HMMT 十一月 2020 · GEN 赛 · 第 9 题
HMMT November 2020 — GEN Round — Problem 9
题目详情
- In the Cartesian plane, a perfectly reflective semicircular room is bounded by the upper half of the unit circle centered at (0 , 0) and the line segment from ( − 1 , 0) to (1 , 0). David stands at the point ◦ ( − 1 , 0) and shines a flashlight into the room at an angle of 46 above the horizontal. How many times does the light beam reflect off the walls before coming back to David at ( − 1 , 0) for the first time?
解析
- In the Cartesian plane, a perfectly reflective semicircular room is bounded by the upper half of the unit circle centered at (0 , 0) and the line segment from ( − 1 , 0) to (1 , 0). David stands at the point ◦ ( − 1 , 0) and shines a flashlight into the room at an angle of 46 above the horizontal. How many times does the light beam reflect off the walls before coming back to David at ( − 1 , 0) for the first time? Proposed by: Kevin Tong Answer: 65 Solution: Note that when the beam reflects off the x -axis, we can reflect the entire room across the x -axis instead. Therefore, the number of times the beam reflects off a circular wall in our semicircular room is equal to the number of times the beam reflects off a circular wall in a room bounded by the unit circle centered at (0 , 0). Furthermore, the number of times the beam reflects off the x -axis wall in our semicircular room is equal to the number of times the beam crosses the x -axis in the room bounded by the unit circle. We will count each of these separately. We first find the number of times the beam reflects off a circular wall. Note that the path of the beam is made up of a series of chords of equal length within the unit circle, each chord connecting the points from two consecutive reflections. Through simple angle chasing, we find that the angle ◦ subtended by each chord is 180 − 2 · 46 = 88 . Therefore, the n th point of reflection in the unit circle is ( − cos(88 n ) , sin(88 n )). The beam returns to ( − 1 , 0) when 88 n ≡ 0 (mod 360) ⇐⇒ 11 n ≡ 0 (mod 45) → n = 45 but since we’re looking for the number of time the beam is reflected before it comes back to David, we only count 45 − 1 = 44 of these reflections. Next, we consider the number of times the beam is reflected off the x -axis. This is simply the number of times the beam crosses the x -axis in the unit circle room before returning to David, which happens 88 · 45 ◦ every 180 around the circle. Thus, we have − 1 = 21 reflections off the x -axis, where we subtract 180 1 to remove the instance when the beam returns to ( − 1 , 0). Thus, the total number of reflections is 44 + 21 = 65.