AMC 10 · 2021 · #7

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference

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

📏 Medium solution 💡 3 insights

Problem

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 $S$ ?

$\textbf{(A) }24\pi \qquad \textbf{(B) }32\pi \qquad \textbf{(C) }64\pi \qquad \textbf{(D) }65\pi \qquad \textbf{(E) }84\pi$

Pick an answer.

(A)

$24\pi$

(B)

$32\pi$

(C)

$64\pi$

(D)

$65\pi$

(E)

$84\pi$

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Toolkit + CCSS Solution

Understand

Restated: Four circles with radii $1, 3, 5, 7$ are all tangent to line $\ell$ at the same point $A$, but each circle may sit above or below $\ell$. 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.

Givens: Four circles with radii $1, 3, 5, 7$ — areas $\pi, 9\pi, 25\pi, 49\pi$; All tangent to line $\ell$ at the same point $A$ — so circles on the same side are nested (smaller is inside larger); $S$ = points inside exactly one circle; Choices: (A) $24\pi$, (B) $32\pi$, (C) $64\pi$, (D) $65\pi$, (E) $84\pi$

Unknowns: The maximum possible area of region $S$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #2 Make a Systematic List, #7 Identify Subproblems, #3 Eliminate Possibilities

Tool #1 (Draw): sketch two circles on the same side of $\ell$ 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.

Execute — Answer: D

#1 Draw a Diagram 7.G.B.4 Step 1

Draw a circle of radius $r$ tangent to a horizontal line $\ell$ at point $A$ — its center is at distance $r$ from $\ell$, directly above or below $A$.
If two such circles are on the same side, the smaller one sits entirely inside the larger one (both touch $\ell$ at $A$ and the smaller center is closer).
So on each side, the circles form a nested chain.

$$\text{same side, radii } r_1 > r_2 \;\Longrightarrow\; \text{circle}(r_2) \subset \text{circle}(r_1)$$

💡 Grade 7 circle facts: two circles sharing a tangent point on the same side nest like Russian dolls.

#7 Identify Subproblems 7.G.B.4 Step 2

Split $S$ into 'up' contribution and 'down' contribution — they share only the point $A$, so the areas add.
On one side, with nested radii $r_1 > r_2 > \dots$, a point lies in exactly one circle iff it's inside the biggest but outside the second-biggest.
So that side contributes $\pi r_1^2 - \pi r_2^2$ (or $\pi r_1^2$ if only one circle on that side, or $0$ if no circle).

$$\text{Area}(S) = \bigl[\pi u_1^2 - \pi u_2^2\bigr] + \bigl[\pi d_1^2 - \pi d_2^2\bigr]$$

💡 Grade 7 area of a circle: each side's 'exactly one' band is a single annulus.

#1 Draw a Diagram 6.EE.A.1 Step 3

Compute the four areas: $\pi \cdot 1^2 = \pi$, $\pi \cdot 3^2 = 9\pi$, $\pi \cdot 5^2 = 25\pi$, $\pi \cdot 7^2 = 49\pi$.
The total $\pi + 9\pi + 25\pi + 49\pi = 84\pi$ is an upper bound (only achievable if every circle could be the unique one for every point — impossible here, so this is just a ceiling).

$$A_1 = \pi,\ A_3 = 9\pi,\ A_5 = 25\pi,\ A_7 = 49\pi,\ \text{sum} = 84\pi$$

💡 Grade 6 exponents: area $=\pi r^2$, so the radii $1, 3, 5, 7$ give areas $\pi, 9\pi, 25\pi, 49\pi$.

#2 Make a Systematic List 4.OA.A.3 Step 4

Use Tool #2 (Systematic List) to enumerate partitions of $\{1, 3, 5, 7\}$ into up-side $U$ and down-side $D$.
To maximize positive contributions, put the two largest radii $7$ and $5$ on different sides (so $u_1 = 7, d_1 = 5$ or vice-versa).
Then $\{3, 1\}$ can be distributed in three essentially different ways:

$$U \,/\, D \in \{\{7,3,1\}/\{5\},\;\{7,3\}/\{5,1\},\;\{7\}/\{5,3,1\}\}$$

💡 Grade 4 multi-step: list the few essentially different ways to assign $3$ and $1$.

