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Difference between revisions of "2007 AIME I Problems/Problem 4"
(Undo revision 109540 by Brudder (talk) Solution 2 (LCM/GCF) is wrong; this method would obtain 105 years even if the periods were 6000, 8400, 14000 for example.)
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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 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?
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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 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?
  
 
== Solution ==
 
== Solution ==
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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.
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== Solution 2 ==
 
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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>
 
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-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}}

Latest revision as of 17:46, 11 March 2021

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?

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.

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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