Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
(→‎Solution 1)
 
(33 intermediate revisions by 20 users not shown)
Line 1: Line 1:
−
==Problem 20==
+
==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>
  
Line 19: Line 19:
 
==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>.
+
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>.
+
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 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{(A) \frac{4}{9}}</math>.
+
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
+
==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>.
 +
 
 +
==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
  
−
==Solution 3==
+
~savannahsolver
  
−
<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{(A) \frac{4}{9}}</math>
+
==Video Solution by SpreadTheMathLove==
 +
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 [mathjax][BEF] = \dfrac{4}{9}[ABC][/mathjax]. Using this information, we get \[[ACFE] = \frac{5}{9}[ABC].\] Then, since $[ADE] = \frac{1}{9}[ABC]$, it follows that the [mathjax][CDEF] = \dfrac{4}{9}[ABC][/mathjax]. Thus, the answer would be $\boxed{\textbf{(A) } \frac{4}{9}}$.

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

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_20&oldid=217775"