Consider an equilateral triangle as the one given in the figure below:
Assume $$\vec{E_1},\vec{E_2},\vec{E_3}$$ are vectors (for instance complex numbers in the complex plane) originating from the centroid (N) and pointing each one to their respective angle (each vector is parallel to a median).
For an equilateral triangle, the property (property 1):
$$\vec{E_1}+\vec{E_2}+\vec{E_3}=0 $$ holds true.
Now consider a generic triangle where the three vectors still originate at the centroid (N). My question is: is the above property (property 1) still true for a generic triangle?
Additional information: this question is related to an electrotechnical problem and to a question I originally asked on Electronics SE. As far as my understanding goes, this geometrical property is the key to answering the electrotechnical question, therefore I thought it could be appropriate to ask it here. If you wish to know more about the original question please let me know.
Best Answer
Here's a proof: By a preexisting formula, if the coordinates of the vertices of a triangle are $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, then the centroid is the point $$(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$$ If this is so, then the horizontal parts of each of the vectors are $\frac{-2x_1+x_2+x_3}{3}$, $\frac{x_1-2x_2+x_3}{3}$, and $\frac{x_1+x_2-2x_3}{3}$, and the vertical parts are $\frac{-2y_1+y_2+y_3}{3}$, $\frac{y_1-2y_2+y_3}{3}$, and $\frac{y_1+y_2-2y_3}{3}$. To sum the vectors, sum the horizontal and vertical parts. The horizontal part will be $$\frac{-2x_1+x_2+x_3}{3}+\frac{x_1-2x_2+x_3}{3}\frac{x_1+x_2-2x_3}{3}$$ Which is $0$, and the vertical part will be $$\frac{-2y_1+y_2+y_3}{3}+\frac{y_1-2y_2+y_3}{3}\frac{y_1+y_2-2y_3}{3}$$ Which is also $0$. All $3$ vectors cancel out. QED.