Difference between revisions of "2023 AMC 10A Problems/Problem 15"
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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>. | 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>. | ||
| − | 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>. | + | 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>. |
~MrThinker | ~MrThinker | ||
Revision as of 20:27, 9 November 2023
Problem
An even number of circles are nested, starting with a radius of 
and increasing by each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius 
but outside the circle of radius 
An example showing 
circles is displayed below. What is the least number of circles needed to make the total shaded area at least 
?
[insert asy of diagram below]
Solution
Notice that the area of the shaded region is 
for any even number 
.
Using the difference of squares, this simplifies to 
. So, we are basically finding the smallest such that 
. Since 
, the only option higher than 
is 
.
~MrThinker
