is it possible to choose points on the graph of $y = x^2$ to form vertices of an equilateral triangle? is it possible to choose points on the graph of $y = x^2$ to form vertices of an equilateral triangle $\Delta ABC$?where three ponit not $(0,0)$ .and find $(S_{\Delta ABC})_{min}$
 A: Summary


*

*Are there other points which can form equilateral triangle?
The answer is yes. In fact, there are infinitely many of them.
The centroids of the triangles lie on another parabola $y = 9x^2 + 2$.

*The minimal area is $3\sqrt{3}$. achieved by the equilateral triangle with vertices $(0,0)$, $(\pm\sqrt{3},3)$.

Identify the euclidean plane $\mathbb{R}^2$ with complex plane $\mathbb{C}$. 
Let $z = x + iy$ and $\bar{z} = x - iy$ be its complex conjugate, the equation of the parabola becomes
$$y = x^2 \iff \frac{z - \bar{z}}{2i} = \left(\frac{z+\bar{z}}{2}\right)^2 \iff
(z + \bar{z})^2 + 2i(z-\bar{z}) = 0$$
Let $\omega = e^{2\pi i/3}$. Given any equilateral triangle $T$, we can always find two complex numbers $\rho = u+iv$ and $a$ such that the vertices of $T$ are $\rho + a\omega^k$ for $k = 0, \pm 1$. 
If they all lie on the parabola above, then for $k = 0, \pm 1$, we have
$$
\begin{align}
 &\; (\rho + \bar{\rho} + a\omega^k + \bar{a}\omega^{-k})^2
+  2i(\rho - \bar{\rho} + a\omega^k - \bar{a}\omega^{-k})\\
= &\; (\rho+\bar{\rho})^2
 +  2(\rho+\bar{\rho})(a\omega^k + \bar{a}\omega^{-k})
 + (a^2\omega^{-k} + 2|a|^2 + \bar{a}^2\omega^k)
 + 2i(\rho - \bar{\rho} + a\omega^k - \bar{a}\omega^{-k})\\
= &\; 0
\end{align}
$$
Comparing coefficients of different powers of $\omega$, we get
$$
\begin{cases}
(\rho + \bar{\rho})^2 + 2i(\rho - \bar{\rho}) + 2|a|^2 &= 0\\
2(\rho + \bar{\rho} + i)a + \bar{a}^2 &= 0\\
2(\rho + \bar{\rho} - i)\bar{a} + a^2 &= 0
\end{cases}
$$
Multiply the $2^{nd}$ equation by $4a^2$ and simplify it using $1^{st}$ equation, 
we get
$$8(\rho + \bar{\rho} +i )a^3 + \left[(\rho + \bar{\rho})^2 + 2i(\rho - \bar{\rho})\right]^2 = 0\tag{*1}$$
Since $\rho + \bar{\rho} + i = 2u + i\ne 0$, we can use this to determine $a$ up to a factor $\omega^k$. 
For this $a$ to be compatible with the $1^{st}$ equation, the condition is
$$\begin{align}
& (\rho + \bar{\rho})^2 + 2i(\rho - \bar{\rho}) = -2|a|^2 = -8|\rho + \bar{\rho} + i|^2
= -8 ((\rho+\bar{\rho})^2 + 1)\\
\iff & 9(\rho + \bar{\rho})^2 + 2i(\rho - \bar{\rho}) + 8 = 36u^2 - 4v + 8 = 0\\
\iff & v = 9u^2 + 2
\end{align}
$$
Base on this, we see start from any point $(u,v)$ from the  parabola $y = 9x^2 + 2$, if one define $\rho = u + iv$ and use $(*1)$ to compute $a$, the 3 points
$\rho + a\omega^k$ will lie on the original parabola
$y = x^2$ and form an equilateral triangle.
Back to question of minimization of area. It is clear it is equivalent to
minimization of $|a|$. Since
$$|a|^2 = -\frac12 \left[ (\rho + \bar{\rho})^2 + 2i(\rho - \bar{\rho}) \right]
= 2(v-u^2) = 2(2+8u^2)$$
the minimal value $|a|$ is achieved at $(u,v) = (0,2)$ with value $2$.
The corresponding triangle has vertices at $(0,0)$, $(\pm\sqrt{3},3)$ 
with side length $2\sqrt{3}$ and area $3\sqrt{3}$.
