2020 AMC 8 Problems/Problem 25

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, $s_{1}+s_{2}+s_{3}=3322, s_{1}-s_{2}+s_{3}=2020.$ If we subtract the second equation from the first we will get $2s_{2}=1302$ or $s_{2}=\boxed{651\textbf{(A)}}$ ~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

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.

Page

Toolbox

Search

2020 AMC 8 Problems/Problem 25

Solution 1

Solution 2

See also