13.432 Two athletes are running on a closed track of a stadium. The speed of each is constant, but the first one takes $a$ seconds less to run the entire track than the second one. If they start running from a common starting point in the same direction, they meet every $b$ seconds. How long will it take for them to meet if they run in opposite directions on the same track at the same speeds?
Solution. Let $x, y$ be the speeds of the athletes, $z$ be the length of the running track. According to the conditions of the problem,

$\left\{\begin{array}{l}\frac{z}{y}-\frac{z}{x}=a, \\ \frac{z}{x-y}=b\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{z}{y}-\frac{z}{x}=a \\ \frac{x}{z}-\frac{y}{z}=\frac{1}{b} .\end{array}\right.\right.$

Let $\frac{x}{z}=w, \frac{y}{z}=t$. Then

$\left\{\begin{array}{l}\frac{1}{b} \cdot \frac{1}{t w}=a, \\ w-t=\frac{1}{b}\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{1}{a b}=t w, \\ \frac{1}{b}=w-t\end{array} \Rightarrow t^{2}+t \cdot \frac{1}{b}-\frac{1}{a b}=0 \Rightarrow\right.\right.$

$\Rightarrow t=\frac{\sqrt{a^{2}+4 a b}-a}{2 a b}, w=\frac{\sqrt{a^{2}+4 a b}+a}{2 a b} \Rightarrow t+w=\frac{\sqrt{a^{2}+4 a b}}{a b} \Rightarrow$

$\Rightarrow \frac{a b}{\sqrt{a^{2}+4 a b}}=\frac{1}{t+w}=\frac{z}{x+y}$, where $\frac{z}{x+y}$ is exactly the time it takes for

the athletes to meet if they run in opposite directions.

Answer: $\frac{a b}{\sqrt{a^{2}+4 a b}} \mathrm{c}$.