2024 AMC 12B Problem 22

Attempt Problem 22 of the 2024 AMC 12B below, then check your answer against the professionally curated solution 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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22.

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

1313

1414

1515

1616

1717

Answer: C
Concepts:law of sinesDiophantine Equationtriangle inequality
Difficulty rating: 2230
Solution:

When B=2A,\angle B = 2\angle A, the side lengths satisfy b2=a(a+c),b^2 = a(a + c), where a=BC,a = BC, b=CA,b = CA, c=AB.c = AB. So c=b2a2ac = \dfrac{b^2 - a^2}{a} must be a positive integer, and the sides must form a valid triangle.

Also b>a,b\gt a, because B=2A>A.\angle B=2\angle A\gt\angle A. The perimeter is a+b+c=b+b2a>2b.a+b+c=b+\dfrac{b^2}{a}\gt2b. Therefore a perimeter below 1515 would require b6.b\le6. For b=2,3,4,5,6,b=2,3,4,5,6, the divisors a<ba\lt b of b2b^2 give the possible pairs (b,a)=(2,1),(3,1),(4,1),(4,2),(5,1),(6,1),(6,2),(6,3),(6,4). \begin{gathered} (b,a)=(2,1),(3,1),\\ (4,1),(4,2),(5,1),\\ (6,1),(6,2),(6,3),(6,4). \end{gathered} Substitution gives a degenerate or invalid triangle in every case except (a,b)=(4,6),(a,b)=(4,6), which gives c=5.c=5. Thus (4,5,6)(4,5,6) is the first valid triangle, and its perimeter is 15.15.

Thus, the correct answer is C.

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