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

(Solution 3)
(Solution 3)
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==Solution 3==
 
==Solution 3==
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 - 2000}{2} = \dfrac{1322}{2} = \boxed{(\text{A}) 651}.</math>
+
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{(\text{A}) 651}.</math>
  
 
-A_MatheMagician
 
-A_MatheMagician

Revision as of 20:40, 19 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

For each square $S_{k}$, let the sidelength of this square be denoted by $s_{k}$.

As the diagram shows, $s_{1}+s_{2}+s_{3}=3322, s_{1}-s_{2}+s_{3}=2020.$ We subtract the second equation from the first, getting $2s_{2}=1302$, and thus $s_{2}=651$, so the answer is $\boxed{\textbf{(A)}\text{ }651}$ ~icematrix, edits by starrynight7210

Solution 2

WLOG, assume that $S_1=S_3$ and $R_1=R_2$. Let the sum of the lengths of $S_1$ and $S_2$ be $x$ and let the length of $S_2$ be $y$. We have the system \[x+y =3322\] \[x-y=2020\]

which we solve to find that $y=\textbf{(A) }651$.

-franzliszt

Solution 3

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

-A_MatheMagician

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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

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