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

(→‎Solution 2: Deleted repetitive solution. Prof. Chen agreed to this.)
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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>
 
<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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==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 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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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 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,</math> or <cmath>A=4s+2\pi.</cmath>
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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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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)

Revision as of 08:53, 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 $a+b+c$?

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

Solution

The side length of the inner square traced out by the disk with radius $1$ is $s-4.$ However, there is a little triangle piece at each corner 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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