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

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<math>\triangle ABC</math> has half the area of the square.
 
<math>\triangle ABC</math> has half the area of the square.
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<math>\triangle FEC</math> has base equal to half the square side length, and by AA Similarity with <math>\triangle FBA</math>, it has <math>\frac{1}{1+2}= \frac{1}{3}</math> the height, so has <math>\dfrac1{12}</math>th area of square(1/2*2*3). Thus, the area of the quadrilateral is <math>1-\frac{1}{2}-\frac{1}{12}=\frac{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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<math>\triangle FEC</math> has base equal to half the square side length, and by AA Similarity with <math>\triangle FBA</math>, it has <math>\frac{1}{1+2}= \frac{1}{3}</math> the height, so has <math>\dfrac1{12}</math>th area of square(<math>\dfrac1{2}</math>*<math>\dfrac1{2}</math>*<math>\dfrac1{3}</math>). Thus, the area of the quadrilateral is <math>1-\frac{1}{2}-\frac{1}{12}=\frac{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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~minor edit by abirgh
  
 
==Solution 3==
 
==Solution 3==
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==Solution 5==
 
==Solution 5==
  
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Since <math>\triangleECF is similar to </math>\triangleBAF and the the base of <math>\triangleECF is half that of </math>\triangleBaF's, we conclude that the area of <math>\triangleBAF is four times that of </math>\triangleECF. Using this fact, we make an equation: <math>\triangleADC + </math>\triangleBEC - <math>\triangleECF + 4 * </math>\triangleECF = area of ABCD.
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Since <math>\triangle BEC  is similar to </math>\triangle BAF  and the the base of <math>\triangle ECF is half that of </math>\triangle BaF's, we conclude that the area of <math>\triangle BAF is four times that of </math>\triangle ECF. Using this fact, we make an equation: <math>\triangle ADC + </math>\triangle BEC - <math>\triangle ECF + 4 * </math>\triangle ECF = area of ABCD.
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We can turn this equation into 3 * <math>\triangleECF=1/4 ABCD, so area of </math>\triangleECF is 1/12 of ABCD.
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We can turn this equation into 3 * <math>\triangle ECF=1/4 ABCD, so area of </math>\triangle ECF is 1/12 of ABCD.
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Since <math>\triangleECF is 1/12 of the area of ABCD and combined with ADFE which has area 45 is half of the area of ABCD, 45=5/12 ABCD.
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Since <math>\triangle ECF is 1/12 of the area of ABCD and combined with ADFE which has area 45 is half of the area of ABCD, 45=5/12 ABCD.
 
Solving this gives an area of \boxed{108}</math>.
 
Solving this gives an area of \boxed{108}</math>.
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can someone help me format this?
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-Sixth Grader
  
 
==Video Solution by OmegaLearn==
 
==Video Solution by OmegaLearn==

Latest revision as of 20:20, 6 January 2025

Problem 22

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

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, we can see that $\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 2

$\triangle ABC$ has half the area of the square. $\triangle FEC$ has base equal to half the square side length, and by AA Similarity with $\triangle FBA$, it has $\frac{1}{1+2}= \frac{1}{3}$ the height, so has $\dfrac1{12}$th area of square($\dfrac1{2}$*$\dfrac1{2}$*$\dfrac1{3}$). Thus, the area of the quadrilateral is $1-\frac{1}{2}-\frac{1}{12}=\frac{5}{12}$ th the area of the square. The area of the square is then $45\cdot\dfrac{12}{5}=\boxed{\textbf{(B) } 108}$.

~minor edit by abirgh

Solution 3

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 4

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 $2(45+9)=\boxed{108}$.

Note: Please do not learn Barycentric Coordinates for the AMC 8.


Solution 5

Since $\triangle BEC is similar to$\triangle BAF and the the base of $\triangle ECF is half that of$\triangle BaF's, we conclude that the area of $\triangle BAF is four times that of$\triangle ECF. Using this fact, we make an equation: $\triangle ADC +$\triangle BEC - $\triangle ECF + 4 *$\triangle ECF = area of ABCD. We can turn this equation into 3 * $\triangle ECF=1/4 ABCD, so area of$\triangle ECF is 1/12 of ABCD. Since $\triangle ECF is 1/12 of the area of ABCD and combined with ADFE which has area 45 is half of the area of ABCD, 45=5/12 ABCD. Solving this gives an area of \boxed{108}$.

can someone help me format this?

-Sixth Grader

Video Solution by OmegaLearn

https://youtu.be/FDgcLW4frg8?t=4038

- pi_is_3.14

Video Solutions

https://youtu.be/c4_-h7DsZFg

- Happytwin

https://youtu.be/EJ-eFP3KHWg

~savannahsolver

Video Solution only problem 22's by SpreadTheMathLove

https://www.youtube.com/watch?v=sOF1Okc0jMc

See Also

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 AJHSME/AMC 8 Problems and Solutions

Set s to be the bottom left triangle. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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