2019 AMC 12A Problems/Problem 19

Problem

In $\triangle ABC$ with integer side lengths, $\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.$ What is the least possible perimeter for $\triangle ABC$ ?

$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$

Solution

We intend to use law of sines, so let's flip all the cosines (Sine is positive for $0\le x \le 180$ , so we're good there).

$\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}$

These are in the ratio $3:2:4$ , so our minimal triangle has side lengths $2$ , $3$ , and $4$ . $\boxed{\text{(A)}\, 9}$ is our answer.

-Rowechen Zhong

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2019 AMC 12A Problems/Problem 19

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