Problem 19 · 2014 AMC 8
Medium
Geometry & Measurement
hide-faces-in-3doptimization
A cube with 3-inch edges is to be constructed from 27 smaller cubes with 1-inch edges. Twenty-one of the cubes are colored red and 6 are colored white. If the 3-inch cube is constructed to have the smallest possible white surface area showing, what fraction of the surface area is white?
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Answer: A — 5/54.
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Hint 1 of 2
Not all 27 positions are equal: a unit cube shows 3 faces at a corner, 2 on an edge, 1 in the middle of a face, and 0 in the dead center. To hide white, fill the lowest-exposure slots first.
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Hint 2 of 2
There's exactly one 0-face slot (center) and six 1-face slots (face-centers) — just enough for the 6 white cubes. Then divide the visible white by the total surface 6 × 9 = 54.
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Approach: greedily put white cubes in the least-exposed positions
- Rank positions by faces showing: center (0), 6 face-centers (1 each), 12 edges (2 each), 8 corners (3 each). White should claim the cheapest slots.
- Put 1 white cube in the dead center (0 faces show) and the other 5 in face-centers (1 face each) ⇒ only 5 white faces are visible — the smallest possible.
- Total surface area = 6 faces × 32 = 54 unit squares.
- White fraction = 5/54.
- Why this transfers: minimizing or maximizing exposure on a cube always comes down to sorting positions by how many faces show — corners are expensive, the center is free — then filling greedily.
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