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.

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.

Dude, bfr, don't forget:

http://chemistry.forumotion.com/Useful-Symbols-h5.htm

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

thanks

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.

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!

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 * 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 x₁= -50cos(5/(2π) * 2π* t); x₂= 60cos(5/(2π) * 2π * t) (I multiply by 2π afterwards because cosine already has a period of 2π) - 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*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.