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. 

So, how about a really simple approach: run through positive integer triples $b<a<c$. If they're sides of a triangle, the angles will be in order $B<A<C$.


*

*$(1,2,3)$: Not a triangle. In fact, we can't have a triangle if the small side is $1$.  

*$(2,3,4)$: Since $4^2=16>13=2^2+3^2$, it's obtuse. Now, the smaller angles: $\cos B = \frac{4^2+3^2-2^2}{2\cdot 4\cdot 3}=\frac{7}{8}$ and $\cos A = \frac{4^2+2^2-3^2}{2\cdot 4\cdot 2}=\frac{11}{16}$. By the double-angle formula, $\cos 2B = 2\cdot\frac{7^2}{8^2}-1=\frac{98-64}{64}=\frac{17}{32}<\frac{11}{16}$. Not this one. 

*$(2,3,5)$: Not a triangle. 

*$(2,4,5)$: $25>20=16+4$, so it's obtuse. $\sin A=2\sin B>\sin 2B$, so it satisfies the second condition as well.
There it is - the second smallest triangle with three different integer sides. Sometimes, the simple approach pays off.
