Three cars depart from city $A$ at the same time and travel a closed path composed of three straight segments $A B, B C$, and $C A$. The speeds of the first car on these segments are 12, 10, and 15 kilometers per hour, respectively. The speeds of the second car are 15, 15, and 10 kilometers per hour, respectively. Finally, the speeds of the third car are 10, 20, and 12 kilometers per hour, respectively. Find the value of the angle $\angle A B C$, knowing that all three cars end up in city $A$ at the same time.
Solution

Let $x, y$ and $z$ be the lengths of $AB, BC$ and $AC$, respectively. The arrival time $t$, common to the three cars, can be found through the equations:

$$
\left\{\begin{array}{l}
\frac{x}{12}+\frac{y}{10}+\frac{z}{15}=t \\
\frac{x}{15}+\frac{y}{15}+\frac{z}{10}=t \\
\frac{x}{10}+\frac{y}{20}+\frac{z}{12}=t
\end{array}\right.
$$

Multiplying all equations by 60, we get

$$
\left\{\begin{aligned}
5 x+6 y+4 z & =60 t \\
4 x+4 y+6 z & =60 t \\
6 x+3 y+5 z & =60 t
\end{aligned}\right.
$$

From the second equation, we have $x+y=(60 t-6 z) / 4=(30 t-3 z) / 2$. Substituting this value into the third equation, we get

$$
x=\frac{60 t-3(x+y)-5 z}{3}=\frac{30 t-z}{6}
$$

Furthermore,

$$
y=\frac{30 t-3 z}{2}-\frac{30 t-z}{6}=\frac{60 t-8 z}{6}
$$

Finally, substituting the two values found into the first equation of the last system, we obtain

$$
\begin{aligned}
5 \cdot \frac{30 t-z}{6}+6 \cdot \frac{60 t-8 z}{6}+4 z & =60 t \\
150 t-5 z+360 t-48 z+24 z & =360 t \\
t & =\frac{29}{150} z
\end{aligned}
$$

Substituting these into the expressions for $x$ and $y$, we can conclude that $(x, y, z)=(4 z / 5,3 z / 5, z)$. Since the sides of triangle $ABC$ are in the ratio $3: 4: 5$, we can conclude that it is similar to the right-angled triangle with sides 3, 4, and 5. Consequently, $\angle ABC=90^{\circ}$.