Propositiones ad Acuendos Juvenes/Problems/29 - De Civitate Rotunda/Solution 1

Propositiones ad Acuendos Juvenes by Alcuin of York: Problem $29$

There is a round town, $8000$ feet in circumference.

I want to build houses there,

each house being $20$ feet long

and $10$ feet wide.

How many houses must it contain,

each house being $30$ feet long

and $20$ feet wide?

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.

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.