13.388 Three swimmers must swim a lane 50 m long in a pool, immediately turn around, and return to the starting point. The first starts first, the second starts after $a$ s, and the third starts after another $a$ s. At some point in time, before reaching the end of the lane, the swimmers found themselves at the same distance from the start. The third swimmer, having reached the end of the lane and turned back, met the second $s$ m from the end of the lane, and the first - $r$ m from the end of the lane. Find the speeds of the first and third swimmers and determine for which $s$ and $r$ the problem has a solution.
Solution. Let $x, y, z$ be the speeds (m/s) of the swimmers. Then $\frac{50+r}{z}$ and $\frac{50-r}{x}$ are the times taken by the third and first swimmers to reach the moment of their second meeting, and $\frac{50+s}{z}$ and $\frac{50-s}{y}$ are the times taken by the third and second swimmers to reach the moment of their second meeting. According to the problem, we have:

\[
\left\{\begin{array}{l}
\frac{50+r}{z}=\frac{50-r}{x}-2 a, \\
\frac{50+s}{z}=\frac{50-s}{y}-a .
\end{array}\right.
\]

Since after some time $t$ from the start, the three swimmers were at the same distance from the start, we have:

\[
t z=(t+a) y=(t+2 a) x \Rightarrow\left\{\begin{array}{l}
\frac{t}{y}=\frac{t+a}{z}, \\
\frac{t+2 a}{y}=\frac{t+a}{x}
\end{array} \Rightarrow \frac{2}{y}=\frac{1}{z}+\frac{1}{x}\right.
\].

By eliminating $\frac{1}{y}$ from the first system, we get:

\[
\left\{\begin{array}{l}
(50+r) \frac{1}{z}=(50-r) \frac{1}{x}-2 a \\
(50+3 s) \frac{1}{z}=(50-s) \frac{1}{x}-2 a
\end{array}\right.
\]

Multiplying the first equation by $50-s$, the second by $50-r$, and subtracting the resulting equations, we get:

\[
\frac{1}{z}=\frac{(r-s) a}{100 s-50 r-r s} \Rightarrow \frac{1}{x}=\frac{(3 s-r) a}{100 s-50 r-r s}
\]

By the meaning of the problem, $r>s$. Therefore, $z>0 \Leftrightarrow 100 s-50 r-r s>0$ or $r<\frac{100 s}{50+s}$. If $r<\frac{100 s}{50+s}$, then $3 s>r$, i.e., $x>0$.

Answer: $x=\frac{100 s-50 r-r s}{(3 s-r) a}, z=\frac{100 s-50 r-r s}{(r-s) a}, s<r<\frac{100 s}{50+s}$.