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

Two circles share the same center. The inner circle has diameter 1 inch and is painted red. The outer circle has diameter 3 inches, and the part inside it but outside the inner circle is painted blue. Find the ratio of the blue area to the red area.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

How to solve

Strategy Identify Subproblems

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.

1STEP 1

Radii from the diameters

Area needs the radius, so halve each diameter: red has radius $\frac{1}{2}$ and the outer circle has radius $\frac{3}{2}$ .

r_small=1/2, r_big=3/2

2STEP 2

Red area (small circle)

The red area is the whole inner circle: radius $\frac{1}{2}$ squared, times π, is $\frac{\pi}{4}$ .

A_red=π(1/2)²=π/4

3STEP 3

Blue area (the ring)

The outer circle has area $\frac{9\pi}{4}$ , and punching the red circle out of it leaves the ring: $2\pi$ .

A_blue=π(3/2)²-π/4=9π/4-π/4=2π

4STEP 4

Form the ratio

Divide blue by red: $2\pi \div \frac{\pi}{4} = 2\pi \cdot \frac{4}{\pi}$ , so π cancels and only 8 is left — choice (D).

A_blue/A_red=2π/(π/4)=2π·4/π=8 → (D)

Answer

Sanity check by scaling: the radii are in ratio 3/2/1/2=3, and areas grow with the square of the radius, so the big circle is 3²=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.

💡Key takeaway

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

Radii from the diameters
Red area (small circle)
Blue area (the ring)
Form the ratio

Radii from the diameters

Red area (small circle)

Blue area (the ring)

Form the ratio

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