Difference between revisions of "2018 AMC 12B Problems/Problem 12"
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* The third inequality can be disregarded, because <math>30/x > 7-x</math> has no real roots. | * The third inequality can be disregarded, because <math>30/x > 7-x</math> has no real roots. | ||
− | Then our interval is simply <math>(3,15)</math> to get <math>18</math> <math>\boxed{C}</math>. | + | Then our interval is simply <math>(3,15)</math> to get <math>18</math> <math>\boxed{C}</math>. |
==See Also== | ==See Also== |
Revision as of 01:07, 19 February 2018
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 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.