Difference between revisions of "2018 AMC 12B Problems/Problem 12"
m (→Solution) |
Rockmanex3 (talk | contribs) (→See Also) |
||
| Line 23: | Line 23: | ||
{{AMC12 box|year=2018|ab=B|num-b=11|num-a=13}} | {{AMC12 box|year=2018|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
Revision as of 16:29, 18 June 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 
?
![\[\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18 \qquad \textbf{(D) }19 \qquad \textbf{(E) }20 \qquad\]](http://latex.artofproblemsolving.com/8/e/7/8e7aabaee0c0bafc06676b5fe9d40cfe4b4d89af.png)
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.
, so 
. Solve this to find that 
, so 
.
, so 
. Solve this to find that 
, so 
.
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. 
