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 "2002 AMC 10P Problems/Problem 14"

(→‎Solution 1)
 
Line 27: Line 27:
  
 
Thus, our answer is <math>\boxed{\textbf{(A) } \frac{1}{4}}.</math>
 
Thus, our answer is <math>\boxed{\textbf{(A) } \frac{1}{4}}.</math>
 +
 +
==Solution 2==
 +
If we draw a diagram as explained by the prompt, <math>\angle EJD = 360^{\circ}-90^{\circ}-90^{\circ}-60^{\circ} = 120^{\circ}</math> as <math>\angle IEJ</math> and <math>\angle JDI</math> are both <math>90^{\circ}</math> because they are angles of the squares, and <math>\angle EID=60^{\circ}</math> given by the question. Similar to solution 1, if we draw <math>EX</math> and <math>EY</math> perpendicular to <math>AD</math> and <math>CD</math> respectively, square <math>EXDY</math> with a side length of <math>\frac{1}{2}</math> will be formed. Looking at <math>\triangle JXE</math>, we will notice <math>\angle EJX = 180^{\circ}-120^{\circ}=60^{\circ}</math>. Therefore <math>\triangle EYI</math> and <math>\triangle EXJ</math> are congruent by <math>A.A.S.</math> as they both have a <math>90^{\circ}</math> and a <math>60^{\circ}</math> angle and <math>EX=EY</math>. Thus, the area of quadrilateral <math>EIDJ</math> is the same as the area of square <math>EXDY</math>, which equals <math>\frac{1}{2} \times \frac{1}{2} = \boxed{\textbf{(A) } \frac{1}{4}}.</math>
  
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2002|ab=P|num-b=13|num-a=15}}
 
{{AMC10 box|year=2002|ab=P|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:16, 18 July 2024

Problem 14

The vertex $E$ of a square $EFGH$ is at the center of square $ABCD.$ The length of a side of $ABCD$ is $1$ and the length of a side of $EFGH$ is $2.$ Side $EF$ intersects $CD$ at $I$ and $EH$ intersects $AD$ at $J.$ If angle $EID=60^{\circ},$ the area of quadrilateral $EIDJ$ is

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

Solution 1

Draw a diagram. Split quadrilateral $EIDJ$ into $\triangle EIJ$ and $\triangle JDI.$ Let the perpendicular from point $E$ intersect $AD$ at $X$, and let the perpendicular from point $E$ intersect $CD$ at $Y.$ We know $\angle EJD=120^{\circ}$ because $\angle JDC=90^{\circ}$ since $ABCD$ is a square, $\angle DCE=60^{\circ}$ as given, and $\angle CEJ = 90^{\circ},$ so $\angle EJD = 360^{\circ}-120^{\circ}-90^{\circ}-90^{\circ}-60^{\circ}=120^{\circ}.$ Since $E$ is at the center of square $ABCD$, $EX=EY=\frac{1}{2}.$ By $30^{\circ}-60^{\circ}-90^{\circ},$ $ED=\frac{EX}{\sqrt{3}}=EC=\frac{EY}{\sqrt{3}}=\frac{1}{3}.$ Additionally, we know $JD=AD-AX-XJ,$ so $JD=1-\frac{1}{2}-\frac{1}{2 \sqrt{3}}=\frac{1}{2}-\frac{1}{2 \sqrt{3}}$ and we know $ID=DY+IY,$ so $ID=\frac{1}{2}+\frac{1}{2 \sqrt{3}}.$ From here, we can sum the areas of $\triangle EIJ$ and $\triangle JDI.$ to get the area of quadrilateral $EIDJ.$ Therefore,

\begin{align*} [EIDJ]&=[EIJ]+[JDI] \\ &=\frac{1}{2}(\frac{1}{\sqrt{3}})(\frac{1}{\sqrt{3}}) + \frac{1}{2} (\frac{1}{2}-\frac{1}{2 \sqrt{3}}) (\frac{1}{2}+\frac{1}{2 \sqrt{3}}) \\ &=\frac{1}{2}(\frac{1}{3})+\frac{1}{2}(\frac{1}{4}-\frac{1}{12}) \\ &=\frac{1}{6}+\frac{1}{12} \\ &=\frac{1}{4} \\ \end{align*}

Thus, our answer is $\boxed{\textbf{(A) } \frac{1}{4}}.$

Solution 2

If we draw a diagram as explained by the prompt, $\angle EJD = 360^{\circ}-90^{\circ}-90^{\circ}-60^{\circ} = 120^{\circ}$ as $\angle IEJ$ and $\angle JDI$ are both $90^{\circ}$ because they are angles of the squares, and $\angle EID=60^{\circ}$ given by the question. Similar to solution 1, if we draw $EX$ and $EY$ perpendicular to $AD$ and $CD$ respectively, square $EXDY$ with a side length of $\frac{1}{2}$ will be formed. Looking at $\triangle JXE$, we will notice $\angle EJX = 180^{\circ}-120^{\circ}=60^{\circ}$. Therefore $\triangle EYI$ and $\triangle EXJ$ are congruent by $A.A.S.$ as they both have a $90^{\circ}$ and a $60^{\circ}$ angle and $EX=EY$. Thus, the area of quadrilateral $EIDJ$ is the same as the area of square $EXDY$, which equals $\frac{1}{2} \times \frac{1}{2} = \boxed{\textbf{(A) } \frac{1}{4}}.$

See also

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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=2002_AMC_10P_Problems/Problem_14&oldid=223051"