Propositiones ad Acuendos Juvenes/Problems/26 - De Campo et Cursu Canis ac Fugo Leporis
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Propositiones ad Acuendos Juvenes by Alcuin of York: Problem $26$
- De Campo et Cursu Canis ac Fugo Leporis
- A Dog Chasing a Hare
- There is a field $150$ feet long.
- At one end is a dog, and at the other a hare.
- The dog chases when the hare runs.
- The dog leaps $9$ feet at a time,
- while the hare travels $7$ feet.
- How many feet will be travelled by the pursuing dog and the fleeing hare before the hare is seized?
Solution
The dog travels $675$ feet.
The hare travels $525$ feet.
Proof
Let $x$ be the length of time it takes the dog to catch the hare.
The dog travels $9 x$ during the time the hare travels $7 x$.
Thus we have:
- $9 x = 150 + 7 x$
which after algebra gives:
- $x = 75$
Hence the result.
$\blacksquare$
Sources
- c. 800: Alcuin of York: Propositiones ad Acuendos Juvenes ... (previous) ... (next)
- 1992: John Hadley/2 and David Singmaster: Problems to Sharpen the Young (Math. Gazette Vol. 76, no. 475: pp. 102 – 126) www.jstor.org/stable/3620384