PUMaC 2012 · 几何（A 组） · 第 6 题
PUMaC 2012 — Geometry (Division A) — Problem 6
题目详情
- [ 6 ] Consider a pool table with the shape of an equilateral triangle. A ball of negligible size is initially placed at the center of the table. After it has been hit, it will keep moving in the direction it was hit towards and bounce off any edges with perfect symmetry. If it eventually reaches the midpoint of any edge, we mark the midpoint of the entire route that the ball has travelled through. Repeating this experiment, how many points can we mark at most?
解析
- [ 6 ] Consider a pool table with the shape of an equilateral triangle. A ball of negligible size is initially placed at the center of the table. After it has been hit, it will keep moving in the direction it was hit towards and bounce off any edges with perfect symmetry. If it eventually reaches the midpoint of any edge, we mark the midpoint of the entire route that the ball has travelled through. Repeating this experiment, how many points can we mark at most? Solution: In the following diagram, the original triangle is the small triangle on the lower left corner of the left diagram, and note that two of its sides have been turned into vectors. 3 A We firstly consider the midpoint A of one edge. We can keep reflecting the triangle along any edge, and the line connecting A and the center of any triangle in the plane represents a route between A and the center in the original triangle. Define two vectors as shown in the diagram on the left: the horizontal one is 1, and the other 1 1 one is v . A can be denoted as + v . For the triangles heading upwards (as in, triangles with 2 2 ( ) 1 1 the same orientation as the lower-left triangle), the center can be denoted as a + + b + v 3 3 ( ) 1 5 1 5 with any integers a, b . Hence the midpoint of the route is a + + b + v . We can remove 2 12 2 12 5 5 11 5 5 the integer parts, so there are four possible vectors starting at any vertex: + v, + v, + 12 12 12 12 12 11 11 11 v, + v . But we need to consider the midpoints of all three different edges, and reflect 12 12 12 them all back to the original triangle. Then we obtain the 12 points above. For the triangles heading downwards (as in, with the same orientation as the middle triangle ( ) 2 2 in the left diagram), the center can be denoted as a + + b + v . Eventually we obtain the 3 3 same 12 points. This conclusion also implies that 12 points are enough to block all such routes. Problem contributed by Xufan Zhang