6.27 In triangle $A B C$, $\angle A=2 \angle B$, $\angle C$ is an obtuse angle, and the three side lengths $a$, $b$, $c$ are all integers. Find the minimum perimeter and provide a proof.
(20th United States of America Mathematical Olympiad, 1991)
[Solution] From the given and the Law of Sines, Law of Cosines, we have
$$
\begin{aligned}
\frac{a}{b} & =\frac{\sin A}{\sin B}=\frac{\sin 2 B}{\sin B}=2 \cos B \\
& =\frac{a^{2}+c^{2}-b^{2}}{a c} .
\end{aligned}
$$

That is
$$
\begin{array}{c}
a^{2} c-a^{2} b-b c^{2}+b^{3}=0, \\
(c-b)\left(a^{2}-b c-b^{2}\right)=0 .
\end{array}
$$

Since $C$ is an obtuse angle, we have $c>b$, thus
$$
a^{2}-b c-b^{2}=0 \text {. }
$$

Therefore,
$$
a^{2}=b(b+c) .
$$

If $(b, c)=d>1$, then by (1), $d \mid a$ must hold, and thus $d^2$ can be canceled out from both sides of (1). Therefore, we can assume $(b, c)=1$.
Thus,
$$
(b, b+c)=1 \text {. }
$$

From (1), it follows that both $b$ and $b+c$ must be perfect squares, so we can set
$$
b=m^{2}, \quad b+c=n^{2} .
$$

At this point,
$$
a=m n, \quad(m, n)=1 .
$$

Since $a+b>c$, we have
$$
a^{2}=b(b+c)b(b+2 b)=3 b^{2} .
$$

Therefore,
$$
\sqrt{3} b1 \text {. }
$$

That is,
$$
\begin{aligned}
m>\frac{1}{2-\sqrt{3}} & =2+\sqrt{3}>3 . \\
m & \geqslant 4 .
\end{aligned}
$$
(1) When $m=4$, from (5)

Then
$$
\begin{array}{c}
4 \sqrt{3}7$.
At this time, $a+b+c=n^{2}+m n>77$.
Therefore, the minimum perimeter is 77, at which the lengths of the three sides $a, b, c$ are $28,16,33$ respectively.