HMMT 二月 2022 · 冲刺赛 · 第 32 题

HMMT February 2022 — Guts Round — Problem 32

[16] An ant starts at the point (0 , 0) in the Cartesian plane. In the first minute, the ant faces towards (1 , 0) and walks one unit. Each subsequent minute, the ant chooses an angle θ uniformly at random ◦ ◦ in the interval [ − 90 , 90 ], and then turns an angle of θ clockwise (negative values of θ correspond to counterclockwise rotations). Then, the ant walks one unit. After n minutes, the ant’s distance from (0 , 0) 2 is d . Let the expected value of d be a . Compute the closest integer to n n n a n 10 lim . n →∞ n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2022, February 19, 2022 — GUTS ROUND Organization Team Team ID#

解析

[16] An ant starts at the point (0 , 0) in the Cartesian plane. In the first minute, the ant faces towards (1 , 0) and walks one unit. Each subsequent minute, the ant chooses an angle θ uniformly at random ◦ ◦ in the interval [ − 90 , 90 ], and then turns an angle of θ clockwise (negative values of θ correspond to counterclockwise rotations). Then, the ant walks one unit. After n minutes, the ant’s distance from 2 (0 , 0) is d . Let the expected value of d be a . Compute the closest integer to n n n a n 10 lim . n →∞ n Proposed by: Carl Schildkraut Answer: 45 Solution: Let α be a random variable that represents the turn made after step k , choosing α k k uniformly at random on the complex plane among the arc of the unit circle containing 1 from − i to i . 2 It is well known that E [ α ] = . We have that k π " # j − 1 n n i − 1 X X Y Y − 1 a = E α α . n k k i =1 j =1 k =1 k =1 Separating the sum based on | i − j | , n − 1 n − 1 t X X 2 a = n + 2 ( n − t ) E [ α · · · α ] = n + 2 ( n − t ) . n 1 t π t =1 t =1 Since terms with large t get very small, we can write ∞ ∞ ∞ t t t X X X a t 2 t 2 2 n lim = lim 1 + 2 1 − = 1 + 2 lim 1 − = 1 + 2 . n →∞ n →∞ n →∞ n n π n π π t =1 t =1 t =1 This gives that π + 2 x = . π − 2 To estimate this quantity we use the approximation π ≈ 22 / 7, which gives us 22 + 14 x ≈ = 4 . 5 . 22 − 14