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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.

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    $\begingroup$ I take it all your $a_i$ are complex numbers, considered as points in the plane. $\endgroup$ – Gerry Myerson Jan 25 '12 at 22:39
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    $\begingroup$ I was a bit confused at first. Only at the end does it appear that these are intended to be complex numbers. $\endgroup$ – Michael Hardy Jan 25 '12 at 22:44
  • $\begingroup$ See also the second half of this blog post on how to solve this problem without explicitly shifting the triangle to have one vertex at the origin: mathematicalgemstones.com/gemstones/opal/… $\endgroup$ – Jonathan Rayner May 14 '18 at 7:34
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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.

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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.

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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,

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