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, 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 |
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All AMC 12 Problems and Solutions |
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