Henry Ernest Dudeney/Modern Puzzles/36 - More Bicycling

by : $36$

 * 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?

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:

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: