Length of triangle madness!

We know that

$\displaystyle ca = \frac{1}{2}\left((c+a)^{2}-(c^{2}+a^{2})\right)$

$\displaystyle = \frac{1}{2}\left((2b)^{2}-(84-b^{2})\right)$

$\displaystyle = \frac{1}{2}(5b^{2}-84)$

Since $c,a$ is the root of the equation

$x^{2}-(c+a)x+ca = 0$

$\displaystyle x^{2}-(2b)x+\frac{1}{2}(5b^{2}-84) = 0$ .

From $a\neq c$ , we know that there are 2 different roots in this equation.

$\Delta = (2b)^{2} - 4(1)(5b^{2}-84) > 0$

Solve the inequality we get $b^{2} < 28$ . (1)

But we know that $ca > 0$ .

$\frac{1}{2}(5b^{2}-84) > 0$

Solve the inequality we get $b^{2} > \displaystyle \frac{84}{5} > 16$ . (2)

From (1),(2) we know that $b$ is an integer, that means $b = 5$ .

We can "guess" that $a = 5 + \sqrt{x}$ and $c = 5 - \sqrt{x}$ for some $x$ .

From $\displaystyle ca = \frac{1}{2}(5b^{2}-84) = \frac{41}{2}$ .

$\displaystyle (5-\sqrt{x})(5+\sqrt{x}) = \frac{41}{2}$

$\displaystyle x^{2} = \frac{9}{2}$

$\displaystyle x = \frac{3}{\sqrt{2}}$

Therefore, $\boxed{(a,b,c) = \left(\displaystyle 5+\frac{3}{\sqrt{2}}, 5, 5 - \frac{3}{\sqrt{2}}\right)}$ . ~~~

We can also check that $a,b,c$ are the length of a triangle.

Substitute for your own, too lazy XD

We denote a as b+x and c as b-x (b>x>0) Hence 3b^2 +2x^2=84 using b>x>0, solving the inequations for integer yields b = 5 and x = 3/ srt(2). a + b - c = b + 2x thus x y z = 5, 3 and 2 respectively