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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    
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Solution Computer Solution | Comment 5 of 15 |

The following Perl program generates all the
Primative Pathagorean Triples (PPTs) with
the hypotenuse listed first and < 100000.
-----------------------------------------------------
LOOP: 
  for ($u=2;;$u++) {
    for ($v=$u-1;$v>0;$v-=2) {
      if (gcd($u,$v)==1) {
        $z = $u*$u+$v*$v;
        if ($z<100000) {
          printf(" %6d %6d %6d\\n",
                  $z,2*$u*$v,$u*$u-$v*$v);
        } else {
          last LOOP;
  }
      }
    }
  }
-----------------------------------------------------
The output is then sorted to make it easy to find
the first groups of 4 and 8 PPTs with the same
hypotenuse. From which we have the following:
-----------------------------------------------------
 Square Side = 1105 cubits
 Triangle Sides (cubits):
           264  1073
           576   943
           744   817
          1104    47
 Square Area = 1221025 square cubits
 Garden Area = 1964113 square cubits
-----------------------------------------------------
 Octagon Side = 32045 cubits
 Triangle Sides (cubits):
             716   32037
            6764   31323
           15916   27813
           22244   23067
           24124   21093
           27004   17253
           30956    8283
           31964    2277
 Octagon Area = 4958225023.4243 square cubits
  Garden Area = 6205484743.4243 square cubits
-----------------------------------------------------
  Ratio Without Triangle Areas = 4060.70721191155
     Ratio With Triangle Areas = 3159.43366976559
-----------------------------------------------------
I look forward to seeing broll's non-computer
solution to the first part of the problem.
 

Edited on August 5, 2010, 9:31 pm

Edited on August 5, 2010, 9:33 pm
  Posted by Bractals on 2010-08-03 22:49:30

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