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Where Will the Runners Meet?Date: 03/29/99 at 05:09:43 From: Jason Mize Subject: Runners and their Intersection Point Two runners, A and B, start 90 degrees away from each other on a circular track and run at the same speed. If Runner B decides to cut across the track, where will they meet? I have been given that the radius of the circular track is one mile. What do I need to do to figure out where on the track they will meet? I understand that C = 2(pi)r, in this case, 2(pi). Therefore, each quarter or quadrant's arc is equal to .5(pi). Can you help me solve this one?
Date: 03/29/99 at 10:37:41
From: Doctor Rob
Subject: Re: Runners and their Intersection Point
I'll start by drawing a picture:
A
_..--+--.._
.-' | `-.
,' | `.
,' | `.
/ | \
/ | \
. | .
| |O |
+--------------+-------------,+ B
| / \ ,-' |
. / \ ,-' .
\ / \ ,-' /
\ / ,-o /
`. / ,-' Q ,'
`. / ,-' .'
`/._ _.-'
P ''-----''
Measure all angles in radians. Then let
<AOP = theta
<BOP = 3*Pi/2 - theta
<BOQ = <POQ = 3*Pi/4 - theta/2
BQ = QP = sin(3*Pi/4 - theta/2)
arc(AP) = theta
BP = 2*sin(3*Pi/4 - theta/2)
According to the conditions of the problem arc(AP) = BP, so
theta = 2*sin(3*Pi/4 - theta/2)
theta/2 = sin(3*Pi/4 - theta/2)
Now I let x = theta/2. So,
x = sin(3*Pi/4 - x)
This is not solvable explicitly for x. One can, of course, solve it
numerically. One pretty quick way is to guess the value of x (I
started with x[1] = 1.), and use the above formula to compute the next
guess:
x[n + 1] = sin(3*Pi/4 - x[n]), n > 0.
This kind of iteration is easy to do on a pocket calculator. Continue
until x[n + 1] - x[n] is smaller than the error you can tolerate. Then
the value of x[n + 1] is equal to the solution x, to within the
tolerance.
This converges pretty rapidly to the solution x, and then theta = 2*x
gives you the distance in miles each traveled. If you put the origin
of your coordinate system at the center of the circle, with A = (0, 1)
and B = (1, 0), then the place they meet is
P = (cos(theta + Pi/2), sin(theta + Pi/2))
= (-sin(theta), cos(theta))
- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
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