Difference between revisions of "2020 AMC 8 Problems/Problem 25"
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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>. | 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>. | ||
| − | ~ | + | ~MrBOSSY |
| − | |||
==Video Solution(🚀Just 1 min🚀)== | ==Video Solution(🚀Just 1 min🚀)== | ||
Revision as of 22:36, 27 November 2023
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (faster version of Solution 1)
- 5 Solution 4
- 6 Video Solution(🚀Just 1 min🚀)
- 7 Video Solution
- 8 Video Solution by OmegaLearn
- 9 Video Solution by WhyMath
- 10 Video Solution by The Learning Royal
- 11 Video Solution by Interstigation
- 12 Video Solution by STEMbreezy
- 13 See also
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?
![[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)); [/asy]](http://latex.artofproblemsolving.com/a/5/5/a55c518a1179d67520c6a03827b4ba25ca1413f2.png)
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 square be 
. We then have the system ![\[\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}\]](http://latex.artofproblemsolving.com/b/5/a/b5ad6916616eee9187590637ee867994d51286aa.png)
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
Let the side length of be s, and the shorter side length of and be 
. 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]](http://latex.artofproblemsolving.com/0/7/a/07a5439c40e473c7c550108923c205e886027cfd.png)
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 
.
~MrBOSSY
Video Solution(🚀Just 1 min🚀)
~Education, the Study of Everything
Video Solution
Please like and subscribe!
https://www.youtube.com/watch?v=gJXMZq2Rbwg ~David
Video Solution by OmegaLearn
https://youtu.be/jhJifWaoUI8?t=441
~ pi_is_3.14
Video Solution by WhyMath
~savannahsolver
Video Solution by The Learning Royal
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=1639
~Interstigation
Video Solution by STEMbreezy
https://youtu.be/wq8EUCe5oQU?t=588
~STEMbreezy
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. 
