Difference between revisions of "2018 AMC 8 Problems/Problem 22"
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==Solution 1== | ==Solution 1== | ||
− | Let the | + | 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>. | ||
− | + | 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. | |
− | < | + | The area of the square is then <math>\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B) } 108}</math>. |
− | |||
==Solution 2== | ==Solution 2== | ||
− | |||
− | + | <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 1/(1+2)= 1/3 the height, so has <math>\dfrac1{12}</math>th area of square. 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>. | ||
+ | |||
+ | ==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 side length of the square is <math>2x</math>, making <math>3h=2x</math>. | ||
− | Now, <math>\triangle</math><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 | |
− | |||
− | |||
==Solution 4== | ==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. | ||
+ | |||
+ | ==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 | |
− | |||
=See Also= | =See Also= |
Latest revision as of 13:28, 14 October 2024
Contents
Problem 22
Point is the midpoint of side in square and meets diagonal at The area of quadrilateral is What is the area of
Solution 1
We can use analytic geometry for this problem.
Let us start by giving the coordinate , the coordinate , and so forth. and can be represented by the equations and , respectively. Solving for their intersection gives point coordinates .
Now, ’s area is simply or . This means that pentagon ’s area is of the entire square, and it follows that quadrilateral ’s area is of the square.
The area of the square is then .
Solution 2
has half the area of the square. has base equal to half the square side length, and by AA Similarity with , it has 1/(1+2)= 1/3 the height, so has th area of square. Thus, the area of the quadrilateral is th the area of the square. The area of the square is then .
Solution 3
Extend and to meet at . Drop an altitude from to and call it . Also, call . As stated before, we have , so the ratio of their heights is in a ratio, making the altitude from to . Note that this means that the side of the square is . In addition, by AA Similarity in a ratio. This means that the side length of the square is , making .
Now, note that . We have and Subtracting makes We are given that so Therefore, so our answer is
- moony_eyed
Solution 4
Solution with Cartesian and Barycentric Coordinates:
We start with the following:
Claim: Given a square , let be the midpoint of and let . Then .
Proof: We use Cartesian coordinates. Let be the origin, . We have that and are governed by the equations and , respectively. Solving, . The result follows.
Now, we apply Barycentric Coordinates w.r.t. . We let . Then .
In the barycentric coordinate system, the area formula is where is a random triangle and is the reference triangle. Using this, we find that Let so that . Then, we have , so the answer is .
Note: Please do not learn Barycentric Coordinates for the AMC 8.
Video Solution by OmegaLearn
https://youtu.be/FDgcLW4frg8?t=4038
- pi_is_3.14
Video Solutions
- Happytwin
~savannahsolver
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.