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

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==Solution==
 
==Solution==
  
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Looking at this diagram, we notice some similar triangles. Since <math>\overline{DE} \parallel \overline{BC}</math>, <math>\angle{ACB}=\angle{ADE}</math>. Since<math>\triangle{ABC}</math> and triangle <math>\triangle{AED}</math> share <math>\angle{A}</math>, <math>\triangle{ABC}</math> is similar to <math>\triangle{AED}</math> by the AA similarity theorem. Using similar logic we can find <math>\triangle{ABC}</math> is similar to <math>\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{A}</math>
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Looking at this diagram, we notice some similar triangles. Since <math>\overline{DE} \parallel \overline{BC}</math>, <math>\angle{ACB}=\angle{ADE}</math>. Since<math>\triangle{ABC}</math> and triangle <math>\triangle{AED}</math> share <math>\angle{A}</math>, <math>\triangle{ABC}</math> is similar to <math>\triangle{AED}</math> by the AA similarity theorem. Using similar logic we can find <math>\triangle{ABC}</math> is similar to <math>\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{(A)}</math>
  
 
{{AMC8 box|year=2018|num-b=19|num-a=21}}
 
{{AMC8 box|year=2018|num-b=19|num-a=21}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:47, 21 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. Since $\overline{DE} \parallel \overline{BC}$, $\angle{ACB}=\angle{ADE}$. Since$\triangle{ABC}$ and triangle $\triangle{AED}$ share $\angle{A}$, $\triangle{ABC}$ is similar to $\triangle{AED}$ by the AA similarity theorem. Using similar logic we can find $\triangle{ABC}$ is similar to $\triangle{EBF}$. The ratio of the areas of two similar triangles is equivalent to the square of the ratio of the lenths, so the area of $\triangle{AED}$ is $\frac{1}{9}$ times the area of $\triangle{ABC}$ and $\triangle{EBF}$ is $\frac{4}{9}$ times the area of $\triangle{ABC}$. This means that the area of quadrilateral $CDEF$ is is $1-(\frac{1}{9}+\frac{4}{9})=\frac{4}{9}$ times the area of $\triangle{ABC}$, so our answer is $\boxed{(A)}$

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