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

## 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 rectangle $S_2$ 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 $2020+2s=3322 \Rightarrow 2s= 1302 \Rightarrow s=\boxed{\textbf{(A) }651}$.

~MrThinker

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~savannahsolver

~Interstigation

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