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

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==Solution 3==
 
==Solution 3==
  
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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{(A) \frac{4}{9}}</math>
+
<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>
  
 
==See Also==
 
==See Also==

Revision as of 13:10, 30 December 2019

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 1

By similar triangles, we have $ADE = \frac{1}{9}ABC$. Similarly, we see that $BEF = \frac{4}{9}ABC.$ Using this information, we get \[ACFE = \frac{5}{9}ABC.\] Then, since $ADE = \frac{1}{9}ABC$, it follows that the $CDEF = \frac{4}{9}ABC$. 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

We can extend it into a parallelogram, so it would equal $3a \cdot 3b$. The smaller parallelogram is 1 a times 2 b. 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 $\boxed{(\text{A}) \frac{4}{9}}$

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 $\boxed{(\textbf{A}) \frac{4}{9}}$

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

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