Propositiones ad Acuendos Juvenes/Problems/29 - De Civitate Rotunda
Propositiones ad Acuendos Juvenes by Alcuin of York: Problem $29$
- De Civitate Rotunda
- A Round Town
- There is a round town, $8000$ feet in circumference.
Solution 1
One circuit of the town is $8000$ feet.
Divided in proportion $3$ to $2$ gives $4800$ and $3200$.
These must contain the length and breadth of the houses.
So take from each one half and there remains:
- of the larger $2400$
- and of the smaller $1600$.
Divide $1600$ by $20$, and there are $80$ $20$s.
Again the greater, that is $2400$, divided into $30$s, gives $80$.
Take $80$ times $80$ and there are $6400$.
This is the number of houses which can be built in the town according to the above instructions.
Solution 2
The boundary of the town is $8000$ feet.
Take a quarter of $8000$, getting $2000$.
Again take a third of $8000$, getting $2666$.
Take the half of $2000$, which is $1000$, and the half of $2666$, which is $1333$.
Not the thirtieth part of $1333$ is $44$.
Likewise the twentieth part of $1000$ is $50$.
$50$ times $44$ is $2200$.
Then form $2200$ times $4$ whch is $8800$.
This is the total number of houses.
Historical Note
The actual area of the town is approximately $8488$ house areas.
Using $\AA = \dfrac {C^2} {16}$ as in $25$: De Campo Rotundo, the area is approximately $6667$ house areas.
Alcuin assumes the circle contains a $1600 \times 2400$ rectangle, but such a circle could hold over $10 \, 000$ house areas.
David Singmaster reports that he managed to fit $8307$ in, but suspects it may be possible to do better.
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