2024 AMC 12B Problems/Problem 22

Problem 22

Let $\triangle{ABC}$ be a triangle with integer side lengths and the property that $\angle{B} = 2\angle{A}$ . What is the least possible perimeter of such a triangle?

$\textbf{(A) }13 \qquad \textbf{(B) }14 \qquad \textbf{(C) }15 \qquad \textbf{(D) }16 \qquad \textbf{(E) }17 \qquad$

Solution 1

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 $a<a+c$ .

$\textbf{Case 1: b=1}$ Clearly, this case yields no valid solutions.

$\textbf{Case 2: b=2}$ 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.

$\textbf{Case 3: b=3}$ For this case, we must have $a=1$ and $c=9$ . However, $(1, 3, 9)$ does not form a triangle. Hence this case yields no valid solutions.

$\textbf{Case 4: b=4}$ 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

$\textbf{Case 5: b=5}$ For this case, we must have $a=1$ and $c=24$ . There are no valid solutions

$\textbf{Case 6: b=6}$ 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 $(4, 6, 5)$ .

It is safe to assume that $(4, 5, 6)$ will be the solution with least perimeter. Hence, the answer is $\fbox{\textbf{(C) }15}$

~tsun26

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2024 AMC 12B Problems/Problem 22

Problem 22

Solution 1