2021 Fall AMC 10A Problems/Problem 19

$19$ . A disk of radius $1$ rolls all the way around in 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 $a + \dfrac{b\pi}{c}$ , 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 1

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 $4$ pieces is $(1+1)^2-\pi\cdot1^2=4-\pi$ . As a result, $A=s^2-(s-4)^2-(4-\pi)=8s-20+\pi$ .

Now, we consider the second disk. The part it sweeps is comprised of $4$ quarter circles with radius $2$ and $4$ rectangles with a side lengths of $2$ and $s$ . When we add it all together, $2A=8s+4\pi\implies A=4s+2\pi$ . $8s-20+\pi=4s+2\pi$ so $s=5+\frac{\pi}{4}$ . Finally, $5+1+4=\boxed{\textbf{(A) } 10}$ .

Solution 2

The area of the region covered by the first disk is $\begin{align*} A & = s^2 - \left( s - 4 \right)^2 - \left( 2^2 - \pi 1^2 \right) \\ & = 8 s - 20 + \pi . \end{align*}$

The area of the region covered by the second disk is $\begin{align*} 2 A & = 4 \cdot 2 s + \pi 2^2 \\ & = 8 s + 4\pi . \end{align*}$

These two equations jointly imply $s = 5 + \frac{1 \cdot \pi}{4}$ .

Therefore, the answer is $\boxed{\textbf{(A) }10}$ .

~Steven Chen (www.professorchenedu.com)

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

Solution 1

Solution 2

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