= Solution
\b[Length functional.] On the unit <sphere>, write the <position vector> in <spherical coordinates> as
$$
r(\theta,\phi)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta).
$$
Its two coordinate <tangent vectors> satisfy $r_\theta\cdot r_\phi=0$, $|r_\theta|=1$ and $|r_\phi|=\sin\theta$. Consequently the induced <Riemannian metric> is $ds^2=d\theta^2+\sin^2\theta\,d\phi^2$. On a segment for which $\theta$ increases and $\phi=\phi(\theta)$,
$$
\boxed{S=\int_{\theta_1}^{\theta_2}\sqrt{1+\sin^2\theta\,(\phi')^2}\,d\theta.}
$$
Segments where $\theta$ is not monotone require another parameter or a subdivision. Endpoint turning points can be handled by a limit.
\b[First integral.] With $F=\sqrt{1+\sin^2\theta\,(\phi')^2}$, there is no explicit $\phi$ dependence. The <Euler-Lagrange equation> therefore says that
$$
\frac{\partial F}{\partial\phi'}=\frac{\sin^2\theta\,\phi'}{\sqrt{1+\sin^2\theta\,(\phi')^2}}=c
$$
is constant. For $c\ne0$, put $k=1/|c|$ and solve:
$$
\boxed{\phi'=\frac{\operatorname{sgn}c}{\sin\theta\sqrt{k^2\sin^2\theta-1}},\qquad\phi(\theta)-\phi(\theta_1)=\operatorname{sgn}c\int_{\theta_1}^{\theta}\frac{d\vartheta}{\sin\vartheta\sqrt{k^2\sin^2\vartheta-1}}.}
$$
The radicand must be nonnegative along the chosen segment. The printed expression with both limits fixed at $\theta_1,\theta_2$ describes a longitude difference between endpoints; a formula for $\phi(\theta)$ needs the variable upper limit and the additive constant shown here. The case $c=0$ gives a <meridian> with constant longitude. The <equator> is also a stationary route, although it cannot be parameterized using changing $\theta$.
\b[Integration and geometric meaning.] Put $t=\cot\theta$, so $d\theta=-dt/(1+t^2)$ and
$$
\sin\theta\sqrt{k^2\sin^2\theta-1}=\frac{\sqrt{k^2-1-t^2}}{1+t^2}.
$$
For a nonconstant segment $k>1$, and writing $K=\sqrt{k^2-1}$, the <integral> becomes
$$
\phi-\phi_0=\mp\arcsin\frac{\cot\theta}K.
$$
Rearranging, and absorbing phase and sign into two constants, gives
$$
\boxed{\cot\theta=A\cos\phi+B\sin\phi.}
$$
Multiplying by $\sin\theta$ shows that the corresponding <Cartesian coordinates> satisfy $z=Ax+By$. Thus every such stationary route lies in a <plane> through the sphere's center: it is a <great circle>, the <geodesic> of the <sphere>. The meridians and equator are the limiting or separately parameterized cases of the same description.
\b[Rome to Singapore.] Use exactly the approximate coordinates supplied in the question: $(\theta,\phi)=(\pi/4,\pi/12)$ at departure and $(\pi/2,7\pi/12)$ at arrival. Inserting them into the preceding relation determines
$$
\boxed{\cot\theta=\cos(\phi-\pi/12),\qquad\frac\pi{12}\leq\phi\leq\frac{7\pi}{12}.}
$$
Indeed the right side is one at Rome and zero at Singapore. This corresponds to $K=1$, $k=\sqrt2$. In latitude $\ell=\pi/2-\theta$, the route is $\ell=\arctan[\cos(\phi-\pi/12)]$: it starts at $45^\circ$ north with an initially eastward heading and then curves southward to the equator. The first-integral expression in $\theta$ has an integrable endpoint singularity at departure, where $\theta$ has a turning point; the longitude parameter is regular there.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-1-great-circle-route.png]
{title=Shortest spherical route for the approximate Rome and Singapore coordinates in the question}
{description=Latitude versus longitude for the derived great circle, with endpoints at 45 degrees north and 15 degrees east, and zero degrees north and 105 degrees east. The quarter-circle parameter increases uniformly in spherical distance.}
For an explicit regular <arc-length parametrization>, let $a=r(\pi/4,\pi/12)$ and $b=r(\pi/2,7\pi/12)$. Their <dot product> is $\sin(\pi/4)\cos(\pi/2)=0$. Hence
$$
\boxed{r(s)=a\cos s+b\sin s,\quad0\leq s\leq\frac\pi2,\qquad S_{\min}=\frac\pi2.}
$$
Both $|r(s)|$ and $|r'(s)|$ equal one. The shorter <great circle> arc is the global shortest path: for any unit-speed curve from $a$, its central angle $\alpha=\arccos(a\cdot r)$ satisfies $|d\alpha/ds|\leq1$ by the <Cauchy-Schwarz inequality>, since the tangential projection of $a$ has length $\sin\alpha$. Integrating gives length at least the endpoint angle $\pi/2$, which this route attains. For a sphere of radius $R$, the corresponding shortest distance is $\pi R/2$.
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