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 1

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

Solution 2

$\angle ACB$ is obtuse since its cosine is negative, so we let the foot of the altitude from $C$ to $AB$ be $H$ . Let $AH=11x$ , $AC=16x$ , $BH=7y$ , and $BC=8y$ . By the Pythagorean Theorem, $CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}$ and $CH=\sqrt{64y^2-49y^2}=y\sqrt{15}$ . Thus, $y=3x$ . The sides of the triangle are then $16x$ , $11x+7(3x)=32x$ , and $24x$ , so for some integers $a,b$ , $16x=a$ and $24x=b$ , where $a$ and $b$ are minimal. Hence, $\frac{a}{16}=\frac{b}{24}$ , or $3a=2b$ . Thus smallest possible positive integers $a$ and $b$ that satisfy this are $a=2$ and $b=3$ , so $x=\frac{1}{8}$ . The sides of the triangle are $2$ , $3$ , and $4$ , so $\boxed{\text{(A)}\, 9}$ is our answer.

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

Contents

Problem

Solution 1

Solution 2

See Also