Given a field $\mathbb k$, an algebraic variety can be defined as the solution set of a system of polynomial equations in $\mathbb k^n$ for some $n\in\mathbb N$. That is, given a finite sequence of $m$ polynomials $F_i,$ we may define a variety $X\subset\mathbb k^n$ as the set of points $(x_1,\cdots,x_n)$ simultaneously satisfying Often, we consider varieties over $\mathbb k=\mathbb R$ or $\mathbb k = \mathbb C$.
Interestingly, the set of all triples of complex numbers $(x,y,z)\in\mathbb C^3$ comprising the vertices of a (possibly degenerate) equilateral triangle is a variety, given by the polynomial equation We can derive this expression by noticing that in an equilateral triangle, the angle between the vectors $y-x$ and $z-x$ must be either $\pi/6$ or $-\pi/6$. This means that the ratio must equal $\zeta_6$ or $\bar\zeta_6$, where $\zeta_6=e^{\pi i/3}$ is a primitive sixth root of unity. This means that or equivalently If we substitute the ratio $(z-x)/(y-x)$ for $\rho$ and simplify this equation, we will find that it becomes the desired polynomial equation. (The above is an exercise in Ahlfors' Complex Analysis.)
Using similar geometric arguments, we can derive the following varieties over $\mathbb C$ as well:
The set of quadruples of points comprising the vertices of a square
The set of $n$-tuples of points comprising a regular $n$-gon
The set of triples of points comprising an isoscles right triangle
However, the following are (somewhat surprisingly) not varieties over $\mathbb C$:
The set of points lying on some fixed straight line
The set of triples of points that are collinear
The set of quadruples of points comprising the vertices of a rectangle
The set of triples of points comprising the vertices of an isoscles triangle
However, if we consider points to be pairs of real numbers, then the above can be expressed as varieties over $\mathbb R^6$.