Henry Ernest Dudeney/Modern Puzzles/35 - Sharing a Bicycle

by : $35$

 * Sharing a Bicycle


 * Two brothers had to go on a journey and arrive at the same time.
 * They had only a single bicycle, which they rode in turns, each rider leaving it in the hedge when he dismounted for the one walking behind to pick up,
 * and walking ahead himself, to be again overtaken.


 * What was the best way of arranging their distances?


 * As their walking and riding speeds were the same, it is extremely easy.
 * Simply divide the route into any even number of equal stages and drop the bicycle at every stage, using the cyclometer.
 * Each man would then walk half-way and ride half-way.


 * But here is a case that will require a little more thought.


 * Anderson and Brown have to go $20$ miles and arrive at exactly the same time.
 * They have only one bicycle.
 * Anderson can only walk $4$ miles an hour,
 * while Brown can walk $5$ miles an hour,
 * but Anderson can ride $10$ miles an hour to Brown's $8$ miles an hour.


 * How are they to arrange the journey?


 * Each man always either walks or rides at the speeds mentioned, without any rests.

Solution
Anderson cycles $11 \tfrac 1 9$ miles, drops the bike, and walks the rest of the way, which is $8 \tfrac 8 9$ miles.

Thus he cycles for $1 \tfrac 1 9$ hours and walks for $2 \tfrac 2 9$ hours.

Brown thus walks $11 \tfrac 1 9$ miles, picks up the bike, and walks the rest of the way, which is $8 \tfrac 8 9$ miles.

Thus he also walks for $2 \tfrac 2 9$ hours and cycles for $1 \tfrac 1 9$ hours.

Proof
Let Anderson and Brown be denoted by $A$ and $B$ respectively.

To keep it simple, we will assume one changeover, and that $A$ starts by cycling.

Let $d$ miles be the distance from the start to where $A$ dismounts to start walking.

Let $t$ hours be the time taken to do the total journey.

Let $t_a$ hours be the time taken by $A$ to travel $d$.

Let $t_b$ hours be the time taken by $B$ to travel $d$.

We have: