Difference between revisions of "2011 AMC 12A Problems/Problem 13"
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| − | Perimeter of <math> \triangle AMN = \frac{12+24+18}{9} \times 5 = 30 </math> | + | Perimeter of <math> \triangle AMN = \frac{12+24+18}{9} \times 5 = 30 \rightarrow \boxed{(B)}</math> |
===Solution 2=== | ===Solution 2=== | ||
Revision as of 21:49, 11 February 2011
Problem
Triangle 
has side-lengths 
and 
The line through the incenter of 
parallel to 
intersects 
at 
and $\overbar{AC}$ (Error compiling LaTeX. Unknown error_msg) at 
What is the perimeter of 
Solution
Solution 1
Let 
be the incenter. 
is the angle bisector of 
. Let the angle bisector of 
meets 
at 
and the angle bisector of 
meets 
at 
. By applying both angle bisector theorem and menelaus' theorem,
Perimeter of 
Solution 2
Using the same notation as in Solution 1, let be the incenter. Because 
and 
is the angle bisector, we have
![\[\angle{MBO} = \angle{CBO} = \angle{MOB} = \frac{1}{2}\angle{MBC}\]](http://latex.artofproblemsolving.com/4/3/0/4305bb6699bacdcba757e231d193c6641d876115.png)
It then follows that 
. Similarly, 
. The perimeter of 
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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 | |
