Difference between revisions of "2018 AMC 8 Problems/Problem 22"

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==Problem 22==
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==Problem==
 
Point <math>E</math> is the midpoint of side <math>\overline{CD}</math> in square <math>ABCD,</math> and <math>\overline{BE}</math> meets diagonal <math>\overline{AC}</math> at <math>F.</math> The area of quadrilateral <math>AFED</math> is <math>45.</math> What is the area of <math>ABCD?</math>
 
Point <math>E</math> is the midpoint of side <math>\overline{CD}</math> in square <math>ABCD,</math> and <math>\overline{BE}</math> meets diagonal <math>\overline{AC}</math> at <math>F.</math> The area of quadrilateral <math>AFED</math> is <math>45.</math> What is the area of <math>ABCD?</math>
  
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The area of the square is then <math>\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}</math>.
 
The area of the square is then <math>\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}</math>.
 
==Solution 3==
 
==Solution 3==
Note that triangle <math>ABC</math> has half the area of the square and triangle <math>FEC</math> has <math>\dfrac1{12}</math>th. This the area of the quadrilateral is <math>1-1/2-1/12=5/12</math>th the area of the square. The area of the square is then <math>45\cdot\dfrac{12}{5}=\boxed{\textbf{(B)}108}</math>.
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Note that triangle <math>ABC</math> has half the area of the square and triangle <math>FEC</math> has <math>\dfrac1{12}</math>th. Thus the area of the quadrilateral is <math>1-1/2-1/12=5/12</math> th the area of the square. The area of the square is then <math>45\cdot\dfrac{12}{5}=\boxed{\textbf{(B.)}108}</math>.
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==Solution 4==
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Extend <math>\overline{AD}</math> and <math>\overline{BE}</math> to meet at <math>X</math>. Drop an altitude from <math>F</math> to <math>\overline{CE}</math> and call it <math>h</math>. Also, call <math>\overline{CE}</math> <math>x</math>. As stated before, we have <math>\triangle ABF \sim \triangle CEF</math>, so the ratio of their heights is in a <math>1:2</math> ratio, making the altitude from <math>F</math> to <math>\overline{AB}</math> <math>2h</math>. Note that this means that the side of the square is <math>3h</math>. In addition, <math>\triangle XDE \sim \triangle XAB</math> by AA Similarity in a <math>1:2</math> ratio. This means that the side length of the square is <math>2x</math>, making <math>3h=2x</math>.
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Now, note that <math>[ADEF]=[XAB]-[XDE]-[ABF]</math>. We have <math>[\triangle XAB]=(4x)(2x)/2=4x^2,</math> <math>[\triangle XDE]=(x)(2x)/2=x^2,</math> and <math>[\triangle ABF]=(2x)(2h)/2=(2x)(4x/3)/2=4x^2/3.</math> Subtracting makes <math>[ADEF]=4x^2-x^2-4x^2/3=5x^2/3.</math> We are given that <math>[ADEF]=45,</math> so <math>5x^2/3=45 \Rightarrow x^2=27.</math> Therefore, <math>x= 3 \sqrt{3},</math> so our answer is <math>(2x)^2=4x^2=4(27)=\boxed{\textbf{(B) }108}.</math> - moony_eyed
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==Solution 5==
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Solution with Cartesian and Barycentric Coordinates:
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We start with the following
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Claim: Given a square <math>ABCD</math>, let <math>E</math> be the midpoint of <math>\overline{DC}</math> and let <math>BE\cap AC = F</math>. Then <math>\frac {AF}{FC}=2</math>.
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Proof. We use Cartesian coordinates. Let <math>D</math> be the origin, <math>A=(0,1),C=(0,1),B=(1,1)</math>. We have that <math>\overline{AC}</math> and <math>\overline{EB}</math> are governed by the equations <math>y=-x+1</math> and <math>y=2x-1</math>, respectively. Solving, <math>F=\left(\frac{2}{3},\frac{1}{3}\right)</math>. The result follows. <math>\square</math>
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Now we apply Barycentric Coordinates w.r.t. <math>\triangle ACD</math>. We let <math>A=(1,0,0),D=(0,1,0),C=(0,0,1)</math>. Then <math>E=(0,\tfrac 12,\tfrac 12),F=(\tfrac 13,0,\tfrac 23)</math>.
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In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix} x_{1} &y_{1}  &z_{1} \\ x_{2} &y_{2}  &z_{2} \\  x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that<cmath>\frac{[FEC]}{[ACD]}=\begin{vmatrix} 0&0&1\\ 0&\tfrac 12&\tfrac 12\\ \tfrac 13&0&\tfrac 23 \end{vmatrix}=\frac16.</cmath> Let <math>[FEC]=x</math> so that <math>[ACD]=45+x</math>. Then we have <math>\frac{x}{x+45}=\frac 16 \Rightarrow x=9</math> so the answer is <math>2(45+9)=108</math>.
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==Video Solution==
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https://youtu.be/c4_-h7DsZFg - Happytwin
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https://youtu.be/veaDx64aX0g
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https://youtu.be/FDgcLW4frg8?t=4038 - pi_is_3.14
  
 
=See Also=
 
=See Also=

Latest revision as of 03:58, 28 January 2021

Problem

Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$

[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); label("$B$",(6,6),NE); label("$C$",(6,0),SE); label("$D$",(0,0),SW); label("$E$",(3,0),S); label("$F$",(4,2),E); [/asy]

$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$

Solution 1

Let the area of $\triangle CEF$ be $x$. Thus, the area of triangle $\triangle ACD$ is $45+x$ and the area of the square is $2(45+x) = 90+2x$.

