AIME 2019 II, Problem 7

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 $200+\frac{425+375}{2}+115=600+115=\fbox{715}$ -ktong

Answer: 715