Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Problem)
Line 1: Line 1:
 
==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>?
 +
 
<math>\textbf{(A) } 46 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 64</math>
 
<math>\textbf{(A) } 46 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 64</math>
−
 
−
 
  
 
==Solution 1==
 
==Solution 1==

Revision as of 20:53, 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 $2023\pi$?

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

Solution 1

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

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10A_Problems/Problem_15&oldid=201342"