Task B-2.7. The tourist ship "Galeb" is sailing straight north. At the bay "Val", two young tourists set off in a fast scooter in a direction that forms a $120^{\circ}$ angle with the ship's direction, heading towards an island $15 \mathrm{~km}$ away from the bay. If the scooter has enough fuel for exactly $40 \mathrm{~km}$ of driving, how many kilometers can the scooter get closest to the island to ensure it can safely return to the ship? In this case, the ship (while the scooter travels $40 \mathrm{~km}$) will travel $16 \mathrm{~km}$ by the time they meet again.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_bde1d6b20cdcb90302c4g-11.jpg?height=816&width=588&top_left_y=660&top_left_x=281)

## Sketch

(2 points) (if angles and the right triangle are not labeled, only 1 point)

Let $x$ be the distance the scooter travels from the shore to point $B$ closest to the island. Let $y$ be the distance from point $B$ to point $C$, where the scooter meets the boat again. Then the total distance the scooter can travel is

$$
|A B|+|B C|
$$

which can be at most $40 \mathrm{~km}$. We write

$$
x+y=40
$$

From the right triangle $\triangle A D B$ it follows that

$$
|A D|=x \cos 60^{\circ}=\frac{x}{2}, \quad|B D|=x \sin 60^{\circ}=\frac{x \sqrt{3}}{2}
$$

From the right triangle $\triangle D B C$ it follows that

$$
\begin{gathered}
y^{2}=\left(16+\frac{x}{2}\right)^{2}+\left(\frac{x \sqrt{3}}{2}\right)^{2} \\
y^{2}=x^{2}+16 x+256 \\
(40-x)^{2}=x^{2}+16 x+256
\end{gathered}
$$

From this, $x=14 \mathrm{~km}$, which means the scooter can get as close as $1 \mathrm{~km}$ to the island.