HMMT 二月 2012 · 冲刺赛 · 第 28 题
HMMT February 2012 — Guts Round — Problem 28
题目详情
- [ 17 ] Alice is sitting in a teacup ride with infinitely many layers of spinning disks. The largest disk has radius 5. Each succeeding disk has its center attached to a point on the circumference of the previous disk and has a radius equal to 2 / 3 of the previous disk. Each disk spins around its center (relative to the disk it is attached to) at a rate of π/ 6 radians per second. Initially, at t = 0, the centers of the disks are aligned on a single line, going outward. Alice is sitting at the limit point of all these disks. After 12 seconds, what is the length of the trajectory that Alice has traced out? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TH 15 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 11 FEBRUARY 2012 — GUTS ROUND
解析
- [ 17 ] Alice is sitting in a teacup ride with infinitely many layers of spinning disks. The largest disk has radius 5. Each succeeding disk has its center attached to a point on the circumference of the previous disk and has a radius equal to 2 / 3 of the previous disk. Each disk spins around its center (relative to the disk it is attached to) at a rate of π/ 6 radians per second. Initially, at t = 0, the centers of the disks are aligned on a single line, going outward. Alice is sitting at the limit point of all these disks. After 12 seconds, what is the length of the trajectory that Alice has traced out? Answer: 18 π Suppose the center of the largest teacup is at the origin in the complex plane, and 2 15 πit/ 6 πit/ 6 let z = e . The center of the second disk is at 5 e at time t ; that is, z . Then the center of 3 2 15 2 the third disk relative to the center of the second disk is at z , and so on. Summing up a geometric 2 series, we get that Alice’s position is 15 15 15 2 3 2 3 ( z + z + z + · · · ) = (1 + z + z + · · · ) − 2 2 2 ( ) 15 1 15 = − . 2 1 − z 2 Guts Now, after 12 seconds, z has made a full circle in the complex plane centered at 0 and of radius 2 / 3. Thus 1 − z is a circle centered at 1 of radius 2 / 3. 1 So 1 − z traces a circle, and now we need to find the path that traces. In the complex plane, 1 − z taking the reciprocal corresponds to a reflection about the real axis followed by a geometric inversion about the unit circle centered at 0. It is well known that geometric inversion maps circles not passing through the center of the inversion to circles. Now, the circle traced by 1 − z contains the points 1 − 2 / 3 = 1 / 3, and 1 + 2 / 3 = 5 / 3. Therefore the 1 circle contains the points 3 and 3 / 5, with the center lying halfway between. So the radius of the 1 − z circle is ( ) 1 3 6 3 − = 2 5 5 and so the perimeter is 2 π (6 / 5) = 12 π/ 5. Scaling by 15 / 2 gives an answer of ( ) 15 12 π = 18 π. 2 5