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2011 AMC 10A Problem 14

2011 AMC 10A Problem 14AMC 10 Step by Step

Name: Date: October 28, 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

With diameter d, the area is less than the circumference exactly when d is less than 4. So the dice must sum to 2 or 3: 3 of 36 rolls.

Solution

First find which diameters work, then count the dice rolls that give them.

Try a few diameters. The radius is half the diameter. Compare area ( $\pi r^2$ ) with circumference ( $\pi \times$ diameter):

Diameter	Radius	Area	Circumference	Area smaller?
$2$	$1$	$\pi$	$2\pi$	yes
$3$	$1.5$	$2.25\pi$	$3\pi$	yes
$4$	$2$	$4\pi$	$4\pi$	no, equal
$5$	$2.5$	$6.25\pi$	$5\pi$	no

2. Why the pattern holds. For diameter $d$ , the area is $\pi \cdot \frac{d}{2} \cdot \frac{d}{2} = \frac{\pi d^2}{4}$ and the circumference is $\pi d$ . Divide both by $\pi d$ (a positive number): the area is smaller exactly when $\frac{d}{4} < 1$ , that is, when $d < 4$ . Larger diameters only make the area pull further ahead.
3. Count the rolls. Two dice can sum to $2$ or $3$ :
- sum $2$ : $(1,1)$ , one way;
- sum $3$ : $(1,2)$ and $(2,1)$ , two ways.

That is $3$ rolls out of $6 \times 6 = 36$ equally likely rolls.
4. Probability: $\dfrac{3}{36} = \dfrac{1}{12}$ .

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

Why this works

Area grows like the diameter times itself, while the circumference grows only like the diameter. So the area is smaller only for small circles, and the cutoff comes from a one-line comparison. After that it is a dice count. Remember that $(1,2)$ and $(2,1)$ are different rolls.

The trap

Using the dice sum as the radius instead of the diameter. That gives the wrong cutoff and only the single roll (1, 1).

Concise 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 $\boxed{\textbf{(B)}\ \frac{1}{12}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Geometry, Chapter 11: Circles
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques
Introduction to Geometry, Chapter 12: Circles and Angles

Techniques

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