Dudeney's "Modern Puzzles" - number 38, The Despatch-Rider

In Henry Ernest Dudeney's "Modern Puzzles" of 1926 (now out of print) we have:

38. The Despatch-Rider

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

My solution is that it seems clear that the rider goes twice as fast as the army, so he goes 80 miles one way and 2/3 of 40 miles the other, being an obvious 19 1/3 miles.

But Dudeney's solution is:

"The answer is the square root of twice the square of 40, added to 40. This is $96.568$ miles, or, roughly $96 \tfrac 1 2$ miles."

Anybody got any idea how he arrives at this?

Solution 1:

Alternative approach:

Assume that the army's length is $(40)$.

Assume that the rider's speed is $(r > 1)$ while the army's speed is $1$.

Assume that the first leg of the trip uses time $t_1.$

Assume that the second leg of the trip uses time $t_2.$

The distance traveled by the rider is $r(t_1 + t_2)$.

Further, from the problem's constraints,
$40 = (1) \times (t_1 + t_2) \implies (t_1 + t_2) = 40.$

Therefore, the problem is reduced to computing $r$.

During $t_1$, the speed at which the rider covers the $(40)$ is (in effect) $(r - 1)$.

During $t_2$, the speed at which the rider covers the $(40)$ is (in effect) $(r + 1)$.

Therefore, $t_1(r - 1) = 40 = t_2(r + 1).$

Therefore $40 = t_1 + t_2 = \displaystyle \frac{40}{r-1} + \frac{40}{r+1}.$

This implies that $1 = \displaystyle \frac{1}{r-1} + \frac{1}{r+1} = \frac{2r}{r^2 - 1}$.

Therefore, $\displaystyle r^2 - 2r - 1 = 0 \implies r = \left(\frac{1}{2}\right) \times \left[2 \pm \sqrt{8}\right] = 1 \pm \sqrt{2}$.

Since $r$ must be $> 0$, (in fact $r$ must be $> 1$), you have that $r = 1 + \sqrt{2}$.

Solution 2:

Let $x_{0}$ be the length of the army in miles (in this problem, $x_{0}=40$). Assume the army is travelling along the $x$-axis, with the rear of the army starting at $0$, the front of the army starting at $x_{0}$. Assume without loss of generality that at time $1$, the despatch-rider has finished (so the front of the army has reached position $2x_{0}$ and the despatch-rider is at position $x_{0}$, the back of the army).

Let $t_{0}$ be the time at which the despatch-rider reached the front of the army (after which point he reverses direction), and let $m$ be the speed of the despatch-rider. Then we have the following facts:

• the position of the front of the army at time $t$ is $x_{0} + x_{0}t$ for $0\le t \le 1$
• position of the despatch-rider at time $t$ is $mt$ for $0\le t \le t_{0}$ and $-mt+b$ for $t_{0}\le t \le 1$, where $m$ is the constant speed of the despatch rider, and $b$ is a constant to be determined.

We can find $m, b, t_{0}$ in terms of known quantities. Then the distance travelled by the despatch-rider will be $m$ (as he travels at speed $m$ for time $1$).

From the second bullet point above, we must have $$mt_{0} = -mt_{0}+b \Rightarrow 2mt_{0} - b = 0. \tag{1}$$

Also, since the despatch-rider is at position $x_{0}$ at time $1$, the second bullet point implies that $$-m+b = x_{0}.\tag{2}$$

Adding $(1)$ and $(2)$ gives $(2t_{0}-1)m = x_{0}\Rightarrow m = \frac{x_{0}}{2t_{0} - 1}$.

So $(1)$ implies that $b = 2t_{0}m = \frac{2t_{0}x_{0}}{2t_{0}-1}$.

Now, the first bullet point above implies that $x_{0}+x_{0}t_{0} = -mt_{0}+b$ (i.e. at time $t_{0}$ the head of the army and the despatch-rider are at the same position). So we have

$$\begin{align} x_{0} + x_{0}t_{0} &= -mt_{0}+b\\ \Rightarrow x_{0} + x_{0}t_{0} &= -\frac{x_{0}t_{0}}{2t_{0} - 1} + \frac{2t_{0}x_{0}}{2t_{0}-1} \\ \Rightarrow 1 + t_{0} &= \frac{t_{0}}{2t_{0} - 1}\\ \Rightarrow (2t_{0} - 1)(1 + t_{0}) &= t_{0} \\ \Rightarrow 2t_{0} + 2t_{0}^{2} - 1 - t_{0} &= t_{0} \\ \Rightarrow 2t_{0}^{2} &= 1 \\ \Rightarrow t_{0} &= \frac{\sqrt{2}}{2}. \end{align}$$

So $m = \frac{x_{0}}{2t_{0} - 1} = \frac{x_{0}}{\sqrt{2}-1} = \left(\sqrt{2} + 1\right)x_{0}$.

Hence the distance travelled by the despatch-rider is $m = \left(\sqrt{2} + 1\right)x_{0}$. In your particular problem, since $x_{0}$ is $40$ miles, the total distance travelled by the despatch-rider is $\left(\sqrt{2} + 1\right)\times 40 \approx 96.57$ miles.