Difference between revisions of "2007 AIME I Problems/Problem 4"
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In order for the planets and the central star to be collinear, <math>a(t)</math>, <math>b(t) </math>, and <math>c(t) </math> must differ by a multiple of <math>\pi </math>. Note that <math> a(t) - b(t) = \frac{t \pi}{105}</math> and <math> b(t) - c(t) = \frac{t \pi}{105}</math>, so <math> a(t) - c(t) = \frac{ 2 t \pi}{105} </math>. These are simultaneously multiples of <math>\pi </math> exactly when <math>t </math> is a multiple of <math>105</math>, so the planets and the star will next be collinear in <math>\boxed{105}</math> years. | In order for the planets and the central star to be collinear, <math>a(t)</math>, <math>b(t) </math>, and <math>c(t) </math> must differ by a multiple of <math>\pi </math>. Note that <math> a(t) - b(t) = \frac{t \pi}{105}</math> and <math> b(t) - c(t) = \frac{t \pi}{105}</math>, so <math> a(t) - c(t) = \frac{ 2 t \pi}{105} </math>. These are simultaneously multiples of <math>\pi </math> exactly when <math>t </math> is a multiple of <math>105</math>, so the planets and the star will next be collinear in <math>\boxed{105}</math> years. | ||
| + | |||
| + | == Solution 2 == | ||
| + | This problem is trivialized since the answer is the LCM/GCF of <math>60,84,140</math>. This is just <math>420/4</math> which is <math>\boxed{105}</math> | ||
| + | -Brudder | ||
== 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}} | {{MAA Notice}} | ||
Revision as of 21:57, 29 August 2019
Contents
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.
Solution 2
This problem is trivialized since the answer is the LCM/GCF of 
. This is just 
which is
-Brudder
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. 
