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2011 AMC 10A Problem 14

2011 AMC 10A Problem 14AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Probability · difficulty 2 of 5 · about 3 min · position P11-15

A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?

A)
$\frac{1}{36}$
B)
$\frac{1}{12}$
C)
$\frac{1}{6}$
D)
$\frac{1}{4}$
E)
$\frac{5}{18}$

Answer

Key insight

Area below circumference means pi d^2/4 < pi d, i.e. d < 4, so the dice sum must be 2 or 3: three of the 36 outcomes.

Solution

Let the dice sum be $d$ , the diameter. The radius is $d/2$ , so the area is $\pi \frac{d^2}{4}$ and the circumference is $\pi d$ . The condition is
$\pi \frac{d^2}{4} < \pi d \quad\Longleftrightarrow\quad d < 4.$

Sums below $4$ are $2$ and $3$ . The sum $2$ occurs one way, $(1,1)$ ; the sum $3$ occurs two ways, $(1,2)$ and $(2,1)$ . That is $3$ of the $36$ equally likely rolls, so the probability is
$\frac{3}{36} = \frac{1}{12}.$

The answer is $\boxed{\textbf{(B)}\ \frac{1}{12}}$ .

Why this works

The geometry collapses to a one-line inequality: area grows like $d^2$ while circumference grows like $d$ , so the area is smaller only for small diameters. After that it is a counting problem over the $36$ ordered dice outcomes, where $(1,2)$ and $(2,1)$ are distinct.

The trap

Using the sum as the radius instead of the diameter, which gives r < 2 and only the single outcome (1,1).

Common mistakes

Using the sum as the radius instead of the diameter, which gives r < 2 and only the single outcome (1,1).
Counting the sum $3$ once instead of twice, or including $d = 4$ where area and circumference are equal, not "less than."

Techniques

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