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P, Q represent complex numbers $\displaystyle \alpha ,\beta $ respectively where O is the origin and O, P, Q are not collinear. In triangle OPQ, the median from O to the midpoint M of PQ meets the median from Q to the midpoint N of OP in the point R, where R represents the complex number z.
Show that $\displaystyle z=\frac{1}{3}(\alpha +\beta )$ and deduce that R is the point of concurrence of the three medians of triangle OPQ
We have $\displaystyle O(0), \ P(\alpha), \ Q(\beta), \ M\left(\frac{\alpha+\beta}{2}\right), \ N\left(\frac{\alpha}{2}\right), \ R(z)$.
Let $\displaystyle \frac{NR}{RQ}=k, \ \frac{MR}{RO}=k_1$. Then
$\displaystyle z=\frac{z_N+kz_Q}{1+k}=\frac{\alpha}{2(1+k)}+\frac {\beta k}{1+k}$
and $\displaystyle z=\frac{z_M+k_1z_O}{1+k_1}=\frac{\alpha}{2(1+k_1)} +\frac{\beta}{2(1+k_1)}$
Then $\displaystyle \left\{\begin{array}{ll}\frac{1}{2(1+k)}=\frac{1}{ 2(1+k_1)}\\\frac{k}{1+k}=\frac{1}{2(1+k_1)}\end{ar ray}\right.$
$\displaystyle \Rightarrow k=k_1=\frac{1}{2}\Rightarrow z=\frac{1}{3}(\alpha+\beta)$.
