Difference between revisions of "2019 AMC 12A Problems/Problem 19"
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==Solution== | ==Solution== | ||
| + | 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 
with integer side lengths,
![\[\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \text{and} \qquad\cos C=-\frac{1}{4}.\]](http://latex.artofproblemsolving.com/3/3/c/33c5cf6fca4eabad219869260b59af5b93d6e5a2.png)
What is the least possible perimeter for ?
Solution
We intend to use law of sines, so let's flip all the cosines (Sine is positive for 
, so we're good there).
These are in the ratio 
, our minimal triangle has side lengths 
, 
, and 
. 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
