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AIME 2007 I · 第 4 题

AIME 2007 I — Problem 4

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Their periods are 60, 84, and 140 years. The three planets and the star are currently collinear. What is the fewest number of years from now that they will all be collinear again?

解析

Solution

Denote the planets A,B,CA, B, C respectively. Let a(t),b(t),c(t)a(t), b(t), c(t) denote the angle which each of the respective planets makes with its initial position after tt years. These are given by a(t)=tπ30a(t) = \frac{t \pi}{30}, b(t)=tπ42b(t) = \frac{t \pi}{42}, c(t)=tπ70c(t) = \frac{t \pi}{70}.

In order for the planets and the central star to be collinear, a(t)a(t), b(t)b(t), and c(t)c(t) must differ by a multiple of π\pi. Note that a(t)b(t)=tπ105a(t) - b(t) = \frac{t \pi}{105} and b(t)c(t)=tπ105b(t) - c(t) = \frac{t \pi}{105}, so a(t)c(t)=2tπ105a(t) - c(t) = \frac{ 2 t \pi}{105}. These are simultaneously multiples of π\pi exactly when tt is a multiple of 105105, so the planets and the star will next be collinear in 105\boxed{105} years.