1. (SWE 2) Two ships sail on the sea with constant speeds and fixed directions. It is known that at $9: 00$ the distance between them was 20 miles; at $9: 35,15$ miles; and at $9: 55,13$ miles. At what moment were the ships the smallest distance from each other, and what was that distance?
1. Since the ships are sailing with constant speeds and directions, the second ship is sailing at a constant speed and direction in reference to the first ship. Let $A$ be the constant position of the first ship in this frame. Let $B_{1}$, $B_{2}, B_{3}$, and $B$ on line $b$ defining the trajectory of the ship be positions of the second ship with respect to the first ship at $9: 00,9: 35,9: 55$, and at the moment the two ships were closest. Then we have the following equations for distances (in miles):
$$
\begin{array}{c}
A B_{1}=20, A B_{2}=15, A B_{3}=13, \\
B_{1} B_{2}: B_{2} B_{3}=7: 4, \quad A B_{i}^{2}=A B^{2}+B B_{i}^{2} .
\end{array}
$$

Since $B B_{1}>B B_{2}>B B_{3}$, it follows that $\mathcal{B}\left(B_{3}, B, B_{2}, B_{1}\right)$ or $B\left(B, B_{3}, B_{2}\right.$, $B_{1}$ ). We get a system of three quadratic equations with three unknowns: $A B, B B_{3}$ and $B_{3} B_{2}$ ( $B B_{3}$ being negative if $B\left(B_{3}, B, B_{1}, B_{2}\right)$, positive otherwise). This can be solved by eliminating $A B$ and then $B B_{3}$. The unique solution ends up being
$$
A B=12, \quad B B_{3}=5, \quad B_{3} B_{2}=4,
$$
and consequently, the two ships are closest at $10: 20$ when they are at a distance of 12 miles.