By AA similarity, $\triangle CEF \sim \triangle ABF$ with a 1:2 ratio, so the area of triangle $\triangle ABF$ is $4x$. Now consider trapezoid $ABED$. Its area is $45+4x$, which is three-fourths the area of the square. We set up an equation in $x$:

\[45+4x = \frac{3}{4}\left(90+2x\right)\] Solving, we get $x = 9$. The area of square $ABCD$ is $90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}$.

Solution 2

We can use analytic geometry for this problem.

Let us start by giving $D$ the coordinate $(0,0)$, $A$ the coordinate $(0,1)$, and so forth. $\overline{AC}$ and $\overline{EB}$ can be represented by the equations $y=-x+1$ and $y=2x-1$, respectively. Solving for their intersection gives point $F$ coordinates $\left(\frac{2}{3},\frac{1}{3}\right)$.

Now, $\triangle$$EFC$’s area is simply $\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}$ or $\frac{1}{12}$. This means that pentagon $ABCEF$’s area is $\frac{1}{2}+\frac{1}{12}=\frac{7}{12}$ of the entire square, and it follows that quadrilateral $AFED$’s area is $\frac{5}{12}$ of the square.

The area of the square is then $\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}$.

Solution 3

Note that triangle $ABC$ has half the area of the square and triangle $FEC$ has $\dfrac1{12}$th. Thus the area of the quadrilateral is $1-1/2-1/12=5/12$ th the area of the square. The area of the square is then $45\cdot\dfrac{12}{5}=\boxed{\textbf{(B.)}108}$.

Solution 4

Extend $\overline{AD}$ and $\overline{BE}$ to meet at $X$. Drop an altitude from $F$ to $\overline{CE}$ and call it $h$. Also, call $\overline{CE}$ $x$. As stated before, we have $\triangle ABF \sim \triangle CEF$, so the ratio of their heights is in a $1:2$ ratio, making the altitude from $F$ to $\overline{AB}$ $2h$. Note that this means that the side of the square is $3h$. In addition, $\triangle XDE \sim \triangle XAB$ by AA Similarity in a $1:2$ ratio. This means that the side length of the square is $2x$, making $3h=2x$.

Now, note that $[ADEF]=[XAB]-[XDE]-[ABF]$. We have $[\triangle XAB]=(4x)(2x)/2=4x^2,$ $[\triangle XDE]=(x)(2x)/2=x^2,$ and $[\triangle ABF]=(2x)(2h)/2=(2x)(4x/3)/2=4x^2/3.$ Subtracting makes $[ADEF]=4x^2-x^2-4x^2/3=5x^2/3.$ We are given that $[ADEF]=45,$ so $5x^2/3=45 \Rightarrow x^2=27.$ Therefore, $x= 3 \sqrt{3},$ so our answer is $(2x)^2=4x^2=4(27)=\boxed{\textbf{(B) }108}.$ - moony_eyed

Solution 5

Solution with Cartesian and Barycentric Coordinates:

We start with the following

Claim: Given a square $ABCD$, let $E$ be the midpoint of $\overline{DC}$ and let $BE\cap AC = F$. Then $\frac {AF}{FC}=2$.

Proof. We use Cartesian coordinates. Let $D$ be the origin, $A=(0,1),C=(0,1),B=(1,1)$. We have that $\overline{AC}$ and $\overline{EB}$ are governed by the equations $y=-x+1$ and $y=2x-1$, respectively. Solving, $F=\left(\frac{2}{3},\frac{1}{3}\right)$. The result follows. $\square$

Now we apply Barycentric Coordinates w.r.t. $\triangle ACD$. We let $A=(1,0,0),D=(0,1,0),C=(0,0,1)$. Then $E=(0,\tfrac 12,\tfrac 12),F=(\tfrac 13,0,\tfrac 23)$.

In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1}  &z_{1} \\ x_{2} &y_{2}  &z_{2} \\   x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that\[\frac{[FEC]}{[ACD]}=\begin{vmatrix} 0&0&1\\ 0&\tfrac 12&\tfrac 12\\ \tfrac 13&0&\tfrac 23 \end{vmatrix}=\frac16.\] Let $[FEC]=x$ so that $[ACD]=45+x$. Then we have $\frac{x}{x+45}=\frac 16 \Rightarrow x=9$ so the answer is $2(45+9)=108$.

Video Solution

https://youtu.be/c4_-h7DsZFg - Happytwin

https://youtu.be/veaDx64aX0g

https://youtu.be/FDgcLW4frg8?t=4038 - pi_is_3.14

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AJHSME/AMC 8 Problems and Solutions

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