Triangle With Random Constraints?

Relevant wiki: Solving Triangles - Problem Solving - Medium

Reciprocal are in HP $\implies$ angles are in AP. Let the angles be $r-y,r,r+y$ . Since they sum up to $\pi$ , we get $3r=\pi$ or $r=\dfrac{\pi}3$ ( i.e one angle of $\Delta$ is $\dfrac{\pi}3$ ). Now this angle $\dfrac{\pi}3$ can be opposite to side $4,5$ or $x$ . I'll be using Cosine Rule in each case:-

$\large\color{#D61F06}{\underline{Case-1}}$

$\dfrac{\pi}3$ is opposite to side of length $4$

$\implies \cos\left(\dfrac{\pi}3\right)=\dfrac{5^2+x^2-4^2}{2\cdot 5\cdot x}$ $\implies x^2-5x+9=0$ Since its discriminant( $=-11<0)$ , hence this quadratic doesn't have any real solutions.

$\large\color{#D61F06}{\underline{Case-2}}$

$\dfrac{\pi}3$ is opposite to side of length $5$ $\implies \cos\left(\dfrac{\pi}3\right)=\dfrac{4^2+x^2-5^2}{2\cdot 4\cdot x}$

$\implies x^2-4x-9=0$

$\implies (x-2)^2=13\implies \boxed{x=2+\sqrt{13}}$

( $2-\sqrt{13}$ is rejected since $x>0$ )

$\large\color{#D61F06}{\underline{Case-3}}$

$\dfrac{\pi}3$ is opposite to side of length $x$ . $\implies \cos\left(\dfrac{\pi}3\right)=\dfrac{5^2+4^2-x^2}{2\cdot 5\cdot 4}$ $\implies x^2=21\implies \boxed{x=\sqrt{21}}$ ( $x=-\sqrt{21}$ is rejected since $x>0$ )

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Hence,

$\mathcal A=(4+5+2+\sqrt{13})+(4+5+\sqrt{21})=\boxed{20+\sqrt{13}+\sqrt{21}}$