AMC 10 · 2009 · #25

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 of a cube's 6 faces gets one stripe joining the midpoint of an edge to the midpoint of the opposite edge. A square face has two pairs of opposite edges, so each face's stripe has 2 equally likely orientations, chosen independently. Find the probability that some stripes join up into a single loop that runs all the way around the cube.

Pick an answer.

(A)
$\frac 18$
(B)
$\frac {3}{16}$
(C)
$\frac 14$
(D)
$\frac 38$
(E)
$\frac 12$

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Each of the 6 faces independently takes 1 of 2 orientations, so the whole cube has 2⁶ = 64 equally likely patterns.

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

Picture an encircling stripe as a band

A loop around the cube crosses 4 faces as a belt and skips one top-and-bottom pair; with 3 such pairs there are exactly 3 belts.

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

Probability one chosen band closes

Fix one belt: each of its 4 faces must take the one orientation running along it, so it closes with probability (1/2)⁴ = 1/16.

(1/2)⁴ = 1/16
4STEP 4

Bands cannot overlap

Any two belts share a face that each would need pointing the other way, so at most one belt closes: the three events are mutually exclusive.

belt_i ∩ belt_j = ∅ (i ≠ j)
5STEP 5

Add the three disjoint bands

Disjoint events add, so the chance some belt closes is 3 × 1/16 = 3/16, matching the direct count 12 of 64. 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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