2007 AIME I Problems/Problem 4

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. 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?

Solution

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

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

Solution 2

This problem is trivialized since the answer is the LCM/GCF of $60,84,140$. This is just $420/4$ which is $\boxed{105}$ -Brudder

See also

2007 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS