Henry Ernest Dudeney/Modern Puzzles/37 - A Side-car Problem

by : $37$

 * A Side-car Problem
 * Atkins, Baldwin and Clarke had to go a journey of $52$ miles across country.
 * Atkins had a motor-bicycle with sidecar 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,
 * so that they should all arrive at their destination at exactly the same time?


 * The motor-bicycle could do $20$ miles per hour,
 * Baldwin could walk $5$ miles per hour,
 * and Clarke could walk $4$ miles per hour.


 * Of course, each went at his proper speed throughout and there was no waiting.


 * I might have complicated the problem by giving more passengers,
 * but I have purposely made it easy,
 * and all the distances are an exact number of miles -- without fractions.

Solution
There are $2$ solutions:


 * $(1): \quad$ Atkins sets off with Clarke and drives $40$ miles, leaving Clarke to walk the remaining $12$ miles
 * Atkins picks up Baldwin $16$ miles from the start and continues to the destination.


 * $(2): \quad$ Atkins sets off with Baldwin and drives $36$ miles, leaving Baldwin to walk the remaining $16$ miles
 * Atkins picks up Clarke $12$ miles from the start and continues to the destination.

The whole journey takes $5$ hours.

Proof
Let Atkins, Baldwin and Clarke be denoted by $A$, $B$ and $C$ respectively.

First let us assume that:
 * $A$ sets off with $C$
 * drops him off $d_1$ miles from the start
 * then returns to pick up $B$ at a point $d_2$ miles from the start.

Let $t_1$ be the time $A$ and $C$ reach $d_1$.

Let $t_2$ be the time $A$ reaches $d_2$ to pick up $B$.

Let $t$ be the time they all arrive at their destination.

We have:

Then we have:

Running the scenario in reverse gives another solution.