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

(Cleaned up the previous solution, fixed wording in some places. also added diagram.)
m (→‎Solution 5 (Similarity))
 
(13 intermediate revisions by 11 users not shown)
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<math> \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 </math>
 
<math> \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 </math>
 +
[[Category: Introductory Geometry Problems]]
  
 
==Solution 1==
 
==Solution 1==
 
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>.
 
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>.
  
−
[asy]
+
<asy>
 
import graph;
 
import graph;
 
size(9cm);
 
size(9cm);
Line 72: Line 73:
  
  
−
label("<math>A\: (0,2)</math>",A,NW);
+
label("$A\: (0,2)$",A,NW);
−
label("<math>B</math>",B,NE);
+
label("$B$",B,NE);
−
label("<math>C</math>",C,SE);
+
label("$C$",C,SE);
−
label("<math>D \: (0,0)</math>",D,SW);
+
label("$D \: (0,0)$",D,SW);
−
label("<math>E</math>",E,E);
+
label("$E$",E,E);
−
label("<math>F\: (\frac12,0)</math>",F,S);
+
label("$F\: (\frac12,0)$",F,S);
−
label("<math>G</math>",G,W);
+
label("$G$",G,W);
−
label("<math>H \: (\frac12,1)</math>",H,N);
+
label("$H \: (\frac12,1)$",H,N);
−
label("<math>Y</math>",Y,E);
+
label("$Y$",Y,E);
−
label("<math>X</math>",X,W);
+
label("$X$",X,W);
  
  
−
label("<math>\displaystyle\frac12</math>",(0.25,0),S);
+
label("$\frac12$",(0.25,0),S);
−
label("<math>\displaystyle\frac12</math>",(0.75,0),S);
+
label("$\frac12$",(0.75,0),S);
−
label("<math>1</math>",(1,0.5),E);
+
label("$1$",(1,0.5),E);
−
label("<math>1</math>",(1,1.5),E);
+
label("$1$",(1,1.5),E);
−
[/asy]
+
</asy>
  
 
==Solution 2==
 
==Solution 2==
  
−
Let the area of the shaded region be <math>x</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>. Note that <math> [ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]</math>. The area of <math>ABF</math> is <math>1</math> and the area of <math>DCH</math> is <math>\dfrac{1}{2}</math>. We will solve for the areas of <math>ADI</math> and <math>BCJ</math> in terms of x by noting that the area of each triangle is the length of the perpendicular from <math>I</math> to <math>AD</math> and <math>J</math> to <math>BC</math> respectively. Because the area of <math>x</math> = <math>\dfrac{1}{2}* IJ</math> based on the area of a kite formula, <math>\dfrac{ab}{2}</math> for diagonals of length <math>a</math> and <math>b</math>, <math>IJ = 2x</math>. So each perpendicular is length <math>\dfrac{1-2x}{2}</math>. So taking our numbers and plugging them into  <math> [ABCD] =[ABF] + [DCH] - x + [ADI] + [BCJ]</math> gives us <math>2 = \dfrac{5}{2} - 3x</math> Solving this equation for <math>x</math> gives us <math> x = \boxed{\textbf{(E)} \: \frac{1}{6}}</math>
+
Let the area of the shaded region be <math>x</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>. Note that <math> [ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]</math>. The area of <math>ABF</math> is <math>1</math> and the area of <math>DCH</math> is <math>\dfrac{1}{2}</math>. We will solve for the areas of <math>ADI</math> and <math>BCJ</math> in terms of x by noting that the area of each triangle is the length of the perpendicular from <math>I</math> to <math>AD</math> and <math>J</math> to <math>BC</math> respectively. Because the area of <math>x</math> = <math>\dfrac{1}{2} \cdot IJ</math> based on the area of a kite formula, <math>\dfrac{ab}{2}</math> for diagonals of length <math>a</math> and <math>b</math>, <math>IJ = 2x</math>. So each perpendicular is length <math>\dfrac{1-2x}{2}</math>. So taking our numbers and plugging them into  <math> [ABCD] =[ABF] + [DCH] - x + [ADI] + [BCJ]</math> gives us <math>2 = \dfrac{5}{2} - 3x</math> Solving this equation for <math>x</math> gives us <math> x = \boxed{\textbf{(E)} \: \frac{1}{6}}</math>
 +
 
