Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AMC 12A Problems/Problem 13"

(→‎See also)
m (→‎Solution)
Line 11: Line 11:
 
== Solution ==
 
== Solution ==
 
===Solution 1===
 
===Solution 1===
−
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,
+
Let <math>O</math> be the incenter. Because <math>MO \parallel BC</math> and <math>BO</math> is the angle bisector, we have  
−
 
 
−
 
 
−
 
 
−
<math>\frac{AO}{OP} \times \frac{BP}{BC} \times \frac{CQ}{QA} = 1 </math>
 
−
 
 
−
 
 
−
<math>\frac{AO}{OP} \times \frac{12}{30} \times \frac{24}{12} = 1 </math>
 
−
 
 
−
 
 
−
<math>\frac{AO}{OP}=\frac{5}{4} </math>
 
−
 
 
−
 
 
−
<math>\frac{AO}{AP}=\frac{5}{9} </math>
 
−
 
 
−
 
 
−
Perimeter of <math> \triangle AMN = \frac{12+24+18}{9} \times 5 = 30  \rightarrow \boxed{(B)}</math>
 
−
 
 
−
===Solution 2===
 
−
Using the same notation as in Solution 1, let <math>O</math> be the incenter. Because <math>MO \parallel BC</math> and <math>BO</math> is the angle bisector, 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>
  
−
It then follows 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>
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}

Revision as of 13:58, 22 September 2013

Problem

Triangle $ABC$ has side-lengths $AB = 12, BC = 24,$ and $AC = 18.$ The line through the incenter of $\triangle ABC$ parallel to $\overline{BC}$ intersects $\overline{AB}$ at $M$ and $\overbar{AC}$ (Error compiling LaTeX. Unknown error_msg) at $N.$ What is the perimeter of $\triangle AMN?$

$\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 33 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 42$

Solution

Solution 1

Let $O$ be the incenter. Because $MO \parallel BC$ and $BO$ is the angle bisector, we have

\[\angle{MBO} = \angle{CBO} = \angle{MOB} = \frac{1}{2}\angle{MBC}\]

It then follows due to alternate interior angles and base angles of isosceles triangles that $MO = MB$. Similarly, $NO = NC$. The perimeter of $\triangle{AMN}$ \begin{align*} AM + MN + NA &= AM + MO + NO + NA \\ &= AM + MB + NC + NA \\ &= AB + AC \\ &= 30 \rightarrow \boxed{(B)} \end{align*}

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_13&oldid=57116"