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

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==Problem==
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== Problem ==
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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>?
  
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In <math>\triangle ABC</math> with integer side lengths,
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<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math>
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<cmath>\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.</cmath>
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What is the least possible perimeter for <math>\triangle ABC</math>?
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== Solutions ==
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=== Solution 1 ===
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Notice that by the Law of Sines, <math>a:b:c = \sin{A}:\sin{B}:\sin{C}</math>, so let's flip all the cosines using <math>\sin^{2}{x} + \cos^{2}{x} = 1</math> (<math>\sin{x}</math> is positive for <math>0^{\circ} < x < 180^{\circ}</math>, so we're good there).
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<math>\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}</math>
  
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<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math>
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These are in the ratio <math>3:2:4</math>, so our minimal triangle has side lengths <math>2</math>, <math>3</math>, and <math>4</math>. <math>\boxed{\textbf{(A) } 9}</math> is our answer.
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=== Solution 2 ===
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<math>\angle ACB</math> is obtuse since its cosine is negative, so we let the foot of the altitude from <math>C</math> to <math>AB</math> be <math>H</math>. Let <math>AH=11x</math>, <math>AC=16x</math>, <math>BH=7y</math>, and <math>BC=8y</math>. By the Pythagorean Theorem, <math>CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}</math> and <math>CH=\sqrt{64y^2-49y^2}=y\sqrt{15}</math>. Thus, <math>y=3x</math>. The sides of the triangle are then <math>16x</math>, <math>11x+7(3x)=32x</math>, and <math>24x</math>, so for some integers <math>a,b</math>, <math>16x=a</math> and <math>24x=b</math>, where <math>a</math> and <math>b</math> are minimal. Hence, <math>\frac{a}{16}=\frac{b}{24}</math>, or <math>3a=2b</math>. Thus the smallest possible positive integers <math>a</math> and <math>b</math> that satisfy this are <math>a=2</math> and <math>b=3</math>, so <math>x=\frac{1}{8}</math>. The sides of the triangle are <math>2</math>, <math>3</math>, and <math>4</math>, so <math>\boxed{\textbf{(A) } 9}</math> is our answer.
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=== Solution 3 ===
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Using the law of cosines, we get the following equations:
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<cmath>c^2=a^2+b^2+\frac{ab}{2}</cmath>
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<cmath>b^2=a^2+c^2-\frac{7ac}{4}</cmath>
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<cmath>a^2=b^2+c^2-\frac{11bc}{8}</cmath>
  
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==Solution==
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Substituting <math>a^2+c^2-\frac{7ac}{4}</math> for <math>b^2</math> in <math>a^2=b^2+c^2-\frac{11bc}{8}</math> and simplifying, we get the following:
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We intend to use law of sines, so let's flip all the cosines (Sine is positive for <math>0\le x \le 180</math>, so we're good there).
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<cmath>14a+11b=16c</cmath>
  
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<math>\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=-\frac{\sqrt{15}}{4}</math>
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Note that since <math>a, b, c</math> are integers, we can solve this for integers. By some trial and error, we get that <math>(a,b,c) = (3,2,4)</math>. Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is <math>3+2+4 = \boxed{\textbf{(A) }9}</math>.
  
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These are in the ratio <math>3:2:4</math>, so our minimal triangle has side lengths <math>2</math>, <math>3</math>, and <math>4</math>. <math>\boxed{(A) 9}</math> is our answer.
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~hiker
  
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-Rowechen Zhong
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== Video Solution1 ==
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https://youtu.be/E8gk7VkLxos
  
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==See Also==
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~ Education, the Study of Everything
  
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== See Also ==
 
{{AMC12 box|year=2019|ab=A|num-b=18|num-a=20}}
 
{{AMC12 box|year=2019|ab=A|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:37, 12 September 2022

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

Video Solution1

https://youtu.be/E8gk7VkLxos

~ Education, the Study of Everything

See Also

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

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