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1998 AHSME Problems/Problem 27

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Problem

A 9 \times 9 \times 9 cube is composed of twenty-seven 3 \times 3 \times 3 cubes. The big cube is ‘tunneled’ as follows: First, the six 3 \times 3 \times 3 cubes which make up the center of each face as well as the center 3 \times 3 \times 3 cube are removed. Second, each of the twenty remaining 3 \times 3 \times 3 cubes is diminished in the same way. That is, the center facial unit cubes as well as each center cube are removed. The surface area of the final figure is:

Image:1998 AHSME num. 27.png

\mathrm{(A)}\ 384\qquad\mathrm{(B)}\ 729\qquad\mathrm{(C)}\ 864\qquad\mathrm{(D)}\ 1024\qquad\mathrm{(E)}\ 1056

Contents

Solution

Solution 1

Each 3 \times 3 \times 3 cube has eight faces on each side, for a surface area of 6 \cdot 8 \cdot (1 \cdot 1) = 48 on the outside. Each face also has to count the surface area in the inside area of the removed cube, for an additional surface area of 6 \cdot 4 \cdot (1 \cdot 1) = 24. Thus the total surface area for each 3 \times 3 \times 3 is 72.

There are 20 of these cubes, for an area of 72 \cdot 20 = 1440. However, many of the cubes share a common face; each corner 3\times 3\times 3 cube has three hidden faces and each edge cube has two hidden faces, for a total of 8\cdot 3 + 12\cdot 2 = 48 hidden faces. Each hidden face has a surface area of 8, so the surface area of the final figure is 1440 - 48 \cdot 8 = 1056 \Rightarrow \mathrm{(E)}.

Solution 2

After the first step, twenty 3 \times 3 \times 3 cubes remain, with 8 corner cubes and 12 edge cubes. Each one of these 3 \times 3 \times 3 corner cubes contributes 27 square units of area, and each edge cube contributes 36 square units of area.

The second stage takes away 3 square units of area (1 for each exposed face) from each of the eight 3 \times 3 \times 3 corner cubes, and adds an additional 24 more units from the center facial cubes removed. Similarly, the twelve 3\times 3\times 3 edge cubes each lose 4 square nits but gain 24 units. Thus, the total surface area is 8 \cdot (27 - 3 + 24) + 12 \cdot (36 - 4 + 24) = 1056

See also

1998 AHSME (Problems)
Preceded by
Problem 26
Followed by
Problem 28
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