2019 AMC 12B Problems/Problem 9
Problem
For how many integral values of can a triangle of positive area be formed having side lengths ?
Solution
For these lengths to form a triangle of positive area, the Triangle Inequality tells us that we need , , and . The second inequality is redundant, as it's always less restrictive than the last inequality.
Let's raise the first inequality to the power of . This gives . Thus, .
Doing the same for the second inequality gives (where we are allowed to divide both sides by since must be positive in order for the logarithms given in the problem statement to even have real values).
Combining our results, is an integer strictly between and , so the number of possible values of is .
Solution 2 (Somewhat Cheating)
Using the triangle inequality, you get . Solving for , you get . Now we need an upper-bound for and since we're dealing with bases of and , we're looking for answer choices close to a power of and . All the answer choices seem to be around , and plugging that into the inequality we see is the correct number. Now we have and the number of integers in between is --OGBooger
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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