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*From*: David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>*Date*: Sun, 15 Aug 2004 14:44:22 -0400

Regarding Carl's solution of my spherical geometry challenge problem:

David Bowman wrote:

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.

Resisting the temptation to look up any references on this

challenge, I have come up with what I believe to be the solution

(where A = interior angle of interest):

cos(A) = tan(s/2)/tan(s)

This satisfied hints 1 through 4, so I suspect it's correct.

Yes, your formula is correct. However I prefer to express it in a

somewhat simplified (yet equivalent) form that doesn't involve the

evaluation of 2 separate trig functions of different arguments on

the RHS. My version is:

cos(A) = 1/(1 + 1/cos(s)) .

I don't remember having seen this formula before, so it was a neat

problem.

I liked it too.

I set up the triangle with its base on the equator and its top

vertex on the Greenwich meridian, to make the derivation simpler.

That is how I attacked it also.

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).

I find the following integral formula but I don't have a symbolic

math program at home to see what it evaluates to:

area = 2 * integral from 0 to MAX of

{arcsin[tan(X)/tan(A)] * sin(X) * dX}

where MAX = arcsec[cos(s/2)/cos(s)]

I don't recognize your formulation. The integral expression I got

was substantially different. I don't know if you version boils down

to mine or not. I'm not about to try transforming your version in

various ways to see if it becomes my solution. When you get the

result in a closed form report back and then we'll compare results.

BTW, I found the extra credit problem *much* more challenging than

the first part.

Carl

David Bowman

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