Shen A.H.

On a highway, a pedestrian and a cyclist are moving in one direction, while a cart and a car are moving in the opposite direction. All participants are moving at constant speeds (each with their own). The cyclist first overtook the pedestrian, then after some time met the cart, and then after the same amount of time met the car. The car first met the cyclist, then after some time met the pedestrian, and then after the same amount of time overtook the cart. The cyclist overtook the pedestrian at 10 o'clock, and the pedestrian met the car at 11 o'clock. When did the pedestrian meet the cart?
Let's draw the graphs of motion and mark their points of intersection. Let points $A, L, C$ correspond to the overtakes and meetings of the cyclist, points $C, K, B$ to the overtakes and meetings of the car, and the meeting of the cart with the pedestrian - point $M$ (see fig). According to the problem, $L$ and $K$ are the midpoints of sides $A C$ and $B C$ of triangle $A B C$, hence $M$ is the point of intersection of its medians. $M$ divides the median $A K$ in the ratio $2: 1$, so the projection of point $M$ divides the time interval from 10 to 11 hours in the same ratio. Therefore, the meeting occurred at 10.40.

![](https://cdn.mathpix.com/cropped/2024_05_06_ed128c2d5b4d71db92ebg-09.jpg?height=894&width=1169&top_left_y=-1&top_left_x=433)

## Answer

At 10.40.