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

Revision as of 15:42, 30 January 2022

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 $S_2$ be $x$ , and let the length of rectangle $R_2$ 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

Assuming that the problem is well-posed, it should be true in the case where $S_1 \cong S_3$ . Let the side length of square $S_1$ be $x$ and the side length of square $S_2$ be $y$ . We then have the system $\begin{dcases}2x-y =2020 \\2x+y =3322\end{dcases}$ and we solve it to determine that $y=\boxed{\textbf{(A) }651}$ .

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 Mathiscool

https://www.youtube.com/watch?v=EKnHXWDA7rw

@@ Line 29: / Line 29: @@
 ==Solution 4==
 Assuming that the problem is well-posed, it should be true in the case where <math>S_1 \cong S_3</math>. Let the side length of square <math>S_1</math> be <math>x</math> and the side length of square <math>S_2</math> be <math>y</math>. We then have the system <cmath>\begin{dcases}2x-y =2020 \\2x+y =3322\end{dcases}</cmath> and we solve it to determine that <math>y=\boxed{\textbf{(A) }651}</math>.
-==Video Solution by Mathiscool==
-https://www.youtube.com/watch?v=EKnHXWDA7rw
 ==Video Solution by WhyMath==
@@ Line 38: / Line 35: @@
 ~savannahsolver
-==Video Solution==
+==Video Solution by The Learning Royal==
 https://youtu.be/JAZXFv1fFGo
@@ Line 45: / Line 42: @@
 ~Interstigation
+==Video Solution by Mathiscool==
+https://www.youtube.com/watch?v=EKnHXWDA7rw
+==See also==
+{{AMC8 box|year=2020|num-b=24|after=Last Problem}}
+[[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

2020 AMC 8 (Problems • Answer Key • Resources)
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