Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3 (faster version of Solution 1))
(→‎Solution 2)
 
(One intermediate revision by one other user not shown)
Line 22: Line 22:
  
 
==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_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>.
+
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>. why? 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>.slayy
+
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>.
  
 
==Solution 4==
 
==Solution 4==

Latest revision as of 16:09, 2 December 2024

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 $R_1 \cong R_2$. why? 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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_25&oldid=236060"