2018 AMC 8 Problems/Problem 20

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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 $\triangle ABC$, it follows that the area of region $CDEF$ is $\frac{4}{9}$$\triangle ABC$. Thus, the answer would be $\boxed {A}.$

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AJHSME/AMC 8 Problems and Solutions

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