<s>[INST] Elegant way of finding the least perimeter of triangle A triangle $ABC$ has positive integer sides, $\angle A = 2\angle B$ and $\angle C > \pi/2$ , then the minimum length of the perimeter of $ABC$ is?
We have $\angle A = 2\angle B$
$\Rightarrow \sin A=\sin 2B=2 \sin B \cos B $
$\sin C=\sin(\pi-3B)=\sin(3B)=3\sin B-4\sin^3B$
Using $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
$a=2b\cos B $
$c=b(3-4\sin^2 B)$
From above two equations $a^2=b(c+b) $
NOTE: similar questions have been asked earlier but I want to do this question analysing with $a^2=b(c+b)$ and no fancy inequalities
 [/INST] Found an another solution:
we have $a^2=b(c+b)$
A triangle of smallest perimeter means $gcd(a,b,c)=1$
In fact $gcd(b,c)=1$ since any common factor of $b,c$ would be a factor of $a$ as well.
A perfect square $a^2$ is being expressed as the product of two relatively prime integers $b$ and $c$.
it must be the case that both $b$ & $b + c$ are perfect squares. Thus for some integer $m$ & $n$ with
$gcd(m,n)= 1$ we have
$b=m^2$ and $b+c=n^2$,$a=mn$
$2\ cosB=\frac{n}{m}=\frac{a}{b}$
As $\angle C >\frac{\pi}{2} \Rightarrow 0<\angle B<\frac{\pi}{6}$
$\Rightarrow \sqrt{3}<2\ cosB=\frac{n}{m}<2$
The smallest value of $(m, n)$ that satisfies the above mentioned conditions are $4$ and $7$ $\blacksquare$
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