AMC 10 · 2023 · #15

Grade 7 geometry-2d

area-circlessequences-arithmeticpattern-recognitiontriangular-numbers pattern-recognitioneasier-related-problemidentify-subproblems ↑ Prerequisites: area-circlessequences-arithmetic

📏 Long solution 💡 3 insights 📊 Diagram

Problem

An even number n of circles are nested with radii 1, 2, 3, …, n and a common point. Shade alternating rings, starting with the ring between radius 2 and radius 1. Find the smallest even n such that the total shaded area is at least 2023π.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Easier Related Problem

Tool #9 (Easier Related Problem) is the lead: instead of fighting the formula at n = 64 directly, compute the total shaded area for small n (n = 2, 4, 6, 8) and read off the pattern. Tool #5 (Look for a Pattern) closes the loop: the partial sums turn out to be π · T_n where T_n = 1 + 2 + … + n is the n-th triangular number — exactly because π[(out)² - (in)²] = π(out+in)(out-in) = π(out+in) when the inner and outer radii differ by 1. Tool #6 (Guess and Check) on the five answer choices finishes the job: plug each n into T_n = $\frac{n(n+1)}{2}$ and find the smallest one that clears 2023. Tool #3 (Eliminate) discards (D) n = 60 etc. quickly along the way.

1STEP 1

Each shaded ring is an outer disk minus an inner disk: area π·outer² − π·inner², with consecutive radii k and k−1.

ring area = π k² - π(k-1)²

2STEP 2

Difference of squares collapses each ring: k² − (k−1)² = (k−1) + k, so one ring contributes π[(k−1)+k].

π[k² - (k-1)²] = π[(k-1) + k]

3STEP 3

Easier version: n=2 gives one ring of area 3π; n=4 gives 10π — each ring adds its inner+outer radii, covering 1,2,…,n.

n=2: 3π; n=4: 10π; n=6: 21π; n=8: 36π

4STEP 4

The pattern is triangular numbers: the shaded total at even n is π· $\frac{n(n+1)}{2}$ = π(1+2+…+n).

S(n) = π ·

\frac{(n(n+1))}{2}

5STEP 5

Need S(n) ≥ 2023π. Drop the positive π and double both sides: n(n+1) ≥ 4046.

n(n+1) ≥ 4046

6STEP 6

Check choices: 60·61=3660 and 62·63=3906 fall short, but 64·65=4160 clears 4046.

60 · 61 = 3660; 62 · 63 = 3906; 64 · 65 = 4160

7STEP 7

Choices 46,48,56,60 all give products below 4046; only n=64 reaches it, so S(64)=2080π ≥ 2023π → (E).

n = 64 → S(64) =

\frac{(64 · 65)}{2}

π = 2080π ≥ 2023π → (E)

Answer

Direction check: S(n) grows like

\frac{n²}{2}

· π, so we want

\frac{n²}{2}

≈ 2023, n ≈ √(4046) ≈ 63.6. The smallest integer above 63.6 is 64, which is conveniently even — no further parity adjustment needed. Magnitude check: S(64) = 2080π, just barely over 2023π (slack of 57π); S(62) = 1953π falls short. The answer sits right at the edge — the problem is well-posed.

💡Key takeaway

Each shaded ring contributes its outer + inner radius (Grade 6 difference of squares!), so the shaded total for the first n circles is just π · (1 + 2 + … + n) = π · $\frac{(n(n+1))}{2}$ . Push this past 2023π and the smallest even n is 64.