Difference between revisions of "2019 AIME II Problems/Problem 7"

Revision as of 12:54, 3 July 2021

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

750px

~MRENTHUSIASM (by Geometry Expressions)

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

@@ Line 1: / Line 1: @@
 ==Problem==
 Triangle <math>ABC</math> has side lengths <math>AB=120,BC=220</math>, and <math>AC=180</math>. Lines <math>\ell_A,\ell_B</math>, and <math>\ell_C</math> are drawn parallel to <math>\overline{BC},\overline{AC}</math>, and <math>\overline{AB}</math>, respectively, such that the intersections of <math>\ell_A,\ell_B</math>, and <math>\ell_C</math> with the interior of <math>\triangle ABC</math> are segments of lengths <math>55,45</math>, and <math>15</math>, respectively. Find the perimeter of the triangle whose sides lie on lines <math>\ell_A,\ell_B</math>, and <math>\ell_C</math>.
+==Diagram==
+[[File:2019 AIME II Problem 7.png|center|750px]]
+~MRENTHUSIASM (by Geometry Expressions)
 ==Solution 1==

2019 AIME II (Problems • Answer Key • Resources)
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