2002 AMC 10A Problem 18

P16-20~4 minPrint

Solid Geometry Basic Counting Set up the equation/formula and compute; no special trick needed Consider the largest/smallest element or boundary case

A $3$ x $3$ x $3$ cube is made of $27$ normal dice. Each die's opposite sides sum to $7$ . What is the smallest possible sum of all of the values visible on the $6$ faces of the large cube?

Show solution without answering

Key insight

Corner dice show 3 faces, edge dice 2, face-center dice 1. Faces 1, 2 and 3 meet at one corner of a die, so each die shows its smallest numbers.

Solution

Each die can be turned any way we like, separately from the others. So make each die's visible total as small as possible, and add them up.

1. Sort the $27$ dice by how many faces show.
- Each of the $8$ corners of the big cube has one die showing $3$ faces.
- Each of the $12$ edges of the big cube has $3$ dice; the middle one (not at a corner) shows $2$ faces. That is $12$ dice.
- Each of the $6$ big faces has $9$ dice; the one in the center shows $1$ face. That is $6$ dice.
- The single die in the very middle shows nothing.

Check: $8 + 12 + 6 + 1 = 27$ .

A face-center die shows one face. Turn it so that face is $1$ .

An edge die shows two faces that touch each other. Two touching faces are never opposite, so $1$ and $2$ can both show (they are not opposite, since $1$ is opposite $6$ ). Best total: $1 + 2 = 3$ .

A corner die shows three faces that meet at one corner. Can those be $1, 2, 3$ ? Their opposite faces are $6, 5, 4$ , so none of $1, 2, 3$ is opposite another. Faces $2$ and $3$ both touch face $1$ and are not opposite each other, so they touch each other too, and all three meet at a corner. Best total: $1 + 2 + 3 = 6$ .

5. Add up.
$8 \times 6 + 12 \times 3 + 6 \times 1 = 48 + 36 + 6 = 90.$

The answer is $\boxed{\textbf{(D)}\ 90}$ .

Why this works

When each part of a total can be chosen on its own, the smallest total comes from making each part as small as possible. The whole problem then rests on counting the corner, edge and face-center cubes of a $3 \times 3 \times 3$ cube ( $8$ , $12$ , $6$ , plus $1$ hidden). The fact that opposite faces add to $7$ is what guarantees $1$ , $2$ and $3$ sit around one corner.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Each die can be oriented independently, so minimize each die's visible total separately and add.

Sort the $27$ dice by how many faces show. The $8$ corner dice show $3$ faces, the $12$ edge dice show $2$ , the $6$ dice at the centers of the big faces show $1$ , and the central die shows none.

Because opposite faces sum to $7$ , the three faces at any corner of a die are pairwise non-opposite; the faces $1, 2, 3$ are pairwise non-opposite (their opposites are $6, 5, 4$ ), so they do meet at a corner and a corner die can show $1 + 2 + 3 = 6$ , which is the least possible. An edge die shows two adjacent faces, at best $1 + 2 = 3$ . A face-center die shows a single face, at best $1$ .

Minimum total:
$8 \cdot 6 + 12 \cdot 3 + 6 \cdot 1 = 48 + 36 + 6 = 90.$

The answer is $\boxed{\textbf{(D)}\ 90}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Geometry, Chapter 15: Curved Surfaces
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques