Henry Ernest Dudeney/Modern Puzzles/37 - A Side-car Problem/Solution
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Modern Puzzles by Henry Ernest Dudeney: $37$
- 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?
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:
\(\text {(1)}: \quad\) | \(\ds t_1\) | \(=\) | \(\ds \dfrac {d_1} {20}\) | $A$ sets off with $C$ | ||||||||||
\(\text {(2)}: \quad\) | \(\ds t_2 - t_1\) | \(=\) | \(\ds \dfrac {d_1 - d_2} {20}\) | $A$ returns to pick up $B$ | ||||||||||
\(\text {(3)}: \quad\) | \(\ds t - t_2\) | \(=\) | \(\ds \dfrac {52 - d_2} {20}\) | $A$ travels to destination with $B$ | ||||||||||
\(\text {(4)}: \quad\) | \(\ds t_2\) | \(=\) | \(\ds \dfrac {d_2} 5\) | $B$ sets off walking to $d_2$ | ||||||||||
\(\text {(5)}: \quad\) | \(\ds t - t_1\) | \(=\) | \(\ds \dfrac {52 - d_1} 4\) | $C$ walks from $d_1$ to the final destination | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {d_2} 5 - \dfrac {d_1} {20}\) | \(=\) | \(\ds \dfrac {d_1 - d_2} {20}\) | substituting for $t_1$ and $t_2$ from $(1)$ and $(4)$ into $(2)$ | ||||||||||
\(\text {(6)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds d_2\) | \(=\) | \(\ds \dfrac {2 d_1} 5\) | simplifying | |||||||||
\(\text {(7)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds t - \dfrac {d_1} {20}\) | \(=\) | \(\ds \dfrac {52 - d_1} 4\) | substituting for $t_1$ from $(1)$ into $(5)$ | |||||||||
\(\text {(8)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds t - \dfrac {d_2} 5\) | \(=\) | \(\ds \dfrac {52 - d_2} {20}\) | substituting for $t_2$ from $(4)$ into $(3)$ | |||||||||
\(\text {(9)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \dfrac {52 - d_1} 4 + \dfrac {d_1} {20}\) | \(=\) | \(\ds \dfrac {52 - d_2} {20} + \dfrac {d_2} 5\) | eliminating $t$ from between $(7)$ and $(8)$ | |||||||||
\(\ds \leadsto \ \ \) | \(\ds 5 \paren {52 - d_1} + d_1\) | \(=\) | \(\ds \paren {52 - d_2} + 4 d_2\) | multiplying by $20$ to clear fractions | ||||||||||
\(\text {(10)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds 4 \times 52 - 4 d_1\) | \(=\) | \(\ds 3 d_2\) | simplifying | |||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {6 d_1} 5\) | substituting for $d_2$ from $(6)$ into $(10)$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds d_1\) | \(=\) | \(\ds 40\) | simplifying |
Then we have:
\(\ds d_2\) | \(=\) | \(\ds \dfrac {2 \times 40} 5 = 16\) | from $(6)$: $d_2 = \dfrac {2 d_1} 5$ | |||||||||||
\(\ds t\) | \(=\) | \(\ds \dfrac {52 - 40} 4 + \dfrac {40} {20} = 5\) | from $(5)$ and $(1)$: $t = \dfrac {52 - d_1} 4 + \dfrac {d_1} {20}$ |
Running the scenario in reverse gives another solution.
$\blacksquare$
Sources
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $37$. -- A Side-car Problem
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $65$. A Sidecar Problem