Last time, I posted a question sent to me by Rob Mount.  So what solutions exist other than the north pole?

You did well: jnelso99 got part of the way there; tanoshimi finished it off.  We can find a set of circles around the south pole with lengths of 1/N miles: 1 mile, 1/2 mile, 1/3 mile, etc.  Any point one mile north of any of these circles will work: you travel south one mile to one of the circles, circle the earth N times, and then head back north to where you started.

Where exactly are these circles?  Well, let’s be a bit mathematical about it.  Let R be the radius of the earth, and let x be the distance of each leg of our path.  Working in radians, let p0 = (lat, long) be our starting point.  (Radians just make the math so much nicer.)

First we walk south x units.  This only changes the latitude, and so we find ourselves at a new point:

p1 = (latx/R, long)

Next we walk x units east, which only impacts the longitude.  How much this affects the longitude depends, however, on our current latitude:

p2 = (latx/R, long + x/cos(latx/R))

Finally we walk x units north.  This brings the latitude back to where we started:

pf = (lat, long + x/(R * cos(latx/R)))

Okay, so what do we make of this?  We want p0 to equal pf.  The latitude is easy, since the latitudes of p0 and pf are the same.  Further, we note that if lat is positive or negative pi/2 then we’re at a pole and the value of long is irrelevant.  Positive pi/2 gives us the north pole solution; –pi/2 ends up being nonsense, since you can’t go south from the south pole.

But if lat is not pi/2, then long matters.  In particular, our starting longitude has to equal our ending, but it can wrap around, so:

long + 2 * pi * N = long + x / (R * cos(latx/R))

2 * pi * N = x / (R * cos(latx/R))

Here N is any integer; the 2 * pi * N factor accounts for the fact that we can circle the globe.  Solving for lat we find:

cos(lat – x/R) = x / ( 2 * pi * N * R )

lat – x/R = acos( x / ( 2 * pi * N * R) )

lat = acos( x / ( 2 * pi * N * R) ) + x/R

The acos term is undefined if N = 0, and if N < 1 then the value lies outside of the range [–pi/2, pi/2].  When N is greater than 0, then the results of the acos term are in our range, but they could be positive or negative; it turns out that the values we want are negative.

So what does this mean?  We have derived a set of latitude values that look like solutions, and the longitudes associated with these solutions are unconstrained.  In other words, we have a set of circles around the south pole corresponding to each N in {1, 2, …}.  These are exactly the circles we’ve already described, but now we can compute exactly where they are.  For  R = 3963 miles and x = 1:

N=1: lat = -acos( 1 / (24900.263) ) + 0.000252 = -1.5705041 = -89.983260 degrees

N=2: lat = -acos( 1 / (49800.527) ) + 0.000252 = -1.5705242 = -89.984410 degrees

N=3: lat = -acos( 1 / (74700.790) ) + 0.000252 = -1.5705309 = -89.984794 degrees

And when N=1, this circle is indeed about 1.16 miles north of the south pole, just as jnelso99 noted.

Cheers,
-Isaac