Difference between revisions of "2019 AMC 12A Problems/Problem 19"

Revision as of 13:24, 19 January 2021

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

@@ Line 1: / Line 1: @@
 == Problem ==
-In <math>\triangle ABC</math> with integer side lengths, <math>\cos A = \frac{11}{16}</math>, <math>\cos B = \frac{7}{8}</math>, and<math>\cos C = -\frac{1}{4}</math>. What is the least possible perimeter for <math>\triangle ABC</math>?
+In <math>\triangle ABC</math> with integer side lengths, <math>\cos A = \frac{11}{16}</math>, <math>\cos B = \frac{7}{8}</math>, and <math>\cos C = -\frac{1}{4}</math>. What is the least possible perimeter for <math>\triangle ABC</math>?
 <math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math>

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search