Henry Ernest Dudeney/Modern Puzzles/36 - More Bicycling/Solution

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Modern Puzzles by Henry Ernest Dudeney: $36$

More Bicycling
Referring to the last puzzle, let us now consider the case where a third rider has to share the same bicycle.
As a matter of fact, I understand that Anderson and Brown have taken a man named Carter into partnership, and the position today is this:
Anderson, Brown and Carter walk respectively $4$, $5$ and $3$ miles per hour,
and ride respectively $10$, $8$ and $12$ miles per hour.
How are they to use that single bicycle so that all shall complete the $20$ miles journey at the same time?


Solution

Anderson:

rides for $7 \tfrac {11} {27}$ miles, taking $44 \tfrac 4 9$ minutes
walks the remaining $12 \tfrac {16} {27}$ miles in $3$ hours and $8 \tfrac 8 9$ minutes.

Brown:

walks those $7 \tfrac {11} {27}$ miles in $1$ hour $28 \tfrac 8 9$ minutes
rides for $1 \tfrac {13} {27}$ miles, taking $11 \tfrac 1 9$ minutes
walks the remaining $11 \tfrac 1 9$ miles in $2$ hours $13 \tfrac 1 3$ minutes.

Carter:

walks $8 \tfrac 8 9$ miles in $2$ hour $57 \tfrac {21} {27}$ minutes
rides the remaining $11 \tfrac 1 9$ miles in $55 \tfrac 5 9$ minutes.

The total journey time is $3$ hours $53 \tfrac 1 3$ minutes.


Proof

Let Anderson, Brown and Carter be denoted by $A$, $B$ and $C$ respectively.

To keep it simple, we will assume:

$A$ starts by cycling, then leaves the bike for $B$
$B$ takes over cycling from $A$, then leaves the bike for $C$
$C$ finishes the journey on the cycle.

Let $d_1$ miles be the distance from the start to where $A$ dismounts to start walking and $B$ starts riding.

Let $d_2$ miles be the distance from $d_1$ to where $B$ dismounts to start walking.

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

Let $t_{a_1}$ hours be the time taken by $A$ to travel $d_1$.

Let $t_{b_1}$ hours be the time taken by $B$ to travel $d_1$.

Let $t_{c_1}$ hours be the time taken by $C$ to travel $d_1$.

Let $t_{a_2}$ hours be the time taken by $A$ to travel $d_2$.

Let $t_{b_2}$ hours be the time taken by $B$ to travel $d_2$.

Let $t_{c_2}$ hours be the time taken by $C$ to travel $d_2$.

We have:

\(\ds d_1\) \(=\) \(\ds 10 t_{a_1}\)
\(\ds \) \(=\) \(\ds 5 t_{b_1}\)
\(\ds \) \(=\) \(\ds 3 t_{c_1}\)
\(\ds d_2\) \(=\) \(\ds 4 t_{a_2}\)
\(\ds \) \(=\) \(\ds 8 t_{b_2}\)
\(\ds \) \(=\) \(\ds 3 t_{c_2}\)
\(\ds 20 - d_1 - d_2\) \(=\) \(\ds 4 \paren {t - t_{a_1} - t_{a_2} }\)
\(\ds \) \(=\) \(\ds 5 \paren {t - t_{b_1}- t_{b_2} }\)
\(\ds \) \(=\) \(\ds 12 \paren {t - t_{c_1} - t_{c_2} }\)


We set up this system of linear simultaneous equations in matrix form as:

$\begin {pmatrix}

1 & 0 & -10 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & -5 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & -3 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & -4 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & -8 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & -3 & 0 \\ 1 & 1 & -4 & 0 & 0 & -4 & 0 & 0 & 4 \\ 1 & 1 & 0 & -5 & 0 & 0 & -5 & 0 & 5 \\ 1 & 1 & 0 & 0 & -12 & 0 & 0 & -12 & 12 \\ \end {pmatrix} \begin {pmatrix} d_1 \\ d_2 \\ t_{a_1} \\ t_{b_1} \\ t_{c_1} \\ t_{a_2} \\ t_{b_2} \\ t_{c_2} \\ t \\ \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 20 \\ 20 \\ 20 \\ \end {pmatrix}$

It remains to solve this matrix equation.


In reduced echelon form, this gives:

$\begin {pmatrix}

1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end {pmatrix} \begin {pmatrix} d_1 \\ d_2 \\ t_{a_1} \\ t_{b_1} \\ t_{c_1} \\ t_{a_2} \\ t_{b_2} \\ t_{c_2} \\ t \\ \end {pmatrix} = \begin {pmatrix} 200/27 \\ 40/27 \\ 20/27 \\ 40/27 \\ 200/81 \\ 10/27 \\ 5/27 \\ 40/81 \\ 35/9 \\ \end {pmatrix}$


The result can be read directly and converted into the appropriate units.

$\blacksquare$


Sources

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