 +
==Solution 3==
 +
 
 +
From the diagram in Solution 1, let <math>e</math> be the height of <math>XHY</math> and <math>f</math> be the height of <math>XFY</math>. It is clear that their sum is <math>1</math> as they are parallel to <math>GD</math>. Let <math>k</math> be the ratio of the sides of the similar triangles <math>XFY</math> and <math>AFB</math>, which are similar because <math>XY</math> is parallel to <math>AB</math> and the triangles share angle <math>F</math>. Then <math>k = f/2</math>, as 2 is the height of <math>AFB</math>. Since <math>XHY</math> and <math>DHC</math> are similar for the same reasons as <math>XFY</math> and <math>AFB</math>, the height of <math>XHY</math> will be equal to the base, like in <math>DHC</math>, making <math>XY = e</math>. However, <math>XY</math> is also the base of <math>XFY</math>, so <math>k = e / AB</math> where <math>AB = 1</math> so <math>k = e</math>. Subbing into <math>k = f/2</math> gives a system of linear equations, <math>e + f = 1</math> and <math>e = f/2</math>. Solving yields <math>e = XY = 1/3</math> and <math>f = \frac{2}{3}</math>, and since the area of the kite is simply the product of the two diagonals over <math>2</math> and <math>HF = 1</math>, our answer is <math>\frac{\frac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>.
 +
 
 +
==Solution 4==
 +
Let the unmarked vertices of the shaded area be labeled <math>I</math> and <math>J</math>, with <math>I</math> being closer to <math>GD</math> than <math>J</math>. Noting that kite <math>HJFI</math> can be split into triangles <math>HJI</math> and <math>JIF</math>.
 +
 
 +
Lemma:  The distance from line segment <math>JI</math> to <math>H</math> is half the distance from <math>JI</math> to <math>F</math> 
 +
 
 +
Proof: Drop perpendiculars of triangles <math>HJI</math> and <math>JIF</math> to line <math>JI</math>, and let the point of intersection be <math>Q</math>. Note that <math>HJI</math> and <math>JIF</math> are similar to <math>HDC</math> and <math>ABF</math>, respectively. Now, the ratio of <math>DC</math> to <math>HF</math> is <math>1:1</math>, which shows that the ratio of <math>JI</math> to <math>HQ</math> is <math>1:1</math>, because of similar triangles as described above. Similarly, the ratio of <math>JI</math> to <math>FQ</math> is <math>1:2</math>. Since these two triangles contain the same base, <math>JI</math>, the ratio of <math>HQ:FQ = 1:2</math>.
 +
 
 +
Because kite <math>HJFI</math> is orthodiagonal, we multiply <math>\frac{1\cdot\tfrac{1}{3}}{2} = \boxed{\textbf{(E)} \: \frac{1}{6}}</math>
 +
 
 +
~Lemma proof by sakshamsethi
 +
 
 +
==Solution 5 (Similarity)==
 +
<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>
 +
 
 +
The area of the shaded area is the area of <math>\triangle DHC</math> minus the two triangles on the side.
 +
Extend <math>\overline{DH}</math> so that it hits point <math>B</math>. Call the intersection of <math>\overline{AF}</math> and <math>\overline{DB}</math> point <math>P</math>. <cmath>\triangle APB \sim \triangle FPD</cmath>
 +
Drop altitudes from <math>P</math> down to <math>\overline{DF}</math> and <math>\overline{AB}</math>; call the intersection points <math>L</math> and <math>M</math> respectively.
 +
<cmath>\frac{DF}{AB}=\frac{LP}{PM}=\frac{\frac{1}{2}}{1}=\frac{1}{2}</cmath>
 +
<cmath>LP=\frac{1}{3}\cdot LM=\frac{2}{3}</cmath>
 +
Thus the two triangles on the side have area <math>\frac{1}{2} \cdot \frac{2}{2} \cdot \frac {1}{3} = \frac{1}{6}</math>. Since there are two, their total area is <math>2 \cdot \frac{1}{6} = \frac{1}{3}</math>. The area of <math>\triangle DHC</math> is <math>\frac{1}{2} \cdot 1 \cdot 1=\frac{1}{2}</math>. The shaded region is <math>\frac{1}{2}-\frac{1}{3} = \frac{1}{6}</math> which is <math>\boxed{\textbf{(E)} \frac{1}{6}}</math>.
 +
 
