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

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m (→‎Solution 4 (Way too short))
 
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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==
−
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>.  
+
<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 <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>.  
 
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>.  
Line 15: Line 59:
  
 
== Solution 2 ==
 
== 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
 +
 +
==Solution 3==
 +
[[File:2019 AIME II 7.png|450px|right]]
 +
Notation shown on diagram. By similar triangles we have
 +
<cmath>k_1 = \frac{EF}{BC} = \frac{AE}{AB} = \frac {AF}{AC} = \frac {1}{4},</cmath>
 +
<cmath>k_2 = \frac{F''E''}{AC} = \frac {BF''}{AB} = \frac{1}{4},</cmath>
 +
<cmath>k_3 = \frac{E'F'}{AB} = \frac{E'C }{AC} = \frac{1}{8}.</cmath>
 +
So, <cmath>\frac{ZE}{BC} = \frac{F''E}{AB} = \frac{AB - AE - BF''}{AB} = 1 - k_1 - k_2,</cmath>
 +
<cmath>\frac{FY}{BC} = \frac{FE'}{AC} = \frac{AC - AF - CE'}{AC} = 1 - k_1 - k_3.</cmath>
 +
<cmath>k = \frac{ZY}{BC} = \frac{ZE + EF + FY}{BC} = (1 - k_1 - k_2) + k_1 + (1 - k_1 - k_3)</cmath>
 +
<cmath>k = 2 -  k_1 - k_2 - k_3 = 2 - \frac{1}{4} - \frac{1}{4}  - \frac{1}{8} = \frac{11}{8}.</cmath>
 +
<cmath>\frac{ZY+YX +XZ}{BC +AB + AC} = k \implies ZY + YX + XZ =\frac{11}{8} (220 + 120 + 180) = \boxed {715}.</cmath>
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 +
==Solution 4 (''Way'' too short, just keep track of which side is which)==
 +
 +
<asy>
 +
pair A,B,C;
 +
B = (0,0);
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C = (1,0);
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A = intersectionpoints(circle(B,3/2),circle(C,11/6))[0];
 +
draw(A--B--C--cycle);
 +
draw((3/2,3/4)--(5/2,3/4));
 +
draw((3/2,1/4)--(5/2,1/4));
 +
draw((9/4,1)--(11/4,1/2)--(9/4,0));
 +
draw(shift(dir(0)*13/4)*shift(dir(30))*polygon(3));
 +
</asy>
 +
 +
<asy>
 +
for (int i=0; i<8; ++i)
 +
{
 +
for (int j=0; j<i+1; ++j)
 +
{
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  draw(shift(dir(30))*shift(dir(0)*i*sqrt(3))*shift(dir(120)*j*sqrt(3))*polygon(3));
 +
}
 +
}
 +
pair A = origin+2*sqrt(3)*dir(0)+7*sqrt(3)*dir(120);
 +
pair B = origin+13*sqrt(3)*dir(0)+7*sqrt(3)*dir(120);
 +
pair C = origin+2*sqrt(3)*dir(0)-4*sqrt(3)*dir(120);
 +
pair D = origin + 2*sqrt(3)*dir(0);
 +
pair E = origin + 2*sqrt(3)*dir(60);
 +
pair F = origin + 7*sqrt(3)*dir(60);
 +
pair G = origin + 7*sqrt(3)*dir(60) + 1*sqrt(3)*dir(0);
 +
pair H = origin + 6*sqrt(3)*dir(0) + 2*sqrt(3)*dir(60);
 +
pair I = origin + 6*sqrt(3)*dir(0);
 +
draw(A--B--C--cycle,linewidth(3));
 +
draw(D--E,linewidth(3)+rgb(3/4,1/4,1/4));
 +
draw(F--G,linewidth(3)+rgb(1/4,3/4,1/4));
 +
draw(H--I,linewidth(3)+rgb(1/4,1/4,3/4));
 +
</asy>
 +
 +
Let's squish a triangle with side lengths 15, 22.5, and 27.5 into a equilateral triangle with side length 1. Then, the original triangle gets turned into a equilateral triangle with side length 8. Since 15 is one eighth of 120, it has a length of one. Since 45 and 55 are one fourth of 180 and 220 respectively, they are both two long. We extend the three segments to form a big equilateral triangle shown in black. Notice it has a side length of 11. Now that our task is done, let's undo the distortion. We get 11(15+22.5+27.5)=11(65)=715.
 +
 +
~ [[User:Afly|Afly]] ([[User talk:Afly|talk]])
  
 
==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}}

Latest revision as of 13:58, 14 January 2024

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

Solution 3

2019 AIME II 7.png

Notation shown on diagram. By similar triangles we have \[k_1 = \frac{EF}{BC} = \frac{AE}{AB} = \frac {AF}{AC} = \frac {1}{4},\] \[k_2 = \frac{F''E''}{AC} = \frac {BF''}{AB} = \frac{1}{4},\] \[k_3 = \frac{E'F'}{AB} = \frac{E'C }{AC} = \frac{1}{8}.\] So, \[\frac{ZE}{BC} = \frac{F''E}{AB} = \frac{AB - AE - BF''}{AB} = 1 - k_1 - k_2,\] \[\frac{FY}{BC} = \frac{FE'}{AC} = \frac{AC - AF - CE'}{AC} = 1 - k_1 - k_3.\] \[k = \frac{ZY}{BC} = \frac{ZE + EF + FY}{BC} = (1 - k_1 - k_2) + k_1 + (1 - k_1 - k_3)\] \[k = 2 - k_1 - k_2 - k_3 = 2 - \frac{1}{4} - \frac{1}{4} - \frac{1}{8} = \frac{11}{8}.\] \[\frac{ZY+YX +XZ}{BC +AB + AC} = k \implies ZY + YX + XZ =\frac{11}{8} (220 + 120 + 180) = \boxed {715}.\] vladimir.shelomovskii@gmail.com, vvsss

Solution 4 (Way too short, just keep track of which side is which)

[asy] pair A,B,C; B = (0,0); C = (1,0); A = intersectionpoints(circle(B,3/2),circle(C,11/6))[0]; draw(A--B--C--cycle); draw((3/2,3/4)--(5/2,3/4)); draw((3/2,1/4)--(5/2,1/4)); draw((9/4,1)--(11/4,1/2)--(9/4,0)); draw(shift(dir(0)*13/4)*shift(dir(30))*polygon(3)); [/asy]

[asy] for (int i=0; i<8; ++i) { for (int j=0; j<i+1; ++j) { draw(shift(dir(30))*shift(dir(0)*i*sqrt(3))*shift(dir(120)*j*sqrt(3))*polygon(3)); } } pair A = origin+2*sqrt(3)*dir(0)+7*sqrt(3)*dir(120); pair B = origin+13*sqrt(3)*dir(0)+7*sqrt(3)*dir(120); pair C = origin+2*sqrt(3)*dir(0)-4*sqrt(3)*dir(120); pair D = origin + 2*sqrt(3)*dir(0); pair E = origin + 2*sqrt(3)*dir(60); pair F = origin + 7*sqrt(3)*dir(60); pair G = origin + 7*sqrt(3)*dir(60) + 1*sqrt(3)*dir(0); pair H = origin + 6*sqrt(3)*dir(0) + 2*sqrt(3)*dir(60); pair I = origin + 6*sqrt(3)*dir(0); draw(A--B--C--cycle,linewidth(3)); draw(D--E,linewidth(3)+rgb(3/4,1/4,1/4)); draw(F--G,linewidth(3)+rgb(1/4,3/4,1/4)); draw(H--I,linewidth(3)+rgb(1/4,1/4,3/4)); [/asy]

Let's squish a triangle with side lengths 15, 22.5, and 27.5 into a equilateral triangle with side length 1. Then, the original triangle gets turned into a equilateral triangle with side length 8. Since 15 is one eighth of 120, it has a length of one. Since 45 and 55 are one fourth of 180 and 220 respectively, they are both two long. We extend the three segments to form a big equilateral triangle shown in black. Notice it has a side length of 11. Now that our task is done, let's undo the distortion. We get 11(15+22.5+27.5)=11(65)=715.

~ Afly (talk)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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