What conditions must hold for a right triangle with legs x and y to have its perimeter numerically equal to its area?

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Answer 1 · 2017-11-28T16:21:08

Condition is $x y + 8 = 4 x + 4 y$ $xy+8=4x+4y$

Area of right angled triangle whose legsare $x$ $x$ and $y$ $y$ is $\frac{1}{2} x y$ $1/2xy$

and its perimeter is $x + y + \sqrt{x^{2} + y^{2}}$ $x+y+sqrt(x^2+y^2)$

if $x + y + \sqrt{x^{2} + y^{2}} = \frac{1}{2} x y$ $x+y+sqrt(x^2+y^2)=1/2xy$

or $\sqrt{x^{2} + y^{2}} = \frac{x y}{2} - (x + y) = \frac{x y - 2 x - 2 y}{2}$ $sqrt(x^2+y^2)=(xy)/2-(x+y)=(xy-2x-2y)/2$

or $x^{2} + y^{2} = \frac{1}{4} (4 x^{2} + 4 y^{2} + x^{2} y^{2} + 8 x y - 4 x^{2} y - 4 x y^{2})$ $x^2+y^2=1/4(4x^2+4y^2+x^2y^2+8xy-4x^2y-4xy^2)$

or $4 x^{2} + 4 y^{2} = 4 x^{2} + 4 y^{2} + x^{2} y^{2} + 8 x y - 4 x^{2} y - 4 x y^{2}$ $4x^2+4y^2=4x^2+4y^2+x^2y^2+8xy-4x^2y-4xy^2$

or $x^{2} y^{2} + 8 x y = 4 x^{2} y + 4 x y^{2}$ $x^2y^2+8xy=4x^2y+4xy^2$

and dividing by $x y$ $xy$ as it is non-zero, we get

$x y + 8 = 4 x + 4 y$ $xy+8=4x+4y$