Difference between revisions of "2011 AMC 12A Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | Let <math>O</math> be the incenter. Because <math>MO \parallel BC</math> and <math>BO</math> is the angle bisector, we have | + | Let <math>O</math> be the incenter of <math>\triangle{ABC}</math>. Because <math>\overline{MO} \parallel \overline{BC}</math> and <math>\overline{BO}</math> is the angle bisector of <math>\angle{ABC}</math>, we have |
<cmath>\angle{MBO} = \angle{CBO} = \angle{MOB} = \frac{1}{2}\angle{MBC}</cmath> | <cmath>\angle{MBO} = \angle{CBO} = \angle{MOB} = \frac{1}{2}\angle{MBC}</cmath> | ||
Revision as of 14:46, 2 November 2019
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 of 
. Because 
and 
is the angle bisector of 
, 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. 
