2019 AMC 12A Problems/Problem 19

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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 $\triangle ABC$?

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


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 $3+2+4 = \boxed{\textbf{(A) }9}$.


See Also

2019 AMC 12A (ProblemsAnswer KeyResources)
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

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