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

Revision as of 08:39, 12 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 $2023\pi$ ?

$[asy] filldraw(circle((0,0),8),gray); filldraw(circle((-1,0),7),white); filldraw(circle((-2,0),6),gray); filldraw(circle((-3,0),5),white); filldraw(circle((-4,0),4),gray); filldraw(circle((-5,0),3),white); filldraw(circle((-6,0),2),gray); filldraw(circle((-7,0),1),white); [/asy]$

$\textbf{(A) } 46 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 64$

Solution 1

Notice that the area of the shaded region is $(2^2\pi-1^2\pi)+(4^2\pi-3^2\pi)+(6^2\pi-5^2\pi)+ \cdots + (n^2\pi-(n-1)^2 \pi)$ 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 $60(61) > 60^2=3600$ , the only option higher than $60$ is $\boxed{\textbf{(E) } 64}$ .

~MrThinker

Solution 2

After first observing the problem, we can work out a few of the areas.

1st area = $4\pi-\pi = 3\pi$

2nd area = $16\pi-9\pi = 7\pi$

3rd area = $36\pi-25\pi = 11\pi$

4th area = $64\pi-49\pi = 15\pi$

We can see that the pattern is an arithmetic sequence with first term $3$ and common difference $4$ . From here, we can start from the bottom of the answer choices and work our way up. As the question asks for the least number of circles needed total, we have to divide the number of total circles by 2.

We can find the sum of the first $32$ terms of the arithmetic sequence by using the formula.

The last term is: $3 + 4\cdot(32-1) = 127$ .

Then, we can find the sum: $(3+127)/2 \cdot 32 = 65\cdot32 = 2080$ . It is clear that $64$ works.

The next answer choice is $60$ , which we have to divide by 2 to get $30$ .

The last term is: $3 + 4\cdot(30-1) = 119$ .

The sum is: $(3+119)/2 \cdot 30 = 61\cdot30 = 1830$ . This does not work.

As answer choice $D$ does not work and $E$ does, we can conclude that the answer is $\boxed{\textbf{(E) } 64}$ .

~zgahzlkw

Solution 3 (Similar to Solution 2)

We can easily see that all of the answer choices are even, which helps us solve this problem a little.

Lets just not consider the $\pi$ , since it is not that important, so let's just cancel that out.

When we plug in 64, we get $64^2-63^2+62^2-61^2+\cdots +4^2-3^2+2^2-1^2$ . By difference of squares, we get $1+2+3+\cdots+62+63+64$ , which by the sum of an arithmetic sequence, is $\frac{64(64+1)}{2}$ , which is $2080$ .

Similarly, we can use this for answer choice $D$ , and we have $\frac{62(62+1)}{2}$ which is $1953$ .

So, we see that answer choice $D$ is too small to satisfy the requirements, so we conclude the answer is $\boxed{\textbf{(E) } 64}$ .

~ESAOPS

Video Solution

https://youtu.be/rZrsrMxA7v0

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/N2lyYRMuZuk?si=o3PKCgxMkvDA9BXi&t=1078 ~Math-X

@@ Line 69: / Line 69: @@
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==Video Solution by Math-X (First understand the problem!!!)==
+https://youtu.be/N2lyYRMuZuk?si=o3PKCgxMkvDA9BXi&t=1078
+~Math-X
 ==See Also==
 {{AMC10 box|year=2023|ab=A|num-b=14|num-a=16}}
 {{MAA Notice}}

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