Question

Find all Pythagorean triangles whose areas are equal to their perimeters.

Solution

Verified

We need to find all Pythagorean triple x, y, zx,\ y,\ z such that

x2+y2=z2\begin{equation} x^2+y^2=z^2 \end{equation}

and

xy2=x+y+z\begin{equation} \dfrac{xy}{2}=x+y+z \end{equation}

From (1)(1) and (2)(2), we obtain that

x2+y2=(x+yxy2)2x2+y2=x2+y2x2y24+2xyx2yxy20=xy4(xy+84y4x)8=(x4)(y4){\color{#c34632}\begin{align*} x^2+y^2&=(x+y-\dfrac{xy}{2})^2\\ \cancel{x^2+y^2}&=\cancel{x^2+y^2}-\dfrac{x^2y^2}{4}+2xy-x^2y-xy^2\\ 0&=\dfrac{xy}{4}(xy+8-4y-4x)\\ 8&= (x-4)(y-4) \end{align*}}

So that, only possible Pythagorean triples are

6, 8, 10{\color{#4257b2}6 ,\ 8 ,\ 10}

5, 12, 13{\color{#4257b2} 5,\ 12,\ 13}

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