Difference between revisions of "2020 AMC 8 Problems/Problem 25"
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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> | ||
+ | |||
+ | ==Solutions== | ||
+ | |||
+ | ===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>. 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)=== | ||
+ | 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>. | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/LebVAuPkpcg | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/JAZXFv1fFGo | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC8 box|year=2020|num-b=24|after=Last Problem}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:25, 26 February 2021
Contents
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?
Solutions
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 . We then have the system which we solve to determine .
Solution 3 (faster version of solution 1)
Since, for each pair of rectangles, the side lengths have a sum of or and a difference of , the answer must be .
Video Solution by WhyMath
~savannahsolver
Video Solution
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.