AIME 2019 II · 第 7 题

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

Solution 4 (Way too short, just keep track of which side is which)

Let's squish a triangle with side lengths 15, 22.5, and 27.5 into a equilateral triangle with side length 1. Then, the original triangle gets turned into a equilateral triangle with side length 8. Since 15 is one eighth of 120, it has a length of one. Since 45 and 55 are one fourth of 180 and 220 respectively, they are both two long. We extend the three segments to form a big equilateral triangle shown in black. Notice it has a side length of 11. Now that our task is done, let's undo the distortion. We get 11(15+22.5+27.5)=11(65)=715.

~ Afly (talk)