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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 19"

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<math>19</math>. A disk of radius <math>1</math> rolls all the way around in 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 + \dfrac{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>
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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>?
  
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<math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14</math>
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<math>\textbf{(A)} ~10\qquad\textbf{(B)} ~11\qquad\textbf{(C)} ~12\qquad\textbf{(D)} ~13\qquad\textbf{(E)} ~14</math>
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==Solution 1==
 
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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 little triangle piece at each corner where the disk never sweeps out. The combined area of these <math>4</math> pieces is <math>(1+1)^2-\pi\cdot1^2=4-\pi</math>. As a result, <math>A=s^2-(s-4)^2-(4-\pi)=8s-20+\pi</math>.
 
  
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Now, we consider the second disk. The part it sweeps is comprised of <math>4</math> quarter circles with radius <math>2</math> and <math>4</math> rectangles with a side lengths of <math>2</math> and <math>s</math>. When we add it all together, <math>2A=8s+4\pi\implies A=4s+2\pi</math>. <math>8s-20+\pi=4s+2\pi</math> so <math>s=5+\frac{\pi}{4}</math>. Finally, <math>5+1+4=\boxed{\textbf{(A) } 10}</math>.
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==Diagram==
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(200);
 +
 
 +
real s = 5 + pi/4;
 +
path p1, p2;
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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;
 +
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);
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fill((0,0)--(s,0)--(s,s)--(0,s)--cycle,white);
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fill(p1,red);
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fill((2,2)--(s-2,2)--(s-2,s-2)--(2,s-2)--cycle,white);
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draw(Circle((2.5,s-1),1)^^Circle((-1,2.5),1),dashed);
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draw((0,0)--(s,0)--(s,s)--(0,s)--cycle,linewidth(1.5));
 +
</asy>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
==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>
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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>
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We equate the expressions for <math>A,</math> and then solve for <math>s:</math> <cmath>8s-20+\pi=4s+2\pi.</cmath>
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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)
  
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== Solution 2 ==
 
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The area of the region covered by the first disk is
 
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<cmath>
 
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\begin{align*}
 
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A & = s^2 - \left( s - 4 \right)^2 - \left( 2^2 - \pi 1^2 \right) \\
 
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& = 8 s - 20 + \pi .
 
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\end{align*}
 
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</cmath>
 
  
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The area of the region covered by the second disk is
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==Video Solution (Under 4 min!)==
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<cmath>
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https://youtu.be/AvgCmcEl5RE  (This solution is pretty straightforward. Just basic geometry.)
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\begin{align*}
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2 A & = 4 \cdot 2 s + \pi 2^2 \\
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<i>~Education, the Study of Everything</i>
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& = 8 s +  4\pi .
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\end{align*}
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==Video Solution by TheBeautyofMath==
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</cmath>
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https://youtu.be/w4w99JBGnYM
  
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These two equations jointly imply <math>s = 5 + \frac{1 \cdot \pi}{4}</math>.
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~IceMatrix
  
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Therefore, the answer is <math>\boxed{\textbf{(A) }10}</math>.
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==Animated Video Solutions==
 +
https://youtu.be/G57mijA4424 (vertical YouTube #Shorts for phone)
  
−
~Steven Chen (www.professorchenedu.com)
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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

Problem

A disk of radius $1$ rolls all the way around the inside of a square of side length $s>4$ and sweeps out a region of area $A$. A second disk of radius $1$ rolls all the way around the outside of the same square and sweeps out a region of area $2A$. The value of $s$ can be written as $a+\frac{b\pi}{c}$, where $a,b$, and $c$ are positive integers and $b$ and $c$ are relatively prime. What is $a+b+c$?

$\textbf{(A)} ~10\qquad\textbf{(B)} ~11\qquad\textbf{(C)} ~12\qquad\textbf{(D)} ~13\qquad\textbf{(E)} ~14$

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

Solution

The side length of the inner square traced out by the disk with radius $1$ is $s-4.$ However, there is a piece at each corner (bounded by two line segments and one $90^\circ$ arc) where the disk never sweeps out. The combined area of these four pieces is $(1+1)^2-\pi\cdot1^2=4-\pi.$ As a result, we have \[A=s^2-(s-4)^2-(4-\pi)=8s-20+\pi.\] Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius $2$ and four rectangles with side lengths of $2$ and $s.$ When we add it all together, we have $2A=8s+4\pi,$ or \[A=4s+2\pi.\] We equate the expressions for $A,$ and then solve for $s:$ \[8s-20+\pi=4s+2\pi.\] We get $s=5+\frac{\pi}{4},$ so the answer is $5+1+4=\boxed{\textbf{(A)} ~10}.$

~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

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

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

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