The minimum value of perimeter of the triangle Let there be $\triangle ABC$ having integral side lengths such that angle $$ \angle A=3 \angle B $$ Then find the minimum value of its perimeter.
I did this with trigonometry and got an equation as: $$ 2p = 4b[2\cos^3( \theta) + \cos^2(\theta) - \cos(\theta)] $$ where $b$ = side opposite to $\angle B$, $\theta$ is the $\angle B $ and $p$ is half of the perimeter. How should I proceed?
Also show any other appropriate method.

My try:
$$\frac{a}{\sin3\beta}=\frac{b}{\sin\beta}$$
$$\sin3\beta=3\sin\beta-4\sin^3\beta=\frac{a}{b}\sin\beta$$
$$3-4\sin^2\beta=\frac{a}{b}$$
$$\sin^2\beta=\frac14(3-\frac ab)\tag{1}$$
$$\frac{b}{\sin\beta}=\frac{c}{\sin4\beta}$$
$$\frac{\sin4\beta}{\sin\beta}=\frac{2\sin2\beta \cos2\beta}{\sin\beta}=4\cos\beta (1-2\sin^2\beta)=\frac cb$$
$$16\cos^2\beta(1-2\sin^2\beta)^2=\frac{c^2}{b^2}$$
$$16(1-\sin^2\beta)(1-2\sin^2\beta)^2=\frac{c^2}{b^2}\tag{2}$$
Replace (1) into (2) and after some simplifications you get:
$$\frac{a^3}{b^3}-\frac{a^2}{b^2}-\frac{a}{b}+1=\frac{c^2}{b^2}$$
$$a^3-a^2 b-a b^2+b^3=c^2b$$
$$(a+b)(a-b)^2=c^2 b$$
However I was not able to proceed and solve this equation, except by guessing. Obviously 10, 8 and 3 fit the equation (I have started with $a=b=1$)
