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

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.

Page

Toolbox

Search

2002 AMC 10P Problems/Problem 14

Problem 14

Solution 1

See also