Propositiones ad Acuendos Juvenes/Problems/28 - De Civitate Triangula
Propositiones ad Acuendos Juvenes by Alcuin of York: Problem $28$
- De Civitate Triangula
- A Triangular Town
- A triangular town measures $100$ feet along one side,
- $100$ feet along another side
- and $90$ feet along the front.
- How many houses can there be?
Solution
Adding the two [[Definition:Side of Polygon|sides] makes $200$ feet.
The half of $200$ is $100$.
The front measures $90$ and the half of $90$ is $45$.
Because each house is $20$ feet long and $20$ feet wide:
- take one $20$th of $100$ which is $5$
- and take one $10$th of $40$ which is $4$.
Then $5$ times $4$ is $20$.
That is the number of houses the town should contain.
Historical Note
According to the Roman formula the area of the triangle is $4500$ square feet.
This is $22 \frac 1 2$ house areas.
The actual area is approximately $4019$ square feet, which is just over $20$ house areas.
The marking down of $45$ to $40$ in the calculation may be an attempt to compensate for the inaccuracy of the formula.
However, fitting them into a triangle is not easy.
David Singmaster reports that he managed to fit $15$ in.
John Hadley, on the other hand, managed $18$, but bent the walls of some of the houses slightly.
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