Latest revision as of 12:14, 16 August 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 of the area of square( $\dfrac1{2}$ * $\dfrac1{2}$ * $\dfrac1{3}$ ). Thus, the area of the quadrilateral is $1-\frac{1}{2}-\frac{1}{12}=\frac{5}{12}$ of 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 square's side length 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

~ Minor Edits by WrenMath

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 (easier)

$\triangle ABF \sim \triangle FEC$ , and the bases $AB$ and $EC$ are in the ratio $2:1$ , so $\frac{[\triangle ABF]}{[\triangle FEC]}=\left(\frac{2}{1}\right)^2=\frac{4}{1}$ , and $[\triangle ABF]=4\cdot[\triangle FEC]$

It is obvious that $[AFED]+[\triangle FEC]=45+[\triangle FEC]=\frac{1}{2}\cdot[ABCD]$

and $[AFED]+[\triangle ABF]=45+4\cdot[\triangle FEC]=\frac{3}{4}\cdot[ABCD]$

Solving the two equations we get $[ABCD]=\boxed{\textbf{(B)}108}$

- bbidev

Solution 6 (Similar to 5)

Since we know that point $E$ is the midpoint of square with area $4x^2$ (Side length of $2x$ ) we can easily tell that $\overline{EC}$ is equal to $\overline{DE}$ . This especially helpful when we realize that $\triangle ECF\sim \triangle ABF$ . Our setup is over, it is time to solve!

We can label the area of $\triangle ECF$ as $a$ , which means that $\triangle ABF$ has an area of $4a$ (This is because the width and height doubles which means the area is four times that of $\triangle ECF$ ). Since we know that the area of quadrilateral $ADEF$ is $45$ , this means that $45+a=2x^2$ (half of the area due to diagonal).

Now we should find the area of trapezoid $ABED$ . Since the bases are $x$ and $2x$ and the height is also $2x$ , we can plug these values into the formula for the area of the trapezoid( $\tfrac{b_{1}+b_{2}}{2}\cdot h$ ), and we will get the area as $3x^2$ which is equal to $45+4a$ . We now have our two equations: $45+a=2x^2$ and $45+4a=3x^3$ .

We can now multiply the first equation by four so we can easily subtract the variable $a$ . Doing this we acquire $180+4a=8x^2$ , and when we subtract both equations, we get $135=5x^2$ . Remember, if we want to find the area, we have to look for $4x^2$ , which is $\frac{135}{5} \cdot 4$ . 135 divided by 5 is 27, and 27 times 4 is $\boxed{\textbf{(B)}108}$ .

~AM24

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

Note

Set s to be the bottom left triangle.

