2011 AMC 12A Problems/Problem 13

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 $\overline{AC}$ 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

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 $\triangle{AMN}$ then becomes \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 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AMC 12 Problems and Solutions

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