 +
~JH. L :)
  
 
==See Also==
 
==See Also==
Line 98: Line 162:
 
{{AMC10 box|year=2014|ab=A|num-b=15|num-a=17}}
 
{{AMC10 box|year=2014|ab=A|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category: Introductory Geometry Problems]]

Latest revision as of 18:51, 25 July 2023

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("$\frac12$",(0.25,0),S); label("$\frac12$",(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 $x$ = $\dfrac{1}{2} \cdot IJ$ 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 $x = \boxed{\textbf{(E)} \: \frac{1}{6}}$

Solution 3

From the diagram in Solution 1, let $e$ be the height of $XHY$ and $f$ be the height of $XFY$. It is clear that their sum is $1$ as they are parallel to $GD$. Let $k$ be the ratio of the sides of the similar triangles $XFY$ and $AFB$, which are similar because $XY$ is parallel to $AB$ and the triangles share angle $F$. Then $k = f/2$, as 2 is the height of $AFB$. Since $XHY$ and $DHC$ are similar for the same reasons as $XFY$ and $AFB$, the height of $XHY$ will be equal to the base, like in $DHC$, making $XY = e$. However, $XY$ is also the base of $XFY$, so $k = e / AB$ where $AB = 1$ so $k = e$. Subbing into $k = f/2$ gives a system of linear equations, $e + f = 1$ and $e = f/2$. Solving yields $e = XY = 1/3$ and $f = \frac{2}{3}$, and since the area of the kite is simply the product of the two diagonals over $2$ and $HF = 1$, our answer is $\frac{\frac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}$.

Solution 4

Let the unmarked vertices of the shaded area be labeled $I$ and $J$, with $I$ being closer to $GD$ than $J$. Noting that kite $HJFI$ can be split into triangles $HJI$ and $JIF$.

Lemma: The distance from line segment $JI$ to $H$ is half the distance from $JI$ to $F$

Proof: Drop perpendiculars of triangles $HJI$ and $JIF$ to line $JI$, and let the point of intersection be $Q$. Note that $HJI$ and $JIF$ are similar to $HDC$ and $ABF$, respectively. Now, the ratio of $DC$ to $HF$ is $1:1$, which shows that the ratio of $JI$ to $HQ$ is $1:1$, because of similar triangles as described above. Similarly, the ratio of $JI$ to $FQ$ is $1:2$. Since these two triangles contain the same base, $JI$, the ratio of $HQ:FQ = 1:2$.

Because kite $HJFI$ is orthodiagonal, we multiply $\frac{1\cdot\tfrac{1}{3}}{2} = \boxed{\textbf{(E)} \: \frac{1}{6}}$

~Lemma proof by sakshamsethi

Solution 5 (Similarity)

[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]

The area of the shaded area is the area of $\triangle DHC$ minus the two triangles on the side. Extend $\overline{DH}$ so that it hits point $B$. Call the intersection of $\overline{AF}$ and $\overline{DB}$ point $P$. \[\triangle APB \sim \triangle FPD\] Drop altitudes from $P$ down to $\overline{DF}$ and $\overline{AB}$; call the intersection points $L$ and $M$ respectively. \[\frac{DF}{AB}=\frac{LP}{PM}=\frac{\frac{1}{2}}{1}=\frac{1}{2}\] \[LP=\frac{1}{3}\cdot LM=\frac{2}{3}\] Thus the two triangles on the side have area $\frac{1}{2} \cdot \frac{2}{2} \cdot \frac {1}{3} = \frac{1}{6}$. Since there are two, their total area is $2 \cdot \frac{1}{6} = \frac{1}{3}$. The area of $\triangle DHC$ is $\frac{1}{2} \cdot 1 \cdot 1=\frac{1}{2}$. The shaded region is $\frac{1}{2}-\frac{1}{3} = \frac{1}{6}$ which is $\boxed{\textbf{(E)} \frac{1}{6}}$.

~JH. L :)

See Also

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

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