2019 AMC 12A Problems/Problem 19

Problem

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

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

Solutions

Solution 1

Notice that by the Law of Sines, $a:b:c = \sin{A}:\sin{B}:\sin{C}$ , so let's flip all the cosines using $\sin^{2}{x} + \cos^{2}{x} = 1$ ( $\sin{x}$ is positive for $0^{\circ} < x < 180^{\circ}$ , 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 $4$ . $\boxed{\textbf{(A) } 9}$ is our answer.

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 the 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 $\boxed{\textbf{(A) } 9}$ is our answer.

Solution 3

Using the law of cosines, we get the following equations:

$c^2=a^2+b^2+\frac{ab}{2}$ $b^2=a^2+c^2-\frac{7ac}{4}$ $a^2=b^2+c^2-\frac{11bc}{8}$

Substituting $a^2+c^2-\frac{7ac}{4}$ for $b^2$ in $a^2=b^2+c^2-\frac{11bc}{8}$ and simplifying, we get the following: $14a+11b=16c$

Note that since $a, b, c$ are integers, we can solve this for integers. By some trial and error, we get that $(a,b,c) = (3,2,4)$ . Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is $3+2+4 = \boxed{\textbf{(A) } 9}$

~hiker

Solution 4

Similar to solution 3, we will get the following by the law of cosines:

$\begin{align} a^2 + b^2 - c^2 = - \frac{1}{2} ab\\ b^2 + c^2 - a^2 = \frac{11}{8} bc\\ a^2 + c^2 - b^2 = \frac{7}{4} ac \end{align}$

By adding equations $(1), (2)$ and $(3)$ together, we will obtain: $\begin{align*} a^2 + b^2 + c^2 &= - \frac{1}{2} ab + \frac{11}{8} bc + \frac{7}{4} ac & \text{(4)} \end{align*}$

Subtracting $(2)$ from $(4)$ will give us: $\begin{align*} (a^2 + b^2 + c^2) - (b^2 + c^2 - a^2) &= (- \frac{1}{2} ab + \frac{11}{8} bc + \frac{7}{4} ac) - \frac{11}{8}\\ 2a^2 &= - \frac{1}{2} ab + \frac{7}{4} ac\\ 2a &= - \frac{1}{2} b + \frac{7}{4} c\\ 16a &= 14c - 4b & \text{(a)} \end{align*}$

Again, we subtract $(3)$ from $(4)$ and obtain: $\begin{align*} 16b &= 11c - 4a & \text{(b)} \end{align*}$

By adding equations $(a)$ and $(b)$ : $\begin{align*} 16a + 16b &= 14c - 4b + 11c - 4a\\ 20(a+b) &= 25c\\ 4(a+b) &= 5c \end{align*}$

Since all $a$ , $b$ and $c$ are integers, the smallest possibility is $a+b=5$ and $c=4$ . Thus, $a+b+c=4+5=\boxed{\textbf{(A) } 9}$

~Andy_li0805

Video Solution1

https://youtu.be/E8gk7VkLxos

~ Education, the Study of Everything

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