1. Along the road from Yolkiino to Palkino, there is an oak tree, from which it is twice as close to Yolkiino as to Palkino. Fyodor, traveling at a constant (and greater than 0) speed from Yolkiino to Palkino, was twice as close to the oak tree as to Yolkiino at 12:00. At 12:40, it was again found that Fyodor was twice as close to the oak tree as to Yolkiino. When will Fyodor arrive in Palkino?
Answer: 13.10. Solution: On the road from Yolkiino to Palkino, there are two points that are twice as close to the oak tree as to Yolkiino. One of them $\left(\Phi_{1}\right)$ is between Yolkiino and the oak tree, and the other ( $\Phi_{2}$ ) is between the oak tree and Palkino (see figure). For

![](https://cdn.mathpix.com/cropped/2024_05_06_2b8477a810387df67742g-1.jpg?height=131&width=1070&top_left_y=577&top_left_x=847)
convenience in calculations, let's assume the entire path from Yolkiino to Palkino is 9. Then ED $=3, \Phi_{1} D=1, E \Phi_{2}=2 ED=ED+D \Phi_{2}$, from which $D \Phi_{2}=ED=3$ and $\Phi_{2} P=EP-E \Phi_{2}=3$. The path $\Phi_{1} \Phi_{2}=\Phi_{1} D + D \Phi_{2}$ of length 4 Fedya traveled in 40 minutes, so the path $\Phi_{2} P$ of length 3 he traveled in 30 minutes, from which we get the answer.