Difference between revisions of "2021 Fall AMC 10A Problems/Problem 19"
Mathfun1000 (talk | contribs) (Adding a solution) |
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<math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14</math> | <math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14</math> | ||
+ | ==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 little triangle piece at each corner where the disk never sweeps out. The combined area of these <math>4</math> pieces is <math>(1+1)^2-\pi\cdot1^2=4-\pi</math>. As a result, <math>A=s^2-(s-4)^2-(4-\pi)=8s-20+\pi</math>. | ||
+ | Now, we consider the second disk. The part it sweeps is comprised of <math>4</math> quarter circles with radius <math>2</math> and <math>4</math> rectangles with a side lengths of <math>2</math> and <math>s</math>. When we add it all together, <math>2A=8s+4\pi\implies A=4s+2\pi</math>. <math>8s-20+\pi=4s+2\pi</math> so <math>s=5+\frac{\pi}{4}</math>. Finally, <math>5+1+4=\boxed{\textbf{(A) } 10}</math>. | ||
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+ | ~MathFun1000 (Inspired by Way Tan) | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=A|num-b=18|num-a=20}} | {{AMC10 box|year=2021 Fall|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:05, 25 November 2021
. A disk of radius rolls all the way around in 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
Solution
The side length of the inner square traced out by the disk with radius is . However, there is a little triangle piece at each corner where the disk never sweeps out. The combined area of these pieces is . As a result, .
Now, we consider the second disk. The part it sweeps is comprised of quarter circles with radius and rectangles with a side lengths of and . When we add it all together, . so . Finally, .
~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 |
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