2023 AMC 10A Problem 15

Even though the circles are tangent at a common point rather than concentric, each circle lies inside the next, so the region between the circles of radii $2k-1$ and $2k$ has area
$$\pi (2k)^2 - \pi (2k-1)^2 = \pi (4k - 1).$$

With $2n$ circles there are $n$ shaded rings, for $k = 1, \ldots, n$, giving total shaded area
$$\pi \sum_{k=1}^{n} (4k - 1) = \pi\left(4 \cdot \frac{n(n+1)}{2} - n\right) = \pi (2n^2 + n).$$

We need $2n^2 + n \ge 2023$. Since $2n^2 \approx 2023$ suggests $n \approx 32$: for $n = 31$, $2 \cdot 961 + 31 = 1953 < 2023$; for $n = 32$, $2 \cdot 1024 + 32 = 2080 \ge 2023$. So $n = 32$ rings, which requires $2n = 64$ circles.

The answer is $\boxed{\textbf{(E)}\ 64}$.

Areas of nested regions are differences of areas, and the difference of consecutive squares is linear, so the shaded total is an arithmetic series with a closed form. The off-center picture is a distraction: only containment matters for subtracting areas. After the algebra, the final trap is a units question: rings versus circles.

Pair the rings differently: the shaded area with $2n$ circles equals $\pi\big[(2^2 - 1^2) + (4^2 - 3^2) + \cdots\big] = \pi(3 + 7 + 11 + \cdots)$, an arithmetic series with $n$ terms, first term $3$ and last term $4n - 1$, sum $n(2n + 1)$. Same inequality.

• Solving for n and reporting n = 32 instead of the number of circles 2n = 64, or using 2n as the ring count.
• Shading the wrong rings (starting with the disk of radius $1$), which gives the sum of $4k - 3$ and a different threshold.