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

m (→‎Solution)
m (→‎Solution: Removed negative, boxed answer prettier)
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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).
 
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).
  
−
<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>
+
<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>
  
−
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.
+
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{\text{(A)}\, 9}</math> is our answer.
  
 
-Rowechen Zhong
 
-Rowechen Zhong

Revision as of 19:10, 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

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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