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

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== Solution 1==
 
== Solution 1==
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Draw a diagram. The perpendicular from point <math>E</math> intersects <math>AD</math> at <math>X</math>, while the perpendicular from point <math>E</math> intersects <math>CD</math> at <math>Y.</math> We know Since <math>E</math> is at the center of square <math>ABCD</math>, <math>EX=EY=\frac{1}{2}.</math>
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Draw a diagram. Split quadrilateral <math>EJDC</math> into <math>\triangle ECJ</math> and <math>\triangle JDC.</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}{2sqrt{3}}</math> and
  
 
== 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 07:39, 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 $EJDC$ into $\triangle ECJ$ and $\triangle JDC.$ 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 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 $JD=1-\frac{1}{2}-\frac{1}{2sqrt{3}}$ and

See also

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AMC 10 Problems and Solutions

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