Triangle with integral side lengths and $\angle A=3\angle B$ 
$ABC$ is a triangle with integral side lengths. Given that $\angle A=3\angle B$, find the minimum possible perimeter of $ABC$.

I got this problem from an old book (which did not provide even a hint). I can think of some approaches, but all of them result in complicated Diophantine equations that would not be solvable without the help of a computer. Any suggestions?

Such triangles exist.  I found one 3:8:10, where the angle opposite 8 is 1/3 of that opposite 10.  The trick here is to pick really small primes in the form of $x+y\sqrt{-n}$, and cube the result.  Note here that we're using $n=7$, and the prime is $1\frac 12 + \frac 12\sqrt{-7}$.  This is a pretty tiny cube, one gets then a matrix
   ( 3  -7 )  (3)  (2)  (36)
   ( 1   3 )  (1)  (6)  (20)
               A         B = A^3

You then divide through by common factors, to get coordinates at $A =0,0$  $B=+6,0$, and $C=-9,5\sqrt{7}$.  The three sides are AB=6, AC=16, and BC=20, which gives the indicated triangle.  
