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Math Help - Napoleon's Theorem & Vectors

  1. #1
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    Napoleon's Theorem & Vectors

    Napoleon's theorem is easy enough to prove using a geometric approach. What I am unable to do is complete a proof using vector methods. For the purposes of discussion, let OAB be an arbitrary triangle. Let H, F and G be the centroids of the equilaterial triangles OAC, ABE and OBD constructed on the sides OA, AB and OB respectively. One must prove that triangle HFG is equilateral.

    I have pursued an approach involving scalar (dot) products but seem to run into a loop and a fair bit of messy algebra. Any thoughts welcome.
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  2. #2
    Senior Member TriKri's Avatar
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    Quote Originally Posted by Ian Moore View Post
    Napoleon's theorem is easy enough to prove using a geometric approach. What I am unable to do is complete a proof using vector methods. For the purposes of discussion, let OAB be an arbitrary triangle. Let H, F and G be the centroids of the equilaterial triangles OAC, ABE and OBD constructed on the sides OA, AB and OB respectively. One must prove that triangle HFG is equilateral.

    I have pursued an approach involving scalar (dot) products but seem to run into a loop and a fair bit of messy algebra. Any thoughts welcome.
    We have:

    H=\frac{O+A}{2}+\frac{F\left( \overrightarrow{OA} \right)\cdot Sin\left( 60^{\circ } \right)}{3}, where F is the linear function that turns a vector 90 degrees counterclockwise.

    F=\frac{A+B}{2}+\frac{F\left( \overrightarrow{AB} \right)\cdot Sin\left( 60^{\circ } \right)}{3}

    G=\frac{B+O}{2}+\frac{F\left( \overrightarrow{BO} \right)\cdot Sin\left( 60^{\circ } \right)}{3}

    I guess what you wantto prove now is that \left|\overrightarrow{HF}\right|=\left|\overrighta  rrow{FG}\right|=\left|\overrightarrow{GH}\right|, since an equilaterial has all sides equally long?
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  3. #3
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    Quote Originally Posted by TriKri View Post
    We have:

    H=\frac{O+A}{2}+\frac{F\left( \overrightarrow{OA} \right)\cdot Sin\left( 60^{\circ } \right)}{3}, where F is the linear function that turns a vector 90 degrees counterclockwise.

    F=\frac{A+B}{2}+\frac{F\left( \overrightarrow{AB} \right)\cdot Sin\left( 60^{\circ } \right)}{3}

    G=\frac{B+O}{2}+\frac{F\left( \overrightarrow{BO} \right)\cdot Sin\left( 60^{\circ } \right)}{3}

    I guess what you wantto prove now is that \left|\overrightarrow{HF}\right|=\left|\overrighta  rrow{FG}\right|=\left|\overrightarrow{GH}\right|, since an equilaterial has all sides equally long?
    Thanks for your response. However, the notation is somewhat unclear to me. What do you mean by the point H equals ... ?
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