

Line 1: 
Line 1: 
−  ==Problem==
 +  yum 
−  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_1s_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 wellposed, 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 \\xy=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>.
 