# Martin Gardner – Problem 1 – The Returning Explorer

In “My Best Mathematical and Logic Puzzles” Martin Gardner’s first problem is titled “The Returning Explorer” in reference of the following riddle.

“”An explorer walks one mile due south, turns and walks one mile due east, turns again and walks one mile due north. He finds himself back where he started. He then shoot [or see (depending of the riddle you wish to tell)] a bear. What color is the bear ?””

The problem proposed by M. Gardner is then to find all points on the globe from which you can walk one mile south, east and north and finally end up at your starting point.

There is an infinite number of points which satisfies this property. Can you find them and give the latitude at which you need to start your journey ?

The answer is the following.  Writing $R$ the earth radius and $\phi_n$ the latitude ($n$ is integer), all latitude on which you can start are given by:

$\phi_n=\frac{1}{R}-arccos\left(\frac{1}{2\pi R n}\right).$

Which at the first order in $1/R$ leads to

$\phi_n=-\frac{\pi}{2}+\frac{1}{R}\frac{2\pi n+1}{2\pi n}$.

Now, starting from any other latitude $\theta_0$, you would not be able to “complete a triangle” by walking south for $X$ miles, east for $Y$ miles and back north for $X$ miles. But how far away from the starting point would you be exactly ?

Well if you were to walk back west from the finish to the starting point, you would have to walk for a distance $\Delta$

$\Delta =Y\frac{\cos(\theta_0)}{\cos(\theta_0-X/R)}.$

Assuming $X/R\ll1$ we have

$\Delta\simeq Y\frac{\cos(\theta_0)}{\cos(\theta_0)+\frac{X}{R}\sin(\theta_0)}.$

However, this is not the shortest distance. The fact is that the shortest path between two points $A$ and $B$ on a sphere is along the circle of radius $R$ passing by $A$ and $B$. Can we find out what the shortest path is ?

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