2002 AMC 10P Problems/Problem 14

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 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 $JD=1-\frac{1}{2}-\frac{1}{2 \sqrt{3}}=\frac{1}{2}-\frac{1}{2 \sqrt{3}}$ 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,

$\begin{aligned} [E I D J] & = [E I J] + [J D I] \\ = \frac{1}{2} (\frac{1}{\sqrt{3}}) (\frac{1}{\sqrt{3}}) + \frac{1}{2} (\frac{1}{2} - \frac{1}{2 \sqrt{3}}) (\frac{1}{2} + \frac{1}{2 \sqrt{3}}) \\ = \frac{1}{2} (\frac{1}{3}) + \frac{\frac{1}{4} - \frac{1}{12}}{2} \\ = \frac{1}{6} + \frac{1}{12} \\ = \frac{1}{4} \end{aligned}$

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

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2002 AMC 10P Problems/Problem 14

Problem 14

Solution 1

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