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Thread: Complex Numbers Q

  1. #1
    Junior Member
    Sep 2009

    Complex Numbers Q

    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
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  2. #2
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    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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