A regular triangle has one vertex at $A(2 \sqrt{3} ; 2 \sqrt{3})$. The centroid of the triangle is at the origin. Determine the coordinates of the other two vertices!
We know that the centroid of a triangle divides each median in a $2: 1$ ratio (the longer part is towards the vertex). Therefore, the coordinates of the midpoint of the side opposite vertex $A$ are: $(-\sqrt{3} ;-\sqrt{3})$. This follows both from the aforementioned ratio and the fact that $F$ lies on the extension of segment $AS$ beyond the origin, in the opposite direction from $A$, thus falling in the fourth quadrant. Since the triangle is equilateral, $A F \perp B C$. The center of the circumcircle of the triangle is the origin, and its radius is

![](https://cdn.mathpix.com/cropped/2024_05_02_0fef593a87c358b06980g-1.jpg?height=767&width=623&top_left_y=332&top_left_x=748)

The equation of the circle is:

$$
x^{2}+y^{2}=24
$$

The slope of line $A F$ is 1, so the slope of the perpendicular line $B C$ is -1, and this line passes through the point $F(-\sqrt{3} ;-\sqrt{3})$, thus its equation is:

$$
y=-x-2 \sqrt{3}
$$

The solutions to the system of equations

$$
x^{2}+y^{2}=24 \quad \text{and} \quad x+y=-2 \sqrt{3}
$$

give the coordinates of the missing vertices, that is,

$$
x_{1}=3-\sqrt{3} \approx 1.268, \quad x_{2}=-3-\sqrt{3} \approx-4.732, \quad y_{1}=-3-\sqrt{3} \approx-4.732 ; \quad y_{2}=3-\sqrt{3} \approx 1.268
$$

According to the labeling on our diagram:

$$
B(-3-\sqrt{3} ; 3-\sqrt{3}), \quad C(3-\sqrt{3} ;-3-\sqrt{3})
$$