is it possible to choose points on the graph of $y = x^2$ to form vertices of an equilateral triangle?

Question

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 $(S_{\Delta ABC})_{min}$

Answer 1

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 $(0,0)$, $(\pm\sqrt{3},3)$.

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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 $3\sqrt{3}$.