2021 Fall AMC 10A Problems/Problem 19

Revision as of 08:53, 3 December 2021 by MRENTHUSIASM (talk | contribs) (Solution 1)

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 (ProblemsAnswer KeyResources)
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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