i think that is 77. I am certain? let (a,b,c) the sides of the triangle with minimum perimeter. suppose that gcd(a,b,c) = 1, (if exist p>1 such that a = px, b = py, c = pz, we have that (x,y,z) is other solution with x+y+z < a+b+c). we have that a² = b²(b+c), because m(A) = 2.m(B). if gcd(b,c)>1, we have gcd(a,b,c)>1. this implies that gcd(b,c) = 1. so, gcd(b,b+c) = 1 and b= m², b+c = n² and a = mn. by the sine law, we have n/m = a/b = 2cos(B). C = 180 - 3B > 90, that implies B < 30°. so, sqrt(3) < n/m < 2. is easy to see that for m = 1,2,3 this inequality no have integer solutions. so, m>=4, n>=7. the pair (m,n) = (4,7) get the solution (a,b,c) = (28,16,33).

I know this is reviving an old thread, but I don't think that answer's right, and AoPS wiki doesn't have the solution, so here's mine: After drawing the triangle, also draw the angle bisector of $ \angle A$, and let it intersect $ \overline{BC}$ at $ D$. Notice that $ \triangle ABC \sim \triangle DAC$, and let $ AD = x$. Now from similarity, \[ \frac {AC}{AD} = \frac {BC}{AB} \Longleftrightarrow \frac {b}{x} = \frac {a}{c}\Longrightarrow x = \frac {bc}{a} \] However, from the angle bisector theorem, we have \[ \frac {AB}{BD} = \frac {AC}{CD}\Longleftrightarrow BD = \frac {ac}{b + c}\text{ and }CD = \frac {ab}{b + c} \] A few steps were skipped to acheive this result. We see that $ \triangle DAB$ is isosceles, so $ BD = AD = x$, which means \[ \frac {ac}{b + c} = \frac {bc}{a}\Longrightarrow \boxed{a^2 = b^2 + bc} \] so all side lengths which satisfy the conditions also meet the boxed condition. We can make a list of all possible values for $ a, b, c$ and impose the condition that $ c$ is the longest side since $ \angle C$ is obtuse. The first set of values that works is $ (a, b, c) = (15, 9, 16)$ which yields a perimeter of $ \boxed{40}$. A thorough check shows that this is the least perimeter, and we can verify that it satisfies the original conditions by using the law of cosines and the fact that $ \cos(2\theta) = 2\cos^2(\theta) - 1$ to show that $ \angle A = 2\angle B$. EDIT: Somehow kalva says it is 77. Dang it! I forgot to show that <C is obtuse by a^2+b^2<c^2. Oops

ZzZzZzZzZzZz wrote: $ \boxed{a^2 = b^2 + bc}$ You can obtain this with the law of sines also. $ \frac{sin2x}{a}=\frac{sin x}{b}=\frac{sin 3x}{c}$ $ \frac{2cosx}{a}=\frac{1}{b}=\frac{-4sin^2x+3}{c}$ So $ cos x=\frac{a}{2b}$, so $ sin^2x=1-cos^2x=\frac{4b^2-a^2}{4b^2}$ Substituting, $ \frac{\frac{a^2-4b^2}{b^2}+3}{c}=\frac{1}{b}$, so $ a^2=b^2+bc$.

Here is a solution, but... http://www.animath.fr/old/tutorat/dossier_02033sol.html

We can also use calculus ...(maxima and minima) and the answer is 77

dgreenb801 wrote: ZzZzZzZzZzZz wrote: $ \boxed{a^2 = b^2 + bc}$ You can obtain this with the law of sines also. $ \frac{sin2x}{a}=\frac{sin x}{b}=\frac{sin 3x}{c}$ $ \frac{2cosx}{a}=\frac{1}{b}=\frac{-4sin^2x+3}{c}$ So $ cos x=\frac{a}{2b}$, so $ sin^2x=1-cos^2x=\frac{4b^2-a^2}{4b^2}$ Substituting, $ \frac{\frac{a^2-4b^2}{b^2}+3}{c}=\frac{1}{b}$, so $ a^2=b^2+bc$. No one said that $C = 3A$

Note that the obtuse condition reduces to $B < 30^{\circ}$. Claim: $a^2=b^2+bc$. Proof. By angle bisector theorem and isosceles triangle $\triangle ABD$ we have $AD=BD=\frac{ac}{b+c}$ and since $\triangle ABC \sim \triangle DAC$ setting ratios gives the desired. $\square$ Now note that $\gcd(a,b,c) = 1$, so $b$ and $b+c$ are perfect squares. Let $a=xy, b=x^2, b+c=y^2$. By LoC on angle $B$ we get $$\sqrt2 < \frac{y}{x} < 2 \rightarrow x \geq 4,$$and checking gives $(x,y)=(4,7)$ as a valid pair, so the minimum perimeter is $77$, achieved at $(a,b,c)=(28,16,33)$. $\blacksquare$

Let the side lengths be $a,b,c$ and let the angle opposite $b$ be $x$ so that the other two angles are $2x$ and $180-3x.$ By Law of Sines, $$a:b:c=\sin(2x):\sin(x):\sin(3x)=2\cos(x):1:3-4\sin^2(x)=2\cos(x):1:4\cos^2(x)-1.$$Let $p=\cos(x)$. Note that by the obtuse condition, $\sqrt{3}/2<p<1.$ Therefore, $$a:b:c=2p:1:4p^2-1.$$Let $p=\frac{m}{n}$ for relatively prime positive integers. This means $$a:b:c=\frac{2m}{n}:1:\frac{4m^2-n^2}{n^2}=2mn:n^2:4m^2-n^2.$$ Case 1: $n$ is even. Then, this can be reduced by a factor of 4 (but no further due to relatively prime $m$ and $n$), so we want to minimize $$\frac{1}{4}(2mn+4m^2)=(1/2)m(n+2m).$$Note that the constraint $\sqrt{3}/2<p<1$ makes it so that $m\geq7, n\geq 8$, so the minimum such value occurs at $m=7,n=8$, which gives $$(a,b,c)=(28,16,33)$$as the smallest solution giving a perimeter of 77. Case 2: $n$ is odd. Clearly, this is not optimal, since we won't be able to reduce by the factor of 4, and $n=9,m=8$ gives a worse value, so we are done.

"It's very easy to $C$ that bashing is not the *right* way of solving this problem" Wow that was a horrible joke but whatever... So we let $B = \alpha, A = 2\alpha, C=180-3\alpha$. By law of sines we have, $$\frac{b}{\sin \alpha} = \frac{a}{\sin 2\alpha} = \frac{a}{2\sin \alpha \cos \alpha} \Rightarrow b = a(2\cos \alpha)$$$$\text{and}$$$$\frac{b}{\sin \alpha} = \frac{c}{\sin(180-3\alpha)} = \frac{c}{\sin 3\alpha} = \frac{c}{\sin \alpha(4\cos^2 \alpha - 1)} \Rightarrow c=a(4\cos^2 \alpha - 1)$$ As $0 < \alpha < 30$, we have that $\sqrt{3}a < b < 2a$. By inspection, a smaller denominator (when reduced) will be most optimal*. Hence, we have $b = \frac{7}{4}a$, and, therefore, $c = \frac{33}{16}$. This gives $(a, b, c) = (16, 28, 33)$ and a minimal perimeter of $\boxed{77}$. *Would I have to prove this in-contest?