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"

m (→‎Solution)
m (→‎Solution)
Line 20: Line 20:
  
 
Looking at this diagram, we notice some similar triangles. <math>\angle{ACB}=\angle{ADE}</math> since <math>\overline{DE} \parallel \overline{BC}</math>. Since<math>\triangle{ABC}</math> and triangle <math>\triangle{AED}</math> share <math>\angle{A}</math>, <math>\triangle{ABC}\sim \triangle{AED}</math> by AA. Using similar logic, <math>\triangle{ABC}\sim \triangle{EBF}</math>.  
 
Looking at this diagram, we notice some similar triangles. <math>\angle{ACB}=\angle{ADE}</math> since <math>\overline{DE} \parallel \overline{BC}</math>. Since<math>\triangle{ABC}</math> and triangle <math>\triangle{AED}</math> share <math>\angle{A}</math>, <math>\triangle{ABC}\sim \triangle{AED}</math> by AA. Using similar logic, <math>\triangle{ABC}\sim \triangle{EBF}</math>.  
−
The ratio of the areas of two similar triangles is equivalent to the square of the ratio of the lenths, so the area of <math>\triangle{AED}</math> is <math>\frac{1}{9}</math> times the area of <math>\triangle{ABC}</math> and <math>\triangle{EBF}</math> is <math>\frac{4}{9}</math> times the area of <math>\triangle{ABC}</math>. This means that the area of quadrilateral <math>CDEF</math> is is <math>1-(\frac{1}{9}+\frac{4}{9})=\frac{4}{9}</math> times the area of <math>\triangle{ABC}</math>, so our answer is <math>\boxed{\textbf{(A) }\frac 49}</math>
+
The ratio of the areas of two similar triangles is equivalent to the square of the ratio of the lenths, so <math>[AED]</math> is <math>\frac{1}{9}</math> the <math>[ABC]</math> and <math>[EBF]</math> is <math>\frac{4}{9}</math> <math>[ABC]</math>. This means that the area of quadrilateral <math>CDEF</math> is is <math>1-(\frac{1}{9}+\frac{4}{9})=\frac{4}{9}</math> <math>[ABC]</math>, so our answer is <math>\boxed{\textbf{(A) }\frac 49}</math>
  
 
==See Also==
 
==See Also==

Revision as of 12:17, 24 November 2018

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

Looking at this diagram, we notice some similar triangles. $\angle{ACB}=\angle{ADE}$ since $\overline{DE} \parallel \overline{BC}$. Since$\triangle{ABC}$ and triangle $\triangle{AED}$ share $\angle{A}$, $\triangle{ABC}\sim \triangle{AED}$ by AA. Using similar logic, $\triangle{ABC}\sim \triangle{EBF}$. The ratio of the areas of two similar triangles is equivalent to the square of the ratio of the lenths, so $[AED]$ is $\frac{1}{9}$ the $[ABC]$ and $[EBF]$ is $\frac{4}{9}$ $[ABC]$. This means that the area of quadrilateral $CDEF$ is is $1-(\frac{1}{9}+\frac{4}{9})=\frac{4}{9}$ $[ABC]$, so our answer is $\boxed{\textbf{(A) }\frac 49}$

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=99026"