Triangle with integral side lengths and $\angle A=3\angle B$
Sophia Terry 
$ABC$ 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?
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$\begingroup$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^3You 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.
$\endgroup$ 1 $\begingroup$Giving it a try using elementary techniques.
In $\triangle ABC$, let $B=\theta$, $A=3\theta$, $C=\pi-4\theta$. By sine-rule$$\frac{a}{\sin 3\theta}=\frac{b}{\sin \theta}=\frac{c}{\sin 4\theta}$$
so that $$a=b(3-4\sin^2\theta) \quad , \quad c=b(4\cos \theta \cos 2\theta)$$
On further simplification,$$a=b(4\cos^2\theta-1) \quad , \quad c=b \cdot 4\cos \theta (2\cos^2 \theta-1)$$
$c>0$ places bound on $\cos \theta$ : $\cos \theta > 1/\sqrt{2} \Rightarrow \theta < 45^{\circ}$ which makes sense since $45^{\circ}+3\cdot 45^{\circ}=180^{\circ}$
We may choose any suitable value of $\cos \theta > 1/\sqrt{2} \approx 0.7071$.
Taking $\cos \theta = 3/4=0.75$, yields $a/b=5/4$, $c/b=3/8$. So one class of triangles would be multiples of$$(a,b,c)=(10,8,3)$$
Taking $\cos \theta = 5/6$, yields $a/b=16/9$, $c/b=35/27$. So another class of triangles would be multiples of$$(a,b,c)=(48,27,35)$$
and so on.
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