Problem 7.4.1. Brothers Lёsha and Sasha decided to get to the skate park. They left at the same time, but Lёsha walked with the skateboard in his hands, while Sasha rode on the skateboard. It is known that Sasha rides the skateboard 3 times faster than Lёsha walks with the skateboard. After some time, they simultaneously switched their modes of transportation: Lёsha started riding the skateboard, and Sasha started walking. As a result, each of their speeds changed by a factor of 2: Lёsha's speed increased, and Sasha's speed decreased. It turned out that they arrived at the skate park at the same time. How many meters did Sasha ride on the skateboard, if the distance from home to the skate park is 3300 meters?
Answer: 1100.

Solution. Let $v$ be the walking speed of Lёsha, then $3v$ is the skateboarding speed of Sasha. After switching their modes of transportation, each of their speeds changed by a factor of 2, so $2v$ is Lёsha's skateboarding speed, and $1.5v$ is Sasha's walking speed. Therefore, on the second part of the journey, Lёsha's speed was $\frac{2v}{1.5v} = \frac{4}{3}$ times greater than Sasha's speed.

Let Lёsha walk $x$ meters, then Sasha skates $3x$ meters. Then Lёsha covers his remaining $(3300 - x)$ meters, and Sasha covers $(3300 - 3x)$ meters. The problem requires finding $3x$. Knowing the ratio of the speeds of the boys after changing their mode of transportation, we set up the proportion:

$$
\begin{gathered}
\frac{3300 - x}{3300 - 3x} = \frac{4}{3} \\
9900 - 3x = 13200 - 12x \\
9x = 3300 \\
3x = 1100
\end{gathered}
$$