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

Revision as of 21:49, 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 $\triangle{AMN}$ $\begin{align*} AM + MN + NA &= AM + MO + NO + NA \\ &= AM + MB + NC + NA \\ &= AB + AC \\ &= 30 \rightarrow \boxed{(B)} \end{align*}$

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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