2021 AMC 10B Problem 7

P1-10~3 minPrint

Circles Geometric Optimization Organized listing / direct enumeration Consider the largest/smallest element or boundary case

In a plane, four circles with radii $1,3,5,$ and $7$ are tangent to line $\ell$ at the same point $A,$ but they may be on either side of $\ell$ . Region $S$ consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region $S$ ?

Show solution without answering

Key insight

Circles on one side sit inside each other, so a side counts only its biggest circle minus its second biggest. Put 7 alone: 49 + 16 = 65.

Solution

Step 1: see how the circles sit.

A circle tangent to line $\ell$ at $A$ has its center straight above (or below) $A$ , at a distance equal to its radius.
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.
Circles on opposite sides only touch at the single point $A$ , so they do not overlap.

Step 2: find the area on one side.

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.)
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.

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

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

One side	Other side	Area
$7$	$5, 3, 1$	$49\pi + 16\pi = 65\pi$
$7, 3, 1$	$5$	$40\pi + 25\pi = 65\pi$
$7, 1$	$5, 3$	$48\pi + 16\pi = 64\pi$
$7, 3$	$5, 1$	$40\pi + 24\pi = 64\pi$
$7, 5$	$3, 1$	$24\pi + 8\pi = 32\pi$
$7, 5, \ldots$	the rest	at most $24\pi + 9\pi = 33\pi$
all four	none	$49\pi - 25\pi = 24\pi$

The largest is $65\pi$ .

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

Why this works

"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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 10: Geometric Inequalities
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Geometry, Chapter 12: Circles and Angles