Problem 32 from the penguin book of puzzles.
A Side-car Problem
Atkins, Baldwin and Clarke had to go a journey of fifty-two miles across country. Atkins had a motor-bicycle with side-car for one passenger. How was he to take one of his companions a certain distance, drop him on the road to walk the remainder of the way, and return to pick up the second friend, who, starting at the same time, was already walking on the road, so that they should all arrive at their destination at exactly the same time? The motor-bicycle could do twenty miles an hour, Baldwin could walk five miles an hour, and Clarke could walk four miles an hour. Of course, each went at his proper speed throughout and there was no waiting.
I found that it’s easier to think about this via a distance time diagram just to get a sense of what the constraints are. Without loss of generality, we can assume A picks up B first.
The constraints here is the speed of A. After dropping off B, it needs to intercept C. If we consider the distance between A and C at time $t_1$, and how long it takes both of them to meet in terms of their relative speed, we can get the equation:
$$ 16 * t_1 = d = 24 * t_2 $$
Similarly, by symmetry,
$$ 15 * t_3 = 25 * t_2 $$
Finally, we get our third equation by considering one of the paths:
$$ 52 = 20 * t_1 + 5 * (t_2 + t_3) $$
Expressing everything in terms of $ t_2 $ gives $ t_2 = 156 / 130 $, and we can then solve for $ t_1 $ and $ t_3 $. Multiplying the speed will then tell us that A took B and travelled 36 miles, then travelled 24 miles back to fetch C and then travelled 40 miles to reach the destination.
I wonder if the condition was relaxed so that a can make multiple trips to fetch the slower person, whether there might be a more optimal strategy.