Let $a$ and $b$ be the two legs of the triangle. We have $\frac{1}{2}ab = 3(a+b+c)$. Then $ab=6 \left(a+b+\sqrt {a^2 + b^2}\right)$. We can complete the square under the root, and we get, $ab=6 \left(a+b+\sqrt {(a+b)^2 - 2ab}\right)$. Let $ab=p$ and $a+b=...
9 years ago