Posted on Feb 13, 2026
From Pierre Bierre comes a planetary puzzle:
A rover is dropped down on a spherical planet with a radius of 1000 miles. The rover has been programmed with a very specific set of motions:
To be picked up, the rover must complete its journey in the same place it was dropped down. What is the minimum value of s, with s > 0, for which this works?
The rover traces three equal great-circle arcs with left turns of 60 degrees at each stop, so the path forms an equilateral spherical triangle whose interior angles are all 120 degrees. Let \(a\) be the angular side length in radians, \(a=s/R\), where \(R=1000\) miles.
Spherical law of cosines for an equilateral triangle gives:
\[ \cos a = \cos^2 a + \sin^2 a \cos A,\quad A=120^\circ. \]
Since \(\cos 120^\circ=-1/2\), this becomes \(\,3\cos^2 a-2\cos a-1=0\), so \(\cos a=1\) (the trivial \(a=0\) solution) or \(\cos a=-1/3\).
Therefore the minimum nonzero solution is:
\(s=1000\arccos(-1/3)\) miles, which is approximately 1910.633236 miles.
Since the equations depend only on \(\cos a\), higher solutions come from other angles with the same cosine. The next one in \((0,2\pi R]\) is:
\(s=1000\left(2\pi-\arccos(-1/3)\right)\) miles, which is approximately 4372.552071 miles.
In general, all positive solutions are:
\(s=1000\left(\arccos(-1/3)+2\pi k\right)\) or \(s=1000\left(2\pi-\arccos(-1/3)+2\pi k\right)\), for integers \(k \ge 0\).
Use the slider to set s (miles) and see the rover's three-step path projected on latitude and longitude lines. The readout shows how far the rover finishes from its starting point.
Finish is 0.00 miles from the start.
There are other values of s for which the rover will end its journey where it was dropped down. How many such positive values of s (including the answer you just found in the Fiddler) are less than 100,000 miles?
Computing solutions in the range (0, 100,000] miles...