2018 AMC 12B Problems/Problem 12
Problem
Side of has length . The bisector of angle meets at , and . The set of all possible values of is an open interval . What is ?
Solution
Let . Then by Angle Bisector Theorem, we have . Now, by the triangle inequality, we have three inequalities.
- , so . Solve this to find that , so .
- , so . Solve this to find that , so .
- The third inequality can be disregarded, because has no real roots.
Then our interval is simply to get . (quinnanyc)
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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All AMC 12 Problems and Solutions |
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