Difference between revisions of "2007 AIME I Problems/Problem 4"
I like pie (talk | contribs) m |
|||
| Line 10: | Line 10: | ||
== See also == | == See also == | ||
{{AIME box|year=2007|n=I|num-b=3|num-a=5}} | {{AIME box|year=2007|n=I|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Revision as of 19:09, 4 July 2013
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 
respectively. Let 
denote the angle which each of the respective planets makes with its initial position after 
years. These are given by 
, 
, 
.
In order for the planets and the central star to be collinear, 
, 
, and 
must differ by a multiple of 
. Note that 
and 
, so 
. These are simultaneously multiples of exactly when is a multiple of 
, so the planets and the star will next be collinear in years.
See also
| 2007 AIME I (Problems • Answer Key • Resources) | ||
| 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. 
