1. Friends Vasya, Petya, and Kolya live in the same house. One day, Vasya and Petya went fishing at the lake on foot. Kolya stayed home, promising to meet his friends on a bicycle on their way back. Vasya was the first to head home, and at the same time, Kolya set out on a bicycle to meet him. Petya, moving at the same speed as Vasya, left the lake for home at the moment Kolya and Vasya met. Kolya, upon meeting Vasya, immediately turned around and took him home, and then immediately set off again on the bicycle towards the lake. Meeting Petya, Kolya turned around again and took his friend home. As a result, the time Petya spent on the journey from the lake to home was $4 / 3$ of the time Vasya spent on the same journey. How many times slower would Vasya have reached home if he had walked the entire way? (8 points)
Solution: Let $x$ be the speed of Vasya and Petya, $v$ be the speed of cyclist Kolya, and $S$ be the distance from home to the lake. Then Vasya and Kolya met after time $t=\frac{S}{x+v}$ from the start of the journey. After Kolya's second departure from home, the meeting between Petya and Kolya occurred after time $t_{1}=\frac{S-x t}{x+v}$. From the lake to home, Vasya was traveling for $2 t$, and Petya for $-t+2 t_{1}$. According to the condition, $t+2 t_{1}=\frac{8 t}{3}$. Therefore, $\frac{2 S-2 x t}{x+v}=\frac{5 S}{3(x+v)} \Rightarrow 6 S-\frac{6 x S}{x+v}=5 S \Rightarrow \frac{6 x}{x+v}=1 \Rightarrow v=5 x$. We need to find the ratio $\frac{S}{x}$ to $2 t$. Thus, the desired value $\frac{S}{2 x t}=\frac{x+v}{2 x}=\frac{x+5 x}{2 x}=3$. Answer: 3 times slower.