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HMMT 二月 2012 · 冲刺赛 · 第 28 题

HMMT February 2012 — Guts Round — Problem 28

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [ 17 ] Alice is sitting in a teacup ride with infinitely many layers of spinning disks. The largest disk hasradius 5. Each succeeding disk has its center attached to a point on the circumference of the previousdisk and has a radius equal to 2 / 3 of the previous disk. Each disk spins around its center (relative tothe disk it is attached to) at a rate of π/ 6 radians per second. Initially, at t = 0, the centers of thedisks 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?
    15 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 11 FEBRUARY 2012 — GUTS ROUNDTH

英文原题

[17]
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15 TH ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 11 FEBRUARY 2012— GUTS ROUND
School Team Team ID#

解析

英文解析

  1. [ 17 ] Alice is sitting in a teacup ride with infinitely many layers of spinning disks. The largest disk hasradius 5. Each succeeding disk has its center attached to a point on the circumference of the previousdisk and has a radius equal to 2 / 3 of the previous disk. Each disk spins around its center (relative tothe disk it is attached to) at a rate of π/ 6 radians per second. Initially, at t = 0, the centers of thedisks 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 geometricseries, we get that Alice’s position is 2
    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.
    So 1 − z traces a circle, and now we need to find the path that traces. In the complex plane,1
    1 − ztaking the reciprocal corresponds to a reflection about the real axis followed by a geometric inversionabout the unit circle centered at 0. It is well known that geometric inversion maps circles not passingthrough 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 circle contains the points 3 and 3 / 5, with the center lying halfway between. So the radius of the 1
    1 − zcircle 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