# AMC 10, question 15

## AMC 10, question 15

Odell and Kershaw run for 30 minutes on a circular track. Odell runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they pass each other?

btw, if anyone knows a good website with full solutions to these kind of contests, please post.

btw, if anyone knows a good website with full solutions to these kind of contests, please post.

**karooomph**- Posts : 74

Join date : 2008-11-19

Age : 22

Location : ubseikastan

## Re: AMC 10, question 15

I found this website, which contains solutions to past AMC 10 problems, but it might not have the solutions to the problems you're looking for (it only has up to 2008, and I'm guessing you took was in 2009).

This turned out to be a pretty cool problem.

First, I thought I'd model model each of their paths, with x1 representing the x-direction path of Odell and x2 representing the x-direction path of Kershaw.

To do this, I needed to convert the rates to appropriate forms. 250m/min * 1lap/(2pi*50m) = 5/(2pi)laps/min; 300m/min * 1lap/(2pi*60m) = 5/(2pi)laps/min (having taken some chemistry helps me to do conversions like this because it makes me think in terms of cancelling and multiplying units). Both of their rates are conveniently 5/(2pi)laps/min, except in opposite directions.

So, I have x1=-50cos(5/(2pi) * 2pi * t); x2=60cos(5/(2pi) * 2pi * t) (I multiply by 2pi afterwards because cosine already has a period of 2pi) - or, x1=-50cos(5t) x2=60cos(5t). At this point, I figured it'd be easier to model the entire thing in polar coordinates, giving: r1=50, θ1=-5t; r2=60, θ2=-5t. We want to find when measurements of the angles are equal (and note that even though the values of θ1 and θ2 won't be equal in this case except when t=0, the angles they represent will still periodically be equal, like how the angle represented by -5*2pi the angle represented by 5*2pi).

Then, by just thinking about the movement of their paths, you can tell that they will pass each other 2 times every lap (once on each opposite side of their paths). Each also runs 5 laps per every multiple of 2pi*t. They run 30 minutes, giving: (30*5*2)/(2pi)~=47.75. You can't really partially pass somebody, so the final answer is 47 times.

This turned out to be a pretty cool problem.

First, I thought I'd model model each of their paths, with x1 representing the x-direction path of Odell and x2 representing the x-direction path of Kershaw.

To do this, I needed to convert the rates to appropriate forms. 250m/min * 1lap/(2pi*50m) = 5/(2pi)laps/min; 300m/min * 1lap/(2pi*60m) = 5/(2pi)laps/min (having taken some chemistry helps me to do conversions like this because it makes me think in terms of cancelling and multiplying units). Both of their rates are conveniently 5/(2pi)laps/min, except in opposite directions.

So, I have x1=-50cos(5/(2pi) * 2pi * t); x2=60cos(5/(2pi) * 2pi * t) (I multiply by 2pi afterwards because cosine already has a period of 2pi) - or, x1=-50cos(5t) x2=60cos(5t). At this point, I figured it'd be easier to model the entire thing in polar coordinates, giving: r1=50, θ1=-5t; r2=60, θ2=-5t. We want to find when measurements of the angles are equal (and note that even though the values of θ1 and θ2 won't be equal in this case except when t=0, the angles they represent will still periodically be equal, like how the angle represented by -5*2pi the angle represented by 5*2pi).

Then, by just thinking about the movement of their paths, you can tell that they will pass each other 2 times every lap (once on each opposite side of their paths). Each also runs 5 laps per every multiple of 2pi*t. They run 30 minutes, giving: (30*5*2)/(2pi)~=47.75. You can't really partially pass somebody, so the final answer is 47 times.

**bfrsoccer**- Administrator
- Posts : 61

Join date : 2008-11-09

## Re: AMC 10, question 15

funion, the pi symbol looks like an n. Any way to change this?

thanks

thanks

**karooomph**- Posts : 74

Join date : 2008-11-19

Age : 22

Location : ubseikastan

## Re: AMC 10, question 15

Well, I guess I could change the default font of the forums, which is the reason that it does look like an n, but I don't think that would work out so well. I forgot about the n thing.

**funion987**- Administrator
- Posts : 146

Join date : 2008-11-09

## Re: AMC 10, question 15

bfrsoccer wrote:I found this website, which contains solutions to past AMC 10 problems, but it might not have the solutions to the problems you're looking for (it only has up to 2008, and I'm guessing you took was in 2009).

thanks, the question was 2006, so I was was able to find it. Plus, you gave a detailed solution, so thanks!

**karooomph**- Posts : 74

Join date : 2008-11-19

Age : 22

Location : ubseikastan

## bfr's solution revised (it's just easier to read for me)

I found this website, which contains solutions to past AMC 10 problems, but it might not have the solutions to the problems you're looking for (it only has up to 2008, and I'm guessing you took was in 2009).

