AMC 10 · 2011 · #10

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
A dice sum becomes a circle's diameter and its two measures are compared as bare numbers. Find the probability.

Pick an answer.

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

Both measures come from the same diameter.

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

Turn the comparison into d ＜ 4

The comparison reduces to a plain bound.

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

Find the qualifying dice sums

Only two dice sums clear it.

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

Compute the probability

The probability is 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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