Integer Triangles

Let $ABC$ be a triangle.

$[ABC]=\sqrt{s(s-a)(s-b)(s-c)}$

And perimeter is given by $2s$ .

So, $\begin{aligned} 2s&=\sqrt{s(s-a)(s-b)(s-c)}\\ \implies 4s^2&=s(s-a)(s-b)(s-c) \\ \implies 4s&=(s-a)(s-b)(s-c) \end{aligned}$

Now let $x=s-a$ , $y=s-b$ , and $z=s-c$ . Hence $x+y+z=s$ .

$\begin{aligned} 4(x+y+z)=xyz \end{aligned}$

Now all you have to do is solve for triplets $(x,y,z)$ which are all positive integers which is quite easy. After going through all the work , you will find out that in total, there are 5 solutions.

The problem and the solution is explained in this video .