2019 AMC 12A Problem 19

Question

In

\triangle ABC

with integer side lengths,

\begin{aligned} \cos A &= \dfrac{11}{16}, \\ \cos B &= \dfrac{7}{8}, \\ \cos C &= -\dfrac{1}{4}. \end{aligned}

What is the least possible perimeter for

\triangle ABC?

9

12

23

27

44

Find

\sin A, \sin B, \sin C

from the given cosines

By the Law of Sines the sides are proportional to

\sin A : \sin B : \sin C = 3 : 2 : 4

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.

Problem 19 in other years