Irrelevant Information.

A.M.-G.M. inequality :

$\dfrac{js^2+t^2}{st}=\dfrac{js}{t}+\dfrac{t}{s}\geq 2\sqrt j$ . So $j\geq 2\sqrt j$ or $j(j-4)\geq 0$ . Since $j>0$ , therefore the minimum value of the given expression is $4$ when $j=\boxed 4$ . This will be true for any positive $j, s, t$ , and this problem is just one example justifying the title!

This problem has nothing to do with the geometry. The aim of the problem is to provide a diagram as a distraction.

Let $0 < s < t$

$\dfrac{js^2 + t^2}{st} = j\dfrac{s}{t} + \dfrac{t}{s}$

Let $u = \dfrac{s}{t} \implies f(u) = ju - u^{-1} \implies f'(u) = \dfrac{ju^2 - 1}{u^2} = 0 \implies u = \dfrac{1}{\sqrt{j}} \implies f(\dfrac{1}{\sqrt{j}}) = 2\sqrt{j} = j \implies 4j = j^2 \implies j(j - 4) = 0$

$j > 0 \implies \boxed{j = 4}$ and $s = \dfrac{t}{2}$ which is the only thing you know about the two triangles.

Note: $f''(\dfrac{1}{\sqrt{j}}) = 2j^{\frac{3}{2}} > 0 \implies$ relative min at $u = \dfrac{1}{\sqrt{j}}$

Let $s = xt$ , where $0 < x < 1$ , since $0 < s < t$ . Then we have:

$\begin{aligned} f(t) & = \frac {js^2+t^2}{st} = \frac {jx^2t^2+t^2}{xt^2} = \frac {jx^2+1}x \\ \frac {df(t)}{dt} & = \frac {2jx^2-jx^2-1}{x^2} = \frac {jx^2 -1}{x^2} \end{aligned}$

As $\dfrac {df(t)}{dt} = 0$ , when $x = \dfrac 1{\sqrt j}$ , and $\dfrac {d^2 f(t)}{dx^2} > 0$ for all $x > 0$ , $\implies \min (f(t) ) = f\left(\dfrac 1{\sqrt j}\right) = \dfrac {j \times \frac 1j + 1}{\frac 1{\sqrt j}} = 2\sqrt j = j$ , $\implies j = \boxed 4$ .