## Task A-4.2.

Let $A B C$ be a right triangle with a right angle at vertex $C$. Let $A^{\prime}, B^{\prime}, C^{\prime}$ be the feet of the perpendiculars dropped from the centroid of triangle $A B C$ to the lines $B C, C A, A B$, respectively.

Determine the ratio of the areas of triangle $A^{\prime} B^{\prime} C^{\prime}$ and $A B C$.
## Second Solution.

Let $B_{1}$ be the midpoint of the leg $\overline{A C}$, and $A_{1}$ be the midpoint of the leg $\overline{B C}$. Triangles $A B^{\prime} T$ and $A C A_{1}$ are similar. Since the centroid divides the median in the ratio 2:1, it follows that $\left|C B^{\prime}\right|:|C A|=1: 3$. Similarly, we prove that $\left|C A^{\prime}\right|:|C B|=1: 3$. Therefore, the right triangles $A^{\prime} C B^{\prime}$ and $B C A$ are similar with a similarity coefficient of $\frac{1}{3}$. Thus, $\left|A^{\prime} B^{\prime}\right|=\frac{1}{3}|A B|$.

Also, $\varangle C A^{\prime} B^{\prime}=\varangle C B A$, which means lines $A^{\prime} B^{\prime}$ and $B A$ are parallel.

![](https://cdn.mathpix.com/cropped/2024_05_30_6f525b6526cae5162cdfg-18.jpg?height=539&width=803&top_left_y=570&top_left_x=632)

Let $v$ be the length of the altitude to side $\overline{A B}$ of triangle $A B C$. The length of the altitude to side $\overline{B^{\prime} A^{\prime}}$ of triangle $B^{\prime} A^{\prime} C$ is three times smaller than $v$. This implies that the distance between lines $A B$ and $B^{\prime} A^{\prime}$ is $\frac{2}{3} v$, i.e., the length of the altitude to side $\overline{B^{\prime} A^{\prime}}$ of triangle $A^{\prime} B^{\prime} C^{\prime}$ is $\frac{2}{3} v$.

Now it follows that

$$
P\left(A^{\prime} B^{\prime} C^{\prime}\right)=\frac{1}{2}\left|B^{\prime} A^{\prime}\right| \cdot \frac{2}{3} v=\frac{1}{2} \cdot \frac{1}{3}|A B| \cdot \frac{2}{3} v=\frac{2}{9} P(A B C)
$$

The desired ratio is $P\left(A^{\prime} B^{\prime} C^{\prime}\right): P(A B C)=2: 9$.