The solution here uses 3D Cartesian
coordinates and the cosine rule. For a similar problem with a
solution that uses scalar products of vectors see Air Routes. A
third method is to draw an altitude of the isosceles triangle
with London, Sydney and the origin as vertices and use the
trigonometry of the rightangled triangle.

This solution comes from Andrei from
Tudor Vianu National College, Bucharest, Romania.
To solve this problem I see that the data are given in
spherical coordinates, but I need to work in Cartesian
coordinates.
I choose the Cartesian system of coordinates with the origin
at the centre of the Earth, with the $xy$ plane  the equatorial
plane, and the $z$ axis directed towards the North Pole. The
spherical system of coordinates has the origin at the same point as
the Cartesian one, i.e. at the centre of the Earth. Let the angle
$p$ measure the latitude (0 at the equator and $\pi/2$ at the North
Pole), i.e. it is the angle between the position vector of the
current point $P$ and the plane $xOy$. Let the angle $q$ measure
the longitude (in respect to the axis $Ox$), i.e. it is the angle
between the projection of the position vector of point $P$ on the
plane $xOy$ and the $x$ axis. So the 3D coordinates in terms of the
angles of latitude and longitude are: $$x = R \cos p \cos q, y = R
\cos p \sin q, z = R \sin p.$$

London has longitude 0 and latitude 51.5 degrees North and
Sydney has longitude 151 degrees East and latitude 34 degrees
South. The corresponding angles for London and Sydney are: $p_L =
51.5, q_L = 0, p_S = 34, q_S = 151$.
The distance between the two cities, measured on the line
joining them and passing through the interior of Earth, is: $$d^2 =
R^2 ((x_L  x_S)^2 + (y_L  y_S)^2 + (z_L  z_S)^2).$$ Both cities
are situated on the sphere, and this means there exists a great
circle of the Earth passing through both of them. Let $\alpha$ be
the angle between the two cities, and having the vertex at the
centre of the Earth (defined on the great circle specified above).
This angle can be found using the cosine rule for the isosceles
triangle defined by the position of London, the centre of Earth and
the position of Sidney: $$d^2 =R^2 + R^2  2R^2 cos \alpha$$ So,
$$\cos \alpha = {2R^2  d^2\over 2R^2}.$$ The distance between the
two cities measured on the surface of the Earth, along the great
circle, is $R\alpha$, where $\alpha$ is measured in radians. Let it
be $L_1$, then: $$L_1 = R\cos^{1}{(2  (x_Lx_S)^2  (y_Ly_S)^2 
(z_Lz_S)^2) \over 2}.$$ The other distance is calculated adding 6
km to the radius of the Earth, and it will be $L_2$.
The numerical results are: $\alpha = 2.66$ radians and, on the
surface of Earth, the distance, to the accuracy of the measures
used, is: $$L_1 = 17000 \ {\rm km}.$$ At the altitude of 6 km the
distance $L_2$ is about 16 kilometres more than $L_1$.
The problem
Over the Pole shows that the great circle distance is shorter
than the path along a line of latitude for two points with the same
latitude.