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Sum of arbitrary vertex to midpoint vectors

  1. May 27, 2014 #1
    I was looking at a homework question posted here requiring proof that the vectors from the vertices of a triangle to the midpoint of the opposite edge sum to zero, and it struck me that there is a more general property:

    Consider a set of points, [itex]\{A_0, A_1, \ldots A_n\}[/itex]. The midpoint of ##A_k## and ##A_{k+1}## is denoted by ##a_k##, with ##a_n## as the midpoint of ##A_n## and ##A_0##.

    Now form a set of ##n## vectors defined by [itex]\stackrel{\rightarrow}{A_ra_s}[/itex] such that each vertex and each midpoint is used once by a vector in the set.

    Now that set of vectors will sum to zero.

    --------------------------

    Sketch of messy proof
    First step - the midpoints are a distraction. ##\stackrel{\rightarrow}{A_ra_s} = \stackrel{\rightarrow}{A_rA_s} + \stackrel{\rightarrow}{A_sa_s} = \stackrel{\rightarrow}{A_rA_s }+ \frac{1}2\stackrel{\rightarrow}{A_sA_{s+1}}##, so the vector sum contains a scaled loop from the midpoint steps which sums to zero.

    Next step - The remaining vectors form a set of closed loops which also sum to zero.

    Improvements on this welcome...
     
  2. jcsd
  3. May 27, 2014 #2

    disregardthat

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    The set of points A_k and a_k may be considered vectors from origon. So the n vectors you describe are on the form a_s-A_r, where s and r ranges from 0 to n with r depending on s via some bijection (<-- I assume this is what you mean).

    a_k may be written as [itex]\frac{1}{2}(A_k+A_{k+1})[/itex], where indices are considered modulo n+1.

    By writing out entire the sum, we get exactly 0.
     
  4. May 27, 2014 #3
    Nice and neat. Thanks.
     
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