Shapovalov A.V.

On the sides $BC$, $CA$, and $AB$ of triangle $ABC$, points $A_1$, $B_1$, and $C_1$ are chosen respectively, such that the medians $A_1A_2$, $B_1B_2$, and $C_1C_2$ of triangle $A_1B_1C_1$ are parallel to the lines $AB$, $BC$, and $CA$ respectively. In what ratio do the points $A_1$, $B_1$, and $C_1$ divide the sides of triangle $ABC$?
Let $O$ be the point of intersection of the medians of triangle $A_{1} B_{1} C_{1}$. Extend the medians $A_{1} A_{2}, B_{1} B_{2}$, and $C_{1} C_{2}$ to intersect segments $A C, A B$, and $B C$ at points $P, Q$, and $R$ respectively. Since segment $O A_{2}$ passes through the midpoint $A_{2}$ of segment $B_{1} C_{1}$ and $O A_{2} \| C_{1} Q$, $O A_{2}$ is the midline of triangle $B_{1} C_{1} Q$. Therefore, $O$ is the midpoint of $B_{1} Q$. Similarly, $O$ is the midpoint of segments $A_{1} P$ and $C_{1} R$. Thus, $A_{1} C_{1} P R$ is a parallelogram. Therefore, $C_{1} P = A_{1} R$ and $C_{1} P \| B C$. Hence, $C R C_{1} P$ and $B A_{1} P C_{1}$ are also parallelograms (the opposite sides of these quadrilaterals are pairwise parallel). Therefore, $B A_{1} = C_{1} P, C R = C_{1} P$. Consequently, $B A_{1} = A_{1} R = C R$ and $B A_{1}: B_{1} A = 1: 2$. Similarly, $C B_{1}: B_{1} A = A C_{1}: C_{1} B = 1: 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_39480ed8b0d60d88e837g-41.jpg?height=509&width=734&top_left_y=1621&top_left_x=685)

## Answer

$B A_{1}: A_{1} C = C B_{1}: B_{1} A = A C_{1}: C_{1} B = 1: 2$.