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2019 AIME II Problems/Problem 7: Difference between revisions

← Older editNewer edit →
Integralarefun (talk | contribs)
editor
149 edits
→Diagram: ; renamed and labeled points
MRENTHUSIASM (talk | contribs)
editor
9,986 edits
I decided to undo: The changes in the diagram section SHOULD NOT label more than the problem statement says. We should leave for the readers to assign points.
Line 7: Line 7:
size(350);
size(350);


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


==Solution 1==
==Solution 1==

Revision as of 21:25, 4 November 2022

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

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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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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