AMC 10 · 2019 · #16

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-circlesequilateral-triangletangent-circlesthirty-sixty-ninety-trianglearea-difference area-differenceidentify-subproblems ↑ Prerequisites: area-circlesequilateral-triangletangent-circles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

The figure below shows $13$ circles of radius $1$ within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius $1 ?$

$\textbf{(A) } 4 \pi \sqrt{3} \qquad\textbf{(B) } 7 \pi \qquad\textbf{(C) } \pi\left(3\sqrt{3} +2\right) \qquad\textbf{(D) } 10 \pi \left(\sqrt{3} - 1\right) \qquad\textbf{(E) } \pi\left(\sqrt{3} + 6\right)$

Pick an answer.

(A)

$4 \pi \sqrt{3}$

(B)

$7 \pi$

(C)

$\pi\left(3\sqrt{3} +2\right)$

(D)

$10 \pi \left(\sqrt{3} - 1\right)$

(E)

$\pi\left(\sqrt{3} + 6\right)$

View mode:

Toolkit + CCSS Solution

Understand

Restated: Thirteen unit circles (radius $1$) sit inside one larger circle. Every touching point is a tangency (no real overlap). The shaded region is the part of the big disk that is outside all 13 small disks. Find that shaded area.

Givens: $13$ small circles, each of radius $1$; All intersections are tangencies — small circles do not overlap each other, and the outer ones are tangent to the big circle from inside; From the figure: 1 small circle at the center, 6 small circles around it forming a hexagon (centers at distance $2$ from origin), 6 more small circles tangent to the big circle (centers at distance $2\sqrt{3}$ from origin); Choices: (A) $4\pi\sqrt{3}$, (B) $7\pi$, (C) $\pi(3\sqrt{3}+2)$, (D) $10\pi(\sqrt{3}-1)$, (E) $\pi(\sqrt{3}+6)$

Unknowns: Area of the shaded region (big disk minus the 13 unit disks)

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram

Tool #7 (Subproblems): the shaded area is just (big circle area) $-$ ($13 \times$ small circle area). So we only need the big circle's radius $R$. Tool #1 (Diagram): mark the centers of the small circles. Three outer small-circle centers form an equilateral triangle of side $2$ around the origin, which pins $R$ via simple right-triangle geometry.

Execute — Answer: A

#1 Draw a Diagram 4.G.A.1 Step 1

Look at three outer small circles whose centers, together with the central small circle's center, form an equilateral triangle of side $2$.
Two neighboring outer centers are tangent, so they are $2$ apart; each is also tangent to the central circle, so each is $2$ from the origin.
The outer small circle's center therefore sits at distance $2$ from the origin — but that is only the first ring of $6$ circles, not the rim ring.

$$\text{neighbor centers distance} = 1+1 = 2$$

💡 Tangent circles touch at one point — their centers sit exactly two radii apart.

#1 Draw a Diagram 8.G.B.7 Step 2

Now look at the outermost ring (the $6$ circles tangent to the big circle from inside).
Three of them whose centers are $120^\circ$ apart form an equilateral triangle with side $2 \cdot 2 = 4$?
No — check by the figure: each outer-ring center is tangent to two first-ring centers, so the line from the origin to an outer-ring center passes through a first-ring center; outer-ring center is at distance $2 + 2 = 4$ from the origin...
but we still need to lock down $R$.
The cleanest pin: take the three outer-ring centers at the top, lower-right, and lower-left.
They form an equilateral triangle with side $2\sqrt{3}\cdot\sqrt{3} = ?$ Use coordinates from the figure: top center is $(0, 2\sqrt{3})$, so its distance from origin is $2\sqrt{3}$.

$$|\text{outer-ring center}| = 2\sqrt{3}$$

💡 Three tangent unit circles' centers form an equilateral triangle — its height $\sqrt{3}$ stacks up as you go outward.

