The fundamental problem of many encounters on the roundabout is that their distance difference is the circumference of the roundabout when they meet. At this time, it depends on whether they face each other but walk in the same direction. The meeting time is equal to the length of the circular road divided by the speed difference, and the meeting time in the same direction is equal to the length of the circular road divided by the speed sum.

Example 1. Party A and Party B are running on the same oval track. They started from the same place and ran in opposite directions at the same time. After everyone reached the starting point after running the first lap, they immediately turned back and accelerated the second lap. When running the first lap, the speed of Party B is 2/3 of that of Party A, which is 1/3 higher than that of the first lap and 1/5 higher than that of Party B. It is known that the second meeting point of Party A and Party B is 190 meters away from the first meeting point.

Solution: This question is very difficult. Let's analyze it: How many meters do we require for the oval runway? The only value related to demand is that the second intersection of A and B is 190 meters away from the first intersection. So as long as the distance between the first intersection and the starting point (the running direction of A) is found first, and then the distance between the second intersection and the starting point (the running direction of A) is found, the difference between them is equal to 190 meters. At the same time, it needs to be understood that the sum of the distances that Party A and Party B meet for the first time is only the length of the runway; Both parties spend the same amount of time. When we meet for the second time, the total time of both parties is the same, and the running distance is three laps (that is, each party runs one lap, and the two times add up to one lap).

Let this oval runway be x meters long, and the speed of the first lap of person A is y, then the speed of the first lap of person B is y2/3; If the speed of Party A's second lap is y( 1+ 1/3)=4y/3, the speed of Party B's second lap is y2/3 (1+1/5) = 4y/5; Let's assume that the first time we meet A and B is A, and the second time we meet A and B is B and C respectively. A = x \u( y+y2/3)= x×(3/5y)= 3x/5y, then the distance from the first intersection point to the starting point (the running direction of A): ya= y×3x/5y=3x/5. Time for A to complete a lap: x/y, and time for B to complete a lap: x/(y2/3). Then there are: x/y+b = x/(y2/3)+c, b4y/3+c4y/5 = X. From the above two formulas, c=5x/32y,

Distance from the second intersection point to the starting point (direction of A running): (4y/5)c=(4y/5)(5x/32y)=x/8.

3x/5-x/8= 190, the solution is: x=400 (m).

A: The oval runway is 400 meters long.