A train departs from $A$ and moves towards $B$ at a constant speed. 11 minutes later, a train departs from $B$ on the same route, also at a constant speed, heading towards $A$. From the moment they meet, it takes 20 and 45 minutes respectively for the trains to reach $B$ and $A$. In what ratio does the meeting point $T$ divide the distance $AB$?
I. Solution: If the speed of the train departing from $A$ is $v_{1}$ and that of the train departing from $B$ is $v_{2}$, then from the data regarding the meeting point,

$$
20 v_{1}=T B, \quad 45 v_{2}=T A, \quad \text { hence } \quad \frac{A T}{T B}=\frac{9}{4} \cdot \frac{v_{2}}{v_{1}}
$$

However, by calculating the time required to travel the distances to $T$, the difference is 11 minutes, i.e.,

$$
\frac{A T}{v_{1}}-\frac{T B}{v_{2}}=11
$$

Substituting the values of $T A$ and $T B$ obtained earlier, and multiplying through by the value of $\frac{v_{2}}{v_{1}}=u$, we get the equation

$$
45 u^{2}-11 u-20=0
$$

From this, taking only the positive root,

$$
u=\frac{4}{5}, \quad \text { and thus } \quad \frac{A T}{T B}=\frac{9}{4} u=\frac{9}{5}
$$

II. Solution: It is easiest to write an equation for the time required for one of the trains to travel from its starting point to $T$ or to complete the entire journey. This can also be aided by visualization, as we can sketch the structure of a graph illustrating the relationships even without specific data (see Figure 1, which represents time on the horizontal axis and distance traveled on the vertical axis).

![](https://cdn.mathpix.com/cropped/2024_05_02_006554322b618288ca93g-1.jpg?height=555&width=696&top_left_y=1083&top_left_x=693)

$$
\text { Figure 1 }
$$

If, for example, we want to calculate the time $t$ required for the train departing from $A$ to reach $T$, then from the similarly shaded similar triangles on the graph,

$$
\frac{t}{20}=\frac{45}{t-11}
$$

The equation holds on both sides (the calculable ratio is the same). Rearranging (if $t \neq 11$),

$$
t^{2}-11 t-900=0
$$

and the positive root of the equation is

$$
t=\frac{11+61}{2}=36
$$

The sought ratio is then

$$
\frac{A T}{T B}=\frac{t}{20}=\frac{9}{5}
$$

Starting from any of the three additional time intervals mentioned, we can proceed in a completely similar manner.

III. Solution: We can also directly set up an equation for the sought ratio $x=A T / B T$. The train departing from $A$ indeed arrives at $T$ from $A$ in $20 x$ minutes, and the train departing from $B$ arrives at $T$ from $B$ in $45 / x$ minutes. According to the first part of the problem,

$$
20 x-\frac{45}{x}=11 \quad \text { or } \quad 20 x^{2}-11 x-45=0
$$

From this, taking only the positive root,

$$
x=\frac{11+61}{40}=\frac{9}{5}
$$