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= (lat–x/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= (lat–x/R,long+x/cos(lat–x/R))

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

pf= (lat,long+x/(R*cos(lat–x/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(lat–x/R))2 *

pi*N=x/ (R * cos(lat–x/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 degreesN=2: lat = -acos( 1

/(49800.527) ) + 0.000252 = -1.5705242 = -89.984410 degreesN=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