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 square <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)== | ||
− | 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== | ||
− | + | 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 | |
− | ==Video Solution by | + | <asy> |
− | https://www.youtube.com/watch?v= | + | 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>. | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/UnVo6jZ3Wnk?si=ryNc6kwFiy7YkEbc&t=6127 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution(🚀Just 1 min🚀)== | ||
+ | https://youtu.be/2yiZ1Mx2P1M | ||
+ | |||
+ | ~<i>Education, the Study of Everything</i> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/wAUam5A-jcA | ||
+ | |||
+ | 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== | ==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 | ||
Line 45: | Line 85: | ||
~Interstigation | ~Interstigation | ||
+ | |||
+ | ==Video Solution by STEMbreezy== | ||
+ | https://youtu.be/wq8EUCe5oQU?t=588 | ||
+ | |||
+ | ~STEMbreezy | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC8 box|year=2020|num-b=24|after=Last Problem}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
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.