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Difference between revisions of "2014 AMC 10A Problems/Problem 16"

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Note that the region is a kite; hence its diagonals are perpendicular and it has area <math>\dfrac{ab}{2}</math> for diagonals of length <math>a</math> and <math>b</math>. Since <math>HF=1</math> as both <math>H</math> and <math>F</math> are midpoints of parallel sides of rectangle <math>GECD</math> and <math>CE=1</math>, we let <math>b=HF=1</math>. Now all we need to do is to find <math>a</math>.
 
Note that the region is a kite; hence its diagonals are perpendicular and it has area <math>\dfrac{ab}{2}</math> for diagonals of length <math>a</math> and <math>b</math>. Since <math>HF=1</math> as both <math>H</math> and <math>F</math> are midpoints of parallel sides of rectangle <math>GECD</math> and <math>CE=1</math>, we let <math>b=HF=1</math>. Now all we need to do is to find <math>a</math>.
  
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Let the other two vertices of the kite be <math>I</math> and <math>J</math> with <math>I</math> closer to <math>AD</math> than <math>J</math>. This gives us <math>a=IJ</math>. Now let <math>D=(0,0)</math>. We thus find that the equation of <math>\overleftrightarrow{AF}</math> is <math>4x+y=2</math> and that of <math>\overleftrightarrow{DH}</math> is <math>2x-y=0</math>. Solving this system gives us <math>x=\dfrac{1}{3}</math>, so the <math>x</math>-coordinate of <math>I</math> is <math>\dfrac{1}{3}</math>; in other words, <math>I</math> is <math>\dfrac{1}{3}</math> from <math>\overline{AD}</math>. By symmetry, <math>J</math> is also the same distance from <math>\overline{BC}</math>, so as <math>CD=1</math> we have <math>a=IJ=1-\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{1}{3}</math>. Hence the area of the kite is <math>\dfrac{ab}{2}=\dfrac{\frac{1}{3}\cdot1}{2}=\dfrac{1}{6}\implies\boxed{\textbf{(C)}\ \dfrac{1}{6}}</math>.
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Let the other two vertices of the kite be <math>I</math> and <math>J</math> with <math>I</math> closer to <math>AD</math> than <math>J</math>. This gives us <math>a=IJ</math>. Now let <math>D=(0,0)</math>. We thus find that the equation of <math>\overleftrightarrow{AF}</math> is <math>4x+y=2</math> and that of <math>\overleftrightarrow{DH}</math> is <math>2x-y=0</math>. Solving this system gives us <math>x=\dfrac{1}{3}</math>, so the <math>x</math>-coordinate of <math>I</math> is <math>\dfrac{1}{3}</math>; in other words, <math>I</math> is <math>\dfrac{1}{3}</math> from <math>\overline{AD}</math>. By symmetry, <math>J</math> is also the same distance from <math>\overline{BC}</math>, so as <math>CD=1</math> we have <math>a=IJ=1-\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{1}{3}</math>. Hence the area of the kite is <math>\dfrac{ab}{2}=\dfrac{\frac{1}{3}\cdot1}{2}=\dfrac{1}{6}\implies\boxed{\textbf{(E)}\ \dfrac{1}{6}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 20:07, 7 February 2014

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

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 $\dfrac{ab}{2}=\dfrac{\frac{1}{3}\cdot1}{2}=\dfrac{1}{6}\implies\boxed{\textbf{(E)}\ \dfrac{1}{6}}$.

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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