Difference between revisions of "2021 Fall AMC 10A Problems/Problem 19"
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− | + | ==Problem== | |
+ | A disk of radius <math>1</math> rolls all the way around the inside of a square of side length <math>s>4</math> and sweeps out a region of area <math>A</math>. A second disk of radius <math>1</math> rolls all the way around the outside of the same square and sweeps out a region of area <math>2A</math>. The value of <math>s</math> can be written as <math>a+\frac{b\pi}{c}</math>, where <math>a,b</math>, and <math>c</math> are positive integers and <math>b</math> and <math>c</math> are relatively prime. What is <math>a+b+c</math>? | ||
− | <math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14</math> | + | <math>\textbf{(A)} ~10\qquad\textbf{(B)} ~11\qquad\textbf{(C)} ~12\qquad\textbf{(D)} ~13\qquad\textbf{(E)} ~14</math> |
− | |||
− | |||
− | + | ==Diagram== | |
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(200); | ||
+ | real s = 5 + pi/4; | ||
+ | path p1, p2; | ||
+ | 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; | ||
+ | fill(p2,green); | ||
+ | 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((-1,2.5),1),dashed); | ||
+ | draw((0,0)--(s,0)--(s,s)--(0,s)--cycle,linewidth(1.5)); | ||
+ | </asy> | ||
− | + | ~MRENTHUSIASM | |
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− | |||
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− | |||
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− | The | + | ==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 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\pi,</math> or <cmath>A=4s+2\pi.</cmath> |
− | 2 | + | 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> | |
− | \ | ||
− | </ | ||
− | + | ~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 17:45, 30 October 2023
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.