2018 AMC 8 Problems/Problem 20

Revision as of 15:46, 23 November 2018 by Xmidnightfirex (talk | contribs) (Solution)

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 $\boxed{\textbf{(A) }\frac 49}$

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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