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

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==Solution==
 
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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 piece at each corner (bounded by two line segments and a <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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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,</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>

Revision as of 20:25, 3 December 2021

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; filldraw(p2,green); fill((0,0)--(s,0)--(s,s)--(0,s)--cycle,white); filldraw(p1,yellow); filldraw((2,2)--(s-2,2)--(s-2,s-2)--(2,s-2)--cycle,white); draw(Circle((2.5,s-1),1)^^Circle((-1,3),1),dashed); draw((0,0)--(s,0)--(s,s)--(0,s)--cycle,linewidth(2.5)); [/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 $2A=8s+4,$ 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)

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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