HMMT 二月 2026 · 冲刺赛 · 第 10 题
HMMT February 2026 — Guts Round — Problem 10
题目详情
- [7] Srinivas picks a uniformly random direction and shoots a laser starting at point (0 , 1) at his chosen direction. The laser bounces off the graph of y = | x | whenever it makes contact. Compute the expected number of times the laser contacts the graph of y = | x | . (When the laser bounces, the angle at which it arrives mirrors the angle at which it departs. See the diagram below.) (0 , 1)
解析
- [7] Srinivas picks a uniformly random direction and shoots a laser starting at point (0 , 1) at his chosen direction. The laser bounces off the graph of y = | x | whenever it makes contact. Compute the expected number of times the laser contacts the graph of y = | x | . (When the laser bounces, the angle at which it arrives mirrors the angle at which it departs. See the diagram below.) (0 , 1) Proposed by: Sam EnMin Huang Answer: 1 ◦ Solution: Let θ be the angle of the laser relative to the x axis. Note that we may assume − 90 ≤ θ ≤ ◦ 90 , since by symmetry, the expected value of the number of bounces given that θ is on the right side of the y axis is the same as the expected value given that θ is on the left side. ©2026 HMMT ◦ ◦ • If 45 ≤ θ ≤ 90 , then the laser never hits the graph of | x | , because the angle of | x | for positive ◦ x is 45 ≤ θ . Thus, there are 0 bounces in this case. ◦ ◦ • If 0 ≤ θ < 45 , the laser bounces once, then stays on the same side of the y -axis but doesn’t hit anything forever: after bouncing once, the new angle of the laser is ◦ ◦ ◦ 45 < 90 − θ ≤ 90 . Thus, there is 1 bounce in this case. ◦ ◦ • If − 45 ≤ θ < 0 , the laser bounces once and eventually gets to the other side of the y axis but never hits anything: after bouncing once, the new angle of the laser is: ◦ ◦ ◦ 90 < 90 − θ ≤ 135 . Thus, there is 1 bounce in this case. ◦ ◦ • If − 90 ≤ θ < − 45 , the laser first bounces off the graph of | x | where x is positive. The new angle is: ◦ ◦ ◦ 135 < 90 − θ ≤ 180 . Thus, there are 2 bounces in this case. Since all cases are equally likely, the expected value is 1 · (0 + 1 + 1 + 2) = 1 . 4