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Difference between revisions of "2011 AMC 12A Problems/Problem 13"
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==Solution 3==
 
==Solution 3==
−
Like in other solutions, let O be the incenter of <math>\triangle ABC</math>. Let <math>AO</math> intersect <math>BC</math> at <math>D</math>. By the angle bisector theorem, <math>\frac{BD}{DC} = \frac{AB}{AC} = \frac{12}{18} = \frac{2}{3}</math>. Since <math>BD+DC = 24</math>, we have <math>\frac{BD}{24-BD} = \frac{2}{3}</math>, so <math>3BD = 48 - 2BD</math>, so <math>BD = \frac{48}{5} = 9.6</math>. By the angle bisector theorem on <math>\triangle ABD</math>, we have <math>\frac{DO}{OA} = \frac{BD}{BA} = \frac{4}{5} = 0.8</math>, so <math>\frac{DA}{OA} = 1 + \frac{DO}{OA} = \frac{9}{5} = 1.8</math>, so <math>\frac{AO}{AD} = \frac{5}{9}</math>. Because <math>\triangle AMN \sim \triangle ABC</math>, the perimeter of <math>\triangle AMN</math> must be <math>\frac{5}{9} (12 + 18 + 24) = 30</math>, so our answer is <math>\boxed{\textbf{(B)}\ 30}</math>.
+
Like in other solutions, let <math>O</math> be the incenter of <math>\triangle ABC</math>. Let <math>AO</math> intersect <math>BC</math> at <math>D</math>. By the angle bisector theorem, <math>\frac{BD}{DC} = \frac{AB}{AC} = \frac{12}{18} = \frac{2}{3}</math>. Since <math>BD+DC = 24</math>, we have <math>\frac{BD}{24-BD} = \frac{2}{3}</math>, so <math>3BD = 48 - 2BD</math>, so <math>BD = \frac{48}{5} = 9.6</math>. By the angle bisector theorem on <math>\triangle ABD</math>, we have <math>\frac{DO}{OA} = \frac{BD}{BA} = \frac{4}{5} = 0.8</math>, so <math>\frac{DA}{OA} = 1 + \frac{DO}{OA} = \frac{9}{5} = 1.8</math>, so <math>\frac{AO}{AD} = \frac{5}{9}</math>. Because <math>\triangle AMN \sim \triangle ABC</math>, the perimeter of <math>\triangle AMN</math> must be <math>\frac{5}{9} (12 + 18 + 24) = 30</math>, so our answer is <math>\boxed{\textbf{(B)}\ 30}</math>.
  
 
Another way to find <math>\frac{AO}{OD}</math> is to use mass points. Assign a mass of 24 to <math>A</math>, a mass of 18 to <math>B</math>, and a mass of 12 to <math>C</math>. Then <math>D</math> has mass 30, so <math>\frac{AO}{OD} = \frac{30}{24} = \frac{5}{4} = 1.25</math>.
 
Another way to find <math>\frac{AO}{OD}</math> is to use mass points. Assign a mass of 24 to <math>A</math>, a mass of 18 to <math>B</math>, and a mass of 12 to <math>C</math>. Then <math>D</math> has mass 30, so <math>\frac{AO}{OD} = \frac{30}{24} = \frac{5}{4} = 1.25</math>.

Revision as of 00:35, 22 July 2020

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 $\frac{AO}{OD} = \frac{30}{24} = \frac{5}{4} = 1.25$.

Video Solution

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

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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