Let's find the coordinates of the centroid of a triangle whose vertices are the midpoints of the sides of the triangle formed by the lines $x+y=11$, $2x=3y-18$, and $x=4y-19$, and let's find the ratio of the areas of these two triangles to each other.

---

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
The coordinates of the vertices of the triangle $ABC$ defined by the given equations are:

$$
A(3,8), B(5,6), C(-3,4)
$$

The coordinates of the midpoints of the sides are

$$
\begin{aligned}
& x_{1}^{\prime}=\frac{5-3}{2}=1, x_{2}^{\prime}=\frac{3-3}{2}=0, x_{3}^{\prime}=\frac{3+5}{2}=4 \\
& y_{1}^{\prime}=\frac{6+4}{2}=5, y_{2}^{\prime}=\frac{8+4}{2}=6, y_{3}^{\prime}=\frac{8+6}{2}=7
\end{aligned}
$$

The centroids of the two triangles coincide; the coordinates of this point are:

$$
x=\frac{x_{1}+x_{2}+x_{3}}{3}=\frac{5}{3}, y=\frac{5+6+7}{3}=6
$$

The triangle defined by the midpoints of the sides is similar to the original triangle; thus, the areas of these triangles are in the ratio of the squares of the corresponding sides, i.e., $1: 4$.

(Irén Juvancz.)

The problem was also solved by: Kohn B., Kornis Ö., Krausz B., Krisztián Gy., Lukhaub Gy., Róth D., Obláth R., Sasvári G., Spitzer Ö., Vida A., Weisz J.