2023 AMC 10A Problem 15

Attempt Problem 15 of the 2023 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 10A solutions, or check the answer key.

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15.

An even number of circles are nested, starting with a radius of 11 and increasing by 11 each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius 22 but outside the circle of radius 1.1. An example showing 88 circles is displayed below. What is the least number of circles needed to make the total shaded area at least 2023π?2023\pi?

4646

4848

5656

6060

6464

Answer: E
Concepts:circle areaarithmetic sequencesummation
Difficulty rating: 1560
Solution:

A circle of radius rr has area πr2.\pi r^2. So the shaded ring between radius 2k2k and 2k12k-1 has area π((2k)2(2k1)2)\pi\big((2k)^2 - (2k-1)^2\big) =(4k1)π.= (4k-1)\pi. With 2n2n circles the shaded total is πk=1n(4k1)=π(2n2+n).\pi\sum_{k=1}^{n}(4k-1) = \pi(2n^2 + n). We want 2n2+n2023.2n^2 + n \ge 2023. At n=31n = 31 it's 1953,1953, at n=32n = 32 it's 2080.2080. So n=32,n = 32, which means 2n=642n = 64 circles. Thus, E is the correct answer.

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