Difference between revisions of "2011 AMC 12A Problems/Problem 13"
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== Solution == | == Solution == | ||
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| + | Let <math> O </math> be the incenter. <math> AO </math> is the angle bisector of <math> \angle MAN </math>. Let the angle bisector of <math> \angle BAC </math> meets <math> BC </math> at <math> P </math> and the angle bisector of <math> \angle ABC </math> meets <math> AC </math> at <math> Q </math>. By applying both angle bisector theorem and menelaus' theorem, | ||
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| + | <math>\frac{AO}{OP} \times \frac{BP}{BC} \times \frac{CQ}{QA} = 1 </math> | ||
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| + | <math>\frac{AO}{OP} \times \frac{12}{30} \times \frac{24}{12} = 1 </math> | ||
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| + | <math>\frac{AO}{OP}=\frac{5}{4} </math> | ||
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| + | <math>\frac{AO}{AP}=\frac{5}{9} </math> | ||
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| + | Perimeter of <math> \triangle AMN = \frac{12+24+18}{9} \times 5 = 30 </math> | ||
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== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=12|num-a=14|ab=A}} | {{AMC12 box|year=2011|num-b=12|num-a=14|ab=A}} | ||
Revision as of 04:25, 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
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 
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 | |
