4. (7 points) Two pedestrians set out at dawn. Each walked at a constant speed. One walked from $A$ to $B$, the other from $B$ to $A$. They met at noon (i.e., exactly at 12 o'clock) and, without stopping, arrived: one at $B$ at 4 PM, and the other at $A$ at 9 PM. At what time was dawn that day?
Answer: at 6 AM.

Solution. Let's denote the meeting point as $C$. Let $x$ be the number of hours from dawn to noon.

The speed of the first pedestrian on segment $A C$ is $A C / x$, and on segment $B C$ it is $B C / 4$. Since his speed is constant, we have $\frac{A C}{x}=\frac{B C}{4}$, which can be rewritten as $\frac{A C}{B C}=\frac{x}{4}$.

Similarly for the second pedestrian: the equality of speeds on segments $B C$ and $A C$ results in the ratio $\frac{B C}{x}=\frac{A C}{9}$, which we can rewrite as $\frac{A C}{B C}=\frac{9}{x}$.

We get $\frac{x}{4}=\frac{9}{x}$, and by the property of proportions, $x^{2}=36, x=6$. Dawn was 6 hours before noon, i.e., at 6 AM.

Criteria. Any correct solution: 7 points.

The time interval from dawn to the meeting is correctly found, but the time of dawn is not found or is found with an error: 5 points.

Only the answer without a solution: 1 point.