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Re: spherical geometry (was Re: navigation riddle)

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.

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

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