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

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==Problem 20==
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==Problem==
 
In <math>\triangle ABC,</math> a point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=1</math> and <math>EB=2.</math> Point <math>D</math> is on <math>\overline{AC}</math> so that <math>\overline{DE} \parallel \overline{BC}</math> and point <math>F</math> is on <math>\overline{BC}</math> so that <math>\overline{EF} \parallel \overline{AC}.</math> What is the ratio of the area of <math>CDEF</math> to the area of <math>\triangle ABC?</math>
 
In <math>\triangle ABC,</math> a point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=1</math> and <math>EB=2.</math> Point <math>D</math> is on <math>\overline{AC}</math> so that <math>\overline{DE} \parallel \overline{BC}</math> and point <math>F</math> is on <math>\overline{BC}</math> so that <math>\overline{EF} \parallel \overline{AC}.</math> What is the ratio of the area of <math>CDEF</math> to the area of <math>\triangle ABC?</math>
  
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==Solution 1==
 
==Solution 1==
  
By similar triangles, we have <math>[ADE] = \frac{1}{9}[ABC]</math>. Similarly, we see that <math>[BEF] = \frac{4}{9}[ABC].</math> Using this information, we get <cmath>[ACFE] = \frac{5}{9}[ABC].</cmath> Then, since <math>[ADE] = \frac{1}{9}[ABC]</math>, it follows that the <math>[CDEF] = \frac{4}{9}[ABC]</math>. Thus, the answer would be <math>\boxed{\textbf{(A) } \frac{4}{9}}</math>.
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By similar triangles, we have <math>[ADE] = \frac{1}{9}[ABC]</math>. Similarly, we see that [mathjax][BEF] = \dfrac{4}{9}[ABC][/mathjax]. Using this information, we get <cmath>[ACFE] = \frac{5}{9}[ABC].</cmath> Then, since <math>[ADE] = \frac{1}{9}[ABC]</math>, it follows that the [mathjax][CDEF] = \dfrac{4}{9}[ABC][/mathjax]. Thus, the answer would be <math>\boxed{\textbf{(A) } \frac{4}{9}}</math>.
  
Sidenote: <math>[ABC]</math> denotes the area of triangle <math>ABC</math>. Similarly, <math>[ABCD]</math> denotes the area of figure <math>ABCD</math>.
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Sidenote: [mathjax][ABC][/mathjax] denotes the area of triangle [mathjax]ABC[/mathjax]. Similarly, [mathjax][ABCD][/mathjax] denotes the area of figure [mathjax]ABCD[/mathjax].
  
 
==Solution 2==
 
==Solution 2==
  
We can extend it into a parallelogram, so it 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
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==Solution 3==
 
==Solution 3==
  
<math>\triangle{ADE} \sim \triangle{ABC} \sim \triangle{EFB}</math>. We can substitute <math>\overline{DA}</math> as <math>\frac{1}{3}x</math> and <math>\overline{CD}</math> as <math>\frac{2}{3}x</math>, where <math>x</math> is <math>\overline{AC}</math>. Side <math>\overline{CB}</math> having, distance <math>y</math>, has <math>2</math> parts also. And <math>\overline{CF}</math> and <math>\overline{FB}</math> are <math>\frac{1}{3}y</math> and <math>\frac{2}{3}y</math> respectfully. You can consider the height of <math>\triangle{ADE}</math> and <math>\triangle{EFB}</math> as <math>z</math> and <math>2z</math> respectfully. The area of <math>\triangle{ADE}</math> is <math>\frac{1\cdot z}{2}=0.5z</math> because the area formula for a triangle is <math>\frac{1}{2}bh</math> or <math>\frac{bh}{2}</math>. The area of <math>\triangle{EFB}</math> will be <math>\frac{2\cdot 2z}{2}=2z</math>. So the area of <math>\triangle{ABC}</math> will be <math>\frac{3\cdot (2z+z)}{2}=\frac{3\cdot 3z}{2}=\frac{9z}{2}=4.5z</math>. The area of parallelogram <math>CDEF</math> will be <math>4.5z-(0.5z+2z)=4.5z-2.5z=2z</math>. Parallelogram <math>CDEF</math> to <math>\triangle{ABC}= \frac{2z}{4.5z}=\frac{2}{4.5}=\frac{4}{9}</math>. The answer is <math>\boxed{(\textbf{A}) \frac{4}{9}}</math>
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<math>\triangle{ADE} \sim \triangle{ABC} \sim \triangle{EFB}</math>. We can substitute <math>\overline{DA}</math> as <math>\frac{1}{3}x</math> and <math>\overline{CD}</math> as <math>\frac{2}{3}x</math>, where <math>x</math> is <math>\overline{AC}</math>. Side <math>\overline{CB}</math> having, distance <math>y</math>, has <math>2</math> parts also. And, <math>\overline{CF}</math> and <math>\overline{FB}</math> are <math>\frac{1}{3}y</math> and <math>\frac{2}{3}y</math> respectfully. You can consider the height of <math>\triangle{ADE}</math> and <math>\triangle{EFB}</math> as <math>z</math> and <math>2z</math> respectfully. The area of <math>\triangle{ADE}</math> is <math>\frac{1\cdot z}{2}=0.5z</math> because the area formula for a triangle is <math>\frac{1}{2}bh</math> or <math>\frac{bh}{2}</math>. The area of <math>\triangle{EFB}</math> will be <math>\frac{2\cdot 2z}{2}=2z</math>. So, the area of <math>\triangle{ABC}</math> will be <math>\frac{3\cdot (2z+z)}{2}=\frac{3\cdot 3z}{2}=\frac{9z}{2}=4.5z</math>. The area of parallelogram <math>CDEF</math> will be <math>4.5z-(0.5z+2z)=4.5z-2.5z=2z</math>. Parallelogram <math>CDEF</math> to <math>\triangle{ABC}= \frac{2z}{4.5z}=\frac{2}{4.5}=\frac{4}{9}</math>. The answer is <math>\boxed{\textbf{(A) } \frac{4}{9}}</math>.
  
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==Video Solution (CREATIVE ANALYSIS!!!)==
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https://youtu.be/ayUmpmgFi3E
  
==Solution 4 (Non-math solution) ==
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~Education, the Study of Everything
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.
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== Video Solution (Meta-Solving Technique) ==
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https://youtu.be/GmUWIXXf_uk?t=1541
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~ pi_is_3.14
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==Video Solution 2==
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https://youtu.be/V_-yIhs_Bps
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~savannahsolver
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==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s
  
 
==See Also==
 
==See Also==

Latest revision as of 00:41, 4 April 2024

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]=49[ABC]. Using this information, we get \[[ACFE] = \frac{5}{9}[ABC].\] Then, since $[ADE] = \frac{1}{9}[ABC]$, it follows that the [CDEF]=49[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 $\boxed{\textbf{(A) } \frac{4}{9}}$.

Video Solution (CREATIVE ANALYSIS!!!)

https://youtu.be/ayUmpmgFi3E

~Education, the Study of Everything

Video Solution (Meta-Solving Technique)

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

~ pi_is_3.14

Video Solution 2

https://youtu.be/V_-yIhs_Bps

~savannahsolver

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
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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