1 = (1-t)z_3+t\frac{z_1+z_2}{2};\quad t\in[0,1].\]

Note that

\[\begin{aligned}\ell_1(t)=\ell_2(t) &\implies (1-t)z_1+t\frac{z_2+z_3}{2}=(1-t)z_2+t\frac{z_1+z_3}{2}\\ &\implies 2z_1-2tz_1+tz_2+tz_3=2z_2-2tz_2+tz_1+tz_3\\&\implies (2-3t)z_1 = (2-3t)z_2\\ &\implies (2-3t)(z_1-z_2)=0\\&\implies 2-3t=0\qquad(\text{since $z_1\neq z_2$})\\ &\implies t=\frac{2}{3}.\end{aligned}\]

Similar calculations show that $\ell_2(t)=\ell_3(t)$ and $\ell_1(t)=\ell_3(t)$ when $t=\frac{2}{3}$. Hence, all medians intersect at the same point $C$ when $t=\frac{2}{3}$; that is, $C=\ell_1(\frac{2}{3})=\ell_2(\frac{2}{3})=\ell_3(\frac{2}{3})=\frac{1}{3}(z_1+z_2+z_3)$.$\hspace{0.25in}\blacksquare$​

​