Re: spherical geometry (was Re: navigation riddle)

[Jack Uretsky <jlu@HEP.ANL.GOV>]

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

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