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

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==Solution==
 
==Solution==
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Let the points of intersection of <math>\ell_a, \ell_b,\ell_c</math> with <math>\triangle ABC</math> divide the sides into consecutive segments <math>BD,DE,EC,CF,FG,GA,AH,HI,IB</math>. Furthermore, let the desired triangle be <math>\triangle XYZ</math>, with <math>X</math> closest to side <math>BC</math>, <math>Y</math> closest to side <math>AC</math>, and <math>Z</math> closest to side <math>AB</math>. Hence, the desired perimeter is <math>XE+EF+FY+YG+GH+HZ+ZI+ID+DX=(DX+XE)+(FY+YG)+(HZ+ZI)+115</math> since <math>HG=55</math>, <math>EF=15</math>, and <math>ID=45</math>.
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Note that <math>\triangle AHG\sim \triangle BID\sim \triangle EFC\sim \triangle ABC</math>, so using similar triangle ratios, we find that <math>BI=HA=30</math>, <math>BD=HG=55</math>, <math>FC=\frac{45}{2}</math>, and <math>EC=\frac{55}{2}</math>.
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We also notice that <math>\triangle EFC\sim \triangle YFG\sim \triangle EXD</math> and <math>\triangle BID\sim \triangle HIZ</math>. Using similar triangles, we get that
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<cmath>FY+YG=\frac{GF}{FC}\cdot \left(EF+EC\right)=\frac{225}{45}\cdot \left(15+\frac{55}{2}\right)=\frac{425}{2}</cmath>
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<cmath>DX+XE=\frac{DE}{EC}\cdot \left(EF+FC\right)=\frac{275}{55}\cdot \left(15+\frac{45}{2}\right)=\frac{375}{2}</cmath>
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<cmath>HZ+ZI=\frac{IH}{BI}\cdot \left(ID+BD\right)=2\cdot \left(45+55\right)=200</cmath>
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Hence, the desired perimeter is <math>200+\frac{425+375}{2}+115=600+115=\boxed{715}</math>
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-ktong
  
 
==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 22:14, 22 March 2019

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

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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