Find all primitive Pythagorean triangles whose areas are equal to their perimeters. Prove that your list is complete. Hint: Use conditions that the equations x^2 + y^2 = z^2 and x +y =1/2xy
Answers
Answer:
(5, 12, 13)
Step-by-step explanation:
The hint says "x+y=(1/2)xy", but to have "perimeter = area" it should say "x+y+z=(1/2)xy".
The conditions are:
- x² + y² = z² [ Pythagorean triple ]
- x, y, z have no common factor greater than 1 [ primitive ]
- x + y + z = xy/2 [ perimeter = area ]
x+y+z = xy/2
2(x+y+z) = xy
=> 4(x+y+z)² = x²y²
=> 4 ( x² + y² + z² + 2xy + 2yz + 2zx ) = x²y²
=> 4 ( 2z² + 2yz + 2zx ) = x²y² - 8xy
=> 8 z ( x + y + z ) = xy ( xy - 8 )
=> 4 xyz = xy ( xy - 8 )
=> 4z = xy - 8
=> 2 ( xy - 2x - 2y ) = xy - 8
=> xy - 4x - 4y + 16 = 8
=> ( x - 4 ) ( y - 4 ) = 8
Since x and y are integers, (x-4)(y-4) is a factorization of 8.
Since (x,y,z) is primitive, x and y are not both even, so the factorization of 8 must be either -1×-8 or 1×8. Since x and y are positive(*), neither x-4 nor y-4 can be equal to -8. So the factorization must be 1×8. Without loss of generality (we can swap x and y if we want to), we therefore have:
x-4 = 1, y-4 = 8
=> x = 5, y = 12.
The corresponding Pythagorean triple is (5, 12, 13).
(*) It is interesting to notice that if we permit negative values, so the problem is purely number theoretic and not associated with triangles anymore, then we also get x-4 = -1, y-4 = -8 => x = 3, y = -4 which together with z = -5 results in:
"perimeter" = 3 - 4 - 5 = -6
and
"area" = (1/2) × 3 × (-4) = -6.