Show that the locus of the centroids of equilateral triangles inscribed in the parabola $y^2=4ax$ is the parabola $9y^2-4ax+32a^2=0.$ Show that the locus of the centroids of equilateral triangles inscribed in the parabola $y^2=4ax$ is the parabola $9y^2-4ax+32a^2=0.$
I tried to solve it.I took three coordinates of the equilateral triangle as $(x_1,y_1),(x_2,y_2),(x_3,y_3)$
let the coordinates of the centroid be $(h,k)$.
$h=\frac{x_1+x_2+x_3}{3},k=\frac{y_1+y_2+y_3}{3}$
$h=\frac{y^2_1+y^2_2+y^2_3}{12a}$
But eliminating $y_1,y_2,y_3$ is difficult.Is my method correct or is there any other better method.Please help me.
 A: If you want to try yourself, translate into algebra the fact the triangle of vertices $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is equilateral. Two equations are enough since congruence is transitive: don't use distance but orthogonality.
Then you need some algebraic manipulations to get the result.
Otherwise the complete solution is in Gelca,Andreescu Putnam and Beyond (2007), pp. 213-214.
A: Writing the solution from Gelca,Andreescu Putnam and Beyond (2007), pp. 213-214 as Tony suggested:


Let us determine first some algebraic conditions that the coordinates $(x_i, y_i)$,
$i = 1, 2, 3$, of the vertices of a triangle should satisfy in order for the triangle to be equilateral. The equation of the median from $(x_3, y_3)$ is $$\dfrac{y-y_3}{x-x_3}=\dfrac{y_1+y_2-2y_3}{x_1+x_2-2x_3}$$
or $$(x_1-x_2)(x_1+x_2-2x_3)+(y_1-y_2)(y_1+y_2-2y_3)=0$$
So this relation along with the two obtained by circular permutations of the indices are necessary and sufficient conditions for the triangle to be equilateral. Of course, the third
condition is redundant. In the case of three points on the parabola, namely $\left(\dfrac{y_i^2}{4a},y_i\right)$ , $i=1,2,3$, after dividing by $y_1-y_2$, respectively, by $y_2-y_3$ (which are both nonzero), we obtain $$(y_1+y_2)(y_1^2+y_2^2-2y_3^2)+16a^2(y_1+y_2-2y_3)=0 \\ (y_2+y_3)(y_2^2+y_3^2-2y_1^2)+16a^2(y_2+y_3-2y_1)=0$$
Subtracting the two gives $$y_1^3-y_3^3+(y_1-y_3)(y_2^2-2y_1y_3)+48a^2(y_1-y_3)=0$$
Divide this by $y_1-y_3 \ne 0$ to transform it into $$y_1^2+y_2^2+y_3^2+3(y_1y_2+y_2y_3+y_3y_1)+48a^2=0$$
This is the condition by the $y$ -coordinates of the vertices of the triangle. Keeping in mind that the coordinates of the centroid of the triangle are $$x=\dfrac{y_1^2+y_2^2+y_3^2}{12a}, y=\dfrac{y_1+y_2+y_3}{3}$$ we rewrite the relations as $$-\dfrac{1}{2}(y_1^2+y_2^2+y_3^2)+\dfrac{3}{2}(y_1+y_2+y_3)+48a^3=0$$ then substitute $12ax=y_1^2+y_2^2+y_3^2$ and $3y=y_1+y_2+y_3$ to obtain the equation of the locus $$-6ax+\dfrac{27}{2}y^2+48a^2=0$$ or $$9y^2-4ax+32a^2=0$$

A: Let the centroid be $G(h,k)$ and the vertices be $P(t_1), Q(t_2), R(t_3)$ in parametric form.
Since $G$ is also the circumcenter, we have that $P,Q,R$ lie on the circle of the form $x^2+y^2-2hx-2ky+c=0$
Setting $x=at^2, y=2at$ we get
$a^2t^4+(4a^2-2ah)t^2-4akt+c=0$
Noting that the sum of roots is zero, the roots of the above quartic are then $t_1,t_2,t_3, -(t_1+t_2+t_3)$
Considering roots taken two at a time,
$t_1t_2+t_2t_3+t_3t_1-(t_1+t_2+t_3)^2 = \dfrac{4a^2-2ah}{a^2}$
We have $3h =a(t_1^2+t_2^2+t_3^2)$ and $3k = 2a(t_1+t_2+t_3)$
The above relation can be written as $t_1^2+t_2^2+t_3^2+(t_1+t_2+t_3)^2 = \dfrac{2(2ah-4a^2)}{a^2}$ or
$\dfrac{3h}{a}+\left(\dfrac{3k}{2a}\right)^2 = \dfrac{2(2ah-4a^2)}{a^2}$
from which we get that $G$ lies on $9y^2-4ax+32a^2=0$
