1994 AMC 12 Problem 11

Question

Three cubes of volume

1,

8

and

27

are glued together at their faces. The smallest possible surface area of the resulting configuration is

36

56

70

72

74

The cube side lengths are

1,2,

and

3

Minimize exposed area by maximizing the total areas of glued face portions

Before gluing, the total surface area is

6(1^2+2^2+3^2)=84.

The side-

2

cube can share area

4

with the side-

3

cube, while the side-

1

cube is placed at their common edge so that it shares area

1

with each larger cube. Thus the total contact area is

6.

Each glued area removes two exposed copies, giving

84-2(6)=72.

No pair can share more than the smaller face, so this is maximal contact and minimal surface area. Thus the correct answer is D.

Problem 11 in other years