This turned out to be a pretty cool problem.

First, I thought I'd model model each of their paths, with x

To do this, I needed to convert the rates to appropriate forms. 250m/min * 1 lap/(2π*50m) = 5/(2π) laps/min; 300m/min * 1 lap/(2π*60m) = 5/(2π) laps/min (having taken some chemistry helps me to do conversions like this because it makes me think in terms of cancelling and multiplying units). Both of their rates are conveniently 5/(2π) laps/min, except in opposite directions.

So, I have x

Then, by just thinking about the movement of their paths, you can tell that they will pass each other 2 times every lap (once on each opposite side of their paths). Each also runs 5 laps per every multiple of 2π*t. They run 30 minutes, giving: (30*5*2)/(2π)~= 47.75. You can't really partially pass somebody, so the final answer is 47 times.

This turned out to be a pretty cool problem.

First, I thought I'd model model each of their paths, with x

_{1}representing the x-direction path of Odell and x_{2}representing the x-direction path of Kershaw.To do this, I needed to convert the rates to appropriate forms. 250m/min * 1 lap/(2π*50m) = 5/(2π) laps/min; 300m/min * 1 lap/(2π*60m) = 5/(2π) laps/min (having taken some chemistry helps me to do conversions like this because it makes me think in terms of cancelling and multiplying units). Both of their rates are conveniently 5/(2π) laps/min, except in opposite directions.

So, I have x

_{1}= -50cos(5/(2π) * 2π* t); x_{2}= 60cos(5/(2π) * 2π * t) (I multiply by 2π afterwards because cosine already has a period of 2π) - or, x_{1}= -50cos(5t) x_{2}=60cos(5t). At this point, I figured it'd be easier to model the entire thing in polar coordinates, giving: r_{1}=50, θ_{1}= -5t; r_{2}= 60, θ_{2}= -5t. We want to find when measurements of the angles are equal (and note that even though the values of θ_{1}and θ_{2}won't be equal in this case except when t= 0, the angles they represent will still periodically be equal, like how the angle represented by -5*2π the angle represented by 5*2π).Then, by just thinking about the movement of their paths, you can tell that they will pass each other 2 times every lap (once on each opposite side of their paths). Each also runs 5 laps per every multiple of 2π*t. They run 30 minutes, giving: (30*5*2)/(2π)~= 47.75. You can't really partially pass somebody, so the final answer is 47 times.

**karooomph**- Posts : 74

Join date : 2008-11-19

Age : 22

Location : ubseikastan

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**monti**- Posts : 10

Join date : 2010-04-28

## Re: AMC 10, question 15

Mathematics Examination (AHSME). The contest was established in 1950.

There are three levels:

the AMC 8 is for students in grades 8 and below

the AMC 10 is for students in grades 10 and below

the AMC 12 is for students in grades 12 and below

Students who perform well on the AMC 10 or AMC 12 exams are invited to participate in the American Invitational Mathematics Examination. Students who perform well on the American Invitational Mathematics Examination (AIME) are then invited to the United States of America Mathematical Olympiad (USAMO). Students who do exceptionally well on the USAMO (typically around 30 students) are invited to go to the Mathematical Olympiad Summer Program (MOSP or more commonly, MOP), and six students are selected from the top twelve scorers on the USAMO (through yet another exam, the Team Selection Test (TST)) to form the United States Math Team.

American Mathematics Competitions is also the name of the organization, based in Lincoln, Nebraska, responsible for creating, distributing and coordinating the American Mathematics Competitions contests, which include the American Mathematics Contest, AIME, and USAMO.

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There are three levels:

the AMC 8 is for students in grades 8 and below

the AMC 10 is for students in grades 10 and below

the AMC 12 is for students in grades 12 and below

Students who perform well on the AMC 10 or AMC 12 exams are invited to participate in the American Invitational Mathematics Examination. Students who perform well on the American Invitational Mathematics Examination (AIME) are then invited to the United States of America Mathematical Olympiad (USAMO). Students who do exceptionally well on the USAMO (typically around 30 students) are invited to go to the Mathematical Olympiad Summer Program (MOSP or more commonly, MOP), and six students are selected from the top twelve scorers on the USAMO (through yet another exam, the Team Selection Test (TST)) to form the United States Math Team.

American Mathematics Competitions is also the name of the organization, based in Lincoln, Nebraska, responsible for creating, distributing and coordinating the American Mathematics Competitions contests, which include the American Mathematics Contest, AIME, and USAMO.

_______________________________________________________________________

link building service

hearing school

**iinfotech01**- Posts : 1

Join date : 2010-07-29

## Re: AMC 10, question 15

I thought I'd model model each of their paths, with x1 representing the x-direction path of Odell and x2 representing the x-direction path of Kershaw.

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**LeeRain**- Posts : 12

Join date : 2010-09-10

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