Difference between revisions of "2018 AMC 8 Problems/Problem 20"

Revision as of 11:06, 18 April 2020

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

We can extend it into a parallelogram, so it would equal $3a \cdot 3b$ . The smaller parallelogram is 1 a times 2 b. 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:).

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 $\boxed{(\textbf{A}) \frac{4}{9}}$

@@ Line 25: / Line 25: @@
 ==Solution 2==
-We can extend it into a parallelogram, so it would equal <math>3a \cdot 3b</math>. The smaller parallelogram is 1 a times 2 b. The smaller parallelogram is <math>\frac{2}{9}</math> of the larger parallelogram, so the answer would be <math>\frac{2}{9} \cdot 2</math>, since the triangle is <math>\frac{1}{2}</math> of the parallelogram, so the answer is <math>\boxed{(\textbf{A}) \frac{4}{9}}</math>
+We can extend it into a parallelogram, so it would equal <math>3a \cdot 3b</math>. The smaller parallelogram is 1 a times 2 b. The smaller parallelogram is <math>\frac{2}{9}</math> of the larger parallelogram, so the answer would be <math>\frac{2}{9} \cdot 2</math>, since the triangle is <math>\frac{1}{2}</math> of the parallelogram, so the answer is <math>\boxed{(\textbf{A}) \frac{4}{9}}</math>.
+By babyzombievillager with credits to many others who helped with the solution:).
 ==Solution 3==

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