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Vectors and medians of a triangle.

  1. Mar 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that the medians of a triangle (the lines joining the vertices of a triangle to the midpoints of the opposite sides) intersect at a point two thirds of the way from each vertex to the opposite sides.


    2. Relevant equations

    If you let the triangle be ABC and the midpoint of AB be Z, the midpoint of BC be X and the midpoint of AC be Y, then:

    x = c/2 +b/2
    y = a/2 + c/2
    z = a/2 + b/2


    3. The attempt at a solution
    Let the intersection of the medians be W (assuming that they intersect in one point)
    Using the above 3 equations, I know that I just need a fourth equation so that I can find a relation between a and x, b and y, and c and z. Then I'd use the formula that if p divides AB in the ratio m:n then p=(na+mb)/(m+n). You'd get two different expressions (eg in terms of a and x in one expression and in terms of b and y in the other expression) which have equal m, n values for 2 different lines. Then you know that since W is the only point lying on both lines, that it must divide them both in the ratio m:n. Similarly do it for the third one and you would know that W is the intersection of all three under the above conditions. I just need to find a fourth equation.
     
  2. jcsd
  3. Mar 8, 2009 #2

    tiny-tim

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    Hi wimma! :smile:
    ooh, so complicated! :cry:

    Stay on the bandwagon :wink:

    use your same method to find the two-thirds point of AX, and you get … ? :smile:
     
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