Difference between revisions of "2011 AMC 12A Problems/Problem 13"
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== Problem == | == Problem == | ||
| − | Triangle <math>ABC</math> has side-lengths <math>AB = 12, BC = 24,</math> and <math>AC = 18.</math> The line through the incenter of <math>\triangle ABC</math> parallel to <math>\overline{BC}</math> intersects <math>\overline{AB}</math> at <math>M</math> and <math>\ | + | Triangle <math>ABC</math> has side-lengths <math>AB = 12, BC = 24,</math> and <math>AC = 18.</math> The line through the incenter of <math>\triangle ABC</math> parallel to <math>\overline{BC}</math> intersects <math>\overline{AB}</math> at <math>M</math> and <math>\overline{AC}</math> at <math>N.</math> What is the perimeter of <math>\triangle AMN?</math> |
<math> | <math> | ||
Revision as of 20:46, 13 March 2015
Problem
Triangle 
has side-lengths 
and 
The line through the incenter of 
parallel to 
intersects 
at 
and 
at 
What is the perimeter of 
Solution
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 due to alternate interior angles and base angles of isosceles triangles that 
. Similarly, 
. The perimeter of 
then becomes
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
