Equilateral triangle and a parabola

Question

We are given an equilateral triangle ABC. Find the coordinates of one of the other apexes. knowing that its apexes lay on the parabola $g(x)=-6x+2x^2$. Point C is the apex of the given parabola.

We can calculate the apex of the parabola by $(-\frac{b}{2a}, -\frac{b^2-4ac}{4a})$ getting (3, -9) as the coordinates of C. How can we move on, though? We only know that the triangle is equilateral. We also know that the coordinates, when plugged to the parabola equation, should output 0=0. How should this be done, though? Could you help?

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(see this meta thread for deletion discussion)

Answer 1

Two generic points are $(a,a^2-6a)$ and $(b,b^2-6b)$. If you write the equations for the distance from these to the point you have and the distance between them, you get three equations in three unknowns. A bit of a cheat is to guess that $a$ and $b$ are symmetric around $x=3$, so write $a=3-c, b=3+c$. That gets rid of one variable for you. So if $d$ is the side of the triangle, $d=2c, d^2=(3-c-3)^2+((3-c)^2-6(3-c)+9)^2$.

Answer 2

In order to find points $A(a,a^2-6a)$ and $B(b,b^2-6b)$ you will need following equalities :

$\sqrt{(3-a)^2+(-9-a^2+6a)^2}=\sqrt{(3-b)^2+(-9-b^2+6b)^2}$ (i)

$\tan \frac{\pi}{3}=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$ (ii)

where $m_1$ is slope of the line that contains points $A$ and $C$ given by : $m_1=\frac{-9-a^2+6a}{3-a}$ , and $m_2$ is slope of the line that contains points $B$ and $C$ given by $m_2=\frac{-9-b^2+6b}{3-b}$

Answer 3

Hint: if one of the triangle's corners is the vertex of the parabola, the two sides of the triangle that meet at the vertex should both be at a $30^\circ$ angle from the parabola's axis; i.e., the two lines should have a slope of $\tan(60^\circ)$ and $\tan(120^\circ)$ (why?). Since you have slopes, and you know that they pass though a vertex, it is a simple matter to find the equations of the lines forming the sides. Find the intersections of those lines with the parabola.