All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
The Garden of Pythagoras (Posted on 2010-08-03) Difficulty: 2 of 5
The god Zeus commanded the Sybarites to furnish his temple with a large piece of land on which to construct a garden or precinct. Not wishing to defy the god, but reluctant to part with so much land, the Sybarites made the donation subject to conditions which they believed could not be fulfilled. They required that the garden be laid out with an open central square, abutted by the hypotenuses of 4 right triangular groves, such that:
1. All dimensions of the square and triangles must be measurable in whole numbers of cubits;
2. No two of the outer sides of the triangles should be of the same length;
3. No two sides of any triangle should have a common divisor.
4. The whole should be of the minimum size permitted by the foregoing requirements.
The priests of Zeus turned to Pythagoras for assistance. To the consternation of the Sybarites, Pythagoras not only immediately produced a plan compliant with these specifications, but into the bargain made proposals for a grand estate, laid out in like manner, but with an octagonal centerpiece!
What was the length (in cubits) of the sides of the central square in the original plan?
Bonus question: approximately how many times larger than the original would the surface area of the larger project proposed by Pythagoras have been?
A cubit is about 50cm.

See The Solution Submitted by broll    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Second Guessing... Solution (part one) | Comment 2 of 15 |

The minimum I have found meeting the conditions seems to be hypoteneuse 1105, with the following four Pythagoreans without a common factor: (squared: 1221025)

47 -- 1104   2209 + 1218816

264 -- 1073   69696 + 1151329

576 -- 943     331776 + 889249

744 -- 817    553536 +  667489

I have used the multiplication/additions above only; perhaps the "whole should be of minimum size" has to be interpreted -- the areas of the center and the four triangles, or an area which surrounds the whole smoothly.)

Pythagoras must have had an ancient Google (a giggle? a gaggle?) to get this solution immediately! OR a better "analytic".  If all four conditions of part one apply to the extended problem, we would need an hypoteneuse which could be derived in eight or more ways, where each pair had no common divisor.

 


  Posted by ed bottemiller on 2010-08-03 17:19:37
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (8)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information