Difference between revisions of "2019 AMC 12A Problems/Problem 19"
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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> | <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>2 | + | 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. |
-Rowechen Zhong | -Rowechen Zhong |
Revision as of 17:40, 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 , so 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 |
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