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 of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$ ?

$\textbf{(A) } 46 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 64$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: An even number $n$ of circles are nested with radii $1, 2, 3, \dots, n$ 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 $2023\pi$.

Givens: Radii: $1, 2, 3, \dots, n$ (consecutive integers); $n$ is even; Shaded rings are between radii $(1, 2), (3, 4), (5, 6), \dots, (n-1, n)$; Each ring's area is $\pi(\text{outer})^2 - \pi(\text{inner})^2$; Total shaded area must be $\ge 2023\pi$; Answer choices: (A) $46$, (B) $48$, (C) $56$, (D) $60$, (E) $64$

Unknowns: The smallest even $n$ that makes the total shaded area $\ge 2023\pi$

Understand

Plan

Primary tool: #9 Easier Related Problem

Secondary: #7 Identify Subproblems, #5 Look for a Pattern, #6 Guess and Check, #3 Eliminate Possibilities

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 $\pi \cdot T_n$ where $T_n = 1 + 2 + \dots + n$ is the $n$-th triangular number — exactly because $\pi[(\text{out})^2 - (\text{in})^2] = \pi(\text{out}+\text{in})(\text{out}-\text{in}) = \pi(\text{out}+\text{in})$ 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 $T_n = n(n+1)/2$ and find the smallest one that clears $2023$. Tool #3 (Eliminate) discards (D) $n = 60$ etc. quickly along the way.

Execute — Answer: E

#7 Identify Subproblems 7.G.B.4 Step 1

Each shaded region is an annulus-like region: outer disk minus inner disk.
Its area is $\pi(\text{outer radius})^2 - \pi(\text{inner radius})^2$.
With consecutive integer radii, every shaded ring has outer radius $k$ and inner radius $k - 1$ for some even $k$.

$$\text{ring area} = \pi k^2 - \pi(k-1)^2$$

💡 Grade 7 introduces $A = \pi r^2$ — and the ring picture splits naturally into "big disk minus little disk".

#5 Look for a Pattern 6.EE.A.3 Step 2

Use the difference-of-squares identity on each ring: $k^2 - (k-1)^2 = (k - (k-1))(k + (k-1)) = 1 \cdot (2k - 1) = (k-1) + k$.
So a single shaded ring contributes $\pi \bigl[(k-1) + k\bigr]$ to the shaded area.

$$\pi[k^2 - (k-1)^2] = \pi[(k-1) + k]$$

💡 Difference of squares ($a^2 - b^2 = (a-b)(a+b)$) is the Grade 6 identity that collapses each ring to a pair-sum.

#9 Easier Related Problem 4.OA.C.5 Step 3

Tool #9 (Easier Related Problem): try $n = 2$ first.
One shaded ring between radii $1$ and $2$.
Area $= \pi(2^2 - 1^2) = \pi(1 + 2) = 3\pi$.
Now $n = 4$: two shaded rings, $(1,2)$ and $(3,4)$.
Sum $= \pi[(1+2) + (3+4)] = \pi(1+2+3+4) = 10\pi$.
Pattern emerging.

$$n=2: 3\pi; \quad n=4: 10\pi; \quad n=6: 21\pi; \quad n=8: 36\pi$$

💡 Small cases reveal: each shaded ring contributes its (inner + outer) radius pair, and the pairs cover $1, 2, 3, \dots, n$ without gaps.

#5 Look for a Pattern 6.EE.A.2 Step 4

Tool #5 (Look for a Pattern).
The shaded total at even $n$ is $\pi(1 + 2 + \dots + n)$ — the $n$-th triangular number times $\pi$.
Recognise $3, 10, 21, 36$ as $T_2, T_4, T_6, T_8$ where $T_n = \dfrac{n(n+1)}{2}$.

$$S(n) = \pi \cdot \dfrac{n(n+1)}{2}$$

💡 Recognising the partial sums as triangular numbers — the Grade 6 named-formula payoff after collecting small cases.

#6 Guess and Check 6.EE.B.8 Step 5

Set up the inequality.
We need $S(n) \ge 2023\pi$, so $\dfrac{n(n+1)}{2} \ge 2023$, i.e.
$n(n+1) \ge 4046$.

$$n(n+1) \ge 4046$$

💡 Strip out $\pi$ (it's positive) and multiply through by $2$ — Grade 6 inequality manipulation.

