# Thread: Napoleon's Theorem & Vectors

1. ## 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.

2. Originally Posted by Ian Moore
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?

3. Originally Posted by TriKri
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 ... ?