Henry Ernest Dudeney/Modern Puzzles/38 - The Despatch-Rider/Solution

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Modern Puzzles by Henry Ernest Dudeney: $38$

The Despatch-Rider
If an army $40$ miles long advances $40$ miles
while a despatch-rider gallops from the rear to the front,
delivers a despatch to the commanding general,
and returns to the rear,
how far has he to travel?


Solution

$40 + \sqrt {2 \times 40^2} \approx 96 \cdotp 568$ miles.


Proof

We have that the length of the army is $40$ miles.

Let $d_1$ miles be the distance travelled by the despatch-rider from the rear to the front.

Let $d_2$ miles be the distance travelled by the despatch-rider returning from the front to the rear.

Let $t_1$ be the time taken to achieve the first leg of the journey.

Let $t_2$ be the time taken to achieve the second leg of the journey.

Let $v_d$ be the speed of the despatch-rider.

Let $v_a$ be the speed of the army.

We have that:

$40 = v_a \paren {t_1 + t_1}$

During the first leg, the speed relative to the army of the despatch-rider is $v_d - v_a$.

During the second leg, the speed relative to the army of the despatch-rider is $v_d + v_a$.

$40 = t_1 \paren {v_d - v_a} = t_2 \paren {v_d + v_a}$

Thus:

$40 = v_a \paren {t_1 + t_1} = v_a \paren {\dfrac {40} {v_d - v_a} + \dfrac {40} {v_d + v_a} }$

Let $r = \dfrac {v_d} {v_a}$.

Hence as $v_d > v_a$ we have that:

$r > 1$

Hence:

$1 = \dfrac 1 {r - 1} + \dfrac 1 {r + 1} = \dfrac {2 r} {r^2 - 1}$

This leads to:

$r^2 - 2 r - 1 = 0$

which, from the Quadratic Formula, gives us:

$r = 1 \pm \sqrt 2$

As $r > 1$:

$r = 1 + \sqrt 2$

which leads to:

$d_1 + d_2 = 40 + 40 \sqrt 2 = 40 + \sqrt {2 \times 40^2}$

$\blacksquare$


Sources

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