AMC10 2010 A

AMC10 2010 A · Q20

AMC10 2010 A · Q20. It mainly tests Permutations, Coordinate geometry.

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?

一只苍蝇被困在一个边长1米的立方体盒子里，它决定通过访问盒子的每个顶点来消磨无聊。它从同一个顶点开始并结束，并恰好访问每个其他顶点一次。从一个顶点到另一个顶点，它要么飞要么爬直线行进。其路径的最大可能长度是多少米？

(A) 4 + 4$\sqrt{2}$ 4 + 4$\sqrt{2}$

(B) 2 + 4$\sqrt{2}$ + 2$\sqrt{3}$ 2 + 4$\sqrt{2}$ + 2$\sqrt{3}$

(D) 4$\sqrt{2}$ + 4$\sqrt{3}$ 4$\sqrt{2}$ + 4$\sqrt{3}$

(E) 3$\sqrt{2}$ + 5$\sqrt{3}$ 3$\sqrt{2}$ + 5$\sqrt{3}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Each of the 8 line segments on the fly’s path is an edge, a face diagonal, or an interior diagonal of the cube. These three type of line segments have lengths $1$, $\sqrt{2}$, and $\sqrt{3}$, respectively. Because each vertex of the cube is visited only once, the two line segments that meet at a vertex have a combined length of at most $\sqrt{2}+\sqrt{3}$. Therefore the sum of the lengths of the 8 segments is at most $4\sqrt{2}+4\sqrt{3}$. This maximum is achieved by the path $A \to G \to B \to H \to C \to E \to D \to F \to A.$

答案（D）：苍蝇路径上的 8 段线段分别是立方体的一条棱、一个面的对角线或立方体的空间对角线。这三类线段的长度分别为 $1$、$\sqrt{2}$ 和 $\sqrt{3}$。由于立方体的每个顶点只被经过一次，在同一顶点相接的两段线段的合长度至多为 $\sqrt{2}+\sqrt{3}$。因此这 8 段线段的总长度至多为 $4\sqrt{2}+4\sqrt{3}$。当路径为 $A \to G \to B \to H \to C \to E \to D \to F \to A$ 时可达到该最大值。

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