U lesa, which had the shape of an isosceles triangle, Ivo and Petr set up camp at one of its vertices. In the middle of the opposite side was a well. The boys decided not to go to it through the forest but along its perimeter. Each took a different direction, but both at a speed of $4 \mathrm{~km} / \mathrm{h}$. Ivo arrived at the well in 15 minutes, and Petr in 12. How long were the sides of the "forest" triangle? (Round the lengths of the sides to whole meters.)

(M. Volfová)
The boys could not camp at the top of the isosceles triangle opposite the base, as they would reach the well at the same time. In the diagram, the locations of the tent and the well are marked, along with the lengths of the legs $b$ and the base $a$.

![](https://cdn.mathpix.com/cropped/2024_04_17_690ff1e527bb42bf625dg-1.jpg?height=409&width=280&top_left_y=972&top_left_x=914)

From the problem, we know that Ivo walked for $\frac{1}{4}$ hour at a speed of $4 \mathrm{~km} / \mathrm{h}$, thus he walked $1 \mathrm{~km}$. Petr walked for 12 minutes, i.e., $\frac{1}{5}$ hour, at a speed of $4 \mathrm{~km} / \mathrm{h}$, thus he walked $\frac{4}{5}=0.8(\mathrm{~km})$. Now we need to discuss the following two possibilities:

1. Ivo walked along the leg, and Petr walked along the base of the triangle.

In this case, Ivo walked $b+\frac{b}{2}=1(\mathrm{~km})$ and Petr walked $a+\frac{1}{2} b=\frac{4}{5}(\mathrm{~km})$. From this, we calculate $\frac{3}{2} b=1(\mathrm{~km})$ and $b=\frac{2}{3}(\mathrm{~km})$, approximately $b \doteq 667 \mathrm{~m}$. Substituting into the second equation, we get $a+\frac{1}{3}=\frac{4}{5}(\mathrm{~km})$, i.e., $a=\frac{4}{5}-\frac{1}{3}=\frac{7}{15}(\mathrm{~km})$, approximately $a \doteq 467 \mathrm{~m}$. The sides of the triangle are approximately 667, 667, and $467(\mathrm{~m})$.

2. Petr walked along the leg, and Ivo walked along the base of the triangle.

In this case, Petr walked $b+\frac{1}{2} b=\frac{4}{5}(\mathrm{~km})$ and Ivo walked $a+\frac{1}{2} b=1(\mathrm{~km})$. We calculate $\frac{3}{2} b=\frac{4}{5}(\mathrm{~km})$ and $b=\frac{8}{15}(\mathrm{~km})$, approximately $b \doteq 533 \mathrm{~m}$. Substituting into the second equation, we get $a+\frac{4}{15}=1(\mathrm{~km})$, i.e., $a=1-\frac{4}{15}=\frac{11}{15}(\mathrm{~km})$, approximately $a \doteq 733 \mathrm{~m}$. The sides of the triangle are approximately 533, 533, and $733(\mathrm{~m})$.

Evaluation. 2 points for analyzing the situation and calculating the distances each boy walked; 2 points for calculating the dimensions of the forest in each of the two scenarios.