2024 AMC 12B Problems/Problem 22
Problem 22
Let be a triangle with integer side lengths and the property that . What is the least possible perimeter of such a triangle?
Solution 1
Let , , . According to the law of sines,
According to the law of cosines,
Hence,
This simplifies to . We want to find the positive integer solution to this equation such that forms a triangle, and is minimized. We proceed by casework on the value of . Remember that .
Clearly, this case yields no valid solutions.
For this case, we must have and . However, does not form a triangle. Hence this case yields no valid solutions.
For this case, we must have and . However, does not form a triangle. Hence this case yields no valid solutions.
For this case, and , or and . As one can check, this case also yields no valid solutions
For this case, we must have and . There are no valid solutions
For this case, and , or and , or and . The only valid solution for this case is .
It is safe to assume that will be the solution with least perimeter. Hence, the answer is
~tsun26
Solution 2 (Similar to Solution 1)
Let , , . Extend to point on such that . This means is isosceles, so . Since is the exterior angle of , we have Thus, is isosceles, so Then, draw the altitude of , from to , and let this point be . Let . Then, by Pythagorean Theorem,
Case : .
This means , so the least possible values are , , but this does not work as it does not satisfy the triangle inequality. Similarly, , also does not satisfy it. Anything larger goes beyond the answer choices, so we stop checking this case.
Case : This means , so the least possible values for and are ,, but this does not satisfy the triangle inequality, and anything larger does not satisfy the answer choices.
Case : This means , and the least possible value for is , which occurs when . Unfortunately, this also does not satisfy the triangle inequality, and similarly, any means the perimeter will get too big.
Case : This means , so we have , so the least possible perimeter so far is .
Case : We have , so least possible value for is , which already does not work as , and the minimum perimeter is already.
Case : We have , so , which already does not work.
Then, notice that when , we also must have and , so , so the least possible perimeter is
~evanhliu2009