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Difference between revisions of "2023 AMC 10A Problems/Problem 15"
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==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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[insert asy of diagram below]
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==Solution==
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Notice that the area of the shaded region is <math>(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)</math> for any even number <math>n</math>.
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Using the difference of squares, this simplifies to <math>(1+2+3+4+\cdots+n)\pi</math>. So, we are basically finding the smallest <math>n</math> such that <math>\frac{n(n+1)}{2}>2023 \Rightarrow n(n+1)>4046</math>. Since <math>60^2=3600</math>, the only option higher than <math>60</math> is <math>\boxed{\textbf{(E) } 64}</math>.
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~MrThinker

Revision as of 20:22, 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 $2023\pi$?

[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

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