Difference between revisions of "2020 AMC 8 Problems/Problem 25"
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==Solution 2== | ==Solution 2== | ||
− | Assuming that the problem is well-posed, it should be true in the particular case where <math>S_1 \cong S_3</math> and <math>R_1 \cong R_2</math>. Let the sum of the side lengths of <math>S_1</math> and <math> | + | Assuming that the problem is well-posed, it should be true in the particular case where <math>S_1 \cong S_3</math> and <math>R_1 \cong R_2</math>. Let the sum of the side lengths of <math>S_1</math> and <math>S_3</math> be <math>x</math>, and let the length of rectangle <math>S_2</math> be <math>y</math>. We then have the system <cmath>\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}</cmath> which we solve to determine <math>y=\boxed{\textbf{(A) }651}</math>. |
==Solution 3 (faster version of Solution 1)== | ==Solution 3 (faster version of Solution 1)== | ||
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Assuming that the problem is well-posed, it should be true in the case where <math>S_1 \cong S_3</math>. Let the side length of square <math>S_1</math> be <math>x</math> and the side length of square <math>S_2</math> be <math>y</math>. We then have the system <cmath>\begin{dcases}2x-y =2020 \\2x+y =3322\end{dcases}</cmath> and we solve it to determine that <math>y=\boxed{\textbf{(A) }651}</math>. | Assuming that the problem is well-posed, it should be true in the case where <math>S_1 \cong S_3</math>. Let the side length of square <math>S_1</math> be <math>x</math> and the side length of square <math>S_2</math> be <math>y</math>. We then have the system <cmath>\begin{dcases}2x-y =2020 \\2x+y =3322\end{dcases}</cmath> and we solve it to determine that <math>y=\boxed{\textbf{(A) }651}</math>. | ||
− | ==Video Solution | + | ==Solution 5== |
− | https://www.youtube.com/watch?v= | + | Let the side length of <math>S_2</math> be s, and the shorter side length of <math>R_1</math> and <math>R_2</math> be <math>r</math>. We have |
+ | |||
+ | <asy> | ||
+ | draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); | ||
+ | draw((3,0)--(3,1)--(0,1)); | ||
+ | draw((3,1)--(3,2)--(5,2)); | ||
+ | draw((3,2)--(2,2)--(2,1)--(2,3)); | ||
+ | label("$R_1$",(3/2,1/2)); | ||
+ | label("$S_3$",(4,1)); | ||
+ | label("$S_2$",(5/2,3/2)); | ||
+ | label("$S_1$",(1,2)); | ||
+ | label("$R_2$",(7/2,5/2)); | ||
+ | label("$r$",(5.2,5/2)); | ||
+ | label("$r$",(3.2,1/2)); | ||
+ | label("$s$",(3.2,3/2)); | ||
+ | </asy> | ||
+ | @ | ||
+ | From this diagram, it is evident that <math>r+s+r=2020</math>. Also, the side length of <math>S_1</math> and <math>S_3</math> is <math>r+s</math>. Then, <math>r+s+s+r+s=3322</math>. Now, we have 2 systems of equations. | ||
+ | |||
+ | <cmath>\begin{align*}r+s+r &= 2020 \\ r+s+r+s+s &= 3322 \\ \end{align*}</cmath> | ||
+ | |||
+ | We can see an <math>r+s+r</math> in the 2nd equation, so substituting that in gives us <math>2020+2s=3322 \Rightarrow 2s= 1302 \Rightarrow s=\boxed{\textbf{(A) }651}</math> | ||
+ | |||
+ | ~MrThinker | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=a3Z7zEc7AXQ | ||
==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
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~savannahsolver | ~savannahsolver | ||
− | ==Video Solution== | + | ==Video Solution by The Learning Royal== |
https://youtu.be/JAZXFv1fFGo | https://youtu.be/JAZXFv1fFGo | ||
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~Interstigation | ~Interstigation | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC8 box|year=2020|num-b=24|after=Last Problem}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 12:29, 29 October 2022
Contents
Problem
Rectangles and and squares and shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of in units?
Solution 1
Let the side length of each square be . Then, from the diagram, we can line up the top horizontal lengths of , , and to cover the top side of the large rectangle, so . Similarly, the short side of will be , and lining this up with the left side of to cover the vertical side of the large rectangle gives . We subtract the second equation from the first to obtain , and thus .
Solution 2
Assuming that the problem is well-posed, it should be true in the particular case where and . Let the sum of the side lengths of and be , and let the length of rectangle be . We then have the system which we solve to determine .
Solution 3 (faster version of Solution 1)
Since, for each pair of rectangles, the side lengths have a sum of or and a difference of , the answer must be .
Solution 4
Assuming that the problem is well-posed, it should be true in the case where . Let the side length of square be and the side length of square be . We then have the system and we solve it to determine that .
Solution 5
Let the side length of be s, and the shorter side length of and be . We have
@ From this diagram, it is evident that . Also, the side length of and is . Then, . Now, we have 2 systems of equations.
We can see an in the 2nd equation, so substituting that in gives us
~MrThinker
Video Solution
https://www.youtube.com/watch?v=a3Z7zEc7AXQ
Video Solution by WhyMath
~savannahsolver
Video Solution by The Learning Royal
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=1639
~Interstigation
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.