AMC 10 · 2009 · #17

Grade 7 probabilitygeometry-3d
Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role
probability-basicspatial-visualizationsystematic-enumeration physical-representation ↑ Prerequisites: probability-basic
📏 Medium solution 💡 3 insights
Problem
Each face of a cube gets one stripe joining opposite edge midpoints, oriented at random. Find the probability some stripes close into a loop around the cube.

Pick an answer.

(A)
$\frac 18$
(B)
$\frac {3}{16}$
(C)
$\frac 14$
(D)
$\frac 38$
(E)
$\frac 12$
How to solve
Strategy Visualize Spatial Relationships

The whole problem turns on seeing, in 3D, what an encircling stripe actually is: a belt of 4 faces wrapping around the cube while skipping one opposite pair. Tool #17 (Visualize Spatial Relationships) is the unlock for that mental picture and for spotting why two belts cannot both close at once. Tool #1 (Draw a Diagram) keeps a labeled cube sketch so 'this face's stripe must run this way' stays concrete. Tool #7 (Identify Subproblems) splits the job into: (i) the chance one chosen belt closes, then (ii) combine the three belts. Tool #2 (Make a Systematic List) both fixes the size of the sample space (2⁶ = 64) and gives a clean favorable-count cross-check (12 of 64).

1STEP 1

Count all equally likely outcomes

There are 64 equally likely outcomes.

2 × 2 × 2 × 2 × 2 × 2 = 2⁶ = 64
2STEP 2

Picture an encircling stripe as a band

A loop is a band of four faces, and there are three such bands.

3 pairs of opposite faces → 3 possible bands
3STEP 3

Probability one chosen band closes

One band closes with probability 1/16.

(1/2)⁴ = 1/16
4STEP 4

Bands cannot overlap

Two bands can never close together.

belt_i ∩ belt_j = ∅ (i ≠ j)
5STEP 5

Add the three disjoint bands

So the three simply add, giving 3/16, choice (B).

3 · 1/16 = 3/16 = 12/64 → (B)
Answer
3/16
The answer 3/16 = 12/64 ≈ 0.19 is a small probability, which fits: getting four specific faces to line up is demanding. Two independent counts agree — the disjoint-sum 3 × 1/16 and the direct favorable count 12/64. The distractors flag the classic slips: 1/16 forgets there are three belts, 3/8 = 6 × 1/16 double-counts as if belts could overlap, and 1/4 or 1/2 overcount the aligned faces. Since the belts are genuinely mutually exclusive (Step 4), no double-counting occurs and 3/16 stands. Answer (B).
💡Key takeaway

A stripe going all the way around the cube is just a belt of 4 faces that skips a top-and-bottom pair; there are 3 such belts, each closes with chance (1/2)⁴ = 1/16, and since two belts can never close at once you add them: 3 × 1/16 = 3/16, choice (B).

  • Count all equally likely outcomes
  • Picture an encircling stripe as a band
  • Probability one chosen band closes
  • Bands cannot overlap
  • Add the three disjoint bands

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