2021 AMC 10B Problem 7

Step 1: see how the circles sit.

1. A circle tangent to line $\ell$ at $A$ has its center straight above (or below) $A$, at a distance equal to its radius.
2. So two circles on the same side both have their centers on the same line through $A$, both touching $\ell$ at $A$. The smaller one sits completely inside the bigger one, like a small coin resting at the bottom of a larger ring.
3. Circles on opposite sides only touch at the single point $A$, so they do not overlap.

Step 2: find the area on one side.

1. On one side the circles are stacked inside each other. A point is inside exactly one of them when it is inside the biggest circle but outside the second biggest. (Anything inside the second biggest is inside at least two circles.)
2. So a side with biggest radius $R$ and second biggest $r$ adds $\pi R^2 - \pi r^2$. A side with just one circle of radius $R$ adds all of $\pi R^2$.

Step 3: choose the sides.

1. We want to take away as little as possible. Put the $7$ alone on one side: it adds $49\pi$.
2. The other side holds $5$, $3$ and $1$. Its biggest is $5$ and second biggest is $3$, so it adds $25\pi - 9\pi = 16\pi$.
3. Total: $49\pi + 16\pi = 65\pi$.

Step 4: check that nothing is bigger. Every other way to split the circles:

The largest is $65\pi$.

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

"Tangent to the same line at the same point" forces the circles on each side to be nested, so only the outer ring on each side counts. After that it is a small puzzle: keep the $7$ and the $5$ on different sides, and make each side's second-biggest circle as small as possible. The easy thing to miss is that a circle alone on its side loses nothing.

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