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

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==Solution 2==
 
==Solution 2==
  
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Let <math>a = DE</math> and <math>b =</math> the height of <math>\triangle ABC</math>. We can extend <math>\triangle ABC</math> To form a parallelogram, which would equal <math>3a \cdot 3b</math>. The smaller parallelogram is <math>a</math> times <math>2b</math>. 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>.
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Let <math>a = DE</math> and <math>b =</math> the height of <math>\triangle ABC</math>. We can extend <math>\triangle ABC</math> to form a parallelogram, which would equal <math>3a \cdot 3b</math>. The smaller parallelogram is <math>a</math> times <math>2b</math>. 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 :D
 
By babyzombievillager with credits to many others who helped with the solution :D

Revision as of 15:00, 17 January 2022

Problem

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 $a = DE$ 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 by drawing lines through points D and F that are parallel to AB and a line through the middle of EF parallel to CB. 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, the answer is $\boxed{\textbf{(A) } \frac{4}{9}}$. (although it could be wrong)

Video Solution (Meta-Solving Technique)

https://youtu.be/GmUWIXXf_uk?t=1541

~ pi_is_3.14

See Also

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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