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

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==Problem==
 
 
Rectangles <math>R_1</math> and <math>R_2,</math> and squares <math>S_1,\,S_2,\,</math> and <math>S_3,</math> shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of <math>S_2</math> in units?
 
Rectangles <math>R_1</math> and <math>R_2,</math> and squares <math>S_1,\,S_2,\,</math> and <math>S_3,</math> shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of <math>S_2</math> in units?
  
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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>S_2</math> be <math>x</math>, and let the length of rectangle <math>R_2</math> be <math>y</math>. Under our assumption, we then have the system <cmath>\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}</cmath> which we solve to find that <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_2</math> be <math>x</math>, and let the length of rectangle <math>R_2</math> be <math>y</math>. We then have the system <cmath>x+y =3322</cmath> <cmath>x-y=2020</cmath> which we solve to find that <math>y=\boxed{\textbf{(A) }651}</math>.
  
 
==Solution 3 (fast)==
 
==Solution 3 (fast)==
Since the sum of the side lengths of each pair of squares/rectangles has a sum of <math>3322</math> or <math>2020</math>, and a difference of <math>S_2</math>, we see that the answer is <math>\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{\textbf{(A) }651}</math>.
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Since each pair of boxes has a sum of <math>3322</math> or <math>2020</math> and a difference of <math>S_2</math>, we see that the answer is <math>\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{\textbf{(A) }651}</math>.
  
 
/* Video Solution */  
 
/* Video Solution */  
  
 
https://www.youtube.com/watch?v=KN441ecLfKM
 
https://www.youtube.com/watch?v=KN441ecLfKM

Revision as of 15:32, 23 November 2020

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 $S_2$ be $x$, and let the length of rectangle $R_2$ be $y$. We then have the system \[x+y =3322\] \[x-y=2020\] which we solve to find that $y=\boxed{\textbf{(A) }651}$.

Solution 3 (fast)

Since each pair of boxes has a sum of $3322$ or $2020$ and a difference of $S_2$, we see that the answer is $\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{\textbf{(A) }651}$.

/* Video Solution */

https://www.youtube.com/watch?v=KN441ecLfKM

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