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Difference between revisions of "2019 AIME II Problems/Problem 7"

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== Solution 2 ==
 
== Solution 2 ==
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Let the diagram be set up like that in Solution 1.
 +
 +
By similar triangles we have
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<cmath>\frac{AH}{AB}=\frac{GH}{BC}\Longrightarrow AH=30</cmath>
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<cmath>\frac{IB}{AB}=\frac{DI}{AC}\Longrightarrow IB=30</cmath>
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Thus <cmath>HI=AB-AH-IB=60</cmath>
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 +
Since <math>\bigtriangleup IHZ\sim\bigtriangleup ABC</math> and <math>\frac{HI}{AB}=\frac{1}{2}</math>, the altitude of <math>\bigtriangleup IHZ</math> from <math>Z</math> is half the altitude of <math>\bigtriangleup ABC</math> from <math>C</math>, say <math>\frac{h}{2}</math>. Also since <math>\frac{EF}{AB}=\frac{1}{8}</math>, the distance from <math>ell_C</math> to <math>AB</math> is <math>\frac{7}{8}h</math>. Therefore the altitude of <math>\bigtriangleup XYZ</math> from <math>Z</math> is
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<cmath>\frac{1}{2}h+\frac{7}{8}h=\frac{11}{8}h</cmath>.
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By triangle scaling, the perimeter of <math>\bigtriangleup XYZ</math> is <math>\frac{11}{8}</math> of that of <math>\bigtriangleup ABC</math>, or
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<cmath>\frac{11}{8}(220+180+120)=\boxed{715}</cmath>
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=II|num-b=6|num-a=8}}
 
{{AIME box|year=2019|n=II|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:16, 4 January 2020

Problem

Triangle $ABC$ has side lengths $AB=120,BC=220$, and $AC=180$. Lines $\ell_A,\ell_B$, and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$, and $\overline{AB}$, respectively, such that the intersections of $\ell_A,\ell_B$, and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,45$, and $15$, respectively. Find the perimeter of the triangle whose sides lie on lines $\ell_A,\ell_B$, and $\ell_C$.

Solution

Let the points of intersection of $\ell_a, \ell_b,\ell_c$ with $\triangle ABC$ divide the sides into consecutive segments $BD,DE,EC,CF,FG,GA,AH,HI,IB$. Furthermore, let the desired triangle be $\triangle XYZ$, with $X$ closest to side $BC$, $Y$ closest to side $AC$, and $Z$ closest to side $AB$. Hence, the desired perimeter is $XE+EF+FY+YG+GH+HZ+ZI+ID+DX=(DX+XE)+(FY+YG)+(HZ+ZI)+115$ since $HG=55$, $EF=15$, and $ID=45$.

Note that $\triangle AHG\sim \triangle BID\sim \triangle EFC\sim \triangle ABC$, so using similar triangle ratios, we find that $BI=HA=30$, $BD=HG=55$, $FC=\frac{45}{2}$, and $EC=\frac{55}{2}$.

We also notice that $\triangle EFC\sim \triangle YFG\sim \triangle EXD$ and $\triangle BID\sim \triangle HIZ$. Using similar triangles, we get that \[FY+YG=\frac{GF}{FC}\cdot \left(EF+EC\right)=\frac{225}{45}\cdot \left(15+\frac{55}{2}\right)=\frac{425}{2}\] \[DX+XE=\frac{DE}{EC}\cdot \left(EF+FC\right)=\frac{275}{55}\cdot \left(15+\frac{45}{2}\right)=\frac{375}{2}\] \[HZ+ZI=\frac{IH}{BI}\cdot \left(ID+BD\right)=2\cdot \left(45+55\right)=200\] Hence, the desired perimeter is $200+\frac{425+375}{2}+115=600+115=\boxed{715}$ -ktong

Solution 2

Let the diagram be set up like that in Solution 1.

By similar triangles we have \[\frac{AH}{AB}=\frac{GH}{BC}\Longrightarrow AH=30\] \[\frac{IB}{AB}=\frac{DI}{AC}\Longrightarrow IB=30\] Thus \[HI=AB-AH-IB=60\]

Since $\bigtriangleup IHZ\sim\bigtriangleup ABC$ and $\frac{HI}{AB}=\frac{1}{2}$, the altitude of $\bigtriangleup IHZ$ from $Z$ is half the altitude of $\bigtriangleup ABC$ from $C$, say $\frac{h}{2}$. Also since $\frac{EF}{AB}=\frac{1}{8}$, the distance from $ell_C$ to $AB$ is $\frac{7}{8}h$. Therefore the altitude of $\bigtriangleup XYZ$ from $Z$ is \[\frac{1}{2}h+\frac{7}{8}h=\frac{11}{8}h\].

By triangle scaling, the perimeter of $\bigtriangleup XYZ$ is $\frac{11}{8}$ of that of $\bigtriangleup ABC$, or \[\frac{11}{8}(220+180+120)=\boxed{715}\]

See Also

2019 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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