I'm standing on the surface of the Earth. I walk ten miles south, ten miles east, ten miles north and ten miles west. I end up exactly where I started.
Where on earth am I?

  • $\begingroup$ Bonus: do you answers work for a flat earth model? (the "disk" one) $\endgroup$ – golimar Jul 23 '20 at 15:48

I think there are three(ish) choices. The simplest one is

5 miles north of the equator. (The lengths of the latitudinal circles change symmetrically on both sides the equator, so the ten miles east and west portions span the same amount of latitude degrees if (and only if) they happen equally far from the equator.)

To find the other solutions, we have to notice that

we don't actually have to span the exact same amount of latitude degrees, it's enough if we span the exact same amount modulo full circles around the planet's axis.

This means we can also start

a bit more than 10 miles from the South Pole: We walk the 10 miles to a carefully chosen point suitably near the pole, and from there the ten miles east go around the pole some appropriate number of times, and then a bit more so that our latitude ends up being a bit under 60 degrees east of the original one, which is what is needed for completing the trip.

The possible paths would look something like this: (nowhere near exact, and not to scale)
enter image description here
(The circle is of course smaller if the eastward ten miles part goes around the pole more than once. The maths needed for figuring out all the (infinitely many) exact distances at which this works get somewhat messy because the angle between 1 and 3 doesn't stay constant; go ahead and have a crack at it if you like challenges.)

and then we can, of course, start

Quite near the North Pole: same pattern as before, but circling the pole happens on the westward portion of the trip.

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    $\begingroup$ Ooh...I didn't think of the multiple transits. Very clever! I wonder how many of these there are? $\endgroup$ – Jeremy Dover Jul 21 '20 at 20:56
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    $\begingroup$ perfect answer! $\endgroup$ – ThomasL Jul 21 '20 at 20:58
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    $\begingroup$ @JeremyDover There are an infinite number of multiple transit solutions, since the transits can become arbitrarily small and result in an arbitrarily large number of transits while still only walking 10 miles. There's no theoretical upper bound for the number of transits, as you can always just start a little closer to the pole, make the circle a little smaller, and go around more times. $\endgroup$ – Nuclear Hoagie Jul 21 '20 at 21:00
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    $\begingroup$ @JeremyDover You can apply the Intermediate Value Theorem. Basically, if you start too far away, you'll not wind around the pole enough, and if you start too close you'll wind around the pole too far. Somewhere in between there must be a starting point that is just right. Calculating where that point is exactly can be very difficult. $\endgroup$ – Jaap Scherphuis Jul 22 '20 at 7:25
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    $\begingroup$ @JaapScherphuis: I did the trig last night, and convinced myself this was true, using a continuity argument as you suggest. It just wasn't obvious to me at the first pass that the dependency was a continuous function :-) $\endgroup$ – Jeremy Dover Jul 22 '20 at 10:35

I think the answer is:

Five miles north of the equator.


This is a riff on the classical North Pole riddle, but the twist is that you seem to make a full circuit, so it is tempting to say "anywhere". However, because the earth is (approximately) a sphere, 10 miles east and 10 miles west are the same distance if and only if they are walked at the same latitude, or equal latitudes in the Northern/Southern hemispheres. Since the east and west segments are walked 10 miles north/south from each other, the only way this can happen is if these two segments are on opposite sides of the equator. Hence you must start five miles north of the equator.

Update: Based on @Bass's answer, I was curious about the actual calculation. Turns out it's really not that hard.

First, let's approximate earth with a sphere of radius 4,000 miles...the number is a bit high, but this is all an approximation anyway. Let $\theta$ be the angle between the axis of the sphere through the south pole, and the axis of the sphere through your starting point, as in this diagram (flagrantly not to scale):

First note that the arc length between the start and end of the walk on this circle is 10 miles, so we know that the angle between the start and end of the walk is $\theta - \phi = \frac{10}{4000} = \frac{1}{400}$ radians.

We can now easily determine the angle of longitude our 10 mile walks take. At the starting point, we are walking around a circle of radius $R = 4000 \sin \theta$ for 10 miles, so the angle of longitude is $\frac{10}{4000 \sin \theta}$. Note the angle equals arclength over radius formulation really helps us here, since it takes care of multiple passes around the circle without any effort on our part. Similarly, the angle of longitude at the bottom walk is $\frac{10}{4000 \sin (\theta - \frac{1}{400})}$.

