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
.
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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