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Difference between revisions of "2002 AMC 10P Problems/Problem 14"

(→‎Solution 1)
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== Solution 1==
 
== Solution 1==
−
Draw a diagram. Split quadrilateral <math>EIDJ</math> into <math>\triangle EIJ</math> and <math>\triangle JDI.</math> Let the perpendicular from point <math>E</math> intersect <math>AD</math> at <math>X</math>, and let the perpendicular from point <math>E</math> intersect <math>CD</math> at <math>Y.</math> We know <math>\angle EJD=120^{\circ}</math> because <math>\angle JDC=90^{\circ}</math> since <math>ABCD</math> is a square, <math>\angle DCE=60^{\circ}</math> as given, and <math>\angle CEJ = 90^{\circ},</math> so \angle EJD = 360^{\circ}-120^{\circ}-90^\{circ}-90^\{circ}-60^\{circ}=120\{circ}.<math> Since </math>E<math> is at the center of square </math>ABCD<math>, </math>EX=EY=\frac{1}{2}.<math> By the </math>30^{\circ}-60^{\circ}-90^{\circ}<math>, </math>ED=\frac{EX}{\sqrt{3}}=EC=\frac{EY}{\sqrt{3}}=\frac{1}{3}. Additionally, we know <math>JD=AD-AX-XJ,</math> so <math>JD=1-\frac{1}{2}-\frac{1}{2 \sqrt{3}}</math> and we know <math>ID=DY+IY,</math> so <math>ID=\frac{1}{2}+\frac{1}{2 \sqrt{3}}.</math> From here, we can sum the areas of <math>\triangle EIJ</math> and <math>\triangle JDI.</math> to get the area of quadrilateral <math>EIDJ.</math> Therefore,
+
Draw a diagram. Split quadrilateral <math>EIDJ</math> into <math>\triangle EIJ</math> and <math>\triangle JDI.</math> Let the perpendicular from point <math>E</math> intersect <math>AD</math> at <math>X</math>, and let the perpendicular from point <math>E</math> intersect <math>CD</math> at <math>Y.</math> We know <math>\angle EJD=120^{\circ}</math> because <math>\angle JDC=90^{\circ}</math> since <math>ABCD</math> is a square, <math>\angle DCE=60^{\circ}</math> as given, and <math>\angle CEJ = 90^{\circ},</math> so <math>\angle EJD = 360^{\circ}-120^{\circ}-90^\{circ}-90^\{circ}-60^\{circ}=120\{circ}.</math> Since <math>E</math> is at the center of square <math>ABCD</math>, <math>EX=EY=\frac{1}{2}.</math> By the <math>30^{\circ}-60^{\circ}-90^{\circ},</math> <math>ED=\frac{EX}{\sqrt{3}}=EC=\frac{EY}{\sqrt{3}}=\frac{1}{3}.</math> Additionally, we know <math>JD=AD-AX-XJ,</math> so <math>JD=1-\frac{1}{2}-\frac{1}{2 \sqrt{3}}</math> and we know <math>ID=DY+IY,</math> so <math>ID=\frac{1}{2}+\frac{1}{2 \sqrt{3}}.</math> From here, we can sum the areas of <math>\triangle EIJ</math> and <math>\triangle JDI.</math> to get the area of quadrilateral <math>EIDJ.</math> Therefore,
  
 
\begin{align*}
 
\begin{align*}
−
[EIDJ]&=[EIJ]+[JDI] //
+
[EIDJ]&=[EIJ]+[JDI] \\
−
&=\frac{1}{2}\frac{1}{\sqrt{3}}\frac{1}{\sqrt{3}} + \frac{1}{2} (1-\frac{1}{2}-\frac{1}{2 \sqrt{3}}) (\frac{1}{2}+\frac{1}{2 \sqrt{3}}) //
+
&=\frac{1}{2}\frac{1}{\sqrt{3}}\frac{1}{\sqrt{3}} + \frac{1}{2} (1-\frac{1}{2}-\frac{1}{2 \sqrt{3}}) (\frac{1}{2}+\frac{1}{2 \sqrt{3}}) \\
−
&=\frac{1}{2}\frac{1}{3}+\frac{\frac{1}{4}-\frac{1}{12}}{2} //
+
&=\frac{1}{2}\frac{1}{3}+\frac{\frac{1}{4}-\frac{1}{12}}{2} \\
−
&=\frac{1}{6}+\frac{1}{12} //
+
&=\frac{1}{6}+\frac{1}{12} \\
−
&=\frac{1}{4} //
+
&=\frac{1}{4} \\
 
\end{align*}
 
\end{align*}
 +
 +
Thus, our answer is \boxed{\textbf{(A) } \frac{1}{4}}.
  
 
== 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}}

Revision as of 08:50, 15 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 $\angle EJD = 360^{\circ}-120^{\circ}-90^\{circ}-90^\{circ}-60^\{circ}=120\{circ}.$ (Error compiling LaTeX. Unknown error_msg) Since $E$ is at the center of square $ABCD$, $EX=EY=\frac{1}{2}.$ By the $30^{\circ}-60^{\circ}-90^{\circ},$ $ED=\frac{EX}{\sqrt{3}}=EC=\frac{EY}{\sqrt{3}}=\frac{1}{3}.$ Additionally, we know $JD=AD-AX-XJ,$ so $JD=1-\frac{1}{2}-\frac{1}{2 \sqrt{3}}$ 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} (1-\frac{1}{2}-\frac{1}{2 \sqrt{3}}) (\frac{1}{2}+\frac{1}{2 \sqrt{3}}) \\ &=\frac{1}{2}\frac{1}{3}+\frac{\frac{1}{4}-\frac{1}{12}}{2} \\ &=\frac{1}{6}+\frac{1}{12} \\ &=\frac{1}{4} \\ \end{align*}

Thus, our answer is \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

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