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 "2014 AMC 10B Problems/Problem 15"
m (→‎Solution 2)
Line 28: Line 28:
  
 
===Solution 2===
 
===Solution 2===
−
WLOG, let <math>AD = 1</math> and <math>BC = 2</math>. Furthermore, drop an an altitude from <math>F</math> to <math>CD</math>, which meets <math>CD</math> at <math>X</math>. Since <math>\angle ADC</math> is right and has been trisected, it follows that <math>\triangle ADE</math> and <math>\triangle DXF</math> are both <math>30-60-90</math> triangles. Therefore, <math>AE = \frac{\sqrt{3}}{3}</math>, and <math>DX = AF = \sqrt{3}</math>. Hence, it follows that <math>EF = \sqrt{3} - \frac{\sqrt{3}}{3}= \frac{2\sqrt{3}}{3}</math>. Since <math>\angle ADE</math> is right, the height and base of <math>\triangle DEF</math> are <math>1</math> and <math> \frac{2\sqrt{3}}{3}</math>, respectively. Thus, the area of <math>\triangle DEF</math> is <math>\frac{\sqrt{3}}{3}</math>, and the area of rectengle <math>ABCD</math> is <math>2</math>, so the ratio beween the area of <math>\triangle DEF</math> and <math>ABCD</math> is <math>\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}</math>.
+
WLOG, let <math>AD = 1</math> and <math>BC = 2</math>. Furthermore, drop an an altitude from <math>F</math> to <math>CD</math>, which meets <math>CD</math> at <math>X</math>. Since <math>\angle ADC</math> is right and has been trisected, it follows that <math>\triangle ADE</math> and <math>\triangle DXF</math> are both <math>30-60-90</math> triangles. Therefore, <math>AE = \frac{\sqrt{3}}{3}</math>, and <math>DX = AF = \sqrt{3}</math>. Hence, it follows that <math>EF = \sqrt{3} - \frac{\sqrt{3}}{3}= \frac{2\sqrt{3}}{3}</math>. Since <math>\angle ADE</math> is right, the height and base of <math>\triangle DEF</math> are <math>1</math> and <math> \frac{2\sqrt{3}}{3}</math>, respectively. Thus, the area of <math>\triangle DEF</math> is <math>\frac{\sqrt{3}}{3}</math>, and the area of rectengle <math>ABCD</math> is <math>2</math>, so the ratio beween the area of <math>\triangle DEF</math> and <math>ABCD</math> is <math>\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}</math>. Note that we are able to assume that <math>AD=1</math> and <math>BC = 2</math> because we were asked to find the ratio between two areas.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2014|ab=B|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:06, 1 January 2018

Problem

In rectangle $ABCD$, $DC = 2CB$ and points $E$ and $F$ lie on $\overline{AB}$ so that $\overline{ED}$ and $\overline{FD}$ trisect $\angle ADC$ as shown. What is the ratio of the area of $\triangle DEF$ to the area of rectangle $ABCD$?

[asy] draw((0, 0)--(0, 1)--(2, 1)--(2, 0)--cycle); draw((0, 0)--(sqrt(3)/3, 1)); draw((0, 0)--(sqrt(3), 1)); label("A", (0, 1), N); label("B", (2, 1), N); label("C", (2, 0), S); label("D", (0, 0), S); label("E", (sqrt(3)/3, 1), N); label("F", (sqrt(3), 1), N); [/asy]

$\textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$

Solution

Solution 1

Let the length of $AD$ be $x$, so that the length of $AB$ is $2x$ and $\text{[}ABCD\text{]}=2x^2$.

Because $ABCD$ is a rectangle, $\angle ADC=90^{\circ}$, and so $\angle ADE=\angle EDF=\angle FDC=30^{\circ}$. Thus $\triangle DAE$ is a $30-60-90$ right triangle; this implies that $\angle DEF=180^{\circ}-60^{\circ}=120^{\circ}$, so $\angle EFD=180^{\circ}-(120^{\circ}+30^{\circ})=30^{\circ}$. Now drop the altitude from $E$ of $\triangle DEF$, forming two $30-60-90$ triangles.

Because the length of $AD$ is $x$, from the properties of a $30-60-90$ triangle the length of $AE$ is $\frac{x\sqrt{3}}{3}$ and the length of $DE$ is thus $\frac{2x\sqrt{3}}{3}$. Thus the altitude of $\triangle DEF$ is $\frac{x\sqrt{3}}{3}$, and its base is $2x$, so its area is $\frac{1}{2}(2x)\left(\frac{x\sqrt{3}}{3}\right)=\frac{x^2\sqrt{3}}{3}$.

To finish, $\frac{\text{[}\triangle DEF\text{]}}{\text{[}ABCD\text{]}}=\frac{\frac{x^2\sqrt{3}}{3}}{2x^2}=\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}$.

Solution 2

WLOG, let $AD = 1$ and $BC = 2$. Furthermore, drop an an altitude from $F$ to $CD$, which meets $CD$ at $X$. Since $\angle ADC$ is right and has been trisected, it follows that $\triangle ADE$ and $\triangle DXF$ are both $30-60-90$ triangles. Therefore, $AE = \frac{\sqrt{3}}{3}$, and $DX = AF = \sqrt{3}$. Hence, it follows that $EF = \sqrt{3} - \frac{\sqrt{3}}{3}= \frac{2\sqrt{3}}{3}$. Since $\angle ADE$ is right, the height and base of $\triangle DEF$ are $1$ and $\frac{2\sqrt{3}}{3}$, respectively. Thus, the area of $\triangle DEF$ is $\frac{\sqrt{3}}{3}$, and the area of rectengle $ABCD$ is $2$, so the ratio beween the area of $\triangle DEF$ and $ABCD$ is $\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}$. Note that we are able to assume that $AD=1$ and $BC = 2$ because we were asked to find the ratio between two areas.

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 AMC 10 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=2014_AMC_10B_Problems/Problem_15&oldid=89327"