Can You Get the Rover Home?

Fiddler

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:

• First, it moves straight forward a fixed distance s, and stops.

• Without moving forward, it turns left 60 degrees. (Importantly, the rover turns 60 degrees, not 120 degrees.)

• Next, the rover moves straight forward in this new direction another distance s, and stops.

• Without moving forward, it again turns left 60 degrees.

• Finally, the rover moves straight forward in this new direction another distance s, and stops.

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?

To help you out, here’s an illustration, courtesy of Pierre, of what such a path might look like:

Solution

This creates an equilateral spherical triangle with $120^\circ$ for each angle.

The law of cosines for the angles of a spherical triangle states

$$\cos A = -\cos B \cos C + \sin B \sin C \cos a$$

For the rover's path :

\begin{align} \cos a &= \dfrac{\cos A + \cos^2 A}{\sin^2 A} \\ \cos a &= \dfrac{\cos 120 \cos^2 120}{\sin^2 120} \\ \cos a &= \dfrac{\frac{-1}{2} + \left(\frac{-1}{2} \right)^2}{\left(\frac{-\sqrt{3}}{2} \right)^2} \\ \cos a &= \dfrac{\frac{-1}{4}}{\frac{3}{4}} = \dfrac{-1}{3}\\ \end{align}

Answer

The minimum solution for the path is

$$p = \boxed{1000\cos^{-1} \dfrac{-1}{3} \approx 1,910.63 \text{ miles}}$$

Extra Credit

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?

Solution

The circumference of the planet is $2000\pi$

Thus,

\begin{align} 1910.63 + 2000\pi q &< 100000 \\ q &< \dfrac{100000-1910.63}{2000\pi} \\ q &< 15.6 \end{align}

This pattern yields 16 solutions.

In addition to $1.91063$, $2\pi - 1.91063 \approx 4.37255 $ is also a solution to $a = \cos^{-1} \dfrac{-1}{3}$.

Thus,

\begin{align} 4372.55 + 2000\pi r &< 100000 \\ r &< \dfrac{100000-4372.55}{2000\pi} \\ r &< 15.2 \end{align}

This pattern also yields 16 solutions.

Answer

$$\boxed{32}$$

Rohan Lewis

2026.02.16