Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've read that if the complex numbers $a_1$, $a_2$ and $a_3$ are the vertices of a triangle in the complex plane such that $$ a_1^2+a_2^2+a_3^2=a_1a_2+a_2a_3+a_1a_3 $$ then the vertices are actually those of an equilateral triangle.

I tried to see why this by first shifting the vertex $a_1$ to the origin, and considering the triangle with vertices $0$, $a_2-a_1$ and $a_3-a_1$. The given equation holds if and only if $$ (a_2-a_1)^2+(a_3-a_1)^2=(a_2-a_1)(a_3-a_1) $$ which is equivalent to saying $$ \frac{a_2-a_1}{a_3-a_1}+\frac{a_3-a_1}{a_2-a_1}=1. $$ Letting $\frac{a_2-a_1}{a_3-a_1}=x$, this gives the quadratic $x+\frac{1}{x}=1$, and so I solve for $x$ to be $$ \frac{a_2-a_1}{a_3-a_1}=x=\frac{1\pm\sqrt{-3}}{2}=e^{\pm i\pi/3}.$$

I'm not sure where I'm going now. How can I conclude that such a relation implies the vertices form an equilateral triangle? Thank you.

share|cite|improve this question
I take it all your $a_i$ are complex numbers, considered as points in the plane. – Gerry Myerson Jan 25 '12 at 22:39
I was a bit confused at first. Only at the end does it appear that these are intended to be complex numbers. – Michael Hardy Jan 25 '12 at 22:44
up vote 2 down vote accepted

If $\frac{a_2 - a_1}{a_3 - a_1} = e^{\pm i \pi/3}$ we have $|a_2 - a_1| = |a_3 - a_1|$. Moreover, $a_3 - a_2 = (a_3 - a_1) + {a_1 - a_2} = (1 - e^{\pm i \pi/3}) (a_3 - a_1)$. But $1 - e^{\pm i \pi/3} = 1/2 \mp i \sqrt{3}/2 = e^{\mp i \pi/3}$ so also $|a_3 - a_2| = |a_3 - a_1|$. Thus all three sides of the triangle have the same length.

share|cite|improve this answer
Thanks everybody. – Hogarth Jan 25 '12 at 23:37

The sides $(a_1,a_2)$ and $(a_1,a_3)$ have the same length (because $|x|=1$) and their nonoriented angle is $\pi/3$ (because $\arg(x)=\pm\pi/3$). Since the triangle $a_1a_2a_3$ is isosceles at vertex $a_1$, its angles at vertices $a_2$ and $a_3$ are equal. The angle at $a_1$ is $\pi/3$ and the sum of the three angles is $\pi$. Hence the angle at each vertex is $\pi/3$. QED.

share|cite|improve this answer

You're very close to the solution. Remember that you shifted everything to the origin, so that $a_2 - a_1$ is the location of the second vertice and $a_3 - a_1$ the location of the third vertice. Your relation says that $$ a_2 - a_1 = e^{\pm i \pi /3} (a_3 - a_1) $$ hence they are at the same distance of your origin (since if you compute modulus in these cases they are the same on both sides) and the angle between them is $\pi/3$, which means the triangle is equilateral (by doing your work with $a_2$ and $a_3$ you actually get the 2 other angles).

In other words, you did all the work, you only needed to interpret what you've found correctly.

Hope that helps,

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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