Difference between revisions of "2020 AMC 8 Problems/Problem 25"
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==Solution 1== | ==Solution 1== | ||
− | Let the side length of each square <math>S_k</math> be <math> | + | Let the side length of each square <math>S_k</math> be <math>s_k</math>. Then, from the diagram, we can line up the top horizontal lengths of <math>S_1</math>, <math>S_2</math>, and <math>S_3</math> to cover the top side of the large rectangle, so <math>s_{1}+s_{2}+s_{3}=3322</math>. Similarly, the short side of <math>R_2</math> will be <math>s_1-s_2</math>, and lining this up with the left side of <math>S_3</math> to cover the vertical side of the large rectangle gives <math>s_{1}-s_{2}+s_{3}=2020</math>. We subtract the second equation from the first to obtain <math>2s_{2}=1302</math>, and thus <math>s_{2}=\boxed{\textbf{(A) }651}</math>. |
==Solution 2== | ==Solution 2== | ||
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==Solution 3 (faster version of Solution 1)== | ==Solution 3 (faster version of Solution 1)== | ||
− | Since, for each pair of rectangles, the side lengths have a sum of <math>3322</math> or <math>2020</math> and a difference of <math>S_2</math>, the answer must be <math>\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{\textbf{(A) }651}</math>. | + | Since, for each pair of rectangles, the side lengths have a sum of <math>3322</math> or <math>2020</math> and a difference of <math>S_2</math>, the answer must be <math>\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{\textbf{(A) }651}</math>.slayy |
==Solution 4== | ==Solution 4== | ||
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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>. | ||
− | + | ==Video Solution by Math-X (First understand the problem!!!)== | |
+ | https://youtu.be/UnVo6jZ3Wnk?si=ryNc6kwFiy7YkEbc&t=6127 | ||
+ | ~Math-X | ||
==Video Solution(🚀Just 1 min🚀)== | ==Video Solution(🚀Just 1 min🚀)== |
Latest revision as of 22:38, 24 August 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (faster version of Solution 1)
- 5 Solution 4
- 6 Video Solution by Math-X (First understand the problem!!!)
- 7 Video Solution(🚀Just 1 min🚀)
- 8 Video Solution
- 9 Video Solution by OmegaLearn
- 10 Video Solution by WhyMath
- 11 Video Solution by The Learning Royal
- 12 Video Solution by Interstigation
- 13 Video Solution by STEMbreezy
- 14 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?
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
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
.slayy
Solution 4
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
.
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=ryNc6kwFiy7YkEbc&t=6127
~Math-X
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.