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

(→‎Solution 2: boxed answer)
(Cleaned up the previous solution, fixed wording in some places. also added diagram.)
Line 44: Line 44:
  
 
==Solution 1==
 
==Solution 1==
−
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>.
+
Denote <math>D=(0,0)</math>. Then <math>A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)</math>. Let the intersection of <math>AF</math> and <math>DH</math> be <math>X</math>, and the intersection of <math>BF</math> and <math>CH</math> be <math>Y</math>. Then we want to find the coordinates of <math>X</math> so we can find <math>XY</math>. From our points, the slope of <math>AF</math> is <math>\bigg(\dfrac{-2}{\tfrac12}\bigg) = -4</math>, and its <math>y</math>-intercept is just <math>2</math>. Thus the equation for <math>AF</math> is <math>y = -4x + 2</math>. We can also quickly find that the equation of <math>DH</math> is <math>y = 2x</math>. Setting the equations equal, we have <math>2x = -4x +2 \implies x = \frac13</math>. Because of symmetry, we can see that the distance from <math>Y</math> to <math>BC</math> is also <math>\frac13</math>, so <math>XY = 1 - 2 \cdot \frac13 = \frac13</math>. Now the area of the kite is simply the product of the two diagonals over <math>2</math>. Since the length <math>HF = 1</math>, our answer is <math>\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>.
  
−
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>.
+
[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);
 +
draw(X--Y,dashed);
 +
 
 +
 
 +
 
 +
label("<math>A\: (0,2)</math>",A,NW);
 +
label("<math>B</math>",B,NE);
 +
label("<math>C</math>",C,SE);
 +
label("<math>D \: (0,0)</math>",D,SW);
 +
label("<math>E</math>",E,E);
 +
label("<math>F\: (\frac12,0)</math>",F,S);
 +
label("<math>G</math>",G,W);
 +
label("<math>H \: (\frac12,1)</math>",H,N);
 +
label("<math>Y</math>",Y,E);
 +
label("<math>X</math>",X,W);
 +
 
 +
 
 +
label("<math>\displaystyle\frac12</math>",(0.25,0),S);
 +
label("<math>\displaystyle\frac12</math>",(0.75,0),S);
 +
label("<math>1</math>",(1,0.5),E);
 +
label("<math>1</math>",(1,1.5),E);
 +
[/asy]
  
 
==Solution 2==
 
==Solution 2==

Revision as of 02:13, 11 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 1

Denote $D=(0,0)$. Then $A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)$. Let the intersection of $AF$ and $DH$ be $X$, and the intersection of $BF$ and $CH$ be $Y$. Then we want to find the coordinates of $X$ so we can find $XY$. From our points, the slope of $AF$ is $\bigg(\dfrac{-2}{\tfrac12}\bigg) = -4$, and its $y$-intercept is just $2$. Thus the equation for $AF$ is $y = -4x + 2$. We can also quickly find that the equation of $DH$ is $y = 2x$. Setting the equations equal, we have $2x = -4x +2 \implies x = \frac13$. Because of symmetry, we can see that the distance from $Y$ to $BC$ is also $\frac13$, so $XY = 1 - 2 \cdot \frac13 = \frac13$. Now the area of the kite is simply the product of the two diagonals over $2$. Since the length $HF = 1$, our answer is $\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}$.

[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); draw(X--Y,dashed);


label("$A\: (0,2)$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D \: (0,0)$",D,SW); label("$E$",E,E); label("$F\: (\frac12,0)$",F,S); label("$G$",G,W); label("$H \: (\frac12,1)$",H,N); label("$Y$",Y,E); label("$X$",X,W);


label("$\displaystyle\frac12$",(0.25,0),S); label("$\displaystyle\frac12$",(0.75,0),S); label("$1$",(1,0.5),E); label("$1$",(1,1.5),E); [/asy]

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 $x = \boxed{\textbf{(E)} \: \frac{1}{6}}$

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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