1998 AHSME Problem 27
A cube is composed of twenty-seven cubes. The big cube is ‘tunneled’ as follows: First the six cubes which make up the center of each face as well as the center cube are removed as shown. Second, each of the twenty remaining 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
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1 solution
First, we find the total surface area of each of the 3x3x3 cubes after they have been “diminished”. We will find it separately as the surface area on the outside and the surface area on the inside. If you visualize this figure, it has eight 1x1x1 cubes on each face, so the surface area of each face is 8. So the total surface area on the outside is 8 x (1x1) per face x 6 faces = 48
Each “hole” at the surface has four faces. The surface area on the inside is 4 x (1x1) = 4 There are six such holes. So total surface area on the inside is 4x6 = 24
Now 20 of these “diminished” 3x3x3 pieces are fitted together to form our final figure. When two pieces are fitted together, some surface area is lost from the outside portion for both pieces. But the area from the inside portion is not lost like this.
We have 8 corner pieces and 12 edge pieces. Corner pieces will lose (3/6) of their outside area because they are connected to other pieces on 3 out of 6 sides, while edge pieces will lose (2/6) of their outside area because they are connected to other pieces on 2 out of 6 sides.
Thus, for corner pieces, area = 48 x (1 - 3/6) + 24 = 48 For edge pieces, area = 48 x (1 - 2/6) + 24 = 56
Total surface area = 48x8 + 56x12 = 1056.