2018 AMC 8 Problems/Problem 20: Difference between revisions

Revision as of 15:24, 1 October 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 $\boxed {A}.$

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 $o0i9u8y7t6r5e4sdrftgyhujik,l.;koiju87y6

by poooooopoooooopopoooopopop

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. Anbe $4.5z-(0.5z+2z)=4.5z-2.5z=2z$ . Parallelogram $CDEF$ to $\triangle{ABC}= \frac{2z}{4.5z}=\frac{2}{4.5}=9iuytrewdrfghnjkmjhuytrdefgbhn \frac{4}{9}}$ (Error compiling LaTeX. Unknown error_msg).

rfedswedfvgb by theultimatepooooooper

@@ Line 31: / Line 31: @@
 ==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. Anbe <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. Anbe <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}=9iuytrewdrfghnjkmjhuytrdefgbhn \frac{4}{9}}</math>.
 rfedswedfvgb   by theultimatepooooooper

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

Page

Toolbox

Search