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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..

Dr. Heinz Doofenshmirtz
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1 Answers1

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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 $\gamma$.

David Quinn
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