AMC 10 · 2006 · #4

Grade 7 geometry-2d

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

area-circlesarea-differenceratio-proportion area-difference ↑ Prerequisites: area-circles

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

Circles of diameter $1$ inch and $3$ inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Two circles share the same center. The inner circle has diameter $1$ inch and is red; the ring between the inner circle and an outer circle of diameter $3$ inches is blue. Find the ratio of the blue area to the red area.

Givens: Inner (red) circle has diameter $1$ inch; Outer circle has diameter $3$ inches, and the two circles are concentric (same center); The blue region is the part inside the outer circle but outside the inner circle — a ring; Answer choices: (A) $2$, (B) $3$, (C) $6$, (D) $8$, (E) $9$

Unknowns: The ratio $\dfrac{\text{blue area}}{\text{red area}}$

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities

The blue region is a ring, which is a compound shape, so Tool #7 (Identify Subproblems) splits the work into three clean pieces: the small circle's area (red), the big circle's area, and then blue = big minus small. Tool #1 (Draw a Diagram) keeps the two concentric circles straight so the ring is not confused with the full outer disk. Tool #3 (Eliminate Possibilities) catches the main trap: the ratio of the two full circles is $9$ (choice (E)), but blue is only the ring, so the true ratio must be one less than that whole-circle ratio.

Execute — Answer: D

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

Radii from the diameters

Area needs the radius, and the radius is half the diameter.
The red circle has diameter $1$, so its radius is $\tfrac{1}{2}$.
The outer circle has diameter $3$, so its radius is $\tfrac{3}{2}$.

$$r_{\text{small}}=\dfrac{1}{2},\qquad r_{\text{big}}=\dfrac{3}{2}$$

💡 Every circle-area calculation starts from the radius, so halve the diameter first.

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

Red area (small circle)

The red area is the whole inner circle.
Use $A=\pi r^2$ with $r=\tfrac{1}{2}$: squaring $\tfrac{1}{2}$ gives $\tfrac{1}{4}$, so the red area is $\tfrac{\pi}{4}$.

$$A_{\text{red}}=\pi\left(\tfrac{1}{2}\right)^2=\dfrac{\pi}{4}$$

💡 Squaring the radius, not the diameter, is what turns a length into an area.

#7 Identify Subproblems 7.NS.A.1 Step 3

Blue area (the ring)

First find the whole outer circle: $A=\pi\left(\tfrac{3}{2}\right)^2=\pi\cdot\tfrac{9}{4}=\tfrac{9\pi}{4}$.
The blue ring is that outer circle with the red circle punched out, so subtract: $\tfrac{9\pi}{4}-\tfrac{\pi}{4}=\tfrac{8\pi}{4}=2\pi$.

$$A_{\text{blue}}=\pi\left(\tfrac{3}{2}\right)^2-\dfrac{\pi}{4}=\dfrac{9\pi}{4}-\dfrac{\pi}{4}=2\pi$$

💡 The ring is everything in the big disk that the small disk does not already cover, so subtract the small area away.

#3 Eliminate Possibilities 6.RP.A.3 Step 4

Form the ratio

Divide the blue area by the red area.
Dividing by $\tfrac{\pi}{4}$ is the same as multiplying by $\tfrac{4}{\pi}$: $2\pi\cdot\tfrac{4}{\pi}=8$.
The $\pi$ cancels, leaving $8$, which is choice (D).
Choice (E) $9$ is the trap for comparing the two whole circles instead of the ring, and $3$ (B) is just the diameter ratio.

$$\dfrac{A_{\text{blue}}}{A_{\text{red}}}=\dfrac{2\pi}{\pi/4}=2\pi\cdot\dfrac{4}{\pi}=8\;\Rightarrow\;\textbf{(D)}$$

💡 A ratio of two areas throws away their shared factor $\pi$, so only the numbers survive.

[1] #7 7.G.B.4 Area needs the radius, and the radius is half the diameter. The red circle has d

[2] #7 7.G.B.4 The red area is the whole inner circle. Use $A=\pi r^2$ with $r=\tfrac{1}{2}$: s

[3] #7 7.NS.A.1 First find the whole outer circle: $A=\pi\left(\tfrac{3}{2}\right)^2=\pi\cdot\tf

[4] #3 6.RP.A.3 Divide the blue area by the red area. Dividing by $\tfrac{\pi}{4}$ is the same a

Review

Reasonableness: Sanity check by scaling: the radii are in ratio $\tfrac{3/2}{1/2}=3$, and areas grow with the square of the radius, so the big circle is $3^2=9$ times the small one. Splitting the big disk into $9$ equal-area parts, the red center is $1$ part and the blue ring is the other $8$ — a ratio of $8$ to $1$. This confirms $8$ and exposes (E) $9$ as the whole-big-to-small ratio, not the ring-to-center ratio.

Alternative: Skip the areas entirely and work in units of the small circle's area. Since area scales as the square of the diameter, the outer circle is $3^2=9$ small-circle-areas and the inner is $1$. Blue is the difference, $9-1=8$ small-circle-areas, and red is $1$, so the ratio is $8:1=8$ without ever writing $\pi$.

CCSS standards used (min grade 7)

7.G.B.4 Know the formulas for the area and circumference of a circle and use them to solve problems (Getting each radius from its diameter and computing the circle areas $\pi(\tfrac12)^2$ and $\pi(\tfrac32)^2$.)
7.NS.A.1 Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers (Subtracting the small circle from the big circle, $\tfrac{9\pi}{4}-\tfrac{\pi}{4}=2\pi$, to get the blue ring.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Dividing the blue area by the red area to form the ratio $\tfrac{2\pi}{\pi/4}=8$.)

⭐ Area depends on the radius squared, so a $3\times$ wider circle is $9\times$ the area; carve out the center and the ring around it is the other $8$ parts.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Radii from the diameters

Red area (small circle)

Blue area (the ring)

Form the ratio

Review

CCSS standards used (min grade 7)

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