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The 40 Mile Army

An old chestnut, dating back to Sam Loyd, asks this question: There is an army marching at a constant speed in a line 40 miles long. A horseman at the back of the army rides to the front with a message, then immediately turns and rides back to the back of the army. By the time he gets there, the army has marched 40 miles since he started his trip. If the horseman also rides at a constant speed, how far did the horse run in the trip to the front and back again?

The solution requires a little algebra. Let

x

be the distance in miles that the army marches during the horseman's run forward. Then the horseman traveled

40 + x

miles while the army marched

x

miles. The return trip for the horseman is 40 miles shorter, since the army has traveled 40 miles by the time he finishes, so it is

x

miles for the horseman, while the army marches

40 - x

miles. The total distances are then 40 miles for the army and

40 + 2 x

miles for the horseman.

Since both the horseman and the army travel at constant rates, the ratio of their rates is also constant. So we have:

$\frac{40 + x}{x} = \frac{40 + 2 x}{40}$

$40 (40 + x) = (40 + 2 x) x$

$1600 + 40 x = 40 x + 2 x^{2}$

$x^{2} = 800$

So $x = \sqrt{800} = 20 \sqrt{2}$ and the horseman has traveled $40 + 40 \sqrt{2}$ miles or about 96.6 miles.