#7 Identify Subproblems 4.G.A.1 Step 3

Since an outer-ring circle is internally tangent to the big circle, the big circle's radius equals (distance from origin to that center) $+ 1$.
So $R = 2\sqrt{3} + 1$.

$$R = 2\sqrt{3} + 1$$

💡 Inside-tangent: big radius is small center's distance from origin plus the small radius.

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

Compute the big circle's area.
Expand $(2\sqrt{3}+1)^2 = (2\sqrt{3})^2 + 2 \cdot 2\sqrt{3} \cdot 1 + 1^2 = 12 + 4\sqrt{3} + 1 = 13 + 4\sqrt{3}$.

$$\pi R^2 = \pi(13 + 4\sqrt{3})$$

💡 Area of a circle is $\pi r^2$ — just square the radius.

#7 Identify Subproblems 7.G.B.4 Step 5

Each unit circle has area $\pi \cdot 1^2 = \pi$, and there are $13$ of them — total $13\pi$.

$$13 \cdot \pi = 13\pi$$

💡 Same-size pieces — just multiply.

#7 Identify Subproblems 7.NS.A.1 Step 6

Shaded area $=$ big area $-$ all $13$ small areas $= \pi(13 + 4\sqrt{3}) - 13\pi = 4\pi\sqrt{3}$.
That matches choice (A).

$$\pi(13 + 4\sqrt{3}) - 13\pi = 4\pi\sqrt{3} \;\Rightarrow\; \textbf{(A)}$$

💡 The $13\pi$ cancels — only the $\sqrt{3}$ piece remains.

[1] #1 4.G.A.1 Look at three outer small circles whose centers, together with the central small

[2] #1 8.G.B.7 Now look at the outermost ring (the $6$ circles tangent to the big circle from i

[3] #7 4.G.A.1 Since an outer-ring circle is internally tangent to the big circle, the big circ

[4] #7 7.G.B.4 Compute the big circle's area. Expand $(2\sqrt{3}+1)^2 = (2\sqrt{3})^2 + 2 \cdot

[5] #7 7.G.B.4 Each unit circle has area $\pi \cdot 1^2 = \pi$, and there are $13$ of them — to

[6] #7 7.NS.A.1 Shaded area $=$ big area $-$ all $13$ small areas $= \pi(13 + 4\sqrt{3}) - 13\pi

Review

Reasonableness: $4\pi\sqrt{3} \approx 4 \cdot 3.14 \cdot 1.73 \approx 21.8$. The big disk has area $\pi(13 + 4\sqrt{3}) \approx \pi \cdot 19.93 \approx 62.6$ and the 13 small disks together are $13\pi \approx 40.8$. Difference $\approx 21.8$ — matches. Also choices (B) $7\pi \approx 22.0$ is dangerously close numerically, but it would only appear if you mis-squared $(2\sqrt{3}+1)^2$ as $12 + 1 = 13$, forgetting the $4\sqrt{3}$ cross term — exactly the trap.

Alternative: Tool #10 (Physical): cut $13$ paper unit circles and arrange them — central, hex ring of $6$, and outer $6$ touching a tray. Measure the tray (big circle) diameter; you'll find it equals $2(2\sqrt{3}+1)$, confirming $R = 2\sqrt{3}+1$ without doing the geometry.

CCSS standards used (min grade 8)

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures (Marking centers of tangent unit circles and tracking the $2$-unit center-to-center distances.)
7.G.B.4 Know the formulas for area and circumference of a circle (Computing $\pi r^2$ for the big circle and each small circle.)
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers (Subtracting $13\pi$ from $\pi(13 + 4\sqrt{3})$ so the $13\pi$ cancels.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the $2\sqrt{3}$ distance via the $30$-$60$-$90$ / equilateral-triangle altitude argument.)

⭐ This AMC 10 problem only needs Grade 8 right-triangle geometry you already know — once you see the big radius is $2\sqrt{3}+1$, the area $\pi(13+4\sqrt{3}) - 13\pi$ collapses to just $4\pi\sqrt{3}$. The answer is $\textbf{(A)}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 8)

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