2024 AMC 12B Problem 22

Below is the professionally curated solution for Problem 22 of the 2024 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12B solutions, or check the answer key.

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Concepts:law of sines Diophantine Equation triangle inequality

Difficulty rating: 2230

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?

$13$

$14$

$15$

$16$

$17$

Solution:

When $\angle B = 2\angle A,$ the side lengths satisfy $b^2 = a(a + c),$ where $a = BC,$ $b = CA,$ $c = AB.$ So $c = \dfrac{b^2 - a^2}{a}$ must be a positive integer, and the sides must form a valid triangle.

Trying small values, $a = 4,$ $b = 6$ gives $c = \dfrac{36 - 16}{4} = 5,$ and the sides $4, 5, 6$ form a valid triangle with $6^2 = 4(4 + 5).$ Its perimeter is $15,$ and a search shows no smaller perimeter works.

Thus, the correct answer is C.

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