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2017 AMC 12A 第 22 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

22.

在笛卡尔坐标平面中画一个正方形，顶点为 (2,2)(2,2)、(−2,2)(-2,2)、(−2,−2)(-2,-2) 和 (2,−2)(2,-2)。一个粒子从 (0,0)(0,0) 出发。每一秒，它等概率地移动到离当前位置最近的八个格点（坐标均为整数的点）之一，且与之前的移动相互独立。换句话说，粒子从 (x,y)(x,y) 移动到 (x,y+1)(x,y+1)、(x+1,y+1)(x+1,y+1)、(x+1,y)(x+1,y)、(x+1,y−1)(x+1,y-1)、(x,y−1)(x,y-1)、(x−1,y−1)(x-1,y-1)、(x−1,y)(x-1,y) 或 (x−1,y+1)(x-1,y+1) 中每一点的概率都是 18\dfrac{1}{8}。粒子最终会第一次碰到这个正方形，碰到的位置要么是正方形的 44 个顶点之一，要么是某条边内部的 1212 个格点之一。它碰到顶点而不是边内部点的概率为 mn\dfrac{m}{n}，其中 mm 和 nn 是互质正整数。m+nm+n 是多少？

A square is drawn in the Cartesian coordinate plane with vertices at (2,2),(2,2), (−2,2),(-2,2), (−2,−2),(-2,-2), and (2,−2).(2,-2). A particle starts at (0,0).(0,0). Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is 18\dfrac{1}{8} that the particle will move from (x,y)(x,y) to each of (x,y+1),(x,y+1), (x+1,y+1),(x+1,y+1), (x+1,y),(x+1,y), (x+1,y−1),(x+1,y-1), (x,y−1),(x,y-1), (x−1,y−1),(x-1,y-1), (x−1,y),(x-1,y), or (x−1,y+1).(x-1,y+1). The particle will eventually hit the square for the first time, either at one of the 44 corners of the square or at one of the 1212 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is mn,\dfrac{m}{n}, where mm and nn are relatively prime positive integers. What is m+n?m+n?

44

55

77

1515

3939

答案：E
知识点：随机游走递推概率对称性
难度评级：2270
小提示：

利用对称性，只追踪三类内部状态：中心、四个轴向邻点和四个对角邻点

By symmetry, track only three interior states: the center, the four edge-midpoint neighbors, and the four diagonal neighbors

大提示：

对每类状态写出碰到顶点的概率方程，并解这个线性方程组求中心处的值

Write hitting-probability equations for each state and solve the linear system for the center’s value

解答：

由对称性，把相关内部点分成三类：C={(0,0)}C=\{(0,0)\}、“轴向”点 A={(±1,0),(0,±1)}A=\{(\pm1,0),(0,\pm1)\} 和“对角”点 I={(±1,±1)}I=\{(\pm1,\pm1)\}。设从 A,C,IA,C,I 三类点出发最终碰到顶点的概率分别为 a,c,ia,c,i。

读出转移概率（例如，AA 中一点以概率 28\tfrac28 到 AA，以 18\tfrac18 到 CC，以 28\tfrac28 到 II，并以 38\tfrac38 到边内部，等等），得到 a=28a+18c+28i,c=48a+48i,i=28a+18c+18。 \begin{aligned} &a=\tfrac28 a+\tfrac18 c+\tfrac28 i,\quad \\ &c=\tfrac48 a+\tfrac48 i,\quad \\ &i=\tfrac28 a+\tfrac18 c+\tfrac18 \end{aligned}\text{。}

解得 a=114a=\dfrac{1}{14}，c=435c=\dfrac{4}{35}，i=1170i=\dfrac{11}{70}。所求概率是 c=435c=\dfrac{4}{35}，所以 m+n=4+35=39m+n=4+35=39。

所以正确答案是 E。

By symmetry, group the relevant interior points into three types: C={(0,0)},C=\{(0,0)\}, the “axis” points A={(±1,0),(0,±1)},A=\{(\pm1,0),(0,\pm1)\}, and the “diagonal” points I={(±1,±1)}.I=\{(\pm1,\pm1)\}. Let a,c,ia,c,i be the probabilities of eventually hitting a corner starting from a point of type A,C,I.A,C,I.

Reading off the transition probabilities (a point in AA goes to AA with prob 28,\tfrac28, to CC with 18,\tfrac18, to II with 28,\tfrac28, and to a side interior with 38,\tfrac38, etc.) gives a=28a+18c+28i,c=48a+48i,i=28a+18c+18. \begin{aligned} &a=\tfrac28 a+\tfrac18 c+\tfrac28 i,\quad \\ &c=\tfrac48 a+\tfrac48 i,\quad \\ &i=\tfrac28 a+\tfrac18 c+\tfrac18. \end{aligned}

Solving yields a=114,a=\dfrac{1}{14}, c=435,c=\dfrac{4}{35}, i=1170.i=\dfrac{11}{70}. The required probability is c=435,c=\dfrac{4}{35}, so m+n=4+35=39.m+n=4+35=39.

Thus, the correct answer is E.

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