Propositiones ad Acuendos Juvenes/Problems/22 - De Campo Fastigioso
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Propositiones ad Acuendos Juvenes by Alcuin of York: Problem $22$
- De Campo Fastigioso
- An Irregular Field
- There is an irregular field,
- measuring $100$ perches along each side,
- and $50$ perches along each end,
- but $60$ perches across the middle.
- How many acres does it contain?
Solution
The area of an acre used here is $144$ square perches.
The length of the field is $100$ perches.
The length of the edges of the field is $50$ perches.
At the middle it is $60$ perches.
Join the length of the enda with the middle, which is $160$.
Take the $3$rd part of that, which is $53$.
Multiply it by $100$, which is $5300$.
Divide this into $12$ equal parts, giving $37$.
This gives the number of acres in the field.
$\blacksquare$
Historical Note
Alcuin's solution to this is suspect in a number of places.
- $(1): \quad$ Fractions are rounded inconsistently:
- At one point $\dfrac {160} 3 = 53 \tfrac 1 3$ is rounded to $53$.
- At another point $\dfrac {5200} {12} = 441 \tfrac 2 3$ is truncated to $441$.
- At yet another point $\dfrac {441} {12} = 36 \tfrac 3 4$ is rounded up to $37$.
- However, the actual value of $\dfrac {16 \, 000} {3 \times 12 \times 12} = 37.037 \ldots$ is actually remarkably accurate.
- $(2): \quad$ The method of averaging the $50$s and $60$s to get $\dfrac {160} 3$ is dubious.
- $(3): \quad$ The length given is measured along the side, which may not be the true length.
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