2019 AMC 12A Problem 19 — Solution | LIVE by Po-Shen Loh

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2019 AMC 12A Problem 19

Below is the professionally curated solution for Problem 19 of the 2019 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 12A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:law of sines ratio and proportion triangle inequality

Difficulty rating: 2000

19.

In $\triangle ABC$ with integer side lengths,

$\cos A = \dfrac{11}{16}, \quad \cos B = \dfrac{7}{8}, \quad \cos C = -\dfrac{1}{4}.$

What is the least possible perimeter for $\triangle ABC?$

$9$

$12$

$23$

$27$

$44$

Solution:

Each sine is $\sqrt{1 - \cos^2}:$ $\sin A = \dfrac{3\sqrt{15}}{16},$ $\sin B = \dfrac{2\sqrt{15}}{16},$ $\sin C = \dfrac{4\sqrt{15}}{16}.$

By the Law of Sines the sides are in ratio $3 : 2 : 4.$ The smallest integer sides are $3, 2, 4,$ which satisfy the triangle inequality.

The least perimeter is $3 + 2 + 4 = 9.$

Thus, the correct answer is A.

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