Let $A, B, C$ be points whose reflections over a line $e$ are $A_{1}, B_{1}, C_{1}$, and whose reflections over a point $P$ are $A_{2}, B_{2}, C_{2}$. What is the reflection of $P$ over the line $e$, if $\overrightarrow{A_{1} A_{2}}+\overrightarrow{B_{1} B_{2}}+\overrightarrow{C_{1} C_{2}}=\overrightarrow{0}$?
In the triangle $A A_{1} A_{2}$, the point $P$ bisects the side $A A_{2}$, and the midpoint of $A A_{1}$ (due to the axial reflection) lies on the line $e$, denoted as $A'$. Thus, $A' P$ is the midline of the triangle $A A_{1} A_{2}$, which is parallel to $A_{1} A_{2}$, and its length is half of that. Similarly, in the triangle $B B_{1} B_{2}$, the midline parallel to $B_{1} B_{2}$ is $B' P$, and in the triangle $C C_{1} C_{2}$, the midline parallel to $C_{1} C_{2}$ is $C' P$.

![](https://cdn.mathpix.com/cropped/2024_05_02_855ffebd2c2d6bde591bg-1.jpg?height=719&width=721&top_left_y=329&top_left_x=710)

We know that

$$
\overrightarrow{A_{1} A_{2}}+\overrightarrow{B_{1} B_{2}}+\overrightarrow{C_{1} C_{2}}=\mathbf{0}
$$

Therefore, the sum of the midlines, which are half the length and parallel to these vectors, $\overrightarrow{A' P}+\overrightarrow{B' P}+\overrightarrow{C' P}$, is also $\mathbf{0}$. Hence, $P$ is the centroid of the points $A', B', C'$. Since $A', B', C'$ lie on the line $e$, $P$ also lies on $e$. Thus, the reflection of the point $P$ over $e$ is itself.

Éva Rácz (Fazekas M. Főv. Gyak. Gimn., 10th grade)