AMC12 2024 B

Let $AB=c$, $BC=a$, $AC=b$. According to the law of sines, \[\frac{b}{a}=\frac{\sin \angle B}{\sin \angle A}=2\cos \angle A\] According to the law of cosines, \[\cos \angle A=\frac{b^2+c^2-a^2}{2bc}\] Hence, \[\frac{b}{a}=\frac{b^2+c^2-a^2}{bc}\] This simplifies to $b^2=a(a+c)$. We want to find the positive integer solution $(a, b, c)$ to this equation such that $a, b, c$ forms a triangle, and $a+b+c$ is minimized. We proceed by casework on the value of $b$. Remember that $a<a+c$. Case $1$: $b=1$ Clearly, this case yields no valid solutions. Case $2$: $b=2$ For this case, we must have $a=1$ and $c=3$. However, $(1, 2, 3)$ does not form a triangle. Hence this case yields no valid solutions. Case $3$: $b=3$ For this case, we must have $a=1$ and $c=8$. However, $(1, 3, 8)$ does not form a triangle. Hence this case yields no valid solutions. Case $4$: $b=4$ For this case, $a=1$ and $c=15$, or $a=2$ and $c=6$. As one can check, this case also yields no valid solutions Case $5$: $b=5$ For this case, we must have $a=1$ and $c=24$. There are no valid solutions Case $6$: $b=6$ For this case, $a=2$ and $c=16$, or $a=4$ and $c=5$, or $a=3$ and $c=9$. The only valid solution for this case is $(4, 6, 5)$, which yields a perimeter of $15$. When $b\ge 7$, it is easy to see that $a+c>7$. Hence $a+b+c>14$, which means $a+b+c\ge15$. Therefore, the answer is $\boxed{\textbf{(C) }15}$