Difference between revisions of "2011 AMC 12A Problems/Problem 13"
Armalite46 (talk | contribs) m (→Solution) |
Armalite46 (talk | contribs) m (→Solution 1) |
||
| Line 15: | Line 15: | ||
<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> | ||
| − | It then follows due to alternate interior angles and base angles of isosceles triangles that <math>MO = MB</math>. Similarly, <math>NO = NC</math>. The perimeter of <math>\triangle{AMN}</math> | + | It then follows due to alternate interior angles and base angles of isosceles triangles that <math>MO = MB</math>. Similarly, <math>NO = NC</math>. The perimeter of <math>\triangle{AMN}</math> then becomes |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
Revision as of 14:58, 22 September 2013
Contents
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. 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. 
