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

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

Page

Toolbox

Search

2019 AIME II Problems/Problem 7

Problem

Solution

See Also