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2006 AMC 10B Problem 4

2006 AMC 10B Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Circles · difficulty 1 of 5 · about 1 min · position P1-10

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?

A)
$2$
B)
$3$
C)
$6$
D)
$8$
E)
$9$

Answer

Key insight

Areas scale with the square of the diameter, so the big circle is 9 times the small one and the ring is 9 - 1 = 8 times.

Solution

The diameters are in ratio $3 : 1$ , so the areas of the two full circles are in ratio $3^2 : 1^2 = 9 : 1$ . Call the red area $R$ ; then the large circle has area $9R$ .

The blue region is everything inside the large circle but outside the small one, so its area is $9R - R = 8R$ .

The ratio of blue to red is $\frac{8R}{R} = 8$ .

The answer is $\boxed{\textbf{(D)}\ 8}$ .

Why this works

Similar figures have areas proportional to the square of any corresponding length, so the actual radii never need to be computed. The ring is a difference of two circles; subtracting the inner one is the step that turns $9$ into $8$ .

Alternative approach

Directly: radii $\tfrac12$ and $\tfrac32$ give areas $\tfrac{\pi}{4}$ and $\tfrac{9\pi}{4}$ . Blue is $\tfrac{9\pi}{4} - \tfrac{\pi}{4} = 2\pi$ , and $2\pi \div \tfrac{\pi}{4} = 8$ .

The trap

Using the diameters as radii, or answering 9 (the ratio of the whole large circle to the small one) instead of ring to small.

Common mistakes

Using the diameters as radii, or answering 9 (the ratio of the whole large circle to the small one) instead of ring to small.
Comparing diameters instead of areas and answering 3 or 2.

Techniques

Set up the equation/formula and compute; no special trick needed