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 1

By similar triangles, we have $[ADE] = \frac{1}{9}[ABC]$ . Similarly, we see that $[BEF] = \frac{4}{9}[ABC].$ Using this information, we get $[ACFE] = \frac{5}{9}[ABC].$ Then, since $[ADE] = \frac{1}{9}[ABC]$ , it follows that the $[CDEF] = \frac{4}{9}[ABC]$ . Thus, the answer would be $\boxed{\textbf{(A) } \frac{4}{9}}$ .

Sidenote: $[ABC]$ denotes the area of triangle $ABC$ . Similarly, $[ABCD]$ denotes the area of figure $ABCD$ .

Solution 2

Let $a = AE$ and $b =$ the height of $\triangle ABC$ . We can extend $\triangle ABC$ To form a parallelogram, which would equal $3a \cdot 3b$ . The smaller parallelogram is $a$ times $2b$ . The smaller parallelogram is $\frac{2}{9}$ of the larger parallelogram, so the answer would be $\frac{2}{9} \cdot 2$ , since the triangle is $\frac{1}{2}$ of the parallelogram, so the answer is $\boxed{(\textbf{A}) \frac{4}{9}}$ .

By babyzombievillager with credits to many others who helped with the solution :D

Solution 3

$\triangle{ADE} \sim \triangle{ABC} \sim \triangle{EFB}$ . We can substitute $\overline{DA}$ as $\frac{1}{3}x$ and $\overline{CD}$ as $\frac{2}{3}x$ , where $x$ is $\overline{AC}$ . Side $\overline{CB}$ having, distance $y$ , has $2$ parts also. And $\overline{CF}$ and $\overline{FB}$ are $\frac{1}{3}y$ and $\frac{2}{3}y$ respectfully. You can consider the height of $\triangle{ADE}$ and $\triangle{EFB}$ as $z$ and $2z$ respectfully. The area of $\triangle{ADE}$ is $\frac{1\cdot z}{2}=0.5z$ because the area formula for a triangle is $\frac{1}{2}bh$ or $\frac{bh}{2}$ . The area of $\triangle{EFB}$ will be $\frac{2\cdot 2z}{2}=2z$ . So the area of $\triangle{ABC}$ will be $\frac{3\cdot (2z+z)}{2}=\frac{3\cdot 3z}{2}=\frac{9z}{2}=4.5z$ . The area of parallelogram $CDEF$ will be $4.5z-(0.5z+2z)=4.5z-2.5z=2z$ . Parallelogram $CDEF$ to $\triangle{ABC}= \frac{2z}{4.5z}=\frac{2}{4.5}=\frac{4}{9}$ . The answer is $\boxed{(\textbf{A}) \frac{4}{9}}$

Solution 4 (Non-math solution)

If you have little time to calculate, divide DEFC into triangles that are equal to dae. Also cut triangle EFB into triangles similar to DAE. We see that there are 9 total triangles, and 4 of those are occupied by DEFC. Thus, 4/9. (although it could be wrong)

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AJHSME/AMC 8 Problems and Solutions

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