Difference between revisions of "2019 AMC 12A Problems/Problem 19"
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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{\textbf{(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{\textbf{(A) } 9}</math> is our answer. | ||
+ | Lmao | ||
==Solution 2== | ==Solution 2== |
Revision as of 00:43, 30 November 2019
Problem
In with integer side lengths, What is the least possible perimeter for ?
Solution 1
Notice that by the Law of Sines, , so let's flip all the cosines using (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. Lmao
Solution 2
is obtuse since its cosine is negative, so we let the foot of the altitude from to be . Let , , , and . By the Pythagorean Theorem, and . Thus, . The sides of the triangle are then , , and , so for some integers , and , where and are minimal. Hence, , or . Thus the smallest possible positive integers and that satisfy this are and , so . The sides of the triangle are , , and , so is our answer.
Solution 3
Using the law of cosines, we get the following equations:
Substituting for in and simplifying, we get the following:
Note that since are integers, we can solve this for integers. By some trial and error, we get that . Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is .
Solution by hiker.
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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