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2023 AMC 10A Problem 15

Problem 15 of 25IntermediateAlgebraGeometry

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?

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Solution

A circle of radius rr has area πr2.\pi r^2. So the shaded ring between radius 2k2k and 2k−12k-1 has area π((2k)2−(2k−1)2)\pi\big((2k)^2 - (2k-1)^2\big) =(4k−1)π.= (4k-1)\pi. With 2n2n circles the shaded total is π∑k=1n(4k−1)=π(2n2+n).\pi\sum_{k=1}^{n}(4k-1) = \pi(2n^2 + n). We want 2n2+n≥2023.2n^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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Tagged: circle area · arithmetic sequence · summation

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