Difference between revisions of "2020 AMC 8 Problems/Problem 25"

Latest revision as of 21:26, 18 November 2024

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 $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ 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$",(7/2,5/2)); [/asy]$

$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$

Solution 1

Let the side length of each square $S_k$ be $s_k$ . Then, from the diagram, we can line up the top horizontal lengths of $S_1$ , $S_2$ , and $S_3$ to cover the top side of the large rectangle, so $s_{1}+s_{2}+s_{3}=3322$ . Similarly, the short side of $R_2$ will be $s_1-s_2$ , and lining this up with the left side of $S_3$ to cover the vertical side of the large rectangle gives $s_{1}-s_{2}+s_{3}=2020$ . We subtract the second equation from the first to obtain $2s_{2}=1302$ , and thus $s_{2}=\boxed{\textbf{(A) }651}$ .

Solution 2

Assuming that the problem is well-posed, it should be true in the particular case where $S_1 \cong S_3$ and $R_1 \cong R_2$ . Let the sum of the side lengths of $S_1$ and $S_3$ be $x$ , and let the length of square $S_2$ be $y$ . We then have the system $\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}$ which we solve to determine $y=\boxed{\textbf{(A) }651}$ .

Solution 3 (faster version of Solution 1)

Since, for each pair of rectangles, the side lengths have a sum of $3322$ or $2020$ and a difference of $S_2$ , the answer must be $\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{\textbf{(A) }651}$ .

Solution 4

Let the side length of $S_2$ be s, and the shorter side length of $R_1$ and $R_2$ be $r$ . 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 $r+s+r=2020$ . Also, the side length of $S_1$ and $S_3$ is $r+s$ . Then, $r+s+s+r+s=3322$ . Now, we have 2 systems of equations.

$\begin{align*}r+s+r &= 2020 \\ r+s+r+s+s &= 3322 \\ \end{align*}$

We can see an $r+s+r$ in the 2nd equation, so substituting that in gives us $2020+2s=3322 \Rightarrow 2s= 1302 \Rightarrow s=\boxed{\textbf{(A) }651}$ .

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

~Education, the Study of Everything

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

https://youtu.be/LebVAuPkpcg

~savannahsolver

Video Solution by The Learning Royal

https://youtu.be/JAZXFv1fFGo

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=1639

~Interstigation

Video Solution by STEMbreezy

https://youtu.be/wq8EUCe5oQU?t=588

~STEMbreezy

@@ Line 22: / Line 22: @@
 ==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>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>.
+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)==
@@ Line 51: / Line 51: @@
 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 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 ==

2020 AMC 8 (Problems • Answer Key • Resources)
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