3. Given a right triangle $ABC$ with area $P$. Express, in terms of $P$, the area of the triangle whose vertices are the orthogonal projections of the centroid onto the sides of the triangle.
Solution. Let $D, E$ and $F$ be the orthogonal projections of the centroid $T$ onto the sides $AC, BC$ and $AB$, respectively.

Quadrilateral $CDTE$ is a rectangle, from which we obtain

$$
\overline{C D}: \overline{C A}=\overline{A_{1} T}: \overline{A_{1} A}=1: 3
$$

and

$\overline{C E}: \overline{C B}=\overline{B_{1} T}: \overline{B_{1} B}=1: 3$.

From this, it follows that $DE$ and $AB$ are parallel and $\overline{D E}=\frac{\overline{A B}}{3}$. If $FX$ is perpendicular to $DE$, i.e., to $AF$, then

![](https://cdn.mathpix.com/cropped/2024_06_05_9b969ba4368c118d4aedg-4.jpg?height=468&width=644&top_left_y=266&top_left_x=839)
$FX$ and $CH$ are parallel and $\overline{F X}=\frac{2}{3} \overline{C H}$. Therefore, the area of triangle $EDF$ is

$$
\frac{\overline{E D} \cdot \overline{F X}}{2}=\frac{\overline{A B} \cdot 2 \overline{C H}}{2 \cdot 3 \cdot 3}=\frac{\overline{A B} \cdot \overline{C H}}{3 \cdot 3}=\frac{2}{9} P
$$