# Distinct points of meeting in a track

$$A$$ and $$B$$ start running simultaneously towards each other from the two ends of a track $$XY$$ and the ratio of their speeds is $$3$$ : $$4$$. Every time they meet, they interchange their speeds and also reverse their directions. At how many distinct points on the track do they meet each other, if they run continuously between $$2$$ ends of the tracks? The length of the track is $$100m.$$

Now I have understood the problem very well and I tried to solve it as well using general approach but after few steps I realized that solving it by general common way will be very lengthy and tedious and I was not getting anywhere.
Then I turned to the solution for this problem and as part of the solution it was given that

It is similar to the case of circular motion where $$2$$ runners are running in opposite direction. Hence number of meeting points is $$(3 +4)=7$$.

Now I am also not able to understand how this case is similar to the $$2$$ runners running in opposite direction around a circular field. I am aware of this formula to find the number of distinct points for the above case. But I am not able to relate the two cases given as part of the problem.

• So the track is nor circular? What happens when a runner reaches one of the ends? Does he turn around? Commented Oct 28, 2021 at 16:08
• Yes, the tracks are not circular. After reaching the end, they turn around. Commented Oct 28, 2021 at 16:12
• Because they turn around, this is equivalent to a loop. Because they interchange speeds, we can interpret this as two runners with the constant speeds who do not change speed or direction. And notice that they always run in the opposite directions. Commented Oct 28, 2021 at 16:17
• @Vasya : Can you please elaborate it a little more? Commented Oct 28, 2021 at 16:18

Let's give each runner a baton, $$A$$ gets baton $$A_1$$ and $$B$$ gets baton $$B_1$$. Every time the runners meet they exchange the batons. Let's track the movement of the batons. At the beginning, baton $$A_1$$ moves with the speed $$3x$$ and baton $$B_1$$ moves with the speed $$4x$$. When the runners meet, runner $$A$$ gets baton $$B_1$$ and runs in the same direction in which $$B_1$$ was moving before the meeting and with the same speed $$4x$$. Similarly, $$A_1$$ continues to move with the speed $$3x$$ in the same direction as before. Thus, we can simplify the problem to track two batons that go with the constant speed in the opposite directions.
Note that the batons start at the opposite ends of the track $$XY$$ and move toward each other. When they meet, they cover the length of the track, baton $$A_1$$ covers $$\frac{3}{7}$$ and baton $$B_1$$ covers $$\frac{4}{7}$$ of the track. Notice that this is equivalent to two batons that move over a circular track of the same length if they start at the same point and move in the opposite directions.
Divide the track into $$7$$ equal parts and name the points as given in the figure. Suppose ratio of the speed of player starting at X and Y be $$3/4$$ then it is easy to see that they will meet each other at point $$3$$. After this player $$X$$ need to cover a distance of $$6$$ units before meeting player $$Y$$ again. $$\begin{array}{c|c|c} \style{font-family:inherit}{\text{Number of times player X and Y met}} & \style{font-family:inherit}{\text{position of X}} & \style{font-family:inherit}{\text{Direction of X}} \\\hline 1 & 3 & \rightarrow \\\hline 2 & 5 & \leftarrow \\\hline 3 & 1 & \rightarrow \\\hline 4 & 7 & \leftarrow\text{(after reflection)} \\\hline 5 & 1 & \leftarrow \\\hline 6 & 5 & \rightarrow \\\hline 7 & 3 & \leftarrow \\\hline 8 & 3 & \rightarrow \end{array}$$
As you can see there are only $$4$$ position in which player $$X$$ and $$Y$$ meet.