May 7th: Some Computations

May 8—2025

Claim: Given $a_1, a_2, a_3 \in \mathbb{C}$ , these 3 values lie on the vertices of an equilateral triangle $\Leftrightarrow$ $a_1^2 + a_2^2 + a_3^2 = a_1a_2 + a_2a_3 + a_1a_3$ .

Proof:

Key: $a_1^2 + a_2^2 + a_3^2 = a_1a_2 + a_2a_3 + a_1a_3$ is invariant under scaling, rotation, and translation. Both scaling and rotation are represented by multiplication of a complex number. Since the equation is homogeneous of degree 2, it follows that it is invariant under multiplication by a constant. Translation is the addition of a complex number. If $(\tilde{a_1}, \tilde{a_2}, \tilde{a_3})$ is a solution to the above, then it follows that for any $\alpha \in \mathbb{C}$ , we have:

$(\tilde{a_1}+ \alpha)^2 + (\tilde{a_2} + \alpha)^2 + (\tilde{a_3} + \alpha)^2 - ((\tilde{a_1} + \alpha)(\tilde{a_2} + \alpha) + (\tilde{a_2} + \alpha)(\tilde{a_3} + \alpha) + (\tilde{a_1} + \alpha)(\tilde{a_3}+ \alpha)$

$= (\tilde{a_1}^2 + 2\tilde{a_1}\alpha + \alpha^2) + (\tilde{a_2}^2 + 2\tilde{a_2}\alpha + \alpha^2) + (\tilde{a_3}^2 + 2a_3\alpha + \alpha^2) - (\tilde{a_1}\tilde{a_2} + \tilde{a_1}\alpha + a_2\alpha + \alpha^2) - (\tilde{a_2}a_3 + \tilde{a_2}\alpha + \tilde{a_3}\alpha + \alpha^2) - (\tilde{a_3}\tilde{a_1} + \tilde{a_3}\alpha + \tilde{a_1}\alpha + \alpha^2)$

$= \tilde{a_1}^2 + \tilde{a_2}^2 + a_3^2 - \tilde{a_1}\tilde{a_2} - \tilde{a_2}\tilde{a_3} - \tilde{a_3}\tilde{a_1}$

$= 0$ .

Hence, $(\tilde{a_1}+\alpha, \tilde{a_2}+\alpha + \tilde{a_3}+\alpha)$ is also a solution.

What does this mean? Given $(a_1, a_2, a_3)$ to be the 3 vertices of an equilateral triangle, we can then apply a sequence of rotation, scaling, and translation to transform it to the equilateral triangle with vertices $(0, 1, \frac{1}{2}+i\frac{sqrt{3}}{2})$ . It suffices to show $(0, 1, \frac{1}{2}+i\frac{sqrt{3}}{2})$ satisfies the above equation.

$1 + 0 + (\frac{1}{2}+i\frac{sqrt{3}}{2})^2 - 1\cdot0 - 0\cdot(\frac{1}{2}+i\frac{sqrt{3}}{2}) - 1\cdot(\frac{1}{2}+i\frac{sqrt{3}}{2})$

$= 1 + (\frac{1}{2}+i\frac{sqrt{3}}{2})^2 - (\frac{1}{2}+i\frac{sqrt{3}}{2})$

$= 0$ , which is what we want.

Now, suppose $(\tilde{a_1}, \tilde{a_2}, \tilde{a_3})$ satisfies the equation. We can apply a sequence of rotation, scaling, and translation to transform $(\tilde{a_1}, \tilde{a_2})$ to $(0, 1)$ . It now suffices to show that $\tilde{a_3}$ must be mapped to $\frac{1}{2}+i\frac{sqrt{3}}{2}$ or $\frac{1}{2}-i\frac{sqrt{3}}{2}$ ; this would mean the original triangle, which is similar to its image, must also be an equilateral triangle. Showing $\tilde{a_3}$ being mapped to $\frac{1}{2}+i\frac{sqrt{3}}{2}$ or $\frac{1}{2}-i\frac{sqrt{3}}{2}$ is equivalent to showing that $\frac{1}{2}+i\frac{sqrt{3}}{2}$ or $\frac{1}{2}-i\frac{sqrt{3}}{2}$ are the only solutions to: