Our three travelers shall be dubbed Adam, Bill and Carl.
! Adam and Bill set off on the horses for a time, then Adam dismounts and proceeds to walk the remainder of the distance. Bill rides his horse, accompanied by the empty horse, and returns to Carl, who has been walking the entire time thus far. Then Carl mounts up and he and Bill proceed to the end, where they arrive at the exact same time as Adam.
! Let us examine the distance that Bill will travel. Bill’s journey can be described as (a) to the point where he will eventually return for Carl; (b) to the point where he drops off Adam; (c) back to the point where he picks up Carl; (d) back to the point where he originally dropped off Adam; and (e) to the end. We can see that b = c = d, since these three legs are all from point 1 to point 2, back to point 1, and back to point 2. Bill then traveled a distance of a + b + c + d + e, but c was backtracking, so his total progress was a + b – c + d + e.
!We now proceed to examine this problem from another angle, leaving Bill of the equations for now. Adam’s journey can be broken into three distances of R1T1, R1T2 and R2T3. Carl, on the other hand, has R2T1, R1T2 and R1T3. [R1 = horse rate, R2 = foot rate] Thus they are walking the same distance for the middle segment, which is intuitive, since they are both on foot. Since they are traveling the same total distance, we have R1T1 + R2T3 = R2T1 + R1T3, or T1(R1-R2) = T3(R1-R2), and T1 = T3. But this means that Carl must now take the same amount of time to get to the end as Adam took to accomplish the first segment – which means that Adam’s first segment is the same distance as Carl’s last segment, since they are both on horseback for that respective segment.
!But we also know that Adam’s first journey of R1T1 = Bill’s a + b, and Carl’s last journey of R1T3 = Bill’s d + e. So the total distance accomplished by Bill is R1T1 – c + R1T3, but R1T1 = R1T3, so we can simplify this to 2xR1T1 – c.
!For calculation purposes, we will state that D1 = R1T1. During this time, Carl has traveled D1 x (R2 / R1), where R2 is the rate of walking. Thus, in order to return to Carl, Bill must travel D1 – D1 x (R2 / R1), or D1 x (1 – R2 / R1), or (D1 x R1 – D1 x R2) / R1, or (D1 / R1) x (R1 – R2). The closing rate of the two is R1 + R2, and he will travel R1 / (R1 + R2) of that distance, or D1 x (R1 – R2) / (R1 + R2). This distance is the same as b = c = d. So the total distance traveled by Adam or Carl is 2D1 – D1 (R1 – R2) / (R1 + R2), or D1(2 - (R1 – R2) / (R1 + R2)). From here it is elementary to determine the distances traveled and their respective times.
!For the given numbers, for instance, we have D1(2 – (6 – 4) / (6 + 4)) = 13.5, so D1 (1.8) = 13.5, and D1 = 7.5. c = 7.5 (0.2) = 1.5. Adam and Bill should ride for 7.5km / 6 kmh = 1.25 hours, and then Bill should go back and get Carl. They will all meet at the end at the same time.