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Show that the triangle in the complex plane whose vertices are $z_1,z_2,z_3$ is equilateral if and only if $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_1 z_3$.

I showed the forward implication, that if those vertices formed an equilateral triangle, it implied the equality. I'm not sure how I would prove the reverse implication.

Twenty-six colours
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  • $$\dfrac{z_2-z_1}{z_3-z_2}=e^{i\pi/3}$$ – lab bhattacharjee Jul 05 '17 at 13:31
  • I used that to show the forward implication. How would I prove that this equality holds from the assumption of the equality? – Twenty-six colours Jul 05 '17 at 13:31
  • Btw, take a look of your forward proof, I am quite sure it could go the other way round and prove the reverse implication. – Sawarnik Jul 05 '17 at 19:58
  • I don't think it does, since it relies on the fact that $z_3 - z_2$ is a rotation of $z_2 - z_1$ by $\frac{\pi}{3}$ radians, to reverse the implication, I'd have to prove that $z_3 - z_2$ is a rotation of $z_2 - z_1$ by $\frac{\pi}{3}$ radians (which assumes the equilateral triangle fact). – Twenty-six colours Jul 07 '17 at 11:06

1 Answers1

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Hint. The equality is equivalent to $(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0$. Let $u=z_1-z_2$, $v=z_2 - z_3$, then $u^2+v^2+(u-v)^2=0$. Now we may view this as a quadratic equation in $u/v$ and solve it easily.

Cave Johnson
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