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# Re: spherical geometry (was Re: navigation riddle)

Regarding Jack's question:

Hi all-
I've skipped some messages, so mebbe the problem changed.
But the last version I saw had an east-directed component (line of
constant latitude). Latitude lines are not -with one exception -
great circles (geodesics).
John Denker wrote, in part:
Here's the outline:

1) The legs of the triangle are by definition geodesically
straight.
2) That means each leg is a piece of a great circle, since
we are dealing with a spherical world.

Have I missed something?

and Clarence's question:

Wait a minute.

Didn't we begin by walking East a mile or so from a pole?

That's not a Great circle.

When did I miss a turn off track?

Yes. Both of you missed something. The history of the problem is
that first John Denker proposed the original (version 1) problem:

Dr. Livingstone starts out at a place which we call
Point A. He then undertakes a journey consisting
of three legs:
-- he travels precisely southward for one mile;
-- then he turns and travels precisely eastward
for one mile;
-- then he turns and travels precisely northward
for one mile.
He discovers that as a result of this journey, he
has returned to Point A.
Note: He travels by airship, at an appropriate constant
altitude, so you don't need to worry about obstructions
or other nonidealities.
The questions are:
1) Where is Point A?
2) Are you sure? How do you know?

As both Jack and Clarence note travelling eastward is not a
geodesically straight path (i.e. a great circle unless one is on the
equator).

Then after the solution set to this problem was posted Joel Rauber
followed it up with a restated version 2 that seemed to forbid the
southern hemisphere solution set (as no bears live in Antarctica,
but I'm not sure if any live within a mile of the north pole either):

If you are old enough, the question is more or less identical to a
seventies. And the description of the problem was followed by the
question: What color are the bears?

After this I proposed the modified problem (version 3a) that *did*
have all three legs of the trip geodesically straight:

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.

and included the extra credit part (version 3b):

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).

After this Brian Whatcott proposed version 4:

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

I wasn't sure what Brian meant by the phrase "turn due East and
continue in that direction" whether "that direction" mean always
east or whether one continued on a geodesically straight path
after making the eastward turn although that direction eventually
ceased being eastward. Brian clarified the situation by saying that
he meant the 2nd leg was supposed to be "a constant Easterly

Now the version to which John Denker was responding above and to
which both Jack and Clarence asked about was version 3 (a & b).
This version and *only* this version *does* have all 3 legs
geodesically straight (segments of great circles). BTW, Carl's
responses also referred to version 3.

Maybe in the future if anyone who responds to any of these problems
that person might want to identify the version number of the problem
they are responding to prevent further mix-ups. Of course it
wouldn't have hurt to give each problem version its own distinct
subject header in the first place.

David Bowman