AMC 12 B, 2024

Let $a$, $b$, and $c$ be the lengths of the sides opposite vertices $A$, $B$, and $C$, respectively. Note that $a < b$. Applying the Law of Sines in $\triangle ABC$, together with the identities $\sin B = \sin(2A) = 2 \sin A \cos A$ and
$$\sin C = \sin(\pi - 3A) = \sin(3A) = (\sin A)(-1 + 4\cos^2 A),$$
give
$$\frac{\sin A}{a} = \frac{2 \sin A \cos A}{b} = \frac{\sin A (-1 + 4 \cos^2 A)}{c}.$$
Thus $\cos A = \frac{b}{2a}$ and
$$c = a(-1 + 4\cos^2 A) = a\left(-1 + \frac{b^2}{a^2}\right),$$
which simplifies to $b^2 = a^2 + ac$.

In looking for the triangle with least perimeter, it can be assumed that $a$ and $c$ are relatively prime, because otherwise a smaller triangle can be obtained by shrinking by a factor of $\gcd(a, c)$. Then $a$ and $a+c$ are relatively prime as well. Because $a(a+c) = b^2$, the numbers $a$ and $a+c$ must be squares, say $a = r^2$ and $a+c = s^2$, where $0 < r < s$ and $\gcd(r, s) = 1$. This gives $a = r^2$, $b = rs$, and $c = s^2 - r^2$.

If $r = 1$, then $b = s \ge 2$ and $c = s^2 - 1 \ge s + 1$, a violation of the Triangle Inequality. If $r = 2$, then the least perimeter will occur when $s = 3$, with $a = 2^2 = 4$, $b = 2 \cdot 3 = 6$, and $c = 3^2 - 2^2 = 5$. Greater values of $r$ lead to greater perimeters. The requested minimum perimeter is $4 + 6 + 5 = 15$.