2021 Fall AMC 10A Problems/Problem 19

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 piece at each corner (bounded by two straight lines and a $90^\circ$ 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)

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2021 Fall AMC 10A Problems/Problem 19

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