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

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-Rowechen Zhong
 
-Rowechen Zhong
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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 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{\text{(A)}\, 9}</math> is our answer.
  
 
==See Also==
 
==See Also==

Revision as of 21:44, 9 February 2019

Problem

In $\triangle ABC$ with integer side lengths, \[\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.\] 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

We intend to use law of sines, so let's flip all the cosines (Sine is positive for $0\le x \le 180$, 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 $4$. $\boxed{\text{(A)}\, 9}$ is our answer.

-Rowechen Zhong

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 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 $\boxed{\text{(A)}\, 9}$ is our answer.

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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