Do Complex Plane Medians Intersect at Triangle Centroid?

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In summary, POTW #346 is a weekly challenge presented by the American Mathematical Society to promote critical thinking and problem-solving skills in mathematics. It was posted on May 9, 2019, and presents a unique and challenging problem that requires creative thinking and mathematical concepts to solve. The purpose of solving POTW #346 is to develop critical thinking skills, practice applying mathematical concepts, and improve problem-solving strategies. The solution to POTW #346 is a mathematical concept or formula that can be used to solve the given problem, and it is important to work through the problem independently or with the help of others. To approach solving POTW #346, one should carefully read and understand the problem statement, identify key information and constraints, brainstorm possible
  • #1
Euge
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Here is this week's POTW, from Chris L T521:

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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 $\frac{1}{3}(z_1+z_2+z_3)$.
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  • #2
No one answered this week's problem. You can read Chris L T521's solution below.
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, $C=\ell_1(\frac{2}{3})=\ell_2(\frac{2}{3})=\ell_3(\frac{2}{3})=\frac{1}{3}(z_1+z_2+z_3)$.$\hspace{0.25in}\blacksquare$​
 

FAQ: Do Complex Plane Medians Intersect at Triangle Centroid?

What is the problem being addressed in POTW #346 - May 09, 2019?

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