13.441 At a distance of $l$ m from bridge $A$ downstream along the river is located bridge $B$. When the athlete was passing by bridge $A$, heading towards bridge $B$, two balls were thrown to him. He caught the first ball and let the second one float downstream. After swimming a certain section of the river with the ball, the athlete left this ball and swam upstream to catch the second ball. After catching the second ball, he turned again towards bridge $B$ and reached it simultaneously with the freely floating first ball. What distance did the athlete have to swim if his own speed was always $\boldsymbol{k}$ times the speed of the current?
Solution. Let $D$ be the point where the swimmer left the first ball and turned back, $C$ be the point where he picked up the second ball, $x$ be the speed (m/min) of the river current, and $y$ be the distance between $C$ and $D$.

Since the first ball freely drifted the distance $D B$, and the second ball drifted the distance $A C$, and they both arrived at $B$ simultaneously, then $D B = A C$ (since the segments $A D$ and $C B$ they "traveled" at the same speed, equal to $(k+1)x$). Therefore, $A C = \frac{l-y}{2}, \frac{l-y}{2x}$ is the time of the free movement of the second ball until it met the swimmer, $\frac{l+y}{2x(k+1)}$ is the time it took the swimmer to swim the segment $A D$, and $\frac{y}{x(k-1)}$ is the time it took him to swim the segment $D C$ (where $x(k-1)$ is the swimmer's speed against the current), hence

$\frac{l-y}{2x} = \frac{l+y}{2x(k+1)} + \frac{y}{x(k-1)} \Rightarrow y = l \cdot \frac{k-1}{k+3} \Rightarrow 2y + l = l \cdot \frac{3k+1}{k+3}$ - the total distance the swimmer swam.

Answer: $l \cdot \frac{3k+1}{k+3} \text{m}$.