Question

一只被困在边长为 $1$ 米的立方体盒子内的苍蝇，决定通过访问盒子的每个角来打发无聊。它从某个角出发并回到同一个角，且其他每个角都恰好访问一次。它从一个角到另一个角时，要么飞行，要么爬行，路径都是直线。它的路径最大可能长度是多少米？

A fly trapped inside a cubical box with side length $1$ meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?

$4+4\sqrt{2}$

$2+4\sqrt{2}+2\sqrt{3}$

$2+3\sqrt{2}+3\sqrt{3}$

$4\sqrt{2}+4\sqrt{3}$

$3\sqrt{2}+5\sqrt{3}$

答案：D

知识点：正方体图论最优化

难度评级：2070

解答：

苍蝇每一步可能的长度只有 $1, \sqrt2$ 或 $\sqrt3$ 。

立方体只有 $4$ 条空间对角线，所以最多 $4$ 步长度为 $\sqrt3$ 。另外 $4$ 步长度最多为 $\sqrt2$ 。

这个上界可以达到，例如在顶点间交替使用空间对角线和面对角线。

因此路径长度为 $4\sqrt2 + 4\sqrt3.$

所以正确答案是 D。

Note that all the paths the fly can take have lengths of $1, \sqrt2,$ or $\sqrt3.$

There are only $4$ space diagonals in the cube, so at most $4$ moves can have length $\sqrt3.$ The other $4$ moves have length at most $\sqrt2.$

This upper bound is attainable, for example by alternating space diagonals and face diagonals around the vertices.

The path has length $4\sqrt2 + 4\sqrt3.$

Thus, D is the correct answer.

2010 AMC 10A 第 20 题

答案：D

难度评级：2070

解答：

其他年份的第 20 题