Source

I got this problem from Rustan Leino, who says he must have heard this
problem ages ago, but as he remembered it, one was always satisfied after
finding just one solution. It was a math professor at the Kaiserslautern
Technical University who brought asking for *all* solutions to his
attention.

I solved it and wrote up my solution.

Problem

Initially, you're somewhere on the surface of the Earth. You travel one
kilometer south, then one kilometer east, then one kilometer north. You
then find yourself back at the initial position. Describe *all*
initial locations from which this is possible.

Solution
Reveal

Let $r$ be the radius of the Earth, in kilometers. The points are the North Pole and, for every positive integer $n,$ every point \[ 1 + r \sin^{-1}\left(\frac{1}{2 \pi n r}\right) \] kilometers north of the South Pole. We'll now explain why this is the complete set of such points.

The North Pole works because if you go a kilometer south, then walk east, you're walking along a circle that keeps you a kilometer south of the North Pole. So, when you then walk a kilometer north, you're back to the North Pole.

From here on, we consider what starting points work other than the North Pole.

The South Pole, and any point less than a kilometer north of it, don't work because you can't go a kilometer south from those locations. Points exactly one kilometer north of the South Pole also don't work because if you go a kilometer south from any of them you're at the South Pole and can't meaningfully walk east.

Thus, your path will never cross either pole and your longitude will always be well-defined. When walking south for the first kilometer, your longitude will remain constant. When walking north for the last kilometer, your longitude will also remain constant. Thus, when walking east one kilometer, you must return to the same longitude that you started at.

This is only possible if the distance you travel is an integer multiple of the circumference of the circle $C$ you walk along as you go east. So $C,$ which is an equi-latitude circle, must have circumference $1/n$ kilometers for some positive integer $n$. This makes its radius $1/(2\pi n).$ If you draw one radius of $C$, then connect both ends of this radius to the center of the Earth, you form a right triangle with the hypotenuse being $r$. Thus, if you define $\theta$ to be the angle opposite the radius of $C$, then $\sin \theta = (1/2\pi n) / r$. In other words, $\theta = \sin^{-1}(1/2\pi n r)$. If you extend the other leg of the right triangle further out from the center of the Earth, it extends to the nearest pole, so we can see that $\theta$ is also the angle between that pole and a point on $C.$ Thus, the distance along the surface of the Earth between the pole and $C$ is $r \theta$.

Since $n \geq 1$, $r \theta \leq r \sin^{-1}(1/2\pi r).$ This latter quantity is approximately 0.16. This means that $C$ is less than 0.17 km from the nearest pole. This means the nearest pole can't be the North Pole since you walked 1 km south to reach $C.$

So the only possibility is that you walk 1 km south to reach a point $r \theta$ km from the South Pole. In this case, you walk east and return to the same point where you started walking east, then walk 1 km north and return to the point where you started.

Thus, besides the North Pole, the only points you can start at are ones $1 + r \theta$ km north of the South Pole. That is, points that are $1 + r \sin^{-1}(1/2\pi n r)$ kilometers north of the South Pole for some positive integer $n$. $\Box$