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Difference between revisions of "2023 AMC 10A Problems/Problem 15"

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==Problem==
 
==Problem==
 
An even number of circles are nested, starting with a radius of <math>1</math> and increasing by <math>1</math> each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius <math>2</math> but outside the circle of radius <math>1.</math> An example showing <math>8</math> circles is displayed below. What is the least number of circles needed to make the total shaded area at least <math>2023\pi</math>?
 
An even number of circles are nested, starting with a radius of <math>1</math> and increasing by <math>1</math> each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius <math>2</math> but outside the circle of radius <math>1.</math> An example showing <math>8</math> circles is displayed below. What is the least number of circles needed to make the total shaded area at least <math>2023\pi</math>?
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<math>\textbf{(A) } 46 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 64</math>
  
 
[insert asy of diagram below]
 
[insert asy of diagram below]
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~MrThinker
 
~MrThinker
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==See Also==
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{{AMC10 box|year=2023|ab=A|num-b=14|num-a=16}}
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{{MAA Notice}}

Revision as of 20:30, 9 November 2023

Problem

An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$? $\textbf{(A) } 46 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 64$

[insert asy of diagram below]

Solution

Notice that the area of the shaded region is $(2^2\pi-1^2\pi)+(4^2\pi-3^2\pi)+(6^2\pi-5^2\pi)+ \cdots + (n^2\pi-(n-1)^2pi)$ for any even number $n$.

Using the difference of squares, this simplifies to $(1+2+3+4+\cdots+n)\pi$. So, we are basically finding the smallest $n$ such that $\frac{n(n+1)}{2}>2023 \Rightarrow n(n+1) > 4046$. Since $60^2=3600$, the only option higher than $60$ is $\boxed{\textbf{(E) } 64}$.

~MrThinker

See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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. AMC logo.png

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