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Let $\Delta ABC$ be a triangle. The vertices $A$, $B$ and $C$ are denoted by the complex numbers $\alpha$, $\beta$ and $\gamma$. Let $\omega=e^{\frac{2\pi}{3}i}$.

Prove that $\alpha+\omega\beta+\omega^2 \gamma =0$ or $\alpha+\omega\gamma+\omega^2\beta=0$ iff $\Delta ABC$ is equilateral.

In previous steps, we need to prove that $1-e^{\frac{\pi}{3}i}+e^{\frac{2\pi}{3}i}=0$ and $1+\omega+\omega^2=0$. I know why this is true, but I don't see the connection to the problem..


marked as duplicate by Martin R, Semiclassical, Joey Zou, JMP, Claude Leibovici Sep 6 '16 at 5:35

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    $\begingroup$ Asked several times ! See as well (math.stackexchange.com/q/481133) with a solution that displays a figure. $\endgroup$ – Jean Marie Sep 5 '16 at 18:46
  • $\begingroup$ It could be a coincidence but this question literally is in the TU Eindhoven HW for linear algebra, in this format. $\endgroup$ – Wesley Strik Sep 6 '18 at 21:26
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    $\begingroup$ @WesleyGroupshaveFeelingsToo slimme gast ben jij $\endgroup$ – Heinz Doofenschmirtz Oct 3 '18 at 13:01

Assuming $\alpha,\beta,\gamma$ are anticlockwise, then

The triangle is equilateral $$\iff \gamma-\beta=\omega(\beta-\alpha)$$

Multiplying by $\omega^2$ and rearranging, noting that $\omega^3=1$, gives $$\alpha+(-1-\omega^2)\beta+\omega^2\gamma=0\iff\alpha+\omega\beta+\omega^2\gamma=0$$

We can conclude the alternative expression when they are clockwise simply by interchanging $\beta$ and $\gamma$.


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