1. What does the geometric place of the centroids of triangles with vertices: $A$ at the vertex of the parabola $y^{2}=2 p x, B$ at the focus of the parabola, and $C$ on the given parabola represent?
Solution. Let $C\left(x_{1}, y_{1}\right)$ be an arbitrary point on the parabola $y^{2}=2 p y$, different from $A(0,0)$. Point $C$ is chosen to be different from $A$ so that $A B C$ forms a triangle. Let $T=(x, y)$ be the centroid of triangle $A B C$. For the vertex $B$ which is at the focus of the parabola $B\left(\frac{p}{2}, 0\right)$, and for the midpoint of side

![](https://cdn.mathpix.com/cropped/2024_06_05_d48bd9f430eb3870413eg-09.jpg?height=488&width=738&top_left_y=1611&top_left_x=722)
$A B$ we have $C_{1}\left(\frac{p}{4}, 0\right)$. Since
![](https://cdn.mathpix.com/cropped/2024_06_05_d48bd9f430eb3870413eg-09.jpg?height=498&width=484&top_left_y=747&top_left_x=978)
$T$ divides the segment $C_{1} C$ in the ratio $1: 2$, we get

$$
x=\frac{p+2 x_{1}}{6} \text { and } y=\frac{y_{1}}{3} \text {, }
$$

from which

$$
x_{1}=\frac{6 x-p}{2} \text { and } y_{1}=3 y
$$

Since $C\left(x_{1}, y_{1}\right)$ lies on the parabola, we get

$$
9 y^{2}=2 p \frac{6 x-p}{2}
$$

from which:

$$
y^{2}=2 p\left(x-\frac{p}{6}\right)
$$

Thus, the required geometric locus of points is a parabola given by (1), but without the vertex $\left(\frac{p}{6}, 0\right)$, because $A \neq C x$.