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*From*: David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>*Date*: Sat, 14 Aug 2004 14:38:41 -0400

Regarding Brian's elaborated explanation of his modified version of

the problem:

Or how about this one:

From a fixed point, head due South, continue in a straight line

for one mile: turn due East and continue in that direction for one

mile, then turn onto a due South heading and continue straight for

one mile, to return to your start point.

Brian Whatcott Altus OK Eureka!

Brian, do you really want two southbound legs? Also, What do you

mean by the phrase "in that direction for one mile" for the 2nd

leg? Do you mean a) along a geodesically straight path, b) always

going due East, c) along a Euclideanly straight path, d) something

else? Do you want the problem performed on the Earth's surface?

David Bowman

A Southbound start, and a constant Easterly heading,

and another Southbound turn on the Earth's surface,

idealized to be spherical, of course.

Brian Whatcott Altus OK Eureka!

Thanks for the clarification. In this case it seems that again

there are an infinite number of solutions starting a fraction of a

mile from the South Pole. The main trick is to realize that once

a person heads South towards the South Pole and continues in a

straight line past the pole one ends up heading North without any

change in direction. Ignoring the tiny curvature effects of the

Earth's surface in the vicinity of the South Pole the solution

set seems to be that one starts about 1 - 1/(2*n*[pi]) miles from

the South pole where n is some positive integer. After traveling

one mile in the same direction and crossing over the pole one

stops 1/(2*n*[pi]) miles from the pole on the other side. One then

travels n times around the pole while traveling East ending up back

at the first turning point for this 2nd 1 mile long leg. One then

travels again one mile back over the pole back to the initial

starting point.

David Bowman

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