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

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== Problem ==
 
== Problem ==
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Triangle <math>ABC</math> has side-lengths <math>AB = 12, BC = 24,</math> and <math>AC = 18.</math> The line through the incenter of <math>\triangle ABC</math> parallel to <math>\overline{BC}</math> intersects <math>\overline{AB}</math> at <math>M</math> and <math>\overbar{AC}</math> at <math>N.</math> What is the perimeter of <math>\triangle AMN?</math>
+
Triangle <math>ABC</math> has side-lengths <math>AB = 12, BC = 24,</math> and <math>AC = 18.</math> The line through the incenter of <math>\triangle ABC</math> parallel to <math>\overline{BC}</math> intersects <math>\overline{AB}</math> at <math>M</math> and <math>\overline{AC}</math> at <math>N.</math> What is the perimeter of <math>\triangle AMN?</math>
  
 
<math>
 
<math>
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\textbf{(E)}\ 42 </math>
 
\textbf{(E)}\ 42 </math>
  
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== Solution ==
+
==Solution 1==
−
===Solution 1===
+
Let <math>O</math> be the incenter of <math>\triangle{ABC}</math>. Because <math>\overline{MO} \parallel \overline{BC}</math> and <math>\overline{BO}</math> is the angle bisector of <math>\angle{ABC}</math>, we have
 +
 
 +
<cmath>\angle{MBO} = \angle{CBO} = \angle{MOB} = \frac{1}{2}\angle{MBC}</cmath>
 +
 
 +
It then follows due to alternate interior angles and base angles of isosceles triangles that <math>MO = MB</math>. Similarly, <math>NO = NC</math>. The perimeter of <math>\triangle{AMN}</math> then becomes
 +
<cmath>
 +
\begin{align*}
 +
AM + MN + NA &= AM + MO + NO + NA \\
 +
&= AM + MB + NC + NA \\
 +
&= AB + AC \\
 +
&= 30 \rightarrow \boxed{(B)}
 +
\end{align*}
 +
</cmath>
 +
 
 +
==Solution 2==
 
Let <math> O </math> be the incenter. <math> AO </math> is the angle bisector of <math> \angle MAN </math>. Let the angle bisector of <math> \angle BAC </math> meets <math> BC </math> at <math> P </math> and the angle bisector of <math> \angle ABC </math> meets <math> AC </math> at <math> Q </math>. By applying both angle bisector theorem and Menelaus' theorem,
 
Let <math> O </math> be the incenter. <math> AO </math> is the angle bisector of <math> \angle MAN </math>. Let the angle bisector of <math> \angle BAC </math> meets <math> BC </math> at <math> P </math> and the angle bisector of <math> \angle ABC </math> meets <math> AC </math> at <math> Q </math>. By applying both angle bisector theorem and Menelaus' theorem,
  
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Perimeter of <math> \triangle AMN = \frac{12+24+18}{9} \times 5 = 30  \rightarrow \boxed{(B)}</math>
 
Perimeter of <math> \triangle AMN = \frac{12+24+18}{9} \times 5 = 30  \rightarrow \boxed{(B)}</math>
  
−
===Solution 2===
+
==Solution 3==
−
Using the same notation as in Solution 1, let <math>O</math> be the incenter. Because <math>MO \parallel BC</math> and <math>BO</math> is the angle bisector, we have  
+
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>.
  
−
<cmath>\angle{MBO} = \angle{CBO} = \angle{MOB} = \frac{1}{2}\angle{MBC}</cmath>
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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>.
 +
 
 +
==Solution 4==
 +
We know that the ratio of the perimeter of <math>\triangle AMN</math> and <math>\triangle ABC</math> is the ratio of their heights, and finding the two heights is pretty easy. Note that the height from <math>A</math> to <math>BC</math> is <math>\frac{9\sqrt{15}}{4}</math> from Herons and then <math>A=\frac{bh}{2}</math> and also that the height from <math>A</math> to <math>MN</math> is simply the height from <math>A</math> to <math>BC</math> minus the inradius. We know the area and the semiperimeter so <math>r=\frac{A}{s}</math> which gives us <math>r=\sqrt{15}</math>. Now we know that the altitude from <math>A</math> to <math>MN</math> is <math>\frac{5\sqrt{15}}{4}</math> so the ratios of the heights from <math>A</math> for <math>\triangle AMN</math> and <math>\triangle ABC</math> is <math>\frac{5}{9}</math>. Thus the perimeter of <math>\triangle AMN</math> is <math>\frac{5}{9} \times 54 = 30</math> so our answer is <math>\boxed{B}</math>
 +
 
 +
-srisainandan6
 +
 
 +
==Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=u23iWcqbJlE
 +
~Shreyas S
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/5jwD5UViZO8?t=1013
  
−
It then follows that <math>MO = MB</math>. Similarly, <math>NO = NC</math>. The perimeter of <math>\triangle{AMN}</math>
+
~ pi_is_3.14
−
<cmath>
 
−
\begin{align*}
 
−
AM + MN + NA &= AM + MO + NO + NA \\
 
−
&= AM + MB + NC + NA \\
 
−
&= AB + AC \\
 
−
&= 30 \rightarrow \boxed{(B)}
 
−
\end{align*}
 
−
</cmath>
 
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=12|num-a=14|ab=A}}
 
{{AMC12 box|year=2011|num-b=12|num-a=14|ab=A}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:12, 23 January 2023

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

Video Solution by OmegaLearn

https://youtu.be/5jwD5UViZO8?t=1013

~ pi_is_3.14

See also

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

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