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

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==Solution==
 
==Solution==
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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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<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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These are in the ratio <math>2:3:4</math>, 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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-Rowechen Zhong
  
 
==See Also==
 
==See Also==

Revision as of 16:40, 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 $2:3:4$, our minimal triangle has side lengths $2$, $3$, and $4$. $\boxed{(A) 9}$ is our answer.

-Rowechen Zhong

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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