1994 AMC 12 第 11 题

Question

三个体积分别为

1

、

8

和

27

的立方体沿着面粘在一起。所得组合体可能达到的最小表面积是

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

三个立方体的边长分别为

1,2

和

3

The cube side lengths are

1,2,

and

3

要使外露面积最小，就要使粘合面的总面积最大

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

粘合前的总表面积为

6(1^2+2^2+3^2)=84\text{。}

边长为

2

的立方体可以共用面积为

4

的面与边长为

3

的立方体粘合；把边长为

1

的立方体放在它们的公共棱处，它就能与两个较大的立方体各共用面积为

1

的面。因此总接触面积为

6

。每块粘合面积会去掉两份外露面，得到

84-2(6)=72

。任意两个立方体共用的面积都不可能超过较小立方体的一个面，所以这里的接触面积已达到最大，表面积也达到最小。因此正确答案是 D。

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.

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