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Difference between revisions of "2007 AIME I Problems/Problem 4"

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== Problem ==
 
== Problem ==
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Three planets orbit a star circularly in the same plane.  Each moves in the same direction and moves at [[constant]] speed.  Their periods are <math>60</math>,<math>84</math>, and <math>140</math>.  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?
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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 ==
 
== Solution ==
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First, take the [[prime factorization]] of all the numbers: <math>\displaystyle 60=2^2*3*5</math>, <math>\displaystyle 84=2^2*3*7</math>, and <math>\displaystyle 140=2^2*5*7</math>. To find out when they are next collinear, you take <math>\displaystyle \frac{lcm}{gcf}</math>, which is <math>\displaystyle \frac{2^2*3*5*7}{2^2}=\frac{420}{4}=105</math>
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Denote the planets <math> \displaystyle A, B, C </math> respectively.  Let <math> \displaystyle a(t), b(t), c(t) </math> denote the angle which each of the respective planets makes with its initial position after <math> \displaystyle t </math> years.  These are given by <math> a(t) = \frac{t \pi}{30} </math>, <math> b(t) = \frac{t \pi}{42} </math>, <math>c(t) = \frac{t \pi}{70}</math>.
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In order for the planets and the central star to be collinear, <math> \displaystyle a(t)</math>, <math> \displaystyle b(t) </math>, and <math> \displaystyle c(t) </math> must differ by a multiple of <math> \displaystyle \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> \displaystyle \pi </math> exactly when <math> \displaystyle t </math> is a multiple of 105, so the planets and the star will next be collinear in 105 years.
  
 
== 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}}

Revision as of 20:27, 18 March 2007

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 $\displaystyle A, B, C$ respectively. Let $\displaystyle a(t), b(t), c(t)$ denote the angle which each of the respective planets makes with its initial position after $\displaystyle t$ 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, $\displaystyle a(t)$, $\displaystyle b(t)$, and $\displaystyle c(t)$ must differ by a multiple of $\displaystyle \pi$. 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 $\displaystyle \pi$ exactly when $\displaystyle t$ is a multiple of 105, so the planets and the star will next be collinear in 105 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
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