# 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 $\ell_C$.

## Solution

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

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