For a point $P=\left(a, a^{2}\right)$ in the coordinate plane, let $\ell(P)$ denote the line passing through $P$ with slope $2 a$. Consider the set of triangles with vertices of the form $P_{1}=\left(a_{1}, a_{1}^{2}\right), P_{2}=\left(a_{2}, a_{2}^{2}\right), P_{3}=\left(a_{3}, a_{3}^{2}\right)$, such that the intersection of the lines $\ell\left(P_{1}\right), \ell\left(P_{2}\right), \ell\left(P_{3}\right)$ form an equilateral triangle $\Delta$. Find the locus of the center of $\Delta$ as $P_{1} P_{2} P_{3}$ ranges over all such triangles.
The answer is the line $y=-1 / 4$. I did not find this problem inspiring, so I will not write out most of the boring calculations since most solutions are just going to be "use Cartesian coordinates and grind all the way through".

The "nice" form of the main claim is as follows (which is certainly overkill for the present task, but is too good to resist including):
Claim (Naoki Sato) - In general, the orthocenter of $\Delta$ lies on the directrix $y=-1 / 4$ of the parabola (even if the triangle $\Delta$ is not equilateral).

Proof. By writing out the equation $y=2 a_{i} x-a_{i}^{2}$ for $\ell\left(P_{i}\right)$, we find the vertices of the triangle are located at
$$
\left(\frac{a_{1}+a_{2}}{2}, a_{1} a_{2}\right) ; \quad\left(\frac{a_{2}+a_{3}}{2}, a_{2} a_{3}\right) ; \quad\left(\frac{a_{3}+a_{1}}{2}, a_{3} a_{1}\right) .
$$

The coordinates of the orthocenter can be checked explicitly to be
$$
H=\left(\frac{a_{1}+a_{2}+a_{3}+4 a_{1} a_{2} a_{3}}{2},-\frac{1}{4}\right) .
$$

An advanced synthetic proof of this fact is given at https://aops.com/community/ p2255814.

This claim already shows that every point lies on $y=-1 / 4$. We now turn to showing that, even when restricted to equilateral triangles, we can achieve every point on $y=-1 / 4$. In what follows $a=a_{1}, b=a_{2}, c=a_{3}$ for legibility.
Claim - Lines $\ell(a), \ell(b), \ell(c)$ form an equilateral triangle if and only if
$$
\begin{aligned}
a+b+c & =-12 a b c \\
a b+b c+c a & =-\frac{3}{4} .
\end{aligned}
$$

Moreover, the $x$-coordinate of the equilateral triangle is $\frac{1}{3}(a+b+c)$.

Proof. The triangle is equilateral if and only if the centroid and orthocenter coincide, i.e.
$$
\left(\frac{a+b+c}{3}, \frac{a b+b c+c a}{3}\right)=G=H=\left(\frac{a+b+c+4 a b c}{2},-\frac{1}{4}\right) .
$$

Setting the $x$ and $y$ coordinates equal, we derive the claimed equations.
Let $\lambda$ be any real number. We are tasked to show that
$$
P(X)=X^{3}-3 \lambda \cdot X^{2}-\frac{3}{4} X+\frac{\lambda}{4}
$$
has three real roots (with multiplicity); then taking those roots as $(a, b, c)$ yields a valid equilateral-triangle triple whose $x$-coordinate is exactly $\lambda$, be the previous claim.
To prove that, pick the values
$$
\begin{aligned}
P(-\sqrt{3} / 2) & =-2 \lambda \\
P(0) & =\frac{1}{4} \lambda \\
P(\sqrt{3} / 2) & =-2 \lambda .
\end{aligned}
$$

The intermediate value theorem (at least for $\lambda \neq 0$ ) implies that $P$ should have at least two real roots now, and since $P$ has degree 3 , it has all real roots. That's all.