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

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Triangle <math>ABC</math> has side lengths <math>AB=120,BC=220</math>, and <math>AC=180</math>. Lines <math>\ell_A,\ell_B</math>, and <math>\ell_C</math> are drawn parallel to <math>\overline{BC},\overline{AC}</math>, and <math>\overline{AB}</math>, respectively, such that the intersections of <math>\ell_A,\ell_B</math>, and <math>\ell_C</math> with the interior of <math>\triangle ABC</math> are segments of lengths <math>55,45</math>, and <math>15</math>, respectively. Find the perimeter of the triangle whose sides lie on lines <math>\ell_A,\ell_B</math>, and <math>\ell_C</math>.
 
Triangle <math>ABC</math> has side lengths <math>AB=120,BC=220</math>, and <math>AC=180</math>. Lines <math>\ell_A,\ell_B</math>, and <math>\ell_C</math> are drawn parallel to <math>\overline{BC},\overline{AC}</math>, and <math>\overline{AB}</math>, respectively, such that the intersections of <math>\ell_A,\ell_B</math>, and <math>\ell_C</math> with the interior of <math>\triangle ABC</math> are segments of lengths <math>55,45</math>, and <math>15</math>, respectively. Find the perimeter of the triangle whose sides lie on lines <math>\ell_A,\ell_B</math>, and <math>\ell_C</math>.
  
==Solution==
+
==Diagram==
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(350);
 +
 
 +
pair A, B, C, D, E, F, G, H, I, J, K, L;
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B = origin;
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C = (220,0);
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A = intersectionpoints(Circle(B,120),Circle(C,180))[0];
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D = A+1/4*(B-A);
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E = A+1/4*(C-A);
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F = B+1/4*(A-B);
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G = B+1/4*(C-B);
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H = C+1/8*(A-C);
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I = C+1/8*(B-C);
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J = extension(D,E,F,G);
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K = extension(F,G,H,I);
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L = extension(H,I,D,E);
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draw(A--B--C--cycle);
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draw(J+9/8*(K-J)--K+9/8*(J-K),dashed);
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draw(L+9/8*(K-L)--K+9/8*(L-K),dashed);
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draw(J+9/8*(L-J)--L+9/8*(J-L),dashed);
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draw(D--E^^F--G^^H--I,red);
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dot("$B$",B,1.5SW,linewidth(4));
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dot("$C$",C,1.5SE,linewidth(4));
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dot("$A$",A,1.5N,linewidth(4));
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dot(D,linewidth(4));
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dot(E,linewidth(4));
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dot(F,linewidth(4));
 +
dot(G,linewidth(4));
 +
dot(H,linewidth(4));
 +
dot(I,linewidth(4));
 +
dot(J,linewidth(4));
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dot(K,linewidth(4));
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dot(L,linewidth(4));
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label("$55$",midpoint(D--E),S,red);
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label("$45$",midpoint(F--G),dir(55),red);
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label("$15$",midpoint(H--I),dir(160),red);
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label("$\ell_A$",J+9/8*(L-J),1.5*dir(B--C));
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label("$\ell_B$",K+9/8*(J-K),1.5*dir(C--A));
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label("$\ell_C$",L+9/8*(K-L),1.5*dir(A--B));
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</asy>
 +
~MRENTHUSIASM
 +
 
 +
==Solution 1==
 +
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>.
 +
 
 +
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>.
 +
 
 +
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
 +
<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>
 +
<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>
 +
<cmath>HZ+ZI=\frac{IH}{BI}\cdot \left(ID+BD\right)=2\cdot \left(45+55\right)=200</cmath>
 +
Hence, the desired perimeter is <math>200+\frac{425+375}{2}+115=600+115=\boxed{715}</math>
 +
-ktong
 +
 
 +
== Solution 2 ==
 +
 
 +
Let the diagram be set up like that in Solution 1.
 +
 
 +
By similar triangles we have
 +
<cmath>\frac{AH}{AB}=\frac{GH}{BC}\Longrightarrow AH=30</cmath>
 +
<cmath>\frac{IB}{AB}=\frac{DI}{AC}\Longrightarrow IB=30</cmath>
 +
Thus <cmath>HI=AB-AH-IB=60</cmath>
 +
 
 +
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
 +
<cmath>\frac{1}{2}h+\frac{7}{8}h=\frac{11}{8}h</cmath>.
 +
 
 +
By triangle scaling, the perimeter of <math>\bigtriangleup XYZ</math> is <math>\frac{11}{8}</math> of that of <math>\bigtriangleup ABC</math>, or
 +
<cmath>\frac{11}{8}(220+180+120)=\boxed{715}</cmath>
 +
 
 +
~ Nafer
  
 
==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}}
 +
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:11, 1 October 2021

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 $\ell_C$.

Diagram

[asy] /* Made by MRENTHUSIASM */ size(350);  pair A, B, C, D, E, F, G, H, I, J, K, L; B = origin; C = (220,0); A = intersectionpoints(Circle(B,120),Circle(C,180))[0]; D = A+1/4*(B-A); E = A+1/4*(C-A); F = B+1/4*(A-B); G = B+1/4*(C-B); H = C+1/8*(A-C); I = C+1/8*(B-C); J = extension(D,E,F,G); K = extension(F,G,H,I); L = extension(H,I,D,E); draw(A--B--C--cycle); draw(J+9/8*(K-J)--K+9/8*(J-K),dashed); draw(L+9/8*(K-L)--K+9/8*(L-K),dashed); draw(J+9/8*(L-J)--L+9/8*(J-L),dashed); draw(D--E^^F--G^^H--I,red); dot("$B$",B,1.5SW,linewidth(4)); dot("$C$",C,1.5SE,linewidth(4)); dot("$A$",A,1.5N,linewidth(4)); dot(D,linewidth(4)); dot(E,linewidth(4)); dot(F,linewidth(4)); dot(G,linewidth(4)); dot(H,linewidth(4)); dot(I,linewidth(4)); dot(J,linewidth(4)); dot(K,linewidth(4)); dot(L,linewidth(4)); label("$55$",midpoint(D--E),S,red); label("$45$",midpoint(F--G),dir(55),red); label("$15$",midpoint(H--I),dir(160),red); label("$\ell_A$",J+9/8*(L-J),1.5*dir(B--C)); label("$\ell_B$",K+9/8*(J-K),1.5*dir(C--A)); label("$\ell_C$",L+9/8*(K-L),1.5*dir(A--B)); [/asy] ~MRENTHUSIASM

Solution 1

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 \[\frac{11}{8}(220+180+120)=\boxed{715}\]

~ Nafer

See Also

2019 AIME II (ProblemsAnswer KeyResources)
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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