2023 AMC 10A Problem 15

Question

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?

46

48

56

60

64

The shaded region between the circles of radius

2k

and

2k-1

has area

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

= (4k-1)\pi

With

2n

circles the shaded area is

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

solve

2n^2 + n \ge 2023

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.

At

n = 31

it’s

1953,

at

n = 32

it’s

2080.

So

n = 32,

which means

2n = 64

circles. Thus, E is the correct answer.

Problem 15 in other years