Complex Numbers Q

**vuze88** · September 4th 2009, 12:08 AM

P, Q represent complex numbers $\alpha ,\beta$ 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 $z=\frac{1}{3}(\alpha +\beta )$ and deduce that R is the point of concurrence of the three medians of triangle OPQ

**red_dog** · September 4th 2009, 02:21 AM

We have $O(0), \ P(\alpha), \ Q(\beta), \ M\left(\frac{\alpha+\beta}{2}\right), \ N\left(\frac{\alpha}{2}\right), \ R(z)$ .

Let $\frac{NR}{RQ}=k, \ \frac{MR}{RO}=k_1$ . Then

$z=\frac{z_N+kz_Q}{1+k}=\frac{\alpha}{2(1+k)}+\frac {\beta k}{1+k}$

and $z=\frac{z_M+k_1z_O}{1+k_1}=\frac{\alpha}{2(1+k_1)} +\frac{\beta}{2(1+k_1)}$

Then $\left\{\begin{array}{ll}\frac{1}{2(1+k)}=\frac{1}{ 2(1+k_1)}\\\frac{k}{1+k}=\frac{1}{2(1+k_1)}\end{ar ray}\right.$

$\Rightarrow k=k_1=\frac{1}{2}\Rightarrow z=\frac{1}{3}(\alpha+\beta)$ .

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