1
$\begingroup$

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.

$\endgroup$

marked as duplicate by dxiv, Daniel W. Farlow, Trevor Gunn, Lord Shark the Unknown, jvdhooft Jul 6 '17 at 6:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $$\dfrac{z_2-z_1}{z_3-z_2}=e^{i\pi/3}$$ $\endgroup$ – lab bhattacharjee Jul 5 '17 at 13:31
  • $\begingroup$ I used that to show the forward implication. How would I prove that this equality holds from the assumption of the equality? $\endgroup$ – Twenty-six colours Jul 5 '17 at 13:31
  • $\begingroup$ Btw, take a look of your forward proof, I am quite sure it could go the other way round and prove the reverse implication. $\endgroup$ – Sawarnik Jul 5 '17 at 19:58
  • $\begingroup$ 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). $\endgroup$ – Twenty-six colours Jul 7 '17 at 11:06
4
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.