AMC 10 · 2011 · #14

Grade 7 probabilitygeometry-2d
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
area-circlesperimeterprobability-basic convert-to-algebra ↑ Prerequisites: probability-basic
📏 Medium solution 💡 2 insights
Problem
Two standard 6-sided dice are rolled once. Their sum is used as the diameter of a circle. Compare the plain number that is the circle's area with the plain number that is its circumference, and find the probability that the area value is smaller than the circumference value.

Pick an answer.

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

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The diameter is the only quantity that changes, so Tool #4 (Introduce a Variable) is the natural anchor: call it d, write area and circumference as formulas in d, and turn the vague word comparison into a clean inequality. Tool #7 (Identify Subproblems) splits the work into two independent pieces — first a pure geometry/algebra step that finds which diameters satisfy the condition, then a pure probability step that counts how often the dice produce those diameters. Tool #2 (Make a Systematic List) handles that counting cleanly by listing the ordered dice results out of 36.

1STEP 1

Write both quantities in terms of d

Let the dice sum be the diameter d: the area is π(d/2)² = (πd²)/4 and the circumference is πd, both in the one letter d.

Area = π(d/2)² = (π d²)/4, Circumference = π d
2STEP 2

Turn the comparison into d ＜ 4

The area is smaller when (πd²)/4 ＜ πd; dividing by the positive πd gives d/4 ＜ 1, so the condition is exactly d ＜ 4.

(π d²)/4 ＜ π d ⟹ d/4 ＜ 1 ⟹ d ＜ 4
3STEP 3

Find the qualifying dice sums

A dice sum is at least 2, so d ＜ 4 allows only d = 2 from (1,1) and d = 3 from (1,2), (2,1): 3 ordered rolls.

d ＜ 4, d ≥ 2 → d ∈ {2,3}; d=2 : (1,1); d=3 : (1,2),(2,1)
4STEP 4

Compute the probability

Those 3 favorable rolls out of 36 equally likely ones give 3/36 = 1/12, choice (B).

P = (1 + 2)/36 = 3/36 = 1/12 → (B)
Answer
1/12
The boundary d = 4 gives equal area and circumference ((π·16)/4 = 4π = π·4), so only diameters below 4 work — a small set, so a small probability is expected. Only the sums 2 and 3 qualify, and these are the rarest sums, worth 1/36 and 2/36. Their total 3/36 = 1/12 is choice (B). The trap answers line up with partial work: (A) 1/36 counts only d=2, (C) 1/6 would wrongly include d=4 as well, and (E) 5/18 ignores the inequality entirely.
💡Key takeaway

Turn a wordy compare-two-things question into one inequality in a single variable, then just count the dice rolls that pass.

  • Write both quantities in terms of d
  • Turn the comparison into d ＜ 4
  • Find the qualifying dice sums
  • Compute the probability

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