2023 AMC 10A Problem 15

Problem 15 of 25IntermediateAlgebra Geometry

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?

Show solution

Solution

A circle of radius

r

has area

\pi r^2.

So the shaded ring between radius

2k

and

2k-1

has area

\pi\big((2k)^2 - (2k-1)^2\big)

= (4k-1)\pi.

With

2n

circles the shaded total is

\pi\sum_{k=1}^{n}(4k-1) = \pi(2n^2 + n).

We want

2n^2 + n \ge 2023.

n = 31

it’s

1953,

n = 32

it’s

2080.

n = 32,

which means

2n = 64

circles. Thus, E is the correct answer.