<s>[INST] Finding the least possible value of perimeter of $\triangle ABC$ with given ranges of angle 
In $\triangle ABC$,$\angle A >2\angle B$ and $\angle C > 90^\circ$. If the length of all side of triangle $\triangle ABC$ are positive integers, then what is the least possible value of perimeter of $\triangle ABC$?

However, I can't think even of the length of the sides related with the possible values for all angle $\angle A, \angle B$ and $\angle C$. How can I construct the triangle and then get all the side having a length belonging to the positive integers? The problem was very weird for me and all of my effort can be hardly shown or described. And how can I get the minimum possible perimeter?
Thanks in advance. 
 [/INST] In the standard notation we have:
$$a+b>c,$$
$$c^2>a^2+b^2$$ and 
$$\frac{\sin\beta}{b}=\frac{\sin\alpha}{a}>\frac{\sin2\beta}{a}=\frac{2\sin\beta\cos\beta}{a},$$ which gives
$$\frac{a}{b}>\frac{a^2+c^2-b^2}{ac}$$ or
$$a^2(c-b)>b(c^2-b^2)$$ or
$$a^2>bc+b^2.$$
If $b=1$ we obtain $$c<a+1$$ or
$$c-a<1,$$ which is impossible.
Thus, $b\geq2$.
Now, $$a<c<\frac{a^2-b^2}{b},$$ which gives
$$a+1\leq c\leq\frac{a^2}{b}-b-1$$ and 
$$a^2-ab-b^2-2b\geq0,$$ which gives
$$a\geq\frac{b+\sqrt{5b^2+8b}}{2}\geq4$$ and since $$c^2>4^2+2^2,$$ we obtain $$c\geq5$$ and $$a+b+c\geq11.$$
Can you end it now?
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