## Task B-1.3.

Ships Bonaca and Galeb are sailing straight and at a constant speed towards the same port. At noon, the positions of the port and the ships form the vertices of an equilateral triangle. At the moment when ship Galeb has traveled $30 \mathrm{~km}$, the positions of the port and the ships form the vertices of a right triangle. When ship Bonaca entered the port, Galeb still had $12.5 \mathrm{~km}$ to go to the port. How far apart were the ships at noon?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_b0f5f68256ba1552c492g-02.jpg?height=486&width=648&top_left_y=2090&top_left_x=704)

Let $L, B$, and $G$ be the vertices of an equilateral triangle, or the positions of the lighthouse and the ships Bonaca and Galeb at noon.

Notice that the speed of the ship Bonaca is greater than the speed of the ship Galeb because Bonaca entered the harbor earlier. This means that when Galeb has traveled $30 \mathrm{~km}$, Bonaca has traveled $y(\mathrm{~km})$ with $y > 30$, and the right angle vertex of the right triangle $L B^{\prime} G^{\prime}$ is at point $B^{\prime}$.

Let $|B G|=x,\left|G G^{\prime}\right|=30 \mathrm{~km}$. Since the triangle $L B^{\prime} G^{\prime}$ is a right triangle, we have

$$
\begin{aligned}
& \left|G^{\prime} L\right|=x-30 \\
& \left|B^{\prime} L\right|=x-y=\frac{x-30}{2} \\
& \left|B B^{\prime}\right|=|B L|-\left|B^{\prime} L\right|=x-\frac{x-30}{2}=\frac{x+30}{2}
\end{aligned}
$$

The ships move at constant speeds, so the ratio of the distances traveled in time $t$ is constant and we have:

$$
\begin{aligned}
& \frac{x+30}{2}: 30=x:(x-12.5) \\
& \frac{x+30}{2} \cdot\left(x-\frac{25}{2}\right)=30 \cdot x \\
& 2 x^{2}-85 x-750=0 \\
& 2 x^{2}-100 x+15 x-750=0 \\
& 2 x(x-50)+15(x-50)=0 \\
& (x-50)(2 x+15)=0 .
\end{aligned}
$$

It follows that $x=50$, so the distance between the ships at noon was $50 \mathrm{~km}$.