Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

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/   
    
Associated Topics:
High School Conic Sections/Circles
High School Coordinate Plane Geometry
High School Geometry
High School Trigonometry

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/