@@ Line 17: / Line 17: @@
 ==Solution 1==
-Let the area of <math>\triangle CEF</math> be <math>x</math>. Thus, the area of triangle <math>\triangle ACD</math> is <math>45+x</math> and the area of the square is <math>2(45+x) = 90+2x</math>.
+We can use analytic geometry for this problem.
-By AAA similarity, <math>\triangle CEF \sim \triangle ABF</math> with a 1:2 ratio, so the area of triangle <math>\triangle ABF</math> is <math>4x</math>. Now consider trapezoid <math>ABED</math>. Its area is <math>45+4x</math>, which is three-fourths the area of the square. We set up an equation in <math>x</math>:
+Let us start by giving <math>D</math> the coordinate <math>(0,0)</math>, <math>A</math> the coordinate <math>(0,1)</math>, and so forth. <math>\overline{AC}</math> and <math>\overline{EB}</math> can be represented by the equations <math>y=-x+1</math> and <math>y=2x-1</math>, respectively. Solving for their intersection gives point <math>F</math> coordinates <math>\left(\frac{2}{3},\frac{1}{3}\right)</math>.
-<cmath> 45+4x = \frac{3}{4}\left(90+2x\right) </cmath>
+Now, we can see that <math>\triangle</math><math>EFC</math>’s area is simply <math>\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}</math> or <math>\frac{1}{12}</math>. This means that pentagon <math>ABCEF</math>’s area is <math>\frac{1}{2}+\frac{1}{12}=\frac{7}{12}</math> of the entire square, and it follows that quadrilateral <math>AFED</math>’s area is <math>\frac{5}{12}</math> of the square.
-Solving, we get <math>x = 9</math>. The area of square <math>ABCD</math> is <math>90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}</math>.
-==Solution 1.5==
+The area of the square is then <math>\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B) } 108}</math>.
-We notice some similar triangles. This is good, but you want to know the ratios in order to begin using them. Say that 45 is the area of half the triangle subtracted by <math>a</math>, so then the area of the whole square would be <math>90+2a</math>.
 ==Solution 2==
-We can use analytic geometry for this problem.
-Let us start by giving <math>D</math> the coordinate <math>(0,0)</math>, <math>A</math> the coordinate <math>(0,1)</math>, and so forth. <math>\overline{AC}</math> and <math>\overline{EB}</math> can be represented by the equations <math>y=-x+1</math> and <math>y=2x-1</math>, respectively. Solving for their intersection gives point <math>F</math> coordinates <math>\left(\frac{2}{3},\frac{1}{3}\right)</math>.
+<math>\triangle ABC</math> has half the area of the square.
+<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 of the 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> of the area of the square. The area of the square is then <math>45\cdot\dfrac{12}{5}=\boxed{\textbf{(B) } 108}</math>.
+~minor edit by abirgh
+==Solution 3==
+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 square's side length is <math>2x</math>, making <math>3h=2x</math>.
+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
+~ Minor Edits by WrenMath
+==Solution 4==
+Solution with Cartesian and Barycentric Coordinates:
+We start with the following:
+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>.
+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>
+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>.
+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)=\boxed{108}</math>.
+Note: Please do not learn Barycentric Coordinates for the AMC 8.
+==Solution 5 (easier)==
+<math>\triangle ABF \sim \triangle FEC</math>, and the bases <math>AB</math> and <math>EC</math> are in the ratio <math>2:1</math>, so <math>\frac{[\triangle ABF]}{[\triangle FEC]}=\left(\frac{2}{1}\right)^2=\frac{4}{1}</math>, and <math>[\triangle ABF]=4\cdot[\triangle FEC]</math>
+It is obvious that <math>[AFED]+[\triangle FEC]=45+[\triangle FEC]=\frac{1}{2}\cdot[ABCD]</math>
+and <math>[AFED]+[\triangle ABF]=45+4\cdot[\triangle FEC]=\frac{3}{4}\cdot[ABCD]</math>
+Solving the two equations we get <math>[ABCD]=\boxed{\textbf{(B)}108}</math>
+- bbidev
+==Solution 6 (Similar to 5)==
+Since we know that point <math>E</math> is the midpoint of square with area <math>4x^2</math> (Side length of <math>2x</math>) we can easily tell that <math>\overline{EC}</math> is equal to <math>\overline{DE}</math>. This especially helpful when we realize that <math>\triangle ECF\sim \triangle ABF</math>. Our setup is over, it is time to solve!
+We can label the area of <math>\triangle ECF</math> as <math>a</math>, which means that <math>\triangle ABF</math> has an area of <math>4a</math> (This is because the width and height doubles which means the area is four times that of <math>\triangle ECF</math>). Since we know that the area of quadrilateral <math>ADEF</math> is <math>45</math>, this means that <math>45+a=2x^2</math>(half of the area due to diagonal).
+Now we should find the area of trapezoid <math>ABED</math>. Since the bases are <math>x</math> and <math>2x</math> and the height is also <math>2x</math>, we can plug these values into the formula for the area of the trapezoid(<math>\tfrac{b_{1}+b_{2}}{2}\cdot h</math>), and we will get the area as <math>3x^2</math> which is equal to <math>45+4a</math>. We now have our two equations: <math>45+a=2x^2</math> and <math>45+4a=3x^3</math>.
+We can now multiply the first equation by four so we can easily subtract the variable <math>a</math>. Doing this we acquire <math>180+4a=8x^2</math>, and when we subtract both equations, we get <math>135=5x^2</math>. Remember, if we want to find the area, we have to look for <math>4x^2</math>, which is <math>\frac{135}{5} \cdot 4</math>. 135 divided by 5 is 27, and 27 times 4 is  <math>\boxed{\textbf{(B)}108}</math>.
+~[https://artofproblemsolving.com/wiki/index.php/User:Am24 AM24]
+==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
-Now, <math>\triangle</math><math>EFC</math>’s area is simply <math>\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}</math> or <math>\frac{1}{12}</math>. This means that pentagon <math>ABCEF</math>’s area is <math>\frac{1}{2}+\frac{1}{12}=\frac{7}{12}</math> of the entire square, and it follows that quadrilateral <math>AFED</math>’s area is <math>\frac{5}{12}</math> of the square.
+== Video Solution only problem 22's by SpreadTheMathLove==
+https://www.youtube.com/watch?v=sOF1Okc0jMc
-The area of the square is then <math>\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}</math>.
+== Note ==
+Set s to be the bottom left triangle.
-==See Also==
+=See Also=
 {{AMC8 box|year=2018|num-b=21|num-a=23}}
 {{MAA Notice}}
+[[Category:Introductory Geometry Problems]]

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

Page

Toolbox

Search

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