Difference between revisions of "2021 Fall AMC 10A Problems/Problem 19"
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==Solution== | ==Solution== | ||
− | The side length of the inner square traced out by the disk with radius <math>1</math> is <math>s-4.</math> However, there is a piece at each corner (bounded by two | + | The side length of the inner square traced out by the disk with radius <math>1</math> is <math>s-4.</math> However, there is a piece at each corner (bounded by two line segments and a <math>90^\circ</math> arc) where the disk never sweeps out. The combined area of these four pieces is <math>(1+1)^2-\pi\cdot1^2=4-\pi.</math> As a result, we have <cmath>A=s^2-(s-4)^2-(4-\pi)=8s-20+\pi.</cmath> |
Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius <math>2</math> and four rectangles with side lengths of <math>2</math> and <math>s.</math> When we add it all together, we have <math>2A=8s+4,</math> or <cmath>A=4s+2\pi.</cmath> | Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius <math>2</math> and four rectangles with side lengths of <math>2</math> and <math>s.</math> When we add it all together, we have <math>2A=8s+4,</math> or <cmath>A=4s+2\pi.</cmath> | ||
We equate the expressions for <math>A,</math> and then solve for <math>s:</math> <cmath>8s-20+\pi=4s+2\pi.</cmath> | We equate the expressions for <math>A,</math> and then solve for <math>s:</math> <cmath>8s-20+\pi=4s+2\pi.</cmath> |
Revision as of 21:25, 3 December 2021
Contents
Problem
A disk of radius rolls all the way around the inside of a square of side length
and sweeps out a region of area
. A second disk of radius
rolls all the way around the outside of the same square and sweeps out a region of area
. The value of
can be written as
, where
, and
are positive integers and
and
are relatively prime. What is
?
Diagram
~MRENTHUSIASM
Solution
The side length of the inner square traced out by the disk with radius is
However, there is a piece at each corner (bounded by two line segments and a
arc) where the disk never sweeps out. The combined area of these four pieces is
As a result, we have
Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius
and four rectangles with side lengths of
and
When we add it all together, we have
or
We equate the expressions for
and then solve for
We get
so the answer is
~MathFun1000 (Inspired by Way Tan)
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.