Question

一个边长为 $3$ 英寸的立方体要由 $27$ 个边长为 $1$ 英寸的小立方体组成。其中二十一个小立方体涂成红色， $6$ 个涂成白色。

如果构造这个 $3$ 英寸立方体时让可见白色表面积尽可能小，那么白色占总表面积的几分之几？

A cube with $3$ -inch edges is to be constructed from $27$ smaller cubes with $1$ -inch edges. Twenty-one of the cubes are colored red and $6$ are colored white.

If the $3$ -inch cube is constructed to have the smallest possible white surface area showing, what fraction of the surface area is white?

$\dfrac{5}{54}$

$\dfrac{1}{9}$

$\dfrac{5}{27}$

$\dfrac{2}{9}$

$\dfrac{1}{3}$

答案：A

知识点：正方体表面积最优化

难度评级：1410

解答：

为了最小化可见白色面积，把一个白色小立方体放在大立方体中心，这样没有任何面可见。把另外五个白色小立方体放在大立方体的面中心位置，每个只贡献一个可见白色小正方形。

可见白色表面积为 $5$ 平方英寸。边长 $3$ 英寸的立方体总表面积为 $6\cdot3^2=54$ 平方英寸，所以白色部分占 $\dfrac{5}{54}$ 。

所以正确答案是 A。

To minimize visible white area, place one white cube in the center of the large cube, where no faces are visible. Place each of the other five white cubes at the center of a face of the large cube, where each contributes only one visible white square.

The visible white surface area is therefore $5$ square inches. The total surface area of the $3$ -inch cube is $6\cdot3^2=54$ square inches, so the fraction that is white is $\dfrac{5}{54}$ .

Thus, A is the correct answer.

2014 AMC 8 第 19 题

答案：A

难度评级：1410

解答：

其他年份的第 19 题