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Finding the centroid of a triangle using complex numbers

  1. Apr 11, 2015 #1
    Hi all,

    I'm preparing for a deferred exam this semester after falling ill last year. Just looking over my course notes and have a question. I understand how this works in the big picture scheme. What I don't understand however is how my instructor simplified the original equation.

    1. The problem statement, all variables and given/known data

    Step 1) (1-t)z1 + t*(z2+z3/2) = (1-s)z1 + s(z2+z3/2)

    Step 2) Simplifies to;
    (2-s)z1+(t-2+2s)z2+(t-s)z3 = 0

    Since z1, z2 and z3 aren't collinear, their coefficients in this equation must be zero. Therefore we have;
    a) 2-s-2t = 0
    b) t-2+2s = 0
    c) t-s = 0

    Then we readily find t = s = 2/3

    Which is then substituted into the original equation for medians to find that it equals (z1+z2+z3)/3

    2. Relevant equations
    I understand how this works in the big picture scheme. What I don't understand however is how my instructor simplified the original equation step 1 to step 2.


    3. The attempt at a solution
    My attempt from equating the equations;
    (1-t)z1 + t*(z2+z3/2) = (1-s)z1 + s(z2+z3/2)

    0 = (2-s)z1+(t-2+2s)z2+(t-s)z3
    = (1-s)z1 + s(z1+z3/2) + s(z2+z3/2) - (1-t)z1 - s(z2+z3/2)
    = z1 - s*z1 - z1 + t*z1 - s*(z2/2) - s*(z3/2) + s*(z2/2) + s*(z3/2)
    = t*z1 - s*z1

    Therefore t*z1 = s*z1 and divide both sides by z1 t = s

    I feel this comes out slightly like my instructors, but its faulty somewhere...
     
  2. jcsd
  3. Apr 11, 2015 #2

    Simon Bridge

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    Why not expand the brackets in the first equation, multiply through by 2, then group like terms?
     
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