2021 Fall AMC 10A Problems/Problem 19

Revision as of 08:45, 3 December 2021 by MRENTHUSIASM (talk | contribs) (Solution 2: Deleted repetitive solution. Prof. Chen agreed to this.)

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 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, $5+1+4=\boxed{\textbf{(A) } 10}$.

~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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