So for this walk to be possible, we must have $$ \frac{10}{4000 \sin (\theta - \frac{1}{400})} - \frac{10}{4000 \sin \theta}$$ be a positive integer multiple of $2\pi$. Since these are both continuous functions of $\theta$ away from $\frac{1}{400}$, it is easy to see by the intermediate value theorem (h/t @JaapScherphuis) that this function takes on all above its minimum.

  • $\begingroup$ Well done. I just spent 2h porting movable-type.co.uk/scripts/latlong.html to python in order to find the possible starting points. Your formula works fine and outputs 2pi for the solution I found with 1 transit and 10pi for 5 transits, just as you described. (I used 3963.2miles for the radius) $\endgroup$ – Eric Duminil Jul 22 '20 at 22:15
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    $\begingroup$ Good job! I'd still say that even with this result, calculating all the exact positions from an equation involving sines of sums of angles isn't quite trivial, but it's very nice that the one obvious solution (the one we get by setting the value to 0, so that $\text{sin}(\theta) = \text{sin}(\theta-\frac{1}{400})$) actually finds the walk across the equator :-) $\endgroup$ – Bass Jul 23 '20 at 18:09
  • $\begingroup$ Oh, I just noticed you said "positive integer multiple". If you drop the "positive", I think your brilliant formula might find the North Pole solutions too! $\endgroup$ – Bass Jul 24 '20 at 19:39

I think the answer depends on what "going east/west" means. That is:

Let us define "going east/west" as "going along a great circle of earth diagonal to the longitude circle you are on". Then going east/west changes latitude (unless you are on the equator). For example, going east from Japan you reach South America as the azimuthal equidistant projection map below indicates.
azimuthal equidistant projection map around Japan

Under this definition, suppose you start from P1, walk south to P2, east to P3, north to P4 and west to P5. (Each walk is 10 miles long.) I check the following three cases.

Case 1:

If P1 and P2 are both in the northern hemisphere, there's no solution. Observe that P2 is 10 miles south of P1, P3 is south of P2, P4 is 10 miles north of P3 and P5 is south of P4. So in total, P5 is south of P1, implying P1 and P5 are different points.

Case 2:

If P1 and P2 are both in the southern hemisphere, there's no solution, by a similar argument to the Case 1.

Case 3:

If P1 is in the northern hemisphere and P2 in the southern, the answer is slightly less than 5 miles north of the equator as in the picture.
circuit near equator
To verify such a solution exists, suppose P1 is $a$ miles north of the equator and let $f(x)$ be the longitudinal distance you move when you are at $x$ miles north/south of the equator and go east/west 10 miles. Then from the picture it must hold $10-a=a+f(10-a)$ or equivalently, $$ 10-2a=f(10-a). $$ When $a$ ranges from 0 to 10, the left side changes from 10 to -10 while the right side is always almost 0 since the latitude is low. So for some $a$ the equality holds and such a circuit exists.

  • $\begingroup$ Interesting. Is there a name for your definition of going east/west? The other answers seem to use rhumb lines (en.wikipedia.org/wiki/Rhumb_line). $\endgroup$ – Eric Duminil Jul 22 '20 at 22:19
  • $\begingroup$ Not sure how it's called, but probably related to the great-circle navigation. Going east/west in such a way uses the shortest path between two points. $\endgroup$ – sj2tpgk Jul 23 '20 at 0:41
  • $\begingroup$ It could be the case, yes. "Go west", with this definition, would be "Go west for the first step, and then just keep on walking, whatever the compass says". Right? $\endgroup$ – Eric Duminil Jul 23 '20 at 8:04
  • $\begingroup$ Yes -- but a more natural interpretation is that "go west 10 miles" means "go to a point which you see 10 miles west from the starting point." $\endgroup$ – sj2tpgk Jul 23 '20 at 12:43
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    $\begingroup$ When I made a much easier puzzle based on projecting straight lines onto a sphere a couple of years back, I used a powerful laser pointer instead of walking, because "go straight west" doesn't really exists on most parts of the planet. $\endgroup$ – Bass Jul 23 '20 at 18:27

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