In an equilateral triangle $A B C$, point $F$ lies on side $B C$. The area of triangle $A B F$ is three times the area of triangle $A C F$, and the difference in the perimeters of these two triangles is $5 \mathrm{~cm}$.

Determine the length of the side of triangle $A B C$.

(E. Semerádová)
Triangles $A B F$ and $A C F$ have a common height from the common vertex $A$, so the areas of these triangles are in the same ratio as the lengths of the sides opposite to vertex $A$. Therefore, it holds that $|B F|=3|C F|$.

![](https://cdn.mathpix.com/cropped/2024_04_17_d9645623b1ef4faf124eg-1.jpg?height=462&width=552&top_left_y=934&top_left_x=752)

Side $A F$ is common to both triangles, and sides $A B$ and $A C$ are equal because triangle $A B C$ is equilateral. From the given relationship for the perimeters of these triangles, it follows that

$$
|B F|=|C F|+5 \mathrm{~cm}
$$

From the previous paragraph, we know the relationship between the lengths of sides $B F$ and $C F$, so we can determine the length of one of these segments:

$$
\begin{aligned}
3|C F| & =|C F|+5 \mathrm{~cm} \\
|C F| & =2.5 \mathrm{~cm} .
\end{aligned}
$$

The length of side $B C$ of triangle $A B C$ is therefore

$$
|B C|=|B F|+|C F|=4|C F|=10 \mathrm{~cm} .
$$

Scoring suggestion. 2 points for determining the ratio of the lengths of segments $B F$ and $C F$; 1 point for recognizing that the lengths of sides $A F, A B$, and $A C$ do not contribute to the difference in perimeters; 3 points for solving the problem.