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 1

Let $O$ be the incenter of $\triangle{ABC}$. Because $\overline{MO} \parallel \overline{BC}$ and $\overline{BO}$ is the angle bisector of $\angle{ABC}$, 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*}

Solution 2

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 $\triangle AMN = \frac{12+24+18}{9} \times 5 = 30  \rightarrow \boxed{(B)}$

Solution 3

Like in other solutions, let $O$ be the incenter of $\triangle ABC$. Let $AO$ intersect $BC$ at $D$. By the angle bisector theorem, $\frac{BD}{DC} = \frac{AB}{AC} = \frac{12}{18} = \frac{2}{3}$. Since $BD+DC = 24$, we have $\frac{BD}{24-BD} = \frac{2}{3}$, so $3BD = 48 - 2BD$, so $BD = \frac{48}{5} = 9.6$. By the angle bisector theorem on $\triangle ABD$, we have $\frac{DO}{OA} = \frac{BD}{BA} = \frac{4}{5} = 0.8$, so $\frac{DA}{OA} = 1 + \frac{DO}{OA} = \frac{9}{5} = 1.8$, so $\frac{AO}{AD} = \frac{5}{9}$. Because $\triangle AMN \sim \triangle ABC$, the perimeter of $\triangle AMN$ must be $\frac{5}{9} (12 + 18 + 24) = 30$, so our answer is $\boxed{\textbf{(B)}\ 30}$.

Another way to find $\frac{AO}{OD}$ is to use mass points. Assign a mass of 24 to $A$, a mass of 18 to $B$, and a mass of 12 to $C$. Then $D$ has mass 30, so $\frac{AO}{OD} = \frac{30}{24} = \frac{5}{4} = 1.25$.

Solution 4

We know that the ratio of the perimeter of $\triangle AMN$ and $\triangle ABC$ is the ratio of their heights, and finding the two heights is pretty easy. Note that the height from $A$ to $BC$ is $\frac{9\sqrt{15}}{4}$ from Herons and then $A=\frac{bh}{2}$ and also that the height from $A$ to $MN$ is simply the height from $A$ to $BC$ minus the inradius. We know the area and the semiperimeter so $r=\frac{A}{s}$ which gives us $r=\sqrt{15}$. Now we know that the altitude from $A$ to $MN$ is $\frac{5\sqrt{15}}{4}$ so the ratios of the heights from $A$ for $\triangle AMN$ and $\triangle ABC$ is $\frac{5}{9}$. Thus the perimeter of $\triangle AMN$ is $\frac{5}{9} \times 54 = 30$ so our answer is $\boxed{B}$

-srisainandan6

Video Solution

https://www.youtube.com/watch?v=u23iWcqbJlE ~Shreyas S

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS