Difference between revisions of "2021 Fall AMC 10A Problems/Problem 19"
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p1 = Arc((1,1),(0,1),(1,0))--(s-1,0)--Arc((s-1,1),(s-1,0),(s,1))--(s,s-1)--Arc((s-1,s-1),(s,s-1),(s-1,s))--(1,s)--Arc((1,s-1),(1,s),(0,s-1))--cycle; | p1 = Arc((1,1),(0,1),(1,0))--(s-1,0)--Arc((s-1,1),(s-1,0),(s,1))--(s,s-1)--Arc((s-1,s-1),(s,s-1),(s-1,s))--(1,s)--Arc((1,s-1),(1,s),(0,s-1))--cycle; | ||
p2 = Arc((0,0),(-2,0),(0,-2))--(s,-2)--Arc((s,0),(s,-2),(s+2,0))--(s+2,s)--Arc((s,s),(s+2,s),(s,s+2))--(0,s+2)--Arc((0,s),(0,s+2),(-2,s))--cycle; | p2 = Arc((0,0),(-2,0),(0,-2))--(s,-2)--Arc((s,0),(s,-2),(s+2,0))--(s+2,s)--Arc((s,s),(s+2,s),(s,s+2))--(0,s+2)--Arc((0,s),(0,s+2),(-2,s))--cycle; | ||
− | + | fill(p2,green); | |
fill((0,0)--(s,0)--(s,s)--(0,s)--cycle,white); | fill((0,0)--(s,0)--(s,s)--(0,s)--cycle,white); | ||
− | + | fill(p1,red); | |
− | + | fill((2,2)--(s-2,2)--(s-2,s-2)--(2,s-2)--cycle,white); | |
− | draw(Circle((2.5,s-1),1)^^Circle((-1, | + | draw(Circle((2.5,s-1),1)^^Circle((-1,2.5),1),dashed); |
− | draw((0,0)--(s,0)--(s,s)--(0,s)--cycle,linewidth( | + | draw((0,0)--(s,0)--(s,s)--(0,s)--cycle,linewidth(1.5)); |
+ | |||
+ | Label L1 = Label("$s$", align=(0,0), position=MidPoint, filltype=Fill(3,0,green)); | ||
+ | draw((0,-0.75)--(s,-0.75), L=L1, arrow=Arrows(),bar=Bars(15)); | ||
</asy> | </asy> | ||
Line 24: | Line 27: | ||
==Solution== | ==Solution== | ||
− | The side length of the inner square traced out by the disk with radius <math>1</math> is <math>s-4.</math> However, there is a piece at each corner (bounded by two | + | The side length of the inner square traced out by the disk with radius <math>1</math> is <math>s-4.</math> However, there is a piece at each corner (bounded by two line segments and one <math>90^\circ</math> arc) where the disk never sweeps out. The combined area of these four pieces is <math>(1+1)^2-\pi\cdot1^2=4-\pi.</math> As a result, we have <cmath>A=s^2-(s-4)^2-(4-\pi)=8s-20+\pi.</cmath> |
− | Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius <math>2</math> and four rectangles with side lengths of <math>2</math> and <math>s.</math> When we add it all together, we have <math>2A=8s+4,</math> or <cmath>A=4s+2\pi.</cmath> | + | Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius <math>2</math> and four rectangles with side lengths of <math>2</math> and <math>s.</math> When we add it all together, we have <math>2A=8s+4\pi,</math> or <cmath>A=4s+2\pi.</cmath> |
We equate the expressions for <math>A,</math> and then solve for <math>s:</math> <cmath>8s-20+\pi=4s+2\pi.</cmath> | We equate the expressions for <math>A,</math> and then solve for <math>s:</math> <cmath>8s-20+\pi=4s+2\pi.</cmath> | ||
We get <math>s=5+\frac{\pi}{4},</math> so the answer is <math>5+1+4=\boxed{\textbf{(A)} ~10}.</math> | We get <math>s=5+\frac{\pi}{4},</math> so the answer is <math>5+1+4=\boxed{\textbf{(A)} ~10}.</math> | ||
~MathFun1000 (Inspired by Way Tan) | ~MathFun1000 (Inspired by Way Tan) | ||
+ | |||
+ | ==Video Solution (Under 4 min!)== | ||
+ | https://youtu.be/AvgCmcEl5RE (This solution is pretty straightforward. Just basic geometry.) | ||
+ | |||
+ | <i>~Education, the Study of Everything</i> | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/w4w99JBGnYM | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Animated Video Solutions== | ||
+ | https://youtu.be/G57mijA4424 (vertical YouTube #Shorts for phone) | ||
+ | |||
+ | https://youtu.be/d5-AluTfxuU (landscape version for desktop) | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=A|num-b=18|num-a=20}} | {{AMC10 box|year=2021 Fall|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:30, 20 October 2024
Contents
Problem
A disk of radius rolls all the way around the inside of a square of side length and sweeps out a region of area . A second disk of radius rolls all the way around the outside of the same square and sweeps out a region of area . The value of can be written as , where , and are positive integers and and are relatively prime. What is ?
Diagram
~MRENTHUSIASM
Solution
The side length of the inner square traced out by the disk with radius is However, there is a piece at each corner (bounded by two line segments and one arc) where the disk never sweeps out. The combined area of these four pieces is As a result, we have Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius and four rectangles with side lengths of and When we add it all together, we have or We equate the expressions for and then solve for We get so the answer is
~MathFun1000 (Inspired by Way Tan)
Video Solution (Under 4 min!)
https://youtu.be/AvgCmcEl5RE (This solution is pretty straightforward. Just basic geometry.)
~Education, the Study of Everything
Video Solution by TheBeautyofMath
~IceMatrix
Animated Video Solutions
https://youtu.be/G57mijA4424 (vertical YouTube #Shorts for phone)
https://youtu.be/d5-AluTfxuU (landscape version for desktop)
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.