The second part: the square problem

Party A and Party B set out from the two vertices A and C opposite to the square pool ABCD with a circumference of 1600 meters and walked along the edge of the pool in the direction of A-B-C-D-A, with the speed of A being 50 meters per minute and the speed of B being 46 meters per minute. When did A and B walk on the same side for the first time? How many minutes did you walk on the same side for the first time?

Analysis:

In order for two people to walk on the same side, the distance between A and B must be less than one side, and A must cross the vertex.

It takes at least 400 ÷ (50-46) = 100 minutes100 = 5000m, 5000 ÷ 400 = 65438 to catch up with B. At this point, A has traveled 50×100 = 5000m.

So it takes 200 ÷ 50 = 4 minutes to walk, that is, 100+4 = 104 minutes after departure, and they walk on the same side for the first time.

At this time, the distance between Party A and Party B is 400× 2- 104× (50-46) = 384 meters, and Party B still has 16 meters to walk the same side for the first time, so it takes 16 ÷ 46 = 8/23 minutes.

The third part: the multi-person tour of the circular runway

Let A, B and C walk around the campus in the same direction at a certain speed for 6, 7, 1 1 min. After starting from the starting point A, B leaves later than A 1 minute, and C leaves later than B, so how many minutes after A leaves, A, B and C pass at the same time for the first time?

Analysis:

According to the conditions, when C set out, A just drove for 5+ 1 = 6 minutes, that is, when A and C passed the starting point again at the same time, it was 6 × 1 1 = 66 minutes later.

Since B takes 7-5 = 2 minutes, it means that the multiple of 66 should be divided by more than 7 minutes. When 66× 3 = 198 minutes, 198 ÷ 7 = 28...2 minutes, the condition is satisfied.

Therefore, the first simultaneous departure place of ABC is 6+ 198 = 204 minutes after the departure of A.