find all fundemental pythagorean triangles whose area are th

find all fundemental pythagorean triangles whose area are three times their perimeters

Solution

Let x and y be the legs and z be the hypotenuse of the right triangle. Since the triangle is a Pythagorean triangle, x,y,& z are integers. Let A be the area and P be the perimeter of the triangle.

A = (1/2)xy
P = x + y + z

We wish to find where (1/2)xy = 3( x + y + z )

By the Pythagorean theorem, we have that:
x^2 + y^2 = z^2
z = sqrt(x² + y²)

Plug into equation:
(1/2)xy = 3( x + y + sqrt(x² + y²))

Multiply both sides by 2:
xy = 6x + 6y + 6sqrt(x² + y²)

Subtract 6x + 6y from both sides:
xy - 6(x + y) = 6sqrt(x^2 + y^2)

Square both sides:
x^2 y^2 + 36(x + y)^2 - 12xy(x + y) = 36 (x^2 + y^2)

Expand .

x^2 y^2 + 36x^2 +72xy +36y^2 - 12xy(x + y) = 36x^2 + 36 y^2

Subtract 36x^2 + 36 y^2 both side on both side,

x^2 y^2 + 72 xy - 12xy(x+y) = 0

factor out this then we get xy ( x y + 72 - 12 (x +y) ) = 0

x y = 0 == > x =0 or y =0
but we want only positive integers.

SO  

x y + 72 - 12 (x +y) = 0

so we get y = 6 and x = 0 here also w e get x = 0

but we want positive values of x , y and z.

So we don\'t have any such type of fundamental Pythagorean triples.