Given parallelograms $ABGH$ and $DEGH$, whose vertices $A, B, D$ and $E$ lie on one line. Point $C$ is the intersection of segments $BG$ and $DH$, point $I$ lies on segment $AH$, and point $F$ lies on segment $EG$. The heptagon $ABCD EGH$ consists of seven triangles, among which one has an area of $3 \, \text{cm}^2$, one has an area of $5 \, \text{cm}^2$, two have an area of $7 \, \text{cm}^2$, and one has an area of $10 \, \text{cm}^2$. Apart from the triangles with areas of $7 \, \text{cm}^2$, no other pair of the mentioned seven triangles has the same area.

Determine whether the triangles with areas of $7 \, \text{cm}^2$ can be identified with certainty. Further, determine the area of the heptagon $ABCD EGH$; find all possibilities.

![](https://cdn.mathpix.com/cropped/2024_04_17_4522893cfc20cd4c648eg-4.jpg?height=500&width=740&top_left_y=681&top_left_x=658)

Note: The image is illustrative only.

(E. Semerádová)
Parallelograms $ABGH$ and $DEGH$ share a common side and have the same height, so they have the same area. This means that for the areas of the corresponding triangles, the following holds:

$$
S_{ABI}+S_{BCI}+S_{CHI}+S_{CGH}=S_{DEF}+S_{CDF}+S_{CFG}+S_{CGH}.
$$

Regardless of the position of point $I$ on segment $AH$, the area of triangle $BCI$ remains the same. If point $I$ coincides with point $A$, this triangle, together with triangle $CGH$, forms half of parallelogram $ABGH$. For the same reason, triangles $CDF$ and $CGH$ form half of the area of parallelogram $DEGH$. Therefore, for the areas of the triangles, the following holds:

$$
S_{BCI}+S_{CGH}=S_{ABI}+S_{CHI}=S_{DEF}+S_{CFG}=S_{CDF}+S_{CGH}.
$$

In particular, triangles $BCI$ and $CDF$ have the same area, which is $7 \text{ cm}^2$.

Three of the four triangles $ABI, CHI, DEF$, and $CFG$ have areas of $3 \text{ cm}^2$, $5 \text{ cm}^2$, and $10 \text{ cm}^2$. We will consider all possible sums (2), from which we will determine the area of the remaining triangle from the given quadruple, the area of triangle $CGH$, and the area of the polygon $ABCDEFGH$:

| sum $(2)$ | $3+5=8$ | $3+10=13$ | $5+10=15$ |
| :--- | :---: | :---: | :---: |
| area of the remaining | $8-10=-2$ | $13-5=8$ | $15-3=12$ |
| area of $CGH$ | - | $13-7=6$ | $15-7=8$ |
| area of the whole | - | $14+26+6=46$ | $14+30+8=52$ |

In both cases, the condition is met that no other pair of triangles, except those with areas of $7 \text{ cm}^2$, have the same area. The polygon $ABCDEFGH$ has an area of either $46 \text{ cm}^2$ or $52 \text{ cm}^2$.

Evaluation. 1 point for deriving the relationship (1); 2 points for the relationships (2) and recognizing the triangles with areas of $7 \text{ cm}^2$; 3 points for analyzing the possibilities, verifying the conditions, and determining the possible areas of the polygon $ABCDEFGH$.