2018 AMC 8 Problems/Problem 20

Problem 20

In $\triangle ABC,$ a point $E$ is on $\overline{AB}$ with $AE=1$ and $EB=2.$ Point $D$ is on $\overline{AC}$ so that $\overline{DE} \parallel \overline{BC}$ and point $F$ is on $\overline{BC}$ so that $\overline{EF} \parallel \overline{AC}.$ What is the ratio of the area of $CDEF$ to the area of $\triangle ABC?$

$[asy] size(7cm); pair A,B,C,DD,EE,FF; A = (0,0); B = (3,0); C = (0.5,2.5); EE = (1,0); DD = intersectionpoint(A--C,EE--EE+(C-B)); FF = intersectionpoint(B--C,EE--EE+(C-A)); draw(A--B--C--A--DD--EE--FF,black+1bp); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",DD,W); label("$E$",EE,S); label("$F$",FF,NE); label("$1$",(A+EE)/2,S); label("$2$",(EE+B)/2,S); [/asy]$

$\textbf{(A) } \frac{4}{9} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{5}{9} \qquad \textbf{(D) } \frac{3}{5} \qquad \textbf{(E) } \frac{2}{3}$

Solution

The $\triangle ADE$ 's area is $\frac{1}{9}$ of $\triangle ABC$ (by similar triangles). We can also get that $\triangle BEF$ is $\frac{4}{9}$ of $\triangle ABC$ (also by similar triangles). Using this information, we can get that the area of region $ACFE$ is $\frac{5}{9}$ of $\triangle ABC$ (because the area of $\triangle BEF$ is $\frac{4}{9}$ of the area of $\triangle ABC$ ). Then, because $\triangle ADE$ is $\frac{1}{9}$ of $\triangle ABC$ , we can get that the area of region $CDEF$ is $\frac{4}{9}=\boxed{\textbf{(B) }\frac{4}{9}}$ .

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2018 AMC 8 Problems/Problem 20

Problem 20

Solution

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