HMMT 二月 2010 · 冲刺赛 · 第 17 题
HMMT February 2010 — Guts Round — Problem 17
题目详情
- [ 9 ] An ant starts at the origin, facing in the positive x -direction. Each second, it moves 1 unit forward, − 1 3 then turns counterclockwise by sin ( ) degrees. What is the least upper bound on the distance 5 between the ant and the origin? (The least upper bound is the smallest real number r that is at least as big as every distance that the ant ever is from the origin.) 1
解析
- [ 9 ] An ant starts at the origin, facing in the positive x -direction. Each second, it moves 1 unit forward, − 1 3 then turns counterclockwise by sin ( ) degrees. What is the least upper bound on the distance 5 between the ant and the origin? (The least upper bound is the smallest real number r that is at least as big as every distance that the ant ever is from the origin.) √ √ 10 Answer: 10 We claim that the points the ant visits lie on a circle of radius . We show this 2 √ 10 1 3 by saying that the ant stays a constant distance from the point ( , ). 2 2 2 ′ Suppose the ant moves on a plane P . Consider a transformation of the plane P such that after the first ′ ′ ′ move, the ant is at the origin of P and facing in the direction of the x axis (on P ). The transformation ′ to get from P to P can be gotten by rotating P about the origin counterclockwise through an angle − 1 3 1 3 sin ( ) and then translating it 1 unit to the right. Observe that the point ( , ) is fixed under this 5 2 2 1 3 4 3 1 3 transformation, which can be shown through the expression ( + i )( + i ) + 1 = + i . It follows 2 2 5 5 2 2 1 3 that at every point the ant stops, it will always be the same distance from ( , ). Since it starts at 2 2 √ 10 (0 , 0), this fixed distance is . 2 Guts Round 1 3 ( , ) 2 2 ′ x x − 1 3 Since sin ( ) is not a rational multiple of π , the points the ant stops at form a dense subset of the 5 circle in question. As a result, the least upper bound on the distance between the ant and the origin √ is the diameter of the circle, which is 10. 1