#6 Guess and Check 6.EE.B.8 Step 6

Tool #6 (Guess and Check) on the answer choices.
Try (D) $n = 60$: $60 \cdot 61 = 3660 < 4046$, so $60$ is too small.
Try (E) $n = 64$: $64 \cdot 65 = 4160 \ge 4046$, so $64$ works.
Try $n = 62$ to make sure the smallest even is $64$: $62 \cdot 63 = 3906 < 4046$, still too small.

$$60 \cdot 61 = 3660; \;\; 62 \cdot 63 = 3906; \;\; 64 \cdot 65 = 4160$$

💡 Marching up from $60$ in even steps and watching the product cross $4046$ — the most direct Grade 6 check.

#3 Eliminate Possibilities 6.EE.B.8 Step 7

Confirm $64$ is the smallest even $n$ satisfying the inequality.
Among the choices: (A) $46 \cdot 47 = 2162$, (B) $48 \cdot 49 = 2352$, (C) $56 \cdot 57 = 3192$, (D) $60 \cdot 61 = 3660$ — all $< 4046$.
Only (E) $n = 64$ gives $\ge 4046$.

$$n = 64 \;\Rightarrow\; S(64) = \dfrac{64 \cdot 65}{2}\,\pi = 2080\pi \ge 2023\pi \;\Rightarrow\; \textbf{(E)}$$

💡 Eliminating (A)-(D) one by one and confirming (E) — clean Grade 6 inequality verification.

[1] #7 7.G.B.4 Each shaded region is an annulus-like region: outer disk minus inner disk. Its a

[2] #5 6.EE.A.3 Use the difference-of-squares identity on each ring: $k^2 - (k-1)^2 = (k - (k-1)

[3] #9 4.OA.C.5 Tool #9 (Easier Related Problem): try $n = 2$ first. One shaded ring between rad

[4] #5 6.EE.A.2 Tool #5 (Look for a Pattern). The shaded total at even $n$ is $\pi(1 + 2 + \dots

[5] #6 6.EE.B.8 Set up the inequality. We need $S(n) \ge 2023\pi$, so $\dfrac{n(n+1)}{2} \ge 202

[6] #6 6.EE.B.8 Tool #6 (Guess and Check) on the answer choices. Try (D) $n = 60$: $60 \cdot 61

[7] #3 6.EE.B.8 Confirm $64$ is the smallest even $n$ satisfying the inequality. Among the choic

Review

Reasonableness: Direction check: $S(n)$ grows like $\tfrac{n^2}{2} \cdot \pi$, so we want $n^2 / 2 \approx 2023$, $n \approx \sqrt{4046} \approx 63.6$. The smallest integer above $63.6$ is $64$, which is conveniently even — no further parity adjustment needed. Magnitude check: $S(64) = 2080\pi$, just barely over $2023\pi$ (slack of $57\pi$); $S(62) = 1953\pi$ falls short. The answer sits right at the edge — the problem is well-posed.

Alternative: Tool #10 (Physical Representation) — sketch the first few rings on paper, add up their areas with simple arithmetic. Ring $1$ (radii $1$-$2$): area $\pi(4 - 1) = 3\pi$. Ring $2$ (radii $3$-$4$): $\pi(16 - 9) = 7\pi$. Ring $3$ (radii $5$-$6$): $\pi(36 - 25) = 11\pi$. Notice the increments $3, 7, 11, 15, \dots$ form an arithmetic sequence with common difference $4$ — sum of first $m$ terms equals $m(2m + 1)$. Setting $m = n/2$ and $m(2m+1) \ge 2023$, again $m = 32$ ($n = 64$) is the smallest even fit.

CCSS standards used (min grade 7)

4.OA.C.5 Generate a number or shape pattern following a given rule (Generating shaded totals for $n = 2, 4, 6, 8$ to see the pattern emerge.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Naming the partial shaded total $S(n) = \pi \cdot \dfrac{n(n+1)}{2}$ in closed form using the triangular-number formula.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Applying the difference-of-squares identity $k^2 - (k-1)^2 = 2k - 1 = (k-1) + k$ to collapse each ring's area.)
6.EE.B.8 Write an inequality of the form x > c or x < c and graph on a number line (Setting up and testing $n(n+1) \ge 4046$ to find the smallest even $n$ that works.)
7.G.B.4 Know the formulas for area and circumference of a circle (Computing the area of each shaded ring as $\pi r_{\text{out}}^2 - \pi r_{\text{in}}^2$ — the Grade 7 circle-area formula.)

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