Question

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 $\triangle ABC?$

$\dfrac{4}{9}$

$\dfrac{1}{2}$

$\dfrac{5}{9}$

$\dfrac{3}{5}$

$\dfrac{2}{3}$

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Written solution:

Let the area of $\triangle ABC$ be equal to $t.$ Since $DE \parallel BC$ and $FE \parallel CA ,$ we can deduce that $ADE \sim ABC\text{ and } EFB \sim ABC.$ Since $AE = \dfrac{AB}3,$ the area of $ADE$ is equal to $\left(\dfrac13\right)^2 t = \dfrac{t}{9} .$ Since $EB = \dfrac{2AB}3,$ the area of $EFB$ is equal to $\left(\dfrac23\right)^2 t = \dfrac{4}{9}t .$ Finally, to find the area of $CDEF,$ we take the area of $ABC =t$ and subtract the areas of $ADE$ and $EFB.$ This is equivalent to the expression $t- \frac{t}{9} - \frac{4t}{9} = \frac{4t}{9} .$ Therefore, the ratio of the area of $CDEF$ and $ABC$ is $\dfrac{\left(\dfrac{4t}{9}\right)}{t} = \dfrac{4}{9} .$

Thus, A is the correct answer.

2018 AMC 8 Problem 20

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Written solution:

Problem 20 in Other Years