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Difference between revisions of "2014 AIME I Problems/Problem 4"

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== Solution ==
 
== Solution ==
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For the purposes of this problem, we will use miles and minutes as our units; thus, the bikers travel at speeds of <math>\dfrac{1}{3}</math> mi/min.
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Let <math>d</math> be the length of the trains, <math>r_1</math> be the speed of train 1 (the faster train), and <math>r_2</math> be the speed of train 2.
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Consider the problem from the bikers' moving frame of reference. In order to pass Jon, the first train has to cover a distance equal to its own length, at a rate of <math>r_1 - \dfrac{1}{3}</math>. Similarly, the second train has to cover a distance equal to its own length, at a rate of <math>r_2 + \dfrac{1}{3}</math>. Since the times are equal and <math>d = rt</math>, we have that <math>\dfrac{d}{r_1 - \dfrac{1}{3}} = \dfrac{d}{r_2 + \dfrac{1}{3}}</math>. Solving for <math>r_1</math> in terms of <math>r_2</math>, we get that <math>r_1 = r_2 + \dfrac{2}{3}</math>.
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Now, let's examine the times it takes the trains to pass Steve. This time, we augment train 1's speed by <math>\dfrac{1}{3}</math>, and decrease train 2's speed by <math>\dfrac{1}{3}</math>. Thus, we have that <math>\dfrac{d}{r_2 - \dfrac{1}{3}} = 10\dfrac{d}{r_1 + \dfrac{1}{3}}</math>.
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Multiplying this out and simplifying, we get that <math>r_1 = 10r_2 - \dfrac{11}{3}</math>. Since we now have 2 expressions for <math>r_1</math> in terms of <math>r_2</math>, we can set them equal to each other:
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<math>r_2 + \dfrac{2}{3} = 10r_2 - \dfrac{11}{3}</math>. Solving for <math>r_2</math>, we get that <math>r_2 = \dfrac{13}{27}. Since we know that it took train 2 1 minute to pass Jon, we know that </math>1 = \dfrac{d}{r_2 + \dfrac{1}{3}}<math>. Plugging in </math>\dfrac{11}{3}<math> for </math>r_2<math> and solving for </math>d<math>, we get that </math>d = \dfrac{22}{27}<math>, and our answer is </math>27 + 22 = 049$.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2014|n=I|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:51, 15 March 2014

Problem 4

Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

For the purposes of this problem, we will use miles and minutes as our units; thus, the bikers travel at speeds of $\dfrac{1}{3}$ mi/min.

Let $d$ be the length of the trains, $r_1$ be the speed of train 1 (the faster train), and $r_2$ be the speed of train 2.

Consider the problem from the bikers' moving frame of reference. In order to pass Jon, the first train has to cover a distance equal to its own length, at a rate of $r_1 - \dfrac{1}{3}$. Similarly, the second train has to cover a distance equal to its own length, at a rate of $r_2 + \dfrac{1}{3}$. Since the times are equal and $d = rt$, we have that $\dfrac{d}{r_1 - \dfrac{1}{3}} = \dfrac{d}{r_2 + \dfrac{1}{3}}$. Solving for $r_1$ in terms of $r_2$, we get that $r_1 = r_2 + \dfrac{2}{3}$.

Now, let's examine the times it takes the trains to pass Steve. This time, we augment train 1's speed by $\dfrac{1}{3}$, and decrease train 2's speed by $\dfrac{1}{3}$. Thus, we have that $\dfrac{d}{r_2 - \dfrac{1}{3}} = 10\dfrac{d}{r_1 + \dfrac{1}{3}}$.

Multiplying this out and simplifying, we get that $r_1 = 10r_2 - \dfrac{11}{3}$. Since we now have 2 expressions for $r_1$ in terms of $r_2$, we can set them equal to each other:

$r_2 + \dfrac{2}{3} = 10r_2 - \dfrac{11}{3}$. Solving for $r_2$, we get that $r_2 = \dfrac{13}{27}. Since we know that it took train 2 1 minute to pass Jon, we know that$1 = \dfrac{d}{r_2 + \dfrac{1}{3}}$. Plugging in$\dfrac{11}{3}$for$r_2$and solving for$d$, we get that$d = \dfrac{22}{27}$, and our answer is$27 + 22 = 049$.

See also

2014 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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