Task 1 - 130911 Two points $A$ and $B$ at the same height above the ground are at the same distance and on the same side of a straight, high wall. The segment $A B$ is $51 \mathrm{~m}$ long. A sound generated at $A$ reaches $B$ directly exactly $\frac{1}{10}$ s earlier than via the reflection off the wall.

Determine the distance of each of the two points $A$ and $B$ from the wall, assuming that the sound travels exactly $340 \mathrm{~m}$ per second.
According to the law of reflection, the angle of incidence is equal to the angle of reflection. Let $C$ be the point on the wall where the sound from $A$ is reflected to $B$. Then, due to the mentioned angle equality and the equality of the distances of points $A$ and $B$ from the wall, the triangle $\triangle A B C$ is isosceles with $|A B|=|C B|$. Since the sound takes exactly $\frac{1}{10}$ s longer on the path from $A$ via $C$ to $B$ than on the direct path, and since the sound travels $34 \mathrm{~m}$ in each tenth of a second according to the problem, we have

$$
|A C|+|C B|=|A B|+34 m=51 m+34 m=85 m
$$

thus $|A C|=|C B|=42.5 \mathrm{~m}$.

The distance of point $A$ from the wall is therefore, according to the Pythagorean theorem,

$$
\sqrt{42.5^{2}-25.5^{2}} m=34 m
$$