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 $\overbar{AC}$ (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

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 $\begin{align*} AM + MN + NA &= AM + MO + NO + NA \\ &= AM + MB + NC + NA \\ &= AB + AC \\ &= 30 \rightarrow \boxed{(B)} \end{align*}$

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.

Page

Toolbox

Search

2011 AMC 12A Problems/Problem 13

Problem

Solution

See also