2019 AMC 12A Problems/Problem 19
Contents
Problem
In with integer side lengths, , , and . What is the least possible perimeter for ?
Solutions
Solution 1
Notice that by the Law of Sines, , so let's flip all the cosines using ( is positive for , so we're good there).
These are in the ratio , so our minimal triangle has side lengths , , and . is our answer.
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 .
~hiker
Video Solution1
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See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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