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2014 AMC 10A Problems/Problem 16

Revision as of 22:30, 7 February 2014 by TheMaskedMagician (talk | contribs) (→‎Solution)

Problem

In rectangle $ABCD$, $AB=1$, $BC=2$, and points $E$, $F$, and $G$ are midpoints of $\overline{BC}$, $\overline{CD}$, and $\overline{AD}$, respectively. Point $H$ is the midpoint of $\overline{GE}$. What is the area of the shaded region?

[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,E); label("$F$",F,S); label("$G$",G,W); label("$H$",H,N); label("$\frac12$",(0.25,0),S); label("$\frac12$",(0.75,0),S); label("$1$",(1,0.5),E); label("$1$",(1,1.5),E); [/asy]

$\textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16$

Solution 1

Note that the region is a kite; hence its diagonals are perpendicular and it has area $\dfrac{ab}{2}$ for diagonals of length $a$ and $b$. Since $HF=1$ as both $H$ and $F$ are midpoints of parallel sides of rectangle $GECD$ and $CE=1$, we let $b=HF=1$. Now all we need to do is to find $a$.

Let the other two vertices of the kite be $I$ and $J$ with $I$ closer to $AD$ than $J$. This gives us $a=IJ$. Now let $D=(0,0)$. We thus find that the equation of $\overleftrightarrow{AF}$ is $4x+y=2$ and that of $\overleftrightarrow{DH}$ is $2x-y=0$. Solving this system gives us $x=\dfrac{1}{3}$, so the $x$-coordinate of $I$ is $\dfrac{1}{3}$; in other words, $I$ is $\dfrac{1}{3}$ from $\overline{AD}$. By symmetry, $J$ is also the same distance from $\overline{BC}$, so as $CD=1$ we have $a=IJ=1-\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{1}{3}$. Hence the area of the kite is $\dfrac{ab}{2}=\dfrac{\frac{1}{3}\cdot1}{2}=\dfrac{1}{6}\implies\boxed{\textbf{(E)}\ \dfrac{1}{6}}$.

Solution 2

Let the area of the shaded region be $x$. Let the other two vertices of the kite be $I$ and $J$ with $I$ closer to $AD$ than $J$. Note that $[ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]$. The area of $ABF$ is $1$ and the area of $DCH$ is $\dfrac{1}{2}$. We will solve for the areas of $ADI$ and $BCJ$ in terms of x by noting that the area of each triangle is the length of the perpendicular from $I$ to $AD$ and $J$ to $BC$ respectively. Because the area of $x$ = $\dfrac{1}{2}* IJ$ based on the area of a kite formula, $\dfrac{ab}{2}$ for diagonals of length $a$ and $b$, $IJ = 2x$. So each perpendicular is length $\dfrac{1-2x}{2}$. So taking our numbers and plugging them into $[ABCD] =[ABF] + [DCH] - x + [ADI] + [BCJ]$ gives us $2 = \dfrac{5}{2} - 3x$ Solving this equation for $x$ gives us $x = {\textbf{(E)}\ \frac{1}{6}}$

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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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