HMMT 二月 2021 · COMB 赛 · 第 6 题

HMMT February 2021 — COMB Round — Problem 6

A light pulse starts at a corner of a reflective square. It bounces around inside the square, reflecting off of the square’s perimeter n times before ending in a different corner. The path of the light pulse, when traced, divides the square into exactly 2021 regions. Compute the smallest possible value of n .

解析

A light pulse starts at a corner of a reflective square. It bounces around inside the square, reflecting off of the square’s perimeter n times before ending in a different corner. The path of the light pulse, when traced, divides the square into exactly 2021 regions. Compute the smallest possible value of n . Proposed by: Krit Boonsiriseth Answer: 129 Solution: The main claim is that if the light pulse reflects vertically (on the left/right edges) a times ( a +2)( b +2) and horizontally b times, then gcd( a + 1 , b + 1) = 1, and the number of regions is . This claim 2 can be conjectured by looking at small values of a and b ; we give a full proof at the end. Assuming the claim, we are trying to find the least possible value of a + b when ( a +2)( b +2) = 2 · 2021 = 2 · 43 · 47. This happens when ( a + 2 , b + 2) = (47 , 86), which also satisfies gcd( a + 1 , b + 1) = 1, and gives a + b = 47 + 86 − 4 = 129. We now prove the claim. Imagine that at each reflection, it is the square that gets reflected instead. Then the path p of the light pulse becomes a straight segment s from (0 , 0) to ( a + 1 , b + 1) of slope a +1

m = . b +1 • The square starts as 1 region; the light pulse hitting a corner at the end creates 1 more region. • Each reflection of the light pulse creates a region. These correspond to intersections of s with a line x = n or y = n for x ∈ [ a ] , y ∈ [ b ]. There are a + b such intersections. • Each self-intersection of p creates a region. An intersection on p corresponds to two on s , and each intersection of s happens with a line of slope − m passing through an even integral point, i.e. a line of the form ( b + 1) x + ( a + 1) y = 2 k . The open segment s intersects these lines for k ∈ [ ab + a + b ]. However, the a + b intersections that happens on a gridline x ∈ Z or y ∈ Z do not count, so here we have an additional ab/ 2 regions. Therefore, the total number of regions is ab ( a + 2)( b + 2) 2 + a + b + = . 2 2