Given a triangle $ABC$. On side $AB$ lies point $X$ and on side $BC$ lies point $Y$ such that $CX$ is a median, $AY$ is an altitude, and $XY$ is a midline of triangle $ABC$. Calculate the area of the shaded triangle in the figure, given that the area of triangle $ABC$ is $24 \, \text{cm}^2$.

(M. Dillingerová)

![](https://cdn.mathpix.com/cropped/2024_04_17_47ababdcc19ff16ad01cg-2.jpg?height=431&width=577&top_left_y=1526&top_left_x=745)
Since $X Y$ is the midline of triangle $A B C$, $Y$ must be the midpoint of side $B C$, and thus segment $A Y$ is not only an altitude but also a median. The intersection of medians $C X$ and $A Y$ is the centroid of triangle $A B C$, which we will denote as $T$.

Each median divides the triangle into two triangles of equal area (the triangles thus formed have a common altitude to the same length sides): $A Y$ is a median of triangle $A B C$, therefore

$$
S_{A B Y}=S_{A C Y}=\frac{1}{2} S_{A B C}=\frac{24}{2}=12\left(\mathrm{~cm}^{2}\right),
$$

and similarly $X Y$ is a median of triangle $A B Y$, therefore

$$
S_{A X Y}=S_{B X Y}=\frac{1}{2} S_{A B Y}=\frac{12}{2}=6\left(\mathrm{~cm}^{2}\right) .
$$

Triangle $A X Y$, whose area we know, consists of triangles $A X T$ and $Y X T$. These two triangles have a common altitude from point $X$, so their areas are in the same ratio as the lengths of sides $A T$ and $T Y$, i.e., in the ratio $2: 1$ (the centroid divides the median in the ratio $2: 1$). The area of triangle $Y X T$ is therefore one-third of the area of triangle $A X Y$, so it is

$$
S_{Y X T}=\frac{1}{3} S_{A X Y}=\frac{6}{3}=2\left(\mathrm{~cm}^{2}\right) .
$$

Evaluation. 2 points for the observation (including justification) that $A Y$ is a median in triangle $A B C$; 1 point each for the areas of triangles $A B Y$ and $A X Y$; 2 points for the final area of triangle $Y X T$ (including its derivation or justification).