# 2019 AIME II Problems/Problem 7

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

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

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