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*From*: Jack Uretsky <jlu@HEP.ANL.GOV>*Date*: Fri, 13 Aug 2004 17:02:37 -0500

I think that the interesting part of the original problem is that it

requires the solver to realize that plane geometry is not involved. The

path has two corners and three equal legs, returning to the starting point.

On a plane surface, this must be an equilateral triangle. But the

problem, stated on a plane surface, would make two of the legs parallel -

impossible on a plane surface. So we need a surface where either a

"southern" path is not parallel to a northern path, or two of the corners

are oincident - impossible on a planar surface.

Regards,

Jack

On Fri, 13 Aug 2004, David Bowman wrote:

Joel R

This discussion reminds me of a somewhat related problem. The

Northern Hemisphere solution can be considered as a closed path that

is, to a good approximation, a closed circular sector whose wedge

angle is 1 radian. The 2nd leg of the path is a circular arc of

approximately 1 radian in turning angle and the 1st and 3rd legs of

the path are two (geodesically straight) radii connecting the arc

ends to the center point of the arc.

Suppose we straightened out the 2nd leg of the path so *all three*

legs of the path are geodesically straight and the length of the 2nd

leg is the same as the length of the 1st and 3rd legs. The whole

closed path is now an equilateral triangle as inscribed onto the

spherical surface. The problem is to find a formula for the measure

of the interior angle of such an equilateral triangle as a function

of the length s of the sides of the triangle (conveniently in units

of the sphere's radius). A few hints are that 1) the value of the

formula must boil down to 60 deg in the limit of s becoming a

zeroth fraction of the sphere's radius, 2) the value of the formula

becomes 90 deg when s is 1/4 of the circumference of the sphere,

3) the maximum size triangle occurs for a great circle with 3

equally-spaced vertices (120 deg apart from each other) on it with

the interior angle at each vertex being 180 deg across the vertex and

each side having a length s of 1/3 of the sphere's circumference, and

4) the messy intermediate math eventually simplifies to a relatively

simplified formula in the general case.

For a lot of extra credit points you can also find the proper formula

for the *area* of this spherical equilateral triangle in terms of the

length s of the sides of the triangle (making sure that the formula

boils down to all the correct values for the variously known special

cases).

David Bowman

--

"Trust me. I have a lot of experience at this."

General Custer's unremembered message to his men,

just before leading them into the Little Big Horn Valley

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