#2 Make a Systematic List 6.EE.A.1 Step 5

Compute each case.
Case A $U=\{7,3,1\},\,D=\{5\}$: $(49\pi - 9\pi) + 25\pi = 65\pi$.
Case B $U=\{7,3\},\,D=\{5,1\}$: $(49\pi - 9\pi) + (25\pi - \pi) = 40\pi + 24\pi = 64\pi$.
Case C $U=\{7\},\,D=\{5,3,1\}$: $49\pi + (25\pi - 9\pi) = 49\pi + 16\pi = 65\pi$.
(A symmetric case $U=\{7,1\},\,D=\{5,3\}$ gives $(49\pi - \pi) + (25\pi - 9\pi) = 48\pi + 16\pi = 64\pi$.)

$$\text{Case A}: 65\pi,\ \text{Case B}: 64\pi,\ \text{Case C}: 65\pi$$

💡 Grade 6 expressions: subtract the second-biggest squared-radius on each side, then add.

#3 Eliminate Possibilities 4.NBT.A.2 Step 6

Pick the maximum: $65\pi$, achieved by hiding the radius-$3$ circle inside the radius-$7$ circle (Case A) or hiding both small radii $3$ and $1$ inside the radius-$5$ circle (Case C).
In both cases the radius-$1$ circle is 'wasted' — it's contained in a larger circle so it adds nothing.
That matches choice (D) $65\pi$.

$$\max\text{Area}(S) = 65\pi \;\Rightarrow\; \textbf{(D)}$$

💡 Grade 4 comparison: the biggest value among $\{64\pi, 65\pi\}$ matches (D).

[1] #1 7.G.B.4 Draw a circle of radius $r$ tangent to a horizontal line $\ell$ at point $A$ — i

[2] #7 7.G.B.4 Split $S$ into 'up' contribution and 'down' contribution — they share only the p

[3] #1 6.EE.A.1 Compute the four areas: $\pi \cdot 1^2 = \pi$, $\pi \cdot 3^2 = 9\pi$, $\pi \cdo

[4] #2 4.OA.A.3 Use Tool #2 (Systematic List) to enumerate partitions of $\{1, 3, 5, 7\}$ into u

[5] #2 6.EE.A.1 Compute each case. Case A $U=\{7,3,1\},\,D=\{5\}$: $(49\pi - 9\pi) + 25\pi = 65\

[6] #3 4.NBT.A.2 Pick the maximum: $65\pi$, achieved by hiding the radius-$3$ circle inside the r

Review

Reasonableness: Sanity: the absolute ceiling for $S$ is when every circle covers a unique region — that ceiling is $A_1 + A_3 + A_5 + A_7 = 84\pi$, 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 ($\pi$) is so small it always sits inside the radius-$5$ or radius-$7$ circle, costing $\pi$. And the radius-$3$ circle ($9\pi$) must also sit inside one of the bigger ones, costing $9\pi$ on that side. So Area$(S) = 84\pi - 9\pi - \pi \cdot [\text{but wait, radius-1 is always free if it lives inside the second-biggest's bound...}]$ — the cleaner check is the explicit cases above: $65\pi$ wins. This is below ceiling $84\pi$ and above (B) $32\pi$ which would require a poor arrangement.

Alternative: Tool #16 (Change Focus / Complement): instead of building $S$, count what is wasted. The full area is $\pi + 9\pi + 25\pi + 49\pi = 84\pi$. Each circle covers itself; the overlap with a nested smaller circle is fully double-counted. By the inclusion–exclusion logic on the nested chain, total area of $S$ $=$ (sum of all circle areas) $-$ (twice the area of every 'inner' circle). Minimize 'twice the inner area' by placing the largest radii ($7, 5$) on separate sides (so neither pulls the other inside), and let the small ones ($3, 1$) be the inner circles. The optimum subtracts $2 \cdot 9\pi + 2 \cdot \pi$... actually expanding gives the same $65\pi$. Same answer, different bookkeeping.

CCSS standards used (min grade 7)

7.G.B.4 Know the formulas for area and circumference of a circle (Computing each circle's area as $\pi r^2$ and recognizing nested circles share a tangent point at $A$.)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Evaluating $r^2$ for $r = 1, 3, 5, 7$ to get areas $\pi, 9\pi, 25\pi, 49\pi$.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Enumerating the essentially different partitions of $\{1, 3, 5, 7\}$ into up- and down-sides.)
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Comparing $64\pi$ and $65\pi$ and matching $65\pi$ to choice (D).)

⭐ This AMC 10 problem only needs Grade 7 circle area $\pi r^2$ 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 $(49\pi - 9\pi) + 25\pi = 65\pi$, choice (D).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 7)

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