2021 AMC 10B Problem 7

Two circles tangent to $\ell$ at the same point $A$ on the same side have their centers on the same perpendicular through $A$, so the smaller one lies entirely inside the larger. Circles on opposite sides only touch at $A$.

So on each side the circles are nested. A point lies inside exactly one circle on that side only if it is inside the largest circle but outside the second largest. Hence each side contributes $\pi(R^2 - r^2)$ where $R, r$ are its two largest radii, or $\pi R^2$ if the side holds a single circle.

To make $S$ large, avoid subtracting big radii. Put the $7$ alone on one side, contributing $49\pi$. The other side holds $5, 3, 1$ and contributes $\pi(25 - 9) = 16\pi$. Total: $65\pi$.

Compare the other splits: $\{7,1\}$ and $\{5,3\}$ give $48\pi + 16\pi = 64\pi$; $\{7,3\}$ and $\{5,1\}$ give $40\pi + 24\pi = 64\pi$; $\{7,5\}$ and $\{3,1\}$ give $24\pi + 8\pi = 32\pi$; $\{7,3,1\}$ and $\{5\}$ give $40\pi + 25\pi = 65\pi$ again. Nothing beats $65\pi$.

The answer is $\boxed{\textbf{(D)}\ 65\pi}$.

"Tangent to the same line at the same point" is a strong condition: it forces nesting on each side. Once you see that only the outermost annulus on each side counts, the problem is a tiny optimization: keep the two largest radii on different sides and let the second-largest on each side be as small as possible. The subtle point is that a side with just one circle subtracts nothing.

• Pairing 7 with 3 and 5 with 1 (or 7 with 1, 5 with 3) to get 64π, missing that a lone circle contributes its full area.
• Treating "inside exactly one circle" as an alternating sum $49 - 9 + 1$ on a side with three circles; the innermost region is inside three circles, not one.