The conditions of the problem are satisfied by a unique arrangement of points $E$ and $F$ on segment $A B$, see the figure. The area of triangle $A B C$ is equal to the sum of the areas of triangles $A E C$, $E F C$, and $F B C$, i.e.,

$$
S_{A B C}=S_{A E C}+S_{E F C}+S_{F B C}=1+3+2=6\left(\mathrm{~cm}^{2}\right) .
$$

The same area can be expressed as the sum of the areas of triangles $A G C$, $G H C$, and $H B C$. The area of the first and third triangles can be derived from the problem statement, after which we can easily determine the area of triangle $G H C$.

![](https://cdn.mathpix.com/cropped/2024_04_17_0263ca193add285924cag-2.jpg?height=394&width=874&top_left_y=997&top_left_x=591)

Segment $C G$ is the median of triangle $A F C$, and it divides this triangle into two triangles of equal area. The area of triangle $A F C$ is the sum of the areas of triangles $A E C$ and $E F C$, which we know. Therefore,

$$
S_{A G C}=\frac{1}{2}\left(S_{A E C}+S_{E F C}\right)=\frac{1}{2}(1+3)=2\left(\mathrm{~cm}^{2}\right)
$$

Similarly, it can be justified that the area of triangle $H B C$ is equal to

$$
S_{H B C}=\frac{1}{2}\left(S_{E F C}+S_{F B C}\right)=\frac{1}{2}(3+2)=2.5\left(\mathrm{~cm}^{2}\right) .
$$

Therefore, the area of triangle $G H C$ is

$$
S_{G H C}=S_{A B C}-S_{A G C}-S_{H B C}=6-2-2.5=1.5\left(\mathrm{~cm}^{2}\right) .
$$

Scoring suggestion. 1 point for arranging points $E$ and $F$ on segment $A B$ (a sketch is sufficient); 2 points for recognizing that segment $C G$, respectively $C H$, divides triangle $A F C$, respectively $E B C$, into two triangles of equal area; 3 points for deriving the desired area.

Note. All discussed triangles share a common height from the common vertex $C$. Therefore, the ratios of the areas of any two triangles are the same as the ratios of the lengths of the sides opposite vertex $C$. To determine the area of triangle $G H C$, it is sufficient to determine the ratio of the length of side $G H$ to the length of a side of a triangle with a known area:

If we denote $|A E|=a$, then $|E F|=3a$ and $|F B|=2a$. Since $C G$ is the median of triangle $A F C$, point $G$ is the midpoint of segment $A F$. The length of this segment is $|A F|=|A E|+|E F|=4a$, so $|A G|=\frac{1}{2}|A F|=2a$. Similarly, it can be justified that $|H B|=\frac{1}{2}(|E F|+|F B|)=2.5a$. The length of segment $G H$ is therefore

$$
|G H|=|A B|-|A G|-|H B|=(6-2-2.5)a=1.5a .
$$

From this, it follows that $S_{G H C}=1.5 \cdot S_{A E C}=1.5 \mathrm{~cm}^{2}$.
In triangle $ABC$, points $E$ and $F$ lie on side $AB$. The area of triangle $AEC$ is $1 \, \text{cm}^2$, the area of triangle $EFC$ is $3 \, \text{cm}^2$, and the area of triangle $FBC$ is $2 \, \text{cm}^2$. Point $T$ is the centroid of triangle $AFC$, and point $G$ is the intersection of line $CT$ and $AB$. Point $R$ is the centroid of triangle $EBC$, and point $H$ is the intersection of lines $CR$ and $AB$.

Determine the area of triangle $GHC$.

(E. Semerádová)