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

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(Undo revision 138245 by A mathemagician (talk)) (Tag: Undo) |
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− | + | ==Problem== | |

+ | Rectangles <math>R_1</math> and <math>R_2,</math> and squares <math>S_1,\,S_2,\,</math> and <math>S_3,</math> shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of <math>S_2</math> 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> | ||

+ | |||

+ | <math>\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666</math> | ||

+ | |||

+ | ==Solution 1== | ||

+ | Let the side length of each square <math>S_k</math> be <math>s_k</math>. Then, from the diagram, we can line up the top horizontal lengths of <math>S_1</math>, <math>S_2</math>, and <math>S_3</math> to cover the top side of the large rectangle, so <math>s_{1}+s_{2}+s_{3}=3322</math>. Similarly, the short side of <math>R_2</math> will be <math>s_1-s_2</math>, and lining this up with the left side of <math>S_3</math> to cover the vertical side of the large rectangle gives <math>s_{1}-s_{2}+s_{3}=2020</math>. We subtract the second equation from the first to obtain <math>2s_{2}=1302</math>, and thus <math>s_{2}=\boxed{\textbf{(A) }651}</math>. | ||

+ | |||

+ | ==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_2</math> be <math>x</math>, and let the length of rectangle <math>R_2</math> be <math>y</math>. Under our assumption, we then have the system <cmath>\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}</cmath> which we solve to find that <math>y=\boxed{\textbf{(A) }651}</math>. | ||

+ | |||

+ | ==Solution 3 (fast)== | ||

+ | Since the sum of the side lengths of each pair of squares/rectangles 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{\textbf{(A) }651}</math>. |

## Revision as of 16:30, 23 November 2020

## Problem

Rectangles and and squares and shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of in units?

## Solution 1

Let the side length of each square be . Then, from the diagram, we can line up the top horizontal lengths of , , and to cover the top side of the large rectangle, so . Similarly, the short side of will be , and lining this up with the left side of to cover the vertical side of the large rectangle gives . We subtract the second equation from the first to obtain , and thus .

## Solution 2

Assuming that the problem is well-posed, it should be true in the particular case where and . Let the sum of the side lengths of and be , and let the length of rectangle be . Under our assumption, we then have the system which we solve to find that .

## Solution 3 (fast)

Since the sum of the side lengths of each pair of squares/rectangles has a sum of or , and a difference of , we see that the answer is .