# Equilateral triangle in complex plane [duplicate]

Prove that the points $a_1,a_2,a_3$ are vertices of an equilateral triangle if and only if $a_1^2+a_2^2+a_3^2=a_1a_2+a_2a_3+a_3a_1$.

I rewrite the equation as $2a_1^2+2a_2^2+2a_3^2-2a_1a_2-2a_2a_3-2a_3a_1=0$, which is $(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_1)^2=0$. This looks quite nice, but I'm not sure how to relate it to the equilateral triangle.

## marked as duplicate by Gerry Myerson, TZakrevskiy, dfeuer, Thomas Andrews, Stefan HansenSep 1 '13 at 8:37

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.

## 1 Answer

I'll prove the 'if' part, leaving the 'only if' part to you:

Denote the points as $A(a), B(b), C(c)$ (easier to type).

Now, rotating $\vec{AC}$ by $\pi/3$ gives you $\vec{AB}$. So:

$$\hat{AC}\cdot e^{i\pi/3} = \hat{AB}$$

Here, $|\vec{AC}| = | \vec{AB} |$. So:

$$\vec{AC}\cdot e^{i\pi/3} = \vec{AB}$$ $$\implies (c-a) \cdot e^{i\pi/3} = (b-a) \ldots...$$

Similarly:

$$(a-b) \cdot e^{i\pi/3} = (c-b) \ldots...$$

Divide equation $1$ by equation $2$:

$$\dfrac{c-a}{a-b} = \dfrac{b-a} {c-b}$$

Cross multiply and you are done! • Check the sign of rotation, not that it matters in the end. – Macavity Sep 1 '13 at 7:48
• It must be counter clockwise and it is. What's wrong? – Parth Thakkar Sep 1 '13 at 7:49
• How does this compare to the proofs given when the question was asked before? – Gerry Myerson Sep 1 '13 at 7:50
• I saw your comment only when I had almost finished the answer. And secondly, I think this is more clear. – Parth Thakkar Sep 1 '13 at 7:53