The brilliant Martin Gardner reviews the old riddle of an explorer who walks one mile due south, turns and walks one mile due east, and finally turns again and walks another one mile due north, being himself at exactly the same point where he started. How is it possible?
This problem is known as the returning explorer and sometimes it is said that the explorer shoots a bear after his walk. Then you are asked for figuring out the color of the bear. After thinking a while, smart people answer “white”, as the explorer must have been right in the North Pole of the Earth (blue point in figure below) when shooting the (polar) bear, what would explain his closed trajectory (red spherical triangle in the figure below). The graph is not to scale.
A mathematician would only partially agree with the answer above and would ask more questions. Is this the only spot on Earth for the returning explorer? Can you find other points, besides the North Pole, from which the explorer could walk a mile south, a mile east, a mile north and then find himself back at his starting position?
The answer is definitely yes. In fact, there are infinitely many points. Let's see it. Imagine a parallel line on the Globe so close to the South pole that its total length is precisely one mile: You walk one mile East along this one-mile parallel, you return to your original place. Therefore, if the explorer is in any point over another parallel located one mile north to the one-mile parallel, it would be a “returning explorer” after going south, east –honestly a waste of time– and north back.
The infinitely many points that ensure a closed trajectory (in red for one of the points) are colored blue in the above figure. (The draw is not to scale.)
Moreover, the same idea applies to a half-mile parallel, a third-mile parallel and so on, with the only difference that the explorer would do 2, 3... full turns in his East trajectory.
Let us calculate the latitude of the blue parallel that ensures going back to the starting point (see previous figure). The length of the red parallel is 1 mile, and the length of a circumference is given by the formula 2πr, thus
2πr = 1,
which gives us the radius r between D and C (see figure):
r = 1/(2π)
Now, remember that the cosine of the angle OĈD (let us call it Φ') is the quotient between DC (that is r) and OC (that is the radius of Earth R):
cos(Φ') = r/R
Combining last two expressions and solving for Φ' with the inverse function of the cosine (that is called arccos), we get
Φ' = arccos(1/(2πR))
Note that this angle Φ'=OĈD equals to the angle AÔC, so it gives the latitude of the one-mile parallel (red line in the figure).
Now let us compute the angle ΔΦ between the red parallel and the blue one (see green triangle in the graph). The blue parallel is where the returning explorer actually is. From the definition of a radian (the standard unit of angle measure) we can calculate this angle as the quotient of the arc length for BC –which is 1 mile– and the radius of Earth R:
ΔΦ = 1/R
Finally, the latitude of the blue parallel is Φ=Φ'–ΔΦ:
Φ = arccos(1/(2πR)) – 1/R
This angle is expressed in radians, but normally latitude is referred in degrees (0 degrees for the Earth's equatorial plane, 90° for the North Pole, and –90° for the South Pole). Then we must multiply by the factor of conversion 180/π and change the global sign to get negative latitudes for the South Hemisphere. Moreover, it is straightforward to generalize the previous formula to include trajectories along red parallels of a 1/n of a mile, being n the number of full turns (n=1, 2, 3,...). We can also consider that the explorer walks an arbitrary distance L in each direction instead of just 1 mile. Then, the general formula for the latitude in the South of the returning explorer would read:
Φ = 180/π·[L/R – arccos(L/(2πnR))] (Latitude, degrees)
This formula is only valid for values of L that give an angle Φ between –90° and 90°. There are always solutions for some n providing that L<πR (although for n=1, L should be less than approximately 2.7R). In the next graph the latitudes in terms of the quotient L/R are shown for some values of n.
The length of the arc between the South Pole and a point along one of the infinitely many blue parallels (one for each valid value of the natural number n) is the product of its subtended angle in radians, π/2 + Φ, and the radius R, which gives:
Arc = L + R·[π/2 – arccos(L/(2πnR))] (distance from the South Pole)
The previous two general exact expressions can be approximated for small values of L/R, as happens in the original example, where L=1 mile and the average Earth's radius R is over 3960 miles. To the first order at L/R, they read:
Φ ≈ –90 + 180L/(πR)·[1+1/(2πn)]
Arc ≈ L·[1+1/(2πn)]
The last expression for L=1 mile and n=1 is the one that Martin Gardner shows us, and gives a distance of around 1.16 miles from the South Pole. For greater values of n, the parallels are even closer to the South Pole.
In any case, there are no bears in the Antarctica, so these infinitely many points near the South Pole that ensure a returning trip for the explorer are not valid and we must admit “white” as the perfect answer for the color of the bear. But it wasn't obvious at all, was it? Now, we know that the explorer could also be shooting black and white penguins (with a camera.)
Martin Gardner, Hexaflexagons and other mathematical diversions, The University of Chicago Press, 1988.
"Penguin vs Bear" by M. A. Cano