Difference between revisions of "2019 AIME II Problems/Problem 7"
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==Problem== | ==Problem== | ||
+ | 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; | ||
+ | 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>. | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==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); | ||
+ | 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. | ||
+ | |||
+ | ~ [[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 14:58, 14 January 2024
Contents
Problem
Triangle has side lengths
, and
. Lines
, and
are drawn parallel to
, and
, respectively, such that the intersections of
, and
with the interior of
are segments of lengths
, and
, respectively. Find the perimeter of the triangle whose sides lie on lines
, and
.
Diagram
~MRENTHUSIASM
Solution 1
Let the points of intersection of with
divide the sides into consecutive segments
. Furthermore, let the desired triangle be
, with
closest to side
,
closest to side
, and
closest to side
. Hence, the desired perimeter is
since
,
, and
.
Note that , so using similar triangle ratios, we find that
,
,
, and
.
We also notice that and
. Using similar triangles, we get that
Hence, the desired perimeter is
-ktong
Solution 2
Let the diagram be set up like that in Solution 1.
By similar triangles we have
Thus
Since and
, the altitude of
from
is half the altitude of
from
, say
. Also since
, the distance from
to
is
. Therefore the altitude of
from
is
.
By triangle scaling, the perimeter of is
of that of
, or
~ Nafer
Solution 3
Notation shown on diagram. By similar triangles we have
So,
vladimir.shelomovskii@gmail.com, vvsss
Solution 4 (Way too short, just keep track of which side is which)
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.
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.