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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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<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>?
 
  
 
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math>
 
<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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==Solution 1==
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== Solutions ==
 +
=== Solution 1 ===
 
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).
 
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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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.
 
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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=== Solution 2 ===
 
<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.
 
<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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==Solution 3==
 
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Using the law of cosines, we get the following equations:
 
Using the law of cosines, we get the following equations:
  
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<cmath>14a+11b=16c</cmath>
 
<cmath>14a+11b=16c</cmath>
  
−
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>.
+
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>
  
 
~hiker
 
~hiker
  
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==See Also==
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=== Solution 4 ===
 +
Similar to solution 3, we will get the following by the law of cosines:
 +
 
 +
<cmath>
 +
\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}
 +
</cmath>
 +
 
 +
By adding equations <math>(1), (2)</math> and <math>(3)</math> together, we will obtain:
 +
<cmath>
 +
\begin{align*}
 +
a^2 + b^2 + c^2 &= - \frac{1}{2} ab + \frac{11}{8} bc + \frac{7}{4} ac & \text{(4)}
 +
\end{align*}
 +
</cmath>
 +
 
 +
Subtracting <math>(2)</math> from <math>(4)</math> will give us:
 +
<cmath>
 +
\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\\
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2a &= - \frac{1}{2} b + \frac{7}{4} c\\
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16a &= 14c - 4b & \text{(a)}
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\end{align*}
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</cmath>
 +
 
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Again, we subtract <math>(3)</math> from <math>(4)</math> and obtain:
 +
<cmath>
 +
\begin{align*}
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16b &= 11c - 4a & \text{(b)}
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\end{align*}
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</cmath>
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By adding equations <math>(a)</math> and <math>(b)</math>:
 +
<cmath>
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\begin{align*}
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16a + 16b &= 14c - 4b + 11c - 4a\\
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20(a+b) &= 25c\\
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4(a+b) &= 5c
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\end{align*}
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</cmath>
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Since all <math>a</math>, <math>b</math> and <math>c</math> are integers, the smallest possibility is <math>a+b=5</math> and <math>c=4</math>. Thus, <math>a+b+c=4+5=\boxed{\textbf{(A) } 9}</math>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Andy_li0805 Andy_li0805]
 +
 
 +
== Video Solution1 ==
 +
https://youtu.be/E8gk7VkLxos
 +
 
 +
~ Education, the Study of Everything
  
 +
== 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 22:36, 19 August 2025

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

See Also

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

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