2021 AMC 12B Fall 第 17 题

先试着解答 2021 AMC 12B Fall 第 17 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2021 AMC 12B Fall 解答,或核对答案

所有题目均经美国数学协会(MAA)官方合法授权使用。

17.

一只虫子从由边长为 11 的等边三角形组成的网格的一个顶点出发。每一步,虫子都沿网格线在 66 个可能方向中随机且独立地等概率移动。经过 55 步后,虫子在整个过程中从未距离起点超过 11 个单位的概率是多少?

A bug starts at a vertex of a grid made of equilateral triangles of side length 1.1. At each step the bug moves in one of the 66 possible directions along the grid lines randomly and independently with equal probability. What is the probability that after 55 moves the bug never will have been more than 11 unit away from the starting position?

13108\dfrac{13}{108}

754\dfrac{7}{54}

29216\dfrac{29}{216}

427\dfrac{4}{27}

116\dfrac{1}{16}

答案:A
知识点:随机游走递推概率
难度评级:2230
小提示:

虫子必须始终在起点或起点的 66 个相邻顶点上。

The bug must always be at the start or one of its 66 neighbors

大提示:

从一个相邻顶点出发,66 个方向中只有 33 个仍保持在距离 11 以内:回到起点或到两个相邻的邻点。

From a neighbor, only 33 of the 66 moves stay within distance 11: back to start or to the two adjacent neighbors

解答:

保持在距离 11 以内意味着虫子始终在原点或其 66 个相邻顶点上。从原点出发,所有 66 个方向都可行;从相邻顶点出发,只有 33 个方向可行:回到原点,或到两个相邻的邻点之一。

aka_kbkb_k 分别表示有效的 kk 步路径中,结束在原点和结束在某个相邻点的路径数。则 ak+1=bka_{k+1} = b_kbk+1=6ak+2bkb_{k+1} = 6a_k + 2b_k,初值为 a0=1a_0 = 1b0=0b_0 = 0

迭代得 b1=6b_1 = 6(a2,b2)=(6,12)(a_2, b_2) = (6, 12)(a3,b3)=(12,60)(a_3, b_3) = (12, 60)(a4,b4)=(60,192)(a_4, b_4) = (60, 192)(a5,b5)=(192,744)(a_5, b_5) = (192, 744)。有效路径总数为 192+744=936192 + 744 = 936

所求概率为 93665=9367776=13108\dfrac{936}{6^5} = \dfrac{936}{7776} = \dfrac{13}{108}

所以正确答案是 A

Staying within distance 11 means the bug is always at the origin or one of its 66 neighbors. From the origin, all 66 moves are allowed. From a neighbor, only 33 moves keep it in range: back to the origin, or to either of the two adjacent neighbors.

Let aka_k and bkb_k count valid kk-step paths ending at the origin and at a neighbor. Then ak+1=bka_{k+1} = b_k and bk+1=6ak+2bk,b_{k+1} = 6a_k + 2b_k, starting from a0=1,a_0 = 1, b0=0.b_0 = 0.

Iterating gives b1=6,b_1 = 6, then (a2,b2)=(6,12),(a_2, b_2) = (6, 12), (a3,b3)=(12,60),(a_3, b_3) = (12, 60), (a4,b4)=(60,192),(a_4, b_4) = (60, 192), (a5,b5)=(192,744).(a_5, b_5) = (192, 744). The total number of valid paths is 192+744=936.192 + 744 = 936.

The probability is 93665=9367776=13108.\dfrac{936}{6^5} = \dfrac{936}{7776} = \dfrac{13}{108}.

Thus, the correct answer is A.

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