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 By Angle Bisector Theorem, we have
from which
Recall that We apply the Triangle Inequality to
We get
We get
We get
Taking the intersection of the solutions gives so the answer is
~quinnanyc ~MRENTHUSIASM
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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