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Simon from Elizabeth College, Guernsey and Andrei from Tudor Vianu National College, Romania have both solved this problem and both solutions are used below.
To solve this problem we shall consider that the point is
situated at latitude $\alpha$. To travel around the line of
latitude, the distance from P to Q would be half the circumference
of the circle at latitude $\alpha$. This circle has a radius $R
\cos \alpha$, where $R$ is the radius of Earth.
So, the distance traveled from P to Q on the line of latitude
is $d_{lat} = \pi R \cos \alpha$.

Traveling over the line of longitude, the circle on which we
have to calculate the distance is a great circle of the sphere, and
the angle of displacement is $2(\pi /2  \alpha)$ radians. The
distance is therefore $4\pi R(90L)/360$ where $L$ is the angle of
latitude in degrees or equivalently $d_{long} = 2R (\pi/2 
\alpha)$.
It is clear that the path on a great circle is always
shorter.

A graph of the ratio of these distances shows that this ratio
seems to tend to $\pi /2$ as the line of latitude approaches the
pole, that is the ratio $${d_{lat}\over d_{long}}= {\pi R \cos
\alpha \over 2R (\pi/2  \alpha)}$$ tends to a limit as $\alpha \to
\pi/2$.
This ratio can also be written as $${d_{lat}\over d_{long}}=
{\pi R \sin (\pi/2 \alpha) \over 2R (\pi/2  \alpha)}.$$ As we
know ${\theta\over \sin \theta} \to 1 $ as $\theta \to 0$ we can
take $\theta = \pi/2  \alpha $ and we see that this limit is $\pi
/2$.
