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

Latest revision as of 23:30, 20 October 2024

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 $a+b+c$ ?

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

Diagram

$[asy] /* Made by MRENTHUSIASM */ size(200); real s = 5 + pi/4; path p1, p2; p1 = Arc((1,1),(0,1),(1,0))--(s-1,0)--Arc((s-1,1),(s-1,0),(s,1))--(s,s-1)--Arc((s-1,s-1),(s,s-1),(s-1,s))--(1,s)--Arc((1,s-1),(1,s),(0,s-1))--cycle; p2 = Arc((0,0),(-2,0),(0,-2))--(s,-2)--Arc((s,0),(s,-2),(s+2,0))--(s+2,s)--Arc((s,s),(s+2,s),(s,s+2))--(0,s+2)--Arc((0,s),(0,s+2),(-2,s))--cycle; fill(p2,green); fill((0,0)--(s,0)--(s,s)--(0,s)--cycle,white); fill(p1,red); fill((2,2)--(s-2,2)--(s-2,s-2)--(2,s-2)--cycle,white); draw(Circle((2.5,s-1),1)^^Circle((-1,2.5),1),dashed); draw((0,0)--(s,0)--(s,s)--(0,s)--cycle,linewidth(1.5)); Label L1 = Label("$s$", align=(0,0), position=MidPoint, filltype=Fill(3,0,green)); draw((0,-0.75)--(s,-0.75), L=L1, arrow=Arrows(),bar=Bars(15)); [/asy]$

~MRENTHUSIASM

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 line segments and one $90^\circ$ 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 $2A=8s+4\pi,$ 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)

Video Solution (Under 4 min!)

https://youtu.be/AvgCmcEl5RE (This solution is pretty straightforward. Just basic geometry.)

~Education, the Study of Everything

Video Solution by TheBeautyofMath

https://youtu.be/w4w99JBGnYM

~IceMatrix

Animated Video Solutions

https://youtu.be/G57mijA4424 (vertical YouTube #Shorts for phone)

https://youtu.be/d5-AluTfxuU (landscape version for desktop)

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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