AMC 10 · 2021 · #7

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Nested Frame / Concentric Difference

area-circlestangent-circlessymmetry-argumentset-partition identify-subproblemscaseworksymmetry-argument ↑ Prerequisites: area-circles

📏 Medium solution 💡 3 insights

Problem

Four circles with radii 1, 3, 5, 7 are all tangent to line ℓ at the same point A, but each circle may sit above or below ℓ. Region S is the set of points lying inside exactly one of the four circles. Each circle can be assigned to the up-side or the down-side; choose the assignment that makes the area of S as large as possible.

Pick an answer.

(A)

$24\pi$

(B)

$32\pi$

(C)

$64\pi$

(D)

$65\pi$

(E)

$84\pi$

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw): sketch two circles on the same side of ℓ tangent at A — they nest, and the 'exactly one' region is the annulus between the largest and second-largest. Tool #7 (Subproblems): split S into 'up contribution' + 'down contribution' — each side acts independently. Tool #2 (Systematic List): partition {1, 3, 5, 7} into two groups and tabulate the area for each partition (up to mirror symmetry only a handful of distinct splits exist). Tool #3 (Eliminate): pick the partition giving the maximum and match to a choice. The big radii 7 and 5 must each be the largest on their own side (separating them maximizes the positive area), and the small radii 3 and 1 should be hidden inside the largest where they cost the least.

1STEP 1

Each circle's center sits distance r from ℓ at A, so two circles on the same side nest — the smaller lies entirely inside the larger.

same side, radii r₁ ＞ r₂ ⟹ circle(r₂) ⊂ circle(r₁)

2STEP 2

Split S at A into an up-side and a down-side; on each side 'exactly one' means only the outer ring, area π r₁² - π r₂².

Area(S) = [π u₁² - π u₂²] + [π d₁² - π d₂²]

3STEP 3

The four areas are π, 9π, 25π, 49π; the sum 84π is only an upper bound, unreachable since all circles share A.

A₁ = π, A₃ = 9π, A₅ = 25π, A₇ = 49π, sum = 84π

4STEP 4

List the splits of {1, 3, 5, 7}: keep the biggest radii 7 and 5 on different sides, then 3 and 1 give three distinct cases.

U / D ∈ {{7,3,1}/{5}, {7,3}/{5,1}, {7}/{5,3,1}}

5STEP 5

Case A {7,3,1}/{5}: 40π+25π=65π. Case B {7,3}/{5,1}: 40π+24π=64π. Case C {7}/{5,3,1}: 49π+16π=65π.

Case A: 65π, Case B: 64π, Case C: 65π

6STEP 6

The maximum is 65π: hide radius 3 inside 7 (Case A) or hide 3 and 1 inside 5 (Case C); the radius-1 circle is always wasted — choice (D).

maxArea(S) = 65π → (D)

Answer

65π

Sanity: the absolute ceiling for S is when every circle covers a unique region — that ceiling is A₁ + A₃ + A₅ + A₇ = 84π, which is choice (E). But circles forced to share the same tangent point cannot all contribute — at least one of {3, 1} is swallowed by {7, 5}. Specifically the radius-1 circle (π) is so small it always sits inside the radius-5 or radius-7 circle, costing π. And the radius-3 circle (9π) must also sit inside one of the bigger ones, costing 9π on that side. So Area(S) = 84π - 9π - π · [but wait, radius-1 is always free if it lives inside the second-biggest's bound...] — the cleaner check is the explicit cases above: 65π wins. This is below ceiling 84π and above (B) 32π which would require a poor arrangement.

💡Key takeaway

This AMC 10 problem only needs Grade 7 circle area π r² you already know — put the big radii 7 and 5 on opposite sides, hide the small ones (3, 1) inside, and on each side the 'exactly-one' band is just the biggest annulus. The best total is (49π - 9π) + 25π = 65π, choice (D).

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