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2011 AMC 12A Problems/Problem 13: Difference between revisions

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== Solution ==
== Solution ==
 
===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. <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,


Line 28: Line 28:


Perimeter of <math> \triangle AMN = \frac{12+24+18}{9} \times 5 = 30 </math>
Perimeter of <math> \triangle AMN = \frac{12+24+18}{9} \times 5 = 30 </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>
It then follows that <math>MO = MB</math>. Similarly, <math>NO = NC</math>. The perimeter of <math>\triangle{AMN}</math>
<cmath>
\begin{align*}
AM + MN + NA &= AM + MO + NO + NA \\
&= AM + MB + NC + NA \\
&= AB + AC \\
&= 30
\end{align*}
</cmath>


== 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 21:27, 11 February 2011

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. $AO$ is the angle bisector of $\angle MAN$. Let the angle bisector of $\angle BAC$ meets $BC$ at $P$ and the angle bisector of $\angle ABC$ meets $AC$ at $Q$. By applying both angle bisector theorem and menelaus' theorem,


$\frac{AO}{OP} \times \frac{BP}{BC} \times \frac{CQ}{QA} = 1$


$\frac{AO}{OP} \times \frac{12}{30} \times \frac{24}{12} = 1$


$\frac{AO}{OP}=\frac{5}{4}$


$\frac{AO}{AP}=\frac{5}{9}$


Perimeter of $\triangle AMN = \frac{12+24+18}{9} \times 5 = 30$

Solution 2

Using the same notation as in 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 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 \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
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