2023 AMC 10A #15

 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 $2023\pi$?

Solution:

Note that the area of the shaded region for $n$ total circles is $$\pi [(n^2 - (n-1)^2) + \dots + (4^2 - 3^2)+ (2^2-1^2)]$$

This must be greater than or equal to $2023\pi$, so we have $$(n^2 - (n-1)^2) + \dots (4^2 - 3^2) + (2^2-1^2) \ge 2023$$

For any $k$, $k^2 - (k-1)^2 = 2k-1$, so we can write this as

$$\implies (2\cdot2 - 1) + (4\cdot2 - 1) + \dots + (2n-1) \ge 2023$$
$$\implies n \ge 64 \longrightarrow \boxed{\textbf{(E